Ryden_IntroCosmo

March 23, 2018 | Author: Rodrigo Carvajal | Category: Redshift, Universe, Electronvolt, Physical Cosmology, Chronology Of The Universe


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Introduction to CosmologyBarbara Ryden Department of Astronomy The Ohio State University January 13, 2006 Contents Preface v 1 Introduction 1 2 Fundamental Observations 7 2.1 Dark night sky . . . . . . . . . . . . . . . . . . . . . . . . . . 7 2.2 Isotropy and homogeneity . . . . . . . . . . . . . . . . . . . . 11 2.3 Redshift proportional to distance . . . . . . . . . . . . . . . . 15 2.4 Types of particles . . . . . . . . . . . . . . . . . . . . . . . . . 22 2.5 Cosmic microwave background . . . . . . . . . . . . . . . . . . 28 3 Newton Versus Einstein 32 3.1 Equivalence principle . . . . . . . . . . . . . . . . . . . . . . . 33 3.2 Describing curvature . . . . . . . . . . . . . . . . . . . . . . . 39 3.3 Robertson-Walker metric . . . . . . . . . . . . . . . . . . . . . 44 3.4 Proper distance . . . . . . . . . . . . . . . . . . . . . . . . . . 47 4 Cosmic Dynamics 55 4.1 Friedmann equation . . . . . . . . . . . . . . . . . . . . . . . . 57 4.2 Fluid and acceleration equations . . . . . . . . . . . . . . . . . 65 4.3 Equations of state . . . . . . . . . . . . . . . . . . . . . . . . . 68 4.4 Learning to love lambda . . . . . . . . . . . . . . . . . . . . . 71 5 Single-Component Universes 79 5.1 Evolution of energy density . . . . . . . . . . . . . . . . . . . 79 5.2 Curvature only . . . . . . . . . . . . . . . . . . . . . . . . . . 86 5.3 Spatially flat universes . . . . . . . . . . . . . . . . . . . . . . 91 5.4 Matter only . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94 ii CONTENTS iii 5.5 Radiation only . . . . . . . . . . . . . . . . . . . . . . . . . . 95 5.6 Lambda only . . . . . . . . . . . . . . . . . . . . . . . . . . . 97 6 Multiple-Component Universes 101 6.1 Matter + curvature . . . . . . . . . . . . . . . . . . . . . . . . 104 6.2 Matter + lambda . . . . . . . . . . . . . . . . . . . . . . . . . 108 6.3 Matter + curvature + lambda . . . . . . . . . . . . . . . . . . 112 6.4 Radiation + matter . . . . . . . . . . . . . . . . . . . . . . . . 116 6.5 Benchmark Model . . . . . . . . . . . . . . . . . . . . . . . . . 118 7 Measuring Cosmological Parameters 126 7.1 “A search for two numbers” . . . . . . . . . . . . . . . . . . . 126 7.2 Luminosity distance . . . . . . . . . . . . . . . . . . . . . . . . 131 7.3 Angular-diameter distance . . . . . . . . . . . . . . . . . . . . 136 7.4 Standard candles & H 0 . . . . . . . . . . . . . . . . . . . . . . 141 7.5 Standard candles & acceleration . . . . . . . . . . . . . . . . . 144 8 Dark Matter 155 8.1 Visible matter . . . . . . . . . . . . . . . . . . . . . . . . . . . 156 8.2 Dark matter in galaxies . . . . . . . . . . . . . . . . . . . . . . 159 8.3 Dark matter in clusters . . . . . . . . . . . . . . . . . . . . . . 164 8.4 Gravitational lensing . . . . . . . . . . . . . . . . . . . . . . . 170 8.5 What’s the matter? . . . . . . . . . . . . . . . . . . . . . . . . 175 9 The Cosmic Microwave Background 179 9.1 Observing the CMB . . . . . . . . . . . . . . . . . . . . . . . 180 9.2 Recombination and decoupling . . . . . . . . . . . . . . . . . . 185 9.3 The physics of recombination . . . . . . . . . . . . . . . . . . 189 9.4 Temperature fluctuations . . . . . . . . . . . . . . . . . . . . . 196 9.5 What causes the fluctuations? . . . . . . . . . . . . . . . . . . 201 10 Nucleosynthesis & the Early Universe 208 10.1 Nuclear physics and cosmology . . . . . . . . . . . . . . . . . 209 10.2 Neutrons and protons . . . . . . . . . . . . . . . . . . . . . . . 213 10.3 Deuterium synthesis . . . . . . . . . . . . . . . . . . . . . . . 218 10.4 Beyond deuterium . . . . . . . . . . . . . . . . . . . . . . . . 222 10.5 Baryon – antibaryon asymmetry . . . . . . . . . . . . . . . . . 228 iv CONTENTS 11 Inflation & the Very Early Universe 233 11.1 The flatness problem . . . . . . . . . . . . . . . . . . . . . . . 234 11.2 The horizon problem . . . . . . . . . . . . . . . . . . . . . . . 236 11.3 The monopole problem . . . . . . . . . . . . . . . . . . . . . . 238 11.4 The inflation solution . . . . . . . . . . . . . . . . . . . . . . . 242 11.5 The physics of inflation . . . . . . . . . . . . . . . . . . . . . . 247 12 The Formation of Structure 255 12.1 Gravitational instability . . . . . . . . . . . . . . . . . . . . . 258 12.2 The Jeans length . . . . . . . . . . . . . . . . . . . . . . . . . 261 12.3 Instability in an expanding universe . . . . . . . . . . . . . . . 266 12.4 The power spectrum . . . . . . . . . . . . . . . . . . . . . . . 272 12.5 Hot versus cold . . . . . . . . . . . . . . . . . . . . . . . . . . 276 Epilogue 284 Annotated Bibliography 286 Preface This book is based on my lecture notes for an upper-level undergraduate cos- mology course at The Ohio State University. The students taking the course were primarily juniors and seniors majoring in physics and astronomy. In my lectures, I assumed that my students, having triumphantly survived fresh- man and sophomore physics, had a basic understanding of electrodynamics, statistical mechanics, classical dynamics, and quantum physics. As far as mathematics was concerned, I assumed that, like modern major generals, they were very good at integral and differential calculus. Readers of this book are assumed to have a similar background in physics and mathemat- ics. In particular, no prior knowledge of general relativity is assumed; the (relatively) small amounts of general relativity needed to understand basic cosmology are introduced as needed. Unfortunately, the National Bureau of Standards has not gotten around to establishing a standard notation for cosmological equations. It seems that every cosmology book has its own notation; this book is no exception. My main motivation was to make the notation as clear as possible for the cosmological novice. I hope that reading this book will inspire students to further explorations in cosmology. The annotated bibliography at the end of the text provides a selection of recommended cosmology books, at the popular, intermediate, and advanced levels. Many people (too many to name individually) helped in the making of this book; I thank them all. I owe particular thanks to the students who took my undergraduate cosmology course at Ohio State University. Their feedback (including nonverbal feedback such as frowns and snores during lectures) greatly improved the lecture notes on which this book is based. Adam Black and Nancy Gee, at Addison Wesley, made possible the great leap from rough lecture notes to polished book. The reviewers of the text, v vi PREFACE both anonymous and onymous, pointed out many errors and omissions. I owe particular thanks to Gerald Newsom, whose careful reading of the entire manuscript improved it greatly. My greatest debt, however, is to Rick Pogge, who acted as my computer maven, graphics guru, and sanity check. (He was also a tireless hunter of creeping fox terrier clones.) As a small sign of my great gratitude, this book is dedicated to him. Chapter 1 Introduction Cosmology is the study of the universe, or cosmos, regarded as a whole. At- tempting to cover the study of the entire universe in a single volume may seem like a megalomaniac’s dream. The universe, after all, is richly tex- tured, with structures on a vast range of scales; planets orbit stars, stars are collected into galaxies, galaxies are gravitationally bound into clusters, and even clusters of galaxies are found within larger superclusters. Given the richness and complexity of the universe, the only way to condense its history into a single book is by a process of ruthless simplification. For much of this book, therefore, we will be considering the properties of an ideal- ized, perfectly smooth, model universe. Only near the end of the book will we consider how relatively small objects, such as galaxies, clusters, and su- perclusters, are formed as the universe evolves. It is amusing to note, in this context, that the words “cosmology” and “cosmetology” come from the same Greek root: the word “kosmos”, meaning harmony or order. Just as cosmetologists try to make a human face more harmonious by smoothing over small blemishes such as pimples and wrinkles, cosmologists sometimes must smooth over small “blemishes” such as galaxies. A science which regards entire galaxies as being small objects might seem, at first glance, very remote from the concerns of humanity. Nevertheless, cos- mology deals with questions which are fundamental to the human condition. The questions which vex humanity are given in the title of a painting by Paul Gauguin (Figure 1.1): “Where do we come from? What are we? Where are we going?” Cosmology grapples with these questions by describing the past, explaining the present, and predicting the future of the universe. Cosmol- ogists ask questions such as “What is the universe made of? Is it finite or 1 2 CHAPTER 1. INTRODUCTION Figure 1.1: Where Do We Come From? What Are We? Where Are We Going? Paul Gauguin, 1897. [Museum of Fine Arts, Boston] infinite in spatial extent? Did it have a beginning some time in the past? Will it come to an end some time in the future?” Cosmology deals with distances that are very large, objects that are very big, and timescales that are very long. Cosmologists frequently find that the standard SI units are not convenient for their purposes: the meter (m) is awkwardly short, the kilogram (kg) is awkwardly tiny, and the second (s) is awkwardly brief. Fortunately, we can adopt the units which have been developed by astronomers for dealing with large distances, masses, and times. One distance unit used by astronomers is the astronomical unit (AU), equal to the mean distance between the Earth and Sun; in metric units, 1 AU = 1.5 10 11 m. Although the astronomical unit is a useful length scale within the Solar System, it is small compared to the distances between stars. To measure interstellar distances, it is useful to use the parsec (pc), equal to the distance at which 1 AU subtends an angle of 1 arcsecond; in metric units, 1 pc = 3.1 10 16 m. For example, we are at a distance of 1.3 pc from Proxima Centauri (the Sun’s nearest neighbor among the stars) and 8500 pc from the center of our Galaxy. Although the parsec is a useful length scale within our Galaxy, it is small compared to the distances between galaxies. To measure intergalactic distances, it is useful to use the megaparsec (Mpc), equal to 10 6 pc, or 3.110 22 m. For example, we are at a distance of 0.7 Mpc from M31 (otherwise known as the Andromeda galaxy) and 15 Mpc from the Virgo cluster (the nearest big cluster of galaxies). The standard unit of mass used by astronomers is the solar mass (M ); in metric units, the Sun’s mass is 1 M = 2.0 10 30 kg. The total mass 3 of our Galaxy is not known as accurately as the mass of the Sun; in round numbers, though, it is ` gal ≈ 10 12 M . The Sun, incidentally, also provides the standard unit of power used in astronomy. The Sun’s luminosity (that is, the rate at which it radiates away energy in the form of light) is 1 L = 3.8 10 26 watts. The total luminosity of our Galaxy is 1 gal = 3.6 10 10 L . For times much longer than a second, astronomers use the year (yr), defined as the time it takes the Earth to go once around the Sun. One year is approximately equal to 3.2 10 7 s. In cosmological context, a year is frequently an inconveniently short period of time, so cosmologists frequently use gigayears (Gyr), equal to 10 9 yr, or 3.2 10 16 s. For example, the age of the Earth is more conveniently written as 4.6 Gyr than as 1.5 10 17 s. In addition to dealing with very large things, cosmology also deals with very small things. Early in its history, as we shall see, the universe was very hot and dense, and some interesting particle physics phenomena were occurring. Consequently, particle physicists have plunged into cosmology, introducing some terminology and units of their own. For instance, particle physicists tend to measure energy units in electron volts (eV) instead of joules (J). The conversion factor between electron volts and joules is 1 eV = 1.6 10 −19 J. The rest energy of an electron, for instance, is : e c 2 = 511.000 eV = 0.511 MeV, and the rest energy of a proton is : P c 2 = 938.3 MeV. When you stop to think of it, you realize that the units of meters, megaparsecs, kilograms, solar masses, seconds, and gigayears could only be devised by ten-fingered Earthlings obsessed with the properties of water. An eighteen-tentacled silicon-based lifeform from a planet orbiting Betel- geuse would devise a different set of units. A more universal, less cultur- ally biased, system of units is the Planck system, based on the universal constants G, c, and ¯ /. Combining the Newtonian gravitational constant, G = 6.7 10 −11 m 3 kg −1 s −2 , the speed of light, c = 3.0 10 8 ms −1 , and the reduced Planck constant, ¯ / = /(2π) = 1.1 10 −34 J s = 6.6 10 −16 eVs, yields a unique length scale, known as the Planck length: / P ≡ G¯ / c 3 1/2 = 1.6 10 −35 m . (1.1) 4 CHAPTER 1. INTRODUCTION The same constants can be combined to yield the Planck mass, 1 ` P ≡ ¯ /c G 1/2 = 2.2 10 −8 kg . (1.2) and the Planck time, t P ≡ G¯ / c 5 1/2 = 5.4 10 −44 s . (1.3) Using Einstein’s relation between mass and energy, we can also define the Planck energy, 1 P = ` P c 2 = 2.0 10 9 J = 1.2 10 28 eV . (1.4) By bringing the Boltzmann constant, / = 8.6 10 −5 eVK −1 , into the act, we can also define the Planck temperature, 1 P = 1 P / = 1.4 10 32 K . (1.5) When distance, mass, time, and temperature are measured in the appropriate Planck units, then c = / = ¯ / = G = 1. This is convenient for individuals who have difficulty in remembering the numerical values of physical constants. However, using Planck units can have potentially confusing side effects. For instance, many cosmology texts, after noting that c = / = ¯ / = G = 1 when Planck units are used, then proceed to omit c, /, ¯ /, and/or G from all equations. For instance, Einstein’s celebrated equation, 1 = :c 2 , becomes 1 = :. The blatant dimensional incorrectness of such an equation is jarring, but it simply means that the rest energy of an object, measured in units of the Planck energy, is equal to its mass, measured in units of the Planck mass. In this book, however, I will retain all factors of c, /, ¯ /, and G, for the sake of clarity. In this book, we will be dealing with distances ranging from the Planck length to 10 4 Mpc or so, a span of some 61 orders of magnitude. Dealing with such a wide range of length scales requires a stretch of the imagination, to be sure. However, cosmologists are not permitted to let their imaginations run totally unfettered. Cosmology, I emphasize strongly, is ultimately based 1 The Planck mass is roughly equal to the mass of a grain of sand a quarter of a millimeter across. 5 Figure 1.2: The ancient Egyptian view of the cosmos: the sky goddess Nut, supported by the air god Shu, arches over the earth god Geb (from the Greenfield Papyrus, ca. 1025 BC). [ c (Copyright The British Museum] on observation of the universe around us. Even in ancient times, cosmology was based on observations; unfortunately, those observations were frequently imperfect and incomplete. Ancient Egyptians, for instance, looked at the desert plains stretching away from the Nile valley and the blue sky overhead. Based on their observations, they developed a model of the universe in which a flat Earth (symbolized by the earth god Geb in Figure 1.2) was covered by a solid dome (symbolized by the sky goddess Nut). Greek cosmology was based on more precise and sophisticated observations. Ancient Greek astronomers deduced, from their observations, that the Earth and Moon are spherical, that the Sun is much farther from the Earth than the Moon is, and that the distance from the Earth to the stars is much greater than the Earth’s diameter. Based on this knowledge, Greek cosmologists devised a “two- sphere” model of the universe, in which the spherical Earth is surrounded by a much larger celestial sphere, a spherical shell to which the stars are attached. Between the Earth and the celestial sphere, in this model, the Sun, Moon, and planets move on their complicated apparatus of epicycles and deferents. Although cosmology is ultimately based on observation, sometimes obser- 6 CHAPTER 1. INTRODUCTION vations temporarily lag behind theory. During periods when data are lacking, cosmologists may adopt a new model for aesthetic or philosophical reasons. For instance, when Copernicus proposed a new Sun-centered model of the universe, to replace the Earth-centered two-sphere model of the Greeks, he didn’t base his model on new observational discoveries. Rather, he believed that putting the Earth in motion around the Sun resulted in a conceptually simpler, more appealing model of the universe. Direct observational evidence didn’t reveal that the Earth revolves around the Sun, rather than vice versa, until the discovery of the aberration of starlight in the year 1728, nearly two centuries after the death of Copernicus. Foucault didn’t demonstrate the ro- tation of the Earth, another prediction of the Copernican model, until 1851, over three centuries after the death of Copernicus. However, although obser- vations sometimes lag behind theory in this way, every cosmological model that isn’t eventually supported by observational evidence must remain pure speculation. The current standard model for the universe is the “Hot Big Bang” model, which states that the universe has expanded from an initially hot and dense state to its current relatively cool and tenuous state, and that the expansion is still going on today. To see why cosmologists have embraced the Hot Big Bang model, let us turn, in the next chapter, to the fundamental observations on which modern cosmology is based. Suggested reading [Full references are given in the “Annotated Bibliography” on page 286.] Cox (2000): Accurate values of physical and astronomical constants Harrison (2000), ch. 1 – 4: A history of early (pre-Einstein) cosmology Chapter 2 Fundamental Observations Some of the observations on which modern cosmology is based are highly complex, requiring elaborate apparatus and sophisticated data analysis. How- ever, other observations are surprisingly simple. Let’s start with an observa- tion which is deceptive in its extreme simplicity. 2.1 The night sky is dark Step outside on a clear, moonless night, far from city lights, and look upward. You will see a dark sky, with roughly two thousand stars scattered across it. The fact that the night sky is dark at visible wavelengths, instead of being uniformly bright with starlight, is known as Olbers’ Paradox, after the astronomer Heinrich Olbers, who wrote a scientific paper on the subject in the year 1826. As it happens, Olbers was not the first person to think about Olbers’ Paradox. As early as 1576, Thomas Digges mentioned how strange it is that the night sky is dark, with only a few pinpoints of light to mark the location of stars. 1 Why should it be paradoxical that the night sky is dark? Most of us simply take for granted the fact that daytime is bright and nighttime is dark. The darkness of the night sky certainly posed no problems to the ancient Egyptians or Greeks, to whom stars were points of light stuck to a dome or sphere. However, the cosmological model of Copernicus required that the 1 The name “Olbers’ Paradox” is thus a prime example of what historians of science jokingly call the law of misonomy: nothing is ever named after the person who really discovers it. 7 8 CHAPTER 2. FUNDAMENTAL OBSERVATIONS dr r Figure 2.1: A star-filled spherical shell, of radius : and thickness d:, centered on the Earth. distance to stars be very much larger than an astronomical unit; otherwise, the parallax of the stars, as the Earth goes around on its orbit, would be large enough to see with the naked eye. Moreover, since the Copernican system no longer requires that the stars be attached to a rotating celestial sphere, the stars can be at different distances from the Sun. These liberating realizations led Thomas Digges, and other post-Copernican astronomers, to embrace a model in which stars are large glowing spheres, like the Sun, scattered throughout infinite space. Let’s compute how bright we expect the night sky to be in an infinite universe. Let : be the average number density of stars in the universe, and let 1 be the average stellar luminosity. The flux received here at Earth from a star of luminosity 1 at a distance : is given by an inverse square law: 1(:) = 1 4π: 2 . (2.1) Now consider a thin spherical shell of stars, with radius : and thickness d:, centered on the Earth (Figure 2.1). The intensity of radiation from the shell of stars (that is, the power per unit area per steradian of the sky) will be dJ(:) = 1 4π: 2 : : 2 d: = :1 4π d: . (2.2) 2.1. DARK NIGHT SKY 9 The total intensity of starlight from a shell thus depends only on its thickness, not on its distance from us. We can compute the total intensity of starlight from all the stars in the universe by integrating over shells of all radii: J = ∞ r=0 dJ = :1 4π ∞ 0 d: = ∞ . (2.3) Thus, I have demonstrated that the night sky is infinitely bright. This is utter nonsense. Therefore, one (or more) of the assumptions that went into the above analysis of the sky brightness must be wrong. Let’s scrutinize some of the assumptions. One assumption that I made is that we have an unobstructed line of sight to every star in the universe. This is not true. In fact, since stars have a finite angular size as seen from Earth, nearby stars will hide more distant stars from our view. Nevertheless, in an infinite distribution of stars, every line of sight should end at the surface of a star; this would imply a surface brightness for the sky equal to the surface brightness of a typical star. This is an improvement on an infinitely bright sky, but is still distinctly different from the dark sky which we actually see. Heinrich Olbers himself tried to resolve Olbers’ Paradox by proposing that distant stars are hidden from view by interstellar matter which absorbs starlight. This resolution will not work, because the interstellar matter will be heated by starlight until it has the same temperature as the surface of a star. At that point, the interstellar matter emits as much light as it absorbs, and is glowing as brightly as the stars themselves. A second assumption I made is that the number density : and mean luminosity 1 of stars are constant throughout the universe; more accurately, the assumption made in equation (2.3) is that the product :1 is constant as a function of :. This might not be true. Distant stars might be less luminous or less numerous than nearby stars. If we are in a clump of stars of finite size, then the absence of stars at large distances will keep the night sky from being bright. Similarly, if distant stars are sufficiently low in luminosity compared to nearby stars, they won’t contribute significantly to the sky brightness. In order for the integrated intensity in equation (2.3) to be finite, the product :1 must fall off more rapidly than :1 ∝ 1: as : →∞. A third assumption is that the universe is infinitely large. This might not be true. If the universe only extends to a maximum distance : max from us, then the total intensity of starlight we see in the night sky will be J ∼ 10 CHAPTER 2. FUNDAMENTAL OBSERVATIONS :1: max (4π). Note that this result will also be found if the universe is infinite in space, but is devoid of stars beyond a distance : max . A fourth assumption, slightly more subtle than the previous ones, is that the universe is infinitely old. This might not be true. Since the speed of light is finite, when we look farther out in space, we are looking farther out in time. Thus, we see the Sun as it was 8.3 minutes ago, Proxima Centauri as it was 4 years ago, and M31 as it was 2 million years ago. If the universe has a finite age t 0 , the intensity of starlight we see at night will be at most J ∼ :1ct 0 (4π). Note that this result will also be found if the universe is infinitely old, but has only contained stars for a finite time t 0 . A fifth assumption is that the flux of light from a distant source is given by the inverse square law of equation (2.1). This might not be true. The assumption that 1 ∝ 1: 2 would have seemed totally innocuous to Olbers and other nineteenth century astronomers; after all, the inverse square law follows directly from Euclid’s laws of geometry. However, in the twentieth century, Albert Einstein, that great questioner of assumptions, demonstrated that the universe might not obey the laws of Euclidean geometry. In addition, the inverse square law assumes that the source of light is stationary relative to the observer. If the universe is systematically expanding or contracting, then the light from distant sources will be redshifted to lower photon energies or blueshifted to higher photon energies. Thus, the infinitely large, eternally old, Euclidean universe which Thomas Digges and his successors pictured simply does not hold up to scrutiny. This is a textbook, not a suspense novel, so I’ll tell you right now: the primary resolution to Olbers’ Paradox comes from the fact that the universe has a finite age. The stars beyond some finite distance, called the horizon distance, are invisible to us because their light hasn’t had time to reach us yet. A particularly amusing bit of cosmological trivia is that the first person to hint at the correct resolution of Olbers’ Paradox was Edgar Allen Poe. 2 In his essay “Eureka: A Prose Poem”, completed in the year 1848, Poe wrote, “Were the succession of stars endless, then the background of the sky would present us an [sic] uniform density . . . since there could be absolutely no point, in all that background, at which would not exist a star. The only mode, therefore, in which, under such a state of affairs, we could comprehend the voids which our telescopes find in innumerable directions, would be by supposing the distance of the invisible background so immense that no ray 2 That’s right, the “Nevermore” guy. 2.2. ISOTROPY AND HOMOGENEITY 11 from it has yet been able to reach us at all.” 2.2 On large scales, the universe is isotropic and homogeneous What does it mean to state that the universe is isotropic and homogeneous? Saying that the universe is isotropic means that there are no preferred direc- tions in the universe; it looks the same no matter which way you point your telescope. Saying that the universe is homogeneous means that there are no preferred locations in the universe; it looks the same no matter where you set up your telescope. Note the very important qualifier: the universe is isotropic and homogeneous on large scales. In this context, “large scales” means that the universe is only isotropic and homogeneous on scales of roughly 100 Mpc or more. The isotropy of the universe is not immediately obvious. In fact, on small scales, the universe is blatantly anisotropic. Consider, for example, a sphere 3 meters in diameter, centered on your navel (Figure 2.2a). Within this sphere, there is a preferred direction; it is the direction commonly referred to as “down”. It is easy to determine the vector pointing down. Just let go of a small dense object. The object doesn’t hover in midair, and it doesn’t move in a random direction; it falls down, toward the center of the Earth. On significantly larger scales, the universe is still anisotropic. Consider, for example, a sphere 3 AU in diameter, centered on your navel (Figure 2.2b). Within this sphere, there is a preferred direction; it is the direction pointing toward the Sun, which is by far the most massive and most luminous object within the sphere. It is easy to determine the vector pointing toward the Sun. Just step outside on a sunny day, and point to that really bright disk of light up in the sky. On still large scales, the universe is still anisotropic. Consider, for ex- ample, a sphere 3 Mpc in diameter, centered on your navel (Figure 2.2c). This sphere contains the Local Group of galaxies, a small cluster of some 40 galaxies. By far the most massive and most luminous galaxies in the Local Group are our own Galaxy and M31, which together contribute about 86% of the total luminosity within the 3 Mpc sphere. Thus, within this sphere, our Galaxy and M31 define a preferred direction. It is fairly easy to determine the vector pointing from our Galaxy to M31; just step outside on a clear 12 CHAPTER 2. FUNDAMENTAL OBSERVATIONS Figure 2.2: (a) A sphere 3 meters in diameter, centered on your navel. (b) A sphere 3 AU in diameter, centered on your navel. (c) A sphere 3 Mpc in diameter, centered on your navel. (d) A sphere 200 Mpc in diameter, centered on your navel. Shown is the number density of galaxies smoothed with a Gaussian of width 17 Mpc. The heavy contour is drawn at the mean density; darker regions represent higher density, lighter regions represent lower density (from Dekel et al. 1999, ApJ, 522, 1, fig. 2) 2.2. ISOTROPY AND HOMOGENEITY 13 night when the constellation Andromeda is above the horizon, and point to the fuzzy oval in the middle of the constellation. It isn’t until you get to considerably larger scales that the universe can be considered as isotropic. Consider a sphere 200 Mpc in diameter, centered on your navel. Figure 2.2d shows a slice through such a sphere, with superclus- ters of galaxies indicated as dark patches. The Perseus-Pisces supercluster is on the right, the Hydra-Centaurus supercluster is on the left, and the edge of the Coma supercluster is just visible at the top of Figure 2.2d. Superclusters are typically ∼ 100 Mpc along their longest dimensions, and are separated by voids (low density regions) which are typically ∼ 100 Mpc across. These are the largest structures in the universe, it seems; surveys of the universe on still larger scales don’t find “superduperclusters”. On small scales, the universe is obviously inhomogeneous, or lumpy, in addition to being anisotropic. For instance, a sphere 3 meters in diameter, centered on your navel, will have an average density of ∼ 100 kg m −3 , in round numbers. However, the average density of the universe as a whole is ρ 0 ∼ 3 10 −27 kg m −3 . Thus, on a scale d ∼ 3 m, the patch of the universe surrounding you is more than 28 orders of magnitude denser than average. On significantly larger scales, the universe is still inhomogeneous. A sphere 3 AU in diameter, centered on your navel, has an average density of 410 −5 kg m −3 ; that’s 22 orders of magnitude denser than the average for the universe. On still larger scales, the universe is still inhomogeneous. A sphere 3 Mpc in diameter, centered on your navel, will have an average density of ∼ 310 −26 kg m −3 , still an order of magnitude denser than the universe as a whole. It’s only when you contemplate a sphere ∼ 100 Mpc in diameter that a sphere centered on your navel is not overdense compared to the universe as a whole. Note that homogeneity does not imply isotropy. A sheet of paper printed with stripes (Figure 2.3a) is homogeneous on scales larger than the stripe width, but it is not isotropic. The direction of the stripes provides a preferred direction by which you can orient yourself. Note also that isotropy around a single point does not imply homogeneity. A sheet of paper printed with a bullseye (Figure 2.3b) is isotropic around the center of the bullseye, but is it not homogeneous. The rings of the bullseye look different far from the center than they do close to the center. You can tell where you are relative to the center by measuring the radius of curvature of the nearest ring. In general, then, saying that something is homogeneous is quite different 14 CHAPTER 2. FUNDAMENTAL OBSERVATIONS Figure 2.3: (a) A pattern which is anisotropic, but which is homogeneous on scales larger than the stripe width. (b) A pattern which is isotropic about the origin, but which is inhomogeneous. from saying it is isotropic. However, modern cosmologists have adopted the cosmological principle, which states “There is nothing special about our location in the universe.” The cosmological principle holds true only on large scales (of 100 Mpc or more). On smaller scales, your navel obviously is in a special location. Most spheres 3 meters across don’t contain a sentient being; most sphere 3 AU across don’t contain a star; most spheres 3 Mpc across don’t contain a pair of bright galaxies. However, most spheres over 100 Mpc across do contain roughly the same pattern of superclusters and voids, statistically speaking. The universe, on scales of 100 Mpc or more, appears to be isotropic around us. Isotropy around any point in the universe, such as your navel, combined with the cosmological principle, implies isotropy around every point in the universe; and isotropy around every point in the universe does imply homogeneity. The cosmological principle has the alternate name of the “Copernican principle” as a tribute to Copernicus, who pointed out that the Earth is not the center of the universe. Later cosmologists also pointed out the Sun is not the center, that our Galaxy is not the center, and that the Local Group is not the center. In fact, there is no center to the universe. 2.3. REDSHIFT PROPORTIONAL TO DISTANCE 15 2.3 Galaxies show a redshift proportional to their distance When we look at a galaxy at visible wavelengths, we are primarily detecting the light from the stars which the galaxy contains. Thus, when we take a galaxy’s spectrum at visible wavelengths, it typically contains absorption lines created in the stars’ relatively cool upper atmospheres. 3 Suppose we consider a particular absorption line whose wavelength, as measured in a laboratory here on Earth, is λ em . The wavelength we measure for the same absorption line in a distant galaxy’s spectrum, λ ob , will not, in general, be the same. We say that the galaxy has a redshift ., given by the formula . ≡ λ ob −λ em λ em . (2.4) Strictly speaking, when . < 0, this quantity is called a blueshift, rather than a redshift. However, the vast majority of galaxies have . 0. The fact that the light from galaxies is generally redshifted to longer wavelengths, rather than blueshifted to shorter wavelengths, was not known until the twentieth century. In 1912, Vesto Slipher, at the Lowell Observatory, measured the shift in wavelength of the light from M31; this galaxy, as it turns out, is one of the few which exhibits a blueshift. By 1925, Slipher had measured the shifts in the spectral lines for approximately 40 galaxies, finding that they were nearly all redshifted; the exceptions were all nearby galaxies within the Local Group. By 1929, enough galaxy redshifts had been measured for the cosmologist Edwin Hubble to make a study of whether a galaxy’s redshift depends on its distance from us. Although measuring a galaxy’s redshift is relatively easy, and can be done with high precision, measuring its distance is difficult. Hubble knew . for nearly 50 galaxies, but had estimated distances for only 20 of them. Nevertheless, from a plot of redshift (.) versus distance (:), reproduced in Figure 2.4, he found the famous linear relation now known as Hubble’s Law: . = H 0 c : . (2.5) where H 0 is a constant (now called the Hubble constant). Hubble interpreted the observed redshift of galaxies as being a Doppler shift due to their radial 3 Galaxies containing active galactic nuclei will also show emission lines from the hot gas in their nuclei. 16 CHAPTER 2. FUNDAMENTAL OBSERVATIONS Figure 2.4: Edwin Hubble’s original plot of the relation between redshift (vertical axis) and distance (horizontal axis). Note that the vertical axis actually plots c. rather than . – and that the units are accidentally written as km rather than km/s. (from Hubble 1929, Proc. Nat. Acad. Sci., 15, 168) 2.3. REDSHIFT PROPORTIONAL TO DISTANCE 17 Figure 2.5: A more modern version of Hubble’s plot, showing c. versus distance. In this case, the galaxy distances have been determined using Cepheid variable stars as standard candles, as described in Chapter 6. (from Freedman, et al. 2001, ApJ, 553, 47) velocity away from Earth. Since the values of . in Hubble’s analysis were all small (. < 0.04), he was able to use the classical, nonrelativistic relation for the Doppler shift, . = ·c, where · is the radial velocity of the light source (in this case, a galaxy). Interpreting the redshifts as Doppler shifts, Hubble’s law takes the form · = H 0 : . (2.6) Since the Hubble constant H 0 can be found by dividing velocity by distance, it is customarily written in the rather baroque units of kms −1 Mpc −1 . When Hubble first discovered Hubble’s Law, he thought that the numerical value of the Hubble constant was H 0 = 500 kms −1 Mpc −1 (see Figure 2.4). However, it turned out that Hubble was severely underestimating the distances to galaxies. Figure 2.5 shows a more recent determination of the Hubble constant from nearby galaxies, using data obtained by (appropriately enough) the 18 CHAPTER 2. FUNDAMENTAL OBSERVATIONS 1 2 3 r 12 r 23 r 31 Figure 2.6: A triangle defined by three galaxies in a uniformly expanding universe. Hubble Space Telescope. The best current estimate of the Hubble constant, combining the results of different research groups, is H 0 = 70 ±7 kms −1 Mpc −1 . (2.7) This is the value for the Hubble constant that I will use in the remainder of this book. Cosmological innocents sometimes exclaim, when first encountering Hub- ble’s Law, “Surely it must be a violation of the cosmological principle to have all those distant galaxies moving away from us! It looks as if we are at a special location in the universe – the point away from which all other galaxies are fleeing.” In fact, what we see here in our Galaxy is exactly what you would expect to see in a universe which is undergoing homogeneous and isotropic expansion. We see distant galaxies moving away from us; but ob- servers in any other galaxy would also see distant galaxies moving away from them. To see on a more mathematical level what we mean by homogeneous, isotropic expansion, consider three galaxies at positions : 1 , : 2 , and : 3 . They define a triangle (Figure 2.6) with sides of length : 12 ≡ [: 1 −: 2 [ (2.8) : 23 ≡ [: 2 −: 3 [ (2.9) : 31 ≡ [: 3 −: 1 [ . (2.10) 2.3. REDSHIFT PROPORTIONAL TO DISTANCE 19 Homogeneous and uniform expansion means that the shape of the triangle is preserved as the galaxies move away from each other. Maintaining the correct relative lengths for the sides of the triangle requires an expansion law of the form : 12 (t) = c(t): 12 (t 0 ) (2.11) : 23 (t) = c(t): 23 (t 0 ) (2.12) : 31 (t) = c(t): 31 (t 0 ) . (2.13) Here the function c(t) is a scale factor which is equal to one at the present moment (t = t 0 ) and which is totally independent of location or direction. The scale factor c(t) tells us how the expansion (or possibly contraction) of the universe depends on time. At any time t, an observer in galaxy 1 will see the other galaxies receding with a speed · 12 (t) = d: 12 dt = ˙ c: 12 (t 0 ) = ˙ c c : 12 (t) (2.14) · 31 (t) = d: 31 dt = ˙ c: 31 (t 0 ) = ˙ c c : 31 (t) . (2.15) You can easily demonstrate that an observer in galaxy 2 or galaxy 3 will find the same linear relation between observed recession speed and distance, with ˙ cc playing the role of the Hubble constant. Since this argument can be applied to any trio of galaxies, it implies that in any universe where the distribution of galaxies is undergoing homogeneous, isotropic expansion, the velocity – distance relation takes the linear form · = H:, with H = ˙ cc. If galaxies are currently moving away from each other, this implies they were closer together in the past. Consider a pair of galaxies currently sep- arated by a distance :, with a velocity · = H 0 : relative to each other. If there are no forces acting to accelerate or decelerate their relative motion, then their velocity is constant, and the time that has elapsed since they were in contact is t 0 = : · = : H 0 : = H −1 0 . (2.16) independent of the current separation :. The time H −1 0 is referred to as the Hubble time. For H 0 = 70 ± 7 kms −1 Mpc −1 , the Hubble time is H −1 0 = 14.0 ±1.4 Gyr. If the relative velocities of galaxies have been constant in the past, then one Hubble time ago, all the galaxies in the universe were crammed together into a small volume. Thus, the observation of galactic redshifts lead 20 CHAPTER 2. FUNDAMENTAL OBSERVATIONS naturally to a Big Bang model for the evolution of the universe. A Big Bang model may be broadly defined as a model in which the universe expands from an initially highly dense state to its current low-density state. The Hubble time of ∼ 14 Gyr is comparable to the ages computed for the oldest known stars in the universe. This rough equivalence is reassuring. However, the age of the universe – that is, the time elapsed since its original highly dense state – is not necessarily exactly equal to the Hubble time. We know that gravity exists, and that galaxies contain matter. If gravity working on matter is the only force at work on large scales, then the attractive force of gravity will act to slow the expansion. In this case, the universe was expanding more rapidly in the past than it is now, and the universe is younger than H −1 0 . On the other hand, if the energy density of the universe is dominated by a cosmological constant (an entity which we’ll examine in more detail in Chapter 4), then the dominant gravitational force is repulsive, and the universe may be older than H −1 0 . Just as the Hubble time provides a natural time scale for our universe, the Hubble distance, cH 0 = 4300 ± 400 Mpc, provides a natural distance scale. Just as the age of the universe is roughly equal to H −1 0 in most Big Bang models, with the exact value depending on the expansion history of the universe, so the horizon distance (the greatest distance a photon can travel during the age of the universe) is roughly equal to cH 0 , with the exact value, again, depending on the expansion history. (Later chapters will deal with computing the exact values of the age and horizon size of our universe.) Note how Hubble’s Law ties in with Olbers’ Paradox. If the universe is of finite age, t 0 ∼ H −1 0 , then the night sky can be dark, even if the universe is infinitely large, because light from distant galaxies has not yet had time to reach us. Galaxy surveys tell us that the luminosity density of galaxies in the local universe is :1 ≈ 2 10 8 L Mpc −3 . (2.17) By terrestrial standards, the universe is not a well-lit place; this luminosity density is equivalent to a single 40 watt light bulb within a sphere 1 AU in radius. If the horizon distance is d hor ∼ cH 0 , then the total flux of light we receive from all the stars from all the galaxies within the horizon will be 1 gal = 4πJ gal ≈ :1 r H 0 d: ∼ :1 c H 0 ∼ 9 10 11 L Mpc −2 ∼ 2 10 −11 L AU −2 . (2.18) 2.3. REDSHIFT PROPORTIONAL TO DISTANCE 21 By the cosmological principle, this is the total flux of starlight you would expect at any randomly located spot in the universe. Comparing this to the flux we receive from the Sun, 1 sun = 1 L 4π AU 2 ≈ 0.08 L AU −2 . (2.19) we find that 1 gal 1 sun ∼ 3 10 −10 . Thus, the total flux of starlight at a randomly selected location in the universe is less than a billionth the flux of light we receive from the Sun here on Earth. For the entire universe to be as well-lit as the Earth, it would have to be over a billion times older than it is; and you’d have to keep the stars shining during all that time. Hubble’s Law occurs naturally in a Big Bang model for the universe, in which homogeneous and isotropic expansion causes the density of the universe to decrease steadily from its initial high value. In a Big Bang model, the properties of the universe evolve with time; the average density decreases, the mean distance between galaxies increases, and so forth. However, Hubble’s Law can also be explained by a Steady State model. The Steady State model was first proposed in the 1940’s by Hermann Bondi, Thomas Gold, and Fred Hoyle, who were proponents of the perfect cosmological principle, which states that not only are there no privileged locations in space, there are no privileged moments in time. Thus, a Steady State universe is one in which the global properties of the universe, such as the mean density ρ 0 and the Hubble constant H 0 , remain constant with time. In a Steady State universe, the velocity – distance relation d: dt = H 0 : (2.20) can be easily integrated, since H 0 is constant with time, to yield an expo- nential law: :(t) ∝ c H 0 t . (2.21) Note that : → 0 only in the limit t → −∞; a Steady State universe is infinitely old. If there existed an instant in time at which the universe started expanding (as in a Big Bang model), that would be a special moment, in violation of the assumed “perfect cosmological principle”. The volume of a spherical region of space, in a Steady State model, increases exponentially with time: \ = 4π 3 : 3 ∝ c 3H 0 t . (2.22) 22 CHAPTER 2. FUNDAMENTAL OBSERVATIONS However, if the universe is in a steady state, the density of the sphere must remain constant. To have a constant density of matter within a growing volume, matter must be continuously created at a rate ˙ ` ss = ρ 0 ˙ \ = ρ 0 3H 0 \ . (2.23) If our own universe, with matter density ρ 0 ∼ 3 10 −27 kg m −3 , happened to be a Steady State universe, then matter would have to be created at a rate ˙ ` ss \ = 3H 0 ρ 0 ∼ 6 10 −28 kg m −3 Gyr −1 . (2.24) This corresponds to creating roughly one hydrogen atom per cubic kilometer per year. During the 1950s and 1960s, the Big Bang and Steady State models bat- tled for supremacy. Critics of the Steady State model pointed out that the continuous creation of matter violates mass-energy conservation. Supporters of the Steady State model pointed out that the continuous creation of matter is no more absurd that the instantaneous creation of the entire universe in a single “Big Bang” billions of years ago. 4 The Steady State model finally fell out of favor when observational evidence increasingly indicated that the per- fect cosmological principle is not true. The properties of the universe do, in fact, change with time. The discovery of the Cosmic Microwave Background, discussed below in section 2.5, is commonly regarded as the observation which decisively tipped the scales in favor of the Big Bang model. 2.4 The universe contains different types of particles It doesn’t take a brilliant observer to confirm that the universe contains a large variety of different things: ships, shoes, sealing wax, cabbages, kings, galaxies, and what have you. From a cosmologist’s viewpoint, though, cab- bages and kings are nearly indistinguishable – the main difference between them is that the mean mass per king is greater than the mean mass per cab- bage. From a cosmological viewpoint, the most significant difference between the different components of the universe is that they are made of different elementary particles. The properties of the most cosmologically important 4 The name “Big Bang” was actually coined by Fred Hoyle, a supporter of the Steady State model. 2.4. TYPES OF PARTICLES 23 Table 2.1: Particle Properties particle symbol rest energy (MeV) charge proton j 938.3 +1 neutron : 939.6 0 electron c − 0.511 -1 neutrino ν e ,ν µ ,ν τ ? 0 photon γ 0 0 dark matter ? ? 0 particles are summarized in Table 2.1. The material objects which surround us during our everyday life are made of protons, neutrons, and electrons. 5 Protons and neutrons are both examples of baryons, where a baryon is defined as a particle made of three quarks. A proton (j) contains two “up” quarks, each with an electrical charge of +23, and a “down” quark, with charge −13. A neutron (:) contains one “up” quark and two “down” quarks. Thus a proton has a net positive charge of +1, while a neutron is electrically neutral. Protons and neutrons also differ in their mass – or equivalently, in their rest energies. The proton mass is : p c 2 = 938.3 MeV, while the neutron mass is : n c 2 = 939.6 MeV, about 0.1% greater. Free neutrons are unstable, decaying into protons with a decay time of τ n = 940 s, about a quarter of an hour. By contrast, experiments have put a lower limit on the decay time of the proton which is very much greater than the Hubble time. Neutrons can be preserved against decay by binding them into an atomic nucleus with one or more protons. Electrons (c − ) are examples of leptons, a class of elementary particles which are not made of quarks. The mass of an electron is much smaller than that of a neutron or proton; the rest energy of an electron is : e c 2 = 0.511 MeV. An electron has an electric charge equal in magnitude to that of a proton, but opposite in sign. On large scales, the universe is electrically neutral; the number of electrons is equal to the number of protons. Since pro- tons outmass electrons by a factor of 1836 to 1, the mass density of electrons is only a small perturbation to the mass density of protons and neutrons. For this reason, the component of the universe made up of ions, atoms, and molecules is generally referred to as baryonic matter, since only the baryons (protons and neutrons) contribute significantly to the mass density. Protons 5 For that matter, we ourselves are made of protons, neutrons, and electrons. 24 CHAPTER 2. FUNDAMENTAL OBSERVATIONS and neutrons are 800-pound gorillas; electrons are only 7-ounce bushbabies. About three-fourths of the baryonic matter in the universe is currently in the form of ordinary hydrogen, the simplest of all elements. In addition, when we look at the remainder of the baryonic matter, it is primarily in the form of helium, the next simplest element. The Sun’s atmosphere, for instance, contains 70% hydrogen by mass, and 28% helium; only 2% is contributed by more massive atoms. When astronomers look at a wide range of astronomical objects – stars and interstellar gas clouds, for instance – they find a minimum helium mass fraction of 24%. The baryonic component of the universe can be described, to lowest order, as a mix of three parts hydrogen to one part helium, with only minor contamination by heavier elements. Another type of lepton, in addition to the electron, is the neutrino (ν). The most poetic summary of the properties of the neutrino was made by John Updike, in his poem “Cosmic Gall” 6 : Neutrinos, they are very small. They have no charge and have no mass And do not interact at all. The earth is just a silly ball To them, through which they simply pass, Like dustmaids down a drafty hall Or photons through a sheet of glass. In truth, Updike was using a bit of poetic license here. It is definitely true that neutrinos have no charge. 7 However, it is not true that neutrinos “do not interact at all”; they actually are able to interact with other particles via the weak nuclear force. The weak nuclear force, though, is very weak indeed; a typical neutrino emitted by the Sun would have to pass through a few parsecs of solid lead before having a 50% chance of interacting with a lead atom. Since neutrinos pass through neutrino detectors with the same facility with which they pass through the Earth, detecting neutrinos from astronomical sources is difficult. There are three types, or “flavors”, of neutrinos: electron neutrinos, muon neutrinos, and tau neutrinos. What Updike didn’t know in 1960, when he wrote his poem, is that some or all of the neutrino types probably have a small 6 From COLLECTED POEMS 1953-1993 by John Updike, c 1993 by John Updike. Used by permission of Alfred A. Knopf, a division of Random House, Inc. 7 Their name, given them by Enrico Fermi, means “little neutral one” in Italian. 2.4. TYPES OF PARTICLES 25 mass. The evidence for massive neutrinos comes indirectly, from the search for neutrino oscillations. An “oscillation” is the transmutation of one flavor of neutrino into another. The rate at which two neutrino flavors oscillate is proportional to the difference of the squares of their masses. Observations of neutrinos from the Sun are most easily explained if electron neutrinos (the flavor emitted by the Sun) oscillate into some other flavor of neutrino, with the difference in the squares of their masses being ∆(: 2 ν c 4 ) ≈ 5 10 −5 eV 2 . Observations of muon neutrinos created by cosmic rays striking the upper atmosphere indicate that muon neutrinos oscillate into tau neutrinos, with ∆(: 2 ν c 4 ) ≈ 3 10 −3 eV 2 for these two flavors. Unfortunately, knowing the differences of the squares of the masses doesn’t tell us the values of the masses themselves. A particle which is known to be massless is the photon. Electromagnetic radiation can be thought of either as a wave or as a stream of particles, called photons. Light, when regarded as a wave, is characterized by its frequency 1 or its wavelength λ = c1. When light is regarded as a stream of photons, each photon is characterized by its energy, 1 γ = /1, where / = 2π¯ / is the Planck constant. Photons of a wide range of energy, from radio to gamma rays, pervade the universe. Unlike neutrinos, photons interact readily with electrons, protons, and neutrons. For instance, photons can ionize an atom by kicking an electron out of its orbit, a process known as photoionization. Higher energy photons can break an atomic nucleus apart, a process known as photodissociation. Photons, in general, are easily created. One way to make photons is to take a dense, opaque object – such as the filament of an incandescent light- bulb – and heat it up. If an object is opaque, then the protons, neutrons, electrons, and photons which it contains frequently interact, and attain ther- mal equilibrium. When a system is in thermal equilibrium, the density of photons in the system, as a function of photon energy, depends only on the temperature 1. It doesn’t matter whether the system is a tungsten filament, or an ingot of steel, or a sphere of ionized hydrogen and helium. The en- ergy density of photons in the frequency range 1 → 1 + d1 is given by the blackbody function ε(1)d1 = 8π/ c 3 1 3 d1 exp(/1/1) −1 . (2.25) illustrated in Figure 2.7. The peak in the blackbody function occurs at /1 peak ≈ 2.82/1. Integrated over all frequencies, equation (2.25) yields a 26 CHAPTER 2. FUNDAMENTAL OBSERVATIONS 0 2 4 6 8 10 12 0 .5 1 1.5 hf/kT ( c 3 h 2 / 8 π k 3 T 3 ) ε Figure 2.7: The energy distribution of a blackbody spectrum. total energy density for blackbody radiation of ε γ = α1 4 . (2.26) where α = π 2 15 / 4 ¯ / 3 c 3 = 7.56 10 −16 J m −3 K −4 . (2.27) The number density of photons in blackbody radiation can be computed from equation (2.25) as : γ = β1 3 . (2.28) where β = 2.404 π 2 / 3 ¯ / 3 c 3 = 2.03 10 7 m −3 K −3 . (2.29) Division of equation (2.26) by equation (2.28) yields a mean photon energy of 1 mean = /1 mean ≈ 2.70/1, close to the peak in the spectrum. You have a tem- perature of 310 K, and you radiate an approximate blackbody spectrum, with a mean photon energy of 1 mean ≈ 0.072 eV, corresponding to a wavelength of λ ≈ 1.7 10 −5 m, in the infrared. By contrast, the Sun produces an approx- imate blackbody spectrum with a temperature 1 ≈ 5800 K. This implies 2.4. TYPES OF PARTICLES 27 a mean photon energy 1 mean ≈ 1.3 eV, corresponding to λ ≈ 9.0 10 −7 m, in the near infrared. Note, however, that although the mean photon energy in a blackbody spectrum is ∼ 3/1, Figure 2.7 shows us that there is a long exponential tail to higher photon energies. A large fraction of the Sun’s out- put is at wavelengths of (4 → 7) 10 −7 m, which our eyes are equipped to detect. The most mysterious component of the universe is dark matter. When observational astronomers refer to dark matter, they usually mean any mas- sive component of the universe which is too dim to be detected readily using current technology. Thus, stellar remnants such as white dwarfs, neutron stars, and black holes are sometimes referred to as dark matter, since an isolated stellar remnant is extremely faint and difficult to detect. Substellar objects such as brown dwarfs are also referred to as dark matter, since brown dwarfs, too low in mass for nuclear fusion to occur in their cores, are very dim. Theoretical astronomers sometimes use a more stringent definition of dark matter than observers do, defining dark matter as any massive compo- nent of the universe which doesn’t emit, absorb, or scatter light at all. 8 If neutrinos have mass, for instance, as the recent neutrino oscillation results indicate, they qualify as dark matter. In some extensions to the Standard Model of particle physics, there exist massive particles which interact, like neutrinos, only through the weak nuclear force and through gravity. These particles, which have not yet been detected in the laboratory, are generically referred to as Weakly Interacting Massive Particles, or WIMPs. In this book, I will generally adopt the broader definition of dark matter as something which is too dim for us to see, even with our best available tech- nology. Detecting dark matter is, naturally, difficult. The standard method of detecting dark matter is by measuring its gravitational effect on luminous matter, just as the planet Neptune was first detected by its gravitational effect on the planet Uranus. Although Neptune no longer qualifies as dark matter, observations of the motions of stars within galaxies and of galaxies within clusters indicate that there’s a significant amount of dark matter in the universe. Exactly how much there is, and what it’s made of, is a topic of great interest to cosmologists. 8 Using this definition, an alternate name for dark matter might be “transparent matter” or “invisible matter”. However, the name “dark matter” has received the sanction of history. 28 CHAPTER 2. FUNDAMENTAL OBSERVATIONS 2.5 The universe is filled with a Cosmic Mi- crowave Background The discovery of the Cosmic Microwave Background (CMB) by Arno Pen- zias and Robert Wilson in 1965 has entered cosmological folklore. Using a microwave antenna at Bell Labs, they found an isotropic background of mi- crowave radiation. More recently, the Cosmic Background Explorer (COBE) satellite has revealed that the Cosmic Microwave Background is exquisitely well fitted by a blackbody spectrum (equation 2.25) with a temperature 1 0 = 2.725 ±0.001 K . (2.30) The energy density of the CMB is, from equation (2.26), ε γ = 4.17 10 −14 J m −3 . (2.31) This is equivalent to roughly a quarter of an MeV per cubic meter of space. The number density of CMB photons is, from equation (2.28), : γ = 4.11 10 8 m −3 . (2.32) Thus, there are about 411 CMB photons in every cubic centimeter of the universe at the present day. The mean energy of CMB photons, however, is quite low, only 1 mean = 6.34 10 −4 eV . (2.33) This is too low in energy to photoionize an atom, much less photodissociate a nucleus. About all they do, from a terrestrial point of view, is cause static on television. The mean CMB photon energy corresponds to a wavelength of 2 millimeters, in the microwave region of the electromagnetic spectrum – hence the name “Cosmic Microwave Background”. The existence of the CMB is a very important cosmological clue. In particular, it is the clue which caused the Big Bang model for the universe to be favored over the Steady State model. In a Steady State universe, the existence of blackbody radiation at 2.725 K is not easily explained. In a Big Bang universe, however, a cosmic background radiation arises naturally if the universe was initially very hot as well as being very dense. If mass is conserved in an expanding universe, then in the past, the universe was denser than it is now. Assume that the early dense universe was very hot (1 10 4 K, or 2.5. COSMIC MICROWAVE BACKGROUND 29 /1 1c\ ). At such high temperatures, the baryonic matter in the universe was completely ionized, and the free electrons rendered the universe opaque. A dense, hot, opaque body, as described in Section 2.4, produces blackbody radiation. So, the early hot dense universe was full of photons, banging off the electrons like balls in a pinball machine, with a spectrum typical of a blackbody (equation 2.25). However, as the universe expanded, it cooled. When the temperature dropped to ∼ 3000 K, ions and electrons combined to form neutral atoms. When the universe no longer contained a significant number of free electrons, the blackbody photons started streaming freely through the universe, without further scattering off free electrons. The blackbody radiation that fills the universe today can be explained as a relic of the time when the universe was sufficiently hot and dense to be opaque. However, at the time the universe became transparent, its tem- perature was ∼ 3000 K. The temperature of the CMB today is 2.725 K, a factor of 1100 lower. The drop in temperature of the blackbody radiation is a direct consequence of the expansion of the universe. Consider a region of volume \ which expands at the same rate as the universe, so that \ ∝ c(t) 3 . The blackbody radiation in the volume can be thought as a photon gas with energy density ε γ = α1 4 . Moreover, since the photons in the volume have momentum as well as energy, the photon gas has a pressure; the pressure of a photon gas is 1 γ = ε γ 3. The photon gas within our imaginary box must follow the laws of thermodynamics; in particular, the boxful of photons must obey the first law dC = d1 +1d\ . (2.34) where dC is the amount of heat flowing into or out of the photon gas in the volume \ , d1 is the change in the internal energy, 1 is the pressure, and d\ is the change in volume of the box. Since, in a homogeneous universe, there is no net flow of heat (everything’s the same temperature, after all), dC = 0. Thus, the first law of thermodynamics, applied to an expanding homogeneous universe, is d1 dt = −1(t) d\ dt . (2.35) Since, for the photons of the CMB, 1 = ε γ \ = α1 4 \ and 1 = 1 γ = α1 4 3, equation (2.35) can be rewritten in the form α 41 3 d1 dt \ +1 4 d\ dt = − 1 3 α1 4 d\ dt . (2.36) 30 CHAPTER 2. FUNDAMENTAL OBSERVATIONS or 1 1 d1 dt = − 1 3\ d\ dt . (2.37) However, since \ ∝ c(t) 3 as the box expands, this means that the rate in change of the photons’ temperature is related to the rate of expansion of the universe by the relation d dt (ln 1) = − d dt (ln c) . (2.38) This implies the simple relation 1(t) ∝ c(t) −1 ; the temperature of the cosmic background radiation has dropped by a factor of 1100 since the universe became transparent because the scale factor c(t) has increased by a factor of 1100 since then. What we now see as a Cosmic Microwave Background was once, at the time the universe became transparent, a Cosmic Near-Infrared Background, with a temperature slightly cooler than the surface of the star Betelgeuse. The evidence cited so far can all be explained within the framework of a Hot Big Bang model, in which the universe was originally very hot and very dense, and since then has been expanding and cooling. The remainder of this book will be devoted to working out the details of the Hot Big Bang model which best fits the universe in which we live. Suggested reading [Full references are given in the “Annotated Bibliography” on page 286.] Bernstein (1995): The “Micropedia” which begins this text is a useful overview of the contents of the universe and the forces which work on them Harrison (1987): The definitive treatment of Olbers’ paradox Problems (2.1) Suppose that in Sherwood Forest, the average radius of a tree is 1 = 1 m and the average number of trees per unit area is Σ = 0.005 m −2 . If Robin Hood shoots an arrow in a random direction, how far, on average, will it travel before it strikes a tree? 2.5. COSMIC MICROWAVE BACKGROUND 31 (2.2) Suppose you are in an infinitely large, infinitely old universe in which the average density of stars is : = 10 9 Mpc −3 and the average stellar radius is equal to the Sun’s radius: 1 = 1 = 7 10 8 m. How far, on average, could you see in any direction before your line of sight struck a star? (Assume standard Euclidean geometry holds true in this universe.) If the stars are clumped into galaxies with a density : g = 1 Mpc −3 and average radius 1 g = 2000 pc, how far, on average, could you see in any direction before your line of sight hit a galaxy? (2.3) Since you are made mostly of water, you are very efficient at absorbing microwave photons. If you were in intergalactic space, approximately how many CMB photons would you absorb per second? (If you like, you may assume you are spherical.) What is the approximate rate, in watts, at which you would absorb radiative energy from the CMB? Ignoring other energy inputs and outputs, how long would it take the CMB to raise your temperature by one nanoKelvin (10 −9 K)? (You may assume your heat capacity is the same as pure water, ( = 4200 J kg −1 K −1 .) (2.4) Suppose that the difference between the square of the mass of the electron neutrino and that of the muon neutrino has the value [:(ν µ ) 2 − :(ν e ) 2 ]c 4 = 5 10 −5 eV 2 , and that the difference between the square of the mass of the muon neutrino and that of the tau neutrino has the value [:(ν τ ) 2 −:(ν µ ) 2 ]c 4 = 3 10 −3 eV 2 . (This is consistent with the observational results discussed in section 2.4.) What values of :(ν e ), :(ν µ ), and :(ν τ ) minimize the sum :(ν e )+:(ν µ )+:(ν τ ), given these constraints? (2.5) A hypothesis once used to explain the Hubble relation is the “tired light hypothesis”. The tired light hypothesis states that the universe is not expanding, but that photons simply lose energy as they move through space (by some unexplained means), with the energy loss per unit distance being given by the law d1 d: = −11 . (2.39) where 1 is a constant. Show that this hypothesis gives a distance- redshift relation which is linear in the limit . <1. What must the value of 1 be in order to yield a Hubble constant of H 0 = 70 kms −1 Mpc −1 ? Chapter 3 Newton Versus Einstein On cosmological scales (that is, on scales greater than 100 Mpc or so), the dominant force determining the evolution of the universe is gravity. The weak and strong nuclear forces are short-range forces; the weak force is effective only on scales of / w ∼ 10 −18 m or less, and the strong force on scales of / s ∼ 10 −15 m or less. Both gravity and electromagnetism are long range forces. On small scales, gravity is negligibly small compared to electromagnetic forces; for instance, the electrostatic repulsion between a pair of protons is larger by a factor ∼ 10 36 than the gravitational attraction between them. However, on large scales, the universe is electrically neutral, so there are no electrostatic forces on large scales. Moreover, intergalactic magnetic fields are sufficiently small that magnetic forces are also negligibly tiny on cosmological scales. Ironically then, gravity – the weakest of all forces from a particle physics standpoint – is the force which determines the evolution of the universe on large scales. Note that in referring to gravity as a force, I am implicitly adopting a Newtonian viewpoint. In physics, there are two useful ways of looking at gravity – the Newtonian, or classical, viewpoint and the Einsteinian, or general relativistic, viewpoint. In Isaac Newton’s view, as formulated by his Laws of Motion and Law of Gravity, gravity is a force which causes massive bodies to be accelerated. By contrast, in Einstein’s view, gravity is a manifestation of the curvature of space-time. Although Newton’s view and Einstein’s view are conceptually very different, in most cosmological contexts they yield the same predictions. The Newtonian predictions differ significantly from the predictions of general relativity only in the limit of deep potential minima (to use Newtonian language) or strong spatial curvature (to 32 3.1. EQUIVALENCE PRINCIPLE 33 use general relativistic language). In these limits, general relativity yields the correct result. In the limit of shallow potential minima and weak spatial curvature, it is permissible to switch back and forth between a Newtonian and a general relativistic viewpoint, adopting whichever one is more convenient. I will frequently adopt the Newtonian view of gravity in this book because, in many contexts, it is mathematically simpler and conceptually more familiar. The question of why it is possible to switch back and forth between the two very different viewpoints of Newton and Einstein is an intriguing one, and deserves closer investigation. 3.1 Equivalence principle In Newton’s view of the universe, space is unchanging and Euclidean. In Euclidean, or “flat”, space, all the axioms and theorems of plane geometry (as codified by Euclid in the third century BC) hold true. In Euclidean space, the shortest distance between two points is a straight line, the angles at the vertices of a triangle sum to π radians, the circumference of a circle is 2π times its radius, and so on, through all the other axioms and theorems you learned in high school geometry. In Newton’s view, moreover, an object with no net force acting on it moves in a straight line at constant speed. However, when we look at objects in the Solar System such as planets, moons, comets, and asteroids, we find that they move on curved lines, with constantly changing speed. Why is this? Newton would tell us, “Their velocities are changing because there is a force acting on them; the force called gravity.” Newton devised a formula for computing the gravitational force between two objects. Every object in the universe, said Newton, has a property which we may call the “gravitational mass”. Let the gravitational masses of two objects be ` g and : g , and let the distance between their centers be :. The gravitational force acting between the two objects (assuming they are both spherical) is 1 = − G` g : g : 2 . (3.1) The negative sign in the above equation indicates that gravity, in the New- tonian view, is always an attractive force, tending to draw two bodies closer together. 34 CHAPTER 3. NEWTON VERSUS EINSTEIN What is the acceleration which results from this gravitational force? New- ton had something to say about that, as well. Every object in the universe, said Newton, has a property which we may call the “inertial mass”. Let the inertial mass of an object be : i . Newton’s second law of motion says that force and acceleration are related by the equation 1 = : i c . (3.2) In equations (3.1) and (3.2) I have distinguished, through the use of different subscripts, between the gravitational mass : g and the inertial mass : i . One of the fundamental principles of physics is that the gravitational mass and the inertial mass of an object are identical: : g = : i . (3.3) When you stop to think about it, this equality is a remarkable fact. The property of an object that determines how strongly it is pulled on by the force of gravity is equal to the property that determines its resistance to acceleration by any force, not just the force of gravity. The equality of gravitational mass and inertial mass is called the equivalence principle, and it is the equivalence principle which led Einstein to devise his theory of general relativity. If the equivalence principle did not hold, then the gravitational accelera- tion of an object toward a mass ` g would be (combining equations 3.1 and 3.2) c = − G` g : 2 : g : i . (3.4) with the ratio : g : i varying from object to object. However, when Galileo dropped objects from towers and slid objects down inclined planes, he found that the acceleration (barring the effects of air resistance and friction) was always the same, regardless of the mass and composition of the object. The magnitude of the gravitational acceleration close to the Earth’s surface is o = G` Earth : 2 Earth = 9.8 ms −2 . Modern tests of the equivalence principle, which are basically more sensitive versions of Galileo’s experiments, reveal that the inertial and gravitational masses are the same to within one part in 10 12 . To see how the equivalence principle led Einstein to devise his theory of general relativity, let’s begin with a thought experiment of the sort Einstein would devise. Suppose you wake up one morning to find that you have been 3.1. EQUIVALENCE PRINCIPLE 35 sealed up (bed and all) within an opaque, soundproof, hermetically sealed box. “Oh no!” you say. “This is what I’ve always feared would happen. I’ve been abducted by space aliens who are taking me away to their home planet.” So startled are you by this realization, you drop your teddy bear. Observing the falling bear, you find that it falls toward the floor of the box with an acceleration c = 9.8 ms −2 . “Whew!” you say, with some relief. “At least I am still on the Earth’s surface; they haven’t taken me away in their spaceship yet.” At that moment, a window in the side of the box opens to reveal (much to your horror) that you are inside an alien spaceship which is being accelerated at 9.8 ms −2 by a rocket engine. When you drop a teddy bear, or any other object, within a sealed box, the equivalence principle permits two possible interpretations, with no way of distinguishing between them. (1) The box is static, or moving with a constant velocity, and the bear is being accelerated downward by a constant gravitational force. (2) The bear is moving at a constant velocity, and the box is being accelerated upward at a constant rate. The behavior of the bear in each case (Figure 3.1) is identical. In each case, a big bear falls at the same rate as a little bear; in each case, a bear stuffed with cotton falls at the same rate as a bear stuffed with lead; and in each case, a sentient anglophone bear would say, “Oh, bother. I’m weightless.” during the interval before it collides with the floor of the box. 1 Einstein’s insight, starting from the equivalence principle, led him to the theory of general relativity. To understand Einstein’s thought processes, imagine yourself back in the sealed box, being accelerated through interplan- etary space at 9.8 ms −2 . You grab the flashlight that you keep on the bedside table and shine a beam of light perpendicular to the acceleration vector (Fig- ure 3.2). Since the box is accelerating upward, the path of the light beam will appear to you to be bent downward, as the floor of the box rushes up to meet the photons. However, thanks to the equivalence principle, we can replace the accelerated box with a stationary box experiencing a constant 1 Note that the equivalence of the two boxes in Figure 3.1 depends on the gravitational acceleration in the left-hand box being constant. In the real universe, though, gravita- tional accelerations are not exactly constant, but vary with position. For instance, the gravitational acceleration near the Earth’s surface is a vector g(r) which varies in direction (always pointing toward the Earth’s center) and in magnitude (decreasing as the inverse square of the distance from the Earth’s center). Thus, in the real universe, the equivalence principle can only be applied to an infinitesimally small box – that is, a box so small that the variation in g is too tiny to be measured. 36 CHAPTER 3. NEWTON VERSUS EINSTEIN Figure 3.1: Equivalence principle (teddy bear version). The behavior of a bear in an accelerated box (left) is identical to that of a bear being accelerated by gravity (right). 3.1. EQUIVALENCE PRINCIPLE 37 Figure 3.2: Equivalence principle (photon version) The path followed by a light beam in an accelerated box (left) is identical to the path followed by a light beam subjected to gravitational acceleration (right). 38 CHAPTER 3. NEWTON VERSUS EINSTEIN gravitational acceleration. Since there’s no way to distinguish between these two cases, we are led to the conclusion that the paths of photons will be curved downward in the presence of a gravitational field. Gravity affects photons, Einstein concluded, even though they have no mass. Contemplat- ing the curved path of the light beam, Einstein had one more insight. One of the fundamental principles of optics is Fermat’s principle, which states that light travels between two points along a path which minimizes the travel time required. 2 In a vacuum, where the speed of light is constant, this translates into the requirement that light takes the shortest path between two points. In Euclidean, or flat, space, the shortest path between two points is a straight line. However, in the presence of gravity, the path taken by light is not a straight line. Thus, Einstein concluded, space is not Euclidean. The presence of mass, in Einstein’s view, causes space to be curved. In fact, in the fully developed theory of general relativity, mass and energy (which Newton thought of as two separate entities) are interchangeable, via the famous equation 1 = :c 2 . Moreover, space and time (which Newton thought of as two separate entities) form a four-dimensional space-time. A more accurate summary of Einstein’s viewpoint, then, is that the presence of mass-energy causes space-time to be curved. We now have a third way of thinking about the motion of the teddy bear in the box: (3) No forces are acting on the bear; it is simply following a geodesic in curved space-time. 3 We now have two ways of describing how gravity works. The Way of Newton: Mass tells gravity how to exert a force (1 = −G`:: 2 ), Force tells mass how to accelerate (1 = :c). The Way of Einstein: Mass-energy tells space-time how to curve, Curved space-time tells mass-energy how to move. 4 Einstein’s description of gravity gives a natural explanation for the equiv- alence principle. In the Newtonian description of gravity, the equality of the 2 More generally, Fermat’s principle requires that the travel time be an extremum – either a minimum or a maximum. In most situations, however, the path taken by light minimizes the travel time rather than maximizing it. 3 In this context, the word “geodesic” is simply a shorter way of saying “the shortest distance between two points”. 4 This pocket summary of general relativity was coined by the physicist John Wheeler, who also popularized the term “black hole”. 3.2. DESCRIBING CURVATURE 39 Figure 3.3: A flat two-dimensional space. gravitational mass and the inertial mass is a remarkable coincidence. How- ever, in Einstein’s theory of general relativity, curvature is a property of space-time itself. It then follows automatically that the gravitational accel- eration of an object should be independent of mass and composition – it’s just following a geodesic, which is dictated by the geometry of space-time. 3.2 Describing curvature In developing a mathematical theory of general relativity, in which space- time curvature is related to the mass-energy density, Einstein needed a way of mathematically describing curvature. Since picturing the curvature of a four-dimensional space-time is, to say the least, difficult, let’s start by considering ways of describing the curvature of two-dimensional spaces, then extend what we have learned to higher dimensions. The simplest of two-dimensional spaces is a plane, on which Euclidean geometry holds (Figure 3.3). On a plane, a geodesic is a straight line. If a triangle is constructed on a plane by connecting three points with geodesics, the angles at its vertices (α, β, and γ in Figure 3.3) obey the relation α +β +γ = π . (3.5) 40 CHAPTER 3. NEWTON VERSUS EINSTEIN where angles are measured in radians. On a plane, we can set up a cartesian coordinate system, and assign to every point a coordinate (r. n). On a plane, the Pythagorean theorem holds, so the distance d: between points (r. n) and (r +dr. n +dn) is given by the relation 5 d: 2 = dr 2 +dn 2 . (3.6) Stating that equation (3.6) holds true everywhere in a two-dimensional space is equivalent to saying that the space is a plane. Of course, other coordinate systems can be used, in place of cartesian coordinates. For instance, in a polar coordinate system, the distance between points (:. θ) and (: +d:. θ +dθ) is d: 2 = d: 2 +: 2 dθ 2 . (3.7) Although equations (3.6) and (3.7) are different in appearance, they both represent the same flat geometry, as you can verify by making the simple coordinate substitution r = : cos θ, n = : sin θ. Now consider another simple two-dimensional space, the surface of a sphere (Figure 3.4). On the surface of a sphere, a geodesic is a portion of a great circle – that is, a circle whose center corresponds to the center of the sphere. If a triangle is constructed on the surface of the sphere by connecting three points with geodesics, the angles at its vertices (α, β, and γ) obey the relation α +β +γ = π +¹1 2 . (3.8) where ¹ is the area of the triangle, and 1 is the radius of the sphere. All spaces in which α + β + γ π are called “positively curved” spaces. The surface of a sphere is a positively curved two-dimensional space. Moreover, it is a space where the curvature is homogeneous and isotropic; no matter where you draw a triangle on the surface of a sphere, or how you orient it, it must always satisfy equation (3.8). On the surface of a sphere, we can set up a polar coordinate system by picking a pair of antipodal points to be the “north pole” and “south pole” and by picking a geodesic from the north to south pole to be the “prime meridian”. If : is the distance from the north pole, and θ is the azimuthal 5 Starting with this equation, I am adopting the convention, commonly used among relativists, that ds 2 = (ds) 2 , and not d(s 2 ). Omitting the parentheses simply makes the equations less cluttered. 3.2. DESCRIBING CURVATURE 41 Figure 3.4: A positively curved two-dimensional space. angle measured relative to the prime meridian, then the distance d: between a point (:. θ) and another nearby point (:+d:. θ+dθ) is given by the relation d: 2 = d: 2 +1 2 sin 2 (:1)dθ 2 . (3.9) Note that the surface of a sphere has a finite area, equal to 4π1 2 , and a maximum possible distance between points; the distance between antipodal points, at the maximum possible separation, is π1. By contrast, a plane has infinite area, and has no upper limits on the possible distance between points. 6 In addition to flat spaces and positively curved spaces, there exist neg- atively curved spaces. An example of a negatively curved two-dimensional space is the hyperboloid, or saddle-shape, shown in Figure 3.5. For illustra- tive purposes, I would like to show you a surface of constant negative cur- vature, just as the surface of a sphere has constant positive curvature. 7 Un- 6 Since the publishers objected to producing a book of infinite size, Figure 3.3 actually shows only a portion of a plane. 7 A space with constant curvature is one where the curvature is homogeneous and isotropic. 42 CHAPTER 3. NEWTON VERSUS EINSTEIN Figure 3.5: A negatively curved two-dimensional space. fortunately, the mathematician David Hilbert proved that a two-dimensional surface of constant negative curvature cannot be constructed in a three- dimensional Euclidean space. The saddle-shape illustrated in Figure 3.5 has constant curvature only in the central region, near the “seat” of the saddle. Despite the difficulties in visualizing a surface of constant negative curva- ture, its properties can easily be written down. Consider a two-dimensional surface of constant negative curvature, with radius of curvature 1. If a trian- gle is constructed on this surface by connecting three points with geodesics, the angles at its vertices (α,β, and γ) obey the relation α +β +γ = π −¹1 2 . (3.10) where ¹ is the area of the triangle. On a surface of constant negative curvature, we can set up a polar coor- dinate system by choosing some point as the pole, and some geodesic leading away from the pole as the prime meridian. If : is the distance from the pole, and θ is the azimuthal angle measured relative to the prime meridian, then the distance d: between a point (:. θ) and a nearby point (: + d:. θ + dθ) is given by d: 2 = d: 2 +1 2 sinh 2 (:1)dθ 2 . (3.11) 3.2. DESCRIBING CURVATURE 43 A surface of constant negative curvature has infinite area, and has no upper limit on the possible distance between points. Relations like those presented in equations (3.7), (3.9), and (3.11), which give the distance d: between two nearby points in space, are known as met- rics. In general, curvature is a local property. A tablecloth can be badly rumpled at one end of the table and smooth at the other end; a bagel (or other toroidal object) is negatively curved on part of its surface and pos- itively curved on other portions. 8 However, if you want a two-dimensional space to be homogeneous and isotropic, there are only three possibilities that fit the bill: the space can be uniformly flat, it can have uniform positive cur- vature, or it can have uniform negative curvature. Thus, if a two-dimensional space has curvature which is homogeneous and isotropic, its geometry can be specified by two quantities, κ, and 1. The number κ, called the curvature constant, is κ = 0 for a flat space, κ = +1 for a positively curved space, and κ = −1 for a negatively curved space. If the space is curved, then the quantity 1, which has dimensions of length, is the radius of curvature. The results for two-dimensional space can be extended straightforwardly to three dimensions. A three-dimensional space, if its curvature is homoge- neous and isotropic, must be flat, or have uniform positive curvature, or have uniform negative curvature. If a three-dimensional space is flat (κ = 0), it has the metric d: 2 = dr 2 +dn 2 +d. 2 . (3.12) expressed in cartesian coordinates, or d: 2 = d: 2 +: 2 [dθ 2 + sin 2 θdφ 2 ] . (3.13) expressed in spherical coordinates. If a three-dimensional space has uniform positive curvature (κ = +1), its metric is d: 2 = d: 2 +1 2 sin 2 (:1)[dθ 2 + sin 2 θdφ 2 ] . (3.14) A positively curved three-dimensional space has finite volume, just as a pos- itively curved two-dimensional space has finite area. The point at : = π1 is the antipodal point to the origin, just as the south pole, at : = π1, is the antipodal point to the north pole, at : = 0, on the surface of a sphere. By traveling a distance ( = 2π1, it is possible to “circumnavigate” a space of uniform positive curvature. 8 You can test this assertion, if you like, by drawing triangles on a bagel. 44 CHAPTER 3. NEWTON VERSUS EINSTEIN Finally, if a three-dimensional space has uniform negative curvature (κ = −1), its metric is d: 2 = d: 2 +1 2 sinh 2 (:1)[dθ 2 + sin 2 θdφ 2 ] . (3.15) Like flat space, negatively curved space has infinite volume. The three possible metrics for a homogeneous, isotropic, three-dimensional space can be written more compactly in the form d: 2 = d: 2 +o κ (:) 2 dΩ 2 . (3.16) where dΩ 2 ≡ dθ 2 + sin 2 θdφ 2 (3.17) and o κ (:) = 1sin(:1) (κ = +1) : (κ = 0) 1sinh(:1) (κ = −1) . (3.18) Note that in the limit : < 1, o κ ≈ :, regardless of the value of κ. When space is flat, or negatively curved, o κ increases monotonically with :, with o κ → ∞ as : → ∞. By contrast, when space is positively curved, o κ increases to a maximum of o max = 1 at :1 = π2, then decreases again to 0 at :1 = π, the antipodal point to the origin. The coordinate system (:. θ. φ) is not the only possible system. For in- stance, if we switch the radial coordinate from : to r ≡ o κ (:), the metric for a homogeneous, isotropic, three-dimensional space can be written in the form d: 2 = dr 2 1 −κr 2 1 2 +r 2 dΩ 2 . (3.19) Although the metrics written in equations (3.16) and (3.19) appear different on the page, they represent the same homogeneous, isotropic spaces. They merely have a different functional form because of the different choice of radial coordinates. 3.3 The Robertson-Walker metric So far, we’ve only considered the metrics for simple two-dimensional and three-dimensional spaces. However, relativity teaches us that space and time 3.3. ROBERTSON-WALKER METRIC 45 together comprise a four-dimensional space-time. Just as we can compute the distance between two points in space using the appropriate metric for that space, so we can compute the four-dimensional distance between two events in space-time. Consider two events, one occurring at the space-time location (t. :. θ. φ), and another occurring at the space-time location (t + dt. : + d:. θ + dθ. φ + dφ). According to the laws of special relativity, the space-time separation between these two events is d: 2 = −c 2 dt 2 +d: 2 +: 2 dΩ 2 . (3.20) The metric given in equation (3.20) is called the Minkowski metric, and the space-time which it describes is called Minkowski space-time. Note, from a comparison with equation (3.16), that the spatial component of Minkowski space-time is Euclidean, or flat. A photon’s path through space-time is a four-dimensional geodesic – and not just any geodesic, mind you, but a special variety called a null geodesic. A null geodesic is one for which, along every infinitesimal segment of the photon’s path, d: = 0. In Minkowski space-time, then, a photon’s trajectory obeys the relation d: 2 = 0 = −c 2 dt 2 +d: 2 +: 2 dΩ 2 . (3.21) If the photon is moving along a radial path, towards or away from the origin, this means, since θ and φ are constant, c 2 dt 2 = d: 2 . (3.22) or d: dt = ±c . (3.23) The Minkowski metric of equation (3.20) applies only within the context of special relativity, so called because it deals with the special case in which space-time is not curved by the presence of mass and energy. Without any gravitational effects, Minkowski space-time is flat and static. When gravity is added, however, the permissible space-times are more interesting. In the 1930’s, the physicists Howard Robertson and Arthur Walker asked “What form can the metric of space-time assume if the universe is spatially homo- geneous and isotropic at all time, and if distances are allowed to expand 46 CHAPTER 3. NEWTON VERSUS EINSTEIN (or contract) as a function of time?” The metric which they derived, inde- pendently of each other, is called the Robertson-Walker metric. It is most generally written in the form d: 2 = −c 2 dt 2 +c(t) 2 ¸ dr 2 1 −κr 2 1 2 0 +r 2 dΩ 2 ¸ . (3.24) Note that the spatial component of the Robertson-Walker metric consists of the spatial metric for a uniformly curved space of radius 1 0 (compare equation 3.19), scaled by the square of the scale factor c(t). The scale fac- tor, first introduced in Section 2.3, describes how distances in a homogeneous, isotropic universe expand or contract with time. The Robertson-Walker met- ric can also be written in the form (see equation 3.16) d: 2 = −c 2 dt 2 +c(t) 2 d: 2 +o κ (:) 2 dΩ 2 . (3.25) with the function o κ (:) for the three different types of curvature given by equation (3.18). The time variable t in the Robertson-Walker metric is the cosmological proper time, called the cosmic time for short, and is the time measured by an observer who sees the universe expanding uniformly around him. The spatial variables (r. θ. φ) or (:. θ. φ) are called the comoving coordinates of a point in space; if the expansion of the universe is perfectly homogeneous and isotropic, the comoving coordinates of any point remain constant with time. The assumption of homogeneity and isotropy is a very powerful one. If the universe is perfectly homogeneous and isotropic, then everything we need to know about its geometry is contained within c(t), κ, and 1 0 . The scale factor c(t) is a dimensionless function of time which describes how distances grow or decrease with time; it is normalized so that c(t 0 ) = 1 at the present moment. The curvature constant κ is a dimensionless number which can take on one of three discrete values: κ = 0 if the universe is spatially flat, κ = −1 if the universe has negative spatial curvature, and κ = +1 if the universe has positive spatial curvature. The radius of curvature 1 0 has dimensions of length, and gives the radius of curvature of the universe at the present moment. Much of modern cosmology, as we’ll see in later chapters, is devoted in one way or another to finding the values of c(t), κ, and 1 0 . The assumption of spatial homogeneity and isotropy is so powerful, Robertson and Walker made it in the 1930’s, long before the available observational evidence gave any support for such an assumption. If homogeneity and isotropy did not 3.4. PROPER DISTANCE 47 exist, as Voltaire might have said, it would be necessary to invent them – at least if your desire is to have a simple, analytically tractable form for the metric of space-time. In truth, the observations reveal that the universe is not homogeneous and isotropic on small scales. Thus, the Robertson-Walker metric is only an approximation which holds good on large scales. On smaller scales, the universe is “lumpy”, and hence does not expand uniformly. Small, dense lumps, such as humans, teddy bears, and interstellar dust grains, are held together by electromagnetic forces, and hence do not expand. Larger lumps, as long as they are sufficiently dense, are held together by their own gravity, and hence do not expand. Examples of such gravitationally bound systems are planetary systems (such as the Solar System in which we live), galaxies (such as the Galaxy in which we live), and clusters of galaxies (such as the Local Group in which we live). It’s only on scales larger than ∼ 100 Mpc that the expansion of the universe can be treated as the ideal, homogeneous, isotropic expansion described by the single scale factor c(t). 3.4 Proper distance Consider a galaxy which is far away from us – sufficiently far away that we may ignore the small scale perturbations of space-time and adopt the Robertson-Walker metric. One question we may ask is, “Exactly how far away is this galaxy?” In an expanding universe, the distance between two objects is increasing with time. Thus, if we want to assign a spatial distance 1 between two objects, we must specify the time t at which the distance is the correct one. Suppose that you are at the origin, and that the galaxy which you are observing is at a comoving coordinate position (:. θ. φ), as illustrated in Figure 3.6. The proper distance d p (t) between two points is equal to the length of the spatial geodesic between them when the scale factor is fixed at the value c(t). The proper distance between the observer and galaxy in Figure 3.6 can be found using the Robertson-Walker metric at a fixed time t: d: 2 = c(t) 2 [d: 2 +o κ (:) 2 dΩ 2 ] . (3.26) Along the spatial geodesic between the observer and galaxy, the angle (θ. φ) is constant, and thus d: = c(t)d: . (3.27) 48 CHAPTER 3. NEWTON VERSUS EINSTEIN Figure 3.6: An observer at the origin observes a galaxy at coordinate position (:. θ. φ). A photon emitted by the galaxy at cosmic time t e reaches the observer at cosmic time t 0 . 3.4. PROPER DISTANCE 49 The proper distance d p is found by integrating over the radial comoving coordinate :: d p (t) = c(t) r 0 d: = c(t): . (3.28) Alternatively, if you wish to use the spatial coordinates (r. θ. φ) instead of (:. θ. φ), where r = o κ (:), you may invert the relations of equation (3.18) to find d p (t) = c(t):(r) = c(t)1 0 sin −1 (r1 0 ) (κ = +1) c(t)r (κ = 0) c(t)1 0 sinh −1 (r1 0 ) (κ = −1) . (3.29) Because the proper distance has the form d p (t) = c(t):, with the comoving coordinate : constant with time, the rate of change for the proper distance between us and a distant galaxy is ˙ d p = ˙ c: = ˙ c c d p . (3.30) Thus, at the current time (t = t 0 ), there is a linear relation between the proper distance to a galaxy and its recession speed: · p (t 0 ) = H 0 d p (t 0 ) . (3.31) where · p (t 0 ) ≡ ˙ d p (t 0 ) (3.32) and H 0 = ˙ c c t=t 0 . (3.33) In a sense, this is just a repetition of what was demonstrated in Section 2.3; if the distance between points is proportional to c(t), there will be a linear relation between the relative velocity of two points and the distance between them. Now, however, we are interpreting the change in distance between widely separated galaxies as being associated with the expansion of space. As the distance between galaxies increases, the radius of curvature of the universe, 1(t) = c(t)1 0 , increases at the same rate. Some cosmology books will contain a statement like “As space expands, it drags galaxies away from each other.” Statements of this sort are mislead- ing, since they make galaxies appear to be entirely passive. On the other hand, a statement like “As galaxies move apart, they drag space along with 50 CHAPTER 3. NEWTON VERSUS EINSTEIN them” would be equally misleading, since it makes space appear to be en- tirely passive. As the theory of general relativity points out, space-time and mass-energy are intimately linked. Yes, the curvature of space-time does tell mass-energy how to move, but then it’s mass-energy which tells space-time how to curve. The linear velocity-distance relation given in equation (3.31) implies that points separated by a proper distance greater than a critical value d H (t 0 ) ≡ cH 0 . (3.34) generally called the Hubble distance, will have · p = ˙ d p c . (3.35) Using the observationally determined value of H 0 = 70±7 kms −1 Mpc −1 , the current value of the Hubble distance in our universe is d H (t 0 ) = 4300 ±400 Mpc . (3.36) Thus, galaxies farther than 4300 megaparsecs from us are currently moving away from us at speeds greater than that of light. Cosmological innocents sometimes exclaim, “Gosh! Doesn’t this violate the law that massive objects can’t travel faster than the speed of light?” Actually, it doesn’t. The speed limit that states that massive objects must travel with · < c relative to each other is one of the results of special relativity, and refers to the relative motion of objects within a static space. In the context of general relativity, there is no objection to having two points moving away from each other at superluminal speed due to the expansion of space. When we observe a distant galaxy, we know its angular position very well, but not its distance. That is, we can point in its direction, but we don’t know its current proper distance d p (t 0 ) – or, for that matter, its comoving coordinate distance :. We can, however, measure the redshift . of the light we receive from the galaxy. Although the redshift doesn’t tell us the proper distance to the galaxy, it does tell us what the scale factor c was at the time the light from that galaxy was emitted. To see the link between c and ., consider the galaxy illustrated in Figure 3.6. Light that was emitted by the galaxy at a time t e is observed by us at a time t 0 . During its travel from the distant galaxy to us, the light traveled along a null geodesic, with 3.4. PROPER DISTANCE 51 d: = 0. The null geodesic has θ and φ constant. 9 Thus, along the light’s null geodesic, c 2 dt 2 = c(t) 2 d: 2 . (3.37) Rearranging this relation, we find c dt c(t) = d: . (3.38) In equation (3.38), the left-hand side is a function only of t, and the right- hand side is independent of t. Suppose the distant galaxy emits light with a wavelength λ e , as measured by an observer in the emitting galaxy. Fix your attention on a single wave crest of the emitted light. The wave crest is emitted at a time t e and observed at a time t 0 , such that c t 0 te dt c(t) = r 0 d: = : . (3.39) The next wave crest of light is emitted at a time t e + λ e c, and is observed at a time t 0 +λ 0 c, where, in general, λ 0 = λ e . For the second wave crest, c t 0 +λ 0 /c te+λe/c dt c(t) = r 0 d: = : . (3.40) Comparing equations (3.39) and (3.40), we find that t 0 te dt c(t) = t 0 +λ 0 /c te+λe/c dt c(t) . (3.41) That is, the integral of dtc(t) between the time of emission and the time of observation is the same for every wave crest in the emitted light. If we subtract the integral t 0 te+λe/c dt c(t) (3.42) from each side of equation (3.41), we find the relation te+λe/c te dt c(t) = t 0 +λ 0 /c t 0 dt c(t) . (3.43) 9 In a homogeneous, isotropic universe there’s no reason for the light to swerve to one side or the other. 52 CHAPTER 3. NEWTON VERSUS EINSTEIN That is, the integral of dtc(t) between the emission of successive wave crests is equal to the integral of dtc(t) between the observation of successive wave crests. This relation becomes still simpler when we realize that during the time between the emission or observation of two wave crests, the universe doesn’t have time to expand by a significant amount. The time scale for expansion of the universe is the Hubble time, H −1 0 ≈ 14 Gyr. The time between wave crests, for visible light, is λc ≈ 2 10 −15 s ≈ 10 −32 H −1 0 . Thus, c(t) is effectively constant in the integrals of equation (3.43). Thus, we may write 1 c(t e ) te+λe/c te dt = 1 c(t 0 ) t 0 +λ 0 /c t 0 dt . (3.44) or λ e c(t e ) = λ 0 c(t 0 ) . (3.45) Using the definition of redshift, . = (λ 0 − λ e )λ e , we find that the redshift of light from a distant object is related to the expansion factor at the time it was emitted via the equation 1 +. = c(t 0 ) c(t e ) = 1 c(t e ) . (3.46) Here, I have used the usual convention that c(t 0 ) = 1. Thus, if we observe a galaxy with a redshift . = 2, we are observing it as it was when the universe had a scale factor c(t e ) = 13. The redshift we observe for a distant object depends only on the relative scale factors at the time of emission and the time of observation. It doesn’t depend on how the transition between c(t e ) and c(t 0 ) was made. It doesn’t matter if the expansion was gradual or abrupt; it doesn’t matter if the transition was monotonic or oscillatory. All that matters is the scale factors at the time of emission and the time of observation. Suggested reading [Full references are given in the “Annotated Bibliography” on page 286.] Harrison (2000), ch. 10 – 12: Curved space and relativity (both special and general) 3.4. PROPER DISTANCE 53 Narlikar (2002), ch. 2 – 3: Delves deeper into general relativity while dis- cussing the Robertson-Walker metric Peacock (2000), ch. 3.1: Derivation and discussion of the Robertson-Walker metric Rich (2001), ch. 3: Coordinates and metrics in the context of general rel- ativity Problems (3.1) What evidence can you provide to support the assertion that the uni- verse is electrically neutral on large scales? (3.2) Suppose you are a two-dimensional being, living on the surface of a sphere with radius 1. An object of width d: < 1 is at a distance : from you (remember, all distances are measured on the surface of the sphere). What angular width dθ will you measure for the object? Explain the behavior of dθ as : →π1. (3.3) Suppose you are still a two-dimensional being, living on the same sphere of radius 1. Show that if you draw a circle of radius :, the circle’s circumference will be ( = 2π1sin(:1) . (3.47) Idealize the Earth as a perfect sphere of radius 1 = 6371 km. If you could measure distances with an error of ±1 meter, how large a circle would you have to draw on the Earth’s surface to convince yourself that the Earth is spherical rather than flat? (3.4) Consider an equilateral triangle, with sides of length 1, drawn on a two-dimensional surface of constant curvature. Can you draw an equilateral triangle of arbitrarily large area ¹ on a surface with κ = +1 and radius of curvature 1? If not, what is the maximum possible value of ¹? Can you draw an equilateral triangle of arbitrarily large area ¹ on a surface with κ = 0? If not, what is the maximum possible value of ¹? Can you draw an equilateral triangle of arbitrarily large area ¹ on a surface with κ = −1 and radius of curvature 1? If not, what is the maximum possible value of ¹? 54 CHAPTER 3. NEWTON VERSUS EINSTEIN (3.5) By making the substitutions r = : sin θ cos φ, n = : sin θ sin φ, and . = : cos θ, demonstrate that equations (3.12) and (3.13) represent the same metric. Chapter 4 Cosmic Dynamics In a universe which is homogeneous and isotropic, but which is allowed to expand or contract with time, everything you need to know about the cur- vature is given by κ, 1 0 , and c(t). The curvature constant κ gives the sign of the curvature: positive (κ = +1), negative (κ = −1), or flat (κ = 0). If κ is non-zero, then 1 0 is the radius of curvature of the universe, as measured at the present moment (t = t 0 ). Finally, the scale factor c(t) tells how dis- tances in the universe increase with time as the universe expands, or decrease with time as the universe contracts. The scale factor is normalized so that c(t 0 ) = 1 at the present moment. The idea that the universe could be curved, or non-Euclidean, actually long predates Einstein’s theory of general relativity. As early as 1829, half a century before Einstein’s birth, Nikolai Ivanovich Lobachevski, one of the founders of non-Euclidean geometry, proposed observational tests to demon- strate whether the universe was curved. In principle, measuring the curvature of the universe is simple; in practice it is much more difficult. In principle, we could determine the curvature by drawing a really, really big triangle, and measuring the angles α, β, and γ at the vertices. Equations (3.5), (3.8), and (3.10) generalize to the equation α +β +γ = π + κ¹ 1 2 0 . (4.1) where ¹ is the area of the triangle. Therefore, if the α +β +γ π radians, the universe is positively curved, and if α + β + γ < π radians, the universe is negatively curved. If, in addition, you measure the area of the triangle, you can determine the radius of curvature 1 0 . Unfortunately for this elegant 55 56 CHAPTER 4. COSMIC DYNAMICS Figure 4.1: A two-dimensional positively curved universe, demonstrating how an observer in such a universe could see multiple images of the same galaxy. geometric plan, the area of the biggest triangle we can draw is much smaller than 1 2 0 , and the deviation of α + β + γ from π radians would be too small to measure. About all we can conclude from geometric arguments is that if the uni- verse is positively curved, it can’t have a radius of curvature 1 0 that is significantly smaller than the current Hubble distance, cH 0 ≈ 4300 Mpc. To understand why this is so, recall that if our universe is positively curved, it has finite size, with a circumference currently equal to ( 0 = 2π1 0 . In the past, since our universe is expanding, its circumference was even smaller. Thus, if the current circumference ( 0 is less than ct 0 , then photons will have had time to circumnavigate the universe. If ( 0 <ct 0 ∼ cH 0 , then photons will have had time to circumnavigate the universe many times. To take an extreme example, suppose the universe were positively curved with a circum- ference of only 10 million light years (roughly 3 Mpc). The two-dimensional analog to such a universe is shown in Figure 4.1. Looking toward the galaxy 4.1. FRIEDMANN EQUATION 57 M31, which is 2 million light years away from us, we would see one image of M31, comprised of photons which had traveled 2 million light years, showing M31 as it was 2 million years ago. We would also see another image, com- prised of photons which had traveled 12 million light years, showing M31 as it was 12 million years ago. And so on. Moreover, looking in the exact oppo- site direction to M31, we would see an image of M31, comprised of photons which had traveled 8 million light years, showing M31 as it was 8 million years ago. We would also see another image, comprised of photons which had traveled 18 million light years, showing M31 as it was 18 million years ago. 1 And so on. Since we don’t see periodicities of this sort, we conclude that if the universe is positively curved, its radius of curvature 1 0 must be very large – comparable to or larger than the current Hubble distance cH 0 . 4.1 The Friedmann equation Although 19th century mathematicians and physicists, such as Lobachevski, were able to conceive of curved space, it wasn’t until Albert Einstein first published his theory of general relativity in 1915 that the curvature of space- time was linked to its mass-energy content. The key equation of general relativity is Einstein’s field equation, which is the relativistic equivalent of Poisson’s equation in Newtonian dynamics. Poisson’s equation, ∇ 2 Φ = 4πGρ . (4.2) gives a mathematical relation between the gravitational potential Φ at a point in space and the mass density ρ at that point. By taking the gradient of the potential, you determine the acceleration, and then can compute the trajectory of objects moving freely through space. Einstein’s field equation, by contrast, gives a mathematical relation between the metric of space-time at a point and the energy and pressure at that space-time point. The tra- jectories of of freely moving objects then correspond to geodesics in curved space-time. In a cosmological context, Einstein’s field equations can be used to find the linkage between c(t), κ, and 1 0 , which describe the curvature of the uni- verse, and the energy density ε(t) and pressure 1(t) of the contents of the 1 This assumes that 2πR 0 /c H −1 0 , and that the universe therefore doesn’t expand significantly as a photon goes once or twice around the universe. 58 CHAPTER 4. COSMIC DYNAMICS universe. The equation which links together c(t), κ, 1 0 , and ε(t) is known as the Friedmann equation, after Alexander Alexandrovich Friedmann, who first derived the equation in 1922. Friedmann actually started his scientific career as a meteorologist. Later, however, he taught himself general relativity, and used Einstein’s field equations to describe how a spatially homogeneous and isotropic universe expands or contracts as a function of time. It is intriguing to note that Friedmann published his first results, implying an expanding or contracting universe, seven years before Hubble published Hubble’s Law in 1929. Unfortunately, Friedmann’s papers received little notice at first. Even Einstein initially dismissed Friedmann’s work as a mathematical curiosity, unrelated to the universe we actually live in. It wasn’t until Hubble’s results were published that Einstein acknowledged the reality of the expanding uni- verse. Alas, Friedmann did not live to see his vindication; he died of typhoid fever in 1925, when he was only 37 years old. Friedmann derived his eponymous equation starting from Einstein’s field equation, using the full power of general relativity. Even without bringing relativity into play, some (though not all) of the aspects of the Friedmann equation can be understood with the use of purely Newtonian dynamics. To see how the expansion or contraction of the universe can be viewed from a Newtonian viewpoint, I will first derive the non-relativistic equivalent of the Friedmann equation, starting from Newton’s Law of Gravity and Second Law of Motion. Then I will state (without proof) the modifications that must be made to find the more correct, general relativistic form of the Friedmann equation. To begin, consider a homogeneous sphere of matter, with total mass ` s constant with time (Figure 4.2). The sphere is expanding or contracting isotropically, so that its radius 1 s (t) is increasing or decreasing with time. Place a test mass, of infinitesimal mass :, at the surface of the sphere. The gravitational force 1 experienced by the test mass will be, from Newton’s Law of Gravity, 1 = − G` s : 1 s (t) 2 . (4.3) The gravitational acceleration at the surface of the sphere will then be, from Newton’s Second Law of Motion, d 2 1 s dt 2 = − G` s 1 s (t) 2 . (4.4) 4.1. FRIEDMANN EQUATION 59 r(t) Figure 4.2: A sphere of radius 1 s (t) and mass ` s , expanding or contracting under its own gravity. Multiply each side of the equation by d1 s dt and integrate to find 1 2 d1 s dt 2 = G` s 1 s (t) +l . (4.5) where l is a constant of integration. Equation (4.5) simply states that the sum of the kinetic energy per unit mass, 1 kin = 1 2 d1 s dt 2 . (4.6) and the gravitational potential energy per unit mass, 1 pot = − G` s 1 s (t) . (4.7) is constant for a bit of matter at the surface of a sphere, as the sphere expands or contracts under its own gravitational influence. Since the mass of the sphere is constant as it expands or contracts, we may write ` s = 4π 3 ρ(t)1 s (t) 3 . (4.8) Since the expansion is isotropic about the sphere’s center, we may write the radius 1 s (t) in the form 1 s (t) = c(t): s . (4.9) 60 CHAPTER 4. COSMIC DYNAMICS where c(t) is the scale factor and : s is the comoving radius of the sphere. In terms of ρ(t) and c(t), the energy conservation equation (4.5) can be rewritten in the form 1 2 : 2 s ˙ c 2 = 4π 3 G: 2 s ρ(t)c(t) 2 +l . (4.10) Dividing each side of equation (4.10) by : 2 s c 2 2 yields the equation ˙ c c 2 = 8πG 3 ρ(t) + 2l : 2 s 1 c(t) 2 . (4.11) Equation (4.11) gives the Friedmann equation in its Newtonian form. Note that the time derivative of the scale factor only enters into equa- tion (4.11) as ˙ c 2 ; a contracting sphere ( ˙ c < 0) is simply the time reversal of an expanding sphere ( ˙ c 0). Let’s concentrate on the case of a sphere which is expanding, analogous to the expanding universe in which we find ourselves. The future of the expanding sphere falls into one of three classes, depending on the sign of l. First, consider the case l 0. In this case, the right hand side of equation (4.11) is always positive. Therefore, ˙ c 2 is always positive, and the expansion of the sphere never stops. Second, consider the case l < 0. In this case, the right hand side of equation (4.11) starts out positive. However, at a maximum scale factor c max = − G` s l: s . (4.12) the right hand side will equal zero, and expansion will stop. Since ¨ c will still be negative, the sphere will then contract. Third, and finally, consider the case l = 0. This is the boundary case in which ˙ c →0 as t →∞ and ρ →0. The three possible fates of an expanding sphere in a Newtonian universe are analogous to the three possible fates of a ball thrown upward from the surface of the Earth. First, the ball can be thrown upward with a speed greater than the escape speed; in this case, the ball continues to go upward forever. Second, the ball can be thrown upward with a speed less than the escape speed; in this case, the ball reaches a maximum altitude, then falls back down. Third, and finally, the ball can be thrown upward with a speed exactly equal to the escape speed; in this case, the speed of the ball approaches zero as t →∞. The Friedmann equation in its Newtonian form (equation 4.11) is useful in picturing how isotropically expanding objects behave under the influence 4.1. FRIEDMANN EQUATION 61 of their self-gravity. However, its application to the real universe must be re- garded with considerable skepticism. First of all, a spherical volume of finite radius 1 s cannot represent a homogeneous, isotropic universe. In a finite spherical volume, there exists a special location (the center of the sphere), violating the principle of homogeneity, and at any point there exists a special direction (the direction pointing toward the center), violating the principle of isotropy. We may instead regard the sphere of radius 1 s as being carved out of an infinite, homogeneous, isotropic universe. In that case, Newtonian dynamics tell us that the gravitational acceleration inside a hollow spheri- cally symmetric shell is equal to zero. We divide up the region outside the sphere into concentric shells, and thus conclude that the test mass : at 1 s experiences no net acceleration from matter at 1 1 s . Unfortunately, a Newtonian argument of this sort assumes that space is Euclidean. A deriva- tion of the correct Friedmann equation, including the possibility of spatial curvature, has to begin with Einstein’s field equations. The correct form of the Friedmann equation, including all general rela- tivistic effects, is ˙ c c 2 = 8πG 3c 2 ε(t) − κc 2 1 2 0 1 c(t) 2 . (4.13) Note the changes made in going from the Newtonian form of the Friedmann equation (equation 4.11) to the correct relativistic form (equation 4.13). The first change is that the mass density ρ has been replaced by an energy density ε divided by the square of the speed of light. One of Einstein’s insights was that in determining the gravitational influence of a particle, the important quantity was not its mass : but its energy, 1 = (: 2 c 4 +j 2 c 2 ) 1/2 . (4.14) Here j is the momentum of the particle as seen by an observer at the parti- cle’s location who sees the universe expanding isotropically around her. Any motion which a particle has, in addition to the motion associated with the expansion or contraction of the universe, is called the particle’s peculiar mo- tion. 2 If a massive particle is non-relativistic – that is, if its peculiar velocity · is much less than c – then its peculiar momentum will be j ≈ :·, and its 2 The adjective “peculiar” comes from the Latin “peculium”, meaning “private prop- erty”. The peculiar motion of a particle is thus the motion which belongs to the particle alone, and not to the global expansion or contraction of the universe. 62 CHAPTER 4. COSMIC DYNAMICS energy will be 1 non−rel ≈ :c 2 (1 +· 2 c 2 ) ≈ :c 2 + 1 2 :· 2 . (4.15) Thus, if the universe contained only massive, slowly moving particles, then the energy density ε would be nearly equal to ρc 2 , with only a small correction for the kinetic energy :· 2 2 of the particles. However, photons and other massless particles also have an energy, 1 rel = jc = /1 . (4.16) which also contributes to the energy density ε. Not only do photons respond to the curvature of space-time, they also contribute to it. The second change that must be made in going from the Newtonian form of the Friedmann equation to the correct relativistic form is making the substitution 2l : 2 s = − κc 2 1 2 0 . (4.17) In the context of general relativity, the curvature κ is related to the Newto- nian energy l of a test mass. The case with l < 0 corresponds to positive curvature (κ = +1), while the case with l 0 corresponds to negative cur- vature (κ = −1). The special case with l = 0 corresponds to the special case where the space is perfectly flat (κ = 0). Although I have not given the derivation of the Friedmann equation in the general relativistic case, it makes sense that the curvature, given by κ and 1 0 , the expansion rate, given by c(t), and the energy density ε should be bound up together in the same equation. After all, in Einstein’s view, the energy density of the universe determines both the curvature of space and the overall dynamics of the expansion. The Friedmann equation is a Very Important Equation in cosmology. 3 However, if we want to apply the Friedmann equation to the real universe, we must have some way of tying it to observable properties. For instance, the Friedmann equation can be tied to the Hubble constant, H 0 . Remember, in a universe whose expansion (or contraction) is described by a scale factor c(t), there’s a linear relation between recession speed · and proper distance d: ·(t) = H(t)d(t) . (4.18) 3 You should consider writing it in reverse on your forehead so that you can see it every morning in the mirror when you comb your hair. 4.1. FRIEDMANN EQUATION 63 where H(t) ≡ ˙ cc. Thus, the Friedmann equation can be rewritten in the form H(t) 2 = 8πG 3c 2 ε(t) − κc 2 1 2 0 c(t) 2 . (4.19) At the present moment, H 0 = H(t 0 ) = ˙ c c t=t 0 = 70 ±7 kms −1 Mpc −1 . (4.20) As an etymological aside, I should point out that the time-varying function H(t) is generally known as the “Hubble parameter”, while H 0 , the value of H(t) at the present day, is known as the “Hubble constant”. The Friedmann equation evaluated at the present moment is H 2 0 = 8πG 3c 2 ε 0 − κc 2 1 2 0 . (4.21) using the convention that a subscript “0” indicates the value of a time- varying quantity evaluated at the present. Thus, the Friedmann equation gives a relation among H 0 , which tells us the current rate of expansion, ε 0 , which tells us the current energy density, and κ1 2 0 , which tells us the current curvature. Due to the difficulty of measuring the curvature directly by geometric means, it is useful to have an indirect method of determining κ and 1 0 . If we were able to measure H 0 and ε 0 with high precision, we could use equation (4.21) to determine the curvature. Even without knowledge of the current density ε 0 , we can use equation (4.21) to place a lower limit on 1 0 in a negatively curved universe. If we assume ε 0 is non-negative, then for a given value of H 0 , the product κ1 2 0 is minimized in the limit ε 0 → 0. In the limit of a totally empty universe, with no energy content, the curvature is negative, with a radius of curvature 1 0 (min) = cH 0 . (4.22) This is the minimum radius of curvature which a negatively curved universe can have, assuming that general relativity correctly describes the curvature. Since we know that the universe contains matter and radiation, and hence that ε 0 0, the radius of curvature must be greater than the Hubble distance if the universe is negatively curved. 4 4 We also know from observations, as discussed earlier, that the radius of curvature must be comparable to or greater than the Hubble distance if the universe is positively curved. 64 CHAPTER 4. COSMIC DYNAMICS As we have seen, the Friedmann equation can generally be written as H(t) 2 = 8πG 3c 2 ε(t) − κc 2 1 2 0 c(t) . (4.23) for all universes with a Robertson-Walker metric whose expansion or contrac- tion is governed by the rules of general relativity. In a spatially flat universe (κ = 0), the Friedmann equation takes a particularly simple form: H(t) 2 = 8πG 3c 2 ε(t) . (4.24) Thus, for a given value of the Hubble parameter, there is a critical density, ε c (t) ≡ 3c 2 8πG H(t) 2 . (4.25) If the energy density ε(t) is greater than this value, the universe is positively curved (κ = +1). If ε(t) is less than this value, the universe is negatively curved (κ = −1). Since we know the current value of the Hubble parameter to within 10%, we can compute the current value of the critical density to within 20%: ε c,0 = 3c 2 8πG H 2 0 = (8.3 ±1.7) 10 −10 J m −3 = 5200 ±1000 MeVm −3 . (4.26) The critical density is frequently written as the equivalent mass density, ρ c,0 ≡ ε c,0 c 2 = (9.2 ±1.8) 10 −27 kg m −3 = (1.4 ±0.3) 10 11 M Mpc −3 . (4.27) Thus, the critical density is currently roughly equivalent to a density of one hydrogen atom per 200 liters, or 140 solar mass stars per cubic kiloparsec. This is definitely not a large density, by terrestrial standards. It’s not even a large density by the standards of interstellar space within our Galaxy, where even the hottest, most tenuous regions have a few protons per liter. However, you must keep in mind that most of the volume of the universe consists of intergalactic voids, where the density is extraordinarily low. When averaged over scales of 100 Mpc or more, the mean density of the universe, as it turns out, is close to the critical density. In discussing the curvature of the universe, it is more convenient to use not the absolute density ε, but the ratio of the density to the critical density 4.2. FLUID AND ACCELERATION EQUATIONS 65 ε c . Thus, when talking about the energy density of the universe, cosmologists often use the dimensionless density parameter Ω(t) ≡ ε(t) ε c (t) . (4.28) The most conservative limits on Ω – that is, limits that even the most bel- ligerent cosmologist will hesitate to quarrel with – state that the current value of the density parameter lies in the range 0.1 < Ω 0 < 2. In terms of the density parameter, the Friedmann equation can be written in yet another form: 1 −Ω(t) = − κc 2 1 2 0 c(t) 2 H(t) 2 . (4.29) Note that since the right hand side of equation (4.29) cannot change sign as the universe expands, neither can the left hand side. If Ω < 1 at any time, it remains less than one for all time; similarly, if Ω 1 at any time, it remains greater than one for all times, and if Ω = 1 at any time, Ω = 1 at all times. A leopard can’t change its spots; a universe governed by the Friedmann equation can’t change the sign of its curvature. At the present moment, the relation among curvature, density, and expansion rate can be written in the form 1 −Ω 0 = − κc 2 1 2 0 H 2 0 . (4.30) or κ 1 2 0 = H 2 0 c 2 (Ω 0 −1) . (4.31) If you know Ω 0 , you know the sign of the curvature (κ). If, in addition, you know the Hubble distance, cH 0 , you can compute the radius of curvature (1 0 ). 4.2 The fluid and acceleration equations Although the Friedmann equation is indeed important, it cannot, all by it- self, tell us how the scale factor c(t) evolves with time. Even if we had accurate boundary conditions (precise values for ε 0 and H 0 , for instance), it still remains a single equation in two unknowns, c(t) and ε(t). 5 5 Or, if we prefer, we may take the unknown functions as H(t) and Ω(t); in any case, there are two of them. 66 CHAPTER 4. COSMIC DYNAMICS We need another equation involving c and ε if we are to solve for c and ε as functions of time. The Friedmann equation, in the Newtonian approximation, is a statement of energy conservation; in particular, it says that the sum of the gravitational potential energy and the kinetic energy of expansion is constant. Energy conservation is a generally useful concept, so let’s look at another manifestation of the same concept – the first law of thermodynamics: dC = d1 +1d\ . (4.32) where dC is the heat flow into or out of a region, d1 is the change in internal energy, 1 is the pressure, and d\ is the change in volume of the region. This equation was applied in Section 2.5 to a comoving volume filled with photons, but it applies equally well to a comoving volume filled with any sort of fluid. If the universe is perfectly homogeneous, then for any volume dC = 0; that is, there is no bulk flow of heat. (Processes for which dC = 0 are known as adiabatic processes. Saying that the expansion of the universe is adiabatic is also a statement about entropy. The change in entropy do within a region is given by the relation do = dC1; thus, an adiabatic process is one in which entropy is not increased. A homogeneous, isotropic expansion of the universe does not increase the universe’s entropy.) Since dC = 0 for a comoving volume as the universe expands, the first law of thermodynamics, as applied to the expanding universe, reduces to the form ˙ 1 +1 ˙ \ = 0 . (4.33) For concreteness, consider a sphere of comoving radius : s expanding along with the universal expansion, so that its proper radius is 1 s (t) = c(t): s . The volume of the sphere is \ (t) = 4π 3 : 3 s c(t) 3 . (4.34) so the rate of change of the sphere’s volume is ˙ \ = 4π 3 : 3 s (3c 2 ˙ c) = \ 3 ˙ c c . (4.35) The internal energy of the sphere is 1(t) = \ (t)ε(t) . (4.36) so the rate of change of the sphere’s internal energy is ˙ 1 = \ ˙ ε + ˙ \ ε = \ ˙ ε + 3 ˙ c c ε . (4.37) 4.2. FLUID AND ACCELERATION EQUATIONS 67 Combining equations (4.33), (4.35), and (4.37), we find that the first law of thermodynamics in an expanding (or contracting) universe takes the form \ ˙ ε + 3 ˙ c c ε + 3 ˙ c c 1 = 0 . (4.38) or ˙ ε + 3 ˙ c c (ε +1) = 0 . (4.39) The above equation is called the fluid equation, and is the second of the key equations describing the expansion of the universe. 6 The Friedmann equation and fluid equation are statements about energy conservation. By combining the two, we can derive an acceleration equation which tells how the expansion of the universe speeds up or slows down with time. The Friedmann equation (equation 4.13), multiplied by c 2 , takes the form ˙ c 2 = 8πG 3c 2 εc 2 − κc 2 1 2 0 . (4.40) Taking the time derivative yields 2˙ c¨ c = 8πG 3c 2 ( ˙ εc 2 + 2εc˙ c) . (4.41) Dividing by 2˙ cc tells us ¨ c c = 4πG 3c 2 ˙ ε c ˙ c + 2ε . (4.42) Using the fluid equation (equation 4.39), we may make the substitution ˙ ε c ˙ c = −3(ε +1) (4.43) to find the usual form of the acceleration equation, ¨ c c = − 4πG 3c 2 (ε + 31) . (4.44) Note that if the energy density ε is positive, it provides a negative acceleration – that is, it decreases the value of ˙ c and reduces the relative velocity of 6 Write it on your forehead just underneath the Friedmann equation. 68 CHAPTER 4. COSMIC DYNAMICS any two points in the universe. The acceleration equation also includes the pressure 1 which is associated with the material filling the universe. 7 A gas made of ordinary baryonic matter has a positive pressure 1, re- sulting from the random thermal motions of the molecules, atoms, or ions of which the gas is made. A gas of photons also has a positive pressure, as does a gas of neutrinos or WIMPs. The positive pressure associated with these components of the universe will cause the expansion to slow down. Suppose, though, that the universe had a component with a pressure 1 < −ε3 . (4.45) Inspection of the acceleration equation (equation 4.44) shows us that such a component will cause the expansion of the universe to speed up rather than slow down. A negative pressure (also called “tension”) is certainly permissible by the laws of physics. Compress a piece of rubber, and its internal pressure will be positive; stretch the same piece of rubber, and its pressure will be negative. In cosmology, the much-discussed cosmological constant is a component of the universe with negative pressure. As we’ll discuss in more detail in Section 4.4, a cosmological constant has 1 = −ε, and thus causes a positive acceleration for the expansion of the universe. 4.3 Equations of state To recap, we now have three key equations which describe how the universe expands. There’s the Friedmann equation, ˙ c c 2 = 8πG 3c 2 ε − κc 2 1 2 0 c 2 . (4.46) the fluid equation, ˙ ε + 3 ˙ c c (ε +1) = 0 . (4.47) and the acceleration equation, ¨ c c = − 4πG 3c 2 (ε + 31) . (4.48) 7 Although we think of ε as an energy per unit volume and P as a force per unit area, they both have the same dimensionality: in SI units, 1 J m −3 = 1 Nm −2 = 1 kg m −1 s −2 . 4.3. EQUATIONS OF STATE 69 Of the above three equations, only two are independent, since equation (4.48), as we’ve just seen, can be derived from equations (4.46) and (4.47). Thus, we have a system of two independent equations in three unknowns – the functions c(t), ε(t), and 1(t). To solve for the scale factor, energy density, and pressure as a function of cosmic time, we need another equation. What we need is an equation of state; that is, a mathematical relation between the pressure and energy density of the stuff that fills up the universe. If only we had a relation of the form 1 = 1(ε) . (4.49) life would be complete – or at least, our set of equations would be complete. We could then, given the appropriate boundary conditions, solve them to find how the universe expanded in the past, and how it will expand (or contract) in the future. In general, equations of state can be dauntingly complicated. Condensed matter physicists frequently deal with substances in which the pressure is a complicated nonlinear function of the density. Fortunately, cosmology usu- ally deals with dilute gases, for which the equation of state is simple. For substances of cosmological importance, the equation of state can be written in a simple linear form: 1 = uε . (4.50) where u is a dimensionless number. Consider, for instance, a low-density gas of non-relativistic massive par- ticles. Non-relativistic, in this case, means that the random thermal motions of the gas particles have peculiar velocities which are tiny compared to the speed of light. Such a non-relativistic gas obeys the perfect gas law, 1 = ρ j /1 . (4.51) where j is the mean mass of the gas particles. The energy density ε of a non- relativistic gas is almost entirely contributed by the mass of the gas particles: ε ≈ ρc 2 . Thus, in terms of ε, the perfect gas law is 1 ≈ /1 jc 2 ε . (4.52) For a non-relativistic gas, the temperature 1 and the root mean square ther- mal velocity '· 2 ` are associated by the relation 3/1 = j'· 2 ` . (4.53) 70 CHAPTER 4. COSMIC DYNAMICS Thus, the equation of state for a non-relativistic gas can be written in the form 1 non−rel = uε non−rel . (4.54) where u ≈ '· 2 ` 3c 2 <1 . (4.55) Most of the gases we encounter in everyday life are non-relativistic. For in- stance, in air at room temperature, the nitrogen molecules are slow-poking along with a root mean square velocity of ∼ 500 ms −1 , yielding u ∼ 10 −12 . Even in astronomical contexts, gases are mainly non-relativistic at the present moment. Within a gas of ionized hydrogen, for instance, the electrons are non-relativistic as long as 1 < 6 10 9 K; the protons are non-relativistic when 1 <10 13 K. A gas of photons, or other massless particles, is guaranteed to be rela- tivistic. Although photons have no mass, they have momentum, and hence exert pressure. The equation of state of photons, or of any other relativistic gas, is 1 rel = 1 3 ε rel . (4.56) (This relation has already been used in Section 2.5, to compute how the Cosmic Microwave Background cools as the universe expands.) A gas of highly relativistic massive particles (with '· 2 ` ∼ c 2 ) will also have u = 13; a gas of mildly relativistic particles (with 0 < '· 2 ` < c 2 ) will have 0 < u < 13. The equation-of-state parameter u can’t take on arbitrary values. Small perturbations in a substance with pressure 1 will travel at the speed of sound. For adiabatic perturbations in a gas with pressure 1 and energy density ε, the sound speed is given by the relation c 2 s = c 2 d1 dε . (4.57) In a substance with u 0, the sound speed is thus c s = √ uc. 8 Sound waves cannot travel faster than the speed of light; if they did, you would be able to send a sound signal into the past, and violate causality. Thus, u is restricted to values u ≤ 1. 8 In a substance with w < 0, the sound speed is an imaginary number; this implies that small pressure perturbations to a substance with negative pressure will not constitute stably propagating sound waves, but will have amplitudes which grow or decay with time. 4.4. LEARNING TO LOVE LAMBDA 71 Some values of u are of particular interest. For instance, the case u = 0 is of interest, because we know that our universe contains non-relativistic matter. The case u = 13 is of interest, because we know that our universe contains photons. For simplicity, I will refer to the component of the universe which consists of non-relativistic particles (and hence has u ≈ 0) as “matter”. I will refer to the component which consists of photons and other relativistic particles (and hence has u = 13) as “radiation”. The case u < −13 is of interest, because a component with u < −13 will provide a positive acceleration (¨ c 0 in equation 4.48). A component of the universe with u < −13 is sometimes referred to generically as “dark energy” (a phrase coined by the cosmologist Michael Turner). One form of “dark energy” is of special interest; some observational evidence, which we’ll review in future chapters, indicates that our universe may contain a cosmological constant. A cosmological constant may be simply defined as a component of the universe which has u = −1, and hence has 1 = −ε. The cosmological constant, also designated by the Greek letter Λ, has had a controversial history, and is still the subject of debate. To learn why cosmologists have had such a long- standing love/hate affair with the cosmological constant Λ, it is necessary to make a brief historical review. 4.4 Learning to love lambda The cosmological constant Λ was first introduced by Albert Einstein. After publishing his first paper on general relativity in 1915, Einstein, naturally enough, wanted to apply his field equation to the real universe. He looked around, and noted that the universe contains both radiation and matter. Since Einstein, along with every other earthling of his time, was unaware of the existence of the Cosmic Microwave Background, he thought that most of the radiation in the universe was in the form of starlight. He also noted, quite correctly, that the energy density of starlight in our Galaxy is much less than the rest energy density of the stars. 9 Thus, Einstein concluded that the primary contribution to the energy density of the universe was from non- relativistic matter, and that he could safely make the approximation that we live in a pressureless universe. So far, Einstein was on the right track. However, in 1915, astronomers were unaware of the existence of the expansion of the universe. In fact, it was 9 That is, stars have only converted a small fraction of their initial mass into light. 72 CHAPTER 4. COSMIC DYNAMICS by no means settled that galaxies besides our own actually existed. After all, the sky is full of faint fuzzy patches of light. It took some time to sort out that some of the faint fuzzy patches are glowing clouds of gas within our Galaxy and that some of them are galaxies in their own right, far beyond our own Galaxy. Thus, when Einstein asked, “Is the universe expanding or contracting?” he looked, not at the motions of galaxies, but at the motions of stars within our Galaxy. Einstein noted that some stars are moving toward us and that others are moving away from us, with no evidence that the Galaxy is expanding or contracting. The incomplete evidence available to Einstein led him to the belief that the universe is static – neither expanding nor contracting – and that it has a positive energy density but negligible pressure. Einstein then had to ask the question, “Can a universe filled with non-relativistic matter, and nothing else, be static?” The answer to this question is “No!” A universe containing nothing but matter must, in general, be either expanding or contracting. The reason why this is true can be illustrated in a Newtonian context. If the mass density of the universe is ρ, then the gravitational potential Φ is given by Poisson’s equation: ∇ 2 Φ = 4πGρ . (4.58) The gravitational acceleration c at any point in space is then found by taking the gradient of the potential: c = − ∇Φ . (4.59) In a static universe, c must vanish everywhere in space. Thus, the poten- tial Φ must be constant in space. However, if Φ is constant, then (from equation 4.58) ρ = 1 4πG ∇ 2 Φ = 0 . (4.60) The only permissible static universe, in this analysis, is a totally empty uni- verse. If you create a matter-filled universe which is initially static, then gravity will cause it to contract. If you create a matter-filled universe which is initially expanding, then it will either expand forever (if the Newtonian en- ergy l is greater than or equal to zero) or reach a maximum radius and then collapse (if l < 0). Trying to make a matter-filled universe which doesn’t expand or collapse is like throwing a ball into the air and expecting it to hover there. 4.4. LEARNING TO LOVE LAMBDA 73 How did Einstein surmount this problem? How did he reconcile the fact that the universe contains matter with his desire for a static universe? Ba- sically, he added a fudge factor to the equations. In Newtonian terms, what he did was analogous to rewriting Poisson’s equation in the form ∇ 2 Φ + Λ = 4πGρ . (4.61) The new term, symbolized by the Greek letter Λ, came to be known as the cosmological constant. Note that it has dimensionality (time) −2 . Introducing Λ into Poisson’s equation allows the universe to be static if you set Λ = 4πGρ. In general relativistic terms, what Einstein did was to add an additional term, involving Λ, to his field equation (the relativistic equivalent of Pois- son’s equation). If the Friedmann equation is re-derived from Einstein’s field equation, with the Λ term added, it becomes ˙ c c 2 = 8πG 3c 2 ε − κc 2 1 2 0 c 2 + Λ 3 . (4.62) The fluid equation is unaffected by the presence of a Λ term, so it still has the form ˙ ε + 3 ˙ c c (ε +1) = 0 . (4.63) With the Λ term present, the acceleration equation becomes ¨ c c = − 4πG 3c 2 (ε + 31) + Λ 3 . (4.64) A look at the Friedmann equation (4.62) tells us that adding the Λ term is equivalent to adding a new component to the universe with energy density ε Λ ≡ c 2 8πG Λ . (4.65) If Λ remains constant with time, then so does its associated energy density ε Λ . The fluid equation (4.63) tells us that to have ε Λ constant with time, the Λ term must have an associated pressure 1 Λ = −ε Λ = − c 2 8πG Λ . (4.66) Thus, we can think of the cosmological constant as a component of the uni- verse which has a constant density ε Λ and a constant pressure 1 Λ = −ε Λ . 74 CHAPTER 4. COSMIC DYNAMICS By introducing a Λ term into his equations, Einstein got the static model universe he wanted. For the universe to remain static, both ˙ c and ¨ c must be equal to zero. If ¨ c = 0, then in a universe with matter density ρ and cosmological constant Λ, the acceleration equation (4.64) reduces to 0 = − 4πG 3 ρ + Λ 3 . (4.67) Thus, Einstein had to set Λ = 4πGρ in order to produce a static universe, just as in the Newtonian case. If ˙ c = 0, the Friedmann equation (4.62) reduces to 0 = 8πG 3 ρ − κc 2 1 2 0 + Λ 3 = 4πGρ − κc 2 1 2 0 . (4.68) Einstein’s static model therefore had to be positively curved (κ = +1), with a radius of curvature 1 0 = c 2(πGρ) 1/2 = c Λ 1/2 . (4.69) Although Einstein published the details of his static, positively curved, matter- filled model in the spring of 1917, he was dissatisfied with the model. He believed that the cosmological constant was “gravely detrimental to the for- mal beauty of the theory”. In addition to its aesthetic shortcomings, the model had a practical defect; it was unstable. Although Einstein’s static model was in equilibrium, with the repulsive force of Λ balancing the attrac- tive force of ρ, it was an unstable equilibrium. Consider expanding Einstein’s universe just a tiny bit. The energy density of Λ remains unchanged, but the energy density of matter drops. Thus, the repulsive force is greater than the attractive force, and the universe expands further. This causes the mat- ter density to drop further, which causes the expansion to accelerate, which causes the matter density to drop further, and so forth. Expanding Einstein’s static universe triggers runaway expansion; similarly, compressing it causes a runaway collapse. Einstein was willing, even eager, to dispose of the “ugly” cosmological constant in his equations. Hubble’s 1929 paper on the redshift – distance re- lation gave Einstein the necessary excuse for tossing Λ onto the rubbish heap. (Einstein later described the cosmological constant Λ as “the greatest blun- der of my career.”) Ironically, however, the same paper that caused Einstein to abandon the cosmological constant caused other scientists to embrace it. In his initial analysis, remember, Hubble badly underestimated the distance 4.4. LEARNING TO LOVE LAMBDA 75 to galaxies, and hence overestimated the Hubble constant. Hubble’s initial value of H 0 = 500 kms −1 Mpc −1 leads to a Hubble time of H −1 0 = 2 Gyr, less than half the age of the Earth, as known from radioactive dating. How could cosmologists reconcile a short Hubble time with an old Earth? Some cosmologists pointed out that one way to increase the age of the universe for a given value of H −1 0 was to introduce a cosmological constant. If the value of Λ is large enough to make ¨ c 0, then ˙ c was smaller in the past than it is now, and consequently the universe is older than H −1 0 . If Λ has a value greater than 4πGρ 0 , then the expansion of the universe is accelerating, and the universe can be arbitrarily old for a given value of H −1 0 . Since 1917, the cosmological constant has gone in and out of fashion, like sideburns or short skirts. It has been particularly fashionable during periods when the favored value of the Hubble time H −1 0 has been embarrassingly short compared to the estimated ages of astronomical objects. Currently, the cosmological constant is popular, thanks to observations, which we will discuss in Chapter 7, that indicate that the expansion of the universe does, indeed, have a positive acceleration. A question which has been asked since the time of Einstein – and one which I’ve assiduously dodged until this moment – is “What is the physical cause of the cosmological constant?” In order to give Λ a real physical meaning, we need to identify some component of the universe whose energy density ε Λ remains constant as the universe expands or contracts. Currently, the leading candidate for this component is the vacuum energy. In classical physics, the idea of a vacuum having energy is nonsense. A vacuum, from the classical viewpoint, contains nothing; and as King Lear would say, “Nothing can come of nothing.” In quantum physics, however, a vacuum is not a sterile void. The Heisenberg uncertainty principle permits particle-antiparticle pairs to spontaneously appear and then annihilate in an otherwise empty vacuum. The total energy ∆1 and the lifetime ∆t of these pairs of virtual particles must satisfy the relation 10 ∆1∆t ≤ / . (4.70) Just as there’s an energy density associated with the real particles in the universe, there is an energy density ε vac associated with the virtual particle- antiparticle pairs. The vacuum density ε vac is a quantum phenomenon which 10 The usual analogy that’s made is to an embezzling bank teller who takes money from the till but who always replaces it before the auditor comes around. Naturally, the more money a teller is entrusted with, the more frequently the auditor checks up on her. 76 CHAPTER 4. COSMIC DYNAMICS doesn’t give a hoot about the expansion of the universe and is independent of time as the universe expands or contracts. Unfortunately, computing the numerical value of ε vac is an exercise in quantum field theory which has not yet been successfully completed. It has been suggested that the natural value for the vacuum energy density is the Planck energy density, ε vac ∼ 1 P / 3 P (???) . (4.71) As we’ve seen in Chapter 1, the Planck energy is large by particle physics standards (1 P = 1.210 28 eV), while the Planck length is small by anybody’s standards (/ P = 1.6 10 −35 m). This gives an energy density ε vac ∼ 3 10 133 eVm −3 (!!!) . (4.72) This is 124 orders of magnitude larger than the current critical density for our universe, and represents a spectacularly bad match between theory and observations. Obviously, we don’t know much yet about the energy density of the vacuum! This is a situation where astronomers can help particle physicists, by deducing the value of ε Λ from observations of the expansion of the universe. By looking at the universe at extremely large scales, we are indirectly examining the structure of the vacuum on extremely small scales. Suggested reading [Full references are given in the “Annotated Bibliography” on page 286.] Bernstein (1995), ch. 2: Friedmann equation, with applications Harrison (2000), ch. 16: Newtonian derivation of Friedmann equation Liddle (1999), ch. 3: Derivation of Friedmann, fluid, and acceleration equa- tions Linder (1997), ch. 2: Dynamics of the universe, including a discussion of equations of state Rich (2001), ch. 4: “Relativistically correct” derivation of the Friedmann equation 4.4. LEARNING TO LOVE LAMBDA 77 Problems (4.1) Suppose the energy density of the cosmological constant is equal to the present critical density ε Λ = ε c,0 = 5200 MeVm −3 . What is the total energy of the cosmological constant within a sphere 1 AU in radius? What is the rest energy of the Sun (1 = ` c 2 )? Comparing these two numbers, do you expect the cosmological constant to have a significant effect on the motion of planets within the Solar System? (4.2) Consider Einstein’s static universe, in which the attractive force of the matter density ρ is exactly balanced by the repulsive force of the cosmological constant, Λ = 4πGρ. Suppose that some of the matter is converted into radiation (by stars, for instance). Will the universe start to expand or contract? Explain your answer. (4.3) If ρ = 310 −27 kg m −3 , what is the radius of curvature 1 0 of Einstein’s static universe? How long would it take a photon to circumnavigate such a universe? (4.4) Suppose that the universe were full of regulation baseballs, each of mass : bb = 0.145 kg and radius : bb = 0.0369 m. If the baseballs were distributed uniformly throughout the universe, what number density of baseballs would be required to make the density equal to the critical density? (Assume non-relativistic baseballs.) Given this density of baseballs, how far would you be able to see, on average, before your line of sight intersected a baseball? In fact, we can see galaxies at a distance ∼ cH 0 ∼ 4000 Mpc; does the transparency of the universe on this length scale place useful limits on the number density of intergalactic baseballs? [Note to readers outside North America or Japan: feel free to substitute regulation cricket balls, with : cr = 0.160 kg and : cr = 0.0360 m.] (4.5) The principle of wave-particle duality tells us that a particle with momentum j has an associated de Broglie wavelength of λ = /j; this wavelength increases as λ ∝ c as the universe expands. The total energy density of a gas of particles can be written as ε = :1, where : is the number density of particles, and 1 is the energy per particle. For simplicity, let’s assume that all the gas particles have the same mass 78 CHAPTER 4. COSMIC DYNAMICS : and momentum j. The energy per particle is then simply 1 = (: 2 c 4 +j 2 c 2 ) 1/2 = (: 2 c 4 +/ 2 c 2 λ 2 ) 1/2 . (4.73) Compute the equation-of-state parameter u for this gas as a function of the scale factor c. Show that u = 13 in the highly relativistic limit (c → 0, j → ∞) and that u = 0 in the highly non-relativistic limit (c →∞, j →0). Chapter 5 Single-Component Universes In a spatially homogeneous and isotropic universe, the relation among the energy density ε(t), the pressure 1(t), and the scale factor c(t) is given by the Friedmann equation, ˙ c c 2 = 8πG 3c 2 ε − κc 2 1 2 0 c 2 . (5.1) the fluid equation, ˙ ε + 3 ˙ c c (ε +1) = 0 . (5.2) and the equation of state, 1 = uε . (5.3) In principle, given the appropriate boundary conditions, we can solve equa- tions (5.1), (5.2), and (5.3) to yield ε(t), 1(t), and c(t) for all times, past and future. 5.1 Evolution of energy density In reality, the evolution of our universe is complicated by the fact that it contains different components with different equations of state. We know that the universe contains non-relativistic matter and radiation – that’s a conclusion as firm as the earth under your feet and as plain as daylight. Thus, the universe contains components with both u = 0 and u = 13. It may well contain a cosmological constant, with u = −1. Moreover, the possibility exists that it may contain still more exotic components, with different values 79 80 CHAPTER 5. SINGLE-COMPONENT UNIVERSES of u. Fortunately for the cause of simplicity, the energy density and pressure for the different components of the universe are additive. We may write the total energy density ε as the sum of the energy density of the different components: ε = ¸ w ε w . (5.4) where ε w represents the energy density of the component with equation-of- state parameter u. The total pressure 1 is the sum of the pressures of the different components: 1 = ¸ w 1 w = ¸ w uε w . (5.5) Because the energy densities and pressures add in this way, the fluid equation (5.2) must hold for each component separately, as long as there is no inter- action between the different components. If this is so, then the component with equation-of-state parameter u obeys the equation ˙ ε w + 3 ˙ c c (ε w +1 w ) = 0 (5.6) or ˙ ε w + 3 ˙ c c (1 +u)ε w = 0 . (5.7) Equation (5.7) can be rearranged to yield dε w ε w = −3(1 +u) dc c . (5.8) If we assume that u is constant, then ε w (c) = ε w,0 c −3(1+w) . (5.9) Here, I’ve used the usual normalization that c 0 = 1 at the present day, when the energy density of the u component is ε w,0 . Note that equation (5.9) is de- rived solely from the fluid equation and the equation of state; the Friedmann equation doesn’t enter into it. Starting from the general result of equation (5.9), we conclude that the energy density ε m associated with non-relativistic matter decreases as the universe expands with the dependence ε m (c) = ε m,0 c 3 . (5.10) 5.1. EVOLUTION OF ENERGY DENSITY 81 Figure 5.1: The dilution of non-relativistic particles (“matter”) and relativis- tic particles (“radiation”) as the universe expands. The energy density in radiation, ε r , drops at the steeper rate ε r (c) = ε r,0 c 4 . (5.11) Why this difference between matter and radiation? We may write the en- ergy density of either component in the form ε = :1, where : is the number density of particles and 1 is the mean energy per particle. For both rela- tivistic and non-relativistic particles, the number density has the dependence : ∝ c −3 as the universe expands, assuming that particles are neither created nor destroyed. Consider, once again, a sphere of comoving radius : s , which expands along with the general expansion of the universe (Figure 5.1). When its 82 CHAPTER 5. SINGLE-COMPONENT UNIVERSES proper radius expands by a factor 2, its volume increases by a factor 8, and the density of particles which it contains falls to 1/8 its previous value. The energy of non-relativistic particles (shown in the upper panel of Figure 5.1) is contributed solely by their rest mass (1 = :c 2 ) and remains constant as the universe expands. Thus, for non-relativistic matter, ε m = :1 = :(:c 2 ) ∝ c −3 . The energy of photons or other massless particles (shown in the lower panel of Figure 5.1) has the dependence 1 = /cλ ∝ c −1 , since, as shown in equation (3.45), λ ∝ c as the universe expands. Thus, for photons and other massless particles, ε r = :1 = :(/cλ) ∝ c −3 c −1 ∝ c −4 . Although I’ve explained why photons have an energy density ε r ∝ c −4 , the explanation required the assumption that photons are neither created nor destroyed. Such an assumption is not, strictly speaking, correct; pho- tons are continuously being created. The Sun, for instance, is emitting out roughly 10 45 photons every second. However, so numerous are the photons of the Cosmic Microwave Background, it turns that the energy density of the CMB is larger than the energy density of all the photons emitted by all the stars in the history of the universe. To see why this is true, remember, from Section 2.5, that the present energy density of the CMB, which has a temperature 1 0 = 2.725 K, is ε CMB,0 = α1 4 0 = 4.17 10 −14 J m −3 = 0.260 MeVm −3 . (5.12) yielding a density parameter for the CMB of Ω CMB,0 ≡ ε CMB,0 ε c,0 = 0.260 MeVm −3 5200 MeVm −3 = 5.0 10 −5 . (5.13) Although the energy density of the CMB is small compared to the critical density, it is large compared to the energy density of starlight. Remember from Section 2.3 that the present luminosity density of galaxies is :1 ≈ 2 10 8 L Mpc −3 ≈ 2.6 10 −33 watts m −3 . (5.14) As a very rough estimate, let’s assume that galaxies have been emitting light at this rate for the entire age of the universe, t 0 ≈ H −1 0 ≈ 14 Gyr ≈ 4.4 10 17 s. This gives an energy density in starlight of ε starlight,0 ∼ :1t 0 ∼ (2.6 10 −33 J s −1 m −3 )(4.4 10 17 s) ∼ 1 10 −15 J m −3 ∼ 0.007 MeVm −3 . (5.15) 5.1. EVOLUTION OF ENERGY DENSITY 83 Thus, the average energy density of starlight is currently only ∼ 3% of the energy density of the CMB. In fact, the estimate given above is a very rough one indeed. Measurements of background radiation from ultraviolet wave- lengths to the near infrared, which includes both direct starlight and starlight absorbed and reradiated by dust, yield the larger value ε starlight ε CMB ≈ 0.1. In the past, the ratio of starlight density to CMB density was even smaller than it is today. For most purposes, it is an acceptable approximation to ig- nore non-CMB photons when computing the mean energy density of photons in the universe. The Cosmic Microwave Background, remember, is a relic of the time when the universe was hot and dense enough to be opaque to photons. If we extrapolate further back in time, we reach a time when the universe was hot and dense enough to be opaque to neutrinos. As a consequence, there should be a Cosmic Neutrino Background today, analogous to the Cosmic Microwave Background. The energy density in neutrinos should be comparable to, but not exactly equal to, the energy density in photons. A detailed calculation indicates that the energy density of each neutrino flavor should be ε = 7 8 4 11 4/3 ε CMB ≈ 0.227 ε CMB . (5.16) (The above result assumes that the neutrinos are relativistic, or, equivalently, that their energy is much greater than their rest energy : ν c 2 .) The density parameter of the Cosmic Neutrino Background, taking into account all three flavors of neutrino, should then be Ω ν = 0.681 Ω CMB . (5.17) as long as all neutrino flavors are relativistic. The mean energy per neutrino will be comparable to, but not exactly equal to, the mean energy per photon: 1 ν ≈ 5 10 −4 eV c . (5.18) as long as 1 ν : ν c 2 . When the mean energy of a particular neutrino species drops to ∼ : ν c 2 , then it makes the transition from being “radiation” to being “matter”. The neutrinos of the Cosmic Neutrino Background, I should note, have not yet been detected. Although neutrinos have been detected from the Sun and from Supernova 1987A, current technology permits the detection only 84 CHAPTER 5. SINGLE-COMPONENT UNIVERSES of neutrinos with energy 1 0.1MeV, far more energetic than the neutrinos of the Cosmic Neutrino Background. If all neutrino species are effectively massless today, with : ν c 2 < 5 10 −4 eV, then the present density parameter in radiation is Ω r,0 = Ω CMB,0 + Ω ν,0 = 5.0 10 −5 + 3.4 10 −5 = 8.4 10 −5 . (5.19) We know the energy density of the Cosmic Microwave Background with high precision. We can calculate theoretically what the energy density of the Cos- mic Neutrino Background should be. Unfortunately, the total energy density of non-relativistic matter is poorly known, as is the energy density of the cosmological constant. As we shall see in future chapters, the available evi- dence favors a universe in which the density parameter for matter is currently Ω m,0 ∼ 0.3, while the density parameter for the cosmological constant is cur- rently Ω Λ,0 ∼ 0.7. Thus, when I want to employ a model which matches the observed properties of the real universe, I will use what I call the “Bench- mark Model”; this model has Ω r,0 = 8.4 10 −5 in radiation, Ω m,0 = 0.3 in non-relativistic matter, and Ω Λ,0 = 1 − Ω r,0 − Ω m,0 ≈ 0.7 in a cosmological constant. 1 In the Benchmark Model, at the present moment, the ratio of the energy density in Λ to the energy density in matter is ε Λ,0 ε m,0 = Ω Λ,0 Ω m,0 ≈ 0.7 0.3 ≈ 2.3 . (5.20) In the language of cosmologists, the cosmological constant is “dominant” over matter today in the Benchmark Model. In the past, however, when the scale factor was smaller, the ratio of densities was ε Λ (c) ε m (c) = ε Λ,0 ε m,0 c 3 = ε Λ,0 ε m,0 c 3 . (5.21) If the universe has been expanding from an initial very dense state, at some moment in the past, the energy density of matter and Λ must have been equal. This moment of matter-Λ equality occurred when the scale factor was c mΛ = Ω m,0 Ω Λ,0 1/3 ≈ 0.3 0.7 1/3 ≈ 0.75 . (5.22) 1 Note that the Benchmark Model is spatially flat. 5.1. EVOLUTION OF ENERGY DENSITY 85 Similarly, the ratio of the energy density in matter to the energy density in radiation is currently ε m,0 ε r,0 = Ω m,0 Ω r,0 ≈ 0.3 8.4 10 −5 ≈ 3600 (5.23) if all three neutrino flavors in the Cosmic Neutrino Background are still relativistic today; it’s even larger if some or all of the neutrino flavors are currently non-relativistic. Thus, matter is now strongly dominant over ra- diation. However, in the past, the ratio of matter density to energy density was ε m (c) ε r (c) = ε m,0 ε r,0 c . (5.24) Thus, the moment of radiation-matter equality took place when the scale factor was c rm = ε m,0 ε r,0 ≈ 1 3600 ≈ 2.8 10 −4 . (5.25) Note that as long as a neutrino’s mass is : ν c 2 <(3600)(510 −4 eV) ∼ 2 eV, then it would have been relativistic at a scale factor c = 13600, and hence would have been “radiation” then even if it is “matter” today. To generalize, if the universe contains different components with different values of u, equation (5.9) tells us that in the limit c → 0, the component with the largest value of u is dominant. If the universe expands forever, then as c → ∞, the component with the smallest value of u is dominant. The evidence indicates we live in a universe where radiation (u = 13) was dominant during the early stages, followed by a period when matter (u = 0) was dominant. If the presently available evidence is correct, and we live in a universe described by the Benchmark Model, we have only recently entered a period when the cosmological constant Λ (u = −1) is dominant. In a universe which is continuously expanding, the scale factor c is a monotonically increasing function of t. Thus, in a continuously expanding universe, the scale factor c can be used as a surrogate for the cosmic time t. We can refer, for instance, to the moment when c = 2.8 10 −4 with the assurance that we are referring to a unique moment in the history of the universe. In addition, because of the simple relation between scale factor and redshift, c = 1 1 +. . (5.26) 86 CHAPTER 5. SINGLE-COMPONENT UNIVERSES cosmologists often use redshift as a surrogate for time. For example, they make statements such as, “Radiation-matter equality took place at a redshift . rm ≈ 3600.” That is, light that was emitted at the time of radiation-matter equality is observed by us with its wavelength increased by a factor of 3600. One reason why cosmologists use scale factor or redshift as a surrogate for time is that the conversion from c to t is not simple to calculate in a multiple- component universe like our own. In a universe with many components, can be written in the form ˙ c 2 = 8πG 3c 2 ¸ w ε w,0 c −1−3w − κc 2 1 2 0 . (5.27) Each term on the right hand side of equation (5.27) has a different depen- dence on scale factor; radiation contributes a term ∝ c −2 , matter contributes a term ∝ c −1 , curvature contributes a term independent of c, and the cos- mological constant Λ contributes a term ∝ c 2 . Solving equation (5.27) for a multiple-component model, such as the Benchmark Model, does not yield a simple analytic form for c(t). However, looking at simplified single- component universes, in which there is only one term on the right hand side of equation (5.27), yields useful insight into the physics of an expanding universe. 5.2 Curvature only A particularly simple universe is one which is empty – no radiation, no mat- ter, no cosmological constant, no contribution to ε of any sort. For this universe, the Friedmann equation (5.27) takes the form ˙ c 2 = − κc 2 1 2 0 . (5.28) One solution to this equation has ˙ c = 0 and κ = 0. An empty, static, spatially flat universe is a permissible solution to the Friedmann equation. This is the universe whose geometry is described by the Minkowski metric of equation (3.20), and in which all the transformations of special relativity hold true. However, equation (5.28) tells us that it is also possible to have an empty universe with κ = −1. (Positively curved empty universes are forbidden, 5.2. CURVATURE ONLY 87 since that would require an imaginary value of ˙ c in equation 5.28.) A nega- tively curved empty universe must be expanding or contracting, with ˙ c = ± c 1 0 . (5.29) In an expanding empty universe, integration of this relation yields a scale factor of the form c(t) = t t 0 . (5.30) where t 0 = 1 0 c. 2 In Newtonian terms, if there’s no gravitational force at work, then the relative velocity of any two points is constant, and thus the scale factor c simply increases linearly with time in an empty universe. The scale factor in an empty, expanding universe is shown as the dashed line in Figure 5.2. Note that in an empty universe, t 0 = H −1 0 ; with nothing to speed or slow the expansion, the age of the universe is exactly equal to the Hubble time. An empty, expanding universe might seem nothing more than a mathe- matical curiosity. 3 However, if a universe has a density ε which is very small compared to the critical density ε c (that is, if Ω < 1), then the linear scale factor of equation (5.30) is a good approximation to the true scale factor. Suppose you were in an expanding universe with a negligibly small value for the density parameter Ω, so that you could reasonably approximate it as an empty, negatively curved universe, with t 0 = H −1 0 = 1 0 c. You observe a distant light source, such as a galaxy, which has a redshift .. The light which you observe now, at t = t 0 , was emitted at an earlier time, t = t e . In an empty expanding universe, 1 +. = 1 c(t e ) = t 0 t e . (5.31) so it is easy to compute the time when the light you observe from the source was emitted: t e = t 0 1 +. = H −1 0 1 +. . (5.32) When observing a galaxy with a redshift ., in addition to asking “When was the light from that galaxy emitted?” you may also ask “How far away 2 Such an empty, negatively curved, expanding universe is sometimes called a Milne universe, after cosmologist to study its properties. 3 If a universe contains nothing, there will be no observers in it to detect the expansion. 88 CHAPTER 5. SINGLE-COMPONENT UNIVERSES −2 0 2 4 6 8 0 2 4 6 8 H 0 (t−t 0 ) a Λ empty matter radiation Figure 5.2: The scale factor as a function of time for an expanding, empty uni- verse (dashed), a flat, matter-dominated universe (dotted), a flat, radiation- dominated universe (solid), and a flat, Λ-dominated universe (dot-dash). 5.2. CURVATURE ONLY 89 is that galaxy?” In Section 3.3 we saw that in any universe described by a Robertson-Walker metric, the current proper distance from an observer at the origin to a galaxy at coordinate location (:. θ. φ) is (see equation 3.28) d p (t 0 ) = c(t 0 ) r 0 d: = : . (5.33) Moreover, if light is emitted by the galaxy at time t e and detected by the observer at time t 0 , the null geodesic followed by the light satisfies equa- tion (3.39): c t 0 te dt c(t) = r 0 d: = : . (5.34) Thus, the current proper distance from you (the observer) to the galaxy (the light source) is d p (t 0 ) = c t 0 te dt c(t) . (5.35) Equation (5.35) holds true in any universe whose geometry is described by a Robertson-Walker metric. In the specific case of an empty expanding uni- verse, c(t) = tt 0 , and thus d p (t 0 ) = ct 0 t 0 te dt t = ct 0 ln(t 0 t e ) . (5.36) Expressed in terms of the redshift . of the observed galaxy, d p (t 0 ) = c H 0 ln(1 +.) . (5.37) This relation is plotted as the dashed line in the upper panel of Figure 5.3. In the limit . < 1, there is a linear relation between d p and ., as seen observationally in Hubble’s law. In the limit . 1, however, d p ∝ ln . in an empty expanding universe. In an empty expanding universe, we can see objects which are currently at an arbitrarily large distance. However, at distances d p (t 0 ) cH 0 , the redshift increases exponentially with distance. At first glance, it may seem counterintuitive that you can see a light source at a proper distance much greater than cH 0 when the age of the universe is only 1H 0 . However, you must remember that d p (t 0 ) is the proper distance to the light source at the time of observation; at the time of emission, the proper distance d p (t e ) was 90 CHAPTER 5. SINGLE-COMPONENT UNIVERSES .01 .1 1 10 100 1000 .01 .1 1 10 100 Observation z ( H 0 / c ) d p ( t 0 ) .01 .1 1 10 100 1000 .01 .1 1 10 100 Emission z ( H 0 / c ) d p ( t e ) Figure 5.3: The proper distance to an object with observed redshift .. The upper panel shows the proper distance at the time the light is observed; the lower panel shows the proper distance at the time the light was emitted. The line types are the same as those of Figure 5.2. 5.3. SPATIALLY FLAT UNIVERSES 91 smaller by a factor c(t e )c(t 0 ) = 1(1 +.). In an empty expanding universe, the proper distance at the time of emission was d p (t e ) = c H 0 ln(1 +.) 1 +. . (5.38) shown as the dashed line in the lower panel of Figure 5.3. In an empty expanding universe, d p (t e ) has a maximum for objects with a redshift . = c − 1 ≈ 1.7, where d p (t e ) = (1c)(cH 0 ) ≈ 0.37(cH 0 ). Objects with much higher redshifts are seen as they were very early in the history of the universe, when their proper distance from the observer was very small. 5.3 Spatially flat universes Setting the energy density ε equal to zero is one way of simplifying the Friedmann equation. Another way is to set κ = 0 and to demand that the universe contain only a single component, with a single value of u. In such a spatially flat, single-component universe, the Friedmann equation takes the simple form ˙ c 2 = 8πGε 0 3c 2 c −(1+3w) . (5.39) To solve this equation, we first make the educated guess that the scale factor has the power law form c ∝ t q . The left hand side of equation (5.39) is then ∝ t 2q−2 , and the right hand side is ∝ t −(1+3w)q , yielding the solution ¡ = 2 3 + 3u . (5.40) with the restriction u = −1. With the proper normalization, the scale factor in a spatially flat, single-component universe is c(t) = (tt 0 ) 2/(3+3w) . (5.41) where the age of the universe, t 0 , is linked to the present energy density by the relation t 0 = 1 1 +u c 2 6πGε 0 1/2 . (5.42) The Hubble constant in such a universe is H 0 ≡ ˙ c c t=t 0 = 2 3(1 +u) t −1 0 . (5.43) 92 CHAPTER 5. SINGLE-COMPONENT UNIVERSES The age of the universe, in terms of the Hubble time, is then t 0 = 2 3(1 +u) H −1 0 . (5.44) In a spatially flat universe, if u −13, the universe is younger than the Hubble time. If u < −13, the universe is older than the Hubble time. As a function of scale factor, the energy density of a component with equation-of-state parameter u is ε(c) = ε 0 c −3(1+w) . (5.45) so in a spatially flat universe with only a single component, the energy density as a function of time is (combining equations 5.41 and 5.45) ε(t) = ε 0 (tt 0 ) −2 . (5.46) regardless of the value of u. Making the substitutions ε 0 = ε c,0 = 3c 2 8πG H 2 0 (5.47) and t 0 = 2 3(1 +u) H −1 0 . (5.48) equation (5.46) can be written in the form ε(t) = 1 6π(1 +u) 2 c 2 G t −2 . (5.49) Expressed in terms of Planck units (introduced in Chapter 1), this relation between energy density and cosmic time is ε(t) = 1 6π(1 +u) 2 1 p / 3 p t t P −2 . (5.50) Suppose yourself to be in a spatially flat, single-component universe. If you see a galaxy, or other distant light source, with a redshift ., you can use the relation 1 +. = c(t 0 ) c(t e ) = t 0 t e 2/(3+3w) (5.51) 5.3. SPATIALLY FLAT UNIVERSES 93 to compute the time t e at which the light from the distant galaxy was emitted: t e = t 0 (1 +.) 3(1+w)/2 = 2 3(1 +u)H 0 1 (1 +.) 3(1+w)/2 . (5.52) The current proper distance to the galaxy is d p (t 0 ) = c t 0 te dt c(t) = ct 0 3(1 +u) 1 + 3u [1 −(t e t 0 ) (1+3w)/(3+3w) ] . (5.53) when u = −13. In terms of H 0 and . rather than t 0 and t e , the current proper distance is d p (t 0 ) = c H 0 2 1 + 3u [1 −(1 +.) −(1+3w)/2 ] . (5.54) The most distant object you can see (in theory) is one for which the light emitted at t = 0 is just now reaching us at t = t 0 . The proper distance (at the time of observation) to such an object is called the horizon distance. 4 Here on Earth, the horizon is a circle centered on you, beyond which you cannot see because of the Earth’s curvature. In the universe, the horizon is a spherical surface centered on you, beyond which you cannot see because light from more distant objects has not had time to reach you. In a universe described by a Robertson-Walker metric, the current horizon distance is d hor (t 0 ) = c t 0 0 dt c(t) . (5.55) In a spatially flat universe, the horizon distance has a finite value if u −13. In such a case, computing the value of d p (t 0 ) in the limit t e →0 (or, equivalently, . →∞) yields d hor (t 0 ) = ct 0 3(1 +u) 1 + 3u = c H 0 2 1 + 3u . (5.56) In a flat universe dominated by matter (u = 0) or by radiation (u = 13), an observer can see only a finite portion of the infinite volume of the universe. The portion of the universe lying within the horizon for a particular observer is referred to as the visible universe for that observer. The visible universe 4 More technically, this is what’s called the particle horizon distance; I’ll continue to call it the horizon distance, for short. 94 CHAPTER 5. SINGLE-COMPONENT UNIVERSES consists of all points in space which have had sufficient time to send informa- tion, in the form of photons or other relativistic particles, to the observer. In other words, the visible universe consists of all points which are causally con- nected to the observer; nothing which happens outside the visible universe can have an effect on the observer. In a flat universe with u ≤ −13, the horizon distance is infinite, and all of space is causally connected to an observer. In such a universe with u ≤ −13, you could see every point in space – assuming the universe was transparent, of course. However, for extremely distant points, you would see extremely redshifted versions of what they looked like extremely early in the history of the universe. 5.4 Matter only Let’s now look at specific examples of spatially flat universes, starting with a universe containing only non-relativistic matter (u = 0). 5 The age of such a universe is t 0 = 2 3H 0 . (5.57) and the horizon distance is d hor (t 0 ) = 3ct 0 = 2cH 0 . (5.58) The scale factor, as a function of time, is c m (t) = (tt 0 ) 2/3 . (5.59) illustrated as the dotted line in Figure 5.2. If you see a galaxy with redshift . in a flat, matter-only universe, the proper distance to that galaxy, at the time of observation, is d p (t 0 ) = c t 0 te dt (tt 0 ) 2/3 = 3ct 0 [1 −(t e t 0 ) 1/3 ] = 2c H 0 ¸ 1 − 1 √ 1 +. ¸ . (5.60) illustrated as the dotted line in the upper panel of Figure 5.3. The proper distance at the time the light was emitted was smaller by a factor 1(1 +.): d p (t e ) = 2c H 0 (1 +.) ¸ 1 − 1 √ 1 +. ¸ . (5.61) 5 Such a universe is sometimes called an Einstein-de Sitter universe. 5.5. RADIATION ONLY 95 illustrated as the dotted line in the lower panel of Figure 5.3. In a flat, matter- only universe, d p (t e ) has a maximum for galaxies with a redshift . = 54, where d p (t e ) = (827)cH 0 . 5.5 Radiation only The case of a spatially flat universe containing only radiation is of particular interest, since early in the history of our own universe, the radiation (u = 13) term dominated the right-hand side of the Friedmann equation (see equation 5.27). Thus, at early times – long before the time of radiation- matter equality – the universe was well described by a spatially flat, radiation- only model. In an expanding, flat universe containing only radiation, the age of the universe is t 0 = 1 2H 0 . (5.62) and the horizon distance at t 0 is d hor (t 0 ) = 2ct 0 = cH 0 . (5.63) In the special case of a flat, radiation-only universe, the horizon distance is exactly equal to the Hubble distance, which is not generally the case. The scale factor of a flat, radiation-only universe is c(t) = (tt 0 ) 1/2 . (5.64) illustrated as the solid line in Figure 5.2. If, at a time t 0 , you observe a distant light source with redshift . in a flat, radiation-only universe, the proper distance to the light source will be d p (t 0 ) = c t 0 te dt (tt 0 ) 1/2 = 2ct 0 [1 −(t e t 0 ) 1/2 ] = c H 0 ¸ 1 − 1 1 +. . (5.65) illustrated as the solid line in the upper panel of Figure 5.3. The proper distance at the time the light was emitted was d p (t e ) = c H 0 (1 +.) ¸ 1 − 1 1 +. = c H 0 . (1 +.) 2 . (5.66) illustrated as the solid line in the lower panel of Figure 5.3. In a flat, radiation-dominated universe, d p (t e ) has a maximum for light sources with a redshift . = 1, where d p (t e ) = (14)cH 0 . 96 CHAPTER 5. SINGLE-COMPONENT UNIVERSES The energy density in a flat, radiation-only universe is ε r (t) = ε 0 (tt 0 ) −2 = 3 32 1 P / 3 P t t P −2 ≈ 0.094 1 P / 3 P t t P −2 . (5.67) Thus, in the early stages of our universe, when radiation was strongly domi- nant, the energy density, measured in units of the Planck density (1 P / 3 P ∼ 310 133 eVm −3 ), was comparable to one over the square of the cosmic time, measured in units of the Planck time (t P ∼ 510 −44 s). Using the blackbody relation between energy density and temperature, given in equations (2.26) and (2.27), we may assign a temperature to a universe dominated by black- body radiation: 1(t) = 45 32π 2 1/4 1 P t t P −1/2 ≈ 0.611 P t t P −1/2 . (5.68) Here 1 P is the Planck temperature, 1 P = 1.410 32 K. The mean energy per photon in a radiation-dominated universe is then 1 mean (t) ≈ 2.70/1(t) ≈ 1.661 P t t P −1/2 . (5.69) and the number density of photons is (combining equations 5.67 and 5.69) :(t) = ε r (t) 1 mean (t) ≈ 0.057 / 3 P t t P −3/2 . (5.70) Note that in a flat, radiation-only universe, as t →0, ε r →∞(equation 5.67). Thus, at the instant t = 0, the energy density of our own universe (well approximated as a flat, radiation-only model in its early stages) was infinite, according to this analysis; this infinite energy density was provided by an infinite number density of photons (equation 5.70), each of infinite energy (equation 5.69). Should we take these infinities seriously? Not really, since the assumptions of general relativity, on which the Friedmann equation is based, break down at t ≈ t P . Thus, extrapolating the results of this chapter to times earlier than the Planck time is not physically justified. Why can’t general relativity be used at times earlier than the Planck time? General relativity is a classical theory; that is, it does not take into account the effects of quantum mechanics. In cosmological contexts, general relativity assumes that the energy content of the universe is smooth down to arbitrarily 5.6. LAMBDA ONLY 97 small scales, instead of being parceled into individual quanta. As long as a radiation-dominated universe has many, many quanta, or photons, within a horizon distance, then the approximation of a smooth, continuous energy density is justifiable, and we may safely use the results of general relativity. However, if there are only a few photons within the visible universe, then quantum mechanical effects must be taken into account, and the classical results of general relativity no longer apply. In a flat, radiation-only universe, the horizon distance grows linearly with time: d hor (t) = 2ct = 2/ P t t P . (5.71) so the volume of the visible universe at time t is \ hor (t) = 4π 3 d 3 hor ≈ 34/ 3 P t t P 3 . (5.72) Combining equations (5.72) and (5.70), we find that the number of photons inside the horizon at time t is `(t) = \ hor (t):(t) ≈ 1.9 t t P 3/2 . (5.73) The quantization of the universe can no longer be ignored when `(t) ≈ 1, equivalent to a time t ≈ 0.7t P . In order to accurately described the universe at its very earliest stages, prior to the Planck time, a theory of quantum gravity is needed. Unfor- tunately, a complete theory of quantum gravity does not yet exist. Conse- quently, in this book, I will not deal with times earlier than the Planck time, t ∼ t P ∼ 5 10 −44 s, when the number density of photons was : ∼ / −3 P ∼ 2 10 104 m −3 , and the mean photon energy was 1 mean ∼ 1 P ∼ 1 10 28 eV. 5.6 Lambda only As seen in Section 5.3, a spatially flat, single-component universe with u = −1 has a power-law dependence of scale factor on time: c ∝ t 2/(3+3w) . (5.74) Now, for the sake of completeness, consider the case with u = −1; that is, a universe in which the energy density is contributed by a cosmological 98 CHAPTER 5. SINGLE-COMPONENT UNIVERSES constant Λ. 6 For such a flat, lambda-dominated universe, the Friedmann equation takes the form ˙ c 2 = 8πGε Λ 3c 2 c 2 . (5.75) where ε Λ is constant with time. This equation can be rewritten in the form ˙ c = H 0 c . (5.76) where H 0 = 8πGε Λ 3c 2 1/2 . (5.77) The solution to equation (5.76) in an expanding universe is c(t) = c H 0 (t−t 0 ) . (5.78) This scale factor is shown as the dot-dashed line in Figure 5.2. A spatially flat universe with nothing but a cosmological constant is exponentially expand- ing; we’ve seen an exponentially expanding universe before, in Section 2.3, under the label “Steady State universe”. In a Steady State universe, the den- sity ε of the universe remains constant because of the continuous creation of real particles. If the cosmological constant Λ is provided by the vacuum energy, then the density ε of a lambda-dominated universe remains con- stant because of the continuous creation and annihilation of virtual particle- antiparticle pairs. A flat universe containing nothing but a cosmological constant is infinitely old, and has an infinite horizon distance d hor . If, in a flat, lambda-only universe, you see a light source with a redshift ., the proper distance to the light source, at the time you observe it, is d p (t 0 ) = c t 0 te c H 0 (t 0 −t) dt = c H 0 [c H 0 (t 0 −te) −1] = c H 0 . . (5.79) shown as the dot-dashed line in the upper panel of Figure 5.3. The proper distance at the time the light was emitted was d p (t e ) = c H 0 . 1 +. . (5.80) shown as the dot-dashed line in the lower panel of Figure 5.3. 6 Such a universe is sometimes called a de Sitter universe. 5.6. LAMBDA ONLY 99 Note that an exponentially growing universe, such as the flat lambda- dominated model, is the only universe for which d p (t 0 ) is linearly proportional to . for all values of .. In other universes, the relation d p (t 0 ) ∝ . only holds true in the limit . < 1. Note also that in the limit . → ∞, d p (t 0 ) → ∞ but d p (t e ) → cH 0 . In a flat, lambda-dominated universe, highly redshifted objects (. 1) are at very large distances (d p (t 0 ) cH 0 ) at the time of observation; the observer sees them as they were just before they reached a proper distance cH 0 . Once a light source is more than a Hubble distance from the observer, their recession velocity is greater than the speed of light, and photons from the light source can no longer reach the observer. The simple models that we’ve examined in this chapter – empty universes, or flat universes with a single component – continue to expand forever if they are expanding at t = t 0 . Is it possible to have universes which stop expanding, then start to collapse? Is it possible to have universes in which the scale factor is not a simple power-law or exponential function of time? The short answer to these questions is “yes”. To study universes with more complicated behavior, however, it is necessary to put aside our simple toy universes, with a single term on the right-hand side of the Friedmann equation, and look at complicated toy universes, with multiple terms on the right-hand side of the Friedmann equation. Suggested reading [Full references are given in the “Annotated Bibiography” on page 286.] Liddle (1999), ch. 4: Flat universes, both matter-only and radiation-only Linder (1997), ch. 2.4,2.5: Evolution of energy density; evolution of scale factor in single-component universes Problems (5.1) The predicted number of neutrinos in the Cosmic Neutrino Back- ground is : ν = (311): γ = 1.12 10 8 m −3 for each of the three species of neutrino. About how many cosmic neutrinos are inside your body right now? What must be the sum of the neutrino masses, :(ν e ) +:(ν µ ) +:(ν τ ), in order for the density of the Cosmic Neutrino Background to be equal to the critical density, ε c,0 ? 100 CHAPTER 5. SINGLE-COMPONENT UNIVERSES (5.2) A light source in a flat, single-component universe has a redshift . when observed at a time t 0 . Show that the observed redshift changes at a rate d. dt 0 = H 0 (1 +.) −H 0 (1 +.) 3(1+w)/2 . (5.81) For what values of u does the redshift decrease with time? For what values of u does the redshift increase with time? (5.3) Suppose you are in a flat, matter-only universe which has a Hubble constant H 0 = 70 kms −1 Mpc −1 . You observe a galaxy with . = 1. How long will you have to keep observing the galaxy to see its redshift change by one part in 10 6 ? [Hint: use the result from the previous problem.] (5.4) In a flat universe with H 0 = 70 kms −1 Mpc −1 , you observe a galaxy at a redshift . = 7. What is the current proper distance to the galaxy, d p (t 0 ), if the universe contains only radiation? What is d p (t 0 ) if the universe contains only matter? What is d p (t 0 ) if the universe contains only a cosmological constant? What was the proper distance at the time the light was emitted, d p (t e ), if the universe contains only radia- tion? What was d p (t e ) if the universe contains only matter? What was d p (t e ) if the universe contains only a cosmological constant? Chapter 6 Multiple-Component Universes The Friedmann equation, in general, can be written in the form H(t) 2 = 8πG 3c 2 ε(t) − κc 2 1 2 0 c(t) 2 . (6.1) where H ≡ ˙ cc, and ε(t) is the energy density contributed by all the compo- nents of the universe, including the cosmological constant. Equation (4.31) tells us the relation among κ, 1 0 , H 0 , and Ω 0 , κ 1 2 0 = H 2 0 c 2 (Ω 0 −1) . (6.2) so we can rewrite the Friedmann equation without explicitly including the curvature: H(t) 2 = 8πG 3c 2 ε(t) − H 2 0 c(t) 2 (Ω 0 −1) . (6.3) Dividing by H 2 0 , this becomes H(t) 2 H 2 0 = ε(t) ε c,0 + 1 −Ω 0 c(t) 2 . (6.4) where the critical density today is ε c,0 ≡ 3c 2 H 2 0 8πG . (6.5) We know that our universe contains matter, for which the energy density ε m has the dependence ε m = ε m,0 c 3 , and radiation, for which the energy 101 102 CHAPTER 6. MULTIPLE-COMPONENT UNIVERSES density has the dependence ε r = ε r,0 c 4 . Current evidence seems to indicate the presence of a cosmological constant, with energy density ε Λ = ε Λ,0 = constant. It is certainly possible that the universe contains other components as well. For instance, as the 21st century began, some cosmologists were investigating the properties of “quintessence”, a component of the universe whose equation-of-state parameter can lie in the range −1 < u < −13, giving a universe with ¨ c 0. However, in the absence of strong evidence for the existence of “quintessence”, I will only consider the contributions of matter (u = 0), radiation (u = 13), and the cosmological constant Λ (u = −1). In our universe, we expect the Friedmann equation (6.4) to take the form H 2 H 2 0 = Ω r,0 c 4 + Ω m,0 c 3 + Ω Λ,0 + 1 −Ω 0 c 2 . (6.6) where Ω r,0 = ε r,0 ε c,0 , Ω m,0 = ε m,0 ε c,0 , Ω Λ,0 = ε Λ,0 ε c,0 , and Ω 0 = Ω r,0 + Ω m,0 +Ω Λ,0 . The Benchmark Model, introduced in the previous chapter as a model consistent with all available data, has Ω 0 = 1, and hence is spatially flat. However, although a perfectly flat universe is consistent with the data, it is not demanded by the data. Thus, prudence dictates that we should keep in mind the possibility that the curvature term, (1−Ω 0 )c 2 in equation (6.6), might be nonzero. Since H = ˙ cc, multiplying equation (6.6) by c 2 , then taking the square root, yields H −1 0 ˙ c = [Ω r,0 c 2 + Ω m,0 c + Ω Λ,0 c 2 + (1 −Ω 0 )] 1/2 . (6.7) The cosmic time t as a function of scale factor c can then be found by performing the integral a 0 dc [Ω r,0 c 2 + Ω m,0 c + Ω Λ,0 c 2 + (1 −Ω 0 )] 1/2 = H 0 t . (6.8) This is not a user-friendly integral: in the general case, it doesn’t have a simple analytic solution. However, for given values of Ω r,0 , Ω m,0 , and Ω Λ,0 , it can be integrated numerically. In many circumstances, the integral in equation (6.8) has a simple analytic approximation to its solution. For instance, as noted in the previous chapter, in a universe with radiation, matter, curvature, and Λ, the radiation term 103 dominates the expansion during the early stages of expansion. In this limit, equation (6.8) simplifies to H 0 t ≈ a 0 cdc Ω r,0 ≈ 1 2 Ω r,0 c 2 . (6.9) or c(t) ≈ (2 Ω r,0 H 0 t) 1/2 . (6.10) In the limit Ω r,0 = 1, this reduces to the solution already found for a flat, radiation-only universe. If the universe continues to expand forever, then in the limit c → ∞, the cosmological constant term will dominate the ex- pansion. For some values of Ω r,0 , Ω m,0 , and Ω Λ,0 , there will be intermediate epochs when the matter or the curvature dominates the expansion. For in- stance, in the Benchmark Model, where radiation-matter equality takes place at a scale factor c rm ≈ 2.8 10 −4 and matter-lambda equality takes place at c mΛ ≈ 0.75, a matter-only universe is a fair approximation to reality when c rm <c <c mΛ . However, during some epochs of the universe’s expansion, two of the components are of comparable density, and provide terms of roughly equal size in the Friedmann equation. During these epochs, a single-component model is a poor description of the universe, and a two-component model must be utilized. For instance, for scale factors near c rm ≈ 2.8 10 −4 , the Benchmark Model is well approximated by a flat universe containing only radiation and matter. Such a universe is examined in Section 6.4. For scale factors near c mΛ ≈ 0.75, the Benchmark Model is well approximated by a flat universe containing only matter and a cosmological constant. Such a universe is examined in Section 6.2. First, however, we will examine a universe which is of great historical in- terest to cosmology; a universe containing both matter and curvature (either negative or positive). After Einstein dismissed the cosmological constant as a blunder, and before astronomers had any clear idea what the value of the density parameter Ω was, they considered the possibility that the universe was negatively curved or positively curved, with the bulk of the density being provided by non-relativistic matter. During the mid-twentieth century, cosmologists concentrated much of their interest on the study of curved, matter-dominated universes. In addition to being of historical inter- est, curved, matter-dominated universes provide useful physical insight into the interplay among curvature, expansion, and density. 104 CHAPTER 6. MULTIPLE-COMPONENT UNIVERSES 6.1 Matter + curvature Consider a universe containing nothing but pressureless matter, with u = 0. If such a universe is spatially flat, then it expands with time, as demonstrated in Section 5.4, with a scale factor c(t) = (tt 0 ) 2/3 . (6.11) Such a flat, matter-only universe expands outward forever. Such a fate is sometimes known as the “Big Chill”, since the temperature of the universe decreases monotonically with time as the universe expands. At this point, it is nearly obligatory for a cosmology text to quote T. S. Eliot: “This is the way the world ends / Not with a bang but a whimper.” 1 In a curved universe containing nothing but matter, the ultimate fate of the cosmos is intimately linked to the density parameter Ω 0 . The Fried- mann equation in a curved, matter-dominated universe (equation 6.6) can be written in the form H(t) 2 H 2 0 = Ω 0 c 3 + 1 −Ω 0 c 2 . (6.12) since Ω m,0 = Ω 0 in such a universe. Suppose you are in a universe which is currently expanding (H 0 0) and which contains nothing but non-relativistic matter. If you ask the question, “Will the universe ever cease to expand?” then equation (6.12) enables you to answer that question. For the universe to cease expanding, there must be some moment at which H(t) = 0. Since the first term on the right hand side of equation (6.12) is always positive, H(t) = 0 requires the second term on the right hand side to be negative. This means that a matter-dominated universe will cease to expand if Ω 0 1, and hence κ = +1. At the time of maximum expansion, H(t) = 0 and thus 0 = Ω 0 c 3 max + 1 −Ω 0 c 2 max . (6.13) The scale factor at the time of maximum expansion will therefore be c max = Ω 0 Ω 0 −1 . (6.14) 1 Interestingly, this quote is from Eliot’s poem The Hollow Men, written, for the most part, in 1924, the year when Friedmann published his second paper on the expansion of the universe. However, this coincidence seems to be just that – a coincidence. Eliot did not keep up to date on the technical literature of cosmology. 6.1. MATTER + CURVATURE 105 where Ω 0 , remember, is the density parameter as measured at a scale factor c = 1. Note that in equation (6.12), the Hubble parameter enters only as H 2 . Thus, the contraction phase, after the universe reaches maximum expansion, is just the time reversal of the expansion phase. 2 Eventually, the Ω 0 1 universe will collapse down to c = 0 (an event sometimes called the “Big Crunch”) after a finite time t = t crunch . A matter-dominated universe with Ω 0 1 not only has finite spatial extent, but also has a finite duration in time; just as it began in a hot, dense state, so it will end in a hot, dense state. When such a universe is in its contracting stage, an observer will see galaxies with a blueshift proportional to their distance. As the universe approaches the Big Crunch, the cosmic microwave background will become a cosmic infrared background, then a cosmic visible background, then a cosmic ultraviolet background, then a cosmic x-ray background, then finally a cosmic gamma-ray background. A matter-dominated universe with Ω 0 1 will expand to a maximum scale factor c max , then collapse in a Big Crunch. What is the ultimate fate of a matter-dominated universe with Ω 0 < 1 and κ = −1? In the Friedmann equation for such a universe (equation 6.12), both terms on the right hand side are positive. Thus if such a universe is expanding at a time t = t 0 , it will continue to expand forever. At early times, when the scale factor is small (c < Ω 0 [1 − Ω 0 ]), the matter term of the Friedmann equation will dominate, and the scale factor will grow at the rate c ∝ t 2/3 . Ultimately, however, the density of matter will be diluted far below the critical density, and the universe will expand like the negatively curved empty universe, with c ∝ t. If a universe contains nothing but matter, its curvature, its density, and its ultimate fate are closely linked, as shown in Table 6.1. At this point, the obligatory quote is from Robert Frost: “Some say the world will end in fire / Some say in ice.” 3 In a matter-dominated universe, if the density is greater 2 The contraction is a perfect time reversal of the expansion only when the universe is perfectly homogeneous and the expansion is perfectly adiabatic, or entropy-conserving. In a real, lumpy universe, entropy is not conserved on small scales. Stars, for instance, generate entropy as they emit photons. During the contraction phase of an Ω 0 > 1 universe, small-scale entropy-producing processes will NOT be reversed. Stars will not absorb the photons which they previously emitted; people will not live backward from grave to cradle. 3 This is from Frost’s poem Fire and Ice, first published in Harper’s Magazine in De- 106 CHAPTER 6. MULTIPLE-COMPONENT UNIVERSES Table 6.1: Curved, matter-dominated universes density curvature ultimate fate Ω 0 < 1 κ = −1 Big Chill (c ∝ t) Ω 0 = 1 κ = 0 Big Chill (c ∝ t 2/3 ) Ω 0 1 κ = +1 Big Crunch than the critical density, the universe will end in a fiery Big Crunch; if the density is less than or equal to the critical density, the universe will end in an icy Big Chill. In a curved universe containing only matter, the scale factor c(t) can be computed explicitly. The Friedmann equation can be written in the form ˙ c 2 H 2 0 = Ω 0 c + (1 −Ω 0 ) . (6.15) so the age t of the universe at a given scale factor c is given by the integral H 0 t = a 0 dc [Ω 0 c + (1 −Ω 0 )] 1/2 . (6.16) When Ω 0 = 1, the solution to this integral is most compactly written in a parametric form. The solution when Ω 0 1 is c(θ) = 1 2 Ω 0 Ω 0 −1 (1 −cos θ) (6.17) and t(θ) = 1 2H 0 Ω 0 (Ω 0 −1) 3/2 (θ −sin θ) . (6.18) where the parameter θ runs from 0 to 2π. Given this parametric form, it is easy to show that the time that elapses between the Big Bang at θ = 0 and the Big Crunch at θ = 2π is t crunch = π H 0 Ω 0 (Ω 0 −1) 3/2 . (6.19) A plot of c versus t in the case Ω 0 = 1.1 is shown as the solid line in Figure 6.1. The c ∝ t 2/3 behavior of an Ω 0 = 1 universe is shown as the dotted line. The cember 1920. Unlike T. S. Eliot, Frost was keenly interested in astronomy, and frequently wrote poems on astronomical themes. 6.1. MATTER + CURVATURE 107 0 20 40 60 80 100 0 10 20 30 40 H 0 (t−t 0 ) a Ω 0 =1.1 Ω 0 =1.0 Ω 0 =0.9 −.5 0 .5 1 1.5 2 0 .5 1 1.5 2 2.5 H 0 (t−t 0 ) a Figure 6.1: The scale factor as a function of time for universes containing only matter. The dotted line is c(t) for a universe with Ω 0 = 1 (flat); the dashed line is c(t) for a universe with Ω 0 = 0.9 (negatively curved); the solid line is c(t) for a universe with Ω 0 = 1.1 (positively curved). The bottom panel is a blow-up of the small rectangle near the lower left corner of the upper panel. 108 CHAPTER 6. MULTIPLE-COMPONENT UNIVERSES solution of equation (6.16) for the case Ω 0 < 1 can be written in parametric form as c(η) = 1 2 Ω 0 1 −Ω 0 (cosh η −1) (6.20) and t(η) = 1 2 Ω 0 (Ω 0 −1) 3/2 (sinh η −η) . (6.21) where the parameter η runs from 0 to infinity. A plot of c versus t in the case Ω 0 = 0.9 is shown as the dashed line in Figure 6.1. Note that although the ultimate fates of an Ω 0 = 0.9 universe is very different from that of an Ω 0 = 1.1 universe, as shown graphically in the upper panel of Figure 6.1, it is very difficult, at t ∼ t 0 , to tell a universe with Ω 0 slightly less than one from that with Ω 0 slightly greater than one. As shown in the lower panel of Figure (6.1), the scale factors of the Ω 0 = 1.1 universe and the Ω 0 = 0.9 universe start to diverge significantly only after a Hubble time or more. Scientists sometimes joke that they are searching for a theory of the uni- verse that is compact enough to fit on the front of a T-shirt. If the energy content of the universe is contributed almost entirely by non-relativistic mat- ter, then an appropriate T-shirt slogan would be: DENSITY IS DESTINY! If the density of matter is less than the critical value, then the destiny of the universe is an ever-expanding Big Chill; if the density is greater than the critical value, then the destiny is a recollapsing Big Crunch. Like all terse summaries of complex concepts, the slogan “Density is Destiny!” requires a qualifying footnote. In this case, the required footnote is “ ∗ if Λ = 0”. If the universe has a cosmological constant (or more generally, any component with u < −13), then the equation Density = Destiny = Curvature no longer applies. 6.2 Matter + lambda Consider a universe which is spatially flat, but which contains both matter and a cosmological constant. 4 If, at a given time t = t 0 , the density parameter 4 Such a universe is of particular interest to us, since it appears to be a close approxi- mation to our own universe at the present day. 6.2. MATTER + LAMBDA 109 in matter is Ω m,0 and the density parameter in a cosmological constant Λ is Ω Λ,0 , the requirement that space be flat tells us that Ω Λ,0 = 1 −Ω m,0 . (6.22) and the Friedmann equation for the flat “matter plus lambda” universe re- duces to H 2 H 2 0 = Ω m,0 c 3 + (1 −Ω m,0 ) . (6.23) The first term on the right hand side of equation (6.23) represents the con- tribution of matter, and is always positive. The second term represents the contribution of a cosmological constant; it is positive if Ω m,0 < 1, implying Ω Λ,0 0, and negative if Ω m,0 1, implying Ω Λ,0 < 0. Thus, a flat universe with Ω Λ,0 0 will continue to expand forever if it is expanding at t = t 0 ; this is another example of a Big Chill universe. In a universe with Ω Λ,0 < 0, however, the negative cosmological constant provides an attractive force, not the repulsive force of a positive cosmological constant. A flat universe with Ω Λ,0 < 0 will cease to expand at a maximum scale factor c max = Ω m,0 Ω m,0 −1 1/3 . (6.24) and will collapse back down to c = 0 at a cosmic time t crunch = 2π 3H 0 1 Ω m,0 −1 . (6.25) For a given value of H 0 , the larger the value of Ω m,0 , the shorter the lifetime of the universe. For a flat, Ω Λ,0 < 0 universe, the Friedmann equation (eq.6.23) can be integrated to yield the analytic solution H 0 t = 2 3 Ω m,0 −1 sin −1 (cc max ) 3/2 . (6.26) A plot of c versus t in the case Ω m,0 = 1.1, Ω Λ,0 = −0.1 is shown as the solid line in Figure 6.2. The c ∝ t 2/3 behavior of a Ω m,0 = 1, Ω Λ,0 = 0 universe is shown, for comparison, as the dotted line. A flat universe with Ω Λ,0 < 0 ends in a Big Crunch, reminiscent of that for a positively curved, matter- only universe. However, with a negative cosmological constant providing an 110 CHAPTER 6. MULTIPLE-COMPONENT UNIVERSES 0 20 40 60 80 100 0 10 20 30 40 H 0 (t−t 0 ) a Ω m,0 =0.9 Ω m,0 =1.0 Ω m,0 =1.1 Figure 6.2: The scale factor as a function of time for flat universes contain- ing both matter and a cosmological constant. The dotted line is c(t) for a universe with Ω m,0 = 1, Ω Λ,0 = 0. The solid line is c(t) for a universe with Ω m,0 = 1.1, Ω Λ,0 = −0.1. The dashed line is c(t) for a universe with Ω m,0 = 0.9, Ω Λ,0 = 0.1. 6.2. MATTER + LAMBDA 111 attractive force, the lifetime of a flat universe with Ω Λ,0 < 0 is exceptionally short. For instance, we have seen that a positively curved universe with Ω m,0 = 1.1 undergoes a Big Crunch after a lifetime t crunch ≈ 110H −1 0 (see Figure 6.1). However, a flat universe with Ω m,0 = 1.1 and Ω Λ,0 = −0.1 has a lifetime t crunch ≈ 7H −1 0 . As soon as the universe becomes lambda- dominated, the negative cosmological constant causes a rapid deceleration of the universe’s expansion. Although a negative cosmological constant is permitted by the laws of physics, it appears that we live in a universe where the cosmological constant is non-negative. In a flat universe with Ω m,0 < 1 and Ω Λ,0 0, the density contributions of matter and the cosmological constant are equal at the scale factor (eq. 5.22): c mΛ = Ω m,0 Ω Λ,0 1/3 = Ω m,0 1 −Ω m,0 1/3 . (6.27) For a flat, Ω Λ,0 0 universe, the Friedmann equation can be integrated to yield the analytic solution H 0 t = 2 3 1 −Ω m,0 ln ¸ (cc mΛ ) 3/2 + 1 + (cc mΛ ) 3 . (6.28) A plot of c versus t in the case Ω m,0 = 0.9, Ω Λ,0 = 0.1 is shown as the dashed line in Figure 6.2. At early times, when c <c mΛ , equation (6.28) reduces to the relation c(t) ≈ 3 2 Ω m,0 H 0 t 2/3 . (6.29) giving the c ∝ t 2/3 dependence required for a flat, matter-dominated universe. At late times, when c c mΛ , equation (6.28) reduces to c(t) ≈ c mΛ exp( 1 −Ω m,0 H 0 t) . (6.30) giving the c ∝ c Kt dependence required for a flat, lambda-dominated uni- verse. Suppose you are in a flat universe containing nothing but matter and a cosmological constant; if you measure H 0 and Ω m,0 , then equation (6.28) tells you that the age of the universe is t 0 = 2H −1 0 3 1 −Ω m,0 ln 1 −Ω m,0 + 1 Ω m,0 ¸ ¸ . (6.31) 112 CHAPTER 6. MULTIPLE-COMPONENT UNIVERSES If we approximate our own universe as having Ω m,0 = 0.3 and Ω Λ,0 = 0.7 (ignoring the contribution of radiation) we find that its current age is t 0 = 0.964H −1 0 = 13.5 ±1.3 Gyr . (6.32) assuming H 0 = 70 ± 7 kms −1 Mpc −1 . (We’ll see in section 6.5 that ignoring the radiation content of the universe has an insignificant effect on our es- timate of t 0 .) The age at which matter and the cosmological constant had equal energy density was t mΛ = 2H −1 0 3 1 −Ω m,0 ln[1 + √ 2] = 0.702H −1 0 = 9.8 ±1.0 Gyr . (6.33) Thus, if our universe is well fit by the Benchmark Model, with Ω m,0 = 0.3 and Ω Λ,0 ≈ 0.7, then the cosmological constant has been the dominant component of the universe for the last four billion years or so. 6.3 Matter + curvature + lambda If a universe contains both matter and a cosmological constant, then the formula “density = destiny = curvature” no longer holds. A flat universe with Ω m,0 1 and Ω Λ,0 < 0, as shown in the previous section, is infinite in spatial extent, but has a finite duration in time. By contrast, a flat universe with Ω m,0 ≤ 1 and Ω Λ,0 ≥ 0 extends to infinity both in space and in time. If a universe containing both matter and lambda is curved (κ = 0) rather than flat, then a wide range of behaviors is possible for the function c(t). For instance, in section 4.4, we encountered Einstein’s static model, in which κ = +1 and ε Λ = ε m 2. A universe described by Einstein’s static model is finite in spatial extent, but has infinite duration in time. By choosing different values of Ω m,0 and Ω Λ,0 , without constraining the universe to be flat, we can create model universes with scale factors c(t) which exhibit very interesting behavior. Start by writing down the Friedmann equation for a curved universe with both matter and a cosmological constant: H 2 H 2 0 = Ω m,0 c 3 + 1 −Ω m,0 −Ω Λ,0 c 2 + Ω Λ,0 . (6.34) If Ω m,0 0 and Ω Λ,0 0, then both the first and last term on the right hand side of equation (6.34) are positive. However, if Ω m,0 +Ω Λ,0 1, so that the 6.3. MATTER + CURVATURE + LAMBDA 113 universe is positively curved, then the central term on the right hand side is negative. As a result, for some choices of Ω m,0 and Ω Λ,0 , the value of H 2 will be positive for small values of c (where matter dominates) and for large values of c (where Λ dominates), but will be negative for intermediate values of c (where the curvature term dominates). Since negative values of H 2 are unphysical, this means that these universes have a forbidden range of scale factors. Suppose such a universe starts out with c 1 and H < 0; that is, it is contracting from a low-density, Λ-dominated state. As the universe contracts, however, the negative curvature term in equation 6.34 becomes dominant, causing the contraction to stop at a minimum scale factor c = c min , and then expand outward again in a “Big Bounce”. Thus, it is possible to have a universe which expands outward at late times, but which never had an initial Big Bang, with c = 0 at t = 0. Another possibility, if the values of Ω m,0 and Ω Λ,0 are chosen just right, is a “loitering” universe. 5 Such a universe starts in a matter-dominated state, expanding outward with c ∝ t 2/3 . Then, however, it enters a stage (called the loitering stage) in which c is very nearly constant for a long period of time. During this time it is almost – but not quite – Einstein’s static universe. After the loitering stage, the cosmological constant takes over, and the universe starts to expand exponentially. Figure 6.3 shows the general behavior of the scale factor c(t) as a function of Ω m,0 and Ω Λ,0 . In the region labeled “Big Crunch”, the universe starts with c = 0 at t = 0, reaches a maximum scale factor c max , then recollapses to c = 0 at a finite time t = t crunch . Note that Big Crunch universes can be positively curved, negatively curved, or flat. In the region labeled “Big Chill”, the universe starts with c = 0 at t = 0, then expands outward forever, with c → ∞ as t → ∞. Like Big Crunch universes, Big Chill universes can have any sign for their curvature. In the region labeled “Big Bounce”, the universe starts in a contracting state, reaches a minimum scale factor c = c min 0 at some time t bounce , then expands outward forever, with c →∞ as t →∞. Universes which fall just below the dividing line between Big Bounce universes and Big Chill universes are loitering universes. The closer such a universe lies to the Big Bounce – Big Chill dividing line in Figure 6.3, the longer its loitering stage lasts. To illustrate the different types of expansion and contraction possible, Figure 6.4 shows c(t) for a set of four model universes. Each of these universes has the same current density parameter for matter: Ω m,0 = 0.3, measured at 5 A loitering universe is sometimes referred to as a Lemaˆıtre universe. 114 CHAPTER 6. MULTIPLE-COMPONENT UNIVERSES 0 .5 1 1.5 2 2.5 −1 0 1 2 3 Ω m,0 Ω Λ , 0 Big Bounce Big Chill Big Crunch κ=−1 κ=+1 l o i t e r i n g Figure 6.3: The curvature and type of expansion for universes containing both matter and a cosmological constant. The dashed line indicates κ = 0; models lying above this line have κ = +1, and those lying below have κ = −1. Also shown are the regions where the universe has a “Big Chill” expansion (c → ∞ as t → ∞), a “Big Crunch” recollapse (c → 0 as t → t crunch ), a loitering phase (c ∼ const for an extended period), or a “Big Bounce” (c = c min 0 at t = t bounce ). 6.3. MATTER + CURVATURE + LAMBDA 115 −4 −2 0 2 4 0 1 2 3 H 0 (t−t 0 ) a Ω Λ,0 =1.8 Ω Λ,0 =1.7134 Ω Λ,0 =0.7 Ω Λ,0 =−0.3 Figure 6.4: The scale factor c as a function of t in four different universes, each with Ω m,0 = 0.3. The dashed line shows a “Big Crunch” universe (Ω Λ,0 = −0.3, κ = −1). The dotted line shows a “Big Chill” universe (Ω Λ,0 = 0.7, κ = 0). The dot-dash line shows a loitering universe (Ω Λ,0 = 1.7134, κ = +1). The solid line shows a “Big Bounce” universe (Ω Λ,0 = 1.8, κ = +1). 116 CHAPTER 6. MULTIPLE-COMPONENT UNIVERSES t = t 0 and c = 1. These universes cannot be distinguished from each other by measuring their current matter density and Hubble constant. Nevertheless, thanks to their different values for the cosmological constant, they have very different pasts and very different futures. The dashed line in Figure 6.4 shows the scale factor for a universe with Ω Λ,0 = −0.3; this universe has negative curvature, and is destined to end in a Big Crunch. The dotted line shows c(t) for a universe with Ω Λ,0 = 0.7; this universe is spatially flat, and is destined to end in an exponentially expanding Big Chill . The dot-dash line shows the scale factor for a universe with Ω Λ,0 = 1.7134; this is a positively curved loitering universe, which spends a long time with a scale factor c ≈ c loiter ≈ 0.44. Finally, the solid line shows a universe with Ω Λ,0 = 1.8. This universe lies above the Big Chill – Big Bounce dividing line in Figure 6.3; it is a positively curved universe which “bounced” at a scale factor c = c bounce ≈ 0.56. There is strong observational evidence that we do not live in a loitering or Big Bounce universe. If we lived in a loitering universe, then as we looked out into space, we would see nearly the same redshift . loiter = 1c loiter −1 for galaxies with a very large range of distances. For instance, with c loiter ≈ 0.44 (the appropriate loitering scale factor for a universe with Ω m,0 = 0.3), this would lead to a large excess of galaxies with . loiter ≈ 1.3. No such excess of galaxies is seen at any redshift in our universe. If we lived in a Big Bounce universe, then the largest redshift we would see for any galaxy would be . max = 1c bounce −1. As we looked further into space, we would see redshifts increase to . max , then see the redshifts decrease until they actually became blueshifts. In our universe, we do not see such distant blueshifted galaxies. Our own universe seems to be a Big Chill universe, fated to eternal expansion. 6.4 Radiation + matter In our universe, radiation-matter equality took place at a scale factor c rm ≡ Ω r,0 Ω m,0 ≈ 2.8 10 −4 . At scale factors c < c rm , the universe is well described by a flat, radiation-only model, as described in Section 5.5. At scale factors c ∼ c rm , the universe is better described by a flat model containing both radiation and matter. The Friedmann equation around the time of radiation-matter equality can be written in the approximate form H 2 H 2 0 = Ω r,0 c 4 + Ω m,0 c 3 . (6.35) 6.4. RADIATION + MATTER 117 This can be rearranged in the form H 0 dt = cdc Ω 1/2 r,0 [1 +cc rm ] −1/2 . (6.36) Integration yields a fairly simple relation for t as a function of c during the epoch when only radiation and matter are significant: H 0 t = 4c 2 rm 3 Ω r,0 ¸ 1 − 1 − c 2c rm 1 + c c rm 1/2 ¸ . (6.37) In the limit c < c rm , this gives the appropriate result for the radiation- dominated phase of evolution (compare equation 6.10), c ≈ 2 Ω r,0 H 0 t 1/2 [c <c rm ] . (6.38) In the limit c c rm (but before curvature or Λ contributes significantly to the Friedmann equation), the approximate result for c(t) becomes c ≈ 3 2 Ω m,0 H 0 t 2/3 [c c rm ] . (6.39) The time of radiation-matter equality, t rm , can be found by setting c = c rm in equation (6.37): t rm = 4 3 (1 − 1 √ 2 ) c 2 rm Ω r,0 H −1 0 ≈ 0.391 Ω 3/2 r,0 Ω 2 m,0 H −1 0 . (6.40) For the Benchmark Model, with Ω r,0 = 8.4 10 −5 , Ω m,0 = 0.3, and H −1 0 = 14 Gyr, the time of radiation-matter equality was t rm = 3.34 10 −6 H −1 0 = 47.000 yr . (6.41) The epoch when the universe was radiation-dominated was only about 47 millennia long. This is sufficiently brief that it justifies our ignoring the effects of radiation when computing the age of the universe. The age t 0 = 0.964H −1 0 = 13.5 Gyr that we computed in Section 6.2 (ignoring radiation) would only be altered by a few parts per million if we included the effects of radiation. This minor correction is dwarfed by the 10% uncertainty in the value of H −1 0 . 118 CHAPTER 6. MULTIPLE-COMPONENT UNIVERSES Table 6.2: Properties of the Benchmark Model List of Ingredients photons: Ω γ,0 = 5.0 10 −5 neutrinos: Ω ν,0 = 3.4 10 −5 total radiation: Ω r,0 = 8.4 10 −5 baryonic matter: Ω bary,0 = 0.04 nonbaryonic dark matter: Ω dm,0 = 0.26 total matter: Ω m,0 = 0.30 cosmological constant: Ω Λ,0 ≈ 0.70 Important Epochs radiation-matter equality: c rm = 2.8 10 −4 t rm = 4.7 10 4 yr matter-lambda equality: c mΛ = 0.75 t mΛ = 9.8 Gyr Now: c 0 = 1 t 0 = 13.5 Gyr 6.5 Benchmark Model The Benchmark Model, which I have adopted as the best fit to the currently available observational data, is spatially flat, and contains radiation, matter, and a cosmological constant. Some of its properties are listed, for ready reference, in Table 6.2. The Hubble constant of the Benchmark Model is assumed to be H 0 = 70 kms −1 Mpc −1 . The radiation in the Benchmark Model consists of photons and neutrinos. The photons are assumed to be provided solely by a Cosmic Microwave Background with current temperature 1 0 = 2.725 K and density parameter Ω γ,0 = 5.010 −5 . The energy density of the cosmic neutrino background is theoretically calculated to be 68% of that of the Cosmic Microwave Background, as long as neutrinos are relativistic. If a neutrino has a non-zero mass : ν , equation (5.18) tells us that it defects from the “radiation” column to the “matter” column when the scale factor is c ∼ 5 10 −4 eV(: ν c 2 ). The matter content of the Benchmark Model consists partly of baryonic matter (that is, matter composed of protons and neutrons, with associated electrons), and partly of nonbaryonic dark matter. As we’ll see in future chapters, the evidence indicates that most of the matter in the universe is nonbaryonic dark matter. The baryonic material that we are familiar with from our everyday existence has a density parameter of Ω bary,0 ≈ 0.04 today. The density parameter of the nonbaryonic dark matter 6.5. BENCHMARK MODEL 119 −10 −8 −6 −4 −2 0 −6 −4 −2 0 2 log(H 0 t) l o g ( a ) t rm t mΛ t 0 a∝t 1/2 a∝t 2/3 a∝e Kt Figure 6.5: The scale factor c as a function of time t (measured in units of the Hubble time), computed for the Benchmark Model. The dotted lines indicate the time of radiation-matter equality, c rm = 2.8 10 −4 , the time of matter-lambda equality, c mΛ = 0.75, and the present moment, c 0 = 1. is roughly six times greater: Ω dm,0 ≈ 0.26. The bulk of the energy density in the Benchmark Model, however, is not provided by radiation or matter, but by a cosmological constant, with Ω Λ,0 = 1 −Ω m,0 −Ω r,0 ≈ 0.70. The Benchmark Model was first radiation-dominated, then matter-dominated, and is now entering into its lambda-dominated phase. As we’ve seen, radi- ation gave way to matter at a scale factor c rm = Ω r,0 Ω m,0 = 2.8 10 −4 , corresponding to a time t rm = 4.7 10 4 yr. Matter, in turn, gave way to the cosmological constant at c mΛ = (Ω m,0 Ω Λ,0 ) 1/3 = 0.75, corresponding to t mΛ = 9.8 Gyr. The current age of the universe, in the Benchmark Model, is t 0 = 13.5 Gyr. With Ω r,0 , Ω m,0 , and Ω Λ,0 known, the scale factor c(t) can be computed numerically using the Friedmann equation, in the form of equation (6.6). Figure 6.5 shows the scale factor, thus computed, for the Benchmark Model. Note that the transition from the c ∝ t 1/2 radiation-dominated phase to the c ∝ t 2/3 matter-dominated phase is not an abrupt one; neither is the later transition from the matter-dominated phase to the exponentially grow- ing lambda-dominated phase. One curious feature of the Benchmark Model 120 CHAPTER 6. MULTIPLE-COMPONENT UNIVERSES which Figure 6.5 illustrates vividly is that we are living very close to the time of matter-lambda equality. Once c(t) is known, other properties of the Benchmark Model can readily be computed. For instance, the upper panel of Figure 6.6 shows the current proper distance to a galaxy with redshift .. The heavy solid line is the result for the Benchmark Model; for purposes of comparison, the result for a flat lambda-only universe is shown as a dot-dash line and the result for a flat matter-only universe is shown as the dotted line. In the limit . → ∞, the proper distance d p (t 0 ) approaches a limiting value d p →3.24cH 0 , in the case of the Benchmark Model. Thus, the Benchmark Model has a finite horizon distance, d hor (t 0 ) = 3.24cH 0 = 3.12ct 0 = 14.000 Mpc . (6.42) If the Benchmark Model is a good description of our own universe, then we can’t see objects more than 14 gigaparsecs away because light from them has not yet had time to reach us. The lower panel of Figure 6.6 shows d p (t e ), the distance to a galaxy with observed redshift . at the time the observed photons were emitted. For the Benchmark Model, d p (t e ) has a maximum for galaxies with redshift . = 1.6, where d p (t e ) = 0.41cH 0 . When astronomers observe a distant galaxy, they ask the related, but not identical, questions, “How far away is that galaxy?” and “How long has the light from that galaxy been traveling?” In the Benchmark Model, or any other model, we can answer the question “How far away is that galaxy?” by computing the proper distance d p (t 0 ). We can answer the question “How long has the light from that galaxy been traveling?” by computing the lookback time. If light emitted at time t e is observed at time t 0 , the lookback time is simply t 0 −t e . In the limits of very small redshifts, t 0 −t e ≈ .H 0 . However, as shown in Figure 6.7, at larger redshifts, the relation between lookback time and redshift becomes nonlinear. The exact dependence of lookback time on redshift depends strongly on the cosmological model used. For example, consider a galaxy with redshift . = 2. In the Benchmark Model, the lookback time to that galaxy is 10.5 Gyr; we are seeing a redshifted image of that galaxy as it was 10.5 billion years ago. In a flat, lambda-only universe, however, the lookback time to a . = 2 galaxy is 15.4 Gyr, assuming H −1 0 = 14 Gyr. In a flat, matter-dominated universe, the lookback time to a . = 2 galaxy is a mere 7.5 Gyr, with the same assumed Hubble constant. Knowing Ω m,0 and Ω Λ,0 thus becomes important to studies of galaxy evolution. How long does it take galaxies at . ≈ 2 to evolve into galaxies similar to those at . ≈ 0? Is 6.5. BENCHMARK MODEL 121 .01 .1 1 10 100 1000 .01 .1 1 10 100 Observation z ( H 0 / c ) d p ( t 0 ) matter−only Benchmark Λ−only .01 .1 1 10 100 1000 .01 .1 1 10 100 Emission z ( H 0 / c ) d p ( t e ) Figure 6.6: The proper distance to a light source with observed redshift .. The upper panel shows the distance at the time of observation; the lower panel shows the distance at the time of emission. The bold solid line indicates the Benchmark Model, the dot-dash line a flat, lambda-only universe, and the dotted line a flat, matter-only universe. 122 CHAPTER 6. MULTIPLE-COMPONENT UNIVERSES 0 2 4 6 0 5 10 15 z t 0 − t e ( G y r ) matter−only Benchmark Λ−only Figure 6.7: The lookback time, t 0 −t e , for galaxies with observed redshift .. The Hubble time is assumed to be H −1 0 = 14 Gyr. The heavy solid line shows the result for the Benchmark Model, the dot-dash line for a flat, lambda-only universe, and the dotted line for a flat, matter-only universe. it 15 billion years, or only half that time, or something in between? In future years, as the Benchmark Model becomes better constrained, our ability to translate observed redshifts into deduced times will become more accurate. The most distant galaxies that have been observed (at the beginning of the 21st century) are at a redshift . ≈ 6. Consider such a high-redshift galaxy. Using the Benchmark Model, we find that the current proper distance to a galaxy with . = 6 is d p (t 0 ) = 1.92cH 0 = 8300 Mpc, about 60% of the current horizon distance. The proper distance at the time the light was emitted was smaller by a multiplicative factor 1(1 + .) = 17. This means that the galaxy was only at a distance d p (t e ) = 0.27cH 0 = 1200 Mpc at the time the light was emitted. The light was emitted when the age of the universe was t e = 0.066H −1 0 = 0.9 Gyr, or less than 7% of the universe’s current age, t 0 = 13.5 Gyr. The lookback time to a . = 6 galaxy in the Benchmark Model is thus t 0 − t e = 12.6 Gyr. Astronomers are fond of saying, “A telescope is a time machine.” As you look further and further out into the universe, to objects with larger and larger values of d p (t 0 ), you are looking back to objects with smaller and smaller values of t e . When you observe a galaxy 6.5. BENCHMARK MODEL 123 with a redshift . = 6, according to the Benchmark Model, you are glimpsing the universe as it was as a youngster, less than a billion years old. Suggested reading [Full references are given in the “Annotated Bibliography” on page 286.] Harrison (2000), ch. 18: A classification of possible universes, by kine- matic and dynamic criteria. Kolb & Turner (1990), ch. 3.2: The scale factor c(t) and current age t 0 for various two-component universes; contains useful formulae and graphs. Linder (1997), ch. 4.5: The fate of the universe, for different values of the equation-of-state parameter u. Problems (6.1) In a positively curved universe containing only matter (Ω 0 1, κ = +1), show that the present age of the universe is given by the formula H 0 t 0 = Ω 0 2(Ω 0 −1) 3/2 cos −1 2 −Ω 0 Ω 0 − 1 Ω 0 −1 . (6.43) Assuming H 0 = 70 kms −1 Mpc −1 , plot t 0 as a function of Ω 0 in the range 1 ≤ Ω 0 ≤ 3. (6.2) In a negatively curved universe containing only matter (Ω 0 < 1, κ = −1), show that the present age of the universe is given by the formula H 0 t 0 = 1 1 −Ω 0 − Ω 0 2(1 −Ω 0 ) 3/2 cosh −1 2 −Ω 0 Ω 0 . (6.44) Assuming H 0 = 70 kms −1 Mpc −1 , plot t 0 as a function of Ω 0 in the range 0 ≤ Ω 0 ≤ 1. The current best estimate for the ages of stars in globular clusters yields an age of t = 13 Gyr for the oldest globular clusters. In a matter-only universe, what is the maximum permissible value of Ω 0 , given the constraints H 0 = 70 kms −1 Mpc −1 and t 0 13 Gyr? 124 CHAPTER 6. MULTIPLE-COMPONENT UNIVERSES (6.3) One of the more recent speculations in cosmology is that the uni- verse may contain a quantum field, called “quintessence”, which has a positive energy density and a negative value of the equation-of-state parameter u. Assume, for the purposes of this problem, that the uni- verse is spatially flat, and contains nothing but matter (u = 0), and quintessence with u = −12. The current density parameter of mat- ter is Ω m,0 ≤ 1, and the current density parameter of quintessence is Ω Q,0 = 1 − Ω m,0 . At what scale factor c mQ will the energy density of quintessence and matter be equal? Solve the Friedmann equation to find c(t) for this universe. What is c(t) in the limit c < c mQ ? What is c(t) in the limit c c mQ ? What is the current age of this universe, expressed in terms of H 0 and Ω m,0 ? (6.4) Suppose you wanted to “pull an Einstein”, and create a static universe ( ˙ c = 0, ¨ c = 0) in which the gravitational attraction of matter is exactly balanced by the gravitational repulsion of quintessence with equation- of-state parameter u Q . Within what range must u Q fall for the effects of quintessence to be repulsive? Is it possible to create a static universe for an arbitrary value of u Q within this range? If so, will the curvature of the universe be negative or positive? What energy density ε Q will be required, for a given value of u Q , to balance an energy density ε m in matter? (6.5) Consider a positively curved universe containing only matter (the “Big Crunch” model discussed in Section 6.1). At some time t 0 t Crunch 2, during the contraction phase of this universe, an astronomer named Elbbuh Niwde discovers that nearby galaxies have blueshifts (−1 ≤ . < 0) proportional to their distance. He then measures H 0 and Ω 0 , finding H 0 < 0 and Ω 0 1. Given H 0 and Ω 0 , how long a time will elapse between Dr. Niwde’s observations at t = t 0 and the final Big Crunch at t = t crunch ? What is the minimum blueshift that Dr. Niwde is able to observe? What is the lookback time to an object with this blueshift? (6.6) Consider an expanding, positively curved universe containing only a cosmological constant (Ω 0 = Ω Λ,0 1). Show that such a universe underwent a “Big Bounce” at a scale factor c bounce = Ω 0 −1 Ω 0 1/2 . (6.45) 6.5. BENCHMARK MODEL 125 and that the scale factor as a function of time is c(t) = c bounce cosh[ Ω 0 H 0 (t −t bounce )] . (6.46) where t bounce is the time at which the Big Bounce occurred. What is the time t 0 −t bounce which has elapsed since the Big Bounce, expressed as a function of H 0 and Ω 0 ? (6.7) A universe is spatially flat, and contains both matter and a cosmolog- ical constant. For what value of Ω m,0 is t 0 exactly equal to H −1 0 ? (6.8) In the Benchmark Model, what is the total mass of all the matter within our horizon? What is the total energy of all the photons within our horizon? How many baryons are within the horizon? Chapter 7 Measuring Cosmological Parameters Cosmologists would like to know the scale factor c(t) for the universe. For a model universe whose contents are known with precision, the scale factor can be computed from the Friedmann equation. Finding c(t) for the real universe, however, is much more difficult. The scale factor is not directly ob- servable; it can only be deduced indirectly from the imperfect and incomplete observations that we make of the universe around us. In the previous three chapters, I’ve pointed out that if we knew the energy density ε for each component of the universe, we could use the Friedmann equation to find the scale factor c(t). The argument works in the other direction, as well; if we could determine c(t) from observations, we could use that knowledge to find ε for each component. Let’s see, then, what constraints we can put on the scale factor by making observations of distant astronomical objects. 7.1 “A search for two numbers” Since determining the exact functional form of c(t) is difficult, it is useful, instead, to do a Taylor series expansion for c(t) around the present moment. The complete Taylor series is c(t) = c(t 0 ) + dc dt t=t 0 (t −t 0 ) + 1 2 d 2 c dt 2 t=t 0 (t −t 0 ) 2 +. . . (7.1) 126 7.1. “A SEARCH FOR TWO NUMBERS” 127 To exactly reproduce an arbitrary function c(t) for all values of t, an infi- nite number of terms is required in the expansion. However, the usefulness of a Taylor series expansion resides in the fact that if c doesn’t fluctuate wildly with t, using only the first few terms of the expansion gives a good approximation in the immediate vicinity of t 0 . The scale factor c(t) is a good candidate for a Taylor expansion. The different model universes examined in the previous two chapters all had smoothly varying scale factors, and there’s no evidence that the real universe has a wildly oscillating scale factor. Keeping the first three terms of the Taylor expansion, the scale factor in the recent past and the near future can be approximated as c(t) ≈ c(t 0 ) + dc dt t=t 0 (t −t 0 ) + 1 2 d 2 c dt 2 t=t 0 (t −t 0 ) 2 . (7.2) Dividing by the current scale factor, c(t 0 ), c(t) c(t 0 ) ≈ 1 + ˙ c c t=t 0 (t −t 0 ) + 1 2 ¨ c c t=t 0 (t −t 0 ) 2 . (7.3) Using the normalization c(t 0 ) = 1, this expansion for the scale factor is customarily written in the form c(t) ≈ 1 +H 0 (t −t 0 ) − 1 2 ¡ 0 H 2 0 (t −t 0 ) 2 . (7.4) In equation (7.4), the parameter H 0 is our old acquaintance the Hubble constant, H 0 ≡ ˙ c c t=t 0 . (7.5) and the parameter ¡ 0 is a dimensionless number called the deceleration pa- rameter, defined as ¡ 0 ≡ − ¨ cc ˙ c 2 t=t 0 = − ¨ c cH 2 t=t 0 . (7.6) Note the choice of sign in defining ¡ 0 . A positive value of ¡ 0 corresponds to ¨ c < 0, meaning that the universe’s expansion is decelerating (that is, the relative velocity of any two points is decreasing). A negative value of ¡ 0 corresponds to ¨ c 0, meaning that the relative velocity of any two points is increasing with time. The choice of sign for ¡ 0 , and the fact that it’s named 128 CHAPTER 7. MEASURING COSMOLOGICAL PARAMETERS the deceleration parameter, is due to the fact that it was first defined during the mid-1950’s, when the limited information available favored a matter- dominated universe with ¨ c < 0. If the universe contains a sufficiently large cosmological constant, however, the deceleration parameter ¡ 0 can have either sign. The Taylor expansion of equation (7.4) is physics-free. It is simply a mathematical description of how the universe expands at times t ∼ t 0 , and says nothing at all about what forces act to accelerate the expansion (to take a Newtonian viewpoint of the physics involved). The parameters H 0 and ¡ 0 are thus purely descriptive of the kinematics, and are free of the the- oretical “baggage” underlying the Friedmann equation and the acceleration equation. 1 In a famous 1970 review article, the observational cosmologist Allan Sandage described all of cosmology as “A Search for Two Numbers”. Those two numbers were H 0 and ¡ 0 . Although the scope of cosmology has widened considerably since Sandage wrote his article, cosmologists are still assiduously searching for H 0 and ¡ 0 . Although H 0 and ¡ 0 are themselves free of the theoretical assumptions underlying the Friedmann and acceleration equations, you can use the accel- eration equation to predict what ¡ 0 will be in a given model universe. If your model universe contains several components, each with a different value of the equation-of-state parameter u, the acceleration equation can be written ¨ c c = − 4πG 3c 2 ¸ w ε w (1 + 3u) . (7.7) Divide each side of the acceleration equation by the square of the Hubble parameter H(t) and change sign: − ¨ c cH 2 = 1 2 ¸ 8πG 3c 2 H 2 ¸ w ε w (1 + 3u) . (7.8) However, the quantity in square brackets in equation (7.8) is just the in- verse of the critical energy density ε c . Thus, we can rewrite the acceleration equation in the form − ¨ c cH 2 = 1 2 ¸ w Ω w (1 + 3u) . (7.9) 1 Remember, the Friedmann equation assumes that the expansion of the universe is controlled by gravity, and that gravity is accurately described by Einstein’s theory of general relativity; although these are reasonable assumptions, they are not 100% iron- clad. 7.1. “A SEARCH FOR TWO NUMBERS” 129 Evaluating equation (7.9) at the present moment, t = t 0 , tells us the relation between the deceleration parameter ¡ 0 and the density parameters of the different components of the universe: ¡ 0 = 1 2 ¸ w Ω w,0 (1 + 3u) . (7.10) For a universe containing radiation, matter, and a cosmological constant, ¡ 0 = Ω r,0 + 1 2 Ω m,0 −Ω Λ,0 . (7.11) Such a universe will currently be accelerating outward (¡ 0 < 0) if Ω Λ,0 Ω r,0 + Ω m,0 2. The Benchmark Model, for instance, has ¡ 0 ≈ −0.55. In principle, determining H 0 should be easy. For small redshifts, the rela- tion between a galaxy’s distance d and its redshift . is linear (equation 2.5): c. = H 0 d . (7.12) Thus, if you measure the distance d and redshift . for a large sample of galax- ies, and fit a straight line to a plot of c. versus d, the slope of the plot gives you the value of H 0 . 2 Measuring the redshift of a galaxy is relatively simple; auto- mated galaxy surveys can find hundreds of galaxy redshifts in a single night. The difficulty is in measuring the distance of a galaxy. Remember, Edwin Hubble was off by a factor of 7 when he estimated H 0 ≈ 500 kms −1 Mpc −1 (see Figure 2.4). This is because he underestimated the distances to galaxies in his sample by a factor of 7. The distance to a galaxy is not only difficult to measure but also, in an expanding universe, somewhat difficult to define. In Section 3.3, the proper distance d p (t) between two points was defined as the length of the spatial geodesic between the points when the scale factor is fixed at the value c(t). The proper distance is perhaps the most straightforward definition of the spatial distance between two points in an expanding universe. Moreover, there is a helpful relation between scale factor and proper distance. If we observe, at time t 0 , light that was emitted by a distant galaxy at time t e , the 2 The peculiar velocities of galaxies cause a significant amount of scatter in the plot, but by using a large number of galaxies, you can beat down the statistical errors. If you use galaxies at d < 100 Mpc, you must also make allowances for the local inhomogeneity and anisotropy. 130 CHAPTER 7. MEASURING COSMOLOGICAL PARAMETERS current proper distance to that galaxy is (equation 5.35): d p (t 0 ) = c t 0 te dt c(t) . (7.13) For the model universes examined in Chapters 5 and 6, we knew the exact functional form of c(t), and hence could exactly compute d p (t 0 ) for a galaxy of any redshift. If we have only partial knowledge of the scale factor, in the form of the Taylor expansion of equation 7.4, we may use the expansion 1 c(t) ≈ 1 −H 0 (t −t 0 ) + (1 +¡ 0 2)H 2 0 (t −t 0 ) 2 (7.14) in equation 7.13. Including the two lowest-order terms in the lookback time, t 0 −t e , we find that the proper distance to the galaxy is d p (t 0 ) ≈ c(t 0 −t e ) + cH 0 2 (t 0 −t e ) 2 . (7.15) The first term in the above equation, c(t 0 −t e ), is what the proper distance would be in a static universe – the lookback time times the speed of light. The second term is a correction due to the expansion of the universe during the time the light was traveling. Equation 7.15 would be extremely useful if the light from distant galaxies carried a stamp telling us the lookback time, t 0 − t e . They don’t; instead, they carry a stamp telling us the scale factor c(t e ) at the time the light was emitted. The observed redshift . of a galaxy, remember, is . = 1 c(t e ) −1 . (7.16) Using equation 7.14, we may write an approximate relation between redshift and lookback time: . ≈ H 0 (t 0 −t e ) + (1 +¡ 0 2)H 2 0 (t 0 −t e ) 2 . (7.17) Inverting equation 7.17 to give the lookback time as a function of redshift, we find t 0 −t e ≈ H −1 0 . −(1 +¡ 0 2). 2 . (7.18) Substituting equation 7.18 into equation 7.15 gives us an approximate rela- tion for the current proper distance to a galaxy with redshift .: d p (t 0 ) ≈ c H 0 . −(1 +¡ 0 2). 2 + cH 0 2 . 2 H 2 0 = c H 0 . ¸ 1 − 1 +¡ 0 2 . . (7.19) 7.2. LUMINOSITY DISTANCE 131 The linear Hubble relation d p ∝ . thus only holds true in the limit . < 2(1 + ¡ 0 ). If ¡ 0 −1, then the proper distance to a galaxy of moderate redshift (. ∼ 0.1, say) is less than would be predicted from the linear Hubble relation. 7.2 Luminosity distance Unfortunately, the current proper distance to a galaxy, d p (t 0 ), is not a mea- surable property. If you tried to measure the distance to a galaxy with a tape measure, for instance, the distance would be continuously increasing as you extended the tape. To measure the proper distance at time t 0 , you would need a tape measure which could be extended with infinite speed; alterna- tively, you would need to stop the expansion of the universe at its current scale factor while you measured the distance at your leisure. Neither of these alternatives is physically possible. Since cosmology is ultimately based on observations, if we want to find the distance to a galaxy, we need some way of computing a distance from that galaxy’s observed properties. In devising ways of computing the dis- tance to galaxies, astronomers have found it useful to adopt and adapt the techniques used to measure shorter distances. Let’s examine, then, the tech- niques used to measure relatively short distances. Within the Solar System, astronomers measure the distance to the Moon and planets by reflecting radar signals from them. If δt is the time taken for a photon to complete the round-trip, then the distance to the reflecting body is d = (c δt)2. 3 The accuracy with which distances have been determined with this technique is impressive; the length of the astronomical unit, for instance, is now known to be 1 AU = 149. 597. 870.61 km. The radar technique is useful only within the Solar System. Beyond ∼ 10 AU, the reflected radio waves are too faint to detect. A favorite method for determining distances to other stars within our Galaxy is the method of trigonometric parallax. When a star is observed from two points separated by a distance /, the star’s apparent position will shift by an angle θ. If the baseline of observation is perpendicular to the line 3 Since the relative speeds of objects within the Solar System are much smaller than c, the corrections due to relative motion during the time δt are minuscule. 132 CHAPTER 7. MEASURING COSMOLOGICAL PARAMETERS of sight to the star, the parallax distance will be d π = 1 pc / 1 AU θ 1 arcsec −1 . (7.20) Measuring the distances to stars using the Earth’s orbit (/ = 2 AU) as a baseline is a standard technique. Since the size of the Earth’s orbit is known with great accuracy from radar measurements, the accuracy with which the parallax distance can be determined is limited by the accuracy with which θ can be measured. The Hipparcos satellite, launched by the European Space Agency in 1989, found the parallax distance for ∼ 10 5 stars, with an accuracy of ∼ 1 milliarcsecond. However, to measure θ for a galaxy ∼ 100 Mpc away, an accuracy of < 10 nanoarcseconds would be required, using the Earth’s orbit as a baseline. The trigonometric parallaxes of galaxies at cosmological distances are too small to be measured with current technology. Let’s focus on the properties that we can measure for objects at cosmo- logical distances. We can measure the flux of light, 1, from the object, in units of watts per square meter. The complete flux, integrated over all wave- lengths of light, is called the bolometric flux. (A bolometer is an extremely sensitive thermometer capable of detecting electromagnetic radiation over a wide range of wavelengths; it was invented in 1881 by the astronomer Samuel Langley, who used it to measure solar radiation. 4 ) More frequently, given the difficulties of measuring the true bolometric flux, the flux over a limited range of wavelengths is measured. If the light from the object has emission or ab- sorption lines, we can measure the redshift, .. If the object is an extended source rather than a point of light, we can measure its angular diameter, δθ. One way of using measured properties to assign a distance is the standard candle method. A standard candle is an object whose luminosity 1 is known. For instance, if some class of astronomical object had luminosities which were the same throughout all of space-time, they would act as excellent standard candles – if their unique luminosity 1 were known. If you know, by some means or other, the luminosity of an object, then you can use its measured flux 1 to define a function called the luminosity distance: d L ≡ 1 4π1 1/2 . (7.21) 4 As expressed more poetically in an anonymous limerick: “Oh, Langley devised the bolometer: / It’s really a kind of thermometer / Which measures the heat / From a polar bear’s feet / At a distance of half a kilometer.” 7.2. LUMINOSITY DISTANCE 133 Figure 7.1: An observer at the origin observes a standard candle, of known luminosity 1, at comoving coordinate location (:. θ. φ). The function d L is called a “distance” because its dimensionality is that of a distance, and because it is what the proper distance to the standard candle would be if the universe were static and Euclidean. In a static Euclidean universe, the propagation of light follows the inverse square law 1 = 1[4πd 2 ]. Suppose, though, that you are in a universe described by a Robertson- Walker metric (equation 3.25): d: 2 = −c 2 dt 2 +c(t) 2 [d: 2 +o κ (:) 2 dΩ 2 ] . (7.22) with o κ (:) = 1 0 sin(:1 0 ) (κ = +1) : (κ = 0) 1 0 sinh(:1 0 ) (κ = −1) . (7.23) You are at the origin. At the present moment, t = t 0 , you see light that was emitted by a standard candle at comoving coordinate location (:. θ. φ) at a time t e (see Figure 7.1). The photons which were emitted at time t e are, at the present moment, spread over a sphere of proper radius d p (t 0 ) = : and proper surface area ¹ p (t 0 ). If space is flat (κ = 0), then the proper area 134 CHAPTER 7. MEASURING COSMOLOGICAL PARAMETERS of the sphere is given by the Euclidean relation ¹ p (t 0 ) = 4πd p (t 0 ) 2 = 4π: 2 . More generally, however, ¹ p (t 0 ) = 4πo κ (:) 2 . (7.24) When space is positively curved, ¹ p (t 0 ) < 4π: 2 , and the photons are spread over a smaller area than they would be in flat space. When space is negatively curved, ¹ p (t 0 ) 4π: 2 , and photons are spread over a larger area than they would be in flat space. In addition to these geometric effects, which would apply even in a static universe, the expansion of the universe causes the observed flux of light from a standard candle of redshift . to be decreased by a factor of (1+.) −2 . First, the expansion of the universe causes the energy of each photon from the standard candle to decrease. If a photon starts with an energy 1 e = /cλ e when the scale factor is c(t e ), by the time we observe it, when the scale factor is c(t 0 ) = 1, the wavelength will have grown to λ 0 = 1 c(t e ) λ e = (1 +.)λ e . (7.25) and the energy will have fallen to 1 0 = 1 e 1 +. . (7.26) Second, thanks to the expansion of the universe, the time between photon detections will be greater. If two photons are emitted in the same direction separated by a time interval δt e , the proper distance between them will ini- tially be c(δt e ); by the time we detect the photons at time t 0 , the proper distance between them will be stretched to c(δt e )(1 + .), and we will detect them separated by a time interval δt 0 = δt e (1 +.). The net result is that in an expanding, spatially curved universe, the relation between the observed flux 1 and the luminosity 1 of a distant light source is 1 = 1 4πo κ (:) 2 (1 +.) 2 . (7.27) and the luminosity distance is d L = o κ (:)(1 +.) . (7.28) 7.2. LUMINOSITY DISTANCE 135 0 2 4 6 0 5 10 15 z ( H 0 / c ) d L matter−only Benchmark Λ−only Figure 7.2: The luminosity distance of a standard candle with observed red- shift .. The bold solid line gives the result for the Benchmark Model, the dot-dash line for a flat, lambda-only universe, and the dotted line for a flat, matter-only universe. The available evidence indicates that our universe is nearly flat, with a radius of curvature 1 0 which is larger than the current horizon distance d hor (t 0 ). Objects with finite redshift are at proper distances smaller than the horizon distance, and hence smaller than the radius of curvature. Thus, it is safe to make the approximation : <1 0 , implying o κ (:) ≈ :. With our assumption that space is very close to being flat, the relation between the luminosity distance and the current proper distance becomes very simple: d L = :(1 +.) = d p (t 0 )(1 +.) [κ = 0] . (7.29) Thus, even if space is perfectly flat, if you estimate the distance to a standard candle by using a na¨ıve inverse square law, you will overestimate the actual proper distance by a factor (1+.), where . is the standard candle’s redshift. Figure 7.2 shows the luminosity distance d L as a function of redshift for the Benchmark Model, and for two other flat universes, one dominated by matter and one dominated by a cosmological constant. When . < 1, the 136 CHAPTER 7. MEASURING COSMOLOGICAL PARAMETERS current proper distance may be approximated as d P (t 0 ) ≈ c H 0 . 1 − 1 +¡ 0 2 . . (7.30) In a universe which is nearly flat, the luminosity distance may thus be ap- proximated as d L ≈ c H 0 . 1 − 1 +¡ 0 2 . (1 +.) ≈ c H 0 . 1 + 1 −¡ 0 2 . . (7.31) Note that in the limit . →0, d p (t 0 ) ≈ d L ≈ c H 0 . . (7.32) In a universe described by the Robertson-Walker metric, the luminosity dis- tance is a good approximation to the current proper distance for objects with small redshifts. 7.3 Angular-diameter distance The luminosity distance d L is not the only distance measure that can be computed using the observable properties of cosmological objects. Suppose that instead of a standard candle, you observed a standard yardstick. A standard yardstick is an object whose proper length / is known. In most cases, it is convenient to chose as your yardstick an object which is tightly bound together, by gravity or duct tape or some other influence, and hence is not expanding along with the universe as a whole. Suppose a yardstick of constant proper length / is aligned perpendicular to your line of sight, as shown in Figure 7.3. You measure an angular distance δθ between the ends of the yardstick, and a redshift . for the light which the yardstick emits. If δθ < 1, and if you know the length / of the yardstick, you can compute a distance to the yardstick using the small-angle formula d A ≡ / δθ . (7.33) This function of / and δθ is called the angular-diameter distance. The angular-diameter distance is equal to the proper distance to the yardstick if the universe is static and Euclidean. 7.3. ANGULAR-DIAMETER DISTANCE 137 Figure 7.3: An observer at the origin observes a standard yardstick, of known proper length /, at comoving coordinate distance :. In general, though, if the universe is expanding or curved, the angular- diameter distance will not be equal to the current proper distance. Suppose you are in a universe described by the Robertson-Walker metric given in equation (7.22). Choose your comoving coordinate system so that you are at the origin. The yardstick is at a comoving coordinate distance :. At a time t e , the yardstick emitted the light which you observe at time t 0 . The comoving coordinates of the two ends of the yardstick, at the time the light was emitted, were (:. θ 1 . φ) and (:. θ 2 . φ). As the light from the yardstick moves toward the origin, it travels along geodesics with θ = constant and φ = constant. Thus, the angular size which you measure for the yardstick will be δθ = θ 2 −θ 1 . The distance d: between the two ends of the yardstick, measured at the time t e when the light was emitted, can be found from the Robertson-Walker metric: d: = c(t e )o κ (:)δθ . (7.34) However, for a standard yardstick whose length / is known, we can set d: = /, and thus find that / = c(t e )o κ (:)δθ = o κ (:)δθ 1 +. . (7.35) 138 CHAPTER 7. MEASURING COSMOLOGICAL PARAMETERS Thus, the angular-diameter distance d A to a standard yardstick is d A ≡ / δθ = o κ (:) 1 +. . (7.36) Comparison with equation (7.28) shows that the relation between the angular- diameter distance and the luminosity distance is d A = d L (1 +.) 2 . (7.37) Thus, if you observe an object which is both a standard candle and a standard yardstick, the angular-diameter distance which you compute for the object will be smaller than the luminosity distance. Moreover, if the universe is spatially flat, d A (1 +.) = d p (t 0 ) = d L 1 +. [κ = 0] . (7.38) In a flat universe, therefore, if you compute the angular-diameter distance d A of a standard yardstick, it isn’t equal to the current proper distance d p (t 0 ); rather, it is equal to the proper distance at the time the light from the object was emitted: d A = d p (t 0 )(1 +.) = d p (t e ). Figure 7.4 shows the angular-diameter distance d A for the Benchmark Model, and for two other spatially flat universes, one dominated by matter and one dominated by a cosmological constant. (Since d A is, for these flat universes, equal to d p (t e ), Figure 7.4 is simply a replotting of of the lower panel in Figure 6.6.) When . < 1, the approximate value of d A is given by the expansion d A ≈ c H 0 . 1 − 3 +¡ 0 2 . . (7.39) Thus, comparing equations (7.30), (7.31), and (7.39), we find that in the limit . → 0, d A ≈ d L ≈ d p (t 0 ) ≈ (cH 0 ).. However, the state of affairs is very different in the limit . → ∞. In models with a finite horizon size, d p (t 0 ) → d hor (t 0 ) as . → ∞. The luminosity distance to highly redshifted objects, in this case, diverges as . →∞, with d L (. →∞) ≈ .d hor (t 0 ) . (7.40) However, the angular-diameter distance to highly redshifted objects approaches zero as . →∞, with d A (. →∞) ≈ d hor (t 0 ) . . (7.41) 7.3. ANGULAR-DIAMETER DISTANCE 139 0 2 4 6 0 .2 .4 .6 .8 1 z ( H 0 / c ) d A matter−only Benchmark Λ−only Figure 7.4: The angular-diameter distance for a standard yardstick with observed redshift .. The bold solid line gives the result for the Benchmark Model, the dot-dash line for a flat, lambda-only universe, and the dotted line for a flat, matter-only universe. 140 CHAPTER 7. MEASURING COSMOLOGICAL PARAMETERS In model universes other than the lambda-only model, the angular-diameter distance d A has a maximum for standard yardsticks at some critical redshift . c . (For the Benchmark Model, . c = 1.6, where d A (max) = 0.41cH 0 = 1800 Mpc.) This means that if the universe were full of glow-in-the-dark yardsticks, all of the same size /, their angular size δθ would decrease with redshift out to . = . c , but then would increase at larger redshifts. The sky would be full of big, faint, redshifted yardsticks. In principle, standard yardsticks can be used to determine H 0 . To begin with, identify a population of standard yardsticks (objects whose physical size / is known). Then, measure the redshift . and angular size δθ of each standard yardstick. Compute the angular-diameter distance d A = /δθ for each standard yardstick. Plot c. versus d A , and the slope of the relation, in the limit . → 0, will give you H 0 . In addition, if you have measured the angular size δθ for standard candles at . ∼ . c , the shape of the c. versus d A plot can be used to determine further cosmological parameters. If you simply want a kinematic description, you can estimate ¡ 0 by fitting equation (7.39) to the data. If you are confident that the universe is dominated by matter and a cosmological constant, you can see which values of Ω m,0 and Ω Λ,0 provide the best fit to the observed data. In practice, the use of standard yardsticks to determine cosmological pa- rameters has long been plagued with observational difficulties. For instance, a standard yardstick must have an angular size large enough to be resolved by your telescope. A yardstick of physical size / will have its angular size δθ minimized when it is at the critical redshift . c . For the Benchmark Model, δθ(min) = / d A (max) = / 1800 Mpc ≈ 0.1 arcsec / 1 kpc . (7.42) Both galaxies and clusters of galaxies are large enough to be useful standard candles. Unfortunately for cosmologists, galaxies and clusters of galaxies do not have sharply defined edges, so assigning a particular angular size δθ, and a corresponding physical size /, to these objects is a somewhat tricky task. Moreover, galaxies and clusters of galaxies are not isolated, rigid yardsticks of fixed length. Galaxies tend to become larger with time as they undergo mergers with their neighbors. Clusters, too, tend to become larger with time, as galaxies fall into them. (Eventually, our Local Group will fall into the Virgo cluster.) Correcting for these evolutionary trends is a difficult task. Given the difficulties involved in using standard yardsticks to determine cosmological parameters, more attention has been focused, in recent years, 7.4. STANDARD CANDLES & H 0 141 on the use of standard candles. Let’s first look, therefore, at how standard candles can be used to determine H 0 , then focus on how they can be used to determine the acceleration of the universe. 7.4 Standard candles & the Hubble constant Using standard candles to determine the Hubble constant has a long and honorable history; it’s the method used by Hubble himself. The recipe for finding the Hubble constant is a simple one: • Identify a population of standard candles with luminosity 1. • Measure the redshift . and flux 1 for each standard candle. • Compute d L = (14π1) 1/2 for each standard candle. • Plot c. versus d L . • Measure the slope of the c. versus d L relation when . < 1; this gives H 0 . As with the apocryphal recipe for rabbit stew which begins “First catch your rabbit,” the hardest step is the first one. A good standard candle is hard to find. For cosmological purposes, a standard candle should be bright enough to be detected at large redshifts. It should also have a luminosity which is well determined. 5 One time-honored variety of standard candle is the class of Cepheid vari- able stars. Cepheids, as they are known, are highly luminous supergiant stars, with mean luminosities lying in the range ¯ 1 = 400 → 40.000 L . Cepheids are pulsationally unstable. As they pulsate radially, their luminosity varies in response, partially due to the change in their surface area, and partially due to the changes in the surface temperature as the star pulsates. The pulsational periods, as reflected in the observed brightness variations of the star, lie in the range 1 = 1.5 →60 days. 5 A useful cautionary tale, in this regard, is the saga of Edwin Hubble. In the 1929 paper which first demonstrated that d L ∝ z when z 1, Hubble underestimated the luminosity distances to galaxies by a factor of ∼ 7 because he underestimated the luminosity of his standard candles by a factor of ∼ 49. 142 CHAPTER 7. MEASURING COSMOLOGICAL PARAMETERS On the face of it, Cepheids don’t seem sufficiently standardized to be standard candles; their mean luminosities range over two orders of magni- tude. How can you tell whether you are looking at an intrinsically faint Cepheid (1 ≈ 400 L ) or at an intrinsically bright Cepheid (1 ≈ 40.000 L ) ten times farther away? The key to calibrating Cepheids was discovered by Henrietta Leavitt, at Harvard College Observatory. In the years prior to World War I, Leavitt was studying variable stars in the Large and Small Magellanic Clouds, a pair of relatively small satellite galaxies orbiting our own Galaxy. For each Cepheid in the Small Magellanic Cloud (SMC), she measured the period 1 by finding the time between maxima in the observed brightness, and found the mean flux ¯ 1, averaged over one complete period. She noted that there was a clear relation between 1 and ¯ 1, with stars having the longest period of variability also having the largest flux. Since the depth of the SMC, front to back, is small compared to its distance from us, she was justified in assuming that the difference in mean flux for the Cepheids was due to differences in their mean luminosity, not differences in their lu- minosity distance. Leavitt had discovered a period – luminosity relation for Cepheid variable stars. If the same period – luminosity relation holds true for all Cepheids, in all galaxies, then Cepheids can act as a standard candle. Suppose, for instance, you find a Cepheid star in the Large Magellanic Cloud (LMC) and another in M31. They both have a pulsational period of 10 days, so you assume, from the period – luminosity relation, that they have the same mean luminosity ¯ 1. By careful measurement, you determine that ¯ 1 LMC ¯ 1 M31 = 230 . (7.43) Thus, you conclude that the luminosity distance to M31 is greater than that to the LMC 6 by a factor d L (M31) d L (LMC) = ¯ 1 LMC ¯ 1 M31 1/2 = √ 230 = 15.2 . (7.44) Note that if you only know the relative fluxes of the two Cepheids, and not their luminosity ¯ 1, you will only know the relative distances of M31 and the 6 In practice, given the intrinsic scatter in the period – luminosity relation, and the inevitable error in measuring fluxes, astronomers would not rely on a single Cepheid in each galaxy. Rather, they would measure ¯ f and P for as many Cepheids as possible in each galaxy, then find the ratio of luminosity distances that would make the period – luminosity relations for the two galaxies coincide. 7.4. STANDARD CANDLES & H 0 143 LMC. To fix an absolute distance to M31, to the LMC, and to other galaxies containing Cepheids, you need to know the luminosity ¯ 1 for a Cepheid of a given period 1. If, for instance, you could measure the parallax distance d π to a Cepheid within our own Galaxy, you could then compute its luminosity ¯ 1 = 4πd 2 π ¯ 1, and use it to normalize the period – luminosity relation for Cepheids. 7 Unfortunately, Cepheids are rare stars; only the very nearest Cepheids in our Galaxy have had their distances measured with even modest accuracy by the Hipparcos satellite. The nearest Cepheid is Polaris, as it turns out, at d π = 130±10 pc. The next nearest is probably δ Cephei (the prototype after which all Cepheids are named), at d π = 300±50 pc. Future space-based astrometric observatories, such as the Space Interferometry Mission (SIM), scheduled for launch in 2009, will allow parallax distances to be measured with an accuracy greater than that provided by Hipparcos. Until the distances (and hence the luminosities) of nearby Cepheids are known with this great accuracy, astronomers must still rely on alternate methods of normalizing the period – luminosity relation for Cepheids. The most usual method involves finding the distance to the Large Magellanic Cloud by secondary methods, 8 then using this distance to compute the mean luminosity of the LMC Cepheids. The current consensus is that the Large Magellanic Cloud has a luminosity distance d L = 50 ±3 kpc, implying a distance to M31 of d L = 760 ±50 kpc. With the Hubble Space Telescope, the fluxes and periods of Cepheids can be accurately measured out to luminosity distances of d L ∼ 20 Mpc. Observation of Cepheid stars in the Virgo cluster of galaxies, for instance, has yielded a distance d L (Virgo) = 300 d L (LMC) = 15 Mpc. One of the motivating reasons for building the Hubble Space Telescope in the first place was to use Cepheids to determine H 0 . The net result of the Hubble Key Project to measure H 0 is displayed in Figure 2.5, showing that the Cepheid data are best fit with a Hubble constant of H 0 = 75 ±8 kms −1 Mpc −1 . There is a hidden difficulty involved in using Cepheid stars to determine H 0 . Cepheids can take you out only to a distance d L ∼ 20 Mpc; on this scale, the universe cannot be assumed to be homogeneous and isotropic. In fact, the Local Group is gravitationally attracted toward the Virgo cluster, causing it to have a peculiar motion in that direction. It is estimated, from dynamical models, that the recession velocity c. which we measure for the Virgo cluster 7 Within our Galaxy, which is not expanding, the parallax distance, the luminosity distance, and the proper distance are identical. 8 A good review of these methods, and the distances they yield, is given by van den Bergh (2000). 144 CHAPTER 7. MEASURING COSMOLOGICAL PARAMETERS is 250 kms −1 less than it would be if the universe were perfectly homogeneous. The plot of c. versus d L given in Figure 2.5 uses recession velocities which are corrected for this “Virgocentric flow”, as it is called. 7.5 Standard candles & the accelerating uni- verse To determine the value of H 0 without having to worry about Virgocentric flow and other peculiar velocities, we need to determine the luminosity distance to standard candles with d L 100 Mpc, or . 0.02. To determine the value of ¡ 0 , we need to view standard candles for which the relation between d L and . deviates significantly from the linear relation which holds true at lower redshifts. In terms of H 0 and ¡ 0 , the luminosity distance at small redshift is d L ≈ c H 0 . ¸ 1 + 1 −¡ 0 2 . . (7.45) At a redshift . = 0.2, for instance, the luminosity distance d L in the Bench- mark Model (with ¡ 0 = −0.55) is 5% larger than d L in an empty universe (with ¡ 0 = 0). 9 For a standard candle to be seen at d L 100 Mpc (to determine H 0 with minimal effects from peculiar velocity) or at d L 1000 Mpc (to determine ¡ 0 ), it must be very luminous. Initial attempts to find a highly luminous standard candle focused on using entire galaxies as standard candles. This attempt foundered on the lack of standardization among galaxies. Not only do galaxies have a wide range of luminosities at the present moment, but any individual galaxy has a luminosity which evolves significantly with time. For instance, an isolated galaxy, after an initial outburst of star formation, will fade gradually with time, as its stars exhaust their nuclear fuel and become dim stellar remnants. A galaxy in a rich cluster, by contrast, can actually become more luminous with time, as it “cannibalizes” smaller galaxies by merging with them. For any particular galaxy, it’s difficult to tell which effect dominates. Since the luminosity evolution of galaxies is imperfectly understood, they aren’t particularly suitable for use as standard candles. 9 If you think, optimistically, that you can determine luminosity distances with an accuracy much better than 5%, then you won’t have to go as deep into space to determine q 0 accurately. 7.5. STANDARD CANDLES & ACCELERATION 145 In recent years, the standard candle of choice among cosmologists has been type Ia supernovae. A supernova may be loosely defined as an exploding star. Early in the history of supernova studies, when little was known about their underlying physics, supernovae were divided into two classes, on the basis of their spectra. Type I supernovae contain no hydrogen absorption lines in their spectra; type II supernovae contain strong hydrogen absorption lines. Gradually, it was realized that all type II supernovae are the same species of beast; they are massive stars (` 8 M ) whose cores collapse to form a black hole or neutron star when their nuclear fuel is exhausted. During the rapid collapse of the core, the outer layers of the star are thrown off into space. Type I supernovae are actually two separate species, which are called type Ia and type Ib. Type Ib supernovae, it is thought, are massive stars whose cores collapse after the hydrogen-rich outer layers of the star have been blown away in strong stellar winds. Thus, type Ib and type II supernovae are driven by very similar mechanisms – their differences are superficial, in the most literal sense. Type Ia supernovae, however, are something completely different. They occur in close binary systems where one of the two stars in the system is a white dwarf; that is, a stellar remnant which is supported against gravity by electron degeneracy pressure. The transfer of mass from the companion star to the white dwarf eventually nudges the white dwarf over the Chandrasekhar limit of 1.4 M ; this is the maximum mass at which the electron degeneracy pressure can support a white dwarf against its own self-gravity. When the Chandrasekhar limit is exceeded, the white dwarf starts to collapse until its increased density triggers a runaway nuclear fusion reaction. The entire white dwarf becomes a fusion bomb, blowing itself to smithereens; unlike type II supernovae, type Ia supernovae do not leave a condensed stellar remnant behind. Within our Galaxy, type Ia supernovae occur roughly once per century, on average. Although type Ia supernovae are not frequent occurrences lo- cally, they are extraordinary luminous, and hence can be seen to large dis- tances. The luminosity of an average type Ia supernova, at peak brightness, is 1 = 410 9 L ; that’s 100,000 times more luminous than even the brightest Cepheid. For a few days, a type Ia supernova in a moderately bright galaxy can outshine all the other stars in the galaxy combined. Since moderately bright galaxies can be seen at . ∼ 1, this means that type Ia supernovae can also be seen at . ∼ 1. Not only are type Ia supernovae bright standard candles, they are also reasonably standardized standard candles. Consider type Ia supernovae in the Virgo cluster. Although there’s only one type Ia 146 CHAPTER 7. MEASURING COSMOLOGICAL PARAMETERS supernova per century in our own Galaxy, the total luminosity of the Virgo cluster is a few hundred times that of our Galaxy. Thus, every year you can expect a few type Ia supernovae to go off in the Virgo cluster. Several type Ia supernovae have been observed in the Virgo cluster in the recent past, and have been found to have similar fluxes at maximum brightness. So far, type Ia supernovae sound like ideal standard candles; very lumi- nous and very standardized. There’s one complication, however. Observa- tion of supernovae in galaxies whose distances have been well determined by Cepheids reveal that type Ia supernovae do not have identical luminosities. Instead of all having 1 = 4 10 9 L , their peak luminosities lie in the fairly broad range 1 ≈ 3 → 5 10 9 L . However, it has also been noted that the peak luminosity of a type Ia supernova is tightly correlated with the shape of its light curve. Type Ia supernovae with luminosities the shoot up rapidly and decline rapidly are less luminous than average at their peak; supernovae with luminosities which rise and fall in a more leisurely manner are more luminous than average. Thus, just as the period of a Cepheid tells you its luminosity, the rise and fall time of a type Ia supernova tells you its peak luminosity. Recently, two research teams, the “Supernova Cosmology Project” and the “High-z Supernova Search Team”, have been conducting searches for supernovae in distant galaxies. They have used the observed light curves and redshifts of type Ia supernovae to measure cosmological parameters. First, by observing type Ia supernovae at . ∼ 0.1, the value of H 0 can be determined. The results of the different groups are in reasonable agreement with each other. If the distance to the Virgo cluster is pegged at d L = 15 Mpc, as indicated by the Cepheid results, then the observed supernovae fluxes and redshifts are consistent with H 0 = 70 ± 7 kms −1 Mpc −1 , the value of the Hubble constant which I have adopted in this text. In addition, the supernova groups have been attempting to measure the acceleration (or deceleration) of the universe by observing type Ia supernovae at higher redshift. To present the most recent supernova results to you, I will have to introduce the “magnitude” system used by astronomers to express fluxes and luminosities. The magnitude system, like much else in astronomy, has its roots in ancient Greece. The Greek astronomer Hipparchus, in the second century BC, divided the stars into six classes, according to their ap- parent brightness. The brightest stars were of “first magnitude”, the faintest stars visible to the naked eye were of “sixth magnitude”, and intermediate stars were ranked as second, third, fourth, and fifth magnitude. Long after 7.5. STANDARD CANDLES & ACCELERATION 147 the time of Hipparchus, it was realized that the response of the human eye is roughly logarithmic, and that stars of the first magnitude have fluxes (at visible wavelengths) about 100 times greater than stars of the sixth magni- tude. On the basis of this realization, the magnitude system was placed on a more rigorous mathematical basis. Nowadays, the bolometric apparent magnitude of a light source is defined in terms of the source’s bolometric flux as : ≡ −2.5 log 10 (11 x ) . (7.46) where the reference flux 1 x is set at the value 1 x = 2.53 10 −8 watt m −2 . Thanks to the negative sign in the definition, a small value of : corresponds to a large flux 1. For instance, the flux of sunlight at the Earth’s location is 1 = 1367 watts m −2 ; the Sun thus has a bolometric apparent magnitude of : = −26.8. The choice of reference flux 1 x constitutes a tip of the hat to Hipparchus, since for stars visible to the naked eye it typically yields 0 < : < 6. The bolometric absolute magnitude of a light source is defined as the apparent magnitude that it would have if it were at a luminosity distance of d L = 10 pc. Thus, a light source with luminosity 1 has a bolometric absolute magnitude ` ≡ −2.5 log 10 (11 x ) . (7.47) where the reference luminosity is 1 x = 78.7 L , since that is the luminosity of an object which produces a flux 1 x = 2.53 10 −8 watt m −2 when viewed from a distance of 10 parsecs. The bolometric absolute magnitude of the Sun is thus ` = 4.74. Although the system of apparent and absolute magnitudes seems strange to the uninitiated, the apparent magnitude is really nothing more than a logarithmic measure of the flux, and the absolute magnitude is a logarithmic measure of the luminosity. Given the definitions of apparent and absolute magnitude, the relation between an object’s apparent magnitude and its absolute magnitude can be written in the form ` = :−5 log 10 d L 10 pc . (7.48) where d L is the luminosity distance to the light source. If the luminosity distance is given in units of megaparsecs, this relation becomes ` = :−5 log 10 d L 1 Mpc −25 . (7.49) 148 CHAPTER 7. MEASURING COSMOLOGICAL PARAMETERS Since astronomers frequently quote fluxes and luminosities in terms of ap- parent and absolute magnitudes, they find it convenient to quote luminosity distances in terms of the distance modulus to a light source. The distance modulus is defined as :− `, and is related to the luminosity distance by the relation :−` = 5 log 10 d L 1 Mpc + 25 . (7.50) The distance modulus of the Large Magellanic Cloud, for instance, at d L = 0.050 Mpc, is :−` = 18.5. The distance modulus of the Virgo cluster, at d L = 15 Mpc, is :− ` = 30.9. When . < 1, the luminosity distance to a light source is d L ≈ c H 0 . 1 + 1 −¡ 0 2 . . (7.51) Substituting this relation into equation (7.50), we have an equation which gives the relation between distance modulus and redshift: :−` ≈ 43.17 −5 log 10 H 0 70 kms −1 Mpc −1 + 5 log 10 . + 1.086(1 −¡ 0 ). . (7.52) For a population of standard candles with known luminosity 1 (and hence of known bolometric absolute magnitude `), you measure the flux 1 (or equivalently, the bolometric apparent magnitude :) and the redshift .. In the limit . → 0, a plot of :− ` versus log 10 . gives a straight line whose amplitude at a given value of . tells you the value of H 0 . At slightly larger values of ., the deviation of the plot from a straight line tells you the value of ¡ 0 . At a given value of ., an accelerating universe (with ¡ 0 < 0) yields standard candles with a smaller flux than would a decelerating universe (with ¡ 0 0). The upper panel of Figure 7.5 shows the plot of distance modulus versus redshift for the combined supernova samples of the High-z Supernova Search Team (given by the filled circles) and the Supernova Cosmology Project (given by the open circles). The observational results are compared to the expected results for three model universes. One universe is flat, and contains nothing but matter (Ω m,0 = 1, ¡ 0 = 0.5). The second is negatively curved, and contains nothing but matter (Ω m,0 = 0.3, ¡ 0 = 0.15). The third is flat, and contains both matter and a cosmological constant (Ω m,0 = 0.3, Ω Λ,0 = 0.7, ¡ 0 = −0.55). The data are best fitted by the third of the models – which is, in fact, our Benchmark Model. The bottom panel of 7.5. STANDARD CANDLES & ACCELERATION 149 Figure 7.5: Distance modulus versus redshift for type Ia supernovae from the Supernova Cosmology Project (Perlmutter et al. 1999, ApJ, 517, 565) and the High-z Supernova Search Team (Riess et al. 1998, AJ, 116, 1009). The bottom panel shows the difference between the data and the predictions of a negatively curved Ω m,0 = 0.3 model (from Riess 2000, PASP, 112, 1284). 150 CHAPTER 7. MEASURING COSMOLOGICAL PARAMETERS Figure 7.5 shows this result more clearly. It shows the difference between the data and the predictions of the negatively curved, matter-only model. The conclusion that the universe is accelerating derives from the observation that the supernovae seen at . ∼ 0.5 are, on average, about 0.25 magnitudes fainter than they would in a decelerating universe with Ω m,0 = 0.3 and no cosmological constant. The supernova data extend out to . ∼ 1; this is beyond the range where an expansion in terms of H 0 and ¡ 0 is adequate to describe the scale factor c(t). Thus, the two supernova teams customarily describe their results in terms of a model universe which contains both matter and a cosmological constant. After choosing values of Ω m,0 and Ω Λ,0 , they compute the expected relation between : − ` and ., and compare it to the observed data. The results of fitting these model universes are given in Figure 7.6. The ovals drawn on Figure 7.6 enclose those values of Ω m,0 and Ω Λ,0 which give the best fit to the supernova data. The results of the two teams (the solid ovals and dotted ovals) give very similar results. Three concentric ovals are shown for each team’s result; they correspond to 1σ, 2σ, and 3σ confidence intervals, with the inner oval representing the highest probability. The best fitting models lie along the line 0.8Ω m,0 −0.6Ω Λ,0 ≈ −0.2. Note that decelerating universes (with ¡ 0 0) can be strongly excluded by the data, as can Big Crunch universes (labeled ‘Recollapses’ in Figure 7.6), and Big Bounce universes (labeled ‘No Big Bang’ in Figure 7.6). The supernova data are consistent with negative curvature (labeled ‘Open’ in Figure 7.6), positive curvature (labeled ‘Closed’ in Figure 7.6), or with a universe which is spatially flat. The results of the supernova teams made headlines when they were first announced; the discovery of the accelerating universe was named by Science magazine as the ‘Scientific Breakthrough of the Year’ for 1998. It is prudent to remember, however, that all the hoopla about the accelerating universe is based on the observation that type Ia supernova at . ∼ 0.5 and beyond have somewhat lower fluxes (by about 25%) than they would have in a decelerating universe. There are other reasons why their fluxes might be low. For instance, if type Ia supernovae were intrinsically less luminous at . ∼ 0.5 than at . ∼ 0, that could explain their low fluxes. (If a typical supernova at . ∼ 0.5 had 1 = 310 9 L rather than 410 9 L , that would explain their observed dimness, without the need to invoke a cosmological constant. Conversely, if the typical supernova at . ∼ 0.5 had 1 = 5 10 9 L rather than 4 10 9 L , that would require an even larger cosmological constant to explain their observed dimness.) However, the other properties of type Ia supernovae, such as their 7.5. STANDARD CANDLES & ACCELERATION 151 Figure 7.6: The values of Ω m,0 (horizontal axis) and Ω Λ,0 (vertical axis) which best fit the data shown in Figure 7.5. The solid ovals show the best-fitting values for the High-z Supernova Search Team data; the dotted ovals show the best-fitting values for the Supernova Cosmology Project data (from Riess 2000, PASP, 112, 1284). 152 CHAPTER 7. MEASURING COSMOLOGICAL PARAMETERS spectra, don’t seem to evolve with time, so why should their luminosity? Perhaps the fluxes of supernovae at . ∼ 0.5 are low because some of their light is scattered or absorbed by intervening dust. However, dust tends to scatter some wavelengths of light more than others. This would change the shape of the spectrum of distant type Ia supernovae, but no dependence of spectral shape on redshift is observed. In sum, the supernova results of Figure 7.6 provide persuasive (but, given the caveats, not absolutely compelling) evidence for an accelerating universe. We will see in future chapters how additional observational evidence inter- locks with the supernova results to suggest that we live in a nearly flat accelerating universe with Ω m,0 ≈ 0.3 and Ω Λ,0 ≈ 0.7. Suggested reading [Full references are given in the “Annotated Bibliography” on page 286.] Liddle (1999), ch. 6: The relation among H 0 , ¡ 0 , Ω 0 , and Λ. Narlikar (2002), ch. 9, 10: Local observations (. < 0.1) and more distant observations (. 0.1) of the universe, and what they tell us about cosmological parameters. Peacock (1999), ch. 5: A review of distance measures used in cosmology. Rich (2001), ch. 5.2: A summary of the supernova Ia results. Problems (7.1) Suppose that a polar bear’s foot has a luminosity of 1 = 10 watts. What is the bolometric absolute magnitude of the bear’s foot? What is the bolometric apparent magnitude of the foot at a luminosity distance of d L = 0.5 km? If a bolometer can detect the bear’s foot at a maximum luminosity distance of d L = 0.5 km, what is the maximum luminosity distance at which it could detect the Sun? What is the maximum luminosity distance at which it could detect a supernova with 1 = 4 10 9 L ? 7.5. STANDARD CANDLES & ACCELERATION 153 (7.2) Suppose that a polar bear’s foot has a diameter of / = 0.16 m. What is the angular size δθ of the foot at an angular-diameter distance of d A = 0.5 km? In the Benchmark Model, what is the minimum possible angular size of the polar bear’s foot? (7.3) Suppose that you are in a spatially flat universe containing a single component with a unique equation-of-state parameter u. What are the current proper distance d P (t 0 ), the luminosity distance d L and the angular-diameter distance d A as a function of . and u? At what redshift will d A have a maximum value? What will this maximum value be, in units of the Hubble distance? (7.4) Verify that equation (7.52) is correct in the limit of small .. (You will probably want to use the relation log 10 (1 + r) ≈ 0.4343 ln(1 + r) ≈ 0.4343r in the limit [r[ <1.) (7.5) The surface brightness Σ of an astronomical object is defined as its observed flux divided by its observed angular area; thus, Σ ∝ 1(δθ) 2 . For a class of objects which are both standard candles and standard yardsticks, what is Σ as a function of redshift? Would observing the surface brightness of this class of objects be a useful way of determining the value of the deceleration parameter ¡ 0 ? Why or why not? (7.6) You observe a quasar at a redshift . = 5.0, and determine that the observed flux of light from the quasar varies on a timescale δt 0 = 3 days. If the observed variation in flux is due to a variation in the intrinsic luminosity of the quasar, what was the variation timescale δt e at the time the light was emitted? For the light from the quasar to vary on a timescale δt e , the bulk of the light must come from a region of physical size 1 ≤ 1 max = c(δt e ). What is 1 max for the observed quasar? What is the angular size of 1 max in the Benchmark Model? (7.7) Derive the relation ¹ p (t 0 ) = 4πo κ (:) 2 , as given in equation (7.24), starting from the Robertson-Walker metric of equation (7.22). (7.8) A spatially flat universe contains a single component with equation- of-state parameter u. In this universe, standard candles of luminosity 1 are distributed homogeneously in space. The number density of the standard candles is : 0 at t = t 0 , and the standard candles are 154 CHAPTER 7. MEASURING COSMOLOGICAL PARAMETERS neither created nor destroyed. Show that the observed flux from a single standard candle at redshift . is 1(.) = 1(1 + 3u) 2 16π(cH 0 ) 2 1 (1 +.) 2 1 −(1 +.) −(1+3w)/2 −2 (7.53) when u = −13. What is the corresponding relation when u = −13? Show that the observed intensity (that is, the power per unit area per steradian of sky) from standard candles with redshifts in the range . →. +d. is dJ(.) = : 0 1(cH 0 ) 4π (1 +.) −(7+3w)/2 d. . (7.54) What will be the total intensity J of all standard candles integrated over all redshifts? Explain why the night sky is of finite brightness even in universes with u ≤ −13, which have an infinite horizon distance. Chapter 8 Dark Matter Cosmologists, over the years, have dedicated a large amount of time and effort to determining the matter density of the universe. There are many reasons for this obsession. First, the density parameter in matter, Ω m,0 , is important in determining the spatial curvature and expansion rate of the universe. Even if the cosmological constant is non-zero, the matter content of the universe is not negligible today, and was the dominant component in the fairly recent past. Another reason for wanting to know the matter density of the universe is to find out what the universe is made of. What fraction of the density is made of stars, and other familiar types of baryonic matter? What fraction of the density is made of dark matter? What constitutes the dark matter – cold stellar remnants, black holes, exotic elementary particles, or some other substance too dim for us to see? These questions, and others, have driven astronomers to make a census of the universe, and find out what types of matter it contains, and in what quantities. We have already seen, in the previous chapter, one method of putting limits on Ω m,0 . The apparent magnitude (or flux) of type Ia supernovae as a function of redshift is consistent with a flat universe having Ω m,0 ≈ 0.3 and Ω Λ,0 ≈ 0.7. However, neither Ω m,0 nor Ω Λ,0 is individually well-constrained by the supernova observations. The supernova data are consistent with Ω m,0 = 0 if Ω Λ,0 ≈ 0.4; they are also consistent with Ω m,0 = 1 if Ω Λ,0 ≈ 1.7. In order to determine Ω m,0 more accurately, we will have to adopt alternate methods of estimating the matter content of the universe. 155 156 CHAPTER 8. DARK MATTER 8.1 Visible matter Some types of matter, such as stars, help astronomers to detect them by broadcasting photons in all directions. Stars primarily emit light in the in- frared, visible, and ultraviolet range of the electromagnetic spectrum. Sup- pose, for instance, you install a 1-band filter on your telescope. Such a filter allows only photons in the wavelength range 4.010 −7 m < λ < 4.910 −7 m to pass through. 1 The “1” in 1-band stands for “blue”; however, in addition to admitting blue light, a 1-band filter also lets through violet light. The Sun’s luminosity in the 1 band is L ,B = 4.7 10 25 watts. 2 In the 1 band, the total luminosity density of stars within a few hundred megaparsecs of our Galaxy is , ,B = 1.2 10 8 L ,B Mpc −3 . (8.1) To convert a luminosity density , ,B into a mass density ρ , we need to know the mass-to-light ratio for the stars. That is, we need to know how many kilograms of star, on average, it takes to produce one watt of starlight in the 1 band. If all stars were identical to the Sun, we could simply say that there is one solar mass of stars for each solar luminosity of output power, or '`1 B ` = 1 M L ,B . However, stars are not uniform in their properties. They have a wide range of masses and a wider range of 1-band luminosities. For main sequence stars, powered by hydrogen fusion in their centers, the mass-to-light ratio ranges from `1 B ∼ 10 −3 M L ,B for the brightest, most massive stars (the O stars in the classic OBAFGKM spectral sequence) to `1 B ∼ 10 3 M L ,B for the dimmest, least massive stars (the M stars). Thus, the mass-to-light ratio of the stars in a galaxy will depend on the mix of stars which it contains. As a first guess, let’s suppose that the mix of stars in the solar neighborhood is not abnormal. Within 1 kiloparsec of the Sun, the mass-to-light ratio of the stars works out to be '`1 B ` ≈ 4 M L ,B ≈ 170.000 kg watt −1 . (8.2) Although a mass-to-light ratio of 170 tons per watt doesn’t seem, at first glance, like a very high efficiency, you must remember that the mass of a star includes all the fuel which it will require during its entire lifetime. 1 For comparison, your eyes detect photons in the wavelength range 4 ×10 −7 m < λ < 7 ×10 −7 m. 2 This is only 12% of the Sun’s total luminosity. About 6% of the luminosity is emitted at ultraviolet wavelengths, and the remaining 82% is emitted at wavelengths too long to pass through the B filter. 8.1. VISIBLE MATTER 157 If the mass-to-light ratio of the stars within a kiloparsec of us is not unusually high or low, then the mass density of stars in the universe is ρ ,0 = '`1 B `, ,B ≈ 5 10 8 M Mpc −3 . (8.3) Since the current critical density of the universe is equivalent to a mass density of ρ c,0 = ε c,0 c 2 = 1.410 11 M Mpc −3 , the current density parameter of stars is Ω ,0 = ρ ,0 ρ c,0 ≈ 5 10 8 M Mpc −3 1.4 10 11 M Mpc −3 ≈ 0.004 . (8.4) Stars make up less than 1/2% of the density necessary to flatten the universe. In truth, the number Ω ,0 ≈ 0.004 is not a precisely determined one, largely because of the uncertainty in the number of low-mass, low-luminosity stars in galaxies. In our Galaxy, for instance, ∼ 95% of the stellar luminosity comes from stars more luminous than the Sun, but ∼ 80% of the stellar mass comes from stars less luminous than the Sun. The density parameter in stars will be further increased if you include in the category of “stars” stellar remnants (such as white dwarfs, neutron stars, and black holes) and brown dwarfs. A brown dwarf is a self-gravitating ball of gas which is too low in mass to sustain nuclear fusion in its interior. Because brown dwarfs and isolated cool stellar remnants are difficult to detect, their number density is not well determined. Galaxies also contain baryonic matter which is not in the form of stars, stellar remnants, or brown dwarfs. The interstellar medium contains signif- icant amounts of gas. In our Galaxy and in M31, for instance, the mass of interstellar gas is roughly equal to 10% of the mass of stars. In irregular galaxies such as the Magellanic Clouds, the ratio of gas to stars is even higher. In addition, there is a significant amount of gas between galaxies. Consider a rich cluster of galaxies such as the Coma cluster, located 100 Mpc from our Galaxy, in the direction of the constellation Coma Berenices. At visible wavelengths, as shown in Figure 8.1, most of the light comes from the stars in the cluster’s galaxies. The Coma cluster contains thousands of galaxies; their summed luminosity in the 1 band comes to 1 Coma,B = 8 10 12 L ,B . If the mass-to-light ratio of the stars in the Coma cluster is '`1 B ` ≈ 4 M L ,B , then the total mass of stars in the Coma cluster is ` Coma, ≈ 3 10 13 M . Although 30 trillion solar masses represents a lot of stars, the stellar mass in the Coma cluster is small compared to the mass of the hot, intracluster gas between the galaxies in the cluster. X-ray images, such as the one shown in 158 CHAPTER 8. DARK MATTER Figure 8.1: The Coma cluster as seen in visible light. The image shown is 35 arcminutes across, equivalent to ∼ 1 Mpc at the distance of the Coma cluster. [From the Digitized Sky Survey, produced at the Space Telescope Science Institute] Figure 8.2: The Coma cluster as seen in x-ray light. The scale is the same as that of the previous image. [From the ROSAT x-ray observatory; cour- tesy Max-Planck-Institut f¨ ur extraterrestriche Physik. This figure and the previous figure were produced by Raymond White, using NASA’s SkyView facility.] 8.2. DARK MATTER IN GALAXIES 159 Figure 8.2, reveal that hot, low-density gas, with a typical temperature of 1 ≈ 110 8 K, fills the space between clusters, emitting x-rays with a typical energy of 1 ∼ /1 gas ∼ 9 keV. The total amount of x-ray emitting gas in the Coma cluster is estimated to be ` Coma,gas ≈ 2 10 14 M , roughly six or seven times the mass in stars. As it turns out, the best current limits on the baryon density of the universe come from the predictions of primordial nucleosynthesis. As we will see in Chapter 10, the efficiency with which fusion takes place in the early universe, converting hydrogen into deuterium, helium, lithium, and other elements, depends on the density of protons and neutrons present. Detailed studies of the amounts of deuterium and other elements present in primordial gas clouds indicate that the density parameter of baryonic matter must be Ω bary,0 = 0.04 ±0.01 . (8.5) an order of magnitude larger than the density parameter for stars. When you stare up at the night sky and marvel at the glory of the stars, you are actually marveling at a minority of the baryonic matter in the universe. Most of the baryons are too cold to be readily visible (the infrared emitting brown dwarfs and cold stellar remnants) or too diffuse to be readily visible (the low density x-ray gas in clusters). 8.2 Dark matter in galaxies The situation, in fact, is even more extreme than stated in the previous section. Not only is most of the baryonic matter undetectable by our eyes, most of the matter is not even baryonic. The majority of the matter in the universe is nonbaryonic dark matter, which doesn’t absorb, emit, or scatter light of any wavelength. One way of detecting dark matter is to look for its gravitational influence on visible matter. A classic method of detecting dark matter involves looking at the orbital speeds of stars in spiral galaxies such as our own Galaxy and M31. Spiral galaxies contain flattened disks of stars; within the disk, stars are on nearly circular orbits around the center of the galaxy. The Sun, for instance, is on such an orbit – it is 1 = 8.5 kpc from the Galactic center, and has an orbital speed of · = 220 kms −1 . Suppose that a star is on a circular orbit around the center of its galaxy. If the radius of the orbit is 1 and the orbital speed is ·, then the star 160 CHAPTER 8. DARK MATTER experiences an acceleration c = · 2 1 . (8.6) directed toward the center of the galaxy. If the acceleration is provided by the gravitational attraction of the galaxy, then c = G`(1) 1 2 . (8.7) where `(1) is the mass contained within a sphere of radius 1 centered on the galactic center. 3 The relation between · and ` is found by setting equation (8.6) equal to equation (8.7): · 2 1 = G`(1) 1 2 . (8.8) or · = G`(1) 1 . (8.9) The surface brightness 1 of the disk of a spiral galaxy typically falls off exponentially with distance from the center: 1(1) = 1(0) exp(−11 s ) . (8.10) with the scale length 1 s typically being a few kiloparsecs. For our Galaxy, 1 s ≈ 4 kpc; for M31, a somewhat larger disk galaxy, 1 s ≈ 6 kpc. Once you are a few scale lengths from the center of the spiral galaxy, the mass of stars inside 1 becomes essentially constant. Thus, if stars contributed all, or most, of the mass in a galaxy, the velocity would fall as · ∝ 1 √ 1 at large radii. This relation between orbital speed and orbital radius, · ∝ 1 √ 1, is referred to as “Keplerian rotation”, since it’s what Kepler found for orbits in the Solar System, where the mass is strongly concentrated toward the center. 4 The orbital speed · of stars within a spiral galaxy can be determined from observations. Consider a galaxy which has the shape of a thin circular disk. In general, we won’t be seeing the disk perfectly face-on or edge-on; 3 Equation (8.7) assumes that the mass distribution of the galaxy is spherically sym- metric. This is not, strictly speaking, true (the stars in the disk obviously have a flattened distribution), but the flattening of the galaxy provides only a small correction to the equation for the gravitational acceleration. 4 99.8% of the Solar System’s mass is contained within the Sun. 8.2. DARK MATTER IN GALAXIES 161 Figure 8.3: An observer sees a disk at an inclination angle i. we’ll see it at an inclination i, where i is the angle between our line of sight to the disk and a line perpendicular to the disk (see Figure 8.3). The disk we see in projection will be elliptical, not circular, with an axis ratio /c = cos i . (8.11) For example, the galaxy M31 looks extremely elongated as seen from Earth, with an observed axis ratio /c = 0.22. This indicates that we are seeing M31 fairly close to edge-on, with an inclination i = cos −1 (0.22) = 77 ◦ . By measuring the redshift of the absorption, or emission, lines in light from the disk, we can find the radial velocity · r (1) = c.(1) along the apparent long axis of the galaxy. Since the redshift contains only the component of the stars’ orbital velocity which lies along the line of sight, the radial velocity which we measure will be · r (1) = · gal +·(1) sin i . (8.12) where · gal is the radial velocity of the galaxy as a whole, resulting from the expansion of the universe, and ·(1) is the orbital speed at a distance 1 from the center of the disk. We can thus compute the orbital speed ·(1) in terms of observable properties as ·(1) = · r (1) −· gal sin i = · r (1) −· gal 1 −/ 2 c 2 . (8.13) 162 CHAPTER 8. DARK MATTER Figure 8.4: The orbital speed · as a function of radius in M31. The open circles show the results of Rubin and Ford (1970, ApJ, 159, 379) at visible wavelengths; the solid dots with error bars show the results of Roberts and Whitehurst (1975, ApJ, 201, 327) at radio wavelengths (figure from van den Bergh, 2000). The first astronomer to detect the rotation of M31 was Vesto Slipher, in 1914. However, given the difficulty of measuring the spectra at low surface brightness, the orbital speed · at 1 31 s = 18 kpc was not accurately measured until more than half a century later. In 1970, Vera Rubin and Kent Ford looked at emission lines from regions of hot ionized gas in M31, and were able to find the orbital speed ·(1) out to a radius 1 = 24 kpc = 41 s . Their results, shown as the open circles in Figure 8.4, give no sign of a Keplerian decrease in the orbital speed. Beyond 1 = 41 s , the visible light from M31 was too faint for Rubin and Ford to measure the redshift; as they wrote in their original paper, “extrapolation beyond that distance is a matter of taste.” At 1 41 s , there is still a small amount of atomic hydrogen in the disk of M31, which can be detected by means of its emission line at λ = 21 cm. By measuring the redshift of this emission line, M. Roberts and R. Whitehurst found that the orbital speed stayed at a nearly constant value of ·(1) ≈ 230 kms −1 out to 1 ≈ 30 kpc ≈ 51 s , as shown by the solid dots in Figure 8.4. Since the orbital speed of the stars and gas at large radii (1 31 s ) is greater than it would be if stars and gas were the only matter present, we deduce the presence of a dark halo within which the visible stellar disk is embedded. The mass of the dark halo provides the necessary 8.2. DARK MATTER IN GALAXIES 163 gravitational “anchor” to keep the high-speed stars and gas from being flung out into intergalactic space. M31 is not a freak; most, if not all, spiral galaxies have comparable dark halos. For instance, our own Galaxy has an orbital speed which actually seems to be rising slightly at 1 15 kpc, instead of decreasing in a Keplerian fashion. Thousands of spiral galaxies have had their orbital velocities ·(1) measured; typically, · is roughly constant at 1 1 s . If we approximate the orbital speed · as being constant with radius, the mass of a spiral galaxy, including both the luminous disk and the dark halo, can be found from equation (8.9): `(1) = · 2 1 G = 9.6 10 10 M · 220 kms −1 2 1 8.5 kpc . (8.14) The values of · and 1 in the above equation are scaled to the Sun’s location in our Galaxy. Since our Galaxy’s luminosity in the 1 band is estimated to be 1 Gal,B = 2.3 10 10 L ,B , this means that the mass-to-light ratio of our Galaxy, taken as a whole, is '`1 B ` Gal ≈ 50 M L ,B 1 halo 100 kpc . (8.15) using · = 220 kms −1 in equation (8.14). The quantity 1 halo is the radius of the dark halo surrounding the luminous disk of our galaxy. The exact value of 1 halo is poorly known. At 1 ≈ 20 kpc, where the last detectable gas exists in the disk of our Galaxy, the orbital speed shows no sign of a Keplerian decrease; thus, 1 halo 20 kpc. A rough estimate of the halo size can be made by looking at the velocities of the globular clusters and satellite galaxies (such as the Magellanic Clouds) which orbit our Galaxy. For these hangers-on to remain gravitationally bound to our Galaxy, the halo must extend as far as 1 halo ≈ 75 kpc, implying a total mass for our Galaxy of ` Gal ≈ 8 10 11 M , and a total mass-to-light ratio '`1 B ` Gal ≈ 40 M L ,B . This mass-to-light ratio is ten times greater than that of the stars in our Galaxy, implying that the dark halo is an order of magnitude more massive than the stellar disk. Some astronomers have speculated that the dark halo is actually four times larger in radius, with 1 halo ≈ 300 kpc; this would mean that our halo stretches nearly halfway to M31. With 1 halo ≈ 300 kpc, the mass of our Galaxy would be ` Gal ≈ 3 10 12 M , and the total mass-to-light ratio would be '`1 B ` Gal ≈ 150 M L ,B . 164 CHAPTER 8. DARK MATTER If our Galaxy is typical in having a dark halo 10 to 40 times more massive than its stellar component, then the density parameter of galaxies (including their dark halos) must be Ω gal,0 = (10 →40)Ω ,0 ≈ 0.04 →0.16 . (8.16) Although the total density of galaxies is poorly known, given the uncertainty in the extent of their dark halos, it is likely to be larger than the density of baryons, Ω bary,0 = 0.04 ±0.01. Thus, some part of the dark halos of galaxies is likely to be comprised of nonbaryonic dark matter. 8.3 Dark matter in clusters The first astronomer to make a compelling case for the existence of large quantities of dark matter was Fritz Zwicky, in the 1930’s. In studying the Coma cluster of galaxies (shown in Figure 8.1), he noted that the disper- sion in the radial velocity of the cluster’s galaxies was very large – around 1000 kms −1 . The stars and gas visible within the galaxies simply did not pro- vide enough gravitational attraction to hold the cluster together. In order to keep the galaxies in the Coma cluster from flying off into the surrounding voids, Zwicky concluded, the cluster must contain a large amount of “dunkle Materie”, or (translated into English) “dark matter”. 5 To follow, at a more mathematical level, Zwicky’s reasoning, let us sup- pose that a cluster of galaxies is comprised of ` galaxies, each of which can be approximated as a point mass, with a mass : i (i = 1. 2. . . . . `), a position r i , and a velocity ˙ r i . Clusters of galaxies are gravitationally bound objects, not expanding with the Hubble flow. They are small compared to the hori- zon size; the radius of the Coma cluster is 1 Coma ≈ 3 Mpc ≈ 0.0002d hor . The galaxies within a cluster are moving at non-relativistic speeds; the velocity dispersion within the Coma cluster is σ Coma ≈ 1000 kms −1 ≈ 0.003c. Because of these considerations, we can treat the dynamics of the Coma cluster, and other clusters of galaxies, in a Newtonian manner. The acceleration of the 5 Although Zwicky’s work popularized the phrase “dark matter”, he was not the first to use it in an astronomical context. For instance, in 1908, Henri Poincar´e discussed the possible existence within our Galaxy of “mati`ere obscure” (translated as “dark matter” in the standard edition of Poincar´e’s works). 8.3. DARK MATTER IN CLUSTERS 165 i th galaxy in the cluster, then, is given by the Newtonian formula ¨ r i = G ¸ j=i : j r j −r i [r j −r i [ 3 . (8.17) Note that equation (8.17) assumes that the cluster is an isolated system, with the gravitational acceleration due to matter outside the cluster being negligibly small. The gravitational potential energy of the system of ` galaxies is \ = − G 2 ¸ i,j j=i : i : j [r j −r i [ . (8.18) This is the energy that would be required to pull the ` galaxies away from each other so that they would all be at infinite distance from each other. (The factor of 12 in front of the double summation ensures that each pair of galaxies is only counted once in computing the potential energy.) The potential energy of the cluster can also be written in the form \ = −α G` 2 : h . (8.19) where ` = ¸ : i is the total mass of all the galaxies in the cluster, α is a numerical factor of order unity which depends on the density profile of the cluster. and : h is the half-mass radius of the cluster – that is, the radius of a sphere centered on the cluster’s center of mass and containing a mass `2. For observed clusters of galaxies, it is found that α ≈ 0.4 gives a good fit to the potential energy. The kinetic energy associated with the relative motion of the galaxies in the cluster is 1 = 1 2 ¸ i : i [ ˙ r i [ 2 . (8.20) The kinetic energy 1 can also be written in the form 1 = 1 2 `'· 2 ` . (8.21) where '· 2 ` ≡ 1 ` ¸ i : i [ ˙ r i [ 2 (8.22) 166 CHAPTER 8. DARK MATTER is the mean square velocity (weighted by galaxy mass) of all the galaxies in the cluster. It is also useful to define the moment of inertia of the cluster as 1 ≡ ¸ i : i [r i [ 2 . (8.23) The moment of inertia 1 can be linked to the kinetic energy and the potential energy if we start by taking the second time derivative of 1: ¨ 1 = 2 ¸ i : i (r i ¨ r i + ˙ r i ˙ r i ) . (8.24) Using equation (8.20), we can rewrite this as ¨ 1 = 2 ¸ i : i (r i ¨ r i ) + 41 . (8.25) To introduce the potential energy \ into the above relation, we can use equation (8.17) to write ¸ i : i (r i ¨ r i ) = G ¸ i,j j=i : i : j r i (r j −r i ) [r j −r i [ 3 . (8.26) However, we could equally well switch around the i and , subscripts to find the equally valid equation ¸ j : j (r j ¨ r j ) = G ¸ j,i i=j : j : i r j (r i −r j ) [r i −r j [ 3 . (8.27) Since ¸ i : i (r i ¨ r i ) = ¸ j : j (r j ¨ r j ) (8.28) (it doesn’t matter whether we call the variable over which we’re summing i or , or / or “Fred”), we can combine equations (8.26) and (8.27) to find ¸ i : i (r i ¨ r i ) = 1 2 ¸ i : i (r i ¨ r i ) + ¸ j : j (r j ¨ r j ) ¸ ¸ = − G 2 ¸ i,j j=i : i : j [r j −r i [ = \ . (8.29) 8.3. DARK MATTER IN CLUSTERS 167 Thus, the first term on the right hand side of equation (8.25) is simply 2\, and we may now write down the simple relation ¨ 1 = 2\ + 41 . (8.30) This relation is known as the virial theorem. It was actually first derived in the nineteenth century in the context of the kinetic theory of gases, but as we have seen, it applies perfectly well to a self-gravitating system of point masses. The virial theorem is particularly useful when it is applied to a system in steady state, with a constant moment of inertia. (This implies, among other things, that the system is neither expanding nor contracting, and that you are using a coordinate system in which the center of mass of the cluster is at rest.) If 1 = constant, then the steady-state virial theorem is 0 = \ + 21 . (8.31) or 1 = −\2 . (8.32) That is, for a self-gravitating system in steady state, the kinetic energy 1 is equal to −12 times the potential energy \. Using equation (8.19) and (8.21) in equation (8.32), we find 1 2 `'· 2 ` = α 2 G` 2 : h . (8.33) This means we can use the virial theorem to estimate the mass of a cluster of galaxies, or any other self-gravitating steady-state system: ` = '· 2 `: h αG . (8.34) Note the similarity between equation (8.14), used to estimate the mass of a rotating spiral galaxy, and equation (8.34), used to estimate the mass of a cluster of galaxies. In either case, we estimate the mass of a self-gravitating system by multiplying the square of a characteristic velocity by a character- istic radius, then dividing by the gravitational constant G. Applying the virial theorem to a real cluster of galaxies, such as the Coma cluster, is complicated by the fact that we have only partial information about the cluster, and thus do not know '· 2 ` and : h exactly. For instance, 168 CHAPTER 8. DARK MATTER we can find the line-of-sight velocity of each galaxy from its redshift, but the velocity perpendicular to the line of sight is unknown. From measurements of the redshifts of hundreds of galaxies in the Coma cluster, the mean redshift of the cluster is found to be '.` = 0.0232 . (8.35) which can be translated into a radial velocity '· r ` = c'.` = 6960 kms −1 (8.36) and a distance d Coma = (cH 0 )'.` = 99 Mpc . (8.37) The velocity dispersion of the cluster along the line of sight is found to be σ r = '(· r −'· r `) 2 ` 1/2 = 880 kms −1 . (8.38) If we assume that the velocity dispersion is isotropic, then the three-dimensional mean square velocity '· 2 ` will be equal to three times the one-dimensional mean square velocity σ 2 r , yielding '· 2 ` = 3(880 kms −1 ) 2 = 2.32 10 12 m 2 s −2 . (8.39) Estimating the half-mass radius : h of the Coma cluster is even more peril- ridden than estimating the mean square velocity '· 2 `. After all, we don’t know the distribution of dark matter in the cluster a priori ; in fact, the total amount of dark matter is what we’re trying to find out. However, if we assume that the mass-to-light ratio is constant with radius, then the sphere containing half the mass of the cluster will be the same as the sphere containing half the luminosity of the cluster. If we further assume that the cluster is intrinsically spherical, then the observed distribution of galaxies within the Coma cluster indicates a half-mass radius : h ≈ 1.5 Mpc ≈ 4.6 10 22 m . (8.40) After all these assumptions and approximations, we may estimate the mass of the Coma cluster to be ` Coma = '· 2 `: h αG ≈ (2.32 10 12 m 2 s −2 )(4.6 10 22 m) (0.4)(6.7 10 −11 m 3 s −2 kg −1 ) (8.41) ≈ 4 10 45 kg ≈ 2 10 15 M . (8.42) 8.3. DARK MATTER IN CLUSTERS 169 Thus, less than two percent of the mass of the Coma cluster consists of stars (` Coma, ≈ 3 10 13 M ), and only ten percent consists of hot intracluster gas (` Coma,gas ≈ 2 10 14 M ). Combined with the luminosity of the Coma cluster, 1 Coma,B = 8 10 12 L ,B , the total mass of the Coma cluster implies a mass-to-light ratio of '`1 B ` Coma ≈ 250 M L ,B . (8.43) greater than the mass-to-light ratio of our Galaxy. The presence of a vast reservoir of dark matter in the Coma cluster is confirmed by the fact that the hot, x-ray emitting intracluster gas, shown in Figure 8.2, is still in place; if there were no dark matter to anchor the gas gravitationally, the hot gas would have expanded beyond the cluster on time scales much shorter than the Hubble time. The temperature and density of the hot gas in the Coma cluster can be used to make yet another estimate of the cluster’s mass. If the hot intracluster gas is supported by its own pressure against gravitational infall, it must obey the equation of hydrostatic equilibrium: d1 d: = − G`(:)ρ(:) : 2 . (8.44) where 1 is the pressure of the gas, ρ is the density of the gas, and ` is the total mass inside a sphere of radius :, including gas, stars, dark matter, lost socks, and anything else. 6 Of course, the gas in Coma isn’t perfectly spherical in shape, as equation (8.44) assumes, but it’s close enough to spherical to give a reasonable approximation to the mass. The pressure of the gas is given by the perfect gas law, 1 = ρ/1 j: p . (8.45) where 1 is the temperature of the gas, and j is its mass in units of the proton mass (: p ). The mass of the cluster, as a function of radius, is found by combining equations (8.44) and (8.45): `(:) = /1(:): Gj: p ¸ − d ln ρ d ln : − d ln 1 d ln : ¸ . (8.46) 6 Equation (8.44) is the same equation which determines the internal structure of a star, where the inward force due to gravity is also exactly balanced by an outward force due to a pressure gradient. 170 CHAPTER 8. DARK MATTER The above equation assumes that j is constant with radius, as you’d ex- pect if the chemical composition and ionization state of the gas is uniform throughout the cluster. The x-rays emitted from the hot intracluster gas are a combination of bremsstrahlung emission (caused by the acceleration of free electrons by pro- tons and helium nuclei) and line emission from highly ionized iron and other heavy elements. Starting from an x-ray spectrum, it is possible to fit models to the emission and thus compute the temperature, density, and chemical composition of the gas. In the Coma cluster, for instance, temperature maps reveal relatively cool regions (at /1 ≈ 5 keV) as well as hotter regions (at /1 ≈ 12 keV), averaging to /1 ≈ 9 keV over the entire cluster. The mass of the Coma cluster, assuming hydrostatic equilibrium, is computed to be (3 →4)10 14 M within 0.7 Mpc of the cluster center and (1 →2)10 15 M within 3.6 Mpc of the center, consistent with the mass estimate of the virial theorem. Other clusters of galaxies besides the Coma cluster have had their masses estimated, using the virial theorem applied to their galaxies or the equation of hydrostatic equilibrium applied to their gas. Typical mass-to-light ratios for clusters lie in the range '`1 B ` = 200 → 300 M L ,B , so the Coma cluster is not unusual in the amount of dark matter which it contains. If the masses of all the clusters of galaxies are added together, it is found that their density parameter is Ω clus,0 ≈ 0.2 . (8.47) This provides a lower limit to the matter density of the universe, since any matter which is smoothly distributed in the intercluster voids will not be included in this number. 8.4 Gravitational lensing So far, I have outlined the classical methods for detecting dark matter via its gravitational effects on luminous matter. 7 We can detect dark matter around spiral galaxies because it affects the motions of stars and interstellar gas. We can detect dark matter in clusters of galaxies because it affects the motions of 7 The roots of these methods can be traced back as far as the year 1846, when Leverrier and Adams deduced the existence of the dim planet Neptune by its effect on the orbit of Uranus. 8.4. GRAVITATIONAL LENSING 171 Figure 8.5: Deflection of light by a massive compact object. galaxies and intracluster gas. However, as Einstein realized, dark matter will affect not only the trajectory of matter, but also the trajectory of photons. Thus, dark matter can bend and focus light, acting as a gravitational lens. The effects of dark matter on photons have been used to search for dark matter within the halo of our own Galaxy, as well as in distant clusters of galaxies. To see how gravitational lensing can be used to detect dark matter, start by considering the dark halo surrounding our Galaxy. Some of the dark matter in the halo might consist of massive compact objects such as brown dwarfs, white dwarfs, neutron stars, and black holes. These objects have been collectively called MACHOs, a slightly strained acronym for MAssive Compact Halo Objects. If a photon passes such a compact massive object at an impact parameter /, as shown in Figure 8.5, the local curvature of space-time will cause the photon to be deflected by an angle α = 4G` c 2 / . (8.48) where ` is the mass of the compact object. For instance, light from a distant star which just grazes the Sun’s surface should be deflected through an angle α = 4GM c 2 R = 1.7 arcsec . (8.49) In 1919, after Einstein predicted a deflection of this magnitude, an eclipse expedition photographed stars in the vicinity of the Sun. Comparison of the eclipse photographs with photographs of the same star field taken six months earlier revealed that the apparent positions of the stars were deflected by the amount which Einstein had predicted. This result brought fame to Einstein and experimental support to the theory of general relativity. Since a star, or a brown dwarf, or a stellar remnant, can deflect light, it can act as a lens. Suppose a MACHO in the halo of our Galaxy passes 172 CHAPTER 8. DARK MATTER Figure 8.6: Light from star in the Large Magellanic Cloud is deflected by a MACHO on its way to an observer in the disk of our Galaxy (seen edge-on in this figure). directly between an observer in our Galaxy and a star in the Large Magellanic Cloud. Figure 8.6 shows such a situation, with a MACHO which happens to be halfway between the observer and the star. As the MACHO deflects the light from the distant star, it produces an image of the star which is both distorted and amplified. If the MACHO is exactly along the line of sight between the observer and the lensed star, the image produced is a perfect ring, with angular radius θ E = 4G` c 2 d 1 −r r 1/2 . (8.50) where ` is the mass of the lensing MACHO, d is the distance from the observer to the lensed star, and rd (where 0 < r < 1) is the distance from the observer to the lensing MACHO. The angle θ E is known as the Einstein radius. If r ≈ 0.5 (that is, if the MACHO is roughly halfway between the observer and the lensed star), then θ E ≈ 4 10 −4 arcsec ` 1 M 1/2 d 50 kpc −1/2 . (8.51) If the MACHO does not lie perfectly along the line of sight to the star, then the image of the star will be distorted into two or more arcs instead of a single unbroken ring. Although the Einstein radius for an LMC star being lensed by a MACHO is too small to be resolved, it is possible, in some cases, to 8.4. GRAVITATIONAL LENSING 173 detect the amplification of the flux from the star. For the amplification to be significant, the angular distance between the MACHO and the lensed star, as seen from Earth, must be comparable to, or smaller than, the Einstein radius. Given the small size of the Einstein radius, the probability of any particular star in the LMC being lensed at any moment is tiny. It has been calculated that if the dark halo of our galaxy were entirely composed of MACHOs, then the probability of any given star in the LMC being lensed at any given time would still only be 1 ∼ 5 10 −7 . To detect lensing by MACHOs, various research groups took up the daunting task of monitoring millions of stars in the Large Magellanic Cloud to watch for changes in their flux. Since the MACHOs in our dark halo and the stars in the LMC are in constant relative motion, the typical signature of a “lensing event” is a star which becomes brighter as the angular distance between star and MACHO decreases, then becomes dimmer as the angular distance increases again. The typical time scale for a lensing event is the time it takes a MACHO to travel through an angular distance equal to θ E as seen from Earth; for a MACHO halfway between here and the LMC, this is ∆t = d θ E 2· ≈ 90 days ` 1 M 1/2 · 200 kms −1 −1 . (8.52) where · is the relative transverse velocity of the MACHO and the lensed star as seen by the observer on Earth. Generally speaking, more massive MACHOs produce larger Einstein rings and thus will amplify the lensed star for a longer time. The research groups which searched for MACHOs found a scarcity of short duration lensing events, suggesting that there is not a significant population of brown dwarfs (with ` < 0.08 M ) in the dark halo of our Galaxy. The total number of lensing events which they detected suggest that as much as 20% of the halo mass could be in the form of MACHOs. The long time scales of the observed lensing events, which have ∆t 35 days, suggest typical MACHO masses of ` 0.15 M . (Perhaps the MACHOs are old, cold white dwarfs, which would have the correct mass.) Alternatively, the observed lensing events could be due, at least in part, to lensing objects within the LMC itself. In any case, the search for MACHOs suggests that most of the matter in the dark halo of our galaxy is due to a smoothly distributed component, instead of being congealed into MACHOs of roughly stellar mass. Gravitational lensing occurs at all mass scales. Suppose, for instance, that a cluster of galaxies, with ` ∼ 10 14 M , at a distance ∼ 500 Mpc from 174 CHAPTER 8. DARK MATTER Figure 8.7: A Hubble Space Telescope picture of the rich cluster Abell 2218, displaying gravitationally lensed arcs. The region shown is roughly 2.4 arcmin by 1.2 arcmin, equivalent to 0.54 Mpc by 0.27 Mpc at the distance of Abell 2218 (courtesy of W. Couch [University of New South Wales] and NASA). our Galaxy, lenses a background galaxy at d ∼ 1000 Mpc. The Einstein radius for this configuration will be θ E ≈ 0.5 arcmin ` 10 14 M 1/2 d 1000 Mpc −1/2 . (8.53) The arc-shaped images into which the background galaxy is distorted by the lensing cluster can thus be resolved. For instance, Figure 8.7 shows an image of the cluster Abell 2218, which has a redshift . = 0.18, and hence is at a proper distance d = 770 Mpc. The elongated arcs seen in Figure 8.7 are not oddly shaped galaxies within the cluster; instead, they are background galaxies, at redshifts . 0.18, which are gravitationally lensed by the cluster mass. The mass of clusters can be estimated by the degree to which they lens background galaxies. The masses calculated in this way are in general agreement with the masses found by applying the virial theorem to the motions of galaxies in the cluster or by applying the equation of hydrostatic equilibrium to the hot intracluster gas. 8.5. WHAT’S THE MATTER? 175 8.5 What’s the matter? I have described how to detect dark matter by its gravitational effects, but I’ve been dodging the essential question: “What is it?” Adding together the masses of clusters of galaxies gives a lower limit on the matter density of the universe, telling us that Ω m,0 ≥ 0.2. However, the density parameter of baryonic matter is only Ω bary,0 ≈ 0.04. Thus, the density of nonbaryonic matter is at least four times the density of the familiar baryonic matter of which people and planets and stars are made. As you might expect, conjecture about the nature of the nonbaryonic dark matter has run rampant (some might even say it has run amok). A component of the universe which is totally invisible is an open invitation to speculation. To give a taste of the variety of speculation, some scientists have proposed that the dark matter might be made of axions, a type of elementary particle with a rest energy of : ax c 2 ∼ 10 −5 eV, equivalent to : ax ∼ 210 −41 kg. This is a rather low mass – it would take some 50 billion axions (if they indeed exist) to equal the mass of one electron. On the other hand, some scientists have conjectured that the dark matter might be made of primordial black holes, with masses up to : BH ∼ 10 5 M , equivalent to : BH ∼ 2 10 35 kg. 8 This is a rather high mass – it would take some 30 billion Earths to equal the mass of one primordial black hole (if they indeed exist). It is a sign of the vast ignorance concerning nonbaryonic dark matter that these two candidates for the role of dark matter differ in mass by 76 orders of magnitude. One nonbaryonic particle which we know exists, and which seems to have a non-zero mass, is the neutrino. As stated in section 5.1, there should exist today a cosmic background of neutrinos. Just as the Cosmic Microwave Background is a relic of the time when the universe was opaque to photons, the Cosmic Neutrino Background is a relic of the time when the universe was hot and dense enough to be opaque to neutrinos. The number density of each of the three flavors of neutrinos (ν e , ν µ , and ν τ ) has been calculated to be 3/11 times the number density of CMB photons, yielding a total number density of neutrinos : ν = 3(311): γ = (911)(4.11 10 8 m −3 ) = 3.36 10 8 m −3 . (8.54) 8 A primordial black hole is one which forms very early in the history of the universe, rather than by the collapse of a massive star later on. 176 CHAPTER 8. DARK MATTER This means that at any moment, about twenty million cosmic neutrinos are zipping through your body, “like photons through a pane of glass”. In order to provide all the nonbaryonic mass in the universe, the average neutrino mass would have to be : ν c 2 = Ω dm,0 ε c,0 : ν . (8.55) Given a density parameter in nonbaryonic dark matter of Ω dm,0 ≈ 0.26, this implies that a mean neutrino mass of : ν c 2 ≈ 0.26(5200 MeVm −3 ) 3.36 10 8 m −3 ≈ 4 eV (8.56) would be necessary to provide all the nonbaryonic dark matter in the uni- verse. Evidence indicates that neutrinos do have some mass. But how much? Enough to contribute significantly to the energy density of the universe? The observations of neutrinos from the Sun, as mentioned in section 2.4, indicate that electron neutrinos oscillate into some other flavor of neutrino, with the difference in the squares of the masses of the two neutrinos being ∆(: 2 ν c 4 ) ≈ 3 10 −5 eV 2 . Observations of muon neutrinos created in the Earth’s atmosphere indicate that muon neutrinos oscillate into tau neutrinos, with ∆(: 2 ν c 4 ) ≈ 3 10 −3 eV 2 . The minimum neutrino masses consistent with these results would have one flavor with : ν c 2 ∼ 0.05 eV, another with : ν c 2 ∼ 0.005 eV, and the third with : ν c 2 <0.005 eV. If the neutrino masses are this small, then the density parameter in neutrinos is only Ω ν ∼ 10 −3 , and neutrinos make up less than 0.5% of the nonbaryonic dark matter. If, on the other hand, neutrinos make up all the nonbaryonic dark matter, the masses of the three species would have to be very nearly identical; for instance, one neutrino flavor with : ν c 2 = 4.0 eV, another with : ν c 2 = 4.0004 eV, and the third with : ν c 2 = 4.000004 eV would be in agreement with the deduced values of ∆(: 2 ν c 4 ). If the masses of all three neutrinos turn out to be significantly less than : ν c 2 ∼ 4 eV, then the bulk of the nonbaryonic dark matter in the universe must be made of some particle other than neutrinos. Particle physicists have provided several possible candidates for the role of dark matter. For instance, consider the extension of the Standard Model of particle physics known as supersymmetry. Various supersymmetric models predict the existence of massive nonbaryonic particles such as photinos, gravitinos, axinos, sneutri- nos, gluinos, and so forth. None of these “inos” have yet been detected in 8.5. WHAT’S THE MATTER? 177 laboratories. The fact that supersymmetric particles such as photinos have not yet been seen in particle accelerator experiments means that they must be massive (if they exist), with :c 2 10 GeV. Like neutrinos, the hypothetical supersymmetric particles interact with other particles only through gravity and through the weak nuclear force, which makes them intrinsically difficult to detect. Particles which interact via the weak nuclear force, but which are much more massive than the upper limit on the neutrino mass, are known generically as Weakly Interacting Massive Particles, or WIMPs. 9 Since WIMPs, like neutrinos, do interact with atomic nuclei on occasion, experimenters have set up WIMP detectors to discover cosmic WIMPs. So far, no convincing detections have been made – but the search goes on. Suggested reading [Full references are given in the “Annotated Bibliography” on page 286.] Liddle (1999), ch. 8: A brief sketch of methods for detecting dark matter. Peacock (1999), ch. 12: Dark matter in the universe, both baryonic and nonbaryonic. Also, chapter 4 gives a good review of gravitational lens- ing. Rich (2001), ch. 2.4: A discussion of the dark matter candidates. Problems (8.1) Suppose it were suggested that black holes of mass 10 −8 M made up all the dark matter in the halo of our Galaxy. How far away would you expect the nearest such black hole to be? How frequently would you expect such a black hole to pass within 1 AU of the Sun? (An order-of-magnitude estimate is sufficient.) Suppose it were suggested that MACHOs of mass 10 −3 M (about the mass of Jupiter) made up all the dark matter in the halo of our Galaxy. How far away would you expect the nearest MACHO to be? How 9 The acronym “MACHO”, encountered in the previous section, was first coined as a humorous riposte to the acronym “WIMP”. 178 CHAPTER 8. DARK MATTER frequently would such a MACHO pass within 1 AU of the Sun? (Again, an order-of-magnitude estimate will suffice.) (8.2) The Draco galaxy is a dwarf galaxy within the Local Group. Its lu- minosity is 1 = (1.8 ± 0.8) 10 5 L and half its total luminosity is contained within a sphere of radius : h = 120 ± 12 pc. The red giant stars in the Draco galaxy are bright enough to have their line-of-sight velocities measured. The measured velocity dispersion of the red giant stars in the Draco galaxy is σ r = 10.5 ±2.2 kms −1 . What is the mass of the Draco galaxy? What is its mass-to-light ratio? Describe the possible sources of error in your mass estimate of this galaxy. (8.3) A light ray just grazes the surface of the Earth (` = 6.0 10 24 kg, 1 = 6.4 10 6 m). Through what angle α is the light ray bent by grav- itational lensing? (Ignore the refractive effects of the Earth’s atmo- sphere.) Repeat your calculation for a white dwarf (` = 2.0 10 30 kg, 1 = 1.5 10 7 m) and for a neutron star (` = 3.0 10 30 kg, 1 = 1.2 10 4 m). (8.4) If the halo of our Galaxy is spherically symmetric, what is the mass density ρ(:) within the halo? If the universe contains a cosmological constant with density parameter Ω Λ,0 = 0.7, would you expect it to significantly affect the dynamics of our Galaxy’s halo? Explain why or why not. (8.5) In the previous chapter, I noted that galaxies in rich clusters are poor standard candles, because they tend to grow brighter with time as they merge with other galaxies. Let’s estimate the galaxy merger rate in the Coma cluster to see whether it’s truly significant. The Coma cluster contains ` ≈ 1000 galaxies within its half-mass radius of : h ≈ 1.5 Mpc. What is the mean number density of galaxies within the half-mass radius? Suppose that the typical cross-section of a galaxy is Σ ≈ 10 −3 Mpc 2 . How far will a galaxy in the Coma cluster travel, on average, before it collides with another galaxy? The velocity dispersion of the Coma cluster is σ ≈ 880 kms −3 . What is the average time between collisions for a galaxy in the Coma cluster? Is this time greater than or less than the Hubble time? Chapter 9 The Cosmic Microwave Background If Heinrich Olbers had lived in intergalactic space and had eyes that operated at millimeter wavelengths (admittedly a very large “if”), he would not have formulated Olbers’ Paradox. At wavelengths of a few millimeters, thousands of times longer than human eyes can detect, most of the light in the universe comes not from the hot balls of gas we call “stars”, but from the Cosmic Microwave Background (CMB). The night sky, unknown to Olbers, actually is uniformly bright – it’s just uniformly bright at a temperature of 1 0 = 2.725 K rather than at a temperature of a few thousand degrees Kelvin. The current energy density of the Cosmic Microwave Background, ε γ,0 = α1 4 0 = 0.261 MeVm −3 . (9.1) is only 5 10 −5 times the current critical density. However, since the energy per CMB photon is small (/1 mean = 6.34 10 −4 eV), the number density of CMB photons in the universe is large: : γ,0 = 4.11 10 8 m −3 . (9.2) It is particularly enlightening to compare the energy density and number density of photons to those of baryons (that is, protons and neutrons). Given a current density parameter for baryons of Ω bary,0 ≈ 0.04, the current energy density of baryons is ε bary,0 = Ω bary,0 ε c,0 ≈ 0.04(5200 MeVm −3 ) ≈ 210 MeVm −3 . (9.3) 179 180 CHAPTER 9. THE COSMIC MICROWAVE BACKGROUND Thus, the energy density in baryons today is about 800 times the energy density in CMB photons. Note, though, that the rest energy of a proton or neutron is 1 bary ≈ 939 MeV; this is more than a trillion times the mean energy of a CMB photon. The number density of baryons, therefore, is much lower than the number density of photons: : bary,0 = ε bary,0 1 bary ≈ 210 MeVm −3 939 MeV ≈ 0.22 m −3 . (9.4) The ratio of baryons to photons in the universe (a number usually designated by the Greek letter η) is, from equations (9.2) and (9.4), η = : bary,0 : γ,0 ≈ 0.22 m −3 4.11 10 8 m −3 ≈ 5 10 −10 . (9.5) Baryons are badly outnumbered by photons in the universe as a whole, by a ratio of roughly two billion to one. 9.1 Observing the CMB Although CMB photons are as common as dirt, 1 Arno Penzias and Robert Wilson were surprised when they serendipitously discovered the Cosmic Mi- crowave Background. At the time of their discovery, Penzias and Wilson were radio astronomers working at Bell Laboratories. The horn-reflector radio an- tenna which they used had previously been utilized to receive microwave signals, of wavelength λ = 7.35 cm, reflected from an orbiting communica- tions satellite. Turning from telecommunications to astronomy, Penzias and Wilson found a slightly stronger signal than they expected when they turned the antenna toward the sky. They did everything they could think of to re- duce “noise” in their system. They even shooed away a pair of pigeons that had roosted in the antenna and cleaned up they later called “the usual white dielectric” generated by pigeons. The excess signal remained. It was isotropic and constant with time, so it couldn’t be associated with an isolated celestial source. Wilson and Penzias were puzzled until they were put in touch with Robert Dicke and his research group at Princeton University. Dicke had deduced that the universe, if it 1 Actually, much commoner than dirt, when you stop to think of it, since dirt is made of baryons. 9.1. OBSERVING THE CMB 181 started in a hot dense state, should now be filled with microwave radiation. 2 In fact, Dicke and his group were in the process of building a microwave antenna when Penzias and Wilson told them that they had already detected the predicted microwave radiation. Penzias and Wilson wrote a paper for The Astrophysical Journal in which they wrote, “Measurements of the effective zenith noise temperature of the 20-foot horn-reflector antenna . . . at 4080 Mc/s have yielded a value about 3.5 K higher than expected. This excess temperature is, within the limits of our observations, isotropic, unpolarized, and free from seasonal variations (July, 1964 – April, 1965). A possible explanation for the observed excess noise temperature is the one given by Dicke, Peebles, Roll, and Wilkinson in a companion letter in this issue.” The companion paper by Dicke and his collaborators points out that the radiation could be a relic of an early, hot, dense, and opaque state of the universe. Measuring the spectrum of the CMB, and confirming that it is indeed a blackbody, is not a simple task, even with modern technology. The current energy per CMB photon, ∼ 6 10 −4 eV, is tiny compared to the energy required to break up an atomic nucleus (∼ 1 MeV) or even the energy required to ionize an atom (∼ 10 eV). However, the mean photon energy is comparable to the energy of vibration or rotation for a small molecule such as H 2 O. Thus, CMB photons can zip along for more than 13 billion years through the tenuous intergalactic medium, then be absorbed a microsecond away from the Earth’s surface by a water molecule in the atmosphere. Microwaves with wavelengths shorter than λ ∼ 3 cm are strongly absorbed by water molecules. Penzias and Wilson observed the CMB at a wavelength λ = 7.35 cm because that was the wavelength of the signals that Bell Labs had been bouncing off orbiting satellites. Thus, Penzias and Wilson were observing at a wavelength 40 times longer than the wavelength (λ ≈ 2 mm) at which the CMB spectrum reaches its peak. The CMB can be measured at wavelengths shorter than 3 cm by observ- ing from high-altitude balloons or from the South Pole, where the combina- tion of cold temperatures and high altitude 3 keeps the atmospheric humidity low. The best way to measure the spectrum of the CMB, however, is to go completely above the damp atmosphere of the Earth. The CMB spectrum 2 To give credit where it’s due, the existence of the cosmic background radiation had actually been predicted by the physicist George Gamow and his colleagues as early as 1948; unfortunately, Gamow’s prediction wasn’t acted on at the time he made it, and had fallen into obscurity during the intervening years. 3 The South Pole is nearly 3 kilometers above sea level. 182 CHAPTER 9. THE COSMIC MICROWAVE BACKGROUND Figure 9.1: The spectrum of the Cosmic Microwave Background, as measured by the FIRAS instrument on the COBE satellite. The uncertainties in the measurement are smaller than the thickness of the line (from Fixsen et al. 1996, ApJ, 473, 576). was first measured accurately over a wide range of wavelengths by the COs- mic Background Explorer (COBE) satellite, launched in 1989, into an orbit 900 km above the Earth’s surface. COBE actually contained three different instruments. The Diffuse InfraRed Background Experiment (DIRBE) was designed to measure radiation at the wavelengths 0.001 mm < λ < 0.24 mm; at these wavelengths, it was primarily detecting stars and dust within our own Galaxy. The second instrument, called the Far InfraRed Absolute Spec- trophotometer (FIRAS), was used to measure the spectrum of the CMB in the range 0.1mm < λ < 10 mm, a wavelength band which includes the peak in the CMB spectrum. The third instrument, called the Differential Microwave Radiometer (DMR), was designed to make full-sky maps of the CMB at three different wavelengths: λ = 3.3 mm, 5.7 mm, and 9.6 mm. Three important results came from the analysis of the COBE data. Result number one: At any angular position (θ. φ) on the sky, the spec- trum of the Cosmic Microwave Background is very close to that of an ideal blackbody, as illustrated in Figure 9.1. How close is very close? FIRAS could have detected fluctuations in the spectrum as small as ∆cc ≈ 10 −4 . No de- viations were found at this level within the wavelength range investigated by 9.1. OBSERVING THE CMB 183 FIRAS. Result number two: The CMB has the dipole distortion in temperature shown in the top panel of Figure 9.2. 4 That is, although each point on the sky has a blackbody spectrum, in one half of the sky the spectrum is slightly blueshifted to higher temperatures, and in the other half the spectrum is slightly redshifted to lower temperatures. 5 This dipole distortion is a simple Doppler shift, caused by the net motion of the COBE satellite relative to a frame of reference in which the CMB is isotropic. After correcting for the orbital motion of COBE around the Earth (· ∼ 8 kms −1 ), for the orbital motion of the Earth around the Sun (· ∼ 30 kms −1 ), for the orbital motion of the Sun around the Galactic center (· ∼ 220 kms −1 ), and for the orbital motion of our Galaxy relative to the center of mass of the Local Group (· ∼ 80 kms −1 ), it is found that the Local Group is moving in the general direction of the constellation Hydra, with a speed · LG = 630 ± 20 kms −1 = 0.0021c. This peculiar velocity for the Local Group is what you’d expect as the result of gravitational acceleration by the largest lumps of matter in the vicinity of the Local Group. The Local Group is being accelerated toward the Virgo cluster, the nearest big cluster to us. In addition, the Virgo cluster is being accelerated toward the Hydra-Centaurus supercluster, the nearest supercluster to us. The combination of these two accelerations, working over the age of the universe, has launched the Local Group in the direction of Hydra, at 0.2% of the speed of light. Result number three: After the dipole distortion of the CMB is subtracted away, the remaining temperature fluctuations, shown in the lower panel of Figure 9.2, are small in amplitude. Let the temperature of the CMB, at a given point on the sky, be 1(θ. φ). The mean temperature, averaging over all locations, is '1` = 1 4π 1(θ. φ) sin θ dθ dφ = 2.725 K . (9.6) The dimensionless temperature fluctuation at a given point on the sky is δ1 1 (θ. φ) ≡ 1(θ. φ) −'1` '1` . (9.7) 4 The dipole distortion of the CMB was first detected in 1977, using aircraft-borne and balloon-borne detectors. The unique contribution of COBE was the precision with which it measured the temperature distortion. 5 The distorted “yin-yang” pattern in the upper panel of Figure 9.2 represents the darker, cooler (yin?) hemisphere of the sky and the hotter, brighter (yang?) hemisphere, distorted by the map projection. 184 CHAPTER 9. THE COSMIC MICROWAVE BACKGROUND Figure 9.2: Top panel: The fluctuations in temperature in the CMB, as mea- sured by COBE. Bottom panel: The fluctuations in temperature remaining after subtraction of the dipole due to the satellite’s proper motion. The band across the middle is due to emission from the disk of our own Galaxy. (Cour- tesy of NASA Goddard Space Flight Center and the COBE Science Working Group.) 9.2. RECOMBINATION AND DECOUPLING 185 From the maps of the sky made by the DMR instrument aboard COBE, it was found that after subtraction of the Doppler dipole, the root mean square temperature fluctuation was δ1 1 2 ¸1/2 = 1.1 10 −5 . (9.8) (This analysis excludes the regions of the sky contaminated by foreground emission from our own Galaxy.) The fact that the temperature of the CMB varies by only 30 microKelvin across the sky represents a remarkably close approach to isotropy. 6 The observations that the CMB has a nearly perfect blackbody spectrum and that it is nearly isotropic (once the Doppler dipole is removed) provide strong support for the Hot Big Bang model of the universe. A background of nearly isotropic blackbody radiation is natural if the universe was once hot, dense, opaque, and nearly homogeneous, as it was in the Hot Big Bang sce- nario. If the universe did not go through such a phase, then any explanation of the Cosmic Microwave Background will have to be much more contrived. 9.2 Recombination and decoupling To understand in more detail the origin of the Cosmic Microwave Back- ground, we’ll have to examine fairly carefully the process by which the bary- onic matter goes from being an ionized plasma to a gas of neutral atoms, and the closely related process by which the universe goes from being opaque to being transparent. To avoid muddle, I will distinguish among three closely related (but not identical) moments in the history of the universe. First, the epoch of recombination is the time at which the baryonic component of the universe goes from being ionized to being neutral. Numerically, you might define it as the instant in time at which the number density of ions is equal to the number density of neutral atoms. 7 Second, the epoch of photon decoupling is the time at which the rate at which photons scatter from elec- trons becomes smaller than the Hubble parameter (which tells us the rate at 6 To make an analogy, if the surface of the Earth were smooth to 11 parts per million, the highest mountains would be just seventy meters above the deepest ocean trenches. 7 Cosmologists sometimes grumble that this should really be called the epoch of “com- bination” rather than the epoch of “recombination”, since this is the very first time when electrons and ions combined to form neutral atoms. 186 CHAPTER 9. THE COSMIC MICROWAVE BACKGROUND Figure 9.3: An observer is surrounded by a spherical last scattering surface. The photons of the CMB travel straight to us from the last scattering surface, being continuously redshifted. which the universe expands). When photons decouple, they cease to interact with the electrons, and the universe becomes transparent. Third, the epoch of last scattering is the time at which a typical CMB photon underwent its last scattering from an electron. Surrounding every observer in the universe is a last scattering surface, illustrated in Figure 9.3, from which the CMB photons have been streaming freely, with no further scattering by electrons. The probability that a photon will scatter from an electron is small once the expansion rate of the universe is faster than the scattering rate; thus, the epoch of last scattering is very close to the epoch of photon decoupling. To keep things from getting too complicated, I will assume that the bary- onic component of the universe consisted entirely of hydrogen at the epoch of recombination. This is not, I concede, a strictly accurate assumption. Even at the time of recombination, before stars had a chance to pollute the universe with heavy elements, there was a significant amount of helium present. 8 However, the presence of helium is merely a complicating factor. All the significant physics of recombination can be studied in a simplified 8 In the next chapter, we will examine how and why this helium was formed in the early universe. 9.2. RECOMBINATION AND DECOUPLING 187 universe containing no elements other than hydrogen. The hydrogen can take the form of a neutral atom (designated by the letter H), or of a naked hydrogen nucleus, otherwise known as a proton (designated by the letter j). To maintain charge neutrality in this hydrogen-only universe, the number density of free electrons must be equal to that of free protons: : e = : p . The degree to which the baryonic content of the universe is ionized can be expressed as the fractional ionization A, defined as A ≡ : p : p +: H = : p : bary = : e : bary . (9.9) The value of A ranges from A = 1 when the baryonic content is fully ionized to A = 0 when it consists entirely of neutral atoms. One useful consequence of assuming that hydrogen is the only element is that there is now a single relevant energy scale in the problem: the ionization energy of hydrogen, C = 13.6 eV. A photon with an energy /1 C is capable of photoionizing a hydrogen atom: H +γ →j +c − . (9.10) This reaction can run in the opposite direction, as well; a proton and an electron can undergo radiative recombination, forming a bound hydrogen atom while a photon carries away the excess energy: j +c − →H +γ . (9.11) In a universe containing protons, electrons, and photons, the fractional ion- ization A will depend on the balance between photoionization and radiative recombination. Let’s travel back in time to a period before the epoch of recombination. For concreteness, let’s choose the moment when c = 10 −5 , corresponding to a redshift . = 10 5 . (In the Benchmark Model, this scale factor was reached when the universe was seventy years old.) The temperature of the background radiation at this time was was 1 ≈ 3 10 5 K, and the average photon energy was /1 mean ≈ 2.7/1 ≈ 60 eV. With such a high energy per photon, and with a ratio of photons to baryons of nearly two billion, any hy- drogen atoms that happened to form by radiative recombination were very short-lived; almost immediately, they were blasted apart into their compo- nent electron and proton by a high-energy photon. At early times, then, the fractional ionization of the universe was very close to A = 1. 188 CHAPTER 9. THE COSMIC MICROWAVE BACKGROUND When the universe was fully ionized, photons interacted primarily with electrons, and the main interaction mechanism was Thomson scattering: γ +c − →γ +c − . (9.12) The scattering interaction is accompanied by a transfer of energy and mo- mentum between the photon and electron. The cross-section for Thomson scattering is the Thomson cross-section of σ e = 6.65 10 −29 m 2 . The mean free path of a photon – that is, the mean distance it travels before scattering from an electron – is λ = 1 : e σ e . (9.13) Since photons travel with a speed c, the rate at which a photon undergoes scattering interactions is Γ = c λ = : e σ e c . (9.14) When the baryonic component of the universe is fully ionized, : e = : p = : bary . Currently, the number density of baryons is : bary,0 = 0.22 m −3 . The number density of conserved particles, such as baryons, goes as 1c 3 , so when the early universe was fully ionized, the free electron density was : e = : bary = : bary,0 c 3 . (9.15) and the scattering rate for photons was Γ = : bary,0 σ e c c 3 = 4.4 10 −21 s −1 c 3 . (9.16) This means, for instance, that at c = 10 −5 , photons would scatter from electrons at a rate Γ = 4.4 10 −6 s −1 , about three times a week. The photons remain coupled to the electrons as long as their scattering rate, Γ, is larger than H, the rate at which the universe expands; this is equivalent to saying that their mean free path λ is shorter than the Hub- ble distance cH. As long as photons scatter frequently from electrons, the photons remain in thermal equilibrium with the electrons (and, indirectly, with the protons as well, thanks to the electrons’ interactions with the pro- tons). The photons, electrons, and protons, as long as they remain in thermal equilibrium, all have the same temperature 1. When the photon scattering rate Γ drops below H, then the electrons are being diluted by expansion 9.3. THE PHYSICS OF RECOMBINATION 189 more rapidly than the photons can interact with them. The photons then decouple from the electrons and the universe becomes transparent. Once the photons are decoupled from the electrons and protons, the baryonic por- tion of the universe is no longer compelled to have the same temperature as the Cosmic Microwave Background. During the early stages of the uni- verse (c < c rm ≈ 3 10 −4 ) the universe was radiation dominated, and the Friedmann equation was H 2 H 2 0 = Ω r,0 c 4 . (9.17) Thus, the Hubble parameter was H = H 0 Ω 1/2 r,0 c 2 = 2.1 10 −20 s −1 c 2 . (9.18) This means, for instance, that at c = 10 −5 , the Hubble parameter was H = 2.1 10 −10 s −1 . Since this is much smaller than the scattering rate Γ = 4.4 10 −6 s −1 at the same scale factor, the photons were well coupled to the electrons and protons. If hydrogen remained ionized (and note the qualifying if ), then photons would have remained coupled to the electrons and protons until a relatively recent time. Taking into account the transition from a radiation-dominated to a matter-dominated universe, and the resulting change in the expansion rate, we can compute that if hydrogen had remained fully ionized, then decoupling would have taken place at a scale factor c ≈ 0.023, corresponding to a redshift . ≈ 42 and a CMB temperature of 1 ≈ 120 K. However, at such a low temperature, the CMB photons are too low in energy to keep the hydrogen ionized. Thus, the decoupling of photons is not a gradual process, caused by the continuous lowering of free electron density as the universe expands. Rather, it is a relatively sudden process, caused by the abrupt plummeting of free electron density during the epoch of recombination, as electrons combine with protons to form hydrogen atoms. 9.3 The physics of recombination When does recombination, and the consequent photon decoupling, take place? It’s easy to do a quick and dirty approximation of the recombination tem- perature. Recombination, one could argue, must take place when the mean 190 CHAPTER 9. THE COSMIC MICROWAVE BACKGROUND energy per photon of the Cosmic Microwave Background falls below the ion- ization energy of hydrogen, C = 13.6 eV. When this happens, the average CMB photon is no longer able to photoionize hydrogen. Since the mean CMB photon energy is ∼ 2.7/1, this line of argument would indicate a re- combination temperature of 1 rec ∼ C 2.7/ ∼ 13.6 eV 2.7(8.6 10 −5 eVK −1 ) ∼ 60.000 K . (9.19) Alas, this crude approximation is a little too crude to be useful. It doesn’t take into account the fact that CMB photons are not of uniform energy – a blackbody spectrum has an exponential tail (see Figure 2.7) trailing off to high energies. Although the mean photon energy is 2.7/1, about one photon in 500 will have 1 10/1, one in 3 million will have 1 20/1, and one in 30 billion will have 1 30/1. Although extremely high energy photons make up only a tiny fraction of the CMB photons, the total number of CMB photons is enormous – nearly 2 billion photons for every baryon. The vast swarms of photons that surround every newly formed hydrogen atom greatly increase the probability that the atom will collide with a photon from the high-energy tail of the blackbody spectrum, and be photoionized. Thus, we expect the recombination temperature to depend on the baryon- to-photon ratio η as well as on the ionization energy C. An exact calculation of the fractional ionization A, as a function of η and 1, requires a smattering of statistical mechanics. Let’s start with the reaction that determines the value of A in the early universe: H +γ ÷ ÷ j +c − . (9.20) While the photons are still coupled to the baryonic component, this reaction will be in statistical equilibrium, with the photoionization rate (going from left to right) balancing the radiative recombination rate (going from right to left). When a reaction is in statistical equilibrium at a temperature 1, the number density : x of particles with mass : x is given by the Maxwell- Boltzmann equation : x = o x : x /1 2π¯ / 2 3/2 exp − : x c 2 /1 . (9.21) as long as the particles are non-relativistic, with /1 < : x c 2 . In equa- tion (9.21), o x is the statistical weight of particle r. For instance, electrons, 9.3. THE PHYSICS OF RECOMBINATION 191 protons, and neutrons (and their anti-particles as well) all have a statistical weight o x = 2, corresponding to their two possible spin states. 9 From the Maxwell-Boltzmann equation for H, j, and c − , we can construct an equation which relates the number densities of these particles: : H : p : e = o H o p o e : H : p : e 3/2 /1 2π¯ / 2 −3/2 exp [: p +: e −: H ]c 2 /1 . (9.22) Equation (9.22) can be simplified further. First, since the mass of an electron is small compared to that of a photon, we can set : H : p = 1. Second, the binding energy C = 13.6 eV is given by the formula (: p +: e −: H )c 2 = C. The statistical weights of the proton and electron, are o p = o e = 2, while the statistical weight of a hydrogen atom is o H = 4. Thus, the factor o H (o p o e ) can be set equal to one. The resulting equation, : H : p : e = : e /1 2π¯ / 2 −3/2 exp C /1 . (9.23) is called the Saha equation. Our next job is to convert the Saha equation into a relation among A, 1, and η. From the definition of A (equation 9.9), we can make the substitution : H = 1 −A A : p . (9.24) and from the requirement of charge neutrality, we can make the substitution : e = : p . This yields 1 −A A = : p : e /1 2π¯ / 2 −3/2 exp C /1 . (9.25) To eliminate : p from the above equation, we recall that η ≡ : bary : γ . In a universe where hydrogen is the only element, and a fraction A of the hydrogen is in the form of naked protons, we may write η = : p A: γ . (9.26) 9 If you are a true thermodynamic maven, you will have noted that equation (9.21) omits the chemical potential term, µ, which appears in the most general form of the Maxwell- Boltzmann equation. In most cosmological contexts, as it turns out, the chemical potential is small enough to be safely neglected. 192 CHAPTER 9. THE COSMIC MICROWAVE BACKGROUND Since the photons have a blackbody spectrum, for which : γ = 2.404 π 2 /1 ¯ /c 3 = 0.243 /1 ¯ /c 3 . (9.27) we can combine equations (9.26) and (9.27) to find : p = 0.243Aη /1 ¯ /c 3 . (9.28) Substituting equation (9.28) back into equation (9.25), we finally find the desired equation for A in terms of 1 and η: 1 −A A 2 = 3.84η /1 : e c 2 3/2 exp C /1 . (9.29) This is a quadratic equation in A, whose positive root is A = −1 + √ 1 + 4o 2o . (9.30) where o(1. η) = 3.84η /1 : e c 2 3/2 exp C /1 . (9.31) If we define the moment of recombination as the exact instant when A = 12, then (assuming η = 5.5 10 −10 ) the recombination temperature is /1 rec = 0.323 eV = C 42 . (9.32) Because of the exponential dependence of o upon the temperature, the exact value of η doesn’t strongly affect the value of 1 rec . In degrees Kelvin, /1 rec = 0.323 eV corresponds to a temperature 1 rec = 3740 K, slightly higher than the melting point of tungsten. 10 The temperature of the universe had a value 1 = 1 rec = 3740 K at a redshift . rec = 1370, when the age of the universe, in the Benchmark Model, was t rec = 240.000 yr. Recombination was not an instantaneous process; however, as shown in Figure 9.4, it proceeded fairly rapidly. The fractional ionization goes from A = 0.9 at a redshift . = 1475 10 Not that there was any tungsten around back then to be melted. 9.3. THE PHYSICS OF RECOMBINATION 193 1200 1300 1400 1500 1600 0 .2 .4 .6 .8 1 z X Figure 9.4: The fractional ionization A as a function of redshift during the epoch of recombination. A baryon-to-photon ratio of η = 5.5 10 −10 is assumed. to A = 0.1 at a redshift . = 1255. In the Benchmark Model, the time that elapses from A = 0.9 to A = 0.1 is ∆t ≈ 70.000 yr. Since the number density of free electrons drops rapidly during the epoch of recombination, the time of photon decoupling comes soon after the time of recombination. The rate of photon scattering, when the hydrogen is partially ionized, is Γ(.) = : e (.)σ e c = A(.)(1 +.) 3 : bary,0 σ e c . (9.33) Using Ω bary,0 = 0.04, the numerical value of the scattering rate is Γ(.) = 4.4 10 −21 s −1 A(.)(1 +.) 3 . (9.34) While recombination is taking place, the universe is matter-dominated, so the Hubble parameter is given by the relation H 2 H 2 0 = Ω m,0 c 3 = Ω m,0 (1 +.) 3 . (9.35) Using Ω m,0 = 0.3, the numerical value of the Hubble parameter during the epoch of recombination is H(.) = 1.24 10 −18 s −1 (1 +.) 3/2 . (9.36) 194 CHAPTER 9. THE COSMIC MICROWAVE BACKGROUND The redshift of photon decoupling is found by setting Γ = H, or (combining equations 9.34 and 9.36), 1 +. rec = 43.0 A(. rec ) 2/3 . (9.37) Using the value of A(.) given by the Saha equation (shown in Figure 9.4), the redshift of photon decoupling is found to be . dec = 1130. In truth, the exact redshift of photon decoupling is somewhat smaller than this value. The Saha equation assumes that the reaction H + γ ÷ ÷ j + c − is in equilibrium. However, when Γ starts to drop below H, the photoionization reaction is no longer in equilibrium. As a consequence, at redshifts smaller than ∼ 1200, the fractional ionization A is larger than would be predicted by the Saha equation, and the decoupling of photons is therefore delayed. Without going into the details of the non-equilibrium physics, let’s content ourselves by saying, in round numbers, . dec ≈ 1100, corresponding to a temperature 1 dec ≈ 3000 K, when the age of the universe was t dec ≈ 350.000 yr in the Benchmark Model. When we examine the CMB with our microwave antennas, the photons we collect have been traveling straight toward us since the last time they scattered from a free electron. During a brief time interval t → t + dt, the probability that a photon undergoes a scattering is d1 = Γ(t)dt, where Γ(t) is the scattering rate at time t. Thus, if we detect a CMB photon at time t 0 , the expected number of scatterings it has undergone since an earlier time t is τ(t) = t 0 t Γ(t)dt . (9.38) The dimensionless number τ is the optical depth. The time t for which τ = 1 is the time of last scattering, and represents the time which has elapsed since a typical CMB photon last scattered from a free electron. If we change the variable of integration in equation (9.38) from t to c, we find that τ(c) = 1 a Γ(c) dc ˙ c = 1 a Γ(c) H(c) dc c . (9.39) using the fact that H = ˙ cc. Alternatively, we can find the optical depth as a function of redshift by making the substitution 1 +. = 1c: τ(.) = z 0 Γ(.) H(.) d. 1 +. = 0.0035 z 0 A(.)(1 +.) 1/2 d. . (9.40) 9.3. THE PHYSICS OF RECOMBINATION 195 Table 9.1: Events in the early universe event redshift temperature (K) time (megayears) radiation-matter equality 3570 9730 0.047 recombination 1370 3740 0.24 photon decoupling 1100 3000 0.35 last scattering 1100 3000 0.35 Here, I have made use of equations (9.34) and (9.36). 11 As it turns out, the last scattering of a typical CMB photon occurs after the photoionization re- action H+γ ÷ ÷ j +c − falls out of equilibrium, so the Saha equation doesn’t strictly apply. To sufficient accuracy for our purposes, we can state that the redshift of last scattering was comparable to the redshift of photon decou- pling: . ls ≈ . dec ≈ 1100. Not all the CMB photons underwent their last scattering simultaneously; the universe doesn’t choreograph its microphysics that well. If you scoop up two photons from the CMB, one may have under- gone its last scattering at . = 1200, while the other may have scattered more recently, at . = 1000. Thus, the “last scattering surface” is really more of a “last scattering layer”; just as we can see a little way into a fog bank here on Earth, we can see a little way into the “electron fog” which hides the early universe from our direct view. The relevant times of various events around the time of recombination are shown in Table 9.1. For purposes of comparison, the table also contains the time of radiation-matter equality, emphasizing the fact that recombina- tion, photon decoupling, and last scattering took place when the universe was matter-dominated. Note that all these times are approximate, and are dependent on the cosmological model you choose. (I have chosen the Bench- mark Model in calculating these numbers.) When we look at the CMB, we are getting an intriguing glimpse of the universe as it was when it was only 140.000 of its present age. The epoch of photon decoupling marked an important change in the state of the universe. Before photon decoupling, there existed a single photon- baryon fluid, consisting of photons, electrons, and protons coupled together. Since the photons traveled about at the speed of light, kicking the electrons 11 By the time the universe becomes Λ dominated, the free electron density has fallen to negligibly small levels, so using the Hubble parameter for a matter-dominated universe is a justifiable approximation in computing τ. 196 CHAPTER 9. THE COSMIC MICROWAVE BACKGROUND before them as they went, they tended to smooth out any density fluctu- ations in the photon-baryon fluid smaller than the horizon. After photon decoupling, however, the photon-baryon fluid became a pair of gases, one of photons and the other of neutral hydrogen. Although the two gases coexisted spatially, they were no longer coupled together. Thus, instead of being kicked to and fro by the photons, the hydrogen gas was free to collapse under its own self-gravity (and the added gravitational attraction of the dark matter). Thus, when we look at the Cosmic Microwave Background, we are looking backward in time to an important epoch in the history of the universe – the epoch when the baryons, free from the harassment of photons, were free to collapse gravitationally, and thus form galaxies, stars, planets, cosmologists, and the other dense knots of baryonic matter which make the universe so rich and strange today. 9.4 Temperature fluctuations The dipole distortion of the Cosmic Microwave Background, shown in the top panel of Figure 9.2, results from the fact that the universe is not perfectly homogeneous today (at . = 0). Because we are gravitationally accelerated towards the nearest large lumps of matter, we see a Doppler shift in the radiation of the CMB. The distortions on a smaller angular scale, shown in the bottom panel of Figure 9.2, tell us that the universe was not perfectly homogeneous at the time of last scattering (at . ≈ 1100). The angular size of the temperature fluctuations reflects in part the physical size of the density and velocity fluctuations at . ≈ 1100. The COBE DMR experiment had lim- ited angular resolution, and was only able to detect temperature fluctuations larger than δθ ≈ 7 ◦ . More recent experiments have provided higher angular resolution. For instance, MAXIMA (a balloon-borne experiment), DASI (an experiment located at the South Pole), and BOOMERANG (a balloon-borne experiment launched from Antarctica), all have provided maps of δ11 down to scales of δθ ∼ 10 arcminutes. The angular size δθ of a temperature fluctuation in the CMB is related to a physical size / on the last scattering surface by the relation d A = / δθ . (9.41) where d A is the angular-diameter distance to the last scattering surface. Since the last scattering surface is at a redshift . ls = 1100 1, a good 9.4. TEMPERATURE FLUCTUATIONS 197 approximation to d A is given by equation (7.41): d A ≈ d hor (t 0 ) . ls . (9.42) In the Benchmark Model, the current horizon distance is d hor (t 0 ) ≈ 14.000 Mpc, so the angular-diameter distance to the surface of last scattering is d A ≈ 14.000 Mpc 1100 ≈ 13 Mpc . (9.43) Thus, fluctuations on the last scattering surface with an observed angular size δθ had a proper size / = d A (δθ) = 13 Mpc δθ 1 rad = 0.22 Mpc δθ 1 ◦ (9.44) at the time of last scattering. Thus, the fluctuations that gave rise to the fluc- tuations seen by COBE (with δθ 7 ◦ ) had a proper size / 1.6 Mpc. How- ever, the fluctuations at the time of last scattering were not gravitationally bound objects; they were expanding along with the universal Hubble expan- sion. Thus, the fluctuations seen by COBE correspond to physical scales of /(1+. ls ) 1700 Mpc today, much larger than the biggest superclusters. The higher-resolution experiments such as MAXIMA, DASI, and BOOMERANG see fluctuations corresponding to scales as small as / ≈ 0.04 Mpc at the time of last scattering, or /(1 + . ls ) ≈ 40 Mpc today, about the size of today’s superclusters. Consider the density fluctuations δ11 observed by a particular exper- iment. Figure 9.5, for instance, shows δ11 as measured by COBE at low resolution over the entire sky, and as measured by BOOMERANG at higher resolution over part of the sky. Since δ11 is defined on the surface of a sphere – the celestial sphere, in this case – it is useful to expand it in spher- ical harmonics: δ1 1 (θ. φ) = ∞ ¸ l=0 l ¸ m=−l c lm ) lm (θ. φ) . (9.45) where ) lm (θ. φ) are the usual spherical harmonic functions. What concerns cosmologists is not the exact pattern of hot spots and cold spots on the sky, but their statistical properties. The most important statistical property of δ11 is the correlation function ((θ). Consider two points on the last 198 CHAPTER 9. THE COSMIC MICROWAVE BACKGROUND Figure 9.5: Upper left: the temperature fluctuations measured by the COBE DMR instrument, after the dipole and galaxy foreground are subtracted. Lower right: the temperature fluctuations measured by the BOOMERANG experiment over a region ∼ 20 ◦ across. The white spots are regions 300 mi- croKelvin warmer than average; the black spots are regions 300 microKelvin cooler than average. (Courtesy of the BOOMERANG collaboration.) 9.4. TEMPERATURE FLUCTUATIONS 199 scattering surface. Relative to an observer, they are in the directions ˆ : and ˆ : , and are separated by an angle θ given by the relation cos θ = ˆ : ˆ : . To find the correlation function ((θ), multiply together the values of δ11 at the two points, then average the product over all points separated by the angle θ: ((θ) = δ1 1 (ˆ :) δ1 1 (ˆ : ) ¸ ˆ n·ˆ n =cos θ . (9.46) If cosmologists knew the precise value of ((θ) for all angles from θ = 0 to θ = 180 ◦ , they would have a complete statistical description of the temperature fluctuations over all angular scales. Unfortunately, the CMB measurements which tell us about ((θ) contain information over only a limited range of angular scales. The limited angular resolution of available observations is what makes the spherical harmonic expansion of δ11, shown in equation (9.45), so useful. Using the expansion of δ11 in spherical harmonics, the correlation function can be written in the form ((θ) = 1 4π ∞ ¸ l=0 (2| + 1)( l 1 l (cos θ) . (9.47) where 1 l are the usual Legendre polynomials: 1 0 (r) = 1 1 1 (r) = r (9.48) 1 2 (r) = 1 2 (3r 2 −1) and so forth. In this way, a measured correlation function ((θ) can be broken down into its multipole moments ( l . For a given experiment, the value of ( l will be nonzero for angular scales larger than the resolution of the experiment and smaller than the patch of sky examined. Generally speaking, a term ( l is a measure of temperature fluctuations on the angular scale θ ∼ 180 ◦ |. Thus, the multipole | is interchangeable, for all practical purposes, with the angular scale θ. The | = 0 (monopole) term of the correlation function vanishes if you’ve defined the mean temperature correctly. The | = 1 (dipole) term results primarily from the Doppler shift due to our motion through space. It is the moments with | ≥ 2 which are of the most interest to cosmologists, since they tell us about the fluctuations present at the time of last scattering. 200 CHAPTER 9. THE COSMIC MICROWAVE BACKGROUND Figure 9.6: The anisotropy of the CMB temperature, ∆ T , expressed as a function of the multipole | (from Wang, Tegmark, & Zaldarriaga, 2002, Phys. Rev. D). In presenting the results of CMB observations, it is customary to plot the function ∆ T ≡ |(| + 1) 2π ( l 1/2 '1` (9.49) since this function tells us the contribution per logarithmic interval in | to the total temperature fluctuation δ1 of the Cosmic Microwave Background. Figure 9.6, which combines data from a large number of CMB experiments, is a plot of ∆ T as a function of the logarithm of |. Note that the temperature fluctuation has a peak at | ∼ 200, corresponding to an angular size of ∼ 1 ◦ . The detailed shape of the ∆ T versus | curve, as shown in Figure 9.6, contains a wealth of information about the universe at the time of photon decoupling. In the next section, we will examine, very briefly, the physics behind the temperature fluctuations, and how we can extract cosmological information from the temperature anisotropy of the Cosmic Microwave Background. 9.5. WHAT CAUSES THE FLUCTUATIONS? 201 9.5 What causes the fluctuations? At the time of last scattering, a particularly interesting length scale, cosmo- logically speaking, is the Hubble distance, cH(. ls ) ≈ 3.0 10 8 ms −1 1.24 10 −18 s −1 (1101) 3/2 ≈ 6.6 10 21 m ≈ 0.2 Mpc . (9.50) where I have used equation (9.36) to compute the Hubble parameter at the redshift of last scattering, . ls ≈ 1100. A patch of the last scattering surface with this physical size will have an angular size, as seen from Earth, of θ H = cH(. ls ) d A ≈ 0.2 Mpc 13 Mpc ≈ 0.015 rad ≈ 1 ◦ . (9.51) It is no coincidence that the peak in the ∆ T versus | curve (Figure 9.6) occurs at an angular scale θ ∼ θ H . The origin of temperature fluctuations with θ θ H (| < 180) is different from those with θ < θ H (| 180). Consider first the large-scale fluctuations – those with angular size θ θ H . These temperature fluctuations arise from the gravitational effect of primor- dial density fluctuations in the distribution of nonbaryonic dark matter. The density of nonbaryonic dark matter at the time of last scattering was ε dm (. ls ) = Ω dm,0 ε c,0 (1 +. ls ) 3 (9.52) since the energy density of matter is ∝ c −3 ∝ (1 + .) 3 . Plugging in the appropriate numbers, we find that ε dm (. ls ) ≈ (0.26)(5200 MeVm −3 )(1101) 3 ≈ 1.8 10 12 MeVm −3 . (9.53) equivalent to a mass density of ∼ 3 10 −18 kg m −3 . The density of baryonic matter at the time of last scattering was ε bary (. ls ) = Ω bary,0 c c,0 (1 +. ls ) 3 ≈ 2.8 10 11 MeVm −3 . (9.54) The density of photons at the time of last scattering, since ε γ ∝ c −4 ∝ (1 +.) 4 , was ε γ (. ls ) = Ω γ,0 c c,0 (1 +. ls ) 4 ≈ 3.8 10 11 MeVm −3 . (9.55) Thus, at the time of last scattering, ε dm ε γ ε bary , with dark matter, photons, and baryons having energy densities in roughly the ratio 6.4 : 1.4 : 1. 202 CHAPTER 9. THE COSMIC MICROWAVE BACKGROUND The nonbaryonic dark matter dominated the energy density c, and hence the gravitational potential, of the universe at the time of last scattering. Suppose that the density of the nonbaryonic dark matter at the time of last scattering was not perfectly homogeneous, but varied as a function of position. Then we could write the energy density of the dark matter as ε(:) = ¯ ε +δε(:) . (9.56) where ¯ ε is the spatially averaged energy density of the nonbaryonic dark matter, and δε is the local deviation from the mean. In the Newtonian approximation, the spatially varying component of the energy density, δε, gives rise to a spatially varying gravitational potential δΦ. The link between δε and δΦ is Poisson’s equation: ∇ 2 (δΦ) = 4πG c 2 δε . (9.57) Unless the distribution of dark matter were perfectly smooth at the time of last scattering, the fluctuations in its density would necessarily have given rise to fluctuations in the gravitational potential. Consider the fate of a CMB photon which happens to be at a local min- imum of the potential at the time of last scattering. (Minima in the gravi- tational potential are known colloquially as “potential wells”.) In climbing out of the potential well, it loses energy, and consequently is redshifted. Conversely, a photon which happens to be at a potential maximum when the universe became transparent gains energy as it falls down the “poten- tial hill”, and thus is blueshifted. The cool (redshifted) spots on the COBE temperature map correspond to minima in δΦ at the time of last scattering; the hot (blueshifted) spots correspond to maxima in δΦ. A detailed general relativistic calculation, first performed by Sachs and Wolfe in 1967, tells us that δ1 1 = 1 3 δΦ c 2 . (9.58) Thus, the temperature fluctuations on large angular scales (θ θ H ≈ 1 ◦ ) give us a map of the potential fluctuations δΦ present at the time of last scattering. The creation of temperature fluctuations by variations in the gravitational potential is known as the Sachs-Wolfe effect, in tribute to the work of Sachs and Wolfe. On smaller scales (θ < θ H ), the origin of the temperature fluctuations in the CMB is complicated by the behavior of the photons and baryons. 9.5. WHAT CAUSES THE FLUCTUATIONS? 203 Consider the situation immediately prior to photon decoupling. The photons, electrons, and protons together make a single photon-baryon fluid, whose energy density is only about a third that of the dark matter. Thus, the photon-baryon fluid moves primarily under the gravitational influence of the dark matter, rather than under its own self-gravity. The equation-of-state parameter u pb of the photon-baryon fluid is intermediate between the value u = 13 expected for a gas containing only photons and the value u = 0 expected for a gas containing only cold baryons and electrons. If the photon- baryon fluid finds itself in a potential well of the dark matter, it will fall to the center of the well. 12 As the photon-baryon fluid is compressed by gravity, however, its pressure starts to rise. Eventually, the pressure is sufficient to cause the fluid to expand outward. As the expansion continues, the pressure drops until gravity causes the photon-baryon fluid to fall inward again. The cycle of compression and expansion which is set up continues until the time of photon decoupling. The inward and outward oscillations of the photon- baryon fluid are called acoustic oscillations, since they represent a type of standing sound wave in the photon-baryon fluid. If the photon-baryon fluid within a potential well is at maximum com- pression at the time of photon decoupling, its density will be higher than average, and the liberated photons, since 1 ∝ ε 1/4 , will be hotter than av- erage. Conversely, if the photon-baryon fluid within a potential well is at maximum expansion at the time of decoupling, the liberated photons will be slightly cooler than average. If the photon-baryon fluid is in the process of expanding or contracting at the time of decoupling, the Doppler effect will cause the liberated photons to be cooler or hotter than average, depending on whether the photon-baryon fluid was moving away from our location or toward it at the time of photon decoupling. Computing the exact shape of the ∆ T versus | curve expected in a particular model universe is a rather complicated chore. Generally speaking, however, the highest peak in the ∆ T curve (at | ∼ 200 or θ ∼ 1 ◦ in Figure 9.6) represents the potential wells within which the photon-baryon fluid had just reached maximum compres- sion at the time of last scattering. These potential wells had proper sizes ∼ cH(. ls ) at the time of last scattering, and hence have angular sizes of 12 Note that if the size of the well is larger than c/H(z ls ), the photon-baryon fluid, which travels at a speed < c, will not have time to fall to the center by the time of last scattering t ls ∼ 1/H(z ls ). This is why the motions of the photons and baryons are irrelevant on scales θ > θ H , and why the temperature fluctuations on these large scales are dictated purely by the distribution of dark matter. 204 CHAPTER 9. THE COSMIC MICROWAVE BACKGROUND ∼ θ H as seen by us at the present day. The location and amplitude of the highest peak in Figure 9.6 is a very useful cosmological diagnostic. As it turns out, the angle θ ≈ 1 ◦ at which it’s located is dependent on the spatial curvature of the universe. In a negatively curved universe (κ = −1), the angular size θ of an object of known proper size at a known redshift is smaller than it is in a positively curved universe (κ = +1). If the universe were negatively curved, the peak in ∆ T would be seen at an angle θ < 1 ◦ , or | 180; if the universe were positively curved, the peak would be seen at an angle θ 1 ◦ , or | < 180. The observed position of the peak is consistent with κ = 0, or Ω 0 = 1. Figure 9.7 shows the values of Ω m,0 and Ω Λ,0 which are permitted by the present CMB data. Note that the shaded areas which show the best fit to the CMB data are roughly parallel to the Ω = 1 line (representing κ = 0 universes) and roughly perpendicular to the region permitted by the type Ia supernova results. The oval curve marked “68%” in Figure 9.7 represents the region consistent with both the CMB results and the supernova results. A spatially flat universe, with Ω m,0 ≈ 0.3 and Ω Λ,0 ≈ 0.7, agrees with the CMB results, the supernova results, and the computed density of matter in clusters (as discussed in Chapter 8). This happy concurrence is the basis for the “Benchmark Model” which I have been using in this book. The angular size corresponding to the highest peak in the ∆ T versus | curve gives useful information about the density parameter Ω and the cur- vature κ. The amplitude of the peak is dependent on the sound speed of the photon-baryon fluid prior to photon decoupling. Since the sound speed of the photon-baryon fluid is c s = √ u pb c . (9.59) and the equation-of-state parameter u pb is in turn dependent on the baryon- to-photon ratio, the amplitude of the peak is a useful diagnostic of the baryon density of the universe. Detailed analysis of the ∆ T curve, with the currently available data, yields Ω bary,0 = 0.04 ±0.02 . (9.60) assuming a Hubble constant of H 0 = 70 kms −1 Mpc −1 . As we’ll see in the next chapter, this baryon density is consistent with that found from the entirely different arguments of primordial nucleosynthesis. 9.5. WHAT CAUSES THE FLUCTUATIONS? 205 Figure 9.7: The values of Ω m,0 and Ω Λ,0 permitted by the MAXIMA and COBE DMR data. The light gray area near the Ω = 1 line shows the best- fitting values of Ω m,0 and Ω Λ,0 . The SN Ia data (as seen earlier in Figure 7.6) are also shown. The curves marked 68%, 95%, and 98% give the best fits, at the designated confidence levels, for the combined CMB and SN Ia data (from Stompor et al. 2001, ApJ, 561, L7). 206 CHAPTER 9. THE COSMIC MICROWAVE BACKGROUND Suggested reading [Full references are given in the “Annotated Bibliography” on page 286.] Coles (1999), Part I, ch. 5: An insightful overview of the origin of the temperature anisotropies in the CMB. Liddle (1999), ch. 9: A simplified overview of photon decoupling and the origin of the CMB. Rich (2001), ch. 7.9: A discussion of the anisotropies in the CMB, and their use as diagnostics of cosmological parameters. Problems (9.1) The purpose of this problem is to determine how the uncertainty in the value of the baryon-to-photon ratio, η, affects the recombination temperature in the early universe. Plot the fractional ionization A as a function of temperature, in the range 3000 K < 1 < 4500 K; first make the plot assuming η = 4 10 −10 , then assuming η = 8 10 −10 . How much does this change in η affect the computed value of the recombination temperature 1 rec , if we define 1 rec as the temperature at which A = 12? (9.2) Suppose the temperature 1 of a blackbody distribution is such that /1 < C, where C = 13.6 eV is the ionization energy of hydrogen. What fraction 1 of the blackbody photons are energetic enough to ionize hydrogen? If 1 = 1 rec = 3740 K, what is the numerical value of 1? (9.3) Imagine that at the time of recombination, the baryonic portion of the universe consisted entirely of 4 He (that is, helium with two protons and two neutrons in its nucleus). The ionization energy of helium (that is, the energy required to convert neutral He to He + ) is C He = 24.6 eV. At what temperature would the fractional ionization of the helium be A = 12? Assume that η = 5.5 10 −10 and that the number density of He ++ is negligibly small. (The relevant statistical weight factor for the ionization of helium is o He (o e o He +) = 14.) 9.5. WHAT CAUSES THE FLUCTUATIONS? 207 (9.4) What is the proper distance d p to the surface of last scattering? What is the luminosity distance d L to the surface of last scattering? Assume that the Benchmark Model is correct, and that the redshift of the last scattering surface is . ls = 1100. (9.5) We know from observations that the intergalactic medium is currently ionized. Thus, at some time between t rec and t 0 , the intergalactic medium must have been reionized. The fact that we can see small fluctuations in the CMB places limits on how early the reionization took place. Assume that the baryonic component of the universe in- stantaneously became completely reionized at some time t ∗ . For what value of t ∗ does the optical depth of reionized material, τ = t 0 t∗ Γ(t)dt = t 0 t∗ : e (t)σ e cdt . (9.61) equal one? For simplicity, assume that the universe is spatially flat and matter-dominated, and that the baryonic component of the universe is pure hydrogen. To what redshift . ∗ does this value of t ∗ correspond? Chapter 10 Nucleosynthesis & the Early Universe The Cosmic Microwave Background tells us a great deal about the state of the universe at the time of last scattering (t ls ≈ 0.35 Myr). However, the opacity of the early universe prevents us from directly seeing what the universe was like at t < t ls . Photons are the “messenger boys” of astronomy; although some information is carried by cosmic rays and by neutrinos, most of what we know about the universe beyond our Solar System has come in the form of photons. Looking at the last scattering surface is like looking at the surface of a cloud, or the surface of the Sun; our curiosity is piqued, and we wish to find out what conditions are like in the opaque regions so tantalizingly hidden from our direct view. From a theoretical viewpoint, many properties of the early universe should be quite simple. For instance, when radiation is strongly dominant over matter, at scale factors c <c rm ≈ 2.8 10 −4 , or times t <t rm ≈ 47.000 yr, the expansion of the universe has the simple power-law form c(t) ∝ t 1/2 . The temperature of the blackbody photons in the early universe, which decreases as 1 ∝ c −1 as the universe expands, is given by the convenient relation 1(t) ≈ 10 10 K t 1 s −1/2 . (10.1) or equivalently /1(t) ≈ 1 MeV t 1 s −1/2 . (10.2) 208 10.1. NUCLEAR PHYSICS AND COSMOLOGY 209 Thus the mean energy per photon was 1 mean (t) ≈ 2.7/1(t) ≈ 3 MeV t 1 s −1/2 . (10.3) The Fermi National Accelerator Laboratory, in Batavia, Illinois, is justly proud of its “Tevatron”, designed to accelerate protons to an energy of 1 TeV; that’s 10 12 eV, or a thousand times the rest energy of a proton. Well, when the universe was one picosecond old (t = 10 −12 s), the entire universe was a Tevatron. Thus, the early universe is referred to as “the poor man’s particle accelerator”, since it provided particles of very high energy without running up an enormous electricity bill or having Congress threaten to cut off its funding. 10.1 Nuclear physics and cosmology As the universe has expanded and cooled, the mean energy per photon has dropped from 1 mean (t P ) ∼ 1 P ∼ 10 28 eV at the Planck time to 1 mean (t 0 ) ≈ 610 −4 eV at the present day. Thus, by studying the universe as it expands, we sample over 31 orders of magnitude in particle energy. Within this wide energy range, some energies are of more interest than others to physicists. For instance, to physicists studying recombination and photoionization, the most interesting energy scale is the ionization energy of an atom. The ionization energy of hydrogen is C = 13.6 eV, as we have already noted. The ionization energies of other elements (that is, the energy required to remove the most loosely bound electron in the neutral atom) are roughly comparable. Thus, atomic physicists, when considering the ionization of atoms, typically deal with energies of ∼ 10 eV, in round numbers. Nuclear physicists are concerned not with ionization and recombination (removing or adding electrons to an atom), but with the much higher energy processes of fission and fusion (splitting or merging atomic nuclei). An atomic nucleus contains 2 protons and ` neutrons, where 2 ≥ 1 and ` ≥ 0. Protons and neutrons are collectively called nucleons. The total number of nucleons within an atomic nucleus is called the mass number, and is given by the formula ¹ = 2 + `. The proton number 2 of a nucleus determines the atomic element to which that nucleus belongs. For instance, hydrogen (H) nuclei all have 2 = 1, helium (He) nuclei have 2 = 2, lithium (Li) nuclei have 2 = 3, beryllium (Be) nuclei have 2 = 4, and so on, through the complete 210 CHAPTER 10. NUCLEOSYNTHESIS & THE EARLY UNIVERSE periodic table. Although all atoms of a given element have the same number of protons in their nuclei, different isotopes of an element can have different numbers of neutrons. A particular isotope of an element is designated by prefixing the mass number ¹ to the symbol for that element. For instance, a standard hydrogen nucleus, with one proton and no neutrons, is symbolized as 1 H. (Since an ordinary hydrogen nucleus is nothing more than a proton, we may also write j in place of 1 H when considering nuclear reactions.) Heavy hydrogen, or deuterium, contains one proton and one neutron, and is symbolized as 2 H. (Since deuterium is very frequently mentioned in the context of nuclear fusion, it has its own special symbol, D.) Ordinary helium contains two protons and two neutrons, and is symbolized as 4 He. The binding energy 1 of a nucleus is the energy required to pull it apart into its component protons and neutrons. Equivalently, it is the energy re- leased when a nucleus is fused together from individual protons and neutrons. For instance, when a neutron and a proton are bound together to form a deu- terium nucleus, an energy of 1 D = 2.22 MeV is released: j +: ÷ ÷ D + 2.22 MeV . (10.4) The deuterium nucleus is not very tightly bound, compared to other atomic nuclei. Figure 10.1 plots the binding energy per nucleon (1¹) for atomic nuclei with different mass numbers. Note that 4 He, with a total binding energy of 1 = 28.30 MeV, and a binding energy per nucleon of 1¹ = 7.07 MeV, is relatively tightly bound, compared to other light nuclei (that is, nuclei with ¹ ≤ 10). The most tightly bound nuclei are those of 56 Fe and 62 Ni, which both have 1¹ ≈ 8.8 MeV. Thus, nuclei more massive than iron or nickel can release energy by fission – splitting into lighter nuclei. Nuclei less massive than iron or nickel can release energy by fusion – merging into heavier nuclei. Thus, just as studies of ionization and recombination deal with an energy scale of ∼ 10 eV (a typical ionization energy), so studies of nuclear fusion and fission deal with an energy scale of ∼ 8 MeV (a typical binding energy per nucleon). Moreover, just as electrons and protons combined to form neutral hydrogen atoms when the temperature dropped sufficiently far below the ion- ization energy of hydrogen (C = 13.6 eV), so protons and neutrons must have fused to form deuterium when the temperature dropped sufficiently far below the binding energy of deuterium (1 D = 2.22 MeV). The epoch of recombina- tion must have been preceded by an epoch of nuclear fusion, commonly called 10.1. NUCLEAR PHYSICS AND COSMOLOGY 211 1 10 100 0 2 4 6 8 D 3 He 3 H 4 He A B / A ( M e V ) Figure 10.1: The binding energy per nucleon (B/A) as a function of the number of nucleons (protons and neutrons) in an atomic nucleus. Note the absence of nuclei at ¹ = 5 and ¹ = 8. 212 CHAPTER 10. NUCLEOSYNTHESIS & THE EARLY UNIVERSE the epoch of Big Bang Nucleosynthesis (BBN). Nucleosynthesis in the early universe starts by the fusion of neutrons and protons to form deuterium, then proceeds to form heavier nuclei by successive acts of fusion. Since the binding energy of deuterium is larger than the ionization energy of hydrogen by a fac- tor 1 D C = 1.610 5 , we would expect, as a rough estimate, the synthesis of deuterium to occur at a temperature 1.610 5 times higher than the recombi- nation temperature 1 rec = 3740 K. That is, deuterium synthesis occurred at a temperature 1 nuc ≈ 1.610 5 (3740 K) ≈ 610 8 K, corresponding to a time t nuc ≈ 300 s. This estimate, as we’ll see when we do the detailed calculations, gives a temperature slightly too low, but it certainly gives the right order of magnitude. As indicated in the title of Steven Weinberg’s classic book, The First Three Minutes, the entire saga of Big Bang Nucleosynthesis takes place when the universe is only a few minutes old. One thing we can say about Big Bang Nucleosynthesis, after taking a look at the present-day universe, is that it was shockingly inefficient. From an energy viewpoint, the preferred universe would be one in which the baryonic matter consisted of an iron-nickel alloy. Obviously, we do not live in such a universe. Currently, three-fourths of the baryonic component (by mass) is still in the form of unbound protons, or 1 H. Moreover, when we look for nuclei heavier than 1 H, we find that they are primarily 4 He, a relatively lightweight nucleus compared to 56 Fe and 62 Ni. The primordial helium fraction of the universe (that is, the helium fraction before nucleosynthesis begins in stars) is usually expressed as the dimensionless number ) p ≡ ρ( 4 He) ρ bary . (10.5) That is, ) p is the mass density of 4 He divided by the mass density of all the baryonic matter. The Sun’s atmosphere has a helium fraction (by mass) of ) = 0.28. However, the Sun is made of recycled interstellar gas, which was contaminated by helium formed in earlier generations of stars. When we look at astronomical objects of different sorts, we find a minimum value of ) = 0.24. That is, baryonic objects such as stars and gas clouds are all at least 24% helium. 1 1 Condensed objects which have undergone chemical or physical fractionation can be much lower in helium than this value. For instance, your helium fraction is 24%. 10.2. NEUTRONS AND PROTONS 213 10.2 Neutrons and protons The basic building blocks for nucleosynthesis are neutrons and protons. The rest energy of a neutron is greater than that of a proton by a factor C n = : n c 2 −: p c 2 = 1.29 MeV . (10.6) A free neutron is unstable, decaying via the reaction : →j +c − + ¯ ν e . (10.7) The decay time for a free neutron is τ n = 890 s. That is, if you start out with a population of free neutrons, after a time t, a fraction 1 = exp(−tτ n ) will remain. 2 Since the energy C n released by the decay of a neutron into a proton is greater than the rest energy of an electron (: e c 2 = 0.51 MeV), the remainder of the energy is carried away by the kinetic energy of the electron and the energy of the electron anti-neutrino. With a decay time of only fifteen minutes, the existence of a free neutron is as fleeting as fame; once the universe was several hours old, it contained essentially no free neutrons. However, a neutron which is bound into a stable atomic nucleus is preserved against decay. There are still neutrons around today, because they’ve been tied up in deuterium, helium, and other atoms. Let’s consider the state of the universe when its age is t = 0.1 s. At that time, the temperature was 1 ≈ 3 10 10 K, and the mean energy per photon was 1 mean ≈ 10 MeV. This energy is much greater than the rest energy of a electron or positron, so there were positrons as well as electrons present at t = 0.1 s, created by pair production: γ +γ ÷ ÷ c − +c + . (10.8) At t = 0.1 s, neutrons and protons were in equilibrium with each other, via the interactions : +ν e ÷ ÷ j +c − (10.9) and : +c + ÷ ÷ j + ¯ ν e . (10.10) As long as neutrons and protons are kept in equilibrium by the reactions shown in equations (10.9) and (10.10), their number density is given by 2 The ‘half-life’, the time it takes for half the neutrons to decay, is related to the decay time by the relation t 1/2 = τ n ln 2 = 617 s. 214 CHAPTER 10. NUCLEOSYNTHESIS & THE EARLY UNIVERSE the Maxwell-Boltzmann equation, as discussed in section 9.3. The number density of neutrons is then : n = o n : n /1 2π¯ / 2 3/2 exp − : n c 2 /1 (10.11) and the number density of protons is : p = o p : p /1 2π¯ / 2 3/2 exp − : p c 2 /1 . (10.12) Since the statistical weights of protons and neutrons are equal, with o p = o n = 2, the neutron-to-proton ratio, from equations (10.11) and (10.12), is : n : p = : n : p 3/2 exp − (: n −: p )c 2 /1 . (10.13) The above equation can be simplified. First, (: n : p ) 3/2 = 1.002; there will be no great loss in accuracy if we set this factor equal to one. Second, the difference in rest energy of the neutron and proton is (: n − : p )c 2 = C n = 1.29 MeV. Thus, the equilibrium neutron-to-proton ratio has the particularly simple form : n : p = exp − C n /1 . (10.14) illustrated as the solid line in Figure 10.2. At temperatures /1 C n = 1.29 MeV, corresponding to 1 1.5 10 10 K and t < 1 s, the number of neutrons is nearly equal to the number of protons. However, as the tempera- ture starts to drop below 1.5 10 10 K, protons begin to be strongly favored, and the neutron-to-proton ratio plummets exponentially. If the neutrons and protons remained in equilibrium, then by the time the universe was six minutes old, there would be only one neutron for every million protons. However, neutrons and protons do not remain in equilibrium for nearly that long. The interactions which mediate between neutrons and protons in the early universe, shown in equations (10.9) and (10.10), involve the interaction of a baryon with a neutrino (or anti-neutrino). Neutrinos interact with baryons via the weak nuclear force. The cross-sections for weak interactions have the temperature dependence σ w ∝ 1 2 ; at the temperatures we are considering, the cross-sections are small. A typical cross-section for 10.2. NEUTRONS AND PROTONS 215 100 10 1 10 −6 10 −5 .0001 .001 .01 .1 1 freezeout T (10 9 K) n n / n p Figure 10.2: Neutron-to-proton ratio in the early universe. The solid line assumes equilibrium; the dotted line gives the value after freezeout. the interaction of a neutrino with any other particle via the weak nuclear force is σ w ∼ 10 −47 m 2 /1 1 MeV 2 . (10.15) (Compare this to the Thomson cross-section for the interaction of electrons via the electromagnetic force: σ e = 6.65 10 −29 m 2 .) In the radiation- dominated universe, the temperature falls at the rate 1 ∝ c(t) −1 ∝ t −1/2 , and thus the cross-sections for weak interactions diminish at the rate σ w ∝ t −1 . The number density of neutrinos falls at the rate : ν ∝ c(t) −3 ∝ t −3/2 , and hence the rate Γ with which neutrons and protons interact with neutrinos via the weak force falls rapidly: Γ = : ν cσ w ∝ t −5/2 . (10.16) Meanwhile, the Hubble parameter is only decreasing at the rate H ∝ t −1 . When Γ ≈ H, the neutrinos decouple from the neutrons and protons, and the ratio of neutrons to protons is “frozen” (at least until the neutrons start to decay, at times t ∼ τ n ). An exact calculation of the temperature 1 freeze at 216 CHAPTER 10. NUCLEOSYNTHESIS & THE EARLY UNIVERSE which Γ = H requires a knowledge of the exact cross-section of the proton and neutron for weak interactions. Using the best available laboratory in- formation, the “freezeout temperature” turns out to be /1 freeze = 0.8 MeV, or 1 freeze = 9 10 9 K. The universe reaches this temperature when its age is t freeze ∼ 1 s. The neutron-to-proton ratio, once the temperature drops below 1 freeze , is frozen at the value : n : p = exp − C n /1 freeze ≈ exp − 1.29 MeV 0.8 MeV ≈ 0.2 (10.17) At times t freeze < t <τ n , there was one neutron for every five protons in the universe. It is the scarcity of neutrons relative to protons that explains why Big Bang Nucleosynthesis was so incomplete, leaving three-fourths of the baryons in the form of unfused protons. A neutron will fuse with a proton much more readily than a proton will fuse with another proton. When two protons fuse to form a deuterium nucleus, a positron must be emitted (to conserve charge); this means that an electron neutrino must also be emitted (to conserve elec- tron quantum number). The proton-proton fusion reaction can be written as j +j →D +c + +ν e . (10.18) The involvement of a neutrino in this reaction tells us that it involves the weak nuclear force, and thus has a minuscule cross-section, of order σ w . By contrast, the neutron-proton fusion reaction is j +: ÷ ÷ D +γ . (10.19) No neutrinos are involved; this is a strong interaction (one involving the strong nuclear force). The cross-section for interactions involving the strong nuclear force are much larger than for those involving the weak nuclear force. The rate of proton-proton fusion is much slower than the rate of neutron- proton fusion, for two reasons. First, the cross-section for proton-proton fusion, since it is a weak interaction, is minuscule compared to the cross- section for neutron-proton fusion. Second, since protons are all positively charged, they must surmount the Coulomb barrier between them in order to fuse. It’s possible, of course, to coax two protons into fusing with each other. It’s happening in the Sun, for instance, even as you read this sentence. How- ever, fusion in the Sun is a very slow process. If you pick out any particular 10.2. NEUTRONS AND PROTONS 217 proton in the Sun’s core, it has only one chance in ten billion of undergo- ing fusion during the next year. Only exceptionally fast protons have any chance of undergoing fusion, and even those high-speed protons have only a tiny probability of quantum tunneling through the Coulomb barrier of an- other proton and fusing with it. The core of the Sun, though, is a stable environment; it’s in hydrostatic equilibrium, and its temperature and den- sity change only slowly with time. In the early universe, by strong contrast, the temperature drops as 1 ∝ t −1/2 and the density of baryons drops as : bary ∝ t −3/2 . Big Bang Nucleosynthesis is a race against time. After less than an hour, the temperature and density have dropped too low for fusion to occur. For the sake of completeness, I should also note that the rate of neutron- neutron fusion in the early universe is negligibly small compared to the rate of neutron-proton fusion. The reaction governing neutron-neutron fusion is : +: →D +c − + ¯ ν e . (10.20) Again, the presence of a neutrino (an electron anti-neutrino, in this case) tells us this is an interaction involving the weak nuclear force. Thus, although their is no Coulomb barrier between neutrons, the neutron-neutron fusion rate is tiny. In part, this is because of the scarcity of neutrons relative to protons, but mainly it is because of the small cross-section for neutron- neutron fusion. Given the alacrity of neutron-proton fusion when compared to the leisurely rate of proton-proton and neutron-neutron fusion, we can state, as a lowest order approximation, that BBN proceeds until every free neutron is bonded into an atomic nucleus, with the leftover protons remaining solitary. In this approximation, we can compute the maximum possible value of ) p , the frac- tion of the baryon mass in the form of 4 He. To compute the maximum possi- ble value of ) p , suppose that every neutron present after the proton-neutron freezeout is incorporated into a 4 He nucleus. Given a neutron-to-proton ratio of : n : p = 15, we can consider a representative group of 2 neutrons and 10 protons. The 2 neutrons can fuse with 2 of the protons to form a single 4 He nucleus. The remaining 8 protons, though, will remain unfused. The mass fraction of 4 He will then be ) max = 4 12 = 1 3 . (10.21) More generally, if 1 ≡ : n : p , with 0 ≤ 1 ≤ 1, then the maximum possible value of ) p is ) max = 21(1 +1). 218 CHAPTER 10. NUCLEOSYNTHESIS & THE EARLY UNIVERSE If the observed value of ) p = 0.24 were greater than the predicted ) max , that would be a cause for worry; it might mean, for example, that we didn’t really understand the process of proton-neutron freezeout. However, the fact that the observed value of ) p is less than ) max is not worrisome; there are various factors which act to reduce the actual value of ) p below its theoret- ical maximum. First, if nucleosynthesis didn’t take place immediately after freezeout at t ≈ 1 s, then the spontaneous decay of neutrons would inevitably lower the neutron-to-proton ratio, and thus reduce the amount of 4 He pro- duced. Next, if some neutrons escape fusion altogether, or end in nuclei lighter than 4 He (such as D or 3 He), they will not contribute to ) p . Finally, if nucleosynthesis goes on long enough to produce nuclei heavier than 4 He, that too will reduce ) p . In order to compute ) p accurately, as well as the abundances of other isotopes, it will be necessary to consider the process of nuclear fusion in more detail. Fortunately, much of the statistical mechanics we will need is just a rehash of what we used when studying recombination. 10.3 Deuterium synthesis Let’s move on to the next stage of Big Bang Nucleosynthesis, just after proton-neutron freezeout is complete. The time is t ≈ 2 s. The neutron-to- proton ratio is : n : p = 0.2. The neutrinos, which ceased to interact with electrons about the same time they stopped interacting with neutrons and protons, are now decoupled from the rest of the universe. The photons, however, are still strongly coupled to the protons and neutrons. Big Bang Nucleosynthesis takes place through a series of two-body reactions, building heavier nuclei step by step. The essential first step in BBN is the fusion of a proton and a neutron to form a deuterium nucleus: j +: ÷ ÷ D +γ . (10.22) When a proton and a neutron fuse, the energy released (and carried away by a gamma ray) is the binding energy of a deuterium nucleus: 1 D = (: n +: p −: D )c 2 = 2.22 MeV . (10.23) Conversely, a photon with energy ≥ 1 D can photodissociate a deuterium nucleus into its component proton and neutron. The reaction shown in equa- tion (10.22) should have a haunting familiarity if you’ve just read Chapter 9; 10.3. DEUTERIUM SYNTHESIS 219 it has the same structural form as the reaction governing the recombination of hydrogen: j +c − ÷ ÷ H +γ . (10.24) A comparison of equation (10.22) with equation (10.24) shows that in each case, two particles become bound together to form a composite object, with the excess energy carried away by a photon. In the case of nucleosynthesis, a proton and neutron are bonded together by the strong nuclear force to form a deuterium nucleus, with a gamma-ray photon being emitted. In the case of photoionization, a proton and electron are bonded together by the electromagnetic force to form a neutral hydrogen atom, with an ultraviolet photon being emitted. A major difference between nucleosynthesis and re- combination, of course, is between the energy scales involved. The binding energy of deuterium, 1 D = 2.200.000 eV, is 160,000 times greater than the ionization energy of hydrogen, C = 13.6 eV. 3 Despite the difference in energy scales, many of the equations used to analyze recombination can be re-used to analyze deuterium nucleosynthesis. Around the time of recombination, for instance, the relative numbers of free protons, free electrons, and neutral hydrogen atoms is given by the Saha equation, : H : p : e = : e /1 2π¯ / 2 −3/2 exp C /1 . (10.25) which tells us that neutral hydrogen is favored in the limit /1 → 0, and that ionized hydrogen is favored in the limit /1 → ∞. Around the time of deuterium synthesis, the relative numbers of free protons, free neutrons, and deuterium nuclei is given by an equation directly analogous to equa- tion (9.22): : D : p : n = o D o p o n : D : p : n 3/2 /1 2π¯ / 2 −3/2 exp [: p +: n −: D ]c 2 /1 . (10.26) From equation (10.23), we can make the substitution [: p +: n −: D ]c 2 = 1 D . The statistical weight factor of the deuterium nucleus is o D = 3, in compar- ison to o p = o n = 2 for a proton or neutron. To acceptable accuracy, we may write : p = : n = : D 2. These substitutions yield the nucleosynthetic 3 As the makers of bombs have long known, you can release much more energy by fusing atomic nuclei than by simply shuffling electrons around. 220 CHAPTER 10. NUCLEOSYNTHESIS & THE EARLY UNIVERSE equivalent of the Saha equation, : D : p : n = 6 : n /1 π¯ / 2 −3/2 exp 1 D /1 . (10.27) which tells us that deuterium is favored in the limit /1 → 0, and that free protons and neutrons are favored in the limit /1 →∞. To define a precise temperature 1 nuc at which the nucleosynthesis of deu- terium takes place, we need to define what we mean by “the nucleosynthesis of deuterium”. Just as recombination is a process which takes a finite length of time, so is nucleosynthesis. It is useful, though, to define 1 nuc as the tem- perature at which : D : n = 1; that is, the temperature at which half the free neutrons have been fused into deuterium nuclei. As long as equation (10.27) holds true, the deuterium-to-neutron ratio can be written as : D : n = 6: p : n /1 π¯ / 2 −3/2 exp 1 D /1 . (10.28) We can write the deuterium-to-neutron ratio as a function of 1 and the baryon-to-photon ratio η if we make some simplifying assumptions. Even today, we know that ∼ 75% of all the baryons in the universe are in the form of unbound protons. Before the start of deuterium synthesis, 5 out of 6 baryons (or ∼ 83%) were in the form of unbound protons. Thus, if we don’t want to be fanatical about accuracy, we can write : p ≈ 0.8: bary = 0.8η: γ = 0.8η 0.243 /1 ¯ /c 3 ¸ ¸ . (10.29) Substituting equation (10.29) into equation (10.28), we find that the deuterium- to-proton ratio is a relatively simple function of temperature: : D : n ≈ 6.5η /1 : n c 2 3/2 exp 1 D /1 . (10.30) This function is plotted in Figure 10.3, assuming a baryon-to-photon ratio of η = 5.5 10 −10 . The temperature 1 nuc of deuterium nucleosynthesis can be found by solving the equation 1 ≈ 6.5η /1 nuc : n c 2 3/2 exp 1 D /1 nuc . (10.31) 10.3. DEUTERIUM SYNTHESIS 221 .9 .8 .7 .6 .0001 .001 .01 .1 1 10 100 1000 10000 T (10 9 K) n D / n n Figure 10.3: The deuterium-to-neutron ratio during the epoch of deuterium synthesis. The nucleosynthetic equivalent of the Saha equation (equa- tion 10.27) is assumed to hold true. 222 CHAPTER 10. NUCLEOSYNTHESIS & THE EARLY UNIVERSE With : n c 2 = 939.6 MeV, 1 D = 2.22 MeV, and η = 5.5 10 −10 , the tem- perature of deuterium synthesis is /1 nuc ≈ 0.066 MeV, corresponding to 1 nuc ≈ 7.6 10 8 K. The temperature drops to this value when the age of the universe is t nuc ≈ 200 s. Note that the time delay until the start of nucleosynthesis, t nuc ≈ 200 s, is not negligible compared to the decay time of the neutron, τ n = 890 s. By the time nucleosynthesis actually gets underway, neutron decay has slightly decreased the neutron-to-proton ratio from : n : p = 15 to : n : p ≈ exp(−200890) 5 + [1 −exp(−200890)] ≈ 0.8 5.2 ≈ 0.15 . (10.32) This in turn lowers the maximum possible 4 He mass fraction from) max ≈ 0.33 to ) max ≈ 0.27. 10.4 Beyond deuterium The deuterium-to-neutron ratio : D : n does not remain indefinitely at the equilibrium value given by equation (10.30). Once a significant amount of deuterium forms, there are many possible nuclear reactions available. For instance, a deuterium nucleus can fuse with a proton to form 3 He: D +j ÷ ÷ 3 He +γ . (10.33) Alternatively, it can fuse with a neutron to form 3 H, also known as “tritium”: D +: ÷ ÷ 3 H +γ . (10.34) Tritium is unstable; it spontaneously decays to 3 He, emitting an electron and an electron anti-neutrino in the process. However, the decay time of tritium is approximately 18 years; during the brief time that Big Bang Nucleosynthesis lasts, tritium can be regarded as effectively stable. Deuterium nuclei can also fuse with each other to form 4 He: D + D ÷ ÷ 4 He +γ . (10.35) However, it is more likely that the interaction of two deuterium nuclei will end in the formation of a tritium nucleus (with the emission of a proton), D + D ÷ ÷ 3 H +j . (10.36) 10.4. BEYOND DEUTERIUM 223 or the formation of a 3 He nucleus (with the emission of a neutron), D + D ÷ ÷ 3 He +: . (10.37) There is never a large amount of 3 H or 3 He present during the time of nu- cleosynthesis. Soon after they are formed, they are converted to 4 He by reactions such as 3 H +j ÷ ÷ 4 He +γ 3 He +: ÷ ÷ 4 He +γ 3 H + D ÷ ÷ 4 He +: 3 He + D ÷ ÷ 4 He +j . (10.38) None of the post-deuterium reactions outlined in equations (10.33) through (10.38) involves neutrinos; they are all involve the strong nuclear force, and have large cross-sections and fast reaction rates. Thus, once nucleosynthesis begins, D, 3 H, and 3 He are all efficiently converted to 4 He. Once 4 He is reached, however, the orderly march of nucleosynthesis to heavier and heavier nuclei reaches a roadblock. For such a light nucleus, 4 He is exceptionally tightly bound, as illustrated in Figure 10.1. By contrast, there are no stable nuclei with ¹ = 5. If you try to fuse a proton or neutron to 4 He, it won’t work; 5 He and 5 Li are not stable nuclei. Thus, 4 He is resistant to fusion with protons and neutrons. Small amounts of 6 Li and 7 Li, the two stable isotopes of lithium, are made by reactions such as 4 He + D ÷ ÷ 6 Li +γ (10.39) and 4 He + 3 H ÷ ÷ 7 Li +γ . (10.40) In addition, small amounts of 7 Be are made by reactions such as 4 He + 3 He ÷ ÷ 7 Be +γ . (10.41) The synthesis of nuclei with ¹ 7 is hindered by the absence of stable nuclei with ¹ = 8. For instance, if 8 Be is made by the reaction 4 He + 4 He → 8 Be . (10.42) the 8 Be nucleus falls back apart into a pair of 4 He nuclei with a decay time of only τ = 3 10 −16 s. 224 CHAPTER 10. NUCLEOSYNTHESIS & THE EARLY UNIVERSE Figure 10.4: Mass fraction of nuclei as a function of time during the epoch of nucleosynthesis. A baryon-to-photon ratio of η = 5.1 10 −10 is assumed (from Tytler et al. (2000) Physica Scripta, T85, 12). The bottom line is that once deuterium begins to be formed, fusion up to the tightly bound 4 He nucleus proceeds very rapidly. Fusion of heavier nuclei occurs much less rapidly. The precise yields of the different isotopes involved in BBN are customarily calculated using a fairly complex computer code. The complexity is necessary because of the large number of possible reactions which can occur once deuterium has been formed, all of which have temperature-dependent cross-sections. Thus, there’s a good deal of bookkeeping involved. The results of a typical BBN code, which follows the mass fraction of different isotopes as the universe expands and cools, is shown in Figure 10.4. Initially, at 1 10 9 K, almost all the baryonic matter is in the form of free protons and neutrons. As the deuterium density climbs upward, however, the point is eventually reached where significant amounts of 3 H, 3 He, and 4 He are formed. By the time the temperature has dropped to 1 ∼ 4 10 8 K, at t ∼ 10 min, Big Bang Nucleosynthesis is essentially over. Nearly all the baryons are in the form of free protons or 4 He nuclei. 10.4. BEYOND DEUTERIUM 225 The small residue of free neutrons decays into protons. Small amounts of D, 3 H, and 3 He are left over, a tribute to the incomplete nature of Big Bang Nucleosynthesis. (The 3 H later decays to 3 He.) Very small amounts of 6 Li, 7 Li, and 7 Be are made. (The 7 Be is later converted to 7 Li by electron capture: 7 Be +c − → 7 Li +ν e .) The yields of D, 3 He, 4 He, 6 Li, and 7 Li depend on various physical pa- rameters. Most importantly, they depend on the baryon-to-photon ratio η. A high baryon-to-photon ratio increases the temperature 1 nuc at which deuterium synthesis occurs, and hence gives an earlier start to Big Bang Nucleosynthesis. Since BBN is a race against the clock as the density and temperature of the universe drop, getting an earlier start means that nu- cleosynthesis is more efficient at producing 4 He, leaving less D and 3 He as “leftovers”. A plot of the mass fraction of various elements produced by Big Bang Nucleosynthesis is shown in Figure 10.5. Note that larger values of η produce larger values for ) p (the 4 He mass fraction) and smaller values for the deuterium density, as explained above. The dependence of the 7 Li density on η is more complicated. Within the range of η plotted in Figure 10.5, the direct production of 7 Li by the fusion of 4 He and 3 H is a decreasing function of η, while the indirect production of 7 Li by 7 Be electron capture is an in- creasing function of η. The net result is a minimum in the predicted density of 7 Li at η ≈ 3 10 −10 . Broadly speaking, we know immediately that the baryon-to-photon ratio can’t be as small as η ∼ 10 −12 . If it were, BBN would be extremely inef- ficient, and we would expect only tiny amounts of helium to be produced () p < 0.01). Conversely, we know that the baryon-to-photon ratio can’t be as large as η ∼ 10 −7 . If it were, nucleosynthesis would have taken place very early (before neutrons had a chance to decay), the universe would be essen- tially deuterium-free, and ) p would be near its maximum permissible value of ) max ≈ 0.33. Pinning down the value of η more accurately requires making accurate observations of the primordial densities of the light elements; that is, the densities before nucleosynthesis in stars started to alter the chemical composition of the universe. In determining the value of η, it is most useful to determine the primordial abundance of deuterium. This is because the deuterium abundance is strongly dependent on η in the range of interest. Thus, determining the deuterium abundance with only modest accuracy will enable us to determine η fairly well. By contrast, the primordial helium frac- tion, ) p , has only a weak dependence on η for the range of interest, as shown in Figure 10.5. Thus, determining η with a fair degree of accuracy would 226 CHAPTER 10. NUCLEOSYNTHESIS & THE EARLY UNIVERSE Figure 10.5: The mass fraction of 4 He, and the number densities of D, D+ 3 He, and 7 Li expressed as a fraction of the H number density. The width of each line represents the 95% confidence interval in the density (from Burles, Nollett, and Turner (2001) ApJ, 552, L1). 10.4. BEYOND DEUTERIUM 227 require measuring ) p with fanatic precision. A major problem in determining the primordial deuterium abundance is that deuterium is very easily destroyed in stars. When an interstellar gas cloud collapses gravitationally to form a star, the first fusion reactions that occur involve the fusion of deuterium into helium. Thus, the abun- dance of deuterium in the universe tends to decrease with time. 4 Deuterium abundances are customarily given as the ratio of the number of deuterium atoms to the number of hydrogen atoms (D/H). For instance, in the local interstellar medium within our Galaxy, astronomers find an average value DH ≈ 1.6 10 −5 ; that is, there’s one deuterium atom for every 60,000 or- dinary hydrogen atoms. However, the Sun and the interstellar medium are contaminated with material that has been cycled through stellar interiors. Thus, we expect the primordial deuterium-to-hydrogen value to have been (DH) p 1.6 10 −5 . Currently, the most promising way to find the primordial value of D/H is to look at the spectra of distant quasars. We don’t care about the deuterium within the quasar itself; instead, we just want to use the quasar as a flashlight to illuminate the intergalactic gas clouds which lie between it and us. If an intergalactic gas cloud contains no detectable stars, and has very low levels of elements heavier than lithium, we can hope that its D/H value is close to the primordial value, and hasn’t been driven downward by the effects of fusion within stars. Neutral hydrogen atoms within these intergalactic clouds will absorb photons whose energy corresponds to the Lyman-α transition; that is, the transition of the atom’s electron from the ground state (: = 1) to the next higher energy level (: = 2). In an ordinary hydrogen atom ( 1 H), the Lyman-α transition corresponds to a wavelength λ H = 121.57 nm. In a deuterium atom, the greater mass of the nucleus causes a small isotopic shift in the electron’s energy levels. As a consequence, the Lyman-α transition in deuterium corresponds to a slightly shorter wavelength, λ D = 121.54 nm. When we look at light from a quasar which has passed through an intergalac- tic cloud at redshift . cl , we will see a strong absorption line at λ H (1 + . cl ), due to absorption from ordinary hydrogen, and a much weaker absorption line at λ D (1 + . cl ), due to absorption from deuterium. Detailed studies of the strength of the absorption lines in the spectra of different quasars give results which are consistent with the ratio (DH) = (3.0 ±0.4) 10 −5 . This 4 There are no mechanisms known that will create deuterium in significant amounts after Big Bang Nucleosynthesis is complete. 228 CHAPTER 10. NUCLEOSYNTHESIS & THE EARLY UNIVERSE translates into a baryon-to-photon ratio of η = (5.5 ±0.5) 10 −10 . The value of η can be converted into a value for the current baryon density by the relation : bary,0 = η: γ,0 = 0.23 ±0.02 m −3 . (10.43) Since most of the baryons are protons, we can write, to acceptable accuracy, ε bary,0 = (: p c 2 ): bary,0 = 210 ±20 MeVm −3 . (10.44) The current density parameter in baryons is thus Ω bary,0 = ε bary,0 ε c,0 = 210 ±20 MeVm −3 5200 ±1000 MeVm −3 = 0.04 ±0.01 . (10.45) Note that the largest source of uncertainty in the value of Ω bary,0 is not the uncertainty in the baryon density, but the uncertainty in the critical density (which in turn results from the fact that H 0 is not particularly well determined). 10.5 Baryon – antibaryon asymmetry The results of Big Bang Nucleosynthesis tell us what the universe was like when it was relatively hot (1 nuc ≈ 7 10 8 K) and dense: ε nuc ≈ α1 4 nuc ≈ 10 33 MeVm −3 . (10.46) This energy density corresponds to a mass density of ε nuc c 2 ≈ 2000 kg m −3 , or about twice the density of water. Remember, though, that the energy density at the time of BBN was almost entirely in the form of radiation. The mass density of baryons at the time of BBN was ρ bary (t nuc ) = Ω bary,0 ρ c,0 1 nuc 1 0 3 (10.47) ≈ (0.04)(9.2 10 −27 kg m −3 ) 7 10 8 2.725 3 ≈ 0.007 kg m −3 . A density of several grams per cubic meter is not outlandishly high, by ev- eryday standards; it’s equal to the density of the Earth’s stratosphere. A mean photon energy of 2.7/1 nuc ≈ 0.2 MeV is not outlandishly high, by ev- eryday standards; you are bombarded with photons of about that energy 10.5. BARYON – ANTIBARYON ASYMMETRY 229 when you have your teeth X-rayed at the dentist. The physics of Big Bang Nucleosynthesis is well understood. Some of the initial conditions for Big Bang Nucleosynthesis, however, are rather puzzling. The baryon-to-photon ratio, η ≈ 5.5 10 −10 , is a remark- ably small number; the universe seems to have a strong preference for photons over baryons. It’s also worthy of remark that the universe seems to have a strong preference for baryons over antibaryons. The laws of physics demand the presence of antiprotons (¯ j), containing two “anti-up” quarks and one “anti-down” quark apiece, as well as antineutrons (¯ :), containing one “anti- up” quark and two “anti-down” quarks apiece. 5 In practice, though, it is found that the universe has an extremely large excess of protons and neu- trons over antiprotons and antineutrons (and hence an excess of quarks over antiquarks). At the time of Big Bang Nucleosynthesis, the number density of antibaryons (¯ : and ¯ j) was tiny compared to the number density of baryons, which in turn was tiny compared to the number density of photons. This imbalance, : antibary < : bary < : γ , has its origin in the physics of the very early universe. When the temperature of the early universe was greater than /1 ≈ 150 MeV, the quarks which it contained were not confined within baryons and other particles, as they are today, but formed a sea of free quarks (some- times referred to by the oddly culinary name of “quark soup”). During the first few microseconds of the universe, when the quark soup was piping hot, quarks and antiquarks were constantly being created by pair production and destroyed by mutual annihilation: γ +γ ÷ ÷ ¡ + ¯ ¡ . (10.48) where ¡ and ¯ ¡ could represent, for instance, an “up” quark and an “anti-up” quark, or a “down” quark and an “anti-down” quark. During this period of quark pair production, the numbers of “up” quarks, “anti-up” quarks, “down” quarks, “anti-down” quarks, and photons were nearly equal to each other. However, suppose there were a very tiny asymmetry between quarks and antiquarks, such that δ q ≡ : q −: ¯ q : q +: ¯ q <1 . (10.49) 5 Note that an “anti-up” quark is not the same as a “down” quark; nor is “anti-down” equivalent to “up”. 230 CHAPTER 10. NUCLEOSYNTHESIS & THE EARLY UNIVERSE As the universe expanded and the quark soup cooled, quark-antiquark pairs would no longer be produced. The existing antiquarks would then annihi- late with the quarks. However, because of the small excess of quarks over antiquarks, there would be a residue of quarks with number density : q : γ ∼ δ q . (10.50) Thus, if there were 1,000,000,000 quarks for every 999,999,997 antiquarks in the early universe, three lucky quarks in a billion would be left over after the others encountered antiquarks and were annihilated. The leftover quarks, however, would be surrounded by roughly 2 billion photons, the product of the annihilations. After the three quarks were bound together into a baryon at /1 ≈ 150 MeV, the resulting baryon-to-photon ratio would be η ∼ 5 10 −10 . Thus, the very strong asymmetry between baryons and antibaryons today and the large number of photons per baryon are both products of a tiny asymmetry between quarks and antiquarks in the early universe. The exact origin of the quark-antiquark asymmetry in the early universe is still not exactly known. The physicist Andrei Sakharov, as far back as the year 1967, was the first to outline the necessary physical conditions for producing a small asymmetry; however, the precise mechanism by which the quarks first developed their few-parts-per-billion advantage over antiquarks still remains to be found. Suggested reading [Full references are given in the “Annotated Bibliography” on page 286.] Bernstein (1995), ch. 4, 5, 6: Puts nucleosynthesis in the larger context of particle physics and thermodynamics. Weinberg (1993) The First Three Minutes: Gives some of the histor- ical background to the development of Big Bang Nucleosynthesis. Problems (10.1) Suppose the neutron decay time were τ n = 89 s instead of τ n = 890 s, with all other physical parameters unchanged. Estimate ) max , 10.5. BARYON – ANTIBARYON ASYMMETRY 231 the maximum possible mass fraction in 4 He, assuming that all available neutrons are incorporated into 4 He nuclei. (10.2) Suppose the difference in rest energy of the neutron and proton were 1 n = (: n − : p )c 2 = 0.129 MeV instead of 1 n = 1.29 MeV, with all other physical parameters unchanged. Estimate ) max , the maximum possible mass fraction in 4 He, assuming that all available neutrons are incorporated into 4 He nuclei. (10.3) A fascinating bit of cosmological history is that of George Gamow’s prediction of the Cosmic Microwave Background in 1948. (Unfortu- nately, his prediction was premature; by the time the CMB was actually discovered in the 1960’s, his prediction had fallen into obscurity.) Let’s see if you can reproduce Gamow’s line of argument. Gamow knew that nucleosynthesis must have taken place at a temperature 1 nuc ≈ 10 9 K, and that the age of the universe is currently t 0 ≈ 10 Gyr. Assume that the universe is flat and contains only radiation. With these assumptions, what was the energy density ε at the time of nucleosyn- thesis? What was the Hubble parameter H at the time of nucleosyn- thesis? What was the time t nuc at which nucleosynthesis took place? What is the current temperature 1 0 of the radiation filling the uni- verse today? If the universe switched from being radiation-dominated to being matter-dominated at a redshift . rm 0, will this increase or decrease 1 0 for fixed values of 1 nuc and t 0 ? Explain your answer. (10.4) The total luminosity of the stars in our Galaxy is 1 ≈ 1.4 10 10 L . Suppose that the luminosity of our Galaxy has been constant for the past 10 Gyr. How much energy has our Galaxy emitted in the form of starlight during that time? Most stars are powered by the fusion of H into 4 He, with the release of 28.4 MeV for every helium nucleus formed. How many helium nuclei have been created within stars in our Galaxy over the course of the past 10 Gyr, assuming that the fusion of H into 4 He is the only significant energy source? If the baryonic mass of our Galaxy is ` ≈ 10 11 M , by what amount has the helium fraction ) of our Galaxy been increased over its primordial value ) 4 = 0.24? (10.5) In section 10.2, it is asserted that the maximum possible value of the 232 CHAPTER 10. NUCLEOSYNTHESIS & THE EARLY UNIVERSE primordial helium fraction is ) max = 21 1 +1 . (10.51) where 1 = : n : p ≤ 1 is the neutron-to-proton ratio at the time of nucleosynthesis. Prove that this assertion is true. (10.6) The typical energy of a neutrino in the Cosmic Neutrino Background, as pointed out in Chapter 5, is 1 ν ∼ /1 ν ∼ 5 10 −4 eV. What is the approximate interaction cross-section σ w for one of these cos- mic neutrinos? Suppose you had a large lump of 56 Fe (with density ρ = 7900 kg m −3 ). What is the number density of protons, neutrons, and electrons within the lump of iron? How far, on average, would a cosmic neutrino travel through the iron before interacting with a proton, neutron, or electron? (Assume that the cross-section for in- teraction is simply σ w , regardless of the type of particle the neutrino interacts with.) Chapter 11 Inflation & the Very Early Universe The observed properties of galaxies, quasars, and supernovae at relatively small redshift (. < 6) tell us about the universe at times t 1 Gyr. The properties of the Cosmic Microwave Background tell us about the universe at the time of last scattering (. ls ≈ 1100, t ls ≈ 0.35 Myr). The abundances of light elements such as deuterium and helium tell us about the universe at the time of Big Bang Nucleosynthesis (. nuc ≈ 3 10 8 , t nuc ≈ 3 min). In fact, the observation that primordial gas clouds are roughly one-fourth helium by mass, rather than being all helium or all hydrogen, tells us that we have a fair understanding of what was happening at the time of neutron-proton freezeout (. freeze ≈ 4 10 9 , t freeze ≈ 1 s). So far, my description of the Hot Big Bang scenario has been a triumphal progress, with only minor details (such as the exact determination of H 0 , Ω m,0 , and Ω Λ,0 ) remaining to be worked out. Whenever a conquering gen- eral made a triumphal progress into ancient Rome, a slave stood behind him, whispering in his ear, “Remember, you are mortal.” Just as every triumphant general must be reminded of his flawed, imperfect nature, lest he become in- sufferably arrogant, so every triumphant theory should be carefully inspected for flaws and imperfections. The standard Hot Big Bang scenario, in which the early universe was radiation-dominated, has three underlying problems, called the flatness problem, the horizon problem, and the monopole problem. The flatness problem can be summarized by the statement, “The universe is nearly flat today, and was even flatter in the past.” The horizon problem can be summarized by the statement, “The universe is nearly isotropic and 233 234 CHAPTER 11. INFLATION & THE VERY EARLY UNIVERSE homogeneous today, and was even more so in the past.” The monopole prob- lem can be summarized by the statement, “The universe is apparently free of magnetic monopoles.” To see why these simple statements pose a problem to the standard Hot Big Bang scenario, it is necessary to go a little deeper into the physics of the expanding universe. 11.1 The flatness problem Let’s start by examining the flatness problem. The spatial curvature of the universe is related to the density parameter Ω by the Friedmann equation: 1 −Ω(t) = − κc 2 1 2 0 c(t) 2 H(t) 2 . (11.1) (Here, I am using the Friedmann equation in the form given by equation 4.29.) At the present moment, the density parameter and curvature are linked by the equation 1 −Ω 0 = − κc 2 1 2 0 H 2 0 . (11.2) The results of the type Ia supernova observations and the measurements of the CMB anisotropy are consistent with the value (see Figure 9.7) [1 −Ω 0 [ ≤ 0.2 . (11.3) Why should the value of Ω 0 be so close to one today? It could have had, for instance, the value Ω 0 = 10 −6 or Ω 0 = 10 6 without violating any laws of physics. We could, of course, invoke coincidence by saying that the initial conditions for the universe just happened to produce Ω 0 ≈ 1 today. After all, Ω 0 = 0.8 or Ω 0 = 1.2 aren’t that close to one. However, when you extrapolate the value of Ω(t) backward into the past, the closeness of Ω to unity becomes more and more difficult to dismiss as a coincidence. By combining equations (11.1) and (11.2), we find the equation that gives the density parameter as a function of time: 1 −Ω(t) = H 2 0 (1 −Ω 0 ) H(t) 2 c(t) 2 . (11.4) When the universe was dominated by radiation and matter, at times t < t mΛ ≈ 10 Gyr, the Hubble parameter was given by equation (6.35): H(t) 2 H 2 0 = Ω r,0 c 4 + Ω m,0 c 3 . (11.5) 11.1. THE FLATNESS PROBLEM 235 Thus, the density parameter evolved at the rate 1 −Ω(t) = (1 −Ω 0 )c 2 Ω r,0 +cΩ m,0 . (11.6) During the period when the universe was dominated by radiation and matter, the deviation of Ω from one was constantly growing. During the radiation- dominated phase, [1 −Ω[ r ∝ c 2 ∝ t . (11.7) During the later matter-dominated phase, [1 −Ω[ m ∝ c ∝ t 2/3 . (11.8) Suppose, as the available evidence indicates, that the universe is described by a model close to the Benchmark Model, with Ω r,0 = 8.4 10 −5 , Ω m,0 = 0.3 ± 0.1, and Ω Λ,0 = 0.7 ± 0.1. At the present, therefore, the density parameter falls within the limits [1−Ω 0 [ ≤ 0.2. At the time of radiation-matter equality, the density parameter Ω rm was equal to one with an accuracy [1 −Ω rm [ ≤ 2 10 −4 . (11.9) If we extrapolate backward to the time of Big Bang Nucleosynthesis, at c nuc ≈ 3.6 10 −8 , the deviation of the density parameter Ω nuc from one was only [1 −Ω nuc [ ≤ 3 10 −14 . (11.10) At the time that deuterium was forming, the density of the universe was equal to the critical density to an accuracy of one part in 30 trillion. If we push our extrapolation as far back as we dare, to the Planck time at t P ≈ 5 10 −44 s, c P ≈ 2 10 −32 , we find that the density parameter Ω P was extraordinarily close to one: [1 −Ω P [ ≤ 1 10 −60 . (11.11) The number 10 −60 is, of course, very tiny. To make an analogy, in order to change the Sun’s mass by one part in 10 60 , you would have to add or subtract two electrons. Our very existence depends on the fanatically close balance between the actual density and the critical density in the early universe. If, for instance, the deviation of Ω from one at the time of nucleosynthesis had been one part in 30 thousand instead of one part in 30 trillion, the universe would have collapsed in a Big Crunch or expanded to a low-density Big Bore 236 CHAPTER 11. INFLATION & THE VERY EARLY UNIVERSE after only a few years. In that case, galaxies, stars, planets, and cosmologists would not have had time to form. You might try to dismiss the extreme flatness of the the early universe as a coincidence, by saying, “Ω P might have had any value, but it just happened to be 1 ± 10 −60 .” However, a coincidence at the level of one part in 10 60 is extremely far-fetched. It would be far more satisfactory if we could find a physical mechanism for flattening the universe early in its history, instead of relying on extremely contrived initial conditions at the Planck time. 11.2 The horizon problem The “flatness problem”, the remarkable closeness of Ω to one in the early universe, is puzzling. It is accompanied, however, by the “horizon problem”, which is, if anything, even more puzzling. The “horizon problem” is simply the statement that the universe is nearly homogeneous and isotropic on very large scales. Why should we regard this as a problem? So far, we’ve treated the homogeneity and isotropy of the universe as a blessing rather than a curse. It’s the homogeneity and isotropy of the universe, after all, which permit us to describe its curvature by the relatively simple Robertson-Walker metric, and its expansion by the relatively simple Friedmann equation. If the universe were inhomogeneous and anisotropic on large scales, it would be much more difficult to describe mathematically. The universe, however, is under no obligation to make things simple for cosmologists. To see why the large scale homogeneity and isotropy of the universe is so unexpected in the standard Hot Big Bang scenario, consider two antipodal points on the last scattering surface, as illustrated in Figure 11.1. The current proper distance to the last scattering surface is d p (t 0 ) = c t 0 t ls dt c(t) . (11.12) Since the last scattering of the CMB photons occurred a long time ago (t ls < t 0 ), the current proper distance to the last scattering surface is only slightly smaller than the current horizon distance. For the Bench- mark Model, the current proper distance to the last scattering surface is d p (t 0 ) = 0.98d hor (t 0 ). Thus, two antipodal points on the last scattering sur- face, separated by 180 ◦ as seen by an observer on Earth, are currently sep- arated by a proper distance of 1.96d hor (t 0 ). Since the two points are farther 11.2. THE HORIZON PROBLEM 237 Figure 11.1: In the standard Hot Big Bang scenario, the current proper distance to the last scattering surface is 98% of the current horizon distance. apart than the horizon distance, they are causally disconnected. That is, they haven’t had time to send messages to each other, and in particular, haven’t had time to come into thermal equilibrium with each other. Nevertheless, the two points have the same temperature (once the dipole distortion is sub- tracted) to within one part in 10 5 . Why? How can two points that haven’t had time to swap information be so nearly identical in their properties? The near-isotropy of the Cosmic Microwave Background is still more re- markable when it is recalled that the temperature fluctuations in the CMB result from the density and velocity fluctuations that existed at the time of last scattering. In the standard Hot Big Bang scenario, the universe was matter-dominated at the time of last scattering, so the horizon distance at that time can be approximated by the value d hor (t ls ) = 2 c H(t ls ) (11.13) appropriate to a flat, matter-only universe (see section 5.4). Since the Hubble distance at the time of last scattering was cH(t ls ) ≈ 0.2 Mpc, the horizon distance at that time was only d hor (t ls ) ≈ 0.4 Mpc. Thus, points more than 238 CHAPTER 11. INFLATION & THE VERY EARLY UNIVERSE 0.4 megaparsecs apart at the time of last scattering were not in causal contact in the standard Hot Big Bang scenario. The angular-diameter distance to the last scattering surface is d A ≈ 13 Mpc, as computed in section 9.4. Thus, points on the last scattering surface that were separated by a horizon distance will have an angular separation equal to θ hor = d hor (t ls ) d A ≈ 0.4 Mpc 13 Mpc ≈ 0.03 rad ≈ 2 ◦ (11.14) as seen from the Earth today. Therefore, points on the last scattering surface separated by an angle as small as ∼ 2 ◦ were out of contact with each other at the time the temperature fluctuations were stamped upon the CMB. Never- theless, we find that δ11 is as small as 10 −5 on scales θ 2 ◦ (corresponding to | < 100 in Figure 9.6). Why should regions which were out of causal contact with each other at t ls have been so nearly homogeneous in their properties? Invoking coinci- dence (“The different patches just happened to have the same temperature”) requires a great stretch of the imagination. The surface of last scattering can be divided into some 20,000 patches, each two degrees across. In the stan- dard Hot Big Bang scenario, the center of each of these patches was out of touch with the other patches at the time of last scattering. Now, if you invite two people to a potluck dinner, and they both bring potato salad, you can dismiss that as coincidence, even if they had 10 5 different dishes to chose from. However, if you invite 20,000 people to a potluck dinner, and they all bring potato salad, it starts to dawn on you that they must have been in contact with each other: “Psst...let’s all bring potato salad. Pass it on.” Similarly, it starts to dawn on you that the different patches of the last scattering surface, in order to be so nearly equal in temperature, must have been in contact with each other: “Psst...let’s all be at 1 = 2.725 K when the universe is 13.5 gigayears old. Pass it on.” 11.3 The monopole problem The monopole problem – that is, the apparent lack of magnetic monopoles in the universe – is not a purely cosmological problem, but one that results from combining the Hot Big Bang scenario with the particle physics concept of a Grand Unified Theory. In particle physics, a Grand Unified Theory, or GUT, is a field theory which attempts to unify the electromagnetic force, the weak 11.3. THE MONOPOLE PROBLEM 239 nuclear force, and the strong nuclear force. Unification of forces has been a goal of scientists since the 1870s, when James Clerk Maxwell demonstrated that electricity and magnetism are both manifestations of a single underlying electromagnetic field. Currently, it is customary to speak of the four funda- mental forces of nature: gravitational, electromagnetic, weak, and strong. In the view of many physicists, though, four forces are three too many; they’ve spent much time and effort to show that two or more of the “fundamental forces” are actually different aspects of a single underlying force. About a century after Maxwell, Steven Weinberg, Abdus Salam, and Sheldon Glashow successfully devised an electroweak theory. They demonstrated that at par- ticle energies greater than 1 ew ∼ 1 TeV, the electromagnetic force and the weak force unite to form a single “electroweak” force. The electroweak en- ergy of 1 ew ∼ 1 TeV corresponds to a temperature 1 ew ∼ 1 ew / ∼ 10 16 K; the universe had this temperature when its age was t ew ∼ 10 −12 s. Thus, when the universe was less than a picosecond old, there were only three fun- damental forces: the gravitational, strong, and electroweak force. When the predictions of the electroweak energy were confirmed experimentally, Wein- berg, Salam, and Glashow toted home their Nobel Prizes, and physicists braced themselves for the next step: unifying the electroweak force with the strong force. By extrapolating the known properties of the strong and electroweak forces to higher particle energies, physicists estimate that at an energy 1 GUT of roughly 10 12 → 10 13 TeV, the strong and electroweak forces should be unified as a single Grand Unified Force. If the GUT energy is 1 GUT ∼ 10 12 TeV, this corresponds to a temperature 1 GUT ∼ 10 28 K and an age for the universe of t GUT ∼ 10 −36 s. The GUT energy is about four orders of magnitude smaller than the Planck energy, 1 P ∼ 10 16 TeV. Physicists are searching for a Theory of Everything (TOE) which describes how the Grand Unified Force and the force of gravity ultimately unite to form a single unified force at the Planck scale. The different unification energy scales, and the corresponding temperatures and times in the early universe, are shown in Figure 11.2. One of the predictions of Grand Unified Theories is that the universe underwent a phase transition as the temperature dropped below the GUT temperature. Generally speaking, phase transitions are associated with a spontaneous loss of symmetry as the temperature of a system is lowered. Take, as an example, the phase transition known as “freezing water”. At temperatures 1 273 K, water is liquid. Individual water molecules are 240 CHAPTER 11. INFLATION & THE VERY EARLY UNIVERSE ?? TOE GUT ew gravity strong weak electromagnetic 10 16 10 32 10 −43 10 12 10 28 10 −36 1 10 16 10 −12 E (TeV) T (K) t (sec) Figure 11.2: The energy, temperature, and time scales at which the different force unifications occur. randomly oriented, and the liquid water thus has rotational symmetry about any point; in other words, it is isotropic. However, when the temperature drops below 1 = 273 K, the water undergoes a phase transition, from liq- uid to solid, and the rotational symmetry of the water is lost. The water molecules are locked into a crystalline structure, and the ice no longer has rotational symmetry about an arbitrary point. In other words, the ice crys- tal is anisotropic, with preferred directions corresponding to the crystal’s axes of symmetry. 1 In a broadly similar vein, there is a loss of symmetry when the universe undergoes the GUT phase transition at t GUT ∼ 10 −36 s. At 1 1 GUT , there was a symmetry between the strong and electroweak forces. At 1 < 1 GUT , the symmetry is spontaneously lost; the strong and electroweak forces begin to behave quite differently from each other. In general, phase transitions associated with a loss of symmetry give rise to flaws known as topological defects. To see how topological defects form, consider a large tub of water which is cooled below 1 = 273 K. Usually, the freezing of the water will start at two or more widely separated nucle- ation sites. The crystal which forms about any given nucleation site is very regular, with well-defined axes of symmetry. However, the axes of symme- 1 Suppose, for instance, that the water freezes in the familiar six-pointed form of a snowflake. It is now only symmetric with respect to rotations of 60 ◦ (or integral multiples of that angle) about the snowflake’s center. 11.3. THE MONOPOLE PROBLEM 241 try of two adjacent ice crystals will be misaligned. At the boundary of two adjacent crystals, there will be a two-dimensional topological defect, called a domain wall, where the axes of symmetry fail to line up. Other types of phase transitions give rise to one-dimensional, or line-like, topological de- fects (in a cosmological context, these linear defects are known as cosmic strings). Still other types of phase transitions give rise to zero-dimensional, or point-like, topological defects. Grand Unified Theories predict that the GUT phase transition creates point-like topological defects which act as mag- netic monopoles. That is, they act as the isolated north pole or south pole of a magnet. The rest energy of the magnetic monopoles created in the GUT phase transition is predicted to be : M c 2 ∼ 1 GUT ∼ 10 12 TeV. This corre- sponds to a mass of over a nanogram (comparable to that of a bacterium), which is a lot of mass for a single particle to be carrying around. At the time of the GUT phase transition, points further apart than the horizon size will be out of causal contact with each other. Thus, we expect roughly one topological defect per horizon volume, due to the mismatch of fields which are not causally linked. The number density of magnetic monopoles, at the time of their creation, would be : M (t GUT ) ∼ 1 (2ct GUT ) 3 ∼ 10 82 m −3 . (11.15) and their energy density would be ε M (t GUT ) ∼ (: M c 2 ): M ∼ 10 94 TeVm −3 . (11.16) This is a large energy density, but it is smaller by ten orders of magnitude than the energy density of radiation at the time of the GUT phase transition: ε γ (t GUT ) ≈ α1 4 GUT ∼ 10 104 TeVm −3 . (11.17) Thus, the magnetic monopoles wouldn’t have kept the universe from being radiation-dominated at the time of the GUT phase transition. However, the magnetic monopoles, being so massive, would soon have become highly non- relativistic, with energy density ε M ∝ c −3 . The energy density in radiation, though, was falling off at the rate ε γ ∝ c −4 . Thus, the magnetic monopoles would have dominated the energy density of the universe when the scale factor had grown by a factor ∼ 10 10 ; that is, when the temperature had fallen to 1 ∼ 10 −10 1 GUT ∼ 10 18 K, and the age of the universe was only t ∼ 10 −16 s. 242 CHAPTER 11. INFLATION & THE VERY EARLY UNIVERSE Obviously, the universe is not dominated by magnetic monopoles today. In fact, there is no strong evidence that they exist at all. Every north magnetic pole which we can find is paired with a south magnetic pole, and vice versa. There are no isolated north poles or isolated south poles. The monopole problem can thus be rephrased as the question, “Where have all the magnetic monopoles gone?” Now, you can always answer the ques- tion, “Where have the monopoles gone?” by saying, “There were never any monopoles to begin with.” There is not yet a single, definitive Grand Uni- fied Theory, and in some variants on the GUT theme, magnetic monopoles are not produced. However, the flatness and horizon problems are not so readily dismissed. When the physicist Alan Guth first proposed the idea of inflation in 1981, he introduced it as a way of resolving the flatness problem, the horizon problem, and the monopole problem with a single cosmological mechanism. 11.4 The inflation solution What is inflation? In a cosmological context, inflation can most generally be defined as the hypothesis that there was a period, early in the history of our universe, when the expansion was accelerating outward; that is, an epoch when ¨ c 0. The acceleration equation, ¨ c c = − 4πG 3c 2 (ε + 31) . (11.18) tells us that ¨ c 0 when 1 < −ε3. Thus, inflation would have taken place if the universe were temporarily dominated by a component with equation-of- state parameter u < −13. The usual implementation of inflation states that the universe was temporarily dominated by a positive cosmological constant Λ i (with u = −1), and thus had an acceleration equation that could be written in the form ¨ c c = Λ i 3 0 . (11.19) In an inflationary phase when the energy density was dominated by a cos- mological constant, the Friedmann equation was ˙ c c 2 = Λ i 3 . (11.20) 11.4. THE INFLATION SOLUTION 243 The Hubble constant H i during the inflationary phase was thus constant, with the value H i = (Λ i 3) 1/2 , and the scale factor grew exponentially with time: c(t) ∝ c H i t . (11.21) Too see how a period of exponential growth can resolve the flatness, horizon, and monopole problems, suppose that the universe had a period of expo- nential expansion sometime in the midst of its early, radiation-dominated phase. For simplicity, suppose the exponential growth was switched on in- stantaneously at a time t i , and lasted until some later time t f , when the exponential growth was switched off instantaneously, and the universe re- verted to its former state of radiation-dominated expansion. In this simple case, we can write the scale factor as c(t) = c i (tt i ) 1/2 t < t i c i c H i (t−t i ) t i < t < t f c i c H i (t f −t i ) (tt f ) 1/2 t t f (11.22) Thus, between the time t i when the exponential inflation began and the time t f when the inflation stopped, the scale factor increased by a factor c(t f ) c(t i ) = c N . (11.23) where `, the number of e-foldings of inflation, was ` ≡ H i (t f −t i ) . (11.24) If the duration of inflation, t f − t i , was long compared to the Hubble time during inflation, H −1 i , then ` was large, and the growth in scale factor during inflation was enormous. For concreteness, let’s take one possible model for inflation. This model states that exponential inflation started around the GUT time, t i ≈ t GUT ≈ 10 −36 s, with a Hubble parameter H i ≈ t −1 GUT ≈ 10 36 s −1 , and lasted for ` ∼ 100 Hubble times. In this particular model, the growth in scale factor during inflation was c(t f ) c(t i ) ∼ c 100 ∼ 10 43 . (11.25) Note that the cosmological constant Λ i present at the time of inflation was very large compared to the cosmological constant which seems to be present 244 CHAPTER 11. INFLATION & THE VERY EARLY UNIVERSE today. Currently, the evidence is consistent with an energy density in Λ of ε Λ,0 ≈ 0.7ε c,0 ≈ 0.004 TeVm −3 . To produce exponential expansion with a Hubble parameter H i ≈ 10 36 s −1 , the cosmological constant during inflation would have had an energy density (see equation 4.65) ε Λ i = c 2 8πG Λ i = 3c 2 8πG H 2 i ∼ 10 105 TeVm −3 . (11.26) over 107 orders of magnitude larger. How does inflation resolve the flatness problem? Equation (11.1), which gives Ω as a function of time, can be written in the form [1 −Ω(t)[ = c 2 1 2 0 c(t) 2 H(t) 2 (11.27) for any universe which is not perfectly flat. If the universe is dominated by a single component with equation-of-state parameter u = −1, then c ∝ t 2/(3+3w) , H ∝ t −1 , and [1 −Ω(t)[ ∝ t 2(1+3w)/(3+3w) . (11.28) Thus, if u < −13, the difference between Ω and one decreases with time. If the universe is expanding exponentially during the inflationary epoch, then [1 −Ω(t)[ ∝ c −2H i t . (11.29) and the difference between Ω and one decreases exponentially with time. If we compare the density parameter at the beginning of exponential inflation (t = t i ) with the density parameter at the end of inflation (t = t f = t i + `H i ), we find [1 −Ω(t f )[ = c −2N [1 −Ω(t i )[ . (11.30) Suppose that prior to inflation, the universe was actually fairly strongly curved, with [1 −Ω(t i )[ ∼ 1 . (11.31) After a hundred e-foldings of inflation, the deviation of Ω from one would be [1 −Ω(t f )[ ∼ c −2N ∼ c −200 ∼ 10 −87 . (11.32) Even if the universe at t i wasn’t particularly close to being flat, a hundred e- foldings of inflation would flatten it like the proverbial pancake. The current 11.4. THE INFLATION SOLUTION 245 limits on the density parameter, [1−Ω 0 [ ≤ 0.2, imply that ` 60, if inflation took place around the GUT time. However, it’s possible that ` may have been very much greater than 60, since the observational data are entirely consistent with [1 −Ω 0 [ <1. How does inflation resolve the horizon problem? At any time t, the horizon distance d hor (t) is given by the relation d hor (t) = c(t)c t 0 dt c(t) . (11.33) Prior to the inflationary period, the universe was radiation-dominated. Thus, the horizon time at the beginning of inflation was d hor (t i ) = c i c t i 0 dt c i (tt i ) 1/2 = 2ct i . (11.34) The horizon size at the end of inflation was d hor (t f ) = c i c N c t i 0 dt c i (tt i ) 1/2 + t f t i dt c i exp[H i (t −t i )] . (11.35) If `, the number of e-foldings of inflation, is large, then the horizon size at the end of inflation was d hor (t f ) = c N c(2t i +H −1 i ) . (11.36) An epoch of exponential inflation causes the horizon size to grow expo- nentially. If inflation started at t i ≈ 10 −36 s, with a Hubble parameter H i ≈ 10 36 s −1 , and lasted for : ≈ 100 e-foldings, then the horizon size imme- diately before inflation was d hor (t i ) = 2ct i ≈ 6 10 −28 m . (11.37) The horizon size immediately after inflation was d hor (t f ) ≈ c N 3ct i ≈ 2 10 16 m ≈ 0.8 pc . (11.38) During the brief period of ∼ 10 −34 s that inflation lasts in this model, the horizon size is boosted exponentially from submicroscopic scales to nearly a parsec. At the end of the inflationary epoch, the horizon size reverts to growing at a sedate linear rate. 246 CHAPTER 11. INFLATION & THE VERY EARLY UNIVERSE The net result of inflation is to increase the horizon length in the post- inflationary universe by a factor ∼ c N over what it would have been without inflation. For instance, we found that, in the absence of inflation, the horizon size at the time of last scattering was d hor (t ls ) ≈ 0.4 Mpc. Given a hundred e-foldings of inflation in the early universe, however, the horizon size at last scattering would have been ∼ 10 43 Mpc, obviously gargantuan enough for the entire last scattering surface to be in causal contact. To look at the resolution of the horizon problem from a slightly different viewpoint, consider the entire universe directly visible to us today; that is, the region bounded by the surface of last scattering. Currently, the proper distance to the surface of last scattering is d p (t 0 ) ≈ 1.4 10 4 Mpc . (11.39) If inflation ended at t f ∼ 10 −34 s, that corresponds to a scale factor c f ∼ 2 10 −27 . Thus, immediately after inflation, the portion of the universe currently visible to us was crammed into a sphere of proper radius d p (t f ) = c f d p (t 0 ) ∼ 3 10 −23 Mpc ∼ 0.9 m . (11.40) Immediately after inflation, in this model, all the mass-energy destined to become the hundreds of billions of galaxies we see today was enclosed within a sphere only six feet across. But, to quote Al Jolson, you ain’t heard nothin’ yet, folks. If there were ` = 100 e-foldings of inflation, then immediately prior to the inflationary epoch, the currently visible universe was enclosed within a sphere of proper radius d p (t i ) = c −N d p (t f ) ∼ 3 10 −44 m . (11.41) Note that this distance is 16 orders of magnitude smaller than the horizon size immediately prior to inflation (d hor (t i ) ∼ 610 −28 m.) Thus, the portion of the universe which we can see today had plenty of time to achieve thermal uniformity before inflation began. How does inflation resolve the monopole problem? If magnetic monopoles were created before or during inflation, then the number density of monopoles was diluted to any undetectably low level. During a period when the universe was expanding exponentially (c ∝ c H i t ), the number density of monopoles, if they were neither created nor destroyed, was decreasing exponentially (: M ∝ c −3H i t ). For instance, if inflation started around the GUT time, 11.5. THE PHYSICS OF INFLATION 247 when the number density of magnetic monopoles was : M (t GUT ) ≈ 10 82 m −3 , then after 100 e-foldings of inflation, the number density would have been : M (t f ) = c −300 : M (t GUT ) ≈ 5 10 −49 m −3 ≈ 15 pc −3 . The number density today, after the additional expansion from c(t f ) ≈ 210 −27 to c 0 = 1, would then be : M (t 0 ) ≈ 1 10 −61 Mpc −3 . The probability of finding even a single monopole within the last scattering surface would be astronomically small. 11.5 The physics of inflation Inflation explains some otherwise puzzling aspects of our universe, by flat- tening it, ensuring its homogeneity over large scales, and driving down the number density of magnetic monopoles which it contains. However, I have not yet answered many crucial questions about the inflationary epoch. What triggers inflation at t = t i , and (just as important) what turns it off at t = t f ? If inflation reduces the number density of monopoles to undetectably low levels, why doesn’t it reduce the number density of photons to undetectably low levels? If inflation is so efficient at flattening the global curvature of the universe, why doesn’t it also flatten out the local curvature due to fluc- tuations in the energy density? We know that the universe wasn’t perfectly homogeneous after inflation, because the Cosmic Microwave Background isn’t perfectly isotropic. To answer these questions, we will have to examine, at least in broad outline, the physics behind inflation. At present, there is not a consensus among cosmologists about the exact mechanism driving inflation. I will re- strict myself to speaking in general terms about one plausible mechanism for bringing about an inflationary epoch. Suppose the universe contains a scalar field φ(:. t) whose value can vary as a function of position and time. (Some early implementations of inflation associated the scalar field φ with the Higgs field which mediates interactions between particles at energies higher than the GUT energy; however, to keep the discussion general, let’s just call the field φ(:. t) the inflaton field.) Gen- erally speaking, a scalar field can have an associated potential energy \ (φ). 2 If φ has units of energy, and its potential \ has units of energy density, 2 As a simple illustrative example, suppose that the scalar field φ is the elevation above sea level at a given point on the Earth’s surface. The associated potential energy, in this case, is the gravitational potential V = gφ, where g = 9.8 kms −1 . 248 CHAPTER 11. INFLATION & THE VERY EARLY UNIVERSE then the energy density of the inflaton field is ε φ = 1 2 1 ¯ /c 3 ˙ φ 2 +\ (φ) (11.42) in a region of space where φ is homogeneous. The pressure of the inflaton field is 1 φ = 1 2 1 ¯ /c 3 ˙ φ 2 −\ (φ) . (11.43) If the inflaton field changes only very slowly as a function of time, with ˙ φ 2 <¯ /c 3 \ (φ) . (11.44) then the inflaton field acts like a cosmological constant, with ε φ ≈ −1 φ ≈ \ (φ) . (11.45) Thus, an inflaton field can drive exponential inflation if there is a temporary period when its rate of change ˙ φ is small (satisfying equation 11.44), and its potential \ (φ) is large enough to dominate the energy density of the universe. Under what circumstances are the conditions for inflation (small ˙ φ and large \ ) met in the early universe? To determine the value of ˙ φ, start with the fluid equation for the energy density of the inflaton field, ˙ ε φ + 3H(t)(ε φ +1 φ ) = 0 . (11.46) where H(t) = ˙ cc. Substituting from equations (11.42) and (11.43), we find the equation which governs the rate of change of φ: ¨ φ + 3H(t) ˙ φ = −¯ /c 3 d\ dφ . (11.47) Note that equation (11.47) mimics the equation of motion for a particle which is being accelerated by a force proportional to −d\dφ and being impeded by a frictional force proportional to the particle’s speed. Thus, the expansion of the universe provides a “Hubble friction” term, 3H ˙ φ, which slows the transition of the inflaton field to a value which will minimize the potential \ . Just as a skydiver reaches terminal velocity when the downward force of gravity is balanced by the upward force of air resistance, so the inflaton field can reach “terminal velocity” (with ¨ φ = 0) when 3H ˙ φ = −¯ /c 3 d\ dφ . (11.48) 11.5. THE PHYSICS OF INFLATION 249 or ˙ φ = − ¯ /c 3 3H d\ dφ . (11.49) If the inflaton field has reached this terminal velocity, then the requirement that ˙ φ 2 <¯ /c 3 \ , necessary if the inflaton field is to play the role of a cosmo- logical constant, translates into d\ dφ 2 < 9H 2 \ ¯ /c 3 . (11.50) If the universe is undergoing exponential inflation driven by the potential energy of the inflaton field, this means that the Hubble parameter is H = 8πGε φ 3c 2 1/2 = 8πG\ 3c 2 1/2 . (11.51) and equation (11.50) becomes d\ dφ 2 < 24πG\ 2 ¯ /c 5 . (11.52) which can also be written as 1 P \ d\ dφ 2 <1 . (11.53) where 1 P is the Planck energy. If the slope of the inflaton’s potential is sufficiently shallow, satisfying equation (11.53), and if the amplitude of the potential is sufficiently large to dominate the energy density of the universe, then the inflaton field is capable of giving rise to exponential expansion. As a concrete example of a potential \ (φ) which can give rise to inflation, consider the potential shown in Figure 11.3. The global minimum in the potential occurs when the value of the inflaton field is φ = φ 0 . Suppose, however, that the inflaton field starts at φ ≈ 0, where the potential is \ (φ) ≈ \ 0 . If d\ dφ 2 < \ 2 0 1 2 P (11.54) on the “plateau” where \ ≈ \ 0 , then while φ is slowly rolling toward φ 0 , the inflaton field contributes an energy density ε φ ≈ \ 0 ≈ constant to the universe. 250 CHAPTER 11. INFLATION & THE VERY EARLY UNIVERSE 0 φ 0 0 V 0 φ V Figure 11.3: A potential which can give rise to an inflationary epoch. The global minimum in \ (or “true vacuum”) is at φ = φ 0 . If the scalar field starts at φ = 0, it is in a “false vacuum” state. 11.5. THE PHYSICS OF INFLATION 251 When an inflaton field has a potential similar to that of Figure 11.3, it is referred to as being in a metastable false vacuum state when it is near the maximum at φ = 0. Such a state is not truly stable; if the inflaton field is nudged from φ = 0 to φ = +dφ, it will continue to slowly roll toward the true vacuum state at φ = φ 0 and \ = 0. However, if the plateau is sufficiently broad as well as sufficiently shallow, it can take many Hubble times for the inflaton field to roll down to the true vacuum state. Whether the inflaton field is dynamically significant during its transition from the false vacuum to the true vacuum depends on the value of \ 0 . As long as ε φ ≈ \ 0 is tiny compared to the energy density of radiation, ε r ∼ α1 4 , the contribution of the inflaton field to the Friedmann equation can be ignored. Exponential inflation, driven by the energy density of the inflaton field, will begin at a temperature 1 i ≈ \ 0 α 1/4 ≈ 2 10 28 K \ 0 10 105 TeVm −3 1/4 (11.55) or /1 i ≈ (¯ / 3 c 3 \ 0 ) 1/4 ≈ 2 10 12 TeV \ 0 10 105 TeVm −3 1/4 . (11.56) This corresponds to a time t i ≈ c 2 G\ 0 1/2 ≈ 3 10 −36 s \ 0 10 105 TeVm −3 −1/2 . (11.57) While the inflaton field is slowly rolling toward the true vacuum state, it produces exponential expansion, with a Hubble parameter H i ≈ 8πG\ 0 3c 2 1/2 ≈ t −1 i . (11.58) The exponential expansion ends as the inflaton field reaches the true vacuum at φ = φ 0 . The duration of inflation thus depends on the exact shape of the potential \ (φ). The number of e-foldings of inflation, for the potential shown in Figure 11.3, should be ` ∼ H i φ 0 ˙ φ ∼ 1 P \ 0 d\ dφ −1 φ 0 1 P . (11.59) Large values of φ 0 and \ 0 (that is, a broad, high plateau) and small values of d\dφ (that is, a shallowly sloped plateau) lead to more e-foldings of inflation. 252 CHAPTER 11. INFLATION & THE VERY EARLY UNIVERSE After rolling off the plateau in Figure 11.3, the inflaton field φ oscillates about the minimum at φ 0 . The amplitude of these oscillations is damped by the “Hubble friction” term proportional to H ˙ φ in equation (11.47). If the inflaton field is coupled to any of the other fields in the universe, however, the oscillations in φ are damped more rapidly, with the energy of the inflaton field being carried away by photons or other relativistic particles. These photons reheat the universe after the precipitous drop in temperature caused by inflation. The energy lost by the inflaton field after its phase transition from the false vacuum to the true vacuum can be thought of as the latent heat of that transition. When water freezes, to use a low-energy analogy, it loses an energy of 3 10 8 J m −3 , which goes to heat its surroundings. 3 Similarly, the transition from false to true vacuum releases an energy \ 0 which goes to reheat the universe. If the scale factor increases by a factor c(t f ) c(t i ) = c N (11.60) during inflation, then the temperature will drop by a factor c −N . If inflation starts around the GUT time, and lasts for ` = 100 e-foldings, then the temperature drops from a toasty 1(t i ) ∼ 1 GUT ∼ 10 28 K to a chilly 1(t f ) ∼ c −100 1 GUT ∼ 10 −15 K. At a temperature of 10 −15 K, you’d expect to find a single photon in a box 25 AU on a side, as compared to the 411 million photons packed into every cubic meter of space today. Not only is inflation very effective at driving down the number density of magnetic monopoles, it is also effective at driving down the number density of every other type of particle, including photons. The chilly post-inflationary period didn’t last, though. As the energy density associated with the inflaton field was converted to relativistic particles such as photons, the temperature of the universe was restored to its pre-inflationary value 1 i . Inflation successfully explains the flatness, homogeneity, and isotropy of the universe. It ensures that we live in a universe with a negligibly low den- sity of magnetic monopoles, while the inclusion of reheating ensures that we don’t live in a universe with a negligibly low density of photons. In some 3 This is why orange growers spray their trees with water when a hard freeze threatens. The energy released by water as it freezes keeps the delicate leaves warm. (The thin layer of ice also cuts down on convective and radiative heat loss, but the release of latent heat is the largest effect.) 11.5. THE PHYSICS OF INFLATION 253 ways, though, inflation seems to be too successful. It makes the universe homogeneous and isotropic all right, but it makes it too homogeneous and isotropic. One hundred e-foldings of inflation not only flattens the global cur- vature of the universe, it also flattens the local curvature due to fluctuations in the energy density. If energy fluctuations prior to inflation were δε¯ ε ∼ 1, a na¨ıve calculation predicts that density fluctuations immediately after 100 e-foldings of inflation would be δε ¯ ε ∼ c −100 ∼ 10 −43 . (11.61) This is a very close approach to homogeneity. Even allowing for the growth in amplitude of density fluctuations prior to the time of last scattering, this would leave the Cosmic Microwave Background much smoother than is ac- tually observed. Remember, however, the saga of how a submicroscopic patch of the uni- verse (d ∼ 3 10 −44 m) was inflated to macroscopic size (d ∼ 1 m), before growing to the size of the currently visible universe. Inflation excels in tak- ing submicroscopic scales and blowing them up to macroscopic scales. On submicroscopic scales, the vacuum, whether true or false, is full of constantly changing quantum fluctuations, with virtual particles popping into and out of existence. On quantum scales, the universe is intrinsically inhomogeneous. Inflation takes the submicroscopic quantum fluctuations in the inflaton field and expands them to macroscopic scales. The energy fluctuations that result are the origin, in the inflationary scenario, of the inhomogeneities in the cur- rent universe. We can replace the old proverb, “Great oaks from tiny acorns grow,” with the yet more amazing proverb, “Great superclusters from tiny quantum fluctuations grow.” Suggested reading [Full references are given in the “Annotated Bibliography” on page 286.] Islam (2002), ch. 9: A general overview of inflation, avoiding technical concepts of particle physics. Liddle (1999), ch. 11: A brief, clear discussion of how inflation solves the horizon, flatness, and monopole problems. 254 CHAPTER 11. INFLATION & THE VERY EARLY UNIVERSE Liddle & Lyth (2000): A thorough, recent review of all aspects of infla- tionary cosmology. Problems (11.1) What upper limit is placed on Ω(t P ) by the requirement that the universe not end in a Big Crunch between the Planck time, t P ≈ 5 10 −44 s, and the start of the inflationary epoch at t i ? Compute the maximum permissible value of Ω(t P ), first assuming t i ≈ 10 −36 s, then assuming t i ≈ 10 −26 s. (Hint: prior to inflation, the Friedmann equation will be dominated by the radiation term and the curvature term.) (11.2) Current observational limits on the density of magnetic monopoles tell us that their density parameter is currently Ω M,0 < 10 −6 . If monopoles formed at the GUT time, with one monopole per hori- zon of mass : M = : GUT , how many e-foldings of inflation would be required to drive the current density of monopoles below the bound Ω M,0 < 10 −6 ? Assume that inflation took place immediately after the formation of monopoles. (11.3) It has been speculated that the present-day acceleration of the uni- verse is due to the existence of a false vacuum which will eventu- ally decay. Suppose that the energy density of the false vacuum is ε Λ = 0.7ε c,0 = 3600 MeVm −3 , and that the current energy density of matter is ε m,0 = 0.3ε c,0 = 1600 MeVm −3 . What will be the value of the Hubble parameter once the false vacuum becomes strongly dominant? Suppose that the false vacuum is fated to decay instantaneously to ra- diation at a time t f = 50t 0 . (Assume, for simplicity, that the radiation takes the form of blackbody photons.) To what temperature will the universe be reheated at t = t f ? What will the energy density of matter be at t = t f ? At what time will the universe again be dominated by matter? Chapter 12 The Formation of Structure The universe can be approximated as being homogeneous and isotropic only if we smooth it with a filter ∼ 100 Mpc across. On smaller scales, the universe contains density fluctuations ranging from subatomic quantum fluctuations up to the large superclusters and voids, ∼ 50 Mpc across, which character- ize the distribution of galaxies in space. If we relax the strict assumption of homogeneity and isotropy which underlies the Robertson-Walker metric and the Friedmann equation, we can ask (and, to some extent, answer) the question, “How do density fluctuations in the universe evolve with time?” The formation of relatively small objects, such as planets, stars, or even galaxies, involves some fairly complicated physics. Consider a galaxy, for instance. As mentioned in Chapter 8, the luminous portions of galaxies are typically much smaller than the dark halos in which they are embedded. In the usual scenario for galaxy formation, this is because the baryonic compo- nent of a galaxy radiates away energy, in the form of photons, and slides to the bottom of the potential well defined by the dark matter. The baryonic gas then fragments to form stars, in a nonlinear magnetohydrodynamical process. In this chapter, however, I will be focusing on the formation of structures larger than galaxies – clusters, superclusters, and voids. Cosmologists use the term “large scale structure of the universe” to refer to all structures bigger than individual galaxies. A map of the large scale structure of the universe, as traced by the positions of galaxies, can be made by measuring the redshifts of a sample of galaxies and using the Hubble relation, d = (cH 0 )., to com- pute their distances from our own Galaxy. For instance, Figure 12.1 shows a redshift map from the 2dF Galaxy Redshift Survey. By plotting redshift as 255 256 CHAPTER 12. THE FORMATION OF STRUCTURE Figure 12.1: A redshift map of ∼ 10 5 galaxies, in a strip ∼ 75 ◦ long, from right ascension α ≈ 10 h to α ≈ 15 h, and ∼ 8 ◦ wide, from declination δ ≈ −5 ◦ to δ ≈ 3 ◦ . (Image courtesy of the 2dF Galaxy Redshift Survey team.) 257 Figure 12.2: The northeastern United States and southeastern Canada at night, as seen by a satellite from the Defense Meteorological Satellite Program (DMSP). a function of angular position for galaxies in a long, narrow strip of the sky, a “slice of the universe” can be mapped. In a slice such as that of Figure 12.1, which reaches to . ≈ 0.3, or d p (t 0 ) ≈ 1300 Mpc, the galaxies obviously do not have a random Poisson distribution. The most prominent structures in Figure 12.1 are superclusters and voids. Superclusters are objects which are just in the process of collapsing under their own self-gravity. Superclusters are typically flattened (roughly planar) or elongated (roughly linear) struc- tures. A supercluster will contain one or more clusters embedded within it; a cluster is a fully collapsed object which has come to equilibrium (more or less), and hence obeys the virial theorem, as discussed in section 8.3. In comparison to the flattened or elongated superclusters, the underdense voids are roughly spherical in shape. When gazing at the large scale structure of the universe, as traced by the distribution of galaxies, cosmologists are likely to call it “bubbly” or “spongy” or “frothy” or “foamy”. Being able to describe the distribution of galaxies in space doesn’t au- tomatically lead to an understanding of the origin of large scale structure. Consider, as an analogy, the distribution of luminous objects shown in Fig- ure 12.2. The distribution of illuminated cities on the Earth’s surface is ob- viously not random. There are “superclusters” of cities, such as the Boswash supercluster stretching from Boston to Washington. There are “voids” such as the Appalachian void. However, the influences which determine the exact 258 CHAPTER 12. THE FORMATION OF STRUCTURE location of cities are often far removed from fundamental physics. 1 Fortunately, the distribution of galaxies in space is more closely tied to fundamental physics than is the distribution of cities on the Earth. The basic mechanism for growing large structures, such as voids and superclusters, is gravitational instability. Suppose that at some time in the past, the density of the universe had slight inhomogeneities. We know, for instance, that such density fluctuations occurred at the time of last scattering, since they left their stamp on the Cosmic Microwave Background. When the universe is matter-dominated, the overdense regions expand less rapidly than the uni- verse as a whole; if their density is sufficiently great, they will collapse and become gravitationally bound objects such as clusters. The dense clusters will, in addition, draw matter to themselves from the surrounding underdense regions. The effect of gravity on density fluctuations is sometimes referred to as the Matthew Effect: “For whosoever hath, to him shall be given, and he shall have more abundance; but whosoever hath not, from him shall be taken away even that he hath” (Matthew 13:12). In less biblical language, the rich get richer and the poor get poorer. 12.1 Gravitational instability To put our study of gravitational instability on a more quantitative basis, consider some component of the universe whose energy density ε(:. t) is a function of position as well as time. At a given time t, the spatially averaged energy density is ¯ ε(t) = 1 \ V ε(:. t)d 3 : . (12.1) To ensure that we have found the true average, the volume \ over which we are averaging must be large compared to the size of the biggest structure in the universe. It is useful to define a dimensionless density fluctuation δ(:. t) ≡ ε(:. t) − ¯ ε(t) ¯ ε(t) . (12.2) The value of δ is thus negative in underdense regions and positive in overdense regions. The minimum possible value of δ is δ = −1, corresponding to ε = 0. 1 Consider, for instance, the complicated politics that went into determining the location of Washington, DC. 12.1. GRAVITATIONAL INSTABILITY 259 R(t) ρ = ρ(1+δ) Figure 12.3: A sphere of radius 1(t) expanding or contracting under the influence of the density fluctuation δ(t). In principle, there is no upper limit on δ. You, for instance, represent a region of space where the baryon density has δ ≈ 2 10 30 . The study of how large scale structure evolves with time requires knowing how a small fluctuation in density, with [δ[ < 1, grows in amplitude under the influence of gravity. This problem is most tractable when [δ[ remains very much smaller than one. In the limit that the amplitude of the fluctuations remains small, we can successfully use linear perturbation theory. To get a feel for how small density contrasts grow with time, consider a particularly simple case. Start with a static, homogeneous, matter-only universe with uniform mass density ¯ ρ. (At this point, we stumble over the inconvenient fact that there’s no such thing as a static, homogeneous, matter- only universe. This is the awkward fact that inspired Einstein to introduce the cosmological constant. However, there are conditions under which we can consider some region of the universe to be approximately static and homogeneous. For instance, the air in a closed room is approximately static and homogeneous; it is stabilized by a pressure gradient with a scale length which is much greater than the height of the ceiling.) In a region of the universe which is approximately static and homogeneous, we add a small amount of mass within a sphere of radius 1, as seen in Figure 12.3, so that the density within the sphere is ¯ ρ(1+δ), with δ <1. If the density excess δ is uniform within the sphere, then the gravitational acceleration at the sphere’s surface, due to the excess mass, will be ¨ 1 = − G(∆`) 1 2 = − G 1 2 4π 3 1 3 ¯ ρδ . (12.3) 260 CHAPTER 12. THE FORMATION OF STRUCTURE or ¨ 1 1 = − 4πG¯ ρ 3 δ(t) . (12.4) Thus, a mass excess (δ 0) will cause the sphere to collapse inward ( ¨ 1 < 0). Equation (12.4) contains two unknowns, 1(t) and δ(t). If we want to find an explicit solution for δ(t), we need a second equation involving 1(t) and δ(t). Conservation of mass tells us that the mass of the sphere, ` = 4π 3 ¯ ρ[1 +δ(t)]1(t) 3 . (12.5) remains constant during the collapse. Thus, we can write another relation between 1(t) and δ(t) which must hold true during the collapse: 1(t) = 1 0 [1 +δ(t)] −1/3 . (12.6) where 1 0 ≡ 3` 4π¯ ρ 1/3 = constant . (12.7) When δ <1, we may make the approximation 1(t) ≈ 1 0 [1 − 1 3 δ(t)] . (12.8) Taking the second time derivative yields ¨ 1 ≈ − 1 3 1 0 ¨ δ ≈ − 1 3 1 ¨ δ . (12.9) Thus, mass conservation tells us that ¨ 1 1 ≈ − 1 3 ¨ δ (12.10) in the limit that δ < 1. Combining equations (12.4) and (12.10), we find a tidy equation which tells us how the small overdensity δ evolves as the sphere collapses: ¨ δ = 4πG¯ ρδ . (12.11) The most general solution of equation (12.11) has the form δ(t) = ¹ 1 c t/t dyn +¹ 2 c −t/t dyn . (12.12) 12.2. THE JEANS LENGTH 261 where the dynamical time for collapse is t dyn = 1 (4πG¯ ρ) 1/2 = c 2 4πG¯ ε 1/2 . (12.13) Note that the dynamical time depends only on ¯ ρ, and not on 1. The con- stants ¹ 1 and ¹ 2 in equation (12.12) depend on the initial conditions of the sphere. For instance, if the overdense sphere starts at rest, with ˙ δ = 0 at t = 0, then ¹ 1 = ¹ 2 = δ(0)2. After a few dynamical times, however, only the exponentially growing term of equation (12.12) is significant. Thus, grav- ity tends to make small density fluctuations in a static, pressureless medium grow exponentially with time. 12.2 The Jeans length The exponential growth of density perturbations is slightly alarming, at first glance. For instance, the density of the air around you is ¯ ρ ≈ 1 kg m −3 , yielding a dynamical time for collapse of t dyn ≈ 9 hours. 2 What keeps small density perturbations in the air from undergoing a runaway collapse over the course of a few days? The answer, of course, is pressure. A non-relativistic gas, as shown in section 4.3, has an equation-of-state parameter u ≈ /1 jc 2 . (12.14) where 1 is the temperature of the gas and j is the mean mass per gas particle. Thus, the pressure of a ideal gas will never totally vanish, but will only approach zero in the limit that the temperature approaches absolute zero. When a sphere of gas is compressed by its own gravity, a pressure gra- dient will build up which tends to counter the effects of gravity. 3 However, hydrostatic equilibrium, the state in which gravity is exactly balanced by a pressure gradient, cannot always be attained. Consider an overdense sphere 2 Slightly longer if you are using this book for recreational reading as you climb Mount Everest. 3 A star is the prime example of a dense sphere of gas in which the inward force of gravity is balanced by the outward force provided by a pressure gradient. 262 CHAPTER 12. THE FORMATION OF STRUCTURE with initial radius 1. If pressure were not present, it would collapse on a timescale t dyn ∼ 1 (G¯ ρ) 1/2 ∼ c 2 G¯ ε 1/2 . (12.15) If the pressure is nonzero, the attempted collapse will be countered by a steepening of the pressure gradient within the perturbation. The steepening of the pressure gradient, however, doesn’t occur instantaneously. Any change in pressure travels at the sound speed. 4 Thus, the time it takes for the pressure gradient to build up in a region of radius 1 will be t pre ∼ 1 c s . (12.16) where c s is the local sound speed. In a medium with equation-of-state pa- rameter u 0, the sound speed is c s = c d1 dε 1/2 = √ uc . (12.17) For hydrostatic equilibrium to be attained, the pressure gradient must build up before the overdense region collapses, implying t pre < t dyn . (12.18) Comparing equation (12.15) with equation (12.16), we find that for a density perturbation to be stabilized by pressure against collapse, it must be smaller than some reference size λ J , given by the relation λ J ∼ c s t dyn ∼ c s c 2 G¯ ε 1/2 . (12.19) The length scale λ J is known as the Jeans length, after the astrophysicist James Jeans, who was among the first to study gravitational instability in a cosmological context. Overdense regions larger than the Jeans length collapse under their own gravity. Overdense regions smaller than the Jeans length merely oscillate in density; they constitute stable sound waves. 4 What is sound, after all, but a traveling change in pressure? 12.2. THE JEANS LENGTH 263 A more precise derivation of the Jeans length, including all the appropri- ate factors of π, yields the result λ J = c s πc 2 G¯ ε 1/2 = 2πc s t dyn . (12.20) The Jeans length of the Earth’s atmosphere, for instance, where the sound speed is a third of a kilometer per second and the dynamical time is nine hours, is λ J ∼ 10 5 km, far longer than the scale height of the Earth’s at- mosphere. You don’t have to worry about density fluctuations in the air undergoing a catastrophic collapse. To consider the behavior of density fluctuations on cosmological scales, consider a spatially flat universe in which the mean density is ¯ ε, but which contains density fluctuations with amplitude [δ[ <1. The characteristic time for expansion of such a universe is the Hubble time, H −1 = 3c 2 8πG¯ ε 1/2 . (12.21) Comparison of equation (12.13) with equation (12.21) reveals that the Hubble time is comparable in magnitude to the dynamical time t dyn for the collapse of an overdense region: H −1 = 3 2 1/2 t dyn ≈ 1.22t dyn . (12.22) The Jeans length in an expanding flat universe will then be λ J = 2πc s t dyn = 2π 3 2 1/2 c s H . (12.23) If we focus on one particular component of the universe, with equation-of- state parameter u and sound speed c s = √ uc, the Jeans length for that component will be λ J = 2π 2 3 1/2 √ u c H . (12.24) Consider, for instance, the “radiation” component of the universe. With u = 13, the sound speed in a gas of photons or other relativistic particles is c s = c √ 3 ≈ 0.58c . (12.25) 264 CHAPTER 12. THE FORMATION OF STRUCTURE The Jeans length for radiation in an expanding universe is then λ J = 2π √ 2 3 c H ≈ 3.0 c H . (12.26) Density fluctuations in the radiative component will be pressure-supported if they are smaller than three times the Hubble distance. Although a universe containing nothing but radiation can have density perturbations smaller than λ J ∼ 3cH, they will be stable sound waves, and will not collapse under their own gravity. In order for a universe to have gravitationally collapsed structures much smaller than the Hubble distance, it must have a non-relativistic compo- nent, with √ u < 1. The gravitational collapse of the baryonic component of the universe is complicated by the fact that it was coupled to photons until a redshift . dec ≈ . ls ≈ 1100. In section 9.5, the Hubble distance at the time of last scattering (effectively equal to the time of decoupling) was shown to be cH(. dec ) ≈ 0.2 Mpc. The energy density of baryons at decoupling was ε bary ≈ 2.8 10 11 MeVm −3 , corresponding to a mass density ρ bary ≈ 5.0 10 −19 kg m −3 , and the energy density of photons was ε γ ≈ 3.8 10 11 MeVm −3 ≈ 1.4ε bary . Prior to decoupling, the photons, electrons, and baryons were all coupled together to form a single photon-baryon fluid. Since the photons were still dominant over the baryons at the time of decoupling, with ε γ ε bary , we can regard the baryons (with only mild exaggeration) as being a dynamically insignificant contaminant in the photon-baryon fluid. Just before decoupling, if we regard the baryons as a minor contaminant, the Jeans length of the photon-baryon fluid was roughly the same as the Jeans length of a pure photon gas: λ J (before) ≈ 3cH(. dec ) ≈ 0.6 Mpc ≈ 1.9 10 22 m . (12.27) The baryonic Jeans mass, ` J , is defined as the mass of baryons contained within a sphere of radius λ J ; ` J ≡ ρ bary 4π 3 λ 3 J . (12.28) Immediately before decoupling, the baryonic Jeans mass was ` J (before) ≈ 5.0 10 −19 kg m −3 4π 3 (1.9 10 22 m) 3 ≈ 1.3 10 49 kg ≈ 7 10 18 M . (12.29) 12.2. THE JEANS LENGTH 265 This is approximately 310 4 times greater than the estimated baryonic mass of the Coma cluster, and represents a mass greater than the baryonic mass of even the largest supercluster seen today. Now consider what happens to the baryonic Jeans mass immediately after decoupling. Once the photons are decoupled, the photons and baryons form two separate gases, instead of a single photon-baryon fluid. The sound speed in the photon gas is c s (photon) = c √ 3 ≈ 0.58c . (12.30) The sound speed in the baryonic gas, by contrast, is c s (baryon) = /1 :c 2 1/2 c . (12.31) At the time of decoupling, the thermal energy per particle was /1 dec ≈ 0.26 eV, and the mean rest energy of the atoms in the baryonic gas was :c 2 = 1.22: p c 2 ≈ 1140 MeV, taking into account the helium mass fraction of ) p = 0.24. Thus, the sound speed of the baryonic gas immediately after decoupling was c s (baryon) ≈ 0.26 eV 1140 10 6 eV 1/2 c ≈ 1.5 10 −5 c . (12.32) only 5 kilometers per second. Thus, once the baryons were decoupled from the photons, their associated Jeans length decreased by a factor 1 = c s (baryon) c s (photon) ≈ 1.5 10 −5 0.58 ≈ 2.6 10 −5 . (12.33) Decoupling causes the baryonic Jeans mass to decrease by a factor 1 3 ≈ 1.8 10 −14 , plummeting from ` J (before) ≈ 7 10 18 M to ` J (after) = 1 3 ` J (before) ≈ 1 10 5 M . (12.34) This is comparable to the baryonic mass of the smallest dwarf galaxies known, and is very much smaller than the baryonic mass of our own Galaxy, which is ∼ 10 11 M . The abrupt decrease of the baryonic Jeans mass at the time of decoupling marks an important epoch in the history of structure formation. Perturba- tions in the baryon density, from supercluster scales down the the size of the 266 CHAPTER 12. THE FORMATION OF STRUCTURE smallest dwarf galaxies, couldn’t grow in amplitude until the time of photon decoupling, when the universe had reached the ripe old age of t dec ≈ 0.35 Myr. After decoupling, the growth of density perturbations in the baryonic com- ponent was off and running. The baryonic Jeans mass, already small by cosmological standards at the time of decoupling, dropped still further with time as the universe expanded and the baryonic component cooled. 12.3 Instability in an expanding universe Density perturbations smaller than the Hubble distance can grow in ampli- tude only when they are no longer pressure-supported. For the baryonic mat- ter, this loss of pressure support happens abruptly at the time of decoupling, when the Jeans length for baryons drops suddenly by a factor 1 ∼ 3 10 −5 . For the dark matter, the loss of pressure support occurs more gradually, as the thermal energy of the dark matter particles drops below their rest energy. When considering the Cosmic Neutrino Background, for instance, which has a temperature comparable to the Cosmic Microwave Background, we found (see equation 5.18) that neutrinos of mass : ν became non-relativistic at a redshift 1 +. = 1 c ≈ : ν c 2 5 10 −4 eV . (12.35) Thus, if the universe contains a Cosmic WIMP Background comparable in temperature to the Cosmic Neutrino Background, the WIMPs, if they have a mass : W c 2 2 eV, would have become non-relativistic long before the time of radiation-matter equality at . rm ≈ 3570. Once the pressure (and hence the Jeans length) of some component be- comes negligibly small, does this imply that the amplitude of density fluc- tuations is free to grow exponentially with time? Not necessarily. The analysis of section 12.1, which yielded δ ∝ exp(tt dyn ), assumed that the universe was static as well as pressureless. In an expanding Friedmann universe, the timescale for the growth of a density perturbation by self- gravity, t dyn ∼ (c 2 G¯ ε) 1/2 , is comparable to the timescale for expansion, H −1 ∼ (c 2 G¯ ε) 1/2 . Self-gravity, in the absence of global expansion, causes overdense regions to become more dense with time. The global expansion of the universe, in the absence of self-gravity, causes overdense regions to be- come less dense with time. Because the timescales for these two competing processes are similar, they must both be taken into account when computing 12.3. INSTABILITY IN AN EXPANDING UNIVERSE 267 the time evolution of a density perturbation. To get a feel how small density perturbations in an expanding universe evolve with time, let’s do a Newtonian analysis of the problem, similar in spirit to the Newtonian derivation of the Friedmann equation given in Chap- ter 4. Suppose you are in a universe filled with pressureless matter which has mass density ¯ ρ(t). As the universe expands, the density decreases at the rate ¯ ρ(t) ∝ c(t) −3 . Within a spherical region of radius 1, a small amount of matter is added, or removed, so that the density within the sphere is ρ(t) = ¯ ρ(t)[1 +δ(t)] . (12.36) with [δ[ < 1. (In performing a Newtonian analysis of this problem, we are implicitly assuming that the radius 1 is small compared to the Hubble distance and large compared to the Jeans length.) The total gravitational acceleration at the surface of the sphere will be ¨ 1 = − G` 1 2 = − G 1 2 4π 3 ρ1 3 = − 4π 3 G¯ ρ1 − 4π 3 G(¯ ρδ)1 . (12.37) The equation of motion for a point at the surface of the sphere can then be written in the form ¨ 1 1 = − 4π 3 G¯ ρ − 4π 3 G¯ ρδ . (12.38) Mass conservation tells us that the mass inside the sphere, ` = 4π 3 ¯ ρ(t)[1 +δ(t)]1(t) 3 . (12.39) remains constant as the sphere expands. Thus, 1(t) ∝ ¯ ρ(t) −1/3 [1 +δ(t)] −1/3 . (12.40) or, since ¯ ρ ∝ c −3 , 1(t) ∝ c(t)[1 +δ(t)] −1/3 . (12.41) That is, if the sphere is slightly overdense, its radius will grow slightly less rapidly than the scale factor c(t). If the sphere is slightly underdense, it will grow slightly more rapidly than the scale factor. Taking two time derivatives of equation (12.41) yields ¨ 1 1 = ¨ c c − 1 3 ¨ δ − 2 3 ˙ c c ˙ δ . (12.42) 268 CHAPTER 12. THE FORMATION OF STRUCTURE when [δ[ <1. Combining equations (12.38) and (12.42), we find ¨ c c − 1 3 ¨ δ − 2 3 ˙ c c ˙ δ = − 4π 3 G¯ ρ − 4π 3 G¯ ρδ . (12.43) When δ = 0, equation (12.43) reduces to ¨ c c = − 4π 3 G¯ ρ . (12.44) which is the correct acceleration equation for a homogeneous, isotropic uni- verse containing nothing but pressureless matter (compare to equation 4.44). By subtracting equation (12.44) from equation (12.43) to leave only the terms linear in the perturbation δ, we find the equation which governs the growth of small perturbations: − 1 3 ¨ δ − 2 3 ˙ c c ˙ δ = − 4π 3 G¯ ρδ . (12.45) or ¨ δ + 2H ˙ δ = 4πG¯ ρδ . (12.46) remembering that H ≡ ˙ cc. In a static universe, with H = 0, equa- tion (12.46) reduces to the result which we have already found in equa- tion (12.11): ¨ δ = 4πG¯ ρδ . (12.47) The additional term, ∝ H ˙ δ, found in an expanding universe, is sometimes called the “Hubble friction” term; it acts to slow the growth of density per- turbations in an expanding universe. A fully relativistic calculation for the growth of density perturbations yields the more general result ¨ δ + 2H ˙ δ = 4πG c 2 ¯ ε m δ . (12.48) This form of the equation can be applied to a universe which contains com- ponents with non-negligible pressure, such as radiation (u = 13) or a cos- mological constant (u = −1). In multiple-component universes, however, it should be remembered that δ represents the fluctuation in the density of matter alone. That is, δ = ε m − ¯ ε m ¯ ε m . (12.49) 12.3. INSTABILITY IN AN EXPANDING UNIVERSE 269 where ¯ ε m (t), the average matter density, might be only a small fraction of ¯ ε(t), the average total density. Rewritten in terms of the density parameter for matter, Ω m = ¯ ε m ε c = 8πG¯ ε m 3c 2 H 2 . (12.50) equation (12.48) takes the form ¨ δ + 2H ˙ δ − 3 2 Ω m H 2 δ = 0 . (12.51) During epochs when the universe is not dominated by matter, density pertur- bations in the matter do not grow rapidly in amplitude. Take, for instance, the early radiation-dominated phase of the universe. During this epoch, Ω m <1 and H = 1(2t), meaning that equation (12.51) takes the form ¨ δ + 1 t ˙ δ ≈ 0 . (12.52) which has a solution of the form δ(t) ≈ 1 1 +1 2 ln t . (12.53) During the radiation-dominated epoch, density fluctuations in the dark mat- ter grew only at a logarithmic rate. In the far future, if the universe is indeed dominated by a cosmological constant, the density parameter for matter will again be negligibly small, the Hubble parameter will have the constant value H = H Λ , and equation (12.51) will take the form ¨ δ + 2H Λ ˙ δ ≈ 0 . (12.54) which has a solution of the form δ(t) ≈ ( 1 +( 2 c −2H Λ t . (12.55) In a lambda-dominated phase, therefore, fluctuations in the matter den- sity reach a constant fractional amplitude, while the average matter density plummets at the rate ¯ ε m ∝ c −3H Λ t . It is only when matter dominates the energy density that fluctuations in the matter density can grow at a significant rate. In a flat, matter-dominated universe, Ω m = 1, H = 2(3t), and equation (12.51) takes the form ¨ δ + 4 3t ˙ δ − 2 3t 2 δ = 0 . (12.56) 270 CHAPTER 12. THE FORMATION OF STRUCTURE If we guess that the solution to the above equation has the power-law form 1t n , plugging this guess into the equation yields :(: −1)1t n−2 + 4 3t :1t n−1 − 2 3t 2 1t n = 0 . (12.57) or :(: −1) + 4 3 : − 2 3 = 0 . (12.58) The two possible solutions for this quadratic equation are : = −1 and : = 23. Thus, the general solution for the time evolution of density pertur- bations in a spatially flat, matter-only universe is δ(t) ≈ 1 1 t 2/3 +1 2 t −1 . (12.59) The values of 1 1 and 1 2 are determined by the initial conditions for δ(t). The decaying mode, ∝ t −1 , eventually becomes negligibly small compared to the growing mode, ∝ t 2/3 . When the growing mode is the only survivor, the density perturbations in a flat, matter-only universe grow at the rate δ ∝ t 2/3 ∝ c(t) ∝ 1 1 +. (12.60) as long as [δ[ <1. When an overdense region attains an overdensity δ ∼ 1, its evolution can no longer be treated with a simple linear perturbation approach. Studies of the nonlinear evolution of structure are commonly made using numerical computer simulations, in which the matter filling the universe is modeled as a distribution of point masses interacting via Newtonian gravity. In these simulations, as in the real universe, when a region reaches an overdensity δ ∼ 1, it breaks away from the Hubble flow and collapses. After one or two oscillations in radius, the overdense region attains virial equilibrium as a gravitationally bound structure. If the baryonic matter within the structure is able to cool efficiently (by bremsstrahlung or some other process) it will ra- diate away energy and fall to the center. The centrally concentrated baryons can then proceed to form stars, becoming the visible portions of galaxies that we see today. The less concentrated nonbaryonic matter forms the dark halo within which the stellar component of the galaxy is embedded. If baryonic matter were the only type of non-relativistic matter in the uni- verse, then density perturbations could have started to grow at . dec ≈ 1100, 12.3. INSTABILITY IN AN EXPANDING UNIVERSE 271 Figure 12.4: A highly schematic drawing of how density fluctuations in dif- ferent components of the universe evolve with time. and they could have grown in amplitude only by a factor ∼ 1100 by the present day. However, the mominant form of non-relativistic matter is dark matter. The density perturbations in the dark matter started to grow effec- tively at . rd ≈ 3570. At the time of decoupling, the baryons fell into the preexisting gravitational wells of the dark matter. The situation is schemat- ically illustrated in Figure 12.4. Having nonbaryonic dark matter allows the universe to get a “head start” on structure formation; perturbations in the matter density can start growing at . rd ≈ 3570 rather than . dec ≈ 1100, as they would in a universe without dark matter. 272 CHAPTER 12. THE FORMATION OF STRUCTURE 12.4 The power spectrum When deriving equation (12.46), which determines the growth rate of density perturbations in a Newtonian universe, I assumed that the perturbation was spherically symmetric. In fact, equation (12.46) and its relativistically correct brother, equation (12.48), both apply to low-amplitude perturbations of any shape. This is fortunate, since the density perturbations in the real universe are not spherically symmetric. The bubbly structure shown in redshift maps of galaxies, such as Figure 12.1, has grown from the density perturbations which were present when the universe became matter dominated. Great oaks from tiny acorns grow – but then, great pine trees from tiny pinenuts grow. By looking at the current large scale structure (the “tree”), we can deduce the properties of the early, low-amplitude, density fluctuations (the “nut”). 5 Let us consider the properties of the early density fluctuations at some time t i when they were still very low in amplitude ([δ[ <1). As long as the density fluctuations are small in amplitude, the expansion of the universe is still nearly isotropic, and the geometry of the universe is still well described by the Robertson-Walker metric (equation 3.25): d: 2 = −c 2 dt 2 +c(t) 2 [d: 2 +o κ (:) 2 dΩ 2 ] . (12.61) Under these circumstances, it is useful to set up a comoving coordinate system. Choose some point as the origin. In a universe described by the Robertson-Walker metric, as shown in section 3.4, the proper distance of any point from the origin can be written in the form d p (t i ) = c(t i ): . (12.62) where the comoving distance : is what the proper distance would be at the present day (c = 1) if the expansion continued to be perfectly isotropic. If we label each bit of matter in the universe with its comoving coordinate position : = (:. θ. φ), then : will remain very nearly constant as long as [δ[ < 1. Thus, when considering the regime where density fluctuations are small, cosmologists typically consider δ(:), the density fluctuation at a comoving location :, at some time t i . (The exact value of t i doesn’t matter, as long as it’s a time after the density perturbations are in place, but before 5 At the risk of carrying the arboreal analogy too far, I should mention that the tem- perature fluctuations of the Cosmic Microwave Background, as shown in Figures 9.2 and 9.5, offer us a look at the “sapling”. 12.4. THE POWER SPECTRUM 273 they reach an amplitude [δ[ ∼ 1. Switching to a different value of t i , under these restrictions, simply changes the amplitude of δ(:), and not its shape.) When discussing the temperature fluctuations of the Cosmic Microwave Background, back in Chapter 9, I pointed out that cosmologists weren’t in- terested in the exact pattern of hot and cold spots on the last scattering surface, but rather in the statistical properties of the field δ11(θ. φ). Sim- ilarly, cosmologists are not interested in the exact locations of the density maxima and minima in the early universe, but rather in the statistical prop- erties of the field δ(:). When studying the temperature fluctuations of the CMB, it is useful to expand δ11(φ. θ) in spherical harmonics. A similar decomposition of δ(:) is also useful. Since δ is defined in three-dimensional space (rather than on the surface of a sphere), a useful expansion of δ is in terms of Fourier components. Within a large comoving box, of comoving volume \ , the density fluctu- ation field δ(:) can be expressed as δ(:) = \ (2π) 3 δ k c −i k·r d 3 / . (12.63) where the individual Fourier components δ k are found by performing the integral δ k = 1 \ δ(:)c i k·r d 3 : . (12.64) In performing the Fourier transform, you are breaking up the function δ(:) into an infinite number of sine waves, each with comoving wavenumber / and comoving wavelength λ = 2π/. If you have complete, uncensored knowledge of δ(:), you can compute all the Fourier components δ k ; conversely, if you know all the Fourier components, you can reconstruct the density field δ(:). Each Fourier component is a complex number, which can be written in the form δ k = [δ k [c iφ k . (12.65) When [δ k [ <1, then each Fourier component obeys equation (12.51), ¨ δ k + 2H ˙ δ k − 3 2 Ω m H 2 δ k = 0 . (12.66) as long as the proper wavelength, c(t)2π/, is large compared to the Jeans 274 CHAPTER 12. THE FORMATION OF STRUCTURE length and small compared to the Hubble distance cH. 6 The phase φ k remains constant as long as the amplitude [δ k [ remains small. Even after fluctuations with a short proper wavelength have reached [δ k [ ∼ 1 and col- lapsed, the growth of the longer wavelength perturbations is still described by equation (12.66). This means, helpfully enough, that we can use linear perturbation theory to study the growth of very large scale structure even af- ter smaller structures, such as galaxies and clusters of galaxies, have already collapsed. The mean square amplitude of the Fourier components defines the power spectrum: 1(/) = '[δ k [ 2 ` . (12.67) where the average is taken over all possible orientations of the wavenumber /. (If δ(:) is isotropic, then no information is lost, statistically speaking, if we average the power spectrum over all angles.) When the phases φ k of the different Fourier components are uncorrelated with each other, then δ(:) is called a Gaussian field. If a Gaussian field is homogeneous and isotropic, then all its statistical properties are summed up in the power spectrum 1(/). If δ(:) is a Gaussian field, then the value of δ at a randomly selected point is drawn from the Gaussian probability distribution j(δ) = 1 √ 2πσ exp − δ 2 2σ 2 . (12.68) where the standard deviation σ can be computed from the power spectrum: σ = \ (2π) 3 1(/)d 3 / = \ 2π 2 ∞ 0 1(/)/ 2 d/ . (12.69) The study of Gaussian density fields is of particular interest to cosmologists because most inflationary scenarios predict that the density fluctuations cre- ated by inflation (see section 11.5) will be an isotropic, homogeneous Gaus- sian field. In addition, the expected power spectrum for the inflationary fluctuations has a scale-invariant, power-law form: 1(/) ∝ / n . (12.70) 6 When a sine wave perturbation has a wavelength large compared to the Hubble dis- tance, its crests are not causally connected to its troughs. As long as the crests remain out of touch with the troughs (that is, as long as a(t)2π/k > c/H(t)), the amplitude of a perturbation grows at the rate δ(t) ∝ a(t). 12.4. THE POWER SPECTRUM 275 with the favored value of the power-law index being : = 1. The preferred power spectrum, 1(/) ∝ /, is often referred to as a Harrison-Zel’dovich spectrum. What would a universe with 1(/) ∝ / n look like? Imagine going through such a universe and marking out randomly located spheres of comoving ra- dius 1. The mean mass of each sphere (considering only the non-relativistic matter which it contains) will be '`` = 4π 3 1 3 ε m,0 c 2 . (12.71) However, the actual mass of each sphere will vary; some spheres will be slightly underdense, and others will be slightly overdense. The mean square density fluctuation of the mass inside each sphere is a function of the power spectrum and of the size of the sphere: ` −'`` '`` 2 ¸ ∝ / 3 1(/) . (12.72) where the comoving wavenumber associated with the sphere is / = 2π1. Thus, if the power spectrum has the form 1(/) ∝ / n , the root mean square mass fluctuation within spheres of comoving radius 1 will be δ` ` ≡ ` −'`` '`` 2 ¸1/2 ∝ 1 −(3+n)/2 . (12.73) This can also be expressed in the form δ`` ∝ ` −(3+n)/6 . For : < −3, the mass fluctuations diverge on large scales, which would be Bad News for our assumption of homogeneity on large scales. (Note that if you scattered point masses randomly throughout the universe, so that they formed a Poisson dis- tribution, you would expect mass fluctuations of amplitude δ`` ∝ ` −1/2 , where ` is the expected number of point masses within the sphere. Since the average mass within a sphere is ` ∝ `, a Poisson point distribu- tion has δ`` ∝ ` −1/2 , or : = 0. The Harrison-Zel’dovich spectrum, with : = 1, thus will produce more power on small length scales than a Poisson distribution of points.) Note that the potential fluctuations asso- ciated with mass fluctuations on a length scale 1 will have an amplitude δΦ ∼ Gδ`1 ∝ δ`` 1/3 ∝ ` (1−n)/6 . Thus, the Harrison-Zel’dovich spec- trum, with : = 1, is the only power law which prevents the divergence of the potential fluctuations on both large and small scales. 276 CHAPTER 12. THE FORMATION OF STRUCTURE 12.5 Hot versus cold Immediately after inflation, the expected power spectrum for density pertur- bations has the form 1(/) ∝ / n , with an index : = 1 being predicted by most inflationary models. However, the shape of the power spectrum will be modified between the end of inflation at t f and the time of radiation-matter equality at t rm ≈ 4.7 10 4 yr. The shape of the power spectrum at t rm , when density perturbations start to grow significantly in amplitude, depends on the properties of the dark matter. More specifically, it depends on whether the dark matter is predominantly cold dark matter or hot dark matter. Cold dark matter consists of particles which were non-relativistic at the time they decoupled from the other components of the universe. For instance, WIMPs would have had thermal velocities much smaller than c at the time they decoupled, and hence qualify as cold dark matter. If any primordial black holes had formed in the early universe, their peculiar velocities would have been much smaller than c at the time they formed; thus primordial black holes would also act as cold dark matter. Axions are a type of elementary particle first proposed by particle physicists for non-cosmological purposes. If they exist, however, they would have formed out of equilibrium in the early universe, with very low thermal velocities. Thus, axions would act as cold dark matter, as well. Hot dark matter, by contrast, consists of particles which were relativistic at the time they decoupled from the other components of the universe, and which remained relativistic until the mass contained within a Hubble volume (a sphere of proper radius cH) was large compared to the mass of a galaxy. In the Benchmark Model, the Hubble distance at the time of radiation-matter equality was c H(t rm ) = c √ 2H 0 Ω 3/2 r,0 Ω 2 m,0 ≈ 1.8ct rm ≈ 0.026 Mpc . (12.74) so the mass within a Hubble volume at that time was 4π 3 c 3 H(t rm ) 3 Ω m,0 ρ c,0 c 3 rm = √ 2π 3 c 3 H 3 0 Ω 3/2 r,0 Ω 2 m,0 ρ c,0 ≈ 1.4 10 17 M . (12.75) much larger than the mass of even a fairly large galaxy such as our own (` gal ≈ 10 12 M ). Thus, a weakly interacting particle which remains rel- ativistic until the universe becomes matter-dominated will act as hot dark 12.5. HOT VERSUS COLD 277 matter. For instance, neutrinos decoupled at t ∼ 1 s, when the universe had a temperature /1 ∼ 1 MeV. Thus, a neutrino with mass : ν c 2 < 1 MeV was hot enough to be relativistic at the time it decoupled. Moreover, as discussed in section 5.1, a neutrino with mass : ν c 2 < 2 eV doesn’t become non-relativistic until after radiation-matter equality, and hence qualifies as hot dark matter. 7 To see how the existence of hot dark matter modifies the spectrum of density perturbations, consider what would happen in a universe filled with weakly interacting particles which are relativistic at the time they decouple. The initially relativistic particles cool as the universe expands, until their thermal velocities drop well below c when 3/1 ∼ : h c 2 . This happens at a temperature 1 h ∼ : h c 2 3/ ∼ 8000 K : h c 2 2 eV . (12.76) In the radiation-dominated universe, this corresponds to a cosmic time (equa- tion 10.2) t h ∼ 2 10 12 s : h c 2 2 eV −2 . (12.77) Prior to the time t h , the hot dark matter particles move freely in random directions with a speed close to that of light. This motion, called free stream- ing, acts to wipe out any density fluctuations present in the hot dark matter. Thus, the net effect of free streaming in the hot dark matter is to wipe out any density fluctuations whose wavelength is smaller than ∼ ct h . When the hot dark matter particles become non-relativistic, there will be no density fluctuations on scales smaller than the physical scale λ min ∼ ct h ∼ 20 kpc : h c 2 2 eV −2 . (12.78) corresponding to a comoving length scale 1 min = λ min c(t h ) ∼ 1 h 2.725 K λ min ∼ 60 Mpc : h c 2 2 eV −1 . (12.79) 7 It may seem odd to refer to neutrinos as “hot” dark matter, when the temperature of the Cosmic Neutrino Background is only two degrees above absolute zero. The label “hot”, in this case, simply means that the neutrinos were hot enough to be relativistic back in the radiation-dominated era. 278 CHAPTER 12. THE FORMATION OF STRUCTURE The total amount of matter within a sphere of comoving radius 1 min is ` min = 4π 3 1 3 min Ω m,0 ρ c,0 ∼ 5 10 16 M : h c 2 2 eV −3 . (12.80) assuming Ω m,0 = 0.3. If the dark matter is contributed by neutrinos with rest energy of a few electron volts, then the free streaming will wipe out all density fluctuations smaller than superclusters. The upper panel of Figure 12.5 shows the power spectrum of density fluc- tuations in hot dark matter, once the hot dark matter has cooled enough to become non-relativistic. Note that for wavenumbers / <2π1 min , the power spectrum of hot dark matter (shown as the dotted line) is indistinguishable from the original 1 ∝ / spectrum (shown as the dashed line). However, the free streaming of the hot dark matter results in a severe loss of power for wavenumbers / 2π1 min . The lower panel of Figure 12.5 shows that the root mean square mass fluctuations in hot dark matter, δ`` ∝ (/ 3 1) 1/2 , have a maximum amplitude at a mass scale ` ∼ 10 16 M . This implies that in a universe filled with hot dark matter, the first structures to collapse are the size of superclusters. Smaller structures, such as clusters and galaxies then form by fragmentation of the superclusters. (This scenario, in which the largest observable structures form first, is called the top-down scenario.) If most of the dark matter in the universe were hot dark matter, such as neutrinos, then we would expect the oldest structures in the universe to be superclusters, and that galaxies would be relatively young. In fact, the opposite seems to be true in our universe. Superclusters are just collapsing today, while galaxies have been around since at least . ∼ 6, when the universe was less than a gigayear old. Thus, most of the dark matter in the universe must be cold dark matter, for which free streaming has been negligible. The evolution of the power spectrum of cold dark matter, given the ab- sence of free streaming, is quite different from the evolution of the power spectrum for hot dark matter. Remember, when the universe is radiation- dominated, density fluctuations δ k in the dark matter do not grow appre- ciably in amplitude, as long as their proper wavelength c(t)2π/ is small compared to the Hubble distance cH(t). However, when the proper wave- length of a density perturbation is large compared to the Hubble distance, its amplitude will be able to increase, regardless of whether the universe is radiation-dominated or matter-dominated. If the cold dark matter consists of WIMPs, they decouple from the radiation at a time t d ∼ 1 s, when the 12.5. HOT VERSUS COLD 279 .001 .01 .1 1 10 10 −7 10 −6 10 −5 .0001 .001 .01 .1 1 k [Mpc −1 ] P HDM CDM P∝k 12 14 16 18 20 10 −5 .0001 .001 .01 .1 1 10 log 10 M ( k 3 P ) 1 / 2 Figure 12.5: Upper panel – The power spectrum at the time of radiation- matter equality for cold dark matter (solid line) and for hot dark matter (dotted line). The initial power spectrum produced by inflation (dashed line) is assumed to have the form 1(/) ∝ /. The normalization of the power spectrum is arbitrary. Lower panel – The root mean square mass fluctuations, δ`` ∝ (/ 3 1) 1/2 , are shown as a function of ` ∝ / −3 (masses are in units of M ). The line types are the same as in the upper panel. 280 CHAPTER 12. THE FORMATION OF STRUCTURE scale factor is c d ∼ 3 10 −10 . At the time of WIMP decoupling, the Hubble distance is cH ∼ 2ct d ∼ 610 8 m, corresponding to a comoving wavenumber / d ∼ 2πc d 2ct d ∼ 10 5 Mpc −1 . (12.81) Thus, density fluctuations with a wavenumber / < / d will have a wavelength greater than the Hubble distance at the time of WIMP decoupling, and will be able to grow freely in amplitude, as long as their wavelength remains longer than the Hubble distance. Density fluctuations with / / d will remain frozen in amplitude until matter starts to dominate the universe at t rm ≈ 4.7 10 4 yr, when the scale factor has grown to c rm ≈ 2.8 10 −4 . At the time of radiation-matter equality, the Hubble distance, as given in equation (12.74), is cH ≈ 1.8ct rm ≈ 0.026 Mpc, corresponding to a comoving wavenumber / rm ≈ 2πc rm 1.8ct rm ≈ 0.07 Mpc −1 . (12.82) Thus, density fluctuations with a wavenumber / < / rm ≈ 0.07 Mpc −1 will grow steadily in amplitude during the entire radiation-dominated era, and for wavenumbers / < / rm ≈ 0.07 Mpc −1 , the power spectrum for cold dark matter retains the original 1(/) ∝ / form which it had immediately after inflation (see the upper panel of Figure 12.5). By contrast, cold dark matter density perturbations with a wavenumber / d / / rm will be able to grow in amplitude only until their physical wave- length c(t)(2π/) ∝ t 1/2 is smaller than the Hubble distance cH(t) ∝ t. At that time, their amplitude will be frozen until the time t rm , when matter dominates, and density perturbations smaller than the Hubble distance are free to grow again. Thus, for wavenumbers / / rm , the power spectrum for cold dark matter is suppressed in amplitude, with the suppression being greatest for the largest wavenumbers (corresponding to shorter wavelengths, which come within the horizon at an earlier time). The top panel of Fig- ure 12.5 shows, as the solid line, the power spectrum for cold dark matter at the time of radiation-matter equality. Note the broad maximum in the power spectrum at / ∼ / rm ≈ 0.07 Mpc −1 . The root mean square mass fluctuations in the cold dark matter, shown in the bottom panel of Figure 12.5 are largest in amplitude for the smallest mass scales. This implies that in a universe filled with cold dark matter, the first objects to form are the smallest, with galaxies forming first, then clusters, then superclusters. This scenario, called 12.5. HOT VERSUS COLD 281 the bottom-up scenario, is consistent with the observed ages of galaxies and superclusters. Assuming that the dark matter consists of nothing but hot dark matter gives a poor fit to the observed large scale structure of the universe. Assuming that the dark matter is purely cold dark matter gives a much better fit. However, there is strong evidence that neutrinos do have some mass, and thus that the universe contains at least some hot dark matter. Cosmologists studying the large scale structure of the universe can adjust the assumed power spectrum of the dark matter, by mixing together hot and cold matter. (It’s a bit like adjusting the temperature of your bath by tweaking the hot and cold water knobs.) Comparison of the assumed power spectrum to the observed large scale structure (as seen, for instance, in figure 12.1) reveals that ∼ 13% or less of the matter in the universe consists of hot dark matter. For Ω m,0 = 0.3, this implies Ω HDM,0 ≤ 0.04. If there were more hot dark matter than this amount, free streaming of the hot dark matter particles would make the universe too smooth on small scales. Some like it hot, but most like it cold – the majority of the dark matter in the universe must be cold dark matter. Suggested reading [Full references are given in the “Annotated Bibliography” on page 286.] Liddle & Lyth (2000): The origin of density perturbations during the in- flationary era, and their growth thereafter. Longair (1998): For those who want to know more about galaxy formation, and how it ties into cosmology. Rich (2001), ch. 7: The origin and evolution of density fluctuations. Problems (12.1) Consider a spatially flat, matter-dominated universe (Ω = Ω m = 1) which is contracting with time. What is the functional form of δ(t) in such a universe? 282 CHAPTER 12. THE FORMATION OF STRUCTURE (12.2) Consider an empty, negatively curved, expanding universe, as de- scribed in section 5.2. If a dynamically insignificant amount of matter (Ω m <1) is present in such a universe, how do density fluctuations in the matter evolve with time? That is, what is the functional form of δ(t)? (12.3) A volume containing a photon-baryon fluid is adiabatically expanded or compressed. The energy density of the fluid is ε = ε γ + ε bary , and the pressure is 1 = 1 γ = ε γ 3. What is d1dε for the photon-baryon fluid? What is the sound speed, c s ? In equation (12.27), how large of an error did I make in my estimate of λ J (before) by ignoring the effect of the baryons on the sound speed of the photon-baryon fluid? (12.4) Suppose that the stars in a disk galaxy have a constant orbital speed · out to the edge of its spherical dark halo, at a distance 1 halo from the galaxy’s center. What is the average density ¯ ρ of the matter in the galaxy, including its dark halo? (Hint: go back to section 8.2.) What is the value of ¯ ρ for our Galaxy, assuming · = 220 kms −1 and 1 halo = 100 kpc? If a bound structure, such as a galaxy, forms by gravitational collapse of an initially small density perturbation, the minimum time for collapse is t min ≈ t dyn ≈ 1 √ G¯ ρ. Show that t min ≈ 1 halo · for a disk galaxy. What is t min for our own Galaxy? What is the maximum possible redshift at which you would expect to see galaxies comparable in · and 1 halo to our own Galaxy? (Assume the Benchmark Model is correct.) (12.5) Within the Coma cluster, as discussed in section 8.3, galaxies have a root mean square velocity of '· 2 ` 1/2 ≈ 1520 kms −1 relative to the center of mass of the cluster; the half-mass radius of the Coma cluster is : h ≈ 1.5 Mpc. Using arguments similar to those of the previous problem, compute the minimum time t min required for the Coma cluster to form by gravitational collapse. (12.6) Derive equation (12.74), giving the Hubble distance at the time of radiation-matter equality. What was the Hubble distance at the time of matter-lambda equality, in the Benchmark Model? How much matter was contained within a Hubble volume at the time of matter-lambda equality? 12.5. HOT VERSUS COLD 283 (12.7) Warm dark matter is defined as matter which became non-relativistic when the amount of matter within a Hubble volume had a mass com- parable to that of a galaxy. In the Benchmark Model, at what time t WDM was the mass contained within a Hubble volume equal to ` gal = 10 12 M ? If the warm dark matter particles have a temperature equal to that of the cosmic neutrino background, what mass must they have in order to have become non-relativistic at t ∼ t WDM ? Epilogue A book dealing with an active field like cosmology can’t really have a neat, tidy ending. Our understanding of the universe is still growing and evolving. During the twentieth century, the growing weight of evidence pointed toward the Hot Big Bang model, in which the universe started in a hot, dense state, but gradually cooled as it expanded. At the end of the twentieth century and the beginning of the twenty-first, cosmological evidence was gathered at an increasing rate, refining our knowledge of the universe. As I write this epilogue, on a sunny spring day in the year 2002, the available evidence is explained by a Benchmark Model which is spatially flat and which has an expansion which is currently accelerating. It seems that 70% of the energy density of the universe is contributed by a cosmological constant (or other form of “dark energy” with negative pressure). Only 30% of the energy density is contributed by matter (and only 4% is contributed by the familiar baryonic matter of which you and I are made). However, many questions about the cosmos remain unanswered. Here are a few of the questions that currently nag at cosmologists: • What are the precise values of cosmological parameters such as H 0 , ¡ 0 , Ω m,0 , and Ω Λ,0 ? Much effort has been invested in determining these parameters, but they are still not pinned down precisely. • What is the dark matter? It can’t be made entirely of baryons. It can’t be made entirely of neutrinos. Most of the dark matter must be in the form of some exotic stuff which has not yet been detected in laboratories. • What is the “dark energy”? Is it vacuum energy which plays the role of a cosmological constant, or is it some other component of the universe with −1 < u < −13? If it is vacuum energy, is it provided by a false 284 285 vacuum, driving a temporary inflationary stage, or are we finally seeing the true vacuum energy? • What drove inflation during the early universe? Our knowledge of the particle physics behind inflation is still sadly incomplete. Indeed, some cosmologists pose the questions, “Did inflation take place at all during the early universe? Is there another way to resolve the flatness, horizon, and monopole problems?” • Why is the universe expanding? At one level, this question is easily answered. The universe is expanding today because it was expanding yesterday. It was expanding yesterday because it was expanding the day before yesterday... However, when you extrapolate back to the Planck time, you find that the universe was expanding then with a Hubble parameter H ∼ 1t P . What determined this set of initial conditions? In other words, “What put the Bang in the Big Bang?” The most interesting questions, however, are those which we are still too ignorant to pose correctly. For instance, in ancient Egypt, a list of unan- swered questions in cosmology might have included “How high is the dome which makes up the sky?” and “What’s the dome made of?” Severely er- roneous models of the universe obviously give rise to irrelevant questions. The exciting, unsettling possibility exists that future observations will ren- der the now-promising Benchmark Model obsolete. I hope, patient reader, that learning about cosmology from this book has encouraged you to become a cosmologist yourself, and to join the scientists who are laboring to make my book a quaint, out-of-date relic from a time when the universe was poorly understood. Annotated Bibliography Works described as ‘Popular’ contain little or no math. Those described as ‘Intermediate’ are at roughly the same level as this book. Those described as ‘Advanced’ have a higher level of mathematical and physical sophistication, appropriate for study at a graduate level. Popular • Harrison, E. R. 1987, Darkness at Night: A Riddle of the Universe (Cambridge: Harvard University Press) A comprehensive discussion of Olbers’ Paradox and its place in the history of cosmology. • Kragh, H. 1996, Cosmology and Controversy (Princeton: Princeton University Press) A well-reseached history of the Big Bang vs. Steady State debate. A fascinating book if you are at all interested in the sociology of science. • Silk, J. 2001, The Big Bang (third edition) (New York: W. H. Freeman & Co.) A broad overview of cosmology. Although aimed at a popular audience (with all mathematical formulas banished to an appendix), it doesn’t skimp on the physics. • Weinberg, S. 1993, The First Three Minutes (revised edition) (Perseus Books) A classic of popular science. Weinberg’s revision has brought the original 1977 version more nearly up-to-date. Intermediate • Bernstein, J. 1995, Introduction to Cosmology (Englewood Cliffs, NJ: Prentice Hall) Has a slightly greater emphasis on particle physics than 286 287 most cosmology texts. • Coles, P. 1999, The Routledge Critical Dictionary of the New Cosmology (New York: Routledge) In addition to a dictionary of cosmology-related terms, from “absorption line” to “Zel’dovich-Sunyaev effect”, this book also contains longer essays on cosmological topics of current interest. • Cox, A. N., ed. 2000, Allen’s Astrophysical Quantities (fourth edition) (New York: Springer-Verlag) A standard reference book of astronom- ically relevant data, from the Euler vectors of the Nazca plate to the intensity of the extragalactic gamma-ray background. • Harrison, E. 2000, Cosmology: The Science of the Universe (second edi- tion) (Cambridge: Cambridge University Press) A wide-ranging book, placing the science of cosmology in its historical context, and discussing its philosophical and religious implications. • Islam, J. N. 2002, An Introduction to Mathematical Cosmology (sec- ond edition) (Cambridge: Cambridge University Press) A book which emphasizes (as its name implies) the mathematical rather than the observational aspects of cosmology. • Liddle, A. 1999, An Introduction to Modern Cosmology (Chichester: John Wiley & Sons) A clear and concise introductory work. • Longair, Malcolm S. 1998, Galaxy Formation (Berlin: Springer-Verlag) A well-written introduction to galaxy formation, approached from a cosmological perspective. • Narlikar, J. V. 2002, Introduction to Cosmology (third edition) (Cam- bridge: Cambridge University Press) Particularly useful for its section on alternative (non-Friedmann) cosmologies. • Peacock, J. A. 1999, Cosmological Physics (Cambridge: Cambridge University Press) A large, well-stuffed grabbag of cosmological top- ics. Contains, among other useful things, a long, detailed discussion of galaxy formation and clustering. • Rich, J. 2001, Fundamentals of Cosmology (Berlin: Springer-Verlag) Aimed primarily at physicists; provides a self-contained introduction 288 ANNOTATED BIBLIOGRAPHY to general relativity, telling you as much as you need to know for cos- mological purposes. • Rowan-Robinson, M. 1996, Cosmology (third edition) (Oxford: Oxford University Press) Has a slightly greater emphasis on astronomical ob- servations than most cosmology texts. • van den Bergh, S. 2000, The Galaxies of the Local Group (Cambridge: Cambridge University Press) Cosmology begins at home: contains in- formation about the distance scale within the Local Group, and its dark matter content. Advanced • Kolb, E. W., and Turner, M. 1990, The Early Universe (Redwood City, CA: Addison-Wesley) A text which helped to define the field currently known as “particle astrophysics”. • Liddle, A. R., and Lyth, D. H. 2000, Cosmological Inflation and Large- Scale Structure (Cambridge: Cambridge University Press) Gives an in-depth treatment of inflation, and how it gives rise to structure in the universe. • Peebles, P. J. E. 1993, Principles of Physical Cosmology (Princeton: Princeton University Press) A classic comprehensive book. Keeping Up to Date Popular astronomy magazines such as Sky and Telescope and Astron- omy provide non-technical news updates on advances in cosmology. Scientific American, from time to time, includes more in-depth arti- cles on cosmological topics. The Annual Review of Astronomy and Astrophysics regularly provides reviews, on a more technical level, of cosmological topics. Index Abell 2218 (cluster of galaxies), 174 accelerating universe, 75, 150 acceleration equation, 67, 68, 128, 242, 268 with cosmological constant, 73 acoustic oscillations, 203 age of universe (t 0 ) Benchmark Model, 112 empty universe, 87 flat universe matter + lambda, 111 matter only, 94 radiation only, 95 single-component, 92 angular-diameter distance (d A ), 136, 138, 139, 196 maximum value, 140 related to luminosity distance, 138 astronomical unit (AU), 2, 131 axions, 276 baryon-antibaryon asymmetry, 229 in early universe, 230 baryon-to-photon ratio (η), 180, 190, 204, 220, 225, 230 smallness of, 229 baryonic Jeans mass (` J ), 264 after decoupling, 265 before decoupling, 264 baryonic matter, 23 baryons, 23 energy density, 179, 228 number density, 180, 228 bear polar, 132, 152–153 teddy, 35, 47 Benchmark Model, 84, 85, 102–103, 118–123, 148, 195, 204, 235, 236 beryllium (Be), 209, 223 Big Bang, 6, 20–22 Big Bang Nucleosynthesis, 212 Big Bounce, 113 evidence against, 116, 150 Big Chill, 104, 106, 109, 113 Big Crunch, 105, 106, 111, 113 evidence against, 150 time of (t crunch ), 106 blackbody radiation, 25 blueshift, 15 bolometric flux (1), 132, 134 Euclidean, 133 Bondi, Hermann, 21 BOOMERANG, 196, 197 bottom-up scenario, 281 Cepheid variable stars, 141 period-luminosity relation, 142 cold dark matter, 276, 278, 281 power spectrum, 278–280 Coma cluster, 157, 164, 167–170 289 290 INDEX distance to, 168 mass of, 168, 170 comoving coordinates, 46 Copernican principle, see cosmolog- ical principle Cosmic Background Explorer (COBE), 28, 182, 197 instruments aboard, 182 Cosmic Microwave Background, 28, 179 blackbody spectrum, 28, 182 compared to starlight, 82–83 cooling of, 30 dipole distortion, 183 discovery of, 180–181 energy density, 28, 179 number density, 28, 179 temperature fluctuations, 183, 196 amplitude, 185 and baryon density, 204 and flatness of universe, 204 correlation function, 197, 199, 200 origin of, 201–204 Cosmic Neutrino Background, 83, 175 energy density, 83 non-detection, 83 number density, 175 cosmic time (t), 46 cosmological constant (Λ), 68, 71– 75, 243 energy density, 73 pressure, 73 cosmological principle, 14 perfect, 21 cosmological proper time, see cos- mic time critical density (ε c ), 64 current value, 64 curvature negative, 41 positive, 40 curvature constant (κ), 43, 46 dark energy, 71 dark halo, 162 dark matter, 27, 164 axions, 175 neutrinos, 175 nonbaryonic, 159, 164 primordial black holes, 175 DASI, 196 de Sitter universe, 98 deceleration parameter (¡ 0 ), 127 Benchmark Model, 129 sign convention, 127 density fluctuations lambda-dominated era, 269 matter-dominated era, 270 power spectrum, 274 radiation-dominated era, 269 density parameter (Ω), 65 baryons (Ω bary ), 159, 228 clusters of galaxies (Ω clus ), 170 cosmological constant (Ω Λ ), 84 galaxies (Ω gal ), 164 matter (Ω m ), 84, 155 radiation (Ω r ), 84 stars (Ω ), 157 deuterium (D), 210 deuterium abundance, 227 determination of, 227 deuterium synthesis, 212, 216, 218 INDEX 291 compared to recombination, 219 temperature of, 222 deuterium-to-neutron ratio, 220 Dicke, Robert, 180 Digges, Thomas, 7, 10 distance modulus (:−`), 148 Einstein radius (θ E ), 172, 174 Einstein’s static universe, 74, 113 instability, 74 radius of curvature, 74 Einstein, Albert, 32–38, 57, 73, 74, 171 cosmological constant, 71 Einstein-de Sitter universe, 94 electron, 23 electron volt (eV), 3 empty universe, 86 expanding, 87 static, 86 energy density (ε), 57 additive, 80 baryons, 179 CMB, 179 flat universe single-component, 92 matter, 80 radiation, 81 entropy, 66 equation of state, 69, 79 equivalence principle, 34 and photons, 38 and teddy bears, 35 expansion adiabatic, 66 superluminal speed, 50 false vacuum, 251 driving inflation, 251 Fermat’s principle, 38 Fermi, Enrico, 24 first law of thermodynamics, 29, 66 flatness problem, 233–236 resolved by inflation, 244–245 fluid equation, 67, 68, 79 Ford, Kent, 162 Fourier transform, 273 fractional ionization (A), 187, 192 free streaming, 277 freezeout, 216 Friedmann equation, 58, 68, 79, 234 during inflation, 242 general relativistic, 61 Newtonian, 60 with cosmological constant, 73 Friedmann, Alexander, 58 fundamental force, 239 electromagnetic, 32 electroweak, 239 Grand Unified Force, 239 gravitational, 32, 33 strong nuclear, 32 weak nuclear, 32, 215 Galaxy (Milky Way), 2, 11, 160 luminosity of, 3 mass of, 163 mass-to-light ratio, 163 Galilei, Galileo, 34 Gamow, George, 181 Gaussian field, 274 geodesic, 38 geometry Euclidean, 33 non-Euclidean, 55 gigayear (Gyr), 3 292 INDEX Glashow, Sheldon, 239 Gold, Thomas, 21 Grand Unified Theory (GUT), 238 phase transition, 240 gravitational instability, 258–261 gravitational lens, 171 cluster of galaxies, 174 MACHO, 171 gravitational potential energy, 165 gravity, Newton vs. Einstein, 38 Guth, Alan, 242 Harrison-Zel’dovich spectrum, 275 helium (He), 209, 223 helium fraction () ), 212, 225 primordial, 217–218 High-z Supernova Search Team, 146 Hilbert, David, 42 Hipparcos satellite, 143 homogeneity, 11, 13 horizon distance (d hor ), 10, 20, 93, 245 at last scattering, 237 Benchmark Model, 120 flat universe matter only, 94 radiation only, 95 single-component, 93 horizon problem, 233, 234, 236–238 and isotropy of CMB, 238 resolved by inflation, 245–246 Hot Big Bang, 6, 30, 233 and production of CMB, 185 problems with, 233 hot dark matter, 276–277 power spectrum, 278 Hoyle, Fred, 21 Hubble constant (H 0 ), 15, 17, 18, 63, 127 Hubble distance (cH), 20, 50 at time of last scattering, 201 Hubble friction, 248, 252, 268 Hubble parameter (H), 63 Hubble time (H −1 0 ), 19, 75 relation to age of universe, 20, 92 Hubble’s Law, 15 consequence of expansion, 18 Hubble, Edwin, 15, 129, 141 Hydra-Centaurus supercluster, 183 hydrogen (H), 209 hydrostatic equilibrium, 169, 261 inflation, 242–253 and density perturbations, 253 and flatness problem, 244–245 and horizon problem, 245–246 and monopole problem, 246–247 inflaton field, 247, 252 ionization energy (C), 187 isotropy, 11–13 Jeans length (λ J ), 262 radiation, 264 Jeans, James, 262 Keplerian rotation, 160 kinetic energy, 165 lambda, see cosmological constant Langley, Samuel, 132 Large Magellanic Cloud, 142, 143, 172 distance to, 143 large scale structure, 255 last scattering, 186 INDEX 293 last scattering surface, 186 angular-diameter distance to, 196, 197, 238 thickness of, 195 Leavitt, Henrietta, 142 Lemaˆıtre universe, 113 lepton, 23 lithium (Li), 209, 223, 225 Lobachevski, Nikolai, 55 loitering universe, 113 evidence against, 116 lookback time, 120 Benchmark Model, 122 luminosity density, 20, 156 luminosity distance (d L ), 132, 136 flat universe, 135 related to angular-diameter dis- tance, 138 M31 (Andromeda galaxy), 2, 11, 15, 160–162 distance to, 143 MACHO, 171–173 magnetic monopole, 241 energy density, 241 number density, 241 magnitude, 146 absolute, 147 apparent, 147 mass gravitational (: g ), 33 inertial (: i ), 34 mass-to-light ratio, 156 cluster of galaxies, 170 Galaxy, 163 stars, 156 matter-dominated universe negative curvature, 105 positive curvature, 104–105 ultimate fate of, 108 matter-lambda equality, 84, 111, 112 Benchmark Model, 118 Matthew Effect, 258 MAXIMA, 196, 205 Maxwell, James Clerk, 239 Maxwell-Boltzmann equation, 190, 214 mean free path, 188 megaparsec (Mpc), 2 metric, 43 curved space, 43, 44 flat space, 43 homogeneous & isotropic, 44 Minkowski, 45 Robertson-Walker, 46 Milne universe, 87 moment of inertia, 166 monopole problem, 233, 234, 238– 242 resolved by inflation, 246–247 neutrino, 24, 175 flavors, 24 massive, 25 oscillation, 25, 176 neutrinos, 277 neutron, 23, 213 decay, 23, 213, 222 neutron-to-proton ratio, 214, 215 after freezeout, 216 Newton, Isaac, 32–38 nuclear binding energy (1), 210, 211 of deuterium, 210, 218 of helium, 210 nucleons, 209 294 INDEX null geodesic, 45 in Minkowski space, 45 Robertson-Walker, 51 Olbers’ Paradox, 7–11, 20–21, 179 Olbers, Heinrich, 7 optical depth (τ), 194 parallax distance, 132 parsec (pc), 2 particle horizon distance, see hori- zon distance peculiar motion, 61 Penzias, Arno, 28, 180 perfect gas law, 69 perturbation theory, 259 phase transition, 239 loss of symmetry, 239 photodissociation, 25, 218 photoionization, 25, 187 photon, 25 photon decoupling, 185, 189, 193 photon decoupling time (t dec ), 194, 195 photon-baryon fluid, 195, 203, 264 Planck energy (1 P ), 4 Planck length (/ P ), 3 Planck mass (` P ), 4 Planck temperature (1 P ), 4 Planck time (t P ), 4, 96 Poe, Edgar Allen, 10 Poincar´e, Henri, 164 Poisson distribution, 275 Poisson’s equation, 57, 72, 202 with cosmological constant, 73 power spectrum, 274, 279 scale invariant, 274 pressure (1), 57 additive, 80 negative, 68 primordial black holes, 276 proper distance Robertson-Walker metric, 49 proper distance (d p ), 47 at time of emission, 91 at time of observation, 89 flat universe lambda only, 98 matter only, 94 radiation only, 95 single-component, 93 time of observation, 130 proton, 23, 213 Pythagorean theorem, 40 quantum gravity, 97 radiation-matter equality, 85, 117 Benchmark Model, 118 radiative recombination, 187 radius of curvature (1), 43, 46 recombination, 185 compared to deuterium synthe- sis, 219 recombination temperature crude approximation, 190 refined calculation, 192 recombination time (t rec ), 192, 195 redshift (.), 15 related to scale factor, 50–52 related to time, 86 redshift survey, 255 reheating, 252 Robertson, Howard, 45 Robertson-Walker metric, 46 Rubin, Vera, 162 INDEX 295 Sachs-Wolfe effect, 202 Saha equation, 191, 219 nucleosynthetic analog, 220 Sakharov, Andrei, 230 Salam, Abdus, 239 Sandage, Allan, 128 scale factor (c), 46 empty universe, 87 flat universe lambda only, 98 matter + lambda, 109, 111 matter only, 94 radiation + matter, 117 radiation only, 95 single-component, 91 negative curvature matter only, 108 positive curvature matter only, 106 Taylor expansion, 126, 128 Slipher, Vesto, 15, 162 Small Magellanic Cloud, 142 solar luminosity (1 ), 3, 156 solar mass (M ), 2 sound speed (c s ), 70, 262 baryonic gas, 265 photon gas, 265 Space Interferometry Mission (SIM), 143 spherical harmonics, 197 standard candle, 132 determining H 0 , 141 standard yardstick, 136 angular resolution, 140 statistical equilibrium, 190 statistical weight (o), 190 Steady State, 21–22 supercluster, 13, 257 Supernova Cosmology Project, 146 supersymmetry, 176 Theory of Everything (TOE), 239 Thomson cross-section (σ e ), 188, 215 Thomson scattering, 188 time of last scattering (t ls ), 194, 195 Hubble distance, 201 top-down scenario, 278 topological defect, 240 cosmic string, 241 domain wall, 241 magnetic monopole, 241 triangle in curved space, 40, 42 in flat space, 39 tritium ( 3 He), 222 Turner, Michael, 71 type Ia supernova, 145 luminosity, 145, 146 vacuum energy density, 75, 76 vs. Planck energy density, 76 Virgo cluster, 2, 143, 183 distance to, 143 Virgocentric flow, 144 virial theorem, 167 visible universe, 93 void, 13, 257 Walker, Arthur, 45 Weinberg, Steven, 212, 239 Wheeler, John, 38 Wilson, Robert, 28, 180 WIMP, 27, 177, 266, 276 year (yr), 3 Zwicky, Fritz, 164 Contents Preface 1 Introduction 2 Fundamental Observations 2.1 Dark night sky . . . . . . . . . . 2.2 Isotropy and homogeneity . . . . 2.3 Redshift proportional to distance 2.4 Types of particles . . . . . . . . . 2.5 Cosmic microwave background . . 3 Newton Versus Einstein 3.1 Equivalence principle . . 3.2 Describing curvature . . 3.3 Robertson-Walker metric 3.4 Proper distance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . v 1 7 7 11 15 22 28 32 33 39 44 47 55 57 65 68 71 79 79 86 91 94 4 Cosmic Dynamics 4.1 Friedmann equation . . . . . . . 4.2 Fluid and acceleration equations 4.3 Equations of state . . . . . . . . 4.4 Learning to love lambda . . . . 5 Single-Component Universes 5.1 Evolution of energy density 5.2 Curvature only . . . . . . . 5.3 Spatially flat universes . . . 5.4 Matter only . . . . . . . . . . . . . ii . . . . CONTENTS 5.5 5.6 iii Radiation only . . . . . . . . . . . . . . . . . . . . . . . . . . 95 Lambda only . . . . . . . . . . . . . . . . . . . . . . . . . . . 97 101 . 104 . 108 . 112 . 116 . 118 126 . 126 . 131 . 136 . 141 . 144 155 . 156 . 159 . 164 . 170 . 175 179 . 180 . 185 . 189 . 196 . 201 208 . 209 . 213 . 218 . 222 . 228 6 Multiple-Component Universes 6.1 Matter + curvature . . . . . . 6.2 Matter + lambda . . . . . . . 6.3 Matter + curvature + lambda 6.4 Radiation + matter . . . . . . 6.5 Benchmark Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 Measuring Cosmological Parameters 7.1 “A search for two numbers” . . . . . 7.2 Luminosity distance . . . . . . . . . . 7.3 Angular-diameter distance . . . . . . 7.4 Standard candles & H0 . . . . . . . . 7.5 Standard candles & acceleration . . . 8 Dark Matter 8.1 Visible matter . . . . . . 8.2 Dark matter in galaxies . 8.3 Dark matter in clusters . 8.4 Gravitational lensing . . 8.5 What’s the matter? . . . 9 The 9.1 9.2 9.3 9.4 9.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Cosmic Microwave Background Observing the CMB . . . . . . . . Recombination and decoupling . . . The physics of recombination . . . Temperature fluctuations . . . . . . What causes the fluctuations? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 Nucleosynthesis & the Early Universe 10.1 Nuclear physics and cosmology . . . . 10.2 Neutrons and protons . . . . . . . . . . 10.3 Deuterium synthesis . . . . . . . . . . 10.4 Beyond deuterium . . . . . . . . . . . 10.5 Baryon – antibaryon asymmetry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iv 11 Inflation & the Very Early 11.1 The flatness problem . . 11.2 The horizon problem . . 11.3 The monopole problem . 11.4 The inflation solution . . 11.5 The physics of inflation . 12 The 12.1 12.2 12.3 12.4 12.5 Universe . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . universe . . . . . . . . . . CONTENTS 233 . 234 . 236 . 238 . 242 . 247 255 . 258 . 261 . 266 . 272 . 276 284 286 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Formation of Structure Gravitational instability . The Jeans length . . . . . Instability in an expanding The power spectrum . . . Hot versus cold . . . . . . Epilogue Annotated Bibliography Preface This book is based on my lecture notes for an upper-level undergraduate cosmology course at The Ohio State University. The students taking the course were primarily juniors and seniors majoring in physics and astronomy. In my lectures, I assumed that my students, having triumphantly survived freshman and sophomore physics, had a basic understanding of electrodynamics, statistical mechanics, classical dynamics, and quantum physics. As far as mathematics was concerned, I assumed that, like modern major generals, they were very good at integral and differential calculus. Readers of this book are assumed to have a similar background in physics and mathematics. In particular, no prior knowledge of general relativity is assumed; the (relatively) small amounts of general relativity needed to understand basic cosmology are introduced as needed. Unfortunately, the National Bureau of Standards has not gotten around to establishing a standard notation for cosmological equations. It seems that every cosmology book has its own notation; this book is no exception. My main motivation was to make the notation as clear as possible for the cosmological novice. I hope that reading this book will inspire students to further explorations in cosmology. The annotated bibliography at the end of the text provides a selection of recommended cosmology books, at the popular, intermediate, and advanced levels. Many people (too many to name individually) helped in the making of this book; I thank them all. I owe particular thanks to the students who took my undergraduate cosmology course at Ohio State University. Their feedback (including nonverbal feedback such as frowns and snores during lectures) greatly improved the lecture notes on which this book is based. Adam Black and Nancy Gee, at Addison Wesley, made possible the great leap from rough lecture notes to polished book. The reviewers of the text, v vi PREFACE both anonymous and onymous, pointed out many errors and omissions. I owe particular thanks to Gerald Newsom, whose careful reading of the entire manuscript improved it greatly. My greatest debt, however, is to Rick Pogge, who acted as my computer maven, graphics guru, and sanity check. (He was also a tireless hunter of creeping fox terrier clones.) As a small sign of my great gratitude, this book is dedicated to him. or cosmos. very remote from the concerns of humanity. planets orbit stars. Attempting to cover the study of the entire universe in a single volume may seem like a megalomaniac’s dream. Only near the end of the book will we consider how relatively small objects.Chapter 1 Introduction Cosmology is the study of the universe. The universe. and superclusters. model universe. A science which regards entire galaxies as being small objects might seem. Nevertheless. cosmologists sometimes must smooth over small “blemishes” such as galaxies. therefore. clusters. at first glance. are formed as the universe evolves. the only way to condense its history into a single book is by a process of ruthless simplification. stars are collected into galaxies. and even clusters of galaxies are found within larger superclusters. regarded as a whole. we will be considering the properties of an idealized. that the words “cosmology” and “cosmetology” come from the same Greek root: the word “kosmos”. with structures on a vast range of scales. is richly textured. such as galaxies. Given the richness and complexity of the universe. and predicting the future of the universe. Cosmologists ask questions such as “What is the universe made of? Is it finite or 1 . meaning harmony or order. Just as cosmetologists try to make a human face more harmonious by smoothing over small blemishes such as pimples and wrinkles. perfectly smooth. after all. It is amusing to note. in this context.1): “Where do we come from? What are we? Where are we going?” Cosmology grapples with these questions by describing the past. For much of this book. galaxies are gravitationally bound into clusters. cosmology deals with questions which are fundamental to the human condition. The questions which vex humanity are given in the title of a painting by Paul Gauguin (Figure 1. explaining the present. One distance unit used by astronomers is the astronomical unit (AU). we can adopt the units which have been developed by astronomers for dealing with large distances. To measure intergalactic distances. masses. The total mass . equal to 106 pc. and the second (s) is awkwardly brief. Fortunately. we are at a distance of 0. 1 pc = 3. it is small compared to the distances between galaxies. the kilogram (kg) is awkwardly tiny. Boston] infinite in spatial extent? Did it have a beginning some time in the past? Will it come to an end some time in the future?” Cosmology deals with distances that are very large. Although the astronomical unit is a useful length scale within the Solar System. Although the parsec is a useful length scale within our Galaxy. For example. 1 AU = 1. in metric units. or 3.2 CHAPTER 1.5 × 1011 m. it is small compared to the distances between stars. objects that are very big. and times. equal to the mean distance between the Earth and Sun. the Sun’s mass is 1 M = 2. we are at a distance of 1. [Museum of Fine Arts. and timescales that are very long. it is useful to use the megaparsec (Mpc). equal to the distance at which 1 AU subtends an angle of 1 arcsecond. To measure interstellar distances. in metric units. Cosmologists frequently find that the standard SI units are not convenient for their purposes: the meter (m) is awkwardly short.1 × 1022 m.0 × 1030 kg. it is useful to use the parsec (pc). 1897.3 pc from Proxima Centauri (the Sun’s nearest neighbor among the stars) and 8500 pc from the center of our Galaxy.1: Where Do We Come From? What Are We? Where Are We Going? Paul Gauguin. For example. in metric units. The standard unit of mass used by astronomers is the solar mass (M ).1 × 1016 m. INTRODUCTION Figure 1.7 Mpc from M31 (otherwise known as the Andromeda galaxy) and 15 Mpc from the Virgo cluster (the nearest big cluster of galaxies). particle physicists have plunged into cosmology. For instance. it is Mgal ≈ 1012 M . and the reduced Planck constant. Combining the Newtonian gravitational constant.511 MeV.7 × 10−11 m3 kg−1 s−2 . and the rest energy of a proton is mP c2 = 938. equal to 109 yr. The Sun’s luminosity (that is. A more universal. based on the universal constants G. or 3.0 × 108 m s−1 . the universe was very hot and dense. cosmology also deals with very small things. and h.2 × 1016 s.5 × 1017 s. less culturally biased. For example. particle physicists tend to measure energy units in electron volts (eV) instead of joules (J).1) . known as the Planck length: P ≡ h G¯ 3 c 1/2 = 1. and some interesting particle physics phenomena were occurring. the speed of light.6 Gyr than as 1.6 × 10−35 m . introducing some terminology and units of their own. the rate at which it radiates away energy in the form of light) is 1 L = 3. kilograms. solar masses. h = h/(2π) = 1. In cosmological context. seconds. In addition to dealing with very large things. and gigayears could only be devised by ten-fingered Earthlings obsessed with the properties of water. For times much longer than a second. defined as the time it takes the Earth to go once around the Sun. incidentally. ¯ yields a unique length scale. The conversion factor between electron volts and joules is 1 eV = 1.1 × 10−34 J s = 6. An eighteen-tentacled silicon-based lifeform from a planet orbiting Betelgeuse would devise a different set of units. The Sun. also provides the standard unit of power used in astronomy. ¯ G = 6. Early in its history. When you stop to think of it. as we shall see.6 × 10−19 J. c.2 × 107 s. The total luminosity of our Galaxy is Lgal = 3. in round numbers. is me c2 = 511.6 × 10−16 eV s.000 eV = 0. a year is frequently an inconveniently short period of time.3 of our Galaxy is not known as accurately as the mass of the Sun. megaparsecs.3 MeV. you realize that the units of meters. (1. One year is approximately equal to 3. The rest energy of an electron.8 × 1026 watts. for instance. though.6 × 1010 L . astronomers use the year (yr). c = 3. system of units is the Planck system. Consequently. so cosmologists frequently use gigayears (Gyr). the age of the Earth is more conveniently written as 4. is equal to its mass. but it simply means that the rest energy of an object. mass. EP = MP c2 = 2. However.4 CHAPTER 1. h. and/or G from all ¯ equations. and G.4 × 1032 K .3) Using Einstein’s relation between mass and energy. (1.0 × 109 J = 1. (1.6 × 10−5 eV K−1 .5) When distance. (1. The blatant dimensional incorrectness of such an equation is jarring. k. This is convenient for individuals who ¯ have difficulty in remembering the numerical values of physical constants. time. is ultimately based The Planck mass is roughly equal to the mass of a grain of sand a quarter of a millimeter across. I will retain all factors of c. E = mc2 .1 MP ≡ and the Planck time. Einstein’s celebrated equation. to be sure. For instance. many cosmology texts. a span of some 61 orders of magnitude. measured in units of the Planck energy. TP = EP /k = 1.4 × 10−44 s . (1. however. we can also define the Planck temperature. into the act. 1 . cosmologists are not permitted to let their imaginations run totally unfettered. k = 8.2) = 5. after noting that c = k = h = G = 1 ¯ when Planck units are used. then c = k = h = G = 1. measured in units of the Planck mass. k. we will be dealing with distances ranging from the Planck length to 104 Mpc or so.2 × 10−8 kg . and temperature are measured in the appropriate Planck units. tP ≡ G¯ h c5 1/2 hc ¯ G 1/2 = 2. In this book. However. In this book. Dealing with such a wide range of length scales requires a stretch of the imagination. For instance.2 × 1028 eV . INTRODUCTION The same constants can be combined to yield the Planck mass. for the sake ¯ of clarity. then proceed to omit c. h. becomes E = m. we can also define the Planck energy. using Planck units can have potentially confusing side effects. I emphasize strongly. Cosmology.4) By bringing the Boltzmann constant. that the Earth and Moon are spherical. ca. Between the Earth and the celestial sphere. Greek cosmology was based on more precise and sophisticated observations. and that the distance from the Earth to the stars is much greater than the Earth’s diameter. Based on this knowledge. Ancient Greek astronomers deduced. Ancient Egyptians. arches over the earth god Geb (from the Greenfield Papyrus. Even in ancient times. that the Sun is much farther from the Earth than the Moon is. unfortunately. Moon.2) was covered by a solid dome (symbolized by the sky goddess Nut). cosmology was based on observations. in this model. they developed a model of the universe in which a flat Earth (symbolized by the earth god Geb in Figure 1. those observations were frequently imperfect and incomplete.5 Figure 1. from their observations. in which the spherical Earth is surrounded by a much larger celestial sphere. looked at the desert plains stretching away from the Nile valley and the blue sky overhead. a spherical shell to which the stars are attached. supported by the air god Shu. the Sun. Although cosmology is ultimately based on observation. and planets move on their complicated apparatus of epicycles and deferents. sometimes obser- . Based on their observations. [ c Copyright The British Museum] on observation of the universe around us. Greek cosmologists devised a “twosphere” model of the universe. for instance. 1025 BC).2: The ancient Egyptian view of the cosmos: the sky goddess Nut. which states that the universe has expanded from an initially hot and dense state to its current relatively cool and tenuous state. when Copernicus proposed a new Sun-centered model of the universe. let us turn. Foucault didn’t demonstrate the rotation of the Earth. in the next chapter. cosmologists may adopt a new model for aesthetic or philosophical reasons. INTRODUCTION vations temporarily lag behind theory.] Cox (2000): Accurate values of physical and astronomical constants Harrison (2000). nearly two centuries after the death of Copernicus. although observations sometimes lag behind theory in this way. rather than vice versa. Direct observational evidence didn’t reveal that the Earth revolves around the Sun. he believed that putting the Earth in motion around the Sun resulted in a conceptually simpler. another prediction of the Copernican model. However. Suggested reading [Full references are given in the “Annotated Bibliography” on page 286. For instance.6 CHAPTER 1. until 1851. To see why cosmologists have embraced the Hot Big Bang model. Rather. over three centuries after the death of Copernicus. until the discovery of the aberration of starlight in the year 1728. ch. more appealing model of the universe. 1 – 4: A history of early (pre-Einstein) cosmology . every cosmological model that isn’t eventually supported by observational evidence must remain pure speculation. The current standard model for the universe is the “Hot Big Bang” model. he didn’t base his model on new observational discoveries. to the fundamental observations on which modern cosmology is based. During periods when data are lacking. and that the expansion is still going on today. to replace the Earth-centered two-sphere model of the Greeks. requiring elaborate apparatus and sophisticated data analysis. to whom stars were points of light stuck to a dome or sphere. 2. who wrote a scientific paper on the subject in the year 1826. Let’s start with an observation which is deceptive in its extreme simplicity.1 Why should it be paradoxical that the night sky is dark? Most of us simply take for granted the fact that daytime is bright and nighttime is dark.Chapter 2 Fundamental Observations Some of the observations on which modern cosmology is based are highly complex. Olbers was not the first person to think about Olbers’ Paradox. is known as Olbers’ Paradox. with roughly two thousand stars scattered across it. after the astronomer Heinrich Olbers. with only a few pinpoints of light to mark the location of stars. other observations are surprisingly simple. 1 7 . instead of being uniformly bright with starlight. The fact that the night sky is dark at visible wavelengths. You will see a dark sky.1 The night sky is dark Step outside on a clear. However. Thomas Digges mentioned how strange it is that the night sky is dark. However. the cosmological model of Copernicus required that the The name “Olbers’ Paradox” is thus a prime example of what historians of science jokingly call the law of misonomy: nothing is ever named after the person who really discovers it. As early as 1576. As it happens. and look upward. far from city lights. The darkness of the night sky certainly posed no problems to the ancient Egyptians or Greeks. moonless night. the power per unit area per steradian of the sky) will be dJ(r) = L nL · n · r2 dr = dr . centered on the Earth (Figure 2. centered on the Earth. Let’s compute how bright we expect the night sky to be in an infinite universe. the parallax of the stars. The flux received here at Earth from a star of luminosity L at a distance r is given by an inverse square law: f (r) = L . 4πr2 (2. since the Copernican system no longer requires that the stars be attached to a rotating celestial sphere. The intensity of radiation from the shell of stars (that is. with radius r and thickness dr. FUNDAMENTAL OBSERVATIONS r dr Figure 2. of radius r and thickness dr. would be large enough to see with the naked eye. like the Sun. Moreover. and let L be the average stellar luminosity. distance to stars be very much larger than an astronomical unit. to embrace a model in which stars are large glowing spheres. as the Earth goes around on its orbit.2) .1) Now consider a thin spherical shell of stars.1). scattered throughout infinite space. Let n be the average number density of stars in the universe. These liberating realizations led Thomas Digges.8 CHAPTER 2.1: A star-filled spherical shell. 4πr2 4π (2. and other post-Copernican astronomers. otherwise. the stars can be at different distances from the Sun. Heinrich Olbers himself tried to resolve Olbers’ Paradox by proposing that distant stars are hidden from view by interstellar matter which absorbs starlight.1. (2. Therefore. one (or more) of the assumptions that went into the above analysis of the sky brightness must be wrong. Distant stars might be less luminous or less numerous than nearby stars. If the universe only extends to a maximum distance rmax from us. This might not be true.2. and is glowing as brightly as the stars themselves. if distant stars are sufficiently low in luminosity compared to nearby stars.3) is that the product nL is constant as a function of r. Similarly. This is not true. the assumption made in equation (2. in an infinite distribution of stars. A second assumption I made is that the number density n and mean luminosity L of stars are constant throughout the universe. the product nL must fall off more rapidly than nL ∝ 1/r as r → ∞. more accurately.3) to be finite.3) Thus. Nevertheless. nearby stars will hide more distant stars from our view. because the interstellar matter will be heated by starlight until it has the same temperature as the surface of a star. In fact. DARK NIGHT SKY 9 The total intensity of starlight from a shell thus depends only on its thickness. not on its distance from us. every line of sight should end at the surface of a star. One assumption that I made is that we have an unobstructed line of sight to every star in the universe. A third assumption is that the universe is infinitely large. This is an improvement on an infinitely bright sky. This might not be true. the interstellar matter emits as much light as it absorbs. but is still distinctly different from the dark sky which we actually see. then the total intensity of starlight we see in the night sky will be J ∼ . they won’t contribute significantly to the sky brightness. this would imply a surface brightness for the sky equal to the surface brightness of a typical star. In order for the integrated intensity in equation (2. I have demonstrated that the night sky is infinitely bright. At that point. then the absence of stars at large distances will keep the night sky from being bright. Let’s scrutinize some of the assumptions. We can compute the total intensity of starlight from all the stars in the universe by integrating over shells of all radii: ∞ J= r=0 dJ = nL 4π ∞ 0 dr = ∞ . since stars have a finite angular size as seen from Earth. This is utter nonsense. If we are in a clump of stars of finite size. This resolution will not work. would be by supposing the distance of the invisible background so immense that no ray 2 That’s right. FUNDAMENTAL OBSERVATIONS nLrmax /(4π). If the universe is systematically expanding or contracting.3 minutes ago. Thus. In addition. The stars beyond some finite distance. completed in the year 1848. Thus. A fourth assumption. Since the speed of light is finite. under such a state of affairs. A particularly amusing bit of cosmological trivia is that the first person to hint at the correct resolution of Olbers’ Paradox was Edgar Allen Poe. Poe wrote. Proxima Centauri as it was 4 years ago. therefore. not a suspense novel. . “Were the succession of stars endless. but is devoid of stars beyond a distance rmax . . when we look farther out in space. This might not be true. The only mode. This is a textbook. The assumption that f ∝ 1/r 2 would have seemed totally innocuous to Olbers and other nineteenth century astronomers. Note that this result will also be found if the universe is infinite in space. Note that this result will also be found if the universe is infinitely old. we could comprehend the voids which our telescopes find in innumerable directions. at which would not exist a star. in the twentieth century. demonstrated that the universe might not obey the laws of Euclidean geometry. However. since there could be absolutely no point.2 In his essay “Eureka: A Prose Poem”. the infinitely large. If the universe has a finite age t0 . that great questioner of assumptions. and M31 as it was 2 million years ago. Albert Einstein. we see the Sun as it was 8. slightly more subtle than the previous ones. so I’ll tell you right now: the primary resolution to Olbers’ Paradox comes from the fact that the universe has a finite age. then the background of the sky would present us an [sic] uniform density . but has only contained stars for a finite time t0 . .1). the “Nevermore” guy. the inverse square law assumes that the source of light is stationary relative to the observer. after all. are invisible to us because their light hasn’t had time to reach us yet. A fifth assumption is that the flux of light from a distant source is given by the inverse square law of equation (2. the intensity of starlight we see at night will be at most J ∼ nLct0 /(4π). Euclidean universe which Thomas Digges and his successors pictured simply does not hold up to scrutiny. This might not be true. the inverse square law follows directly from Euclid’s laws of geometry. we are looking farther out in time. eternally old. in which. then the light from distant sources will be redshifted to lower photon energies or blueshifted to higher photon energies. called the horizon distance.10 CHAPTER 2. is that the universe is infinitely old. in all that background. 2c). just step outside on a clear . for example. In this context. our Galaxy and M31 define a preferred direction. The object doesn’t hover in midair. the universe is still anisotropic. the universe is blatantly anisotropic. Within this sphere. This sphere contains the Local Group of galaxies. it is the direction commonly referred to as “down”. for example. ISOTROPY AND HOMOGENEITY from it has yet been able to reach us at all. and point to that really bright disk of light up in the sky. there is a preferred direction. it looks the same no matter which way you point your telescope. Consider. Within this sphere. for example. the universe is still anisotropic. a sphere 3 meters in diameter. centered on your navel (Figure 2. a sphere 3 AU in diameter. Consider. the universe is isotropic and homogeneous What does it mean to state that the universe is isotropic and homogeneous? Saying that the universe is isotropic means that there are no preferred directions in the universe. it falls down. It is fairly easy to determine the vector pointing from our Galaxy to M31.2b). which is by far the most massive and most luminous object within the sphere. It is easy to determine the vector pointing down. it looks the same no matter where you set up your telescope.2.2 On large scales. Thus. and it doesn’t move in a random direction. “large scales” means that the universe is only isotropic and homogeneous on scales of roughly 100 Mpc or more. It is easy to determine the vector pointing toward the Sun.” 11 2. The isotropy of the universe is not immediately obvious. centered on your navel (Figure 2. which together contribute about 86% of the total luminosity within the 3 Mpc sphere. toward the center of the Earth. By far the most massive and most luminous galaxies in the Local Group are our own Galaxy and M31. it is the direction pointing toward the Sun. on small scales. centered on your navel (Figure 2. Consider. On significantly larger scales.2a). In fact. Saying that the universe is homogeneous means that there are no preferred locations in the universe. within this sphere.2. Just let go of a small dense object. On still large scales. a small cluster of some 40 galaxies. there is a preferred direction. a sphere 3 Mpc in diameter. Just step outside on a sunny day. Note the very important qualifier: the universe is isotropic and homogeneous on large scales. 522. 1. centered on your navel. lighter regions represent lower density (from Dekel et al. fig. centered on your navel. (c) A sphere 3 Mpc in diameter. ApJ. Shown is the number density of galaxies smoothed with a Gaussian of width 17 Mpc. centered on your navel. centered on your navel. The heavy contour is drawn at the mean density. darker regions represent higher density. (b) A sphere 3 AU in diameter.2: (a) A sphere 3 meters in diameter. FUNDAMENTAL OBSERVATIONS Figure 2.12 CHAPTER 2. (d) A sphere 200 Mpc in diameter. 1999. 2) . Note that homogeneity does not imply isotropy. You can tell where you are relative to the center by measuring the radius of curvature of the nearest ring. ISOTROPY AND HOMOGENEITY 13 night when the constellation Andromeda is above the horizon. the average density of the universe as a whole is ρ0 ∼ 3 × 10−27 kg m−3 . the patch of the universe surrounding you is more than 28 orders of magnitude denser than average. saying that something is homogeneous is quite different . On significantly larger scales. In general. the universe is still inhomogeneous. will have an average density of ∼ 3 × 10−26 kg m−3 . surveys of the universe on still larger scales don’t find “superduperclusters”.2. centered on your navel.2d shows a slice through such a sphere. it seems. the universe is obviously inhomogeneous. centered on your navel.2d. Note also that isotropy around a single point does not imply homogeneity. Consider a sphere 200 Mpc in diameter. For instance. However. will have an average density of ∼ 100 kg m−3 . These are the largest structures in the universe.3b) is isotropic around the center of the bullseye.3a) is homogeneous on scales larger than the stripe width. The Perseus-Pisces supercluster is on the right. and point to the fuzzy oval in the middle of the constellation. The direction of the stripes provides a preferred direction by which you can orient yourself. the Hydra-Centaurus supercluster is on the left. and are separated by voids (low density regions) which are typically ∼ 100 Mpc across. then. still an order of magnitude denser than the universe as a whole. It isn’t until you get to considerably larger scales that the universe can be considered as isotropic. the universe is still inhomogeneous. A sphere 3 AU in diameter. centered on your navel.2. On still larger scales. or lumpy. and the edge of the Coma supercluster is just visible at the top of Figure 2. has an average density of 4 × 10−5 kg m−3 . The rings of the bullseye look different far from the center than they do close to the center. centered on your navel. A sphere 3 Mpc in diameter. in addition to being anisotropic. a sphere 3 meters in diameter. that’s 22 orders of magnitude denser than the average for the universe. Superclusters are typically ∼ 100 Mpc along their longest dimensions. On small scales. Figure 2. A sheet of paper printed with stripes (Figure 2. with superclusters of galaxies indicated as dark patches. A sheet of paper printed with a bullseye (Figure 2. but it is not isotropic. on a scale d ∼ 3 m. It’s only when you contemplate a sphere ∼ 100 Mpc in diameter that a sphere centered on your navel is not overdense compared to the universe as a whole. Thus. but is it not homogeneous. in round numbers. most spheres 3 Mpc across don’t contain a pair of bright galaxies. On smaller scales.” The cosmological principle holds true only on large scales (of 100 Mpc or more). modern cosmologists have adopted the cosmological principle. your navel obviously is in a special location. . statistically speaking. In fact. such as your navel. who pointed out that the Earth is not the center of the universe. appears to be isotropic around us. implies isotropy around every point in the universe. but which is inhomogeneous. most spheres over 100 Mpc across do contain roughly the same pattern of superclusters and voids. from saying it is isotropic. most sphere 3 AU across don’t contain a star. Isotropy around any point in the universe. but which is homogeneous on scales larger than the stripe width.14 CHAPTER 2. Most spheres 3 meters across don’t contain a sentient being. FUNDAMENTAL OBSERVATIONS Figure 2. there is no center to the universe.3: (a) A pattern which is anisotropic. Later cosmologists also pointed out the Sun is not the center. that our Galaxy is not the center. (b) A pattern which is isotropic about the origin. on scales of 100 Mpc or more. However. However. The universe. and that the Local Group is not the center. which states “There is nothing special about our location in the universe. The cosmological principle has the alternate name of the “Copernican principle” as a tribute to Copernicus. combined with the cosmological principle. and isotropy around every point in the universe does imply homogeneity. this quantity is called a blueshift. when we take a galaxy’s spectrum at visible wavelengths. from a plot of redshift (z) versus distance (r). We say that the galaxy has a redshift z. In 1912. this galaxy. (2. measured the shift in wavelength of the light from M31.5) c where H0 is a constant (now called the Hubble constant). in general. By 1929. it typically contains absorption lines created in the stars’ relatively cool upper atmospheres. the vast majority of galaxies have z > 0. rather than a redshift. is one of the few which exhibits a blueshift. REDSHIFT PROPORTIONAL TO DISTANCE 15 2. Thus.3 Suppose we consider a particular absorption line whose wavelength.2. measuring its distance is difficult. be the same. λob .4) Strictly speaking. Slipher had measured the shifts in the spectral lines for approximately 40 galaxies. By 1925. The wavelength we measure for the same absorption line in a distant galaxy’s spectrum. as it turns out. Hubble knew z for nearly 50 galaxies. enough galaxy redshifts had been measured for the cosmologist Edwin Hubble to make a study of whether a galaxy’s redshift depends on its distance from us. we are primarily detecting the light from the stars which the galaxy contains. will not. but had estimated distances for only 20 of them. the exceptions were all nearby galaxies within the Local Group. when z < 0. given by the formula z≡ λob − λem . Hubble interpreted the observed redshift of galaxies as being a Doppler shift due to their radial Galaxies containing active galactic nuclei will also show emission lines from the hot gas in their nuclei. 3 .3 Galaxies show a redshift proportional to their distance When we look at a galaxy at visible wavelengths. Nevertheless. finding that they were nearly all redshifted. However. was not known until the twentieth century. as measured in a laboratory here on Earth. and can be done with high precision. is λem . he found the famous linear relation now known as Hubble’s Law: H0 z= r. rather than blueshifted to shorter wavelengths.4. Although measuring a galaxy’s redshift is relatively easy.3. Vesto Slipher. reproduced in Figure 2. λem (2. at the Lowell Observatory. The fact that the light from galaxies is generally redshifted to longer wavelengths. 168) . Proc.. Note that the vertical axis actually plots cz rather than z – and that the units are accidentally written as km rather than km/s. FUNDAMENTAL OBSERVATIONS Figure 2.16 CHAPTER 2. Sci. 15. Nat.4: Edwin Hubble’s original plot of the relation between redshift (vertical axis) and distance (horizontal axis). Acad. (from Hubble 1929. 2001. he was able to use the classical. as described in Chapter 6. it is customarily written in the rather baroque units of km s−1 Mpc−1 .2. (from Freedman. In this case.4). where v is the radial velocity of the light source (in this case. z = v/c. a galaxy). using data obtained by (appropriately enough) the .04). the galaxy distances have been determined using Cepheid variable stars as standard candles. nonrelativistic relation for the Doppler shift.3. 47) velocity away from Earth. he thought that the numerical value of the Hubble constant was H0 = 500 km s−1 Mpc−1 (see Figure 2. showing cz versus distance. When Hubble first discovered Hubble’s Law. Figure 2. However. REDSHIFT PROPORTIONAL TO DISTANCE 17 Figure 2.5: A more modern version of Hubble’s plot. Since the values of z in Hubble’s analysis were all small (z < 0. ApJ. et al.5 shows a more recent determination of the Hubble constant from nearby galaxies. Hubble’s law takes the form v = H0 r . Interpreting the redshifts as Doppler shifts. 553. it turned out that Hubble was severely underestimating the distances to galaxies.6) Since the Hubble constant H0 can be found by dividing velocity by distance. (2. Hubble Space Telescope. We see distant galaxies moving away from us.7) This is the value for the Hubble constant that I will use in the remainder of this book.6) with sides of length r12 ≡ |r1 − r2 | r23 ≡ |r2 − r3 | r31 ≡ |r3 − r1 | . “Surely it must be a violation of the cosmological principle to have all those distant galaxies moving away from us! It looks as if we are at a special location in the universe – the point away from which all other galaxies are fleeing. (2.9) (2. but observers in any other galaxy would also see distant galaxies moving away from them.6: A triangle defined by three galaxies in a uniformly expanding universe.” In fact. consider three galaxies at positions r1 .10) . r2 .8) (2. The best current estimate of the Hubble constant. combining the results of different research groups. is H0 = 70 ± 7 km s−1 Mpc−1 . (2. when first encountering Hubble’s Law. and r3 .18 CHAPTER 2. They define a triangle (Figure 2. To see on a more mathematical level what we mean by homogeneous. what we see here in our Galaxy is exactly what you would expect to see in a universe which is undergoing homogeneous and isotropic expansion. Cosmological innocents sometimes exclaim. FUNDAMENTAL OBSERVATIONS 2 r12 r31 r23 3 1 Figure 2. isotropic expansion. with a/a playing the role of the Hubble constant. Consider a pair of galaxies currently separated by a distance r. the velocity – distance relation takes the linear form v = Hr. Maintaining the correct relative lengths for the sides of the triangle requires an expansion law of the form r12 (t) = a(t)r12 (t0 ) r23 (t) = a(t)r23 (t0 ) r31 (t) = a(t)r31 (t0 ) .13) Here the function a(t) is a scale factor which is equal to one at the present moment (t = t0 ) and which is totally independent of location or direction. If there are no forces acting to accelerate or decelerate their relative motion. (2. For H0 = 70 ± 7 km s−1 Mpc−1 . The scale factor a(t) tells us how the expansion (or possibly contraction) of the universe depends on time. At any time t. then one Hubble time ago. the observation of galactic redshifts lead . If the relative velocities of galaxies have been constant in the past.16) v H0 r −1 independent of the current separation r. an observer in galaxy 1 will see the other galaxies receding with a speed v12 (t) = dr12 = ar12 (t0 ) = ˙ dt dr31 v31 (t) = = ar31 (t0 ) = ˙ dt a ˙ r12 (t) a a ˙ r31 (t) . isotropic expansion. ˙ If galaxies are currently moving away from each other.14) (2.2. (2. with H = a/a. and the time that has elapsed since they were in contact is r r −1 t0 = = = H0 . all the galaxies in the universe were crammed together into a small volume.15) You can easily demonstrate that an observer in galaxy 2 or galaxy 3 will find the same linear relation between observed recession speed and distance. The time H0 is referred to as the −1 Hubble time.12) (2. then their velocity is constant. this implies they were closer together in the past. Since this argument can ˙ be applied to any trio of galaxies.3. the Hubble time is H0 = 14. it implies that in any universe where the distribution of galaxies is undergoing homogeneous. Thus.11) (2.0 ± 1. with a velocity v = H0 r relative to each other.4 Gyr. a (2. REDSHIFT PROPORTIONAL TO DISTANCE 19 Homogeneous and uniform expansion means that the shape of the triangle is preserved as the galaxies move away from each other. This rough equivalence is reassuring.) Note how Hubble’s Law ties in with Olbers’ Paradox. −1 and the universe may be older than H0 . Just as the age of the universe is roughly equal to H0 in most Big Bang models. If the universe is −1 of finite age. because light from distant galaxies has not yet had time to reach us. If gravity working on matter is the only force at work on large scales. the age of the universe – that is. again. The Hubble time of ∼ 14 Gyr is comparable to the ages computed for the oldest known stars in the universe. so the horizon distance (the greatest distance a photon can travel during the age of the universe) is roughly equal to c/H0 . and the universe is −1 younger than H0 . depending on the expansion history. FUNDAMENTAL OBSERVATIONS naturally to a Big Bang model for the evolution of the universe. By terrestrial standards. On the other hand. (2. the universe was expanding more rapidly in the past than it is now. with the exact value. However. t0 ∼ H0 . If the horizon distance is dhor ∼ c/H0 . c/H0 = 4300 ± 400 Mpc. the universe is not a well-lit place. Just as the Hubble time provides a natural time scale for our universe. then the night sky can be dark. this luminosity density is equivalent to a single 40 watt light bulb within a sphere 1 AU in radius. the Hubble distance. then the dominant gravitational force is repulsive. provides a natural distance −1 scale. then the total flux of light we receive from all the stars from all the galaxies within the horizon will be Fgal = 4πJgal ≈ nL 11 rH 0 dr ∼ nL ∼ 9 × 10 L Mpc−2 c H0 ∼ 2 × 10−11 L AU−2 .17) nL ≈ 2 × 108 L Mpc−3 . (Later chapters will deal with computing the exact values of the age and horizon size of our universe. and that galaxies contain matter. We know that gravity exists. the time elapsed since its original highly dense state – is not necessarily exactly equal to the Hubble time. if the energy density of the universe is dominated by a cosmological constant (an entity which we’ll examine in more detail in Chapter 4). even if the universe is infinitely large. then the attractive force of gravity will act to slow the expansion.20 CHAPTER 2. with the exact value depending on the expansion history of the universe. Galaxy surveys tell us that the luminosity density of galaxies in the local universe is (2. In this case.18) . A Big Bang model may be broadly defined as a model in which the universe expands from an initially highly dense state to its current low-density state. In a Big Bang model.08 L AU−2 . The Steady State model was first proposed in the 1940’s by Hermann Bondi. 4π AU2 (2. in violation of the assumed “perfect cosmological principle”. remain constant with time. REDSHIFT PROPORTIONAL TO DISTANCE 21 By the cosmological principle. The volume of a spherical region of space. who were proponents of the perfect cosmological principle. Thomas Gold. If there existed an instant in time at which the universe started expanding (as in a Big Bang model). in which homogeneous and isotropic expansion causes the density of the universe to decrease steadily from its initial high value. In a Steady State universe. there are no privileged moments in time.3. the average density decreases. However. in a Steady State model. such as the mean density ρ0 and the Hubble constant H0 .22) 3 . Comparing this to the flux we receive from the Sun. Fsun = 1L ≈ 0. the velocity – distance relation dr = H0 r dt (2. which states that not only are there no privileged locations in space. the properties of the universe evolve with time. the total flux of starlight at a randomly selected location in the universe is less than a billionth the flux of light we receive from the Sun here on Earth. For the entire universe to be as well-lit as the Earth. and Fred Hoyle.19) we find that Fgal /Fsun ∼ 3 × 10−10 .2. a Steady State universe is infinitely old. that would be a special moment. (2. (2. it would have to be over a billion times older than it is. this is the total flux of starlight you would expect at any randomly located spot in the universe.20) can be easily integrated. a Steady State universe is one in which the global properties of the universe. Hubble’s Law occurs naturally in a Big Bang model for the universe. the mean distance between galaxies increases. since H0 is constant with time. to yield an exponential law: r(t) ∝ eH0 t .21) Note that r → 0 only in the limit t → −∞. Thus. Thus. and you’d have to keep the stars shining during all that time. increases exponentially with time: 4π 3 V = r ∝ e3H0 t . Hubble’s Law can also be explained by a Steady State model. and so forth. shoes. and what have you. The properties of the universe do. FUNDAMENTAL OBSERVATIONS However. in fact.4 The Steady State model finally fell out of favor when observational evidence increasingly indicated that the perfect cosmological principle is not true. the most significant difference between the different components of the universe is that they are made of different elementary particles. (2. the density of the sphere must remain constant.24) V This corresponds to creating roughly one hydrogen atom per cubic kilometer per year. matter must be continuously created at a rate ˙ ˙ Mss = ρ0 V = ρ0 3H0 V . cabbages and kings are nearly indistinguishable – the main difference between them is that the mean mass per king is greater than the mean mass per cabbage. The properties of the most cosmologically important The name “Big Bang” was actually coined by Fred Hoyle. 2. then matter would have to be created at a rate ˙ Mss = 3H0 ρ0 ∼ 6 × 10−28 kg m−3 Gyr−1 . though.22 CHAPTER 2. with matter density ρ0 ∼ 3 × 10−27 kg m−3 . (2.5. During the 1950s and 1960s. discussed below in section 2. From a cosmological viewpoint. The discovery of the Cosmic Microwave Background. sealing wax.4 The universe contains different types of particles It doesn’t take a brilliant observer to confirm that the universe contains a large variety of different things: ships. a supporter of the Steady State model. cabbages. is commonly regarded as the observation which decisively tipped the scales in favor of the Big Bang model. 4 . From a cosmologist’s viewpoint. kings. To have a constant density of matter within a growing volume. the Big Bang and Steady State models battled for supremacy. if the universe is in a steady state. change with time. Supporters of the Steady State model pointed out that the continuous creation of matter is no more absurd that the instantaneous creation of the entire universe in a single “Big Bang” billions of years ago.23) If our own universe. Critics of the Steady State model pointed out that the continuous creation of matter violates mass-energy conservation. happened to be a Steady State universe. galaxies. 3 +1 neutron n 939. By contrast. a class of elementary particles which are not made of quarks. the universe is electrically neutral. A neutron (n) contains one “up” quark and two “down” quarks.νµ . The mass of an electron is much smaller than that of a neutron or proton. and electrons. For this reason. experiments have put a lower limit on the decay time of the proton which is very much greater than the Hubble time.4.ντ ? 0 photon γ 0 0 dark matter ? ? 0 23 particles are summarized in Table 2. about a quarter of an hour. and molecules is generally referred to as baryonic matter.6 MeV. The material objects which surround us during our everyday life are made of protons. Thus a proton has a net positive charge of +1. the mass density of electrons is only a small perturbation to the mass density of protons and neutrons. An electron has an electric charge equal in magnitude to that of a proton. in their rest energies. and electrons. decaying into protons with a decay time of τn = 940 s. with charge −1/3. while the neutron mass is mn c2 = 939. while a neutron is electrically neutral. On large scales. Protons and neutrons also differ in their mass – or equivalently.1.2.1% greater. where a baryon is defined as a particle made of three quarks. the number of electrons is equal to the number of protons. TYPES OF PARTICLES Table 2.511 -1 neutrino νe . the rest energy of an electron is me c2 = 0. but opposite in sign. The proton mass is mp c2 = 938.1: Particle Properties particle symbol rest energy (MeV) charge proton p 938.5 Protons and neutrons are both examples of baryons. Protons 5 For that matter. we ourselves are made of protons. each with an electrical charge of +2/3. the component of the universe made up of ions. . neutrons. since only the baryons (protons and neutrons) contribute significantly to the mass density. A proton (p) contains two “up” quarks. Electrons (e− ) are examples of leptons.6 0 electron e− 0. about 0. Neutrons can be preserved against decay by binding them into an atomic nucleus with one or more protons. atoms. and a “down” quark.511 MeV. neutrons.3 MeV. Free neutrons are unstable. Since protons outmass electrons by a factor of 1836 to 1. Knopf. detecting neutrinos from astronomical sources is difficult. is very weak indeed. Inc.24 CHAPTER 2. In truth. FUNDAMENTAL OBSERVATIONS and neutrons are 800-pound gorillas. Updike was using a bit of poetic license here. electrons are only 7-ounce bushbabies. given them by Enrico Fermi. in his poem “Cosmic Gall”6 : Neutrinos. they are very small. 7 Their name. In addition. About three-fourths of the baryonic matter in the universe is currently in the form of ordinary hydrogen. means “little neutral one” in Italian. is that some or all of the neutrino types probably have a small From COLLECTED POEMS 1953-1993 by John Updike. What Updike didn’t know in 1960. a division of Random House. it is not true that neutrinos “do not interact at all”. There are three types. for instance – they find a minimum helium mass fraction of 24%. is the neutrino (ν). they actually are able to interact with other particles via the weak nuclear force. muon neutrinos. It is definitely true that neutrinos have no charge. only 2% is contributed by more massive atoms.7 However. as a mix of three parts hydrogen to one part helium. Like dustmaids down a drafty hall Or photons through a sheet of glass. through which they simply pass. The most poetic summary of the properties of the neutrino was made by John Updike. When astronomers look at a wide range of astronomical objects – stars and interstellar gas clouds. in addition to the electron. 6 . Used by permission of Alfred A. for instance. The earth is just a silly ball To them. with only minor contamination by heavier elements. and tau neutrinos. it is primarily in the form of helium. of neutrinos: electron neutrinos. the simplest of all elements. contains 70% hydrogen by mass. Another type of lepton. to lowest order. Since neutrinos pass through neutrino detectors with the same facility with which they pass through the Earth. or “flavors”. c 1993 by John Updike. The Sun’s atmosphere. a typical neutrino emitted by the Sun would have to pass through a few parsecs of solid lead before having a 50% chance of interacting with a lead atom. when we look at the remainder of the baryonic matter. The weak nuclear force. the next simplest element. and 28% helium. though. when he wrote his poem. The baryonic component of the universe can be described. They have no charge and have no mass And do not interact at all. as a function of photon energy. from radio to gamma rays. photons can ionize an atom by kicking an electron out of its orbit. electrons. ν Observations of muon neutrinos created by cosmic rays striking the upper atmosphere indicate that muon neutrinos oscillate into tau neutrinos. when regarded as a wave.4. where h = 2π¯ is the h Planck constant. When a system is in thermal equilibrium.7. or a sphere of ionized hydrogen and helium. called photons. knowing the ν differences of the squares of the masses doesn’t tell us the values of the masses themselves. The peak in the blackbody function occurs at hfpeak ≈ 2. are easily created. depends only on the temperature T . from the search for neutrino oscillations. TYPES OF PARTICLES 25 mass. If an object is opaque. Higher energy photons can break an atomic nucleus apart.82kT . The energy density of photons in the frequency range f → f + df is given by the blackbody function ε(f )df = 8πh f 3 df . It doesn’t matter whether the system is a tungsten filament. is characterized by its frequency f or its wavelength λ = c/f . A particle which is known to be massless is the photon. and attain thermal equilibrium. a process known as photoionization. neutrons. and photons which it contains frequently interact. Photons of a wide range of energy. photons interact readily with electrons. or an ingot of steel. the density of photons in the system. with the difference in the squares of their masses being ∆(m2 c4 ) ≈ 5 × 10−5 eV2 . Observations of neutrinos from the Sun are most easily explained if electron neutrinos (the flavor emitted by the Sun) oscillate into some other flavor of neutrino. An “oscillation” is the transmutation of one flavor of neutrino into another. protons. pervade the universe. equation (2. and neutrons. then the protons. The rate at which two neutrino flavors oscillate is proportional to the difference of the squares of their masses. Unfortunately. One way to make photons is to take a dense. a process known as photodissociation.2. When light is regarded as a stream of photons. Eγ = hf .25) yields a .25) illustrated in Figure 2. in general. Integrated over all frequencies. Unlike neutrinos. For instance. opaque object – such as the filament of an incandescent lightbulb – and heat it up. each photon is characterized by its energy. Light. Electromagnetic radiation can be thought of either as a wave or as a stream of particles. c3 exp(hf /kT ) − 1 (2. The evidence for massive neutrinos comes indirectly. with ∆(m2 c4 ) ≈ 3 × 10−3 eV2 for these two flavors. Photons. 5 0 0 2 4 6 8 hf/kT 10 12 Figure 2.7 × 10−5 m. You have a temperature of 310 K.7: The energy distribution of a blackbody spectrum.28) yields a mean photon energy of Emean = hfmean ≈ 2. the Sun produces an approximate blackbody spectrum with a temperature T ≈ 5800 K.26) π2 k4 −16 α= J m−3 K−4 .5 (c h /8πk T ) ε 1 .29) 2 h 3 c3 π ¯ Division of equation (2. FUNDAMENTAL OBSERVATIONS 1. (2. This implies β= where 3 2 3 3 .27) 3 3 = 7.404 k 3 = 2.072 eV. and you radiate an approximate blackbody spectrum.26 CHAPTER 2. By contrast. corresponding to a wavelength of λ ≈ 1. (2.28) 2. (2. in the infrared. close to the peak in the spectrum.03 × 107 m−3 K−3 . with a mean photon energy of Emean ≈ 0.56 × 10 15 h c ¯ The number density of photons in blackbody radiation can be computed from equation (2.26) by equation (2.25) as nγ = βT 3 .70kT . total energy density for blackbody radiation of εγ = αT 4 . where (2. they qualify as dark matter.4. Using this definition. corresponding to λ ≈ 9. even with our best available technology. difficult. are very dim. and what it’s made of. In this book. Thus. 8 . defining dark matter as any massive component of the universe which doesn’t emit. too low in mass for nuclear fusion to occur in their cores. there exist massive particles which interact. since an isolated stellar remnant is extremely faint and difficult to detect. which our eyes are equipped to detect. they usually mean any massive component of the universe which is too dim to be detected readily using current technology. that although the mean photon energy in a blackbody spectrum is ∼ 3kT . absorb. like neutrinos. stellar remnants such as white dwarfs. neutron stars. Substellar objects such as brown dwarfs are also referred to as dark matter. A large fraction of the Sun’s output is at wavelengths of (4 → 7) × 10−7 m. When observational astronomers refer to dark matter. Exactly how much there is. Although Neptune no longer qualifies as dark matter. Note. These particles.3 eV. However. The standard method of detecting dark matter is by measuring its gravitational effect on luminous matter. for instance. which have not yet been detected in the laboratory. however. Detecting dark matter is.7 shows us that there is a long exponential tail to higher photon energies. only through the weak nuclear force and through gravity. naturally. Figure 2. In some extensions to the Standard Model of particle physics. 8 If neutrinos have mass.0 × 10−7 m. in the near infrared. or scatter light at all. observations of the motions of stars within galaxies and of galaxies within clusters indicate that there’s a significant amount of dark matter in the universe. as the recent neutrino oscillation results indicate. since brown dwarfs. and black holes are sometimes referred to as dark matter. is a topic of great interest to cosmologists. I will generally adopt the broader definition of dark matter as something which is too dim for us to see. TYPES OF PARTICLES 27 a mean photon energy Emean ≈ 1.2. an alternate name for dark matter might be “transparent matter” or “invisible matter”. just as the planet Neptune was first detected by its gravitational effect on the planet Uranus. are generically referred to as Weakly Interacting Massive Particles. Theoretical astronomers sometimes use a more stringent definition of dark matter than observers do. the name “dark matter” has received the sanction of history. The most mysterious component of the universe is dark matter. or WIMPs. then in the past. the universe was denser than it is now. The existence of the CMB is a very important cosmological clue. nγ = 4. is quite low. (2. there are about 411 CMB photons in every cubic centimeter of the universe at the present day. they found an isotropic background of microwave radiation. (2.25) with a temperature T0 = 2. only Emean = 6. it is the clue which caused the Big Bang model for the universe to be favored over the Steady State model. from equation (2.33) This is too low in energy to photoionize an atom.34 × 10−4 eV .11 × 108 m−3 .725 ± 0. About all they do.32) Thus. or . The mean CMB photon energy corresponds to a wavelength of 2 millimeters. (2.001 K . from equation (2. much less photodissociate a nucleus. εγ = 4. More recently. from a terrestrial point of view. Using a microwave antenna at Bell Labs. In particular. If mass is conserved in an expanding universe. is cause static on television. however. the Cosmic Background Explorer (COBE) satellite has revealed that the Cosmic Microwave Background is exquisitely well fitted by a blackbody spectrum (equation 2. Assume that the early dense universe was very hot (T 104 K.31) (2. The number density of CMB photons is.26). In a Big Bang universe. a cosmic background radiation arises naturally if the universe was initially very hot as well as being very dense.30) This is equivalent to roughly a quarter of an MeV per cubic meter of space. FUNDAMENTAL OBSERVATIONS 2.28).28 CHAPTER 2.17 × 10−14 J m−3 . The energy density of the CMB is.5 The universe is filled with a Cosmic Microwave Background The discovery of the Cosmic Microwave Background (CMB) by Arno Penzias and Robert Wilson in 1965 has entered cosmological folklore. The mean energy of CMB photons. in the microwave region of the electromagnetic spectrum – hence the name “Cosmic Microwave Background”.725 K is not easily explained. the existence of blackbody radiation at 2. however. In a Steady State universe. the blackbody photons started streaming freely through the universe. equation (2. The blackbody radiation in the volume can be thought as a photon gas with energy density εγ = αT 4 . The temperature of the CMB today is 2. dE is the change in the internal energy. (2. the boxful of photons must obey the first law dQ = dE + P dV . its temperature was ∼ 3000 K. The blackbody radiation that fills the universe today can be explained as a relic of the time when the universe was sufficiently hot and dense to be opaque.4. dt dt (2. and dV is the change in volume of the box.2. Consider a region of volume V which expands at the same rate as the universe. hot. the baryonic matter in the universe was completely ionized. and the free electrons rendered the universe opaque. 3 dt (2. at the time the universe became transparent. without further scattering off free electrons. dQ = 0. after all). the pressure of a photon gas is Pγ = εγ /3. P is the pressure. applied to an expanding homogeneous universe. COSMIC MICROWAVE BACKGROUND 29 kT 1eV ). Moreover. the first law of thermodynamics.36) . The photon gas within our imaginary box must follow the laws of thermodynamics. there is no net flow of heat (everything’s the same temperature. the photon gas has a pressure. it cooled.35) can be rewritten in the form α 4T 3 dT dV V + T4 dt dt 1 dV = − αT 4 .5. Since. When the universe no longer contained a significant number of free electrons. since the photons in the volume have momentum as well as energy.35) Since.25). in particular. for the photons of the CMB.34) where dQ is the amount of heat flowing into or out of the photon gas in the volume V . However. with a spectrum typical of a blackbody (equation 2. is dE dV = −P (t) . opaque body. E = εγ V = αT 4 V and P = Pγ = αT 4 /3. When the temperature dropped to ∼ 3000 K. the early hot dense universe was full of photons.725 K. A dense. as described in Section 2. as the universe expanded. The drop in temperature of the blackbody radiation is a direct consequence of the expansion of the universe. At such high temperatures. in a homogeneous universe. So. produces blackbody radiation. a factor of 1100 lower. However. ions and electrons combined to form neutral atoms. so that V ∝ a(t)3 . banging off the electrons like balls in a pinball machine. Thus. how far. on average. the temperature of the cosmic background radiation has dropped by a factor of 1100 since the universe became transparent because the scale factor a(t) has increased by a factor of 1100 since then. with a temperature slightly cooler than the surface of the star Betelgeuse. The evidence cited so far can all be explained within the framework of a Hot Big Bang model. and since then has been expanding and cooling. the average radius of a tree is R = 1 m and the average number of trees per unit area is Σ = 0. If Robin Hood shoots an arrow in a random direction.38) This implies the simple relation T (t) ∝ a(t)−1 .] Bernstein (1995): The “Micropedia” which begins this text is a useful overview of the contents of the universe and the forces which work on them Harrison (1987): The definitive treatment of Olbers’ paradox Problems (2.005 m−2 . FUNDAMENTAL OBSERVATIONS 1 dV 1 dT =− .1) Suppose that in Sherwood Forest.37) T dt 3V dt However. at the time the universe became transparent. (2.30 or CHAPTER 2. in which the universe was originally very hot and very dense. since V ∝ a(t)3 as the box expands. will it travel before it strikes a tree? . What we now see as a Cosmic Microwave Background was once. dt dt (2. this means that the rate in change of the photons’ temperature is related to the rate of expansion of the universe by the relation d d (ln T ) = − (ln a) . The remainder of this book will be devoted to working out the details of the Hot Big Bang model which best fits the universe in which we live. a Cosmic Near-Infrared Background. Suggested reading [Full references are given in the “Annotated Bibliography” on page 286. and m(ντ ) minimize the sum m(νe )+m(νµ )+m(ντ ).) What is the approximate rate.5.3) Since you are made mostly of water. and that the difference between the square of the mass of the muon neutrino and that of the tau neutrino has the value [m(ντ )2 − m(νµ )2 ]c4 = 3 × 10−3 eV2 . you may assume you are spherical. How far. in watts. could you see in any direction before your line of sight hit a galaxy? (2.5) A hypothesis once used to explain the Hubble relation is the “tired light hypothesis”. m(νµ ). how far. Show that this hypothesis gives a distanceredshift relation which is linear in the limit z 1.2) Suppose you are in an infinitely large. how long would it take the CMB to raise your temperature by one nanoKelvin (10−9 K)? (You may assume your heat capacity is the same as pure water. COSMIC MICROWAVE BACKGROUND 31 (2. could you see in any direction before your line of sight struck a star? (Assume standard Euclidean geometry holds true in this universe.39) where K is a constant. What must the value of K be in order to yield a Hubble constant of H0 = 70 km s−1 Mpc−1 ? . at which you would absorb radiative energy from the CMB? Ignoring other energy inputs and outputs.4) Suppose that the difference between the square of the mass of the electron neutrino and that of the muon neutrino has the value [m(νµ )2 − m(νe )2 ]c4 = 5 × 10−5 eV2 .) If the stars are clumped into galaxies with a density ng = 1 Mpc−3 and average radius Rg = 2000 pc. If you were in intergalactic space. C = 4200 J kg −1 K−1 . (This is consistent with the observational results discussed in section 2. dr (2. approximately how many CMB photons would you absorb per second? (If you like.4. with the energy loss per unit distance being given by the law dE = −KE . on average. on average.) (2. you are very efficient at absorbing microwave photons. infinitely old universe in which the average density of stars is n = 109 Mpc−3 and the average stellar radius is equal to the Sun’s radius: R = R = 7 × 108 m.2. The tired light hypothesis states that the universe is not expanding.) What values of m(νe ). given these constraints? (2. but that photons simply lose energy as they move through space (by some unexplained means). In Isaac Newton’s view. gravity is negligibly small compared to electromagnetic forces. By contrast. Ironically then. Both gravity and electromagnetism are long range forces. there are two useful ways of looking at gravity – the Newtonian. the dominant force determining the evolution of the universe is gravity. or classical. as formulated by his Laws of Motion and Law of Gravity. so there are no electrostatic forces on large scales. gravity – the weakest of all forces from a particle physics standpoint – is the force which determines the evolution of the universe on large scales. viewpoint. However. in Einstein’s view. On small scales. The weak and strong nuclear forces are short-range forces. Although Newton’s view and Einstein’s view are conceptually very different. for instance. the universe is electrically neutral. the weak force is effective only on scales of w ∼ 10−18 m or less. The Newtonian predictions differ significantly from the predictions of general relativity only in the limit of deep potential minima (to use Newtonian language) or strong spatial curvature (to 32 . Moreover. Note that in referring to gravity as a force. viewpoint and the Einsteinian. on scales greater than 100 Mpc or so). intergalactic magnetic fields are sufficiently small that magnetic forces are also negligibly tiny on cosmological scales. and the strong force on scales of s ∼ 10−15 m or less. the electrostatic repulsion between a pair of protons is larger by a factor ∼ 1036 than the gravitational attraction between them. I am implicitly adopting a Newtonian viewpoint.Chapter 3 Newton Versus Einstein On cosmological scales (that is. gravity is a manifestation of the curvature of space-time. on large scales. gravity is a force which causes massive bodies to be accelerated. or general relativistic. in most cosmological contexts they yield the same predictions. In physics. has a property which we may call the “gravitational mass”. when we look at objects in the Solar System such as planets. Every object in the universe. is always an attractive force. In Newton’s view. with constantly changing speed. (3. space. in the Newtonian view. 3.1. The question of why it is possible to switch back and forth between the two very different viewpoints of Newton and Einstein is an intriguing one. Let the gravitational masses of two objects be Mg and mg . the circumference of a circle is 2π times its radius. In the limit of shallow potential minima and weak spatial curvature. an object with no net force acting on it moves in a straight line at constant speed. However. Why is this? Newton would tell us. the shortest distance between two points is a straight line. tending to draw two bodies closer together. we find that they move on curved lines. through all the other axioms and theorems you learned in high school geometry.1) r2 The negative sign in the above equation indicates that gravity. all the axioms and theorems of plane geometry (as codified by Euclid in the third century BC) hold true. comets. and asteroids. said Newton. and let the distance between their centers be r. moreover. in many contexts. The gravitational force acting between the two objects (assuming they are both spherical) is GMg mg F =− . . In Euclidean. EQUIVALENCE PRINCIPLE 33 use general relativistic language). it is permissible to switch back and forth between a Newtonian and a general relativistic viewpoint. general relativity yields the correct result. the force called gravity.3. adopting whichever one is more convenient. In these limits. and deserves closer investigation. I will frequently adopt the Newtonian view of gravity in this book because. “Their velocities are changing because there is a force acting on them. space is unchanging and Euclidean.1 Equivalence principle In Newton’s view of the universe.” Newton devised a formula for computing the gravitational force between two objects. and so on. it is mathematically simpler and conceptually more familiar. the angles at the vertices of a triangle sum to π radians. moons. or “flat”. In Euclidean space. Newton’s second law of motion says that force and acceleration are related by the equation F = mi a .8 m s−2 . not just the force of gravity. then the gravitational acceleration of an object toward a mass Mg would be (combining equations 3.34 CHAPTER 3. when Galileo dropped objects from towers and slid objects down inclined planes.2) GMg mg . The equality of gravitational mass and inertial mass is called the equivalence principle. Let the inertial mass of an object be mi . said Newton. between the gravitational mass mg and the inertial mass mi . and it is the equivalence principle which led Einstein to devise his theory of general relativity. he found that the acceleration (barring the effects of air resistance and friction) was always the same.4) a=− 2 r mi with the ratio mg /mi varying from object to object. reveal that the inertial and gravitational masses are the same to within one part in 1012 .1) and (3.2) In equations (3. Suppose you wake up one morning to find that you have been . The magnitude of the gravitational acceleration close to the Earth’s surface is 2 g = GMEarth /rEarth = 9. which are basically more sensitive versions of Galileo’s experiments. (3. NEWTON VERSUS EINSTEIN What is the acceleration which results from this gravitational force? Newton had something to say about that. let’s begin with a thought experiment of the sort Einstein would devise. through the use of different subscripts.1 and 3. One of the fundamental principles of physics is that the gravitational mass and the inertial mass of an object are identical: mg = m i . Every object in the universe.2) I have distinguished. Modern tests of the equivalence principle. (3. The property of an object that determines how strongly it is pulled on by the force of gravity is equal to the property that determines its resistance to acceleration by any force. regardless of the mass and composition of the object. as well. To see how the equivalence principle led Einstein to devise his theory of general relativity. However. has a property which we may call the “inertial mass”. (3.3) When you stop to think about it. this equality is a remarkable fact. If the equivalence principle did not hold. “Oh. Observing the falling bear. we can replace the accelerated box with a stationary box experiencing a constant Note that the equivalence of the two boxes in Figure 3. starting from the equivalence principle. and the box is being accelerated upward at a constant rate. the equivalence principle permits two possible interpretations. a sentient anglophone bear would say. 1 . imagine yourself back in the sealed box. bother. Thus.1 Einstein’s insight. with some relief. a big bear falls at the same rate as a little bear. they haven’t taken me away in their spaceship yet. in the real universe. or moving with a constant velocity. The behavior of the bear in each case (Figure 3. “Oh no!” you say. you drop your teddy bear. In the real universe. as the floor of the box rushes up to meet the photons. “This is what I’ve always feared would happen. the path of the light beam will appear to you to be bent downward. hermetically sealed box. In each case. but vary with position. EQUIVALENCE PRINCIPLE 35 sealed up (bed and all) within an opaque.8 m s−2 .” At that moment. For instance. being accelerated through interplanetary space at 9. the gravitational acceleration near the Earth’s surface is a vector g(r) which varies in direction (always pointing toward the Earth’s center) and in magnitude (decreasing as the inverse square of the distance from the Earth’s center). in each case.8 m s−2 by a rocket engine. within a sealed box.3. I’ve been abducted by space aliens who are taking me away to their home planet. thanks to the equivalence principle. gravitational accelerations are not exactly constant. with no way of distinguishing between them. Since the box is accelerating upward.1) is identical. (2) The bear is moving at a constant velocity. When you drop a teddy bear. To understand Einstein’s thought processes. led him to the theory of general relativity.” during the interval before it collides with the floor of the box. and in each case. “At least I am still on the Earth’s surface. and the bear is being accelerated downward by a constant gravitational force. I’m weightless. a bear stuffed with cotton falls at the same rate as a bear stuffed with lead. you find that it falls toward the floor of the box with an acceleration a = 9.1. “Whew!” you say. However. a window in the side of the box opens to reveal (much to your horror) that you are inside an alien spaceship which is being accelerated at 9. (1) The box is static. a box so small that the variation in g is too tiny to be measured. You grab the flashlight that you keep on the bedside table and shine a beam of light perpendicular to the acceleration vector (Figure 3. soundproof. though.1 depends on the gravitational acceleration in the left-hand box being constant.8 m s−2 . or any other object.” So startled are you by this realization.2). the equivalence principle can only be applied to an infinitesimally small box – that is. The behavior of a bear in an accelerated box (left) is identical to that of a bear being accelerated by gravity (right).36 CHAPTER 3.1: Equivalence principle (teddy bear version). NEWTON VERSUS EINSTEIN Figure 3. . 2: Equivalence principle (photon version) The path followed by a light beam in an accelerated box (left) is identical to the path followed by a light beam subjected to gravitational acceleration (right). .1. EQUIVALENCE PRINCIPLE 37 Figure 3.3. the path taken by light minimizes the travel time rather than maximizing it. which states that light travels between two points along a path which minimizes the travel time required. Einstein had one more insight. We now have a third way of thinking about the motion of the teddy bear in the box: (3) No forces are acting on the bear. it is simply following a geodesic in curved space-time. However. Curved space-time tells mass-energy how to move. The presence of mass. In the Newtonian description of gravity. Contemplating the curved path of the light beam. Einstein concluded. in the fully developed theory of general relativity. Fermat’s principle requires that the travel time be an extremum – either a minimum or a maximum. this translates into the requirement that light takes the shortest path between two points.2 In a vacuum. the shortest path between two points is a straight line. In fact. In Euclidean. In most situations. the word “geodesic” is simply a shorter way of saying “the shortest distance between two points”. we are led to the conclusion that the paths of photons will be curved downward in the presence of a gravitational field. in the presence of gravity. who also popularized the term “black hole”. or flat. 2 .38 CHAPTER 3. One of the fundamental principles of optics is Fermat’s principle. A more accurate summary of Einstein’s viewpoint. Thus. The Way of Einstein: Mass-energy tells space-time how to curve. however. space is not Euclidean. 3 In this context. mass and energy (which Newton thought of as two separate entities) are interchangeable. 4 This pocket summary of general relativity was coined by the physicist John Wheeler. then. Gravity affects photons. in Einstein’s view. Moreover. via the famous equation E = mc2 . Since there’s no way to distinguish between these two cases. the path taken by light is not a straight line. 4 Einstein’s description of gravity gives a natural explanation for the equivalence principle. even though they have no mass. Einstein concluded. The Way of Newton: Mass tells gravity how to exert a force (F = −GM m/r 2 ).3 We now have two ways of describing how gravity works. where the speed of light is constant. Force tells mass how to accelerate (F = ma). space and time (which Newton thought of as two separate entities) form a four-dimensional space-time. space. is that the presence of mass-energy causes space-time to be curved. causes space to be curved. the equality of the More generally. NEWTON VERSUS EINSTEIN gravitational acceleration. in which spacetime curvature is related to the mass-energy density.2. and γ in Figure 3.5) . let’s start by considering ways of describing the curvature of two-dimensional spaces. in Einstein’s theory of general relativity. a geodesic is a straight line. on which Euclidean geometry holds (Figure 3. then extend what we have learned to higher dimensions.2 Describing curvature In developing a mathematical theory of general relativity. If a triangle is constructed on a plane by connecting three points with geodesics.3). However. the angles at its vertices (α. Since picturing the curvature of a four-dimensional space-time is.3: A flat two-dimensional space. (3. Einstein needed a way of mathematically describing curvature. to say the least.3) obey the relation α+β+γ =π . which is dictated by the geometry of space-time. 3. curvature is a property of space-time itself. It then follows automatically that the gravitational acceleration of an object should be independent of mass and composition – it’s just following a geodesic. On a plane. The simplest of two-dimensional spaces is a plane. gravitational mass and the inertial mass is a remarkable coincidence.3. β. difficult. DESCRIBING CURVATURE 39 Figure 3. 6) Stating that equation (3. and θ is the azimuthal Starting with this equation. If r is the distance from the north pole. and assign to every point a coordinate (x.8). the angles at its vertices (α. y). 5 . a circle whose center corresponds to the center of the sphere. in place of cartesian coordinates.8) where A is the area of the triangle. no matter where you draw a triangle on the surface of a sphere. it is a space where the curvature is homogeneous and isotropic.6) and (3. NEWTON VERSUS EINSTEIN where angles are measured in radians. On the surface of a sphere. as you can verify by making the simple coordinate substitution x = r cos θ. the surface of a sphere (Figure 3. the Pythagorean theorem holds.7) Although equations (3. it must always satisfy equation (3. θ + dθ) is ds2 = dr2 + r2 dθ2 . θ) and (r + dr. in a polar coordinate system. Moreover.4). Now consider another simple two-dimensional space. β. The surface of a sphere is a positively curved two-dimensional space. For instance. or how you orient it. If a triangle is constructed on the surface of the sphere by connecting three points with geodesics. they both represent the same flat geometry.40 CHAPTER 3. y) and (x + dx. I am adopting the convention. other coordinate systems can be used. a geodesic is a portion of a great circle – that is. and R is the radius of the sphere. y = r sin θ. the distance between points (r. commonly used among relativists. (3. On the surface of a sphere.7) are different in appearance. Omitting the parentheses simply makes the equations less cluttered.6) holds true everywhere in a two-dimensional space is equivalent to saying that the space is a plane. (3. we can set up a polar coordinate system by picking a pair of antipodal points to be the “north pole” and “south pole” and by picking a geodesic from the north to south pole to be the “prime meridian”. and not d(s2 ). so the distance ds between points (x. All spaces in which α + β + γ > π are called “positively curved” spaces. Of course. y + dy) is given by the relation5 ds2 = dx2 + dy 2 . On a plane. On a plane. we can set up a cartesian coordinate system. that ds2 = (ds)2 . and γ) obey the relation α + β + γ = π + A/R2 . (3. (3. θ) and another nearby point (r +dr. For illustrative purposes.3.2.5.9) Note that the surface of a sphere has a finite area. the distance between antipodal points. angle measured relative to the prime meridian. is πR. and a maximum possible distance between points. shown in Figure 3. DESCRIBING CURVATURE 41 Figure 3. I would like to show you a surface of constant negative curvature. 7 A space with constant curvature is one where the curvature is homogeneous and isotropic. An example of a negatively curved two-dimensional space is the hyperboloid. 6 . θ +dθ) is given by the relation ds2 = dr2 + R2 sin2 (r/R)dθ 2 . Figure 3. then the distance ds between a point (r.7 UnSince the publishers objected to producing a book of infinite size. equal to 4πR 2 . there exist negatively curved spaces. and has no upper limits on the possible distance between points.4: A positively curved two-dimensional space. By contrast. just as the surface of a sphere has constant positive curvature.6 In addition to flat spaces and positively curved spaces.3 actually shows only a portion of a plane. a plane has infinite area. or saddle-shape. at the maximum possible separation. (3. The saddle-shape illustrated in Figure 3. and θ is the azimuthal angle measured relative to the prime meridian.10) where A is the area of the triangle.5 has constant curvature only in the central region. If a triangle is constructed on this surface by connecting three points with geodesics.5: A negatively curved two-dimensional space. θ) and a nearby point (r + dr. NEWTON VERSUS EINSTEIN Figure 3.β. Despite the difficulties in visualizing a surface of constant negative curvature. (3. fortunately. θ + dθ) is given by ds2 = dr2 + R2 sinh2 (r/R)dθ 2 . If r is the distance from the pole.11) . the angles at its vertices (α. and some geodesic leading away from the pole as the prime meridian. then the distance ds between a point (r. its properties can easily be written down. and γ) obey the relation α + β + γ = π − A/R2 . Consider a two-dimensional surface of constant negative curvature. On a surface of constant negative curvature. with radius of curvature R. the mathematician David Hilbert proved that a two-dimensional surface of constant negative curvature cannot be constructed in a threedimensional Euclidean space. near the “seat” of the saddle. we can set up a polar coordinate system by choosing some point as the pole.42 CHAPTER 3. 3. or ds2 = dr2 + r2 [dθ2 + sin2 θdφ2 ] . (3. and (3. its metric is ds2 = dr2 + R2 sin2 (r/R)[dθ 2 + sin2 θdφ2 ] . if a two-dimensional space has curvature which is homogeneous and isotropic.12) expressed in cartesian coordinates.9). at r = 0. it can have uniform positive curvature. it has the metric ds2 = dx2 + dy 2 + dz 2 . is κ = 0 for a flat space. or have uniform positive curvature.8 However. In general. there are only three possibilities that fit the bill: the space can be uniformly flat.2. by drawing triangles on a bagel. κ. and has no upper limit on the possible distance between points. The number κ. A tablecloth can be badly rumpled at one end of the table and smooth at the other end. If a three-dimensional space has uniform positive curvature (κ = +1). The results for two-dimensional space can be extended straightforwardly to three dimensions. (3. If the space is curved. if you like. Relations like those presented in equations (3. (3. . A three-dimensional space. its geometry can be specified by two quantities. is the radius of curvature. 8 You can test this assertion. Thus. DESCRIBING CURVATURE 43 A surface of constant negative curvature has infinite area. are known as metrics. and R. or have uniform negative curvature. By traveling a distance C = 2πR. must be flat.11). then the quantity R. just as a positively curved two-dimensional space has finite area. and κ = −1 for a negatively curved space.13) expressed in spherical coordinates.14) A positively curved three-dimensional space has finite volume. If a three-dimensional space is flat (κ = 0). on the surface of a sphere. curvature is a local property. which has dimensions of length. or it can have uniform negative curvature. (3. is the antipodal point to the north pole. κ = +1 for a positively curved space. called the curvature constant. it is possible to “circumnavigate” a space of uniform positive curvature. The point at r = πR is the antipodal point to the origin. if you want a two-dimensional space to be homogeneous and isotropic. if its curvature is homogeneous and isotropic.7). just as the south pole. a bagel (or other toroidal object) is negatively curved on part of its surface and positively curved on other portions. at r = πR. which give the distance ds between two nearby points in space. The three possible metrics for a homogeneous. The coordinate system (r. φ) is not the only possible system. They merely have a different functional form because of the different choice of radial coordinates. then decreases again to 0 at r/R = π. isotropic. Sκ ≈ r. where dΩ2 ≡ dθ2 + sin2 θdφ2 and (κ = +1) (κ = 0) Sκ (r) =  r  R sinh(r/R) (κ = −1) . 3. its metric is ds2 = dr2 + R2 sinh2 (r/R)[dθ 2 + sin2 θdφ2 ] . negatively curved space has infinite volume.17) (3.19) 1 − κx2 /R2 Although the metrics written in equations (3. By contrast. Sκ increases to a maximum of Smax = R at r/R = π/2. regardless of the value of κ.16) and (3. they represent the same homogeneous. isotropic spaces. (3.16) (3. the metric for a homogeneous. For instance. When space is flat. Sκ increases monotonically with r.18) Note that in the limit r R. when space is positively curved.   R sin(r/R)  (3. if we switch the radial coordinate from r to x ≡ Sκ (r). if a three-dimensional space has uniform negative curvature (κ = −1). θ. the antipodal point to the origin.19) appear different on the page. However. with Sκ → ∞ as r → ∞. (3. relativity teaches us that space and time . NEWTON VERSUS EINSTEIN Finally. we’ve only considered the metrics for simple two-dimensional and three-dimensional spaces. or negatively curved.3 The Robertson-Walker metric So far. isotropic.15) Like flat space. three-dimensional space can be written more compactly in the form ds2 = dr2 + Sκ (r)2 dΩ2 .44 CHAPTER 3. three-dimensional space can be written in the form dx2 ds2 = + x2 dΩ2 . that the spatial component of Minkowski space-time is Euclidean. According to the laws of special relativity.22) (3.3. φ + dφ).20) is called the Minkowski metric. Note. this means. φ). Consider two events. the physicists Howard Robertson and Arthur Walker asked “What form can the metric of space-time assume if the universe is spatially homogeneous and isotropic at all time. ds = 0. θ + dθ.21) If the photon is moving along a radial path.3. towards or away from the origin. from a comparison with equation (3. however. along every infinitesimal segment of the photon’s path. but a special variety called a null geodesic. and if distances are allowed to expand .16). Without any gravitational effects. or flat. (3. c2 dt2 = dr2 . and the space-time which it describes is called Minkowski space-time. the permissible space-times are more interesting. r + dr. A null geodesic is one for which. r. or dr = ±c . since θ and φ are constant. A photon’s path through space-time is a four-dimensional geodesic – and not just any geodesic. When gravity is added. one occurring at the space-time location (t.20) The metric given in equation (3. so we can compute the four-dimensional distance between two events in space-time. then. In Minkowski space-time. Minkowski space-time is flat and static. the space-time separation between these two events is ds2 = −c2 dt2 + dr2 + r2 dΩ2 . ROBERTSON-WALKER METRIC 45 together comprise a four-dimensional space-time. dt (3.23) The Minkowski metric of equation (3. so called because it deals with the special case in which space-time is not curved by the presence of mass and energy. a photon’s trajectory obeys the relation ds2 = 0 = −c2 dt2 + dr2 + r2 dΩ2 . θ. In the 1930’s. and another occurring at the space-time location (t + dt. Just as we can compute the distance between two points in space using the appropriate metric for that space. mind you.20) applies only within the context of special relativity. (3. (3. The radius of curvature R0 has dimensions of length. NEWTON VERSUS EINSTEIN (or contract) as a function of time?” The metric which they derived.19).18). it is normalized so that a(t0 ) = 1 at the present moment. The scale factor. and κ = +1 if the universe has positive spatial curvature. κ = −1 if the universe has negative spatial curvature. first introduced in Section 2.16) ds2 = −c2 dt2 + a(t)2 dr2 + Sκ (r)2 dΩ2 .3. θ. It is most generally written in the form ds = −c dt + a(t) 2 2 2 2 dx2 + x2 dΩ2 2 /R2 1 − κx 0 .24) Note that the spatial component of the Robertson-Walker metric consists of the spatial metric for a uniformly curved space of radius R0 (compare equation 3. κ. and R0 . The curvature constant κ is a dimensionless number which can take on one of three discrete values: κ = 0 if the universe is spatially flat.25) with the function Sκ (r) for the three different types of curvature given by equation (3. φ) or (r. The Robertson-Walker metric can also be written in the form (see equation 3. the comoving coordinates of any point remain constant with time. called the cosmic time for short. independently of each other. is devoted in one way or another to finding the values of a(t). If homogeneity and isotropy did not . The assumption of homogeneity and isotropy is a very powerful one. If the universe is perfectly homogeneous and isotropic. The time variable t in the Robertson-Walker metric is the cosmological proper time. long before the available observational evidence gave any support for such an assumption. φ) are called the comoving coordinates of a point in space. describes how distances in a homogeneous. The scale factor a(t) is a dimensionless function of time which describes how distances grow or decrease with time. κ. if the expansion of the universe is perfectly homogeneous and isotropic. as we’ll see in later chapters. Robertson and Walker made it in the 1930’s. and is the time measured by an observer who sees the universe expanding uniformly around him. and gives the radius of curvature of the universe at the present moment. isotropic universe expand or contract with time. and R0 . θ. The spatial variables (x. then everything we need to know about its geometry is contained within a(t). The assumption of spatial homogeneity and isotropy is so powerful. scaled by the square of the scale factor a(t).46 CHAPTER 3. Much of modern cosmology. is called the Robertson-Walker metric. (3. the observations reveal that the universe is not homogeneous and isotropic on small scales. the Robertson-Walker metric is only an approximation which holds good on large scales.26) Along the spatial geodesic between the observer and galaxy. are held together by electromagnetic forces. as illustrated in Figure 3. as Voltaire might have said.4 Proper distance Consider a galaxy which is far away from us – sufficiently far away that we may ignore the small scale perturbations of space-time and adopt the Robertson-Walker metric. φ) is constant. (3. Thus. and that the galaxy which you are observing is at a comoving coordinate position (r. teddy bears. φ). isotropic expansion described by the single scale factor a(t). PROPER DISTANCE 47 exist. Small. 3.6 can be found using the Robertson-Walker metric at a fixed time t: ds2 = a(t)2 [dr2 + Sκ (r)2 dΩ2 ] . One question we may ask is. θ. galaxies (such as the Galaxy in which we live). The proper distance between the observer and galaxy in Figure 3.27) . we must specify the time t at which the distance is the correct one. “Exactly how far away is this galaxy?” In an expanding universe. the universe is “lumpy”. On smaller scales. as long as they are sufficiently dense. and interstellar dust grains. In truth. the distance between two objects is increasing with time.3. dense lumps. The proper distance dp (t) between two points is equal to the length of the spatial geodesic between them when the scale factor is fixed at the value a(t). Thus. homogeneous. it would be necessary to invent them – at least if your desire is to have a simple. and clusters of galaxies (such as the Local Group in which we live). the angle (θ.4. Suppose that you are at the origin.6. and thus ds = a(t)dr . Larger lumps. and hence do not expand. and hence does not expand uniformly. It’s only on scales larger than ∼ 100 Mpc that the expansion of the universe can be treated as the ideal. are held together by their own gravity. and hence do not expand. (3. analytically tractable form for the metric of space-time. such as humans. Examples of such gravitationally bound systems are planetary systems (such as the Solar System in which we live). if we want to assign a spatial distance D between two objects. A photon emitted by the galaxy at cosmic time te reaches the observer at cosmic time t0 .48 CHAPTER 3.6: An observer at the origin observes a galaxy at coordinate position (r. θ. φ). . NEWTON VERSUS EINSTEIN Figure 3. where and ˙ vp (t0 ) ≡ dp (t0 ) H0 = (3. PROPER DISTANCE 49 The proper distance dp is found by integrating over the radial comoving coordinate r: r dp (t) = a(t) dr = a(t)r . where x = Sκ (r). the rate of change for the proper distance between us and a distant galaxy is a ˙ ˙ dp = ar = dp . there is a linear relation between the proper distance to a galaxy and its recession speed: vp (t0 ) = H0 dp (t0 ) . since they make galaxies appear to be entirely passive. Some cosmology books will contain a statement like “As space expands. you may invert the relations of equation (3. if you wish to use the spatial coordinates (x. (3.29) Because the proper distance has the form dp (t) = a(t)r.   a(t)R0 sin−1 (x/R0 )  (3. this is just a repetition of what was demonstrated in Section 2.33) a t=t0 In a sense. a statement like “As galaxies move apart. they drag space along with .31) (3. θ. we are interpreting the change in distance between widely separated galaxies as being associated with the expansion of space.18) to find (κ = +1) (κ = 0) dp (t) = a(t)r(x) =  a(t)x  a(t)R0 sinh−1 (x/R0 ) (κ = −1) . φ). (3. it drags galaxies away from each other. Now.32) a ˙ .4.” Statements of this sort are misleading. the radius of curvature of the universe. there will be a linear relation between the relative velocity of two points and the distance between them. On the other hand.3.3. however. increases at the same rate.28) 0 Alternatively. R(t) = a(t)R0 . θ. ˙ a (3. As the distance between galaxies increases. if the distance between points is proportional to a(t). φ) instead of (r. with the comoving coordinate r constant with time.30) Thus. at the current time (t = t0 ). When we observe a distant galaxy.36) Thus. since it makes space appear to be entirely passive. its comoving coordinate distance r. but not its distance. As the theory of general relativity points out. but then it’s mass-energy which tells space-time how to curve. it doesn’t. consider the galaxy illustrated in Figure 3. galaxies farther than 4300 megaparsecs from us are currently moving away from us at speeds greater than that of light. with . however.34) Using the observationally determined value of H0 = 70 ± 7 km s−1 Mpc−1 .50 CHAPTER 3.31) implies that points separated by a proper distance greater than a critical value dH (t0 ) ≡ c/H0 . generally called the Hubble distance. We can. space-time and mass-energy are intimately linked. we know its angular position very well. “Gosh! Doesn’t this violate the law that massive objects can’t travel faster than the speed of light?” Actually. the curvature of space-time does tell mass-energy how to move. and refers to the relative motion of objects within a static space.6. Yes. the light traveled along a null geodesic. we can point in its direction. the current value of the Hubble distance in our universe is dH (t0 ) = 4300 ± 400 Mpc . The linear velocity-distance relation given in equation (3. Light that was emitted by the galaxy at a time te is observed by us at a time t0 . Cosmological innocents sometimes exclaim. measure the redshift z of the light we receive from the galaxy. During its travel from the distant galaxy to us. (3. To see the link between a and z. NEWTON VERSUS EINSTEIN them” would be equally misleading. (3. Although the redshift doesn’t tell us the proper distance to the galaxy. will have ˙ vp = d p > c .35) (3. there is no objection to having two points moving away from each other at superluminal speed due to the expansion of space. That is. The speed limit that states that massive objects must travel with v < c relative to each other is one of the results of special relativity. it does tell us what the scale factor a was at the time the light from that galaxy was emitted. but we don’t know its current proper distance dp (t0 ) – or. In the context of general relativity. for that matter. we find c dt = dr . c2 dt2 = a(t)2 dr2 . The wave crest is emitted at a time te and observed at a time t0 . a(t) (3. The null geodesic has θ and φ constant.40) Comparing equations (3.41) That is.42) te +λe /c a(t) from each side of equation (3. (3.3. along the light’s null geodesic. Suppose the distant galaxy emits light with a wavelength λe . and is observed at a time t0 + λ0 /c. where. a(t) (3. For the second wave crest. If we subtract the integral t0 dt (3.9 Thus. the left-hand side is a function only of t.43) In a homogeneous. we find the relation te +λe /c te 9 dt = a(t) t0 +λ0 /c t0 dt . the integral of dt/a(t) between the time of emission and the time of observation is the same for every wave crest in the emitted light. in general. a(t) (3. PROPER DISTANCE 51 ds = 0. (3.39) and (3. we find that t0 te dt = a(t) t0 +λ0 /c te +λe /c dt .37) Rearranging this relation.38).40). (3. isotropic universe there’s no reason for the light to swerve to one side or the other. Fix your attention on a single wave crest of the emitted light. such that c t0 te dt = a(t) r 0 dr = r . λ0 = λe . c t0 +λ0 /c te +λe /c dt = a(t) r 0 dr = r .4.41). as measured by an observer in the emitting galaxy. . and the righthand side is independent of t.38) In equation (3.39) The next wave crest of light is emitted at a time te + λe /c. Thus. The time −1 between wave crests. It doesn’t matter if the expansion was gradual or abrupt. if we observe a galaxy with a redshift z = 2. This relation becomes still simpler when we realize that during the time between the emission or observation of two wave crests. H0 ≈ 14 Gyr. the integral of dt/a(t) between the emission of successive wave crests is equal to the integral of dt/a(t) between the observation of successive wave crests. it doesn’t matter if the transition was monotonic or oscillatory. we find that the redshift of light from a distant object is related to the expansion factor at the time it was emitted via the equation 1+z = 1 a(t0 ) = . It doesn’t depend on how the transition between a(te ) and a(t0 ) was made.] Harrison (2000). The redshift we observe for a distant object depends only on the relative scale factors at the time of emission and the time of observation.45) a(te ) a(t0 ) Using the definition of redshift. Thus. The time scale for −1 expansion of the universe is the Hubble time. Suggested reading [Full references are given in the “Annotated Bibliography” on page 286. a(te ) a(te ) (3. NEWTON VERSUS EINSTEIN That is. z = (λ0 − λe )/λe . ch. a(t) is effectively constant in the integrals of equation (3. Thus.46) Here. for visible light.52 CHAPTER 3. (3. (3. we may write te +λe /c t0 +λ0 /c 1 1 dt = dt . is λ/c ≈ 2 × 10−15 s ≈ 10−32 H0 . All that matters is the scale factors at the time of emission and the time of observation. I have used the usual convention that a(t0 ) = 1. we are observing it as it was when the universe had a scale factor a(te ) = 1/3. 10 – 12: Curved space and relativity (both special and general) .44) a(te ) te a(t0 ) t0 or λ0 λe = . the universe doesn’t have time to expand by a significant amount.43). Can you draw an equilateral triangle of arbitrarily large area A on a surface with κ = +1 and radius of curvature R? If not. the circle’s circumference will be C = 2πR sin(r/R) .3) Suppose you are still a two-dimensional being. (3. 3: Coordinates and metrics in the context of general relativity Problems (3. Show that if you draw a circle of radius r.2) Suppose you are a two-dimensional being.4) Consider an equilateral triangle. all distances are measured on the surface of the sphere). An object of width ds R is at a distance r from you (remember. what is the maximum possible value of A? . (3. ch. What angular width dθ will you measure for the object? Explain the behavior of dθ as r → πR. If you could measure distances with an error of ±1 meter.1) What evidence can you provide to support the assertion that the universe is electrically neutral on large scales? (3.47) Idealize the Earth as a perfect sphere of radius R = 6371 km. how large a circle would you have to draw on the Earth’s surface to convince yourself that the Earth is spherical rather than flat? (3.3. with sides of length L. PROPER DISTANCE 53 Narlikar (2002). ch. what is the maximum possible value of A? Can you draw an equilateral triangle of arbitrarily large area A on a surface with κ = −1 and radius of curvature R? If not. 3. ch.1: Derivation and discussion of the Robertson-Walker metric Rich (2001). what is the maximum possible value of A? Can you draw an equilateral triangle of arbitrarily large area A on a surface with κ = 0? If not. living on the same sphere of radius R. living on the surface of a sphere with radius R. drawn on a two-dimensional surface of constant curvature.4. 2 – 3: Delves deeper into general relativity while discussing the Robertson-Walker metric Peacock (2000). y = r sin θ sin φ.5) By making the substitutions x = r sin θ cos φ.54 CHAPTER 3.12) and (3. NEWTON VERSUS EINSTEIN (3. demonstrate that equations (3.13) represent the same metric. . and z = r cos θ. in addition. or non-Euclidean. but which is allowed to expand or contract with time. The idea that the universe could be curved. we could determine the curvature by drawing a really. you can determine the radius of curvature R0 . measuring the curvature of the universe is simple. As early as 1829. Unfortunately for this elegant 55 . actually long predates Einstein’s theory of general relativity. and (3. 2 R0 (4. negative (κ = −1). one of the founders of non-Euclidean geometry.8). as measured at the present moment (t = t0 ).5). In principle. really big triangle. In principle.10) generalize to the equation α+β+γ =π+ κA . if the α + β + γ > π radians. half a century before Einstein’s birth. everything you need to know about the curvature is given by κ. The scale factor is normalized so that a(t0 ) = 1 at the present moment. Nikolai Ivanovich Lobachevski. β. the universe is negatively curved. and if α + β + γ < π radians. and measuring the angles α.Chapter 4 Cosmic Dynamics In a universe which is homogeneous and isotropic. (3. proposed observational tests to demonstrate whether the universe was curved. then R0 is the radius of curvature of the universe. or decrease with time as the universe contracts.1) where A is the area of the triangle. Finally. Equations (3. and γ at the vertices. The curvature constant κ gives the sign of the curvature: positive (κ = +1). you measure the area of the triangle. If. or flat (κ = 0). Therefore. If κ is non-zero. the scale factor a(t) tells how distances in the universe increase with time as the universe expands. and a(t). R0 . the universe is positively curved. in practice it is much more difficult. then photons will have had time to circumnavigate the universe many times.1. About all we can conclude from geometric arguments is that if the universe is positively curved. it can’t have a radius of curvature R0 that is significantly smaller than the current Hubble distance. the area of the biggest triangle we can draw is much smaller 2 than R0 . Looking toward the galaxy . demonstrating how an observer in such a universe could see multiple images of the same galaxy. since our universe is expanding. if the current circumference C0 is less than ct0 . geometric plan. COSMIC DYNAMICS Figure 4. and the deviation of α + β + γ from π radians would be too small to measure. with a circumference currently equal to C0 = 2πR0 .1: A two-dimensional positively curved universe. c/H0 ≈ 4300 Mpc. To take an extreme example. recall that if our universe is positively curved. To understand why this is so. it has finite size.56 CHAPTER 4. its circumference was even smaller. then photons will have had time to circumnavigate the universe. Thus. If C0 ct0 ∼ c/H0 . suppose the universe were positively curved with a circumference of only 10 million light years (roughly 3 Mpc). The two-dimensional analog to such a universe is shown in Figure 4. In the past. which is 2 million light years away from us. showing M31 as it was 18 million years ago.2) gives a mathematical relation between the gravitational potential Φ at a point in space and the mass density ρ at that point. showing M31 as it was 8 million years ago. comprised of photons which had traveled 18 million light years. Einstein’s field equation. such as Lobachevski. Poisson’s equation.1. you determine the acceleration. which is the relativistic equivalent of Poisson’s equation in Newtonian dynamics. its radius of curvature R0 must be very large – comparable to or larger than the current Hubble distance c/H0 . and that the universe therefore doesn’t expand significantly as a photon goes once or twice around the universe. gives a mathematical relation between the metric of space-time at a point and the energy and pressure at that space-time point. 1 . Since we don’t see periodicities of this sort. 2 Φ = 4πGρ . comprised of photons which had traveled 2 million light years.4. Moreover. comprised of photons which had traveled 12 million light years. By taking the gradient of the potential. it wasn’t until Albert Einstein first published his theory of general relativity in 1915 that the curvature of spacetime was linked to its mass-energy content.1 The Friedmann equation Although 19th century mathematicians and physicists. Einstein’s field equations can be used to find the linkage between a(t). showing M31 as it was 2 million years ago. We would also see another image. The key equation of general relativity is Einstein’s field equation. FRIEDMANN EQUATION 57 M31. In a cosmological context. by contrast. The trajectories of of freely moving objects then correspond to geodesics in curved space-time. were able to conceive of curved space. κ. 4. and R0 . showing M31 as it was 12 million years ago. (4. we conclude that if the universe is positively curved. which describe the curvature of the universe. And so on. we would see one image of M31. and the energy density ε(t) and pressure P (t) of the contents of the −1 This assumes that 2πR0 /c H0 .1 And so on. we would see an image of M31. and then can compute the trajectory of objects moving freely through space. We would also see another image. comprised of photons which had traveled 8 million light years. looking in the exact opposite direction to M31. The gravitational force F experienced by the test mass will be. consider a homogeneous sphere of matter. It wasn’t until Hubble’s results were published that Einstein acknowledged the reality of the expanding universe. when he was only 37 years old. The sphere is expanding or contracting isotropically. who first derived the equation in 1922. Alas. Unfortunately. Friedmann actually started his scientific career as a meteorologist. with total mass Ms constant with time (Figure 4. d 2 Rs GMs =− . κ. To begin. Friedmann derived his eponymous equation starting from Einstein’s field equation. Later. R0 . some (though not all) of the aspects of the Friedmann equation can be understood with the use of purely Newtonian dynamics. The equation which links together a(t). from Newton’s Second Law of Motion. Friedmann did not live to see his vindication. Then I will state (without proof) the modifications that must be made to find the more correct. and ε(t) is known as the Friedmann equation.2). (4. and used Einstein’s field equations to describe how a spatially homogeneous and isotropic universe expands or contracts as a function of time. implying an expanding or contracting universe. 2 dt Rs (t)2 (4. Place a test mass. Even without bringing relativity into play. he died of typhoid fever in 1925.3) F =− Rs (t)2 The gravitational acceleration at the surface of the sphere will then be.58 CHAPTER 4. Friedmann’s papers received little notice at first.4) . however. It is intriguing to note that Friedmann published his first results. GMs m . of infinitesimal mass m. unrelated to the universe we actually live in. after Alexander Alexandrovich Friedmann. at the surface of the sphere. starting from Newton’s Law of Gravity and Second Law of Motion. general relativistic form of the Friedmann equation. from Newton’s Law of Gravity. he taught himself general relativity. seven years before Hubble published Hubble’s Law in 1929. so that its radius Rs (t) is increasing or decreasing with time. using the full power of general relativity. COSMIC DYNAMICS universe. To see how the expansion or contraction of the universe can be viewed from a Newtonian viewpoint. Even Einstein initially dismissed Friedmann’s work as a mathematical curiosity. I will first derive the non-relativistic equivalent of the Friedmann equation. 5) simply states that the sum of the kinetic energy per unit mass. Rs (t) (4. Epot = − GMs . we may write 4π ρ(t)Rs (t)3 . Multiply each side of the equation by dRs /dt and integrate to find 1 2 dRs dt 2 = GMs +U .8) Ms = 3 Since the expansion is isotropic about the sphere’s center. FRIEDMANN EQUATION 59 r(t) Figure 4. (4.2: A sphere of radius Rs (t) and mass Ms . Ekin 1 = 2 dRs dt 2 .5) where U is a constant of integration.1. Rs (t) (4. (4. Since the mass of the sphere is constant as it expands or contracts.9) . Equation (4.6) and the gravitational potential energy per unit mass. expanding or contracting under its own gravity. as the sphere expands or contracts under its own gravitational influence.4.7) is constant for a bit of matter at the surface of a sphere. we may write the radius Rs (t) in the form Rs (t) = a(t)rs . (4. 11) as a2 .12) the right hand side will equal zero.10) by rs a2 /2 yields the equation 2 (4. Note that the time derivative of the scale factor only enters into equation (4. then falls back down. In terms of ρ(t) and a(t). U rs (4. in this case.10) a ˙ a = 8πG 2U 1 ρ(t) + 2 . the energy conservation equation (4. Let’s concentrate on the case of a sphere ˙ which is expanding. a2 is always ˙ positive. in this case. the right hand side of equation (4. the ball can be thrown upward with a speed less than the escape speed. the speed of the ball approaches zero as t → ∞. at a maximum scale factor amax = − GMs .11) is always positive. In this case. This is the boundary case in which a → 0 as t → ∞ and ρ → 0. a contracting sphere (a < 0) is simply the time reversal ˙ ˙ of an expanding sphere (a > 0). consider the case U < 0.11) is useful in picturing how isotropically expanding objects behave under the influence . and finally. Since a will still ¨ be negative.11) Equation (4. and finally. Therefore. First. However.11) starts out positive. Third. In this case. in this case. The future of the expanding sphere falls into one of three classes. COSMIC DYNAMICS where a(t) is the scale factor and rs is the comoving radius of the sphere. and the expansion of the sphere never stops. and expansion will stop. consider the case U > 0. The Friedmann equation in its Newtonian form (equation 4. analogous to the expanding universe in which we find ourselves. depending on the sign of U . 2 s 3 2 Dividing each side of equation (4. the ball reaches a maximum altitude. the ball can be thrown upward with a speed exactly equal to the escape speed. Second. First. Second. ˙ The three possible fates of an expanding sphere in a Newtonian universe are analogous to the three possible fates of a ball thrown upward from the surface of the Earth. the sphere will then contract.60 CHAPTER 4. consider the case U = 0.5) can be rewritten in the form 1 2 2 4π 2 r a = ˙ Grs ρ(t)a(t)2 + U . Third. the ball continues to go upward forever.11) gives the Friedmann equation in its Newtonian form. the ball can be thrown upward with a speed greater than the escape speed. 3 rs a(t)2 (4. the right hand side of equation (4. Newtonian dynamics tell us that the gravitational acceleration inside a hollow spherically symmetric shell is equal to zero. We may instead regard the sphere of radius Rs as being carved out of an infinite. violating the principle of isotropy. homogeneous.4. One of Einstein’s insights was that in determining the gravitational influence of a particle. FRIEDMANN EQUATION 61 of their self-gravity. in addition to the motion associated with the expansion or contraction of the universe. the important quantity was not its mass m but its energy. However. including all general relativistic effects. if its peculiar velocity v is much less than c – then its peculiar momentum will be p ≈ mv. Any motion which a particle has. and thus conclude that the test mass m at Rs experiences no net acceleration from matter at R > Rs .1. The peculiar motion of a particle is thus the motion which belongs to the particle alone.13). a Newtonian argument of this sort assumes that space is Euclidean. We divide up the region outside the sphere into concentric shells. and its The adjective “peculiar” comes from the Latin “peculium”. The correct form of the Friedmann equation. is called the particle’s peculiar motion. 2 . violating the principle of homogeneity. isotropic universe.13) = a 3c2 R0 a(t)2 Note the changes made in going from the Newtonian form of the Friedmann equation (equation 4. including the possibility of spatial curvature. and not to the global expansion or contraction of the universe. there exists a special location (the center of the sphere). has to begin with Einstein’s field equations. First of all. In a finite spherical volume. Unfortunately.11) to the correct relativistic form (equation 4. The first change is that the mass density ρ has been replaced by an energy density ε divided by the square of the speed of light.2 If a massive particle is non-relativistic – that is. isotropic universe. (4. is a 2 8πG ˙ κc2 1 ε(t) − 2 . a spherical volume of finite radius Rs cannot represent a homogeneous. (4. and at any point there exists a special direction (the direction pointing toward the center).14) Here p is the momentum of the particle as seen by an observer at the particle’s location who sees the universe expanding isotropically around her. E = (m2 c4 + p2 c2 )1/2 . meaning “private property”. A derivation of the correct Friedmann equation. In that case. its application to the real universe must be regarded with considerable skepticism. 62 energy will be CHAPTER 4. we must have some way of tying it to observable properties. the Friedmann equation can be tied to the Hubble constant. However. The Friedmann equation is a Very Important Equation in cosmology. given by a(t). (4. the expansion rate. Not only do photons respond to the curvature of space-time. Erel = pc = hf . they also contribute to it. it makes sense that the curvature. if we want to apply the Friedmann equation to the real universe. there’s a linear relation between recession speed v and proper distance d: v(t) = H(t)d(t) .18) You should consider writing it in reverse on your forehead so that you can see it every morning in the mirror when you comb your hair. Although I have not given the derivation of the Friedmann equation in the general relativistic case. (4. After all. COSMIC DYNAMICS 1 Enon−rel ≈ mc2 (1 + v 2 /c2 ) ≈ mc2 + mv 2 . in a universe whose expansion (or contraction) is described by a scale factor a(t).3 However. The special case with U = 0 corresponds to the special case where the space is perfectly flat (κ = 0). (4. slowly moving particles. with only a small correction for the kinetic energy mv 2 /2 of the particles.15) Thus. given by κ and R0 . the energy density of the universe determines both the curvature of space and the overall dynamics of the expansion. if the universe contained only massive. For instance.16) which also contributes to the energy density ε. Remember. in Einstein’s view. H0 . 2 (4. The second change that must be made in going from the Newtonian form of the Friedmann equation to the correct relativistic form is making the substitution 2U κc2 =− 2 . then the energy density ε would be nearly equal to ρc2 . and the energy density ε should be bound up together in the same equation. while the case with U > 0 corresponds to negative curvature (κ = −1).17) 2 rs R0 In the context of general relativity. photons and other massless particles also have an energy. The case with U < 0 corresponds to positive curvature (κ = +1). 3 . the curvature κ is related to the Newtonian energy U of a test mass. 19) H(t)2 = 3c2 R0 a(t)2 At the present moment.21) to determine the curvature.21) to place a lower limit on R0 in a negatively curved universe. 3c2 R0 (4. (4. and κ/R0 . is known as the “Hubble constant”. If we assume ε0 is non-negative. FRIEDMANN EQUATION 63 where H(t) ≡ a/a.22) This is the minimum radius of curvature which a negatively curved universe can have. the radius of curvature must be greater than the Hubble distance if the universe is negatively curved.20) As an etymological aside.4. the Friedmann equation can be rewritten in the ˙ form κc2 8πG ε(t) − 2 .1. Thus. that the radius of curvature must be comparable to or greater than the Hubble distance if the universe is positively curved. 2 ε0 . assuming that general relativity correctly describes the curvature. Since we know that the universe contains matter and radiation. which tells us the current curvature. and hence that ε0 > 0. the value of H(t) at the present day. as discussed earlier. we could use equation (4. which tells us the current rate of expansion. H0 = H(t0 ) = a ˙ a t=t0 = 70 ± 7 km s−1 Mpc−1 . (4. The Friedmann equation evaluated at the present moment is 2 H0 = 8πG κc2 ε0 − 2 .4 We also know from observations. while H0 . we can use equation (4. with no energy content.21) using the convention that a subscript “0” indicates the value of a timevarying quantity evaluated at the present. the Friedmann equation gives a relation among H0 . Thus. the curvature is negative. which tells us the current energy density. 4 . I should point out that the time-varying function H(t) is generally known as the “Hubble parameter”. Even without knowledge of the current density ε0 . Due to the difficulty of measuring the curvature directly by geometric means. In the limit of a totally empty universe. If we were able to measure H0 and ε0 with high precision. it is useful to have an indirect method of determining κ and R0 . the product κ/R0 is minimized in the limit ε0 → 0. with a radius of curvature R0 (min) = c/H0 . then for 2 a given value of H0 . (4. most tenuous regions have a few protons per liter. is close to the critical density. 8πG (4. but the ratio of the density to the critical density . εc (t) ≡ 3c2 H(t)2 . where the density is extraordinarily low.3) × 1011 M Mpc−3 . as it turns out. (4.2 ± 1. ρc. the Friedmann equation can generally be written as κc2 8πG ε(t) − 2 . for a given value of the Hubble parameter.0 /c2 = (9. In discussing the curvature of the universe.3 ± 1.0 ≡ εc. However.7) × 10−10 J m−3 = 5200 ± 1000 MeV m−3 . the universe is positively curved (κ = +1).4 ± 0. the Friedmann equation takes a particularly simple form: H(t)2 = 8πG ε(t) . H(t) = 3c2 R0 a(t) 2 (4.23) for all universes with a Robertson-Walker metric whose expansion or contraction is governed by the rules of general relativity. In a spatially flat universe (κ = 0). where even the hottest. COSMIC DYNAMICS As we have seen.27) Thus. or 140 solar mass stars per cubic kiloparsec. we can compute the current value of the critical density to within 20%: εc. Since we know the current value of the Hubble parameter to within 10%.25) If the energy density ε(t) is greater than this value. the critical density is currently roughly equivalent to a density of one hydrogen atom per 200 liters. the universe is negatively curved (κ = −1). the mean density of the universe. it is more convenient to use not the absolute density ε. When averaged over scales of 100 Mpc or more. you must keep in mind that most of the volume of the universe consists of intergalactic voids.24) Thus.0 = 3c2 2 H = (8.64 CHAPTER 4.26) 8πG 0 The critical density is frequently written as the equivalent mass density. by terrestrial standards. 3c2 (4. there is a critical density. It’s not even a large density by the standards of interstellar space within our Galaxy. If ε(t) is less than this value. (4.8) × 10−27 kg m−3 = (1. This is definitely not a large density. you know the Hubble distance. Thus. limits that even the most belligerent cosmologist will hesitate to quarrel with – state that the current value of the density parameter lies in the range 0. in addition. tell us how the scale factor a(t) evolves with time. neither can the left hand side. for instance). and expansion rate can be written in the form κc2 1 − Ω0 = − 2 2 .2. and if Ω = 1 at any time. 2 R0 a(t)2 H(t)2 (4. (4. there are two of them. cosmologists often use the dimensionless density parameter Ω(t) ≡ ε(t) . εc (t) (4. you can compute the radius of curvature (R0 ). we may take the unknown functions as H(t) and Ω(t). it remains less than one for all time. 5 .1 < Ω0 < 2.29) Note that since the right hand side of equation (4. a(t) and ε(t). A leopard can’t change its spots. when talking about the energy density of the universe.5 Or. If Ω < 1 at any time. the relation among curvature. a universe governed by the Friedmann equation can’t change the sign of its curvature. c/H0 . If. in any case. it still remains a single equation in two unknowns. In terms of the density parameter. Even if we had accurate boundary conditions (precise values for ε0 and H0 .31) 2 R0 c If you know Ω0 . it cannot. it remains greater than one for all times. if Ω > 1 at any time. At the present moment.2 The fluid and acceleration equations Although the Friedmann equation is indeed important. FLUID AND ACCELERATION EQUATIONS 65 εc .28) The most conservative limits on Ω – that is. (4.30) R0 H 0 or H2 κ = 20 (Ω0 − 1) . Ω = 1 at all times. 4. density. similarly. you know the sign of the curvature (κ).29) cannot change sign as the universe expands.4. if we prefer. all by itself. the Friedmann equation can be written in yet another form: 1 − Ω(t) = − κc2 . 5 to a comoving volume filled with photons. so the rate of change of the sphere’s internal energy is ˙ E =Vε+Vε=V ˙ ˙ a ˙ ε+3 ε ˙ a . so let’s look at another manifestation of the same concept – the first law of thermodynamics: dQ = dE + P dV . (4. thus. The Friedmann equation. the first law of thermodynamics. A homogeneous. (4. then for any volume dQ = 0. it says that the sum of the gravitational potential energy and the kinetic energy of expansion is constant. Energy conservation is a generally useful concept.32) where dQ is the heat flow into or out of a region. reduces to the form ˙ ˙ E + PV = 0 . and dV is the change in volume of the region.33) For concreteness.34) 3 s so the rate of change of the sphere’s volume is 4π 3 2 ˙ ˙ r (3a a) = V V = 3 s The internal energy of the sphere is 3 a ˙ a . This equation was applied in Section 2. Saying that the expansion of the universe is adiabatic is also a statement about entropy. there is no bulk flow of heat. isotropic expansion of the universe does not increase the universe’s entropy. P is the pressure.35) E(t) = V (t)ε(t) . an adiabatic process is one in which entropy is not increased. (4.37) . (Processes for which dQ = 0 are known as adiabatic processes. The volume of the sphere is 4π 3 V (t) = r a(t)3 .66 CHAPTER 4. is a statement of energy conservation. so that its proper radius is Rs (t) = a(t)rs . (4. (4.36) (4. but it applies equally well to a comoving volume filled with any sort of fluid. consider a sphere of comoving radius rs expanding along with the universal expansion. dE is the change in internal energy. in the Newtonian approximation. in particular. that is. as applied to the expanding universe. If the universe is perfectly homogeneous. COSMIC DYNAMICS We need another equation involving a and ε if we are to solve for a and ε as functions of time. The change in entropy dS within a region is given by the relation dS = dQ/T .) Since dQ = 0 for a comoving volume as the universe expands. it decreases the value of a and reduces the relative velocity of ˙ 6 Write it on your forehead just underneath the Friedmann equation. 3c2 R0 Taking the time derivative yields 2a¨ = ˙a Dividing by 2aa tells us ˙ 4πG a a ¨ = ε + 2ε ˙ a 3c2 a ˙ .39).40) εa − 2 . By combining the two. FLUID AND ACCELERATION EQUATIONS 67 Combining equations (4. it provides a negative acceleration – that is.13).37). multiplied by a2 .44) (4. 4πG a ¨ = − 2 (ε + 3P ) .33). a 3c (4. takes the form 8πG 2 κc2 a2 = ˙ (4. (4.35). and is the second of the key equations describing the expansion of the universe. . ˙ a The above equation is called the fluid equation.6 The Friedmann equation and fluid equation are statements about energy conservation.2. we may make the substitution a ε = −3(ε + P ) ˙ a ˙ to find the usual form of the acceleration equation. ˙ ˙ 3c2 (4. (4. we can derive an acceleration equation which tells how the expansion of the universe speeds up or slows down with time.4.38) a ˙ (4. The Friedmann equation (equation 4.42) 8πG 2 (εa + 2εaa) . we find that the first law of thermodynamics in an expanding (or contracting) universe takes the form a ˙ a ˙ V ε+3 ε+3 P ˙ a a or =0.39) ε + 3 (ε + P ) = 0 .41) Using the fluid equation (equation 4. and (4. (4.43) Note that if the energy density ε is positive. 44) shows us that such a component will cause the expansion of the universe to speed up rather than slow down. that the universe had a component with a pressure P < −ε/3 . as does a gas of neutrinos or WIMPs. The acceleration equation also includes the pressure P which is associated with the material filling the universe. resulting from the random thermal motions of the molecules. COSMIC DYNAMICS any two points in the universe. stretch the same piece of rubber.4. a ˙ a the fluid equation. (4. In cosmology. a ¨ 4πG = − 2 (ε + 3P ) . A negative pressure (also called “tension”) is certainly permissible by the laws of physics. ˙ a and the acceleration equation.47) (4.48) Although we think of ε as an energy per unit volume and P as a force per unit area. The positive pressure associated with these components of the universe will cause the expansion to slow down.46) a ˙ ε + 3 (ε + P ) = 0 . and its internal pressure will be positive. they both have the same dimensionality: in SI units. though. and its pressure will be negative.3 Equations of state To recap. There’s the Friedmann equation. 3c2 R0 a (4. 1 J m−3 = 1 N m−2 = 1 kg m−1 s−2 . 2 = κc2 8πG ε− 2 2 . Compress a piece of rubber. As we’ll discuss in more detail in Section 4. a cosmological constant has P = −ε. Suppose. a 3c 7 (4. . 4. A gas of photons also has a positive pressure. atoms.7 A gas made of ordinary baryonic matter has a positive pressure P .68 CHAPTER 4. or ions of which the gas is made. we now have three key equations which describe how the universe expands.45) Inspection of the acceleration equation (equation 4. the much-discussed cosmological constant is a component of the universe with negative pressure. and thus causes a positive acceleration for the expansion of the universe. 4.3. EQUATIONS OF STATE 69 Of the above three equations, only two are independent, since equation (4.48), as we’ve just seen, can be derived from equations (4.46) and (4.47). Thus, we have a system of two independent equations in three unknowns – the functions a(t), ε(t), and P (t). To solve for the scale factor, energy density, and pressure as a function of cosmic time, we need another equation. What we need is an equation of state; that is, a mathematical relation between the pressure and energy density of the stuff that fills up the universe. If only we had a relation of the form P = P (ε) , (4.49) life would be complete – or at least, our set of equations would be complete. We could then, given the appropriate boundary conditions, solve them to find how the universe expanded in the past, and how it will expand (or contract) in the future. In general, equations of state can be dauntingly complicated. Condensed matter physicists frequently deal with substances in which the pressure is a complicated nonlinear function of the density. Fortunately, cosmology usually deals with dilute gases, for which the equation of state is simple. For substances of cosmological importance, the equation of state can be written in a simple linear form: P = wε , (4.50) where w is a dimensionless number. Consider, for instance, a low-density gas of non-relativistic massive particles. Non-relativistic, in this case, means that the random thermal motions of the gas particles have peculiar velocities which are tiny compared to the speed of light. Such a non-relativistic gas obeys the perfect gas law, ρ (4.51) P = kT , µ where µ is the mean mass of the gas particles. The energy density ε of a nonrelativistic gas is almost entirely contributed by the mass of the gas particles: ε ≈ ρc2 . Thus, in terms of ε, the perfect gas law is P ≈ kT ε. µc2 (4.52) For a non-relativistic gas, the temperature T and the root mean square thermal velocity v 2 are associated by the relation 3kT = µ v 2 . (4.53) 70 CHAPTER 4. COSMIC DYNAMICS Thus, the equation of state for a non-relativistic gas can be written in the form Pnon−rel = wεnon−rel , (4.54) v2 1. (4.55) 3c2 Most of the gases we encounter in everyday life are non-relativistic. For instance, in air at room temperature, the nitrogen molecules are slow-poking along with a root mean square velocity of ∼ 500 m s−1 , yielding w ∼ 10−12 . Even in astronomical contexts, gases are mainly non-relativistic at the present moment. Within a gas of ionized hydrogen, for instance, the electrons are non-relativistic as long as T 6 × 109 K; the protons are non-relativistic when T 1013 K. A gas of photons, or other massless particles, is guaranteed to be relativistic. Although photons have no mass, they have momentum, and hence exert pressure. The equation of state of photons, or of any other relativistic gas, is 1 (4.56) Prel = εrel . 3 (This relation has already been used in Section 2.5, to compute how the Cosmic Microwave Background cools as the universe expands.) A gas of highly relativistic massive particles (with v 2 ∼ c2 ) will also have w = 1/3; a gas of mildly relativistic particles (with 0 < v 2 < c2 ) will have 0 < w < 1/3. The equation-of-state parameter w can’t take on arbitrary values. Small perturbations in a substance with pressure P will travel at the speed of sound. For adiabatic perturbations in a gas with pressure P and energy density ε, the sound speed is given by the relation w≈ c2 = c 2 s dP dε . (4.57) where √ In a substance with w > 0, the sound speed is thus cs = wc.8 Sound waves cannot travel faster than the speed of light; if they did, you would be able to send a sound signal into the past, and violate causality. Thus, w is restricted to values w ≤ 1. In a substance with w < 0, the sound speed is an imaginary number; this implies that small pressure perturbations to a substance with negative pressure will not constitute stably propagating sound waves, but will have amplitudes which grow or decay with time. 8 4.4. LEARNING TO LOVE LAMBDA 71 Some values of w are of particular interest. For instance, the case w = 0 is of interest, because we know that our universe contains non-relativistic matter. The case w = 1/3 is of interest, because we know that our universe contains photons. For simplicity, I will refer to the component of the universe which consists of non-relativistic particles (and hence has w ≈ 0) as “matter”. I will refer to the component which consists of photons and other relativistic particles (and hence has w = 1/3) as “radiation”. The case w < −1/3 is of interest, because a component with w < −1/3 will provide a positive acceleration (¨ > 0 in equation 4.48). A component of the universe with a w < −1/3 is sometimes referred to generically as “dark energy” (a phrase coined by the cosmologist Michael Turner). One form of “dark energy” is of special interest; some observational evidence, which we’ll review in future chapters, indicates that our universe may contain a cosmological constant. A cosmological constant may be simply defined as a component of the universe which has w = −1, and hence has P = −ε. The cosmological constant, also designated by the Greek letter Λ, has had a controversial history, and is still the subject of debate. To learn why cosmologists have had such a longstanding love/hate affair with the cosmological constant Λ, it is necessary to make a brief historical review. 4.4 Learning to love lambda The cosmological constant Λ was first introduced by Albert Einstein. After publishing his first paper on general relativity in 1915, Einstein, naturally enough, wanted to apply his field equation to the real universe. He looked around, and noted that the universe contains both radiation and matter. Since Einstein, along with every other earthling of his time, was unaware of the existence of the Cosmic Microwave Background, he thought that most of the radiation in the universe was in the form of starlight. He also noted, quite correctly, that the energy density of starlight in our Galaxy is much less than the rest energy density of the stars.9 Thus, Einstein concluded that the primary contribution to the energy density of the universe was from nonrelativistic matter, and that he could safely make the approximation that we live in a pressureless universe. So far, Einstein was on the right track. However, in 1915, astronomers were unaware of the existence of the expansion of the universe. In fact, it was 9 That is, stars have only converted a small fraction of their initial mass into light. 72 CHAPTER 4. COSMIC DYNAMICS by no means settled that galaxies besides our own actually existed. After all, the sky is full of faint fuzzy patches of light. It took some time to sort out that some of the faint fuzzy patches are glowing clouds of gas within our Galaxy and that some of them are galaxies in their own right, far beyond our own Galaxy. Thus, when Einstein asked, “Is the universe expanding or contracting?” he looked, not at the motions of galaxies, but at the motions of stars within our Galaxy. Einstein noted that some stars are moving toward us and that others are moving away from us, with no evidence that the Galaxy is expanding or contracting. The incomplete evidence available to Einstein led him to the belief that the universe is static – neither expanding nor contracting – and that it has a positive energy density but negligible pressure. Einstein then had to ask the question, “Can a universe filled with non-relativistic matter, and nothing else, be static?” The answer to this question is “No!” A universe containing nothing but matter must, in general, be either expanding or contracting. The reason why this is true can be illustrated in a Newtonian context. If the mass density of the universe is ρ, then the gravitational potential Φ is given by Poisson’s equation: 2 Φ = 4πGρ . (4.58) The gravitational acceleration a at any point in space is then found by taking the gradient of the potential: a=− Φ. (4.59) In a static universe, a must vanish everywhere in space. Thus, the potential Φ must be constant in space. However, if Φ is constant, then (from equation 4.58) 1 2 Φ=0. (4.60) ρ= 4πG The only permissible static universe, in this analysis, is a totally empty universe. If you create a matter-filled universe which is initially static, then gravity will cause it to contract. If you create a matter-filled universe which is initially expanding, then it will either expand forever (if the Newtonian energy U is greater than or equal to zero) or reach a maximum radius and then collapse (if U < 0). Trying to make a matter-filled universe which doesn’t expand or collapse is like throwing a ball into the air and expecting it to hover there. 4.4. LEARNING TO LOVE LAMBDA 73 How did Einstein surmount this problem? How did he reconcile the fact that the universe contains matter with his desire for a static universe? Basically, he added a fudge factor to the equations. In Newtonian terms, what he did was analogous to rewriting Poisson’s equation in the form 2 Φ + Λ = 4πGρ . (4.61) The new term, symbolized by the Greek letter Λ, came to be known as the cosmological constant. Note that it has dimensionality (time)−2 . Introducing Λ into Poisson’s equation allows the universe to be static if you set Λ = 4πGρ. In general relativistic terms, what Einstein did was to add an additional term, involving Λ, to his field equation (the relativistic equivalent of Poisson’s equation). If the Friedmann equation is re-derived from Einstein’s field equation, with the Λ term added, it becomes a ˙ a 2 = 8πG κc2 Λ ε− 2 2 + . 3c2 R0 a 3 (4.62) The fluid equation is unaffected by the presence of a Λ term, so it still has the form a ˙ ε + 3 (ε + P ) = 0 . ˙ (4.63) a With the Λ term present, the acceleration equation becomes Λ a ¨ 4πG = − 2 (ε + 3P ) + . a 3c 3 (4.64) A look at the Friedmann equation (4.62) tells us that adding the Λ term is equivalent to adding a new component to the universe with energy density εΛ ≡ c2 Λ. 8πG (4.65) If Λ remains constant with time, then so does its associated energy density εΛ . The fluid equation (4.63) tells us that to have εΛ constant with time, the Λ term must have an associated pressure PΛ = −εΛ = − c2 Λ. 8πG (4.66) Thus, we can think of the cosmological constant as a component of the universe which has a constant density εΛ and a constant pressure PΛ = −εΛ . 74 CHAPTER 4. COSMIC DYNAMICS By introducing a Λ term into his equations, Einstein got the static model universe he wanted. For the universe to remain static, both a and a must ˙ ¨ be equal to zero. If a = 0, then in a universe with matter density ρ and ¨ cosmological constant Λ, the acceleration equation (4.64) reduces to 0=− Λ 4πG ρ+ . 3 3 (4.67) Thus, Einstein had to set Λ = 4πGρ in order to produce a static universe, just as in the Newtonian case. If a = 0, the Friedmann equation (4.62) ˙ reduces to 8πG κc2 Λ κc2 0= (4.68) ρ − 2 + = 4πGρ − 2 . 3 R0 3 R0 Einstein’s static model therefore had to be positively curved (κ = +1), with a radius of curvature c c R0 = = 1/2 . (4.69) 1/2 2(πGρ) Λ Although Einstein published the details of his static, positively curved, matterfilled model in the spring of 1917, he was dissatisfied with the model. He believed that the cosmological constant was “gravely detrimental to the formal beauty of the theory”. In addition to its aesthetic shortcomings, the model had a practical defect; it was unstable. Although Einstein’s static model was in equilibrium, with the repulsive force of Λ balancing the attractive force of ρ, it was an unstable equilibrium. Consider expanding Einstein’s universe just a tiny bit. The energy density of Λ remains unchanged, but the energy density of matter drops. Thus, the repulsive force is greater than the attractive force, and the universe expands further. This causes the matter density to drop further, which causes the expansion to accelerate, which causes the matter density to drop further, and so forth. Expanding Einstein’s static universe triggers runaway expansion; similarly, compressing it causes a runaway collapse. Einstein was willing, even eager, to dispose of the “ugly” cosmological constant in his equations. Hubble’s 1929 paper on the redshift – distance relation gave Einstein the necessary excuse for tossing Λ onto the rubbish heap. (Einstein later described the cosmological constant Λ as “the greatest blunder of my career.”) Ironically, however, the same paper that caused Einstein to abandon the cosmological constant caused other scientists to embrace it. In his initial analysis, remember, Hubble badly underestimated the distance In classical physics. a vacuum is not a sterile void.4. and the universe can be arbitrarily old for a given value of H0 . like sideburns or short skirts.4. as known from radioactive dating. It has been particularly fashionable during periods −1 when the favored value of the Hubble time H0 has been embarrassingly short compared to the estimated ages of astronomical objects.” In quantum physics. The Heisenberg uncertainty principle permits particle-antiparticle pairs to spontaneously appear and then annihilate in an otherwise empty vacuum. indeed. the more frequently the auditor checks up on her. have a positive acceleration.70) Just as there’s an energy density associated with the real particles in the universe. If the value ¨ ˙ of Λ is large enough to make a > 0. A vacuum. Since 1917. If Λ has a value greater than 4πGρ0 . and as King Lear would say. less than half the age of the Earth. and hence overestimated the Hubble constant. the idea of a vacuum having energy is nonsense. “Nothing can come of nothing. the cosmological constant is popular. A question which has been asked since the time of Einstein – and one which I’ve assiduously dodged until this moment – is “What is the physical cause of the cosmological constant?” In order to give Λ a real physical meaning. however. The total energy ∆E and the lifetime ∆t of these pairs of virtual particles must satisfy the relation10 ∆E∆t ≤ h . contains nothing. that indicate that the expansion of the universe does. which we will discuss in Chapter 7. Currently. Currently. (4. LEARNING TO LOVE LAMBDA 75 to galaxies. we need to identify some component of the universe whose energy density εΛ remains constant as the universe expands or contracts. 10 . then a was smaller in the past than it is −1 now. The vacuum density εvac is a quantum phenomenon which The usual analogy that’s made is to an embezzling bank teller who takes money from the till but who always replaces it before the auditor comes around. from the classical viewpoint. there is an energy density εvac associated with the virtual particleantiparticle pairs. the leading candidate for this component is the vacuum energy. and consequently the universe is older than H0 . the more money a teller is entrusted with. the cosmological constant has gone in and out of fashion. Naturally. then the expansion of the universe is −1 accelerating. How could cosmologists reconcile a short Hubble time with an old Earth? Some cosmologists pointed out that one way to increase the age of the universe for −1 a given value of H0 was to introduce a cosmological constant. Hubble’s initial −1 value of H0 = 500 km s−1 Mpc−1 leads to a Hubble time of H0 = 2 Gyr. thanks to observations. 4: “Relativistically correct” derivation of the Friedmann equation . (4.72) This is 124 orders of magnitude larger than the current critical density for our universe. fluid. and represents a spectacularly bad match between theory and observations. we are indirectly examining the structure of the vacuum on extremely small scales. It has been suggested that the natural value for the vacuum energy density is the Planck energy density. ch. ch. computing the numerical value of εvac is an exercise in quantum field theory which has not yet been successfully completed. with applications Harrison (2000). Suggested reading [Full references are given in the “Annotated Bibliography” on page 286. εvac ∼ EP 3 P (???) . 16: Newtonian derivation of Friedmann equation Liddle (1999). By looking at the universe at extremely large scales. This gives an energy density εvac ∼ 3 × 10133 eV m−3 (!!!) . ch. ch. Unfortunately.2×1028 eV). by deducing the value of εΛ from observations of the expansion of the universe. and acceleration equations Linder (1997).6 × 10−35 m).] Bernstein (1995). Obviously.76 CHAPTER 4. the Planck energy is large by particle physics standards (EP = 1. we don’t know much yet about the energy density of the vacuum! This is a situation where astronomers can help particle physicists. while the Planck length is small by anybody’s standards ( P = 1. COSMIC DYNAMICS doesn’t give a hoot about the expansion of the universe and is independent of time as the universe expands or contracts. including a discussion of equations of state Rich (2001).71) As we’ve seen in Chapter 1. 3: Derivation of Friedmann. (4. 2: Friedmann equation. 2: Dynamics of the universe. ch. 3) If ρ = 3×10−27 kg m−3 . Suppose that some of the matter is converted into radiation (by stars.4) Suppose that the universe were full of regulation baseballs. If the baseballs were distributed uniformly throughout the universe. with mcr = 0. let’s assume that all the gas particles have the same mass . where n is the number density of particles. The total energy density of a gas of particles can be written as ε = nE. (4.0360 m. this wavelength increases as λ ∝ a as the universe expands. What is the total energy of the cosmological constant within a sphere 1 AU in radius? What is the rest energy of the Sun (E = M c2 )? Comparing these two numbers.1) Suppose the energy density of the cosmological constant is equal to the present critical density εΛ = εc.2) Consider Einstein’s static universe. Λ = 4πGρ. we can see galaxies at a distance ∼ c/H0 ∼ 4000 Mpc. and E is the energy per particle. For simplicity. in which the attractive force of the matter density ρ is exactly balanced by the repulsive force of the cosmological constant.0 = 5200 MeV m−3 .145 kg and radius rbb = 0. on average. what is the radius of curvature R0 of Einstein’s static universe? How long would it take a photon to circumnavigate such a universe? (4. do you expect the cosmological constant to have a significant effect on the motion of planets within the Solar System? (4. for instance). does the transparency of the universe on this length scale place useful limits on the number density of intergalactic baseballs? [Note to readers outside North America or Japan: feel free to substitute regulation cricket balls. Will the universe start to expand or contract? Explain your answer. what number density of baseballs would be required to make the density equal to the critical density? (Assume non-relativistic baseballs.4.4.5) The principle of wave-particle duality tells us that a particle with momentum p has an associated de Broglie wavelength of λ = h/p.] (4.0369 m.160 kg and rcr = 0. each of mass mbb = 0. before your line of sight intersected a baseball? In fact. LEARNING TO LOVE LAMBDA 77 Problems (4.) Given this density of baseballs. how far would you be able to see. 78 CHAPTER 4. p → 0). COSMIC DYNAMICS m and momentum p. p → ∞) and that w = 0 in the highly non-relativistic limit (a → ∞. Show that w = 1/3 in the highly relativistic limit (a → 0. . (4.73) Compute the equation-of-state parameter w for this gas as a function of the scale factor a. The energy per particle is then simply E = (m2 c4 + p2 c2 )1/2 = (m2 c4 + h2 c2 /λ2 )1/2 . the universe contains components with both w = 0 and w = 1/3.3) to yield ε(t). the pressure P (t).2). past and future. we can solve equations (5.Chapter 5 Single-Component Universes In a spatially homogeneous and isotropic universe. Thus. (5. the evolution of our universe is complicated by the fact that it contains different components with different equations of state. the relation among the energy density ε(t). It may well contain a cosmological constant. 3c2 R0 a (5. Moreover. with different values 79 . the possibility exists that it may contain still more exotic components. P (t). and the scale factor a(t) is given by the Friedmann equation. In principle. and (5. with w = −1. a ˙ a the fluid equation.1 Evolution of energy density In reality.3) and the equation of state. and a(t) for all times.1). 2 8πG κc2 = ε− 2 2 . (5. We know that the universe contains non-relativistic matter and radiation – that’s a conclusion as firm as the earth under your feet and as plain as daylight.1) a ˙ ε + 3 (ε + P ) = 0 . 5. given the appropriate boundary conditions.2) (5. ˙ a P = wε . 10) . the fluid equation (5. the energy density and pressure for the different components of the universe are additive. as long as there is no interaction between the different components.9). the Friedmann equation doesn’t enter into it. Starting from the general result of equation (5. We may write the total energy density ε as the sum of the energy density of the different components: ε= εw .5) w w Because the energy densities and pressures add in this way. (5. (5.7) (5. If this is so. The total pressure P is the sum of the pressures of the different components: P = Pw = wεw . (5. ˙ a Equation (5. SINGLE-COMPONENT UNIVERSES of w.0 . εw a If we assume that w is constant.0 /a3 . then the component with equation-of-state parameter w obeys the equation a ˙ εw + 3 (εw + Pw ) = 0 ˙ a or a ˙ εw + 3 (1 + w)εw = 0 . I’ve used the usual normalization that a0 = 1 at the present day.2) must hold for each component separately.6) (5.4) w where εw represents the energy density of the component with equation-ofstate parameter w.9) is derived solely from the fluid equation and the equation of state.0 a−3(1+w) .8) Here. we conclude that the energy density εm associated with non-relativistic matter decreases as the universe expands with the dependence εm (a) = εm. (5.7) can be rearranged to yield da dεw = −3(1 + w) . Note that equation (5. Fortunately for the cause of simplicity. then εw (a) = εw. when the energy density of the w component is εw.9) (5.80 CHAPTER 5. The energy density in radiation. When its .0 /a4 . EVOLUTION OF ENERGY DENSITY 81 Figure 5. εr .1. once again. (5. where n is the number density of particles and E is the mean energy per particle. For both relativistic and non-relativistic particles.11) Why this difference between matter and radiation? We may write the energy density of either component in the form ε = nE.5. a sphere of comoving radius rs . assuming that particles are neither created nor destroyed. the number density has the dependence n ∝ a−3 as the universe expands.1: The dilution of non-relativistic particles (“matter”) and relativistic particles (“radiation”) as the universe expands. which expands along with the general expansion of the universe (Figure 5. drops at the steeper rate εr (a) = εr. Consider.1). for photons and other massless particles.0 × 10−5 . let’s assume that galaxies have been emitting −1 light at this rate for the entire age of the universe.15) . (5.1) is contributed solely by their rest mass (E = mc2 ) and remains constant as the universe expands. correct. However. its volume increases by a factor 8.5.82 CHAPTER 5. The Sun.3 that the present luminosity density of galaxies is nL ≈ 2 × 108 L Mpc−3 ≈ 2. which has a temperature T0 = 2. Remember from Section 2. for non-relativistic matter. it is large compared to the energy density of starlight. εc. from Section 2. is emitting out roughly 1045 photons every second. εm = nE = n(mc2 ) ∝ a−3 . as shown in equation (3. The energy of non-relativistic particles (shown in the upper panel of Figure 5.0 ≡ εCMB. Although I’ve explained why photons have an energy density εr ∝ a−4 . Thus. strictly speaking. is 4 εCMB.260 MeV m−3 = = 5. (5. λ ∝ a as the universe expands. SINGLE-COMPONENT UNIVERSES proper radius expands by a factor 2. photons are continuously being created. the explanation required the assumption that photons are neither created nor destroyed.4 × 1017 s. This gives an energy density in starlight of εstarlight. that the present energy density of the CMB.0 5200 MeV m−3 (5.6 × 10−33 J s−1 m−3 )(4.14) As a very rough estimate. t0 ≈ H0 ≈ 14 Gyr ≈ 4.260 MeV m−3 . (5. εr = nE = n(hc/λ) ∝ a−3 a−1 ∝ a−4 . The energy of photons or other massless particles (shown in the lower panel of Figure 5.725 K.45). since.0 = αT0 = 4.12) yielding a density parameter for the CMB of ΩCMB.17 × 10−14 J m−3 = 0.6 × 10−33 watts m−3 .1) has the dependence E = hc/λ ∝ a−1 .0 0. and the density of particles which it contains falls to 1/8 its previous value. so numerous are the photons of the Cosmic Microwave Background. it turns that the energy density of the CMB is larger than the energy density of all the photons emitted by all the stars in the history of the universe. remember.0 ∼ nLt0 ∼ (2.007 MeV m−3 . To see why this is true. for instance. Such an assumption is not.13) Although the energy density of the CMB is small compared to the critical density.4 × 1017 s) ∼ 1 × 10−15 J m−3 ∼ 0. Thus. 17) as long as all neutrino flavors are relativistic. A detailed calculation indicates that the energy density of each neutrino flavor should be ε= 7 4 8 11 4/3 εCMB ≈ 0. the mean energy per photon: Eν ≈ 5 × 10−4 eV . The energy density in neutrinos should be comparable to. have not yet been detected. the average energy density of starlight is currently only ∼ 3% of the energy density of the CMB. remember. When the mean energy of a particular neutrino species drops to ∼ mν c2 . the estimate given above is a very rough one indeed. analogous to the Cosmic Microwave Background. Although neutrinos have been detected from the Sun and from Supernova 1987A. equivalently. the ratio of starlight density to CMB density was even smaller than it is today. then it makes the transition from being “radiation” to being “matter”.) The density parameter of the Cosmic Neutrino Background. yield the larger value εstarlight /εCMB ≈ 0. but not exactly equal to.681 ΩCMB . should then be Ων = 0.18) as long as Eν > mν c2 . taking into account all three flavors of neutrino. As a consequence. we reach a time when the universe was hot and dense enough to be opaque to neutrinos. it is an acceptable approximation to ignore non-CMB photons when computing the mean energy density of photons in the universe.227 εCMB . is a relic of the time when the universe was hot and dense enough to be opaque to photons. (5.16) (The above result assumes that the neutrinos are relativistic. I should note. a (5. The mean energy per neutrino will be comparable to. which includes both direct starlight and starlight absorbed and reradiated by dust.1. The Cosmic Microwave Background. If we extrapolate further back in time. the energy density in photons. current technology permits the detection only . In fact. but not exactly equal to. The neutrinos of the Cosmic Neutrino Background.5. there should be a Cosmic Neutrino Background today. EVOLUTION OF ENERGY DENSITY 83 Thus. or.1. (5. For most purposes. that their energy is much greater than their rest energy mν c2 . In the past. Measurements of background radiation from ultraviolet wavelengths to the near infrared. As we shall see in future chapters. I will use what I call the “Benchmark Model”.3 in non-relativistic matter.0 − Ωm.3 . (5. 3 εm (a) εm. Thus. the energy density of matter and Λ must have been equal. and ΩΛ.0 + Ων. = ≈ εm.4 × 10−5 = 8. In the past.0 ∼ 0.7 1/3 ≈ 0.0 1/3 ≈ 0.0 = 5. at the present moment.0 Ωm.0 3 = = a . (5. however.4 × 10−5 in radiation. when I want to employ a model which matches the observed properties of the real universe.0 = 1 − Ωr.20) In the language of cosmologists.0 ≈ 0. .0 ΩΛ.0 /a εm.21) If the universe has been expanding from an initial very dense state.0 0.0 = 0.22) Note that the Benchmark Model is spatially flat.0 ∼ 0.3. the ratio of densities was εΛ (a) εΛ. when the scale factor was smaller.7. with mν c2 5× −4 10 eV. Unfortunately. This moment of matter-Λ equality occurred when the scale factor was amΛ = 1 Ωm.7 in a cosmological constant. at some moment in the past.0 = ΩCMB.1 In the Benchmark Model. while the density parameter for the cosmological constant is currently ΩΛ.75 .0 = 8. this model has Ωr. Ωm.0 × 10−5 + 3. the available evidence favors a universe in which the density parameter for matter is currently Ωm.3 0.19) We know the energy density of the Cosmic Microwave Background with high precision. the total energy density of non-relativistic matter is poorly known. the cosmological constant is “dominant” over matter today in the Benchmark Model.0 εΛ. then the present density parameter in radiation is Ωr. as is the energy density of the cosmological constant. SINGLE-COMPONENT UNIVERSES of neutrinos with energy E > 0.4 × 10−5 .84 CHAPTER 5.0 ΩΛ. far more energetic than the neutrinos of the Cosmic Neutrino Background.3 (5.1MeV.7 ≈ 2. the ratio of the energy density in Λ to the energy density in matter is εΛ.0 (5. If all neutrino species are effectively massless today.0 0. We can calculate theoretically what the energy density of the Cosmic Neutrino Background should be. 8 × 10−4 . if the universe contains different components with different values of w.1. and we live in a universe described by the Benchmark Model. In addition. followed by a period when matter (w = 0) was dominant. If the presently available evidence is correct. To generalize. the moment of radiation-matter equality took place when the scale factor was 1 εm. In a universe which is continuously expanding. the component with the smallest value of w is dominant.24) εr (a) εr.4 × 10−5 (5. The evidence indicates we live in a universe where radiation (w = 1/3) was dominant during the early stages. we have only recently entered a period when the cosmological constant Λ (w = −1) is dominant.0 8. (5. because of the simple relation between scale factor and redshift. the ratio of the energy density in matter to the energy density in radiation is currently Ωm. to the moment when a = 2. the scale factor a can be used as a surrogate for the cosmic time t.25) arm = εr.0 = ≈ ≈ 3600 εr.0 Thus. for instance. the ratio of matter density to energy density was εm (a) εm. in the past. EVOLUTION OF ENERGY DENSITY 85 Similarly. and hence would have been “radiation” then even if it is “matter” today.26) 1+z . If the universe expands forever.0 = a. then as a → ∞.0 ≈ ≈ 2. Thus.5. equation (5.0 0. it’s even larger if some or all of the neutrino flavors are currently non-relativistic. However.9) tells us that in the limit a → 0.8 × 10−4 with the assurance that we are referring to a unique moment in the history of the universe. Thus. then it would have been relativistic at a scale factor a = 1/3600. matter is now strongly dominant over radiation.23) if all three neutrino flavors in the Cosmic Neutrino Background are still relativistic today.3 εm. in a continuously expanding universe. 1 a= . (5. (5. the scale factor a is a monotonically increasing function of t.0 Ωr. We can refer. the component with the largest value of w is dominant.0 3600 Note that as long as a neutrino’s mass is mν c2 (3600)(5×10−4 eV) ∼ 2 eV. and in which all the transformations of special relativity hold true. 2 R0 (5. equation (5. the Friedmann equation (5.27) Each term on the right hand side of equation (5. matter contributes a term ∝ a−1 . R0 (5. in which there is only one term on the right hand side of equation (5. and the cosmological constant Λ contributes a term ∝ a2 .28) tells us that it is also possible to have an empty universe with κ = −1. ˙ spatially flat universe is a permissible solution to the Friedmann equation. yields useful insight into the physics of an expanding universe. .27) has a different dependence on scale factor. curvature contributes a term independent of a. (Positively curved empty universes are forbidden. no cosmological constant. An empty. SINGLE-COMPONENT UNIVERSES cosmologists often use redshift as a surrogate for time. In a universe with many components.28) One solution to this equation has a = 0 and κ = 0. 5. However. “Radiation-matter equality took place at a redshift zrm ≈ 3600. radiation contributes a term ∝ a−2 . For example. For this universe. However. does not yield a simple analytic form for a(t).27) for a multiple-component model. no contribution to ε of any sort. This is the universe whose geometry is described by the Minkowski metric of equation (3. no matter. light that was emitted at the time of radiation-matter equality is observed by us with its wavelength increased by a factor of 3600.20). static. such as the Benchmark Model.0 a w −1−3w κc2 − 2 . Solving equation (5.27) takes the form a2 = − ˙ κc2 . One reason why cosmologists use scale factor or redshift as a surrogate for time is that the conversion from a to t is not simple to calculate in a multiplecomponent universe like our own.2 Curvature only A particularly simple universe is one which is empty – no radiation. can be written in the form 8πG a = ˙ 3c2 2 εw.” That is.86 CHAPTER 5.27). they make statements such as. looking at simplified singlecomponent universes. You observe a distant light source. a(te ) te (5. after cosmologist to study its properties. negatively curved universe.2 In Newtonian terms. Note that in an empty universe. if there’s no gravitational force at work. so that you could reasonably approximate it as −1 an empty. (5.30) is a good approximation to the true scale factor. at t = t0 .) A nega˙ tively curved empty universe must be expanding or contracting. 2 . 3 If a universe contains nothing. with nothing to speed or slow the expansion. and thus the scale factor a simply increases linearly with time in an empty universe. In an empty expanding universe.2.5. The light which you observe now. there will be no observers in it to detect the expansion. expanding universe is sometimes called a Milne universe.3 However. then the linear scale compared to the critical density εc (that is. t = te . if Ω factor of equation (5. t0 = H0 . (5. CURVATURE ONLY 87 since that would require an imaginary value of a in equation 5.32) te = 1+z 1+z When observing a galaxy with a redshift z. R0 (5. expanding universe might seem nothing more than a mathematical curiosity. An empty. expanding universe is shown as the dashed line in −1 Figure 5. integration of this relation yields a scale factor of the form t a(t) = . The scale factor in an empty. such as a galaxy.2.28. negatively curved. with t0 = H0 = R0 /c.29) In an expanding empty universe. Suppose you were in an expanding universe with a negligibly small value for the density parameter Ω. in addition to asking “When was the light from that galaxy emitted?” you may also ask “How far away Such an empty. the age of the universe is exactly equal to the Hubble time.31) so it is easy to compute the time when the light you observe from the source was emitted: H −1 t0 = 0 . with a=± ˙ c . was emitted at an earlier time. if a universe has a density ε which is very small 1).30) t0 where t0 = R0 /c. 1+z = t0 1 = . which has a redshift z. then the relative velocity of any two points is constant. empty universe (dashed). and a flat. SINGLE-COMPONENT UNIVERSES 8 6 a 4 2 0 −2 0 Λ empty matter radiation 2 4 H0(t−t0) 6 8 Figure 5. radiationdominated universe (solid). . Λ-dominated universe (dot-dash). matter-dominated universe (dotted). a flat.88 CHAPTER 5. a flat.2: The scale factor as a function of time for an expanding. In the limit z 1.5.3 we saw that in any universe described by a Robertson-Walker metric. if light is emitted by the galaxy at time te and detected by the observer at time t0 . at distances dp (t0 ) c/H0 . However. In the specific case of an empty expanding universe.34) te a(t) 0 Thus. CURVATURE ONLY 89 is that galaxy?” In Section 3.28) dp (t0 ) = a(t0 ) r 0 dr = r .39): r t0 dt = c dr = r . φ) is (see equation 3. and thus dp (t0 ) = ct0 t0 te dt = ct0 ln(t0 /te ) . dp (t0 ) = c ln(1 + z) . θ.3. (5.35) te a(t) Equation (5. there is a linear relation between dp and z. you must remember that dp (t0 ) is the proper distance to the light source at the time of observation. H0 (5. At first glance. as seen observationally in Hubble’s law. at the time of emission. However.33) Moreover. the redshift increases exponentially with distance. the current proper distance from an observer at the origin to a galaxy at coordinate location (r. the current proper distance from you (the observer) to the galaxy (the light source) is t0 dt dp (t0 ) = c . however. the proper distance dp (te ) was . the null geodesic followed by the light satisfies equation (3.35) holds true in any universe whose geometry is described by a Robertson-Walker metric.37) This relation is plotted as the dashed line in the upper panel of Figure 5. we can see objects which are currently at an arbitrarily large distance. In an empty expanding universe. dp ∝ ln z in an empty expanding universe.36) Expressed in terms of the redshift z of the observed galaxy. (5. t (5. In the limit z 1.2. it may seem counterintuitive that you can see a light source at a proper distance much greater than c/H0 when the age of the universe is only 1/H0 . a(t) = t/t0 . (5. 2. the lower panel shows the proper distance at the time the light was emitted.1 1 z 10 100 1000 100 (H0/c) dp(te) 10 1 . The upper panel shows the proper distance at the time the light is observed.90 CHAPTER 5.01 Emission . SINGLE-COMPONENT UNIVERSES 100 (H0/c) dp(t0) 10 1 .1 .1 1 z 10 100 1000 Figure 5. .01 .3: The proper distance to an object with observed redshift z. The line types are the same as those of Figure 5.1 .01 Observation .01 . SPATIALLY FLAT UNIVERSES 91 smaller by a factor a(te )/a(t0 ) = 1/(1 + z). t0 . yielding the solution q= 2 . H0 1 + z (5.42) 1 + w 6πGε0 The Hubble constant in such a universe is 2 a ˙ t−1 . where dp (te ) = (1/e)(c/H0 ) ≈ 0.3 Spatially flat universes Setting the energy density ε equal to zero is one way of simplifying the Friedmann equation. single-component universe.5. the proper distance at the time of emission was dp (te ) = c ln(1 + z) . The left hand side of equation (5.43) . single-component universe is a(t) = (t/t0 )2/(3+3w) . the scale factor in a spatially flat.40) with the restriction w = −1. (5. = H0 ≡ a t=t0 3(1 + w) 0 (5. In such a spatially flat. (5.7. and the right hand side is ∝ t−(1+3w)q . 3 + 3w (5.38) shown as the dashed line in the lower panel of Figure 5. In an empty expanding universe. 5.39) 3c2 To solve this equation. when their proper distance from the observer was very small. In an empty expanding universe.41) where the age of the universe. we first make the educated guess that the scale factor has the power law form a ∝ tq .37(c/H0 ). dp (te ) has a maximum for objects with a redshift z = e − 1 ≈ 1. (5. with a single value of w. is linked to the present energy density by the relation 1/2 1 c2 t0 = . the Friedmann equation takes the simple form 8πGε0 −(1+3w) a2 = ˙ a .39) is then ∝ t2q−2 .3.3. Another way is to set κ = 0 and to demand that the universe contain only a single component. Objects with much higher redshifts are seen as they were very early in the history of the universe. With the proper normalization. 46) can be written in the form ε(t) = c2 −2 1 t .47) (5. the universe is older than the Hubble time. or other distant light source. 3(1 + w) 0 (5. the universe is younger than the Hubble time.50) Suppose yourself to be in a spatially flat.46) 2 H −1 . the energy density as a function of time is (combining equations 5. the energy density of a component with equation-of-state parameter w is ε(a) = ε0 a−3(1+w) .45) ε(t) = ε0 (t/t0 )−2 . (5. if w > −1/3. 6π(1 + w)2 G (5.51) a(te ) te .49) Expressed in terms of Planck units (introduced in Chapter 1). this relation between energy density and cosmic time is ε(t) = 1 Ep 2 3 6π(1 + w) p t tP −2 . (5.0 = and t0 = 3c2 2 H 8πG 0 (5. As a function of scale factor. single-component universe. 3(1 + w) 0 (5.44) In a spatially flat universe.45) so in a spatially flat universe with only a single component. regardless of the value of w.41 and 5.92 CHAPTER 5. in terms of the Hubble time. Making the substitutions ε0 = εc. If w < −1/3. SINGLE-COMPONENT UNIVERSES The age of the universe. is then t0 = 2 H −1 . If you see a galaxy. you can use the relation a(t0 ) t0 2/(3+3w) 1+z = = (5.48) equation (5. with a redshift z. In terms of H0 and z rather than t0 and te . In the universe. The visible universe More technically. this is what’s called the particle horizon distance. The proper distance (at the time of observation) to such an object is called the horizon distance.54) The most distant object you can see (in theory) is one for which the light emitted at t = 0 is just now reaching us at t = t0 . beyond which you cannot see because light from more distant objects has not had time to reach you. 3(1+w)/2 (1 + z) 3(1 + w)H0 (1 + z)3(1+w)/2 (5. computing the value of dp (t0 ) in the limit te → 0 (or.3. In such a case. 1 + 3w H0 1 + 3w (5. SPATIALLY FLAT UNIVERSES 93 to compute the time te at which the light from the distant galaxy was emitted: te = 2 1 t0 = . for short. the current horizon distance is dhor (t0 ) = c t0 0 dt . In a universe described by a Robertson-Walker metric.53) when w = −1/3. beyond which you cannot see because of the Earth’s curvature. the horizon is a spherical surface centered on you. equivalently. 4 Here on Earth. a(t) (5. a(t) 1 + 3w (5.5. the current proper distance is dp (t0 ) = 2 c [1 − (1 + z)−(1+3w)/2 ] .52) The current proper distance to the galaxy is dp (t0 ) = c t0 te dt 3(1 + w) = ct0 [1 − (te /t0 )(1+3w)/(3+3w) ] . the horizon is a circle centered on you. I’ll continue to call it the horizon distance. H0 1 + 3w (5. The portion of the universe lying within the horizon for a particular observer is referred to as the visible universe for that observer. the horizon distance has a finite value if w > −1/3. z → ∞) yields dhor (t0 ) = ct0 c 3(1 + w) 2 = .55) In a spatially flat universe.56) In a flat universe dominated by matter (w = 0) or by radiation (w = 1/3). 4 . an observer can see only a finite portion of the infinite volume of the universe. starting with a universe containing only non-relativistic matter (w = 0).61) Such a universe is sometimes called an Einstein-de Sitter universe. SINGLE-COMPONENT UNIVERSES consists of all points in space which have had sufficient time to send information. matter-only universe.3. If you see a galaxy with redshift z in a flat. the horizon distance is infinite.4 Matter only Let’s now look at specific examples of spatially flat universes.57) t0 = 3H0 and the horizon distance is dhor (t0 ) = 3ct0 = 2c/H0 . The proper distance at the time the light was emitted was smaller by a factor 1/(1 + z): dp (te ) = 5 2c 1 1− √ H0 (1 + z) 1+z . is dp (t0 ) = c t0 te 2c 1 dt = 3ct0 [1 − (te /t0 )1/3 ] = 1− √ 2/3 (t/t0 ) H0 1+z .58) illustrated as the dotted line in Figure 5. (5. the visible universe consists of all points which are causally connected to the observer. The scale factor. the proper distance to that galaxy. at the time of observation.59) (5. to the observer. In a flat universe with w ≤ −1/3.94 CHAPTER 5. for extremely distant points. nothing which happens outside the visible universe can have an effect on the observer. In other words. and all of space is causally connected to an observer. in the form of photons or other relativistic particles. as a function of time. you could see every point in space – assuming the universe was transparent. In such a universe with w ≤ −1/3.2. .60) illustrated as the dotted line in the upper panel of Figure 5. you would see extremely redshifted versions of what they looked like extremely early in the history of the universe. (5. of course. 5.5 The age of such a universe is 2 . (5. (5. is am (t) = (t/t0 )2/3 . However. radiation-only universe. Thus.3. dp (te ) has a maximum for galaxies with a redshift z = 5/4. In an expanding. In a flat. 5. at early times – long before the time of radiationmatter equality – the universe was well described by a spatially flat. If. The proper distance at the time the light was emitted was dp (te ) = c 1 c z . flat universe containing only radiation. The scale factor of a flat.5. In a flat. (5. (5. .66) illustrated as the solid line in the lower panel of Figure 5. dp (te ) has a maximum for light sources with a redshift z = 1. since early in the history of our own universe. RADIATION ONLY 95 illustrated as the dotted line in the lower panel of Figure 5. the proper distance to the light source will be dp (t0 ) = c t0 te c 1 dt = 2ct0 [1 − (te /t0 )1/2 ] = 1− (t/t0 )1/2 H0 1+z . at a time t0 . matteronly universe. (5. radiationonly model.27). where dp (te ) = (8/27)c/H0 . which is not generally the case. where dp (te ) = (1/4)c/H0 .63) In the special case of a flat.3. the horizon distance is exactly equal to the Hubble distance.3.5.64) illustrated as the solid line in Figure 5. the age of the universe is 1 t0 = . radiation-only universe is a(t) = (t/t0 )1/2 . radiation-dominated universe. radiation-only universe.5 Radiation only The case of a spatially flat universe containing only radiation is of particular interest.65) illustrated as the solid line in the upper panel of Figure 5. 1− = H0 (1 + z) 1+z H0 (1 + z)2 (5. you observe a distant light source with redshift z in a flat. the radiation (w = 1/3) term dominated the right-hand side of the Friedmann equation (see equation 5.62) 2H0 and the horizon distance at t0 is dhor (t0 ) = 2ct0 = c/H0 .2. (5. radiation-only universe.094 EP 3 P t tP −2 . measured in units of the Planck density (EP / 3 ∼ P 3 × 10133 eV m−3 ). the energy density of our own universe (well approximated as a flat. given in equations (2. in the early stages of our universe. (5.26) and (2.70). each of infinite energy (equation 5. since the assumptions of general relativity.61TP t tP −1/2 .67) Thus.69). that is. In cosmological contexts. (5.70kT (t) ≈ 1.67 and 5. Using the blackbody relation between energy density and temperature. The mean energy per photon in a radiation-dominated universe is then Emean (t) ≈ 2. Thus. at the instant t = 0. we may assign a temperature to a universe dominated by blackbody radiation: T (t) = 45 32π 2 1/4 TP t tP −1/2 ≈ 0. radiation-only model in its early stages) was infinite. (5. Thus. radiation-only universe is εr (t) = ε0 (t/t0 )−2 = 3 EP 32 3 P t tP −2 ≈ 0.4 × 1032 K. εr → ∞ (equation 5. on which the Friedmann equation is based. it does not take into account the effects of quantum mechanics. (5. Why can’t general relativity be used at times earlier than the Planck time? General relativity is a classical theory. according to this analysis. TP = 1. SINGLE-COMPONENT UNIVERSES The energy density in a flat.67).96 CHAPTER 5.66EP t tP −1/2 . when radiation was strongly dominant.27).70) Note that in a flat. measured in units of the Planck time (tP ∼ 5×10−44 s).69) and the number density of photons is (combining equations 5.68) Here TP is the Planck temperature. was comparable to one over the square of the cosmic time.057 t n(t) = ≈ 3 Emean (t) tP P −3/2 .69) εr (t) 0. break down at t ≈ tP . this infinite energy density was provided by an infinite number density of photons (equation 5. extrapolating the results of this chapter to times earlier than the Planck time is not physically justified. the energy density. Should we take these infinities seriously? Not really. general relativity assumes that the energy content of the universe is smooth down to arbitrarily . as t → 0. (5. consider the case with w = −1. and we may safely use the results of general relativity.6 Lambda only As seen in Section 5.73) The quantization of the universe can no longer be ignored when N (t) ≈ 1. In a flat. that is.74) Now. and the mean photon energy was Emean ∼ EP ∼ 1 × 1028 eV.3.5. In order to accurately described the universe at its very earliest stages.71) so the volume of the visible universe at time t is Vhor (t) = 4π 3 d ≈ 34 3 hor 3 P 3 . a theory of quantum gravity is needed. when the number density of photons was n ∼ −3 ∼ P 2 × 10104 m−3 . for the sake of completeness. a complete theory of quantum gravity does not yet exist.9 t tP 3/2 . then quantum mechanical effects must be taken into account. Consequently. (5.72) and (5. LAMBDA ONLY 97 small scales. we find that the number of photons inside the horizon at time t is N (t) = Vhor (t)n(t) ≈ 1. (5. instead of being parceled into individual quanta. 5. and the classical results of general relativity no longer apply. the horizon distance grows linearly with time: dhor (t) = 2ct = 2 P t tP t tP . Unfortunately. radiation-only universe. a spatially flat. in this book. within a horizon distance.7tP . equivalent to a time t ≈ 0. (5. continuous energy density is justifiable. prior to the Planck time. or photons. t ∼ tP ∼ 5 × 10−44 s. single-component universe with w = −1 has a power-law dependence of scale factor on time: a ∝ t2/(3+3w) . if there are only a few photons within the visible universe. a universe in which the energy density is contributed by a cosmological . many quanta.6. then the approximation of a smooth.72) Combining equations (5. I will not deal with times earlier than the Planck time. As long as a radiation-dominated universe has many.70). However. (5.3. lambda-only universe. If the cosmological constant Λ is provided by the vacuum energy.2.80) shown as the dot-dashed line in the lower panel of Figure 5. (5. ˙ where 8πGεΛ 1/2 .79) shown as the dot-dashed line in the upper panel of Figure 5. lambda-dominated universe. is dp (t0 ) = c t0 te eH0 (t0 −t) dt = c H0 (t0 −te ) c [e − 1] = z. in a flat.3. under the label “Steady State universe”. the Friedmann equation takes the form 8πGεΛ 2 a . the density ε of the universe remains constant because of the continuous creation of real particles. H0 H0 (5. A flat universe containing nothing but a cosmological constant is infinitely old.77) (5. This equation can be rewritten in the form a = H0 a . H0 1 + z (5. at the time you observe it.3. and has an infinite horizon distance dhor . then the density ε of a lambda-dominated universe remains constant because of the continuous creation and annihilation of virtual particleantiparticle pairs. A spatially flat universe with nothing but a cosmological constant is exponentially expanding. In a Steady State universe. SINGLE-COMPONENT UNIVERSES constant Λ.98 CHAPTER 5. in Section 2. we’ve seen an exponentially expanding universe before.76) (5. The proper distance at the time the light was emitted was dp (te ) = c z . the proper distance to the light source. .76) in an expanding universe is a(t) = eH0 (t−t0 ) . H0 = 3c2 The solution to equation (5.6 For such a flat. If.75) a2 = ˙ 3c2 where εΛ is constant with time. 6 Such a universe is sometimes called a de Sitter universe. you see a light source with a redshift z.78) This scale factor is shown as the dot-dashed line in Figure 5. or flat universes with a single component – continue to expand forever if they are expanding at t = t0 . In a flat. such as the flat lambdadominated model. their recession velocity is greater than the speed of light. evolution of scale factor in single-component universes Problems (5. About how many cosmic neutrinos are inside your body right now? What must be the sum of the neutrino masses. The simple models that we’ve examined in this chapter – empty universes. is the only universe for which dp (t0 ) is linearly proportional to z for all values of z. and look at complicated toy universes. To study universes with more complicated behavior. it is necessary to put aside our simple toy universes. then start to collapse? Is it possible to have universes in which the scale factor is not a simple power-law or exponential function of time? The short answer to these questions is “yes”. LAMBDA ONLY 99 Note that an exponentially growing universe. and photons from the light source can no longer reach the observer. with a single term on the right-hand side of the Friedmann equation. both matter-only and radiation-only Linder (1997). Is it possible to have universes which stop expanding.12 × 108 m−3 for each of the three species of neutrino. the relation dp (t0 ) ∝ z only holds true in the limit z 1. ch.6.] Liddle (1999). the observer sees them as they were just before they reached a proper distance c/H0 .0 ? . dp (t0 ) → ∞ but dp (te ) → c/H0 . 2.5. ch. Suggested reading [Full references are given in the “Annotated Bibiography” on page 286. In other universes.1) The predicted number of neutrinos in the Cosmic Neutrino Background is nν = (3/11)nγ = 1. Once a light source is more than a Hubble distance from the observer. however. εc.5: Evolution of energy density. lambda-dominated universe. highly redshifted objects (z 1) are at very large distances (dp (t0 ) c/H0 ) at the time of observation. m(νe ) + m(νµ ) + m(ντ ). Note also that in the limit z → ∞.4.2. 4: Flat universes. with multiple terms on the right-hand side of the Friedmann equation. in order for the density of the Cosmic Neutrino Background to be equal to the critical density. SINGLE-COMPONENT UNIVERSES (5. What is the current proper distance to the galaxy.] (5.3) Suppose you are in a flat. you observe a galaxy at a redshift z = 7. matter-only universe which has a Hubble constant H0 = 70 km s−1 Mpc−1 . if the universe contains only radiation? What was dp (te ) if the universe contains only matter? What was dp (te ) if the universe contains only a cosmological constant? . How long will you have to keep observing the galaxy to see its redshift change by one part in 106 ? [Hint: use the result from the previous problem. dp (te ). if the universe contains only radiation? What is dp (t0 ) if the universe contains only matter? What is dp (t0 ) if the universe contains only a cosmological constant? What was the proper distance at the time the light was emitted.100 CHAPTER 5. (5. You observe a galaxy with z = 1. Show that the observed redshift changes at a rate dz = H0 (1 + z) − H0 (1 + z)3(1+w)/2 . dp (t0 ).2) A light source in a flat.81) dt0 For what values of w does the redshift decrease with time? For what values of w does the redshift increase with time? (5. single-component universe has a redshift z when observed at a time t0 .4) In a flat universe with H0 = 70 km s−1 Mpc−1 . (6.2) so we can rewrite the Friedmann equation without explicitly including the curvature: 2 8πG H0 H(t)2 = ε(t) − (Ω0 − 1) .Chapter 6 Multiple-Component Universes The Friedmann equation. in general. for which the energy 101 . and Ω0 .3) 3c2 a(t)2 2 Dividing by H0 . and radiation. 3c2 R0 a(t)2 (6. 2 R0 c (6.4) (6.0 a(t)2 where the critical density today is εc. including the cosmological constant. and ε(t) is the energy density contributed by all the compo˙ nents of the universe.0 ≡ εm 2 3c2 H0 .1) where H ≡ a/a. H0 . 8πG (6.31) tells us the relation among κ.5) We know that our universe contains matter. H2 κ = 20 (Ω0 − 1) . this becomes ε(t) 1 − Ω0 H(t)2 = + . can be written in the form H(t)2 = 8πG κc2 ε(t) − 2 . for which the energy density has the dependence εm = εm. 2 H0 εc.0 /a3 . R0 . Equation (4. Thus. Current evidence seems to indicate the presence of a cosmological constant. For instance. However. (1−Ω0 )/a2 in equation (6. with energy density εΛ = εΛ.0 /a2 da = H0 t . + Ωm. For instance. Ωm.0 = εm. it can be integrated numerically.6) by a2 .0 /a4 .0 + Ωm.0 = εΛ. in a universe with radiation. we expect the Friedmann equation (6. In many circumstances.0 .0 + 2 H0 a a a2 (6. = 4 + 3 + ΩΛ.0 . However.8) This is not a user-friendly integral: in the general case.0 /εc.0 . radiation (w = 1/3). as the 21st century began. then taking the square ˙ root. ¨ giving a universe with a > 0.0 .0 /a + ΩΛ.0 a2 + (1 − Ω0 )]1/2 . it is not demanded by the data.0 H2 . the integral in equation (6. it doesn’t have a simple analytic solution. might be nonzero.7) The cosmic time t as a function of scale factor a can then be found by performing the integral a 0 [Ωr. I will only consider the contributions of matter (w = 0).0 = constant. It is certainly possible that the universe contains other components as well.0 . prudence dictates that we should keep in mind the possibility that the curvature term.102 CHAPTER 6. a component of the universe whose equation-of-state parameter can lie in the range −1 < w < −1/3.0 /a + ΩΛ. and hence is spatially flat.0 Ωm. some cosmologists were investigating the properties of “quintessence”. and Ω0 = Ωr.0 /εc.0 a2 + (1 − Ω0 )]1/2 (6. introduced in the previous chapter as a model consistent with all available data.0 /εc. as noted in the previous chapter.0 . MULTIPLE-COMPONENT UNIVERSES density has the dependence εr = εr. the radiation term . Ωm.0 .0 /a2 + Ωm.0 = εr. In our universe.8) has a simple analytic approximation to its solution. yields −1 H0 a = [Ωr. and ΩΛ. The Benchmark Model. in the absence of strong evidence for the existence of “quintessence”. ΩΛ. and Λ. and the cosmological constant Λ (w = −1). However. has Ω0 = 1. although a perfectly flat universe is consistent with the data. curvature.0 + ΩΛ. Since H = a/a.6). ˙ (6. for given values of Ωr.4) to take the form 1 − Ω0 Ωr. matter. multiplying equation (6.6) where Ωr. During these epochs. and ΩΛ. and density.0 a2 . however. matter-dominated universes. After Einstein dismissed the cosmological constant as a blunder.8 × 10−4 and matter-lambda equality takes place at amΛ ≈ 0. However. Such a universe is examined in Section 6. they considered the possibility that the universe was negatively curved or positively curved. the Benchmark Model is well approximated by a flat universe containing only matter and a cosmological constant. In this limit. matter-dominated universes provide useful physical insight into the interplay among curvature. For instance. a universe containing both matter and curvature (either negative or positive). Ωm. In addition to being of historical interest. During the mid-twentieth century. the Benchmark Model is well approximated by a flat universe containing only radiation and matter. this reduces to the solution already found for a flat. two of the components are of comparable density. a 0 ada Ωr.0 ≈ 1 2 Ωr.2.103 dominates the expansion during the early stages of expansion.0 .0 . For some values of Ωr. expansion. equation (6. For instance. in the Benchmark Model.0 = 1. with the bulk of the density being provided by non-relativistic matter.75. (6.0 . If the universe continues to expand forever.9) . there will be intermediate epochs when the matter or the curvature dominates the expansion.8) simplifies to H0 t ≈ or a(t) ≈ (2 Ωr.4. a single-component model is a poor description of the universe. then in the limit a → ∞. the cosmological constant term will dominate the expansion. (6. For scale factors near amΛ ≈ 0. radiation-only universe.10) In the limit Ωr.0 H0 t)1/2 . First. for scale factors near arm ≈ 2. and before astronomers had any clear idea what the value of the density parameter Ω was.75. we will examine a universe which is of great historical interest to cosmology. curved. and a two-component model must be utilized.8 × 10−4 . where radiation-matter equality takes place at a scale factor arm ≈ 2. during some epochs of the universe’s expansion. cosmologists concentrated much of their interest on the study of curved. Such a universe is examined in Section 6. and provide terms of roughly equal size in the Friedmann equation. a matter-only universe is a fair approximation to reality when arm a amΛ . The Friedmann equation in a curved. Suppose you are in a universe which is currently expanding (H0 > 0) and which contains nothing but non-relativistic matter.”1 In a curved universe containing nothing but matter. it is nearly obligatory for a cosmology text to quote T.11) Such a flat. For the universe to cease expanding. then it expands with time.12) enables you to answer that question. Ω0 − 1 (6.1 Matter + curvature Consider a universe containing nothing but pressureless matter. Eliot: “This is the way the world ends / Not with a bang but a whimper. matter-dominated universe (equation 6. and hence κ = +1. as demonstrated in Section 5. this quote is from Eliot’s poem The Hollow Men. this coincidence seems to be just that – a coincidence. with w = 0. Since the first term on the right hand side of equation (6. “Will the universe ever cease to expand?” then equation (6. . MULTIPLE-COMPONENT UNIVERSES 6. = 3 + 2 H0 a a2 (6. Eliot did not keep up to date on the technical literature of cosmology. the year when Friedmann published his second paper on the expansion of the universe. However.12) since Ωm.0 = Ω0 in such a universe. for the most part. At the time of maximum expansion. If you ask the question.6) can be written in the form Ω0 1 − Ω 0 H(t)2 .104 CHAPTER 6. H(t) = 0 requires the second term on the right hand side to be negative. matter-only universe expands outward forever. Such a fate is sometimes known as the “Big Chill”. S. (6. there must be some moment at which H(t) = 0.13) The scale factor at the time of maximum expansion will therefore be amax = 1 (6. since the temperature of the universe decreases monotonically with time as the universe expands. in 1924.4. If such a universe is spatially flat. the ultimate fate of the cosmos is intimately linked to the density parameter Ω0 . H(t) = 0 and thus 0= Ω0 1 − Ω0 + 2 . This means that a matter-dominated universe will cease to expand if Ω0 > 1.12) is always positive. 3 amax amax Ω0 . written. At this point.14) Interestingly. with a scale factor a(t) = (t/t0 )2/3 . lumpy universe. Stars will not absorb the photons which they previously emitted.12). As the universe approaches the Big Crunch. Thus if such a universe is expanding at a time t = t0 . then a cosmic ultraviolet background. When such a universe is in its contracting stage. In a real. generate entropy as they emit photons. it will continue to expand forever. At early times. A matter-dominated universe with Ω0 > 1 not only has finite spatial extent. however.1. and the scale factor will grow at the rate a ∝ t2/3 . entropy is not conserved on small scales. dense state. then finally a cosmic gamma-ray background. remember. A matter-dominated universe with Ω0 > 1 will expand to a maximum scale factor amax . just as it began in a hot.6. and its ultimate fate are closely linked. after the universe reaches maximum expansion. is the density parameter as measured at a scale factor a = 1. or entropy-conserving. the matter term of the Friedmann equation will dominate. 3 This is from Frost’s poem Fire and Ice. If a universe contains nothing but matter. small-scale entropy-producing processes will NOT be reversed. for instance. What is the ultimate fate of a matter-dominated universe with Ω0 < 1 and κ = −1? In the Friedmann equation for such a universe (equation 6. Note that in equation (6. and the universe will expand like the negatively curved empty universe. MATTER + CURVATURE 105 where Ω0 . Stars.2 Eventually. its density. its curvature. both terms on the right hand side are positive.1. dense state. Ultimately. if the density is greater The contraction is a perfect time reversal of the expansion only when the universe is perfectly homogeneous and the expansion is perfectly adiabatic. the Ω0 > 1 universe will collapse down to a = 0 (an event sometimes called the “Big Crunch”) after a finite time t = tcrunch . people will not live backward from grave to cradle. the Hubble parameter enters only as H 2 . then a cosmic visible background. the contraction phase. an observer will see galaxies with a blueshift proportional to their distance. when the scale factor is small (a Ω0 /[1 − Ω0 ]). first published in Harper’s Magazine in De2 . During the contraction phase of an Ω 0 > 1 universe. the density of matter will be diluted far below the critical density. then collapse in a Big Crunch. At this point. the obligatory quote is from Robert Frost: “Some say the world will end in fire / Some say in ice. so it will end in a hot. then a cosmic x-ray background. with a ∝ t.12). but also has a finite duration in time. as shown in Table 6. is just the time reversal of the expansion phase. Thus. the cosmic microwave background will become a cosmic infrared background.”3 In a matter-dominated universe. Given this parametric form. it is easy to show that the time that elapses between the Big Bang at θ = 0 and the Big Crunch at θ = 2π is tcrunch = Ω0 π . In a curved universe containing only matter.18) A plot of a versus t in the case Ω0 = 1. matter-dominated universes density curvature ultimate fate Ω0 < 1 κ = −1 Big Chill (a ∝ t) Ω0 = 1 κ=0 Big Chill (a ∝ t2/3 ) Ω0 > 1 κ = +1 Big Crunch than the critical density. The Friedmann equation can be written in the form a2 ˙ = Ω0 /a + (1 − Ω0 ) .1. The solution when Ω0 > 1 is a(θ) = and t(θ) = 1 Ω0 (1 − cos θ) 2 Ω0 − 1 (6. the scale factor a(t) can be computed explicitly. the solution to this integral is most compactly written in a parametric form. MULTIPLE-COMPONENT UNIVERSES Table 6. The a ∝ t2/3 behavior of an Ω0 = 1 universe is shown as the dotted line. if the density is less than or equal to the critical density. 2H0 (Ω0 − 1)3/2 (6.19) 1 Ω0 (θ − sin θ) . [Ω0 /a + (1 − Ω0 )]1/2 (6. the universe will end in a fiery Big Crunch. Eliot.1: Curved. 2 H0 (6.16) When Ω0 = 1.1 is shown as the solid line in Figure 6. the universe will end in an icy Big Chill. S. . H0 (Ω0 − 1)3/2 (6.17) where the parameter θ runs from 0 to 2π. Unlike T. and frequently wrote poems on astronomical themes.106 CHAPTER 6. The cember 1920.15) so the age t of the universe at a given scale factor a is given by the integral H0 t = a 0 da . Frost was keenly interested in astronomy. The dotted line is a(t) for a universe with Ω0 = 1 (flat).5 0 −.9 (negatively curved).5 a 1 .1: The scale factor as a function of time for universes containing only matter.5 2 1.5 0 . The bottom panel is a blow-up of the small rectangle near the lower left corner of the upper panel.1 (positively curved).0 Ω0=1.1 80 100 2.5 2 Figure 6. the solid line is a(t) for a universe with Ω0 = 1. the dashed line is a(t) for a universe with Ω0 = 0.1. MATTER + CURVATURE 107 40 30 a 20 10 0 0 20 40 60 H0(t−t0) Ω0=0.6.5 1 H0(t−t0) 1.9 Ω0=1. . 21) t(η) = 2 (Ω0 − 1)3/2 where the parameter η runs from 0 to infinity. any component with w < −1/3). If the universe has a cosmological constant (or more generally. it is very difficult. as shown graphically in the upper panel of Figure 6.9 is shown as the dashed line in Figure 6.1 universe and the Ω0 = 0. then the equation Density = Destiny = Curvature no longer applies. at t ∼ t0 . MULTIPLE-COMPONENT UNIVERSES solution of equation (6. at a given time t = t0 . Note that although the ultimate fates of an Ω0 = 0.1. to tell a universe with Ω0 slightly less than one from that with Ω0 slightly greater than one. 6. but which contains both matter and a cosmological constant.2 Matter + lambda Consider a universe which is spatially flat.1 universe.108 CHAPTER 6.16) for the case Ω0 < 1 can be written in parametric form as 1 Ω0 a(η) = (cosh η − 1) (6. Like all terse summaries of complex concepts. then an appropriate T-shirt slogan would be: DENSITY IS DESTINY! If the density of matter is less than the critical value. A plot of a versus t in the case Ω0 = 0.4 If. In this case.1.9 universe is very different from that of an Ω0 = 1. As shown in the lower panel of Figure (6. since it appears to be a close approximation to our own universe at the present day. Scientists sometimes joke that they are searching for a theory of the universe that is compact enough to fit on the front of a T-shirt. 4 .20) 2 1 − Ω0 and Ω0 1 (sinh η − η) .1). the density parameter Such a universe is of particular interest to us.9 universe start to diverge significantly only after a Hubble time or more. the required footnote is “∗ if Λ = 0”. the scale factors of the Ω0 = 1. then the destiny of the universe is an ever-expanding Big Chill. if the density is greater than the critical value. (6. the slogan “Density is Destiny!” requires a qualifying footnote. If the energy content of the universe is contributed almost entirely by non-relativistic matter. then the destiny is a recollapsing Big Crunch. matteronly universe.0 = 0 universe is shown.0 < 0.26) A plot of a versus t in the case Ωm. and negative if Ωm. In a universe with ΩΛ. ΩΛ.6. not the repulsive force of a positive cosmological constant. as the dotted line.0 = 1.2. ΩΛ.0 − 1 sin−1 (a/amax )3/2 . the shorter the lifetime of the universe. For a flat. and is always positive. the requirement that space be flat tells us that ΩΛ.0 .0 < 1. MATTER + LAMBDA 109 in matter is Ωm. reminiscent of that for a positively curved.0 − 1 1/3 .6.0 − 1 2π 3H0 1 Ωm. the negative cosmological constant provides an attractive force. A flat universe with ΩΛ.0 > 0 will continue to expand forever if it is expanding at t = t0 . The a ∝ t2/3 behavior of a Ωm. with a negative cosmological constant providing an .0 < 0. implying ΩΛ. The second term represents the contribution of a cosmological constant.0 H2 = 3 + (1 − Ωm.2. however.23) represents the contribution of matter.0 > 1.0 < 0 universe.0 < 0 will cease to expand at a maximum scale factor amax = Ωm. (6.0 . the Friedmann equation (eq.0 ) .23) can be integrated to yield the analytic solution H0 t = 2 3 Ωm.0 < 0 ends in a Big Crunch. for comparison.25) For a given value of H0 .0 = 1 − Ωm.0 . (6.0 = −0.0 = 1. implying ΩΛ. Thus. ΩΛ.23) 2 H0 a The first term on the right hand side of equation (6. (6. (6.22) and the Friedmann equation for the flat “matter plus lambda” universe reduces to Ωm.0 > 0. the larger the value of Ωm. A flat universe with ΩΛ.0 and the density parameter in a cosmological constant Λ is ΩΛ. it is positive if Ωm. (6.1.24) and will collapse back down to a = 0 at a cosmic time tcrunch = . this is another example of a Big Chill universe. a flat universe with ΩΛ. However.1 is shown as the solid line in Figure 6.0 Ωm. ΩΛ. .0=1. The dashed line is a(t) for a universe with Ωm. ΩΛ.0 = 1.9 Ωm.1.2: The scale factor as a function of time for flat universes containing both matter and a cosmological constant.1. ΩΛ.9.1.110 CHAPTER 6. MULTIPLE-COMPONENT UNIVERSES 40 30 20 10 0 Ωm.1 0 20 40 60 H0(t−t0) 80 100 Figure 6.0 = −0.0 = 0.0 = 1.0=0.0 = 0. The dotted line is a(t) for a universe with Ωm.0 = 0. The solid line is a(t) for a universe with Ωm.0 a Ωm.0=1. the negative cosmological constant causes a rapid deceleration of the universe’s expansion. if you measure H0 and Ωm.31) . At late times.0 > 0 universe.0 ΩΛ.0 ln (a/amΛ )3/2 + 1 + (a/amΛ )3 . MATTER + LAMBDA 111 attractive force. In a flat universe with Ωm. As soon as the universe becomes lambdadominated. (6.0 > 0.28) reduces to a(t) ≈ amΛ exp( 1 − Ωm.28) A plot of a versus t in the case Ωm.0 1/3 .1 is shown as the dashed line in Figure 6. (6.1 −1 has a lifetime tcrunch ≈ 7H0 .0 = 0. ΩΛ.0   .0 + 1 Ωm. At early times.29) a(t) ≈ 2 giving the a ∝ t2/3 dependence required for a flat.0 = 0.0 1/3 = Ωm. when a amΛ . Suppose you are in a flat universe containing nothing but matter and a cosmological constant. lambda-dominated universe. then equation (6. Although a negative cosmological constant is permitted by the laws of physics.0 1 − Ωm.30) giving the a ∝ eKt dependence required for a flat.0 = 1. the density contributions of matter and the cosmological constant are equal at the scale factor (eq.6. the Friedmann equation can be integrated to yield the analytic solution H0 t = 2 3 1 − Ωm.0 . we have seen that a positively curved universe with −1 Ωm. the lifetime of a flat universe with ΩΛ.27) For a flat. (6. matter-dominated universe. For instance.0 < 0 is exceptionally short. equation (6.0 H0 t .2.0 = 1.0 H0 t) . (6.28) tells you that the age of the universe is t0 = −1 2H0 3 1 − Ωm.28) reduces to the relation 2/3 3 Ωm. a flat universe with Ωm.2. ΩΛ.0 ln   1 − Ωm.9.0 = −0.1).1 undergoes a Big Crunch after a lifetime tcrunch ≈ 110H0 (see Figure 6. However. it appears that we live in a universe where the cosmological constant is non-negative. when a amΛ .0 < 1 and ΩΛ. 5. equation (6.22): amΛ = Ωm.1 and ΩΛ. (6. if Ωm.0 < 0.0 .3 Matter + curvature + lambda If a universe contains both matter and a cosmological constant. we encountered Einstein’s static model.34) are positive. a flat universe with Ωm. but has infinite duration in time. we can create model universes with scale factors a(t) which exhibit very interesting behavior.33) Thus.0 − ΩΛ.0 = 0.8 ± 1.0 ≈ 0. MULTIPLE-COMPONENT UNIVERSES If we approximate our own universe as having Ωm. but has a finite duration in time.4.0 + ΩΛ.0 > 1. By contrast.0 ≥ 0 extends to infinity both in space and in time.0 ln[1 + √ −1 2] = 0. (6.0 > 0. then the formula “density = destiny = curvature” no longer holds.0 . A flat universe with Ωm.32) assuming H0 = 70 ± 7 km s−1 Mpc−1 . so that the .0 + ΩΛ.3 and ΩΛ.0 = 0.0 > 0 and ΩΛ. without constraining the universe to be flat.0 ≤ 1 and ΩΛ. = 3 + 2 H0 a a2 (6. 6.5 that ignoring the radiation content of the universe has an insignificant effect on our estimate of t0 . (6.) The age at which matter and the cosmological constant had equal energy density was tmΛ = −1 2H0 3 1 − Ωm. For instance.964H0 = 13. A universe described by Einstein’s static model is finite in spatial extent. in which κ = +1 and εΛ = εm /2. is infinite in spatial extent.0 Gyr .0 1 − Ωm. (We’ll see in section 6.7 (ignoring the contribution of radiation) we find that its current age is −1 t0 = 0.3 Gyr .3 and ΩΛ. then a wide range of behaviors is possible for the function a(t). with Ωm. if our universe is well fit by the Benchmark Model.0 = 0. then the cosmological constant has been the dominant component of the universe for the last four billion years or so.7. However. If a universe containing both matter and lambda is curved (κ = 0) rather than flat. then both the first and last term on the right hand side of equation (6.702H0 = 9.112 CHAPTER 6.34) If Ωm. Start by writing down the Friedmann equation for a curved universe with both matter and a cosmological constant: H2 Ωm. By choosing different values of Ωm. as shown in the previous section.0 > 1 and ΩΛ.0 and ΩΛ.5 ± 1. in section 4. reaches a minimum scale factor a = amin > 0 at some time tbounce .0 and ΩΛ. The closer such a universe lies to the Big Bounce – Big Chill dividing line in Figure 6. the longer its loitering stage lasts. or flat. the negative curvature term in equation 6. the cosmological constant takes over. then the central term on the right hand side is negative.34 becomes dominant. Each of these universes has the same current density parameter for matter: Ωm. Note that Big Crunch universes can be positively curved. then expands outward forever. As a result. In the region labeled “Big Chill”. Big Chill universes can have any sign for their curvature. with a → ∞ as t → ∞. To illustrate the different types of expansion and contraction possible. ıtre . however. Figure 6. In the region labeled “Big Crunch”.3. then expands outward forever. Universes which fall just below the dividing line between Big Bounce universes and Big Chill universes are loitering universes. and the universe starts to expand exponentially. with a → ∞ as t → ∞. Thus. measured at 5 A loitering universe is sometimes referred to as a Lemaˆ universe. the universe starts with a = 0 at t = 0. MATTER + CURVATURE + LAMBDA 113 universe is positively curved. expanding outward with a ∝ t2/3 .0 and ΩΛ. In the region labeled “Big Bounce”. the universe starts in a contracting state. if the values of Ω m. it is possible to have a universe which expands outward at late times. negatively curved. Like Big Crunch universes. this means that these universes have a forbidden range of scale factors. Figure 6. Λ-dominated state.0 = 0. the universe starts with a = 0 at t = 0.0 . with a = 0 at t = 0. then recollapses to a = 0 at a finite time t = tcrunch .5 Such a universe starts in a matter-dominated state. After the loitering stage.6. Since negative values of H 2 are unphysical.0 are chosen just right.3. Another possibility. and then expand outward again in a “Big Bounce”. Then.3. Suppose such a universe starts out with a 1 and H < 0. it is contracting from a low-density. it enters a stage (called the loitering stage) in which a is very nearly constant for a long period of time. causing the contraction to stop at a minimum scale factor a = amin . but which never had an initial Big Bang. however.0 . As the universe contracts. is a “loitering” universe. that is. the value of H 2 will be positive for small values of a (where matter dominates) and for large values of a (where Λ dominates).0 and ΩΛ.4 shows a(t) for a set of four model universes. During this time it is almost – but not quite – Einstein’s static universe. but will be negative for intermediate values of a (where the curvature term dominates). for some choices of Ωm.3 shows the general behavior of the scale factor a(t) as a function of Ωm. reaches a maximum scale factor amax . or a “Big Bounce” (a = amin > 0 at t = tbounce ).0 1 Big Chill 0 Big Crunch κ=−1 κ=+1 −1 0 . MULTIPLE-COMPONENT UNIVERSES 3 Big Bounce loi te rin g 2 ΩΛ. a loitering phase (a ∼ const for an extended period). and those lying below have κ = −1.5 Ωm. . models lying above this line have κ = +1. a “Big Crunch” recollapse (a → 0 as t → tcrunch ). Also shown are the regions where the universe has a “Big Chill” expansion (a → ∞ as t → ∞).0 2 2.5 1 1.114 CHAPTER 6.5 Figure 6.3: The curvature and type of expansion for universes containing both matter and a cosmological constant. The dashed line indicates κ = 0. 7.0 = 0.3. The dot-dash line shows a loitering universe (ΩΛ. κ = +1).0 = −0.0=−0.8.3.7 2 ΩΛ. .3. κ = 0).0=0. κ = −1). MATTER + CURVATURE + LAMBDA 115 3 ΩΛ.6.0 = 0. κ = +1). The solid line shows a “Big Bounce” universe (ΩΛ.0=1.8 ΩΛ. The dashed line shows a “Big Crunch” universe (ΩΛ.0 = 1.3 a 1 ΩΛ.7134 0 −4 −2 0 H0(t−t0) 2 4 Figure 6.0=1.4: The scale factor a as a function of t in four different universes.0 = 1. The dotted line shows a “Big Chill” universe (ΩΛ.7134. each with Ωm. If we lived in a loitering universe. 6. Finally.8.0 H2 = 4 + 3 . MULTIPLE-COMPONENT UNIVERSES t = t0 and a = 1. The dashed line in Figure 6.0 = 1.8 × 10−4 .0 /Ωm. This universe lies above the Big Chill – Big Bounce dividing line in Figure 6.7134. this universe has negative curvature.0 = 0.5.0 = −0. radiation-only model. and is destined to end in a Big Crunch. thanks to their different values for the cosmological constant.0 Ωm. then see the redshifts decrease until they actually became blueshifts. These universes cannot be distinguished from each other by measuring their current matter density and Hubble constant. the universe is better described by a flat model containing both radiation and matter.4 Radiation + matter In our universe. fated to eternal expansion.3).44 (the appropriate loitering scale factor for a universe with Ωm. which spends a long time with a scale factor a ≈ aloiter ≈ 0. the solid line shows a universe with ΩΛ. At scale factors a ∼ arm . they have very different pasts and very different futures. Nevertheless.44. this would lead to a large excess of galaxies with zloiter ≈ 1. 2 H0 a a (6. Our own universe seems to be a Big Chill universe.0 = 1.3. then the largest redshift we would see for any galaxy would be zmax = 1/abounce − 1. and is destined to end in an exponentially expanding Big Chill . The dot-dash line shows the scale factor for a universe with ΩΛ. As we looked further into space.56. it is a positively curved universe which “bounced” at a scale factor a = abounce ≈ 0. this is a positively curved loitering universe.35) .3. we would see nearly the same redshift zloiter = 1/aloiter − 1 for galaxies with a very large range of distances. with aloiter ≈ 0. this universe is spatially flat.3.7. we do not see such distant blueshifted galaxies. radiation-matter equality took place at a scale factor arm ≡ Ωr. we would see redshifts increase to zmax . The dotted line shows a(t) for a universe with ΩΛ. then as we looked out into space. No such excess of galaxies is seen at any redshift in our universe. as described in Section 5.0 = 0.0 ≈ 2.4 shows the scale factor for a universe with ΩΛ. In our universe. There is strong observational evidence that we do not live in a loitering or Big Bounce universe. At scale factors a arm . If we lived in a Big Bounce universe.116 CHAPTER 6. For instance. the universe is well described by a flat. The Friedmann equation around the time of radiation-matter equality can be written in the approximate form Ωr. 000 yr .39) The time of radiation-matter equality.0 −1 1 a2 4 −1 = (1 − √ ) rm H0 ≈ 0.41) The epoch when the universe was radiation-dominated was only about 47 millennia long. The age t0 = −1 0.0 1− 1− a 2arm 1+ a arm 1/2 .0 H0 t 1/2 [a arm ] . (6.0 = 0.3. . This is sufficiently brief that it justifies our ignoring the effects of radiation when computing the age of the universe. (6. a ≈ 2 Ωr.2 (ignoring radiation) would only be altered by a few parts per million if we included the effects of radiation.0 117 [1 + a/arm ]−1/2 . and H0 = 14 Gyr. the approximate result for a(t) becomes a≈ 3 Ωm. 3 Ωm.0 = 8.37): trm Ωr.0 3/2 (6.10). trm . can be found by setting a = arm in equation (6.6.0 2 Ωr. (6.5 Gyr that we computed in Section 6. (6.37) In the limit a arm .4.0 H0 t 2 2/3 [a arm ] .4 × 10−5 . Ωm. the time of radiation-matter equality was −1 trm = 3. RADIATION + MATTER This can be rearranged in the form H0 dt = ada 1/2 Ωr.964H0 = 13.38) In the limit a arm (but before curvature or Λ contributes significantly to the Friedmann equation). this gives the appropriate result for the radiationdominated phase of evolution (compare equation 6.34 × 10−6 H0 = 47. This minor correction is dwarfed by the 10% uncertainty in the −1 value of H0 . (6.391 2 H0 . with Ωr.40) −1 For the Benchmark Model.36) Integration yields a fairly simple relation for t as a function of a during the epoch when only radiation and matter are significant: H0 t = 4a2 rm 3 Ωr. The Hubble constant of the Benchmark Model is assumed to be H0 = 70 km s−1 Mpc−1 .04 today. Some of its properties are listed. in Table 6. The radiation in the Benchmark Model consists of photons and neutrinos. as long as neutrinos are relativistic. is spatially flat.0 × 10−5 .4 × 10−5 Ωbary. The photons are assumed to be provided solely by a Cosmic Microwave Background with current temperature T0 = 2.0 = 3. with associated electrons).0 = 0. and a cosmological constant. which I have adopted as the best fit to the currently available observational data.0 ≈ 0. The baryonic material that we are familiar with from our everyday existence has a density parameter of Ωbary.0 = 0.725 K and density parameter Ωγ.0 = 0.5 Gyr photons: neutrinos: total radiation: baryonic matter: nonbaryonic dark matter: total matter: cosmological constant: radiation-matter equality: matter-lambda equality: Now: 6.30 ΩΛ.118 CHAPTER 6.0 = 5. The matter content of the Benchmark Model consists partly of baryonic matter (that is. matter composed of protons and neutrons.8 × 10−4 amΛ = 0.26 Ωm. equation (5. As we’ll see in future chapters. MULTIPLE-COMPONENT UNIVERSES Table 6.18) tells us that it defects from the “radiation” column to the “matter” column when the scale factor is a ∼ 5 × 10−4 eV/(mν c2 ).2. The energy density of the cosmic neutrino background is theoretically calculated to be 68% of that of the Cosmic Microwave Background. The density parameter of the nonbaryonic dark matter .75 a0 = 1 trm = 4. and partly of nonbaryonic dark matter. and contains radiation.7 × 104 yr tmΛ = 9. If a neutrino has a non-zero mass mν .5 Benchmark Model The Benchmark Model. the evidence indicates that most of the matter in the universe is nonbaryonic dark matter.2: Properties of the Benchmark Model List of Ingredients Ωγ.0 = 8. for ready reference. matter.04 Ωdm.8 Gyr t0 = 13.0 ≈ 0.0 = 5.0 × 10−5 Ων.4 × 10−5 Ωr.70 Important Epochs arm = 2. 8 Gyr.5. corresponding to tmΛ = 9. One curious feature of the Benchmark Model .0 known. is roughly six times greater: Ωdm. is t0 = 13. the scale factor a(t) can be computed numerically using the Friedmann equation. with ΩΛ. for the Benchmark Model.0 . Ωm. The Benchmark Model was first radiation-dominated.0 = 2. is not provided by radiation or matter.0 ≈ 0. computed for the Benchmark Model. arm = 2. The bulk of the energy density in the Benchmark Model. The current age of the universe.5 shows the scale factor. Matter. thus computed. then matter-dominated. radiation gave way to matter at a scale factor arm = Ωr. neither is the later transition from the matter-dominated phase to the exponentially growing lambda-dominated phase.6. a0 = 1. in turn. the time of matter-lambda equality.0 . With Ωr. BENCHMARK MODEL 119 2 0 log(a) −2 −4 −6 −10 a∝t 1/2 a∝e Kt a∝t 2/3 tmΛ trm t0 −8 −6 −4 log(H0t) −2 0 Figure 6. and is now entering into its lambda-dominated phase. in the Benchmark Model.75.0 )1/3 = 0. Note that the transition from the a ∝ t1/2 radiation-dominated phase to the a ∝ t2/3 matter-dominated phase is not an abrupt one.0 − Ωr.0 = 1 − Ωm.6).5: The scale factor a as a function of time t (measured in units of the Hubble time).5 Gyr. in the form of equation (6.7 × 104 yr. but by a cosmological constant.26. and the present moment.8 × 10−4 . amΛ = 0.8 × 10−4 . gave way to the cosmological constant at amΛ = (Ωm. and ΩΛ.75. however.0 /ΩΛ.0 /Ωm. As we’ve seen.0 ≈ 0. Figure 6. corresponding to a time trm = 4. The dotted lines indicate the time of radiation-matter equality.70. lambda-only universe. other properties of the Benchmark Model can readily be computed. In the limit z → ∞. or any other model. dhor (t0 ) = 3. In a flat. How long does it take galaxies at z ≈ 2 to evolve into galaxies similar to those at z ≈ 0? Is . consider a galaxy with redshift z = 2. the lookback time to that galaxy is 10. at larger redshifts. assuming H0 = 14 Gyr. the lookback time to a z = 2 galaxy is a mere 7. as shown in Figure 6.41c/H0 . In the limits of very small redshifts.7. Thus. for purposes of comparison. questions. but not identical. then we can’t see objects more than 14 gigaparsecs away because light from them has not yet had time to reach us. the proper distance dp (t0 ) approaches a limiting value dp → 3. In a flat.5 Gyr. In the Benchmark Model. The heavy solid line is the result for the Benchmark Model. we are seeing a redshifted image of that galaxy as it was 10. We can answer the question “How long has the light from that galaxy been traveling?” by computing the lookback time.24c/H0 = 3.24c/H0 .4 Gyr. For example.120 CHAPTER 6. However. in the case of the Benchmark Model. MULTIPLE-COMPONENT UNIVERSES which Figure 6.6. the Benchmark Model has a finite horizon distance.000 Mpc . The exact dependence of lookback time on redshift depends strongly on the cosmological model used. the −1 lookback time to a z = 2 galaxy is 15. the result for a flat lambda-only universe is shown as a dot-dash line and the result for a flat matter-only universe is shown as the dotted line. Knowing Ωm. The lower panel of Figure 6.5 illustrates vividly is that we are living very close to the time of matter-lambda equality. the lookback time is simply t0 −te .6 shows the current proper distance to a galaxy with redshift z.5 Gyr. with the same assumed Hubble constant. the distance to a galaxy with observed redshift z at the time the observed photons were emitted.0 and ΩΛ.0 thus becomes important to studies of galaxy evolution.5 billion years ago. dp (te ) has a maximum for galaxies with redshift z = 1. For instance. we can answer the question “How far away is that galaxy?” by computing the proper distance dp (t0 ). Once a(t) is known. matter-dominated universe.12ct0 = 14. t0 −te ≈ z/H0 . For the Benchmark Model. the upper panel of Figure 6. the relation between lookback time and redshift becomes nonlinear. If light emitted at time te is observed at time t0 . however. “How far away is that galaxy?” and “How long has the light from that galaxy been traveling?” In the Benchmark Model. they ask the related. When astronomers observe a distant galaxy.42) If the Benchmark Model is a good description of our own universe.6 shows dp (te ). (6. where dp (te ) = 0. and the dotted line a flat.01 Observation Λ−only Benchmark matter−only .1 .01 .01 Emission . The bold solid line indicates the Benchmark Model.6. the lower panel shows the distance at the time of emission. matter-only universe.1 1 z 10 100 1000 Figure 6. lambda-only universe. The upper panel shows the distance at the time of observation.1 1 z 10 100 1000 100 (H0/c) dp(te) 10 1 . .6: The proper distance to a light source with observed redshift z. BENCHMARK MODEL 121 100 (H0/c) dp(t0) 10 1 .5.1 . the dot-dash line a flat.01 . you are looking back to objects with smaller and smaller values of te .” As you look further and further out into the universe. The heavy solid line shows the result for the Benchmark Model. t0 − te . MULTIPLE-COMPONENT UNIVERSES 15 t0−te (Gyr) 10 Λ−only Benchmark matter−only 5 0 0 2 z 4 6 Figure 6.122 CHAPTER 6. “A telescope is a time machine. Consider such a high-redshift galaxy.5 Gyr. it 15 billion years. The proper distance at the time the light was emitted was smaller by a multiplicative factor 1/(1 + z) = 1/7. This means that the galaxy was only at a distance dp (te ) = 0. Using the Benchmark Model. as the Benchmark Model becomes better constrained. Astronomers are fond of saying. matter-only universe. The light was emitted when the age of the universe −1 was te = 0. When you observe a galaxy . or less than 7% of the universe’s current age. The most distant galaxies that have been observed (at the beginning of the 21st century) are at a redshift z ≈ 6.7: The lookback time. for galaxies with observed redshift z.066H0 = 0. our ability to translate observed redshifts into deduced times will become more accurate. the dot-dash line for a flat. to objects with larger and larger values of dp (t0 ).27c/H0 = 1200 Mpc at the time the light was emitted.9 Gyr. lambda-only universe. about 60% of the current horizon distance.92c/H0 = 8300 Mpc.6 Gyr. t0 = 13. and the dotted line for a flat. or something in between? In future years. The lookback time to a z = 6 galaxy in the Benchmark Model is thus t0 − te = 12. or only half that time. −1 The Hubble time is assumed to be H0 = 14 Gyr. we find that the current proper distance to a galaxy with z = 6 is dp (t0 ) = 1. plot t0 as a function of Ω0 in the range 1 ≤ Ω0 ≤ 3. Linder (1997). ch. for different values of the equation-of-state parameter w. by kinematic and dynamic criteria. you are glimpsing the universe as it was as a youngster. The current best estimate for the ages of stars in globular clusters yields an age of t = 13 Gyr for the oldest globular clusters. less than a billion years old.2) In a negatively curved universe containing only matter (Ω0 < 1. 3.1) In a positively curved universe containing only matter (Ω0 > 1. Problems (6. given the constraints H0 = 70 km s−1 Mpc−1 and t0 > 13 Gyr? . In a matter-only universe.5: The fate of the universe. show that the present age of the universe is given by the formula H0 t 0 = 1 2 − Ω0 Ω0 cosh−1 − 3/2 1 − Ω0 2(1 − Ω0 ) Ω0 .43) Assuming H0 = 70 km s−1 Mpc−1 . what is the maximum permissible value of Ω0 . 4.44) Assuming H0 = 70 km s−1 Mpc−1 . ch. (6.6. − 3/2 2(Ω0 − 1) Ω0 Ω0 − 1 (6. show that the present age of the universe is given by the formula H0 t 0 = Ω0 2 − Ω0 1 cos−1 . contains useful formulae and graphs. Kolb & Turner (1990).] Harrison (2000). κ = −1). 18: A classification of possible universes.5. Suggested reading [Full references are given in the “Annotated Bibliography” on page 286. plot t0 as a function of Ω0 in the range 0 ≤ Ω0 ≤ 1.2: The scale factor a(t) and current age t0 for various two-component universes. according to the Benchmark Model. ch. (6. κ = +1). BENCHMARK MODEL 123 with a redshift z = 6. 124 CHAPTER 6. At some time t0 > tCrunch /2. for the purposes of this problem. Within what range must wQ fall for the effects of quintessence to be repulsive? Is it possible to create a static universe for an arbitrary value of wQ within this range? If so. What is a(t) in the limit a amQ ? What is a(t) in the limit a amQ ? What is the current age of this universe. and the current density parameter of quintessence is ΩQ. and quintessence with w = −1/2. He then measures H0 and Ω0 . At what scale factor amQ will the energy density of quintessence and matter be equal? Solve the Friedmann equation to find a(t) for this universe.3) One of the more recent speculations in cosmology is that the universe may contain a quantum field. and contains nothing but matter (w = 0). The current density parameter of matter is Ωm.1). how long a time will elapse between Dr. that the universe is spatially flat. Niwde is able to observe? What is the lookback time to an object with this blueshift? (6.0 = 1 − Ωm.0 . Given H0 and Ω0 . a = 0) in which the gravitational attraction of matter is exactly balanced by the gravitational repulsion of quintessence with equationof-state parameter wQ . and create a static universe ˙ ¨ (a = 0. for a given value of wQ .4) Suppose you wanted to “pull an Einstein”.6) Consider an expanding.0 ≤ 1. which has a positive energy density and a negative value of the equation-of-state parameter w.0 > 1). Niwde’s observations at t = t0 and the final Big Crunch at t = tcrunch ? What is the minimum blueshift that Dr. (6.5) Consider a positively curved universe containing only matter (the “Big Crunch” model discussed in Section 6.0 ? (6. will the curvature of the universe be negative or positive? What energy density εQ will be required. Show that such a universe underwent a “Big Bounce” at a scale factor abounce = Ω0 − 1 Ω0 1/2 . an astronomer named Elbbuh Niwde discovers that nearby galaxies have blueshifts (−1 ≤ z < 0) proportional to their distance.45) . to balance an energy density εm in matter? (6. finding H0 < 0 and Ω0 > 1. called “quintessence”. expressed in terms of H0 and Ωm. positively curved universe containing only a cosmological constant (Ω0 = ΩΛ. MULTIPLE-COMPONENT UNIVERSES (6. during the contraction phase of this universe. Assume. and contains both matter and a cosmolog−1 ical constant.8) In the Benchmark Model. BENCHMARK MODEL and that the scale factor as a function of time is a(t) = abounce cosh[ Ω0 H0 (t − tbounce )] . 125 (6.5.6. what is the total mass of all the matter within our horizon? What is the total energy of all the photons within our horizon? How many baryons are within the horizon? .46) where tbounce is the time at which the Big Bounce occurred. What is the time t0 − tbounce which has elapsed since the Big Bounce.0 is t0 exactly equal to H0 ? (6. For what value of Ωm. expressed as a function of H0 and Ω0 ? (6.7) A universe is spatially flat. Let’s see. I’ve pointed out that if we knew the energy density ε for each component of the universe. to do a Taylor series expansion for a(t) around the present moment. it is useful. the scale factor can be computed from the Friedmann equation. if we could determine a(t) from observations. is much more difficult. it can only be deduced indirectly from the imperfect and incomplete observations that we make of the universe around us. . we could use that knowledge to find ε for each component. The complete Taylor series is a(t) = a(t0 ) + da dt (t − t0 ) + 126 1 d2 a 2 dt2 (t − t0 )2 + . Finding a(t) for the real universe. then.1) t=t0 t=t0 . The argument works in the other direction. however. we could use the Friedmann equation to find the scale factor a(t). The scale factor is not directly observable. what constraints we can put on the scale factor by making observations of distant astronomical objects. 7. (7. . In the previous three chapters. as well. instead.1 “A search for two numbers” Since determining the exact functional form of a(t) is difficult.Chapter 7 Measuring Cosmological Parameters Cosmologists would like to know the scale factor a(t) for the universe. For a model universe whose contents are known with precision. 4) In equation (7. the ¨ relative velocity of any two points is decreasing).7. Keeping the first three terms of the Taylor expansion. the parameter H0 is our old acquaintance the Hubble constant. “A SEARCH FOR TWO NUMBERS” 127 To exactly reproduce an arbitrary function a(t) for all values of t. The choice of sign for q0 . a ˙ . (7. and there’s no evidence that the real universe has a wildly oscillating scale factor. meaning that the universe’s expansion is decelerating (that is. and the fact that it’s named . A negative value of q0 corresponds to a > 0. this expansion for the scale factor is customarily written in the form 1 2 a(t) ≈ 1 + H0 (t − t0 ) − q0 H0 (t − t0 )2 . However.3) t=t0 t=t0 Using the normalization a(t0 ) = 1. the scale factor in the recent past and the near future can be approximated as a(t) ≈ a(t0 ) + da dt (t − t0 ) + 1 d2 a 2 dt2 (t − t0 )2 . defined as q0 ≡ − aa ¨ a2 ˙ =− a ¨ aH 2 . an infinite number of terms is required in the expansion. (7. meaning that the relative velocity of any two points is ¨ increasing with time. a(t0 ). 2 (7. using only the first few terms of the expansion gives a good approximation in the immediate vicinity of t0 .1. the usefulness of a Taylor series expansion resides in the fact that if a doesn’t fluctuate wildly with t. A positive value of q0 corresponds to a < 0.2) t=t0 t=t0 Dividing by the current scale factor. The scale factor a(t) is a good candidate for a Taylor expansion. The different model universes examined in the previous two chapters all had smoothly varying scale factors. (7.4). t=t0 (7.6) t=t0 Note the choice of sign in defining q0 . a(t) a ˙ ≈1+ a(t0 ) a (t − t0 ) + ¨ 1 a 2 a (t − t0 )2 .5) H0 ≡ a t=t0 and the parameter q0 is a dimensionless number called the deceleration parameter. 8) However. If your model universe contains several components. w (7. MEASURING COSMOLOGICAL PARAMETERS the deceleration parameter. w (7.128 CHAPTER 7. w (7. Although the scope of cosmology has widened considerably since Sandage wrote his article. although these are reasonable assumptions. If the universe contains a sufficiently large ¨ cosmological constant. the Friedmann equation assumes that the expansion of the universe is controlled by gravity. and says nothing at all about what forces act to accelerate the expansion (to take a Newtonian viewpoint of the physics involved).9) Remember. It is simply a mathematical description of how the universe expands at times t ∼ t0 .4) is physics-free. they are not 100% ironclad.1 In a famous 1970 review article. Those two numbers were H0 and q0 . you can use the acceleration equation to predict what q0 will be in a given model universe. is due to the fact that it was first defined during the mid-1950’s.8) is just the inverse of the critical energy density εc . The Taylor expansion of equation (7. when the limited information available favored a matterdominated universe with a < 0. the acceleration equation can be written 4πG a ¨ =− 2 a 3c εw (1 + 3w) . the deceleration parameter q0 can have either sign. we can rewrite the acceleration equation in the form − 1 1 a ¨ = 2 aH 2 Ωw (1 + 3w) . the quantity in square brackets in equation (7. and are free of the theoretical “baggage” underlying the Friedmann equation and the acceleration equation. Although H0 and q0 are themselves free of the theoretical assumptions underlying the Friedmann and acceleration equations. . however. Thus. each with a different value of the equation-of-state parameter w.7) Divide each side of the acceleration equation by the square of the Hubble parameter H(t) and change sign: − a ¨ 1 8πG = 2 aH 2 3c2 H 2 εw (1 + 3w) . The parameters H0 and q0 are thus purely descriptive of the kinematics. and that gravity is accurately described by Einstein’s theory of general relativity. cosmologists are still assiduously searching for H0 and q0 . the observational cosmologist Allan Sandage described all of cosmology as “A Search for Two Numbers”. 0 + Ωm. (7. and a cosmological constant.2 Measuring the redshift of a galaxy is relatively simple. the The peculiar velocities of galaxies cause a significant amount of scatter in the plot. The distance to a galaxy is not only difficult to measure but also. determining H0 should be easy. In Section 3. The Benchmark Model. there is a helpful relation between scale factor and proper distance. the relation between a galaxy’s distance d and its redshift z is linear (equation 2. 2 . “A SEARCH FOR TWO NUMBERS” 129 Evaluating equation (7. w (7.0 .11) Such a universe will currently be accelerating outward (q0 < 0) if ΩΛ.3. Remember. 2 (7. This is because he underestimated the distances to galaxies in his sample by a factor of 7.4).0 + Ωm.9) at the present moment. t = t0 . automated galaxy surveys can find hundreds of galaxy redshifts in a single night.7.12) Thus. somewhat difficult to define.0 (1 + 3w) . the slope of the plot gives you the value of H0 . matter.0 /2. but by using a large number of galaxies. at time t0 . in an expanding universe.1. if you measure the distance d and redshift z for a large sample of galaxies. for instance. has q0 ≈ −0. 1 q0 = Ωr. the proper distance dp (t) between two points was defined as the length of the spatial geodesic between the points when the scale factor is fixed at the value a(t). The difficulty is in measuring the distance of a galaxy. and fit a straight line to a plot of cz versus d. tells us the relation between the deceleration parameter q0 and the density parameters of the different components of the universe: q0 = 1 2 Ωw. you can beat down the statistical errors. For small redshifts.0 > Ωr. In principle.10) For a universe containing radiation.55. Moreover.5): cz = H0 d .0 − ΩΛ. light that was emitted by a distant galaxy at time te . The proper distance is perhaps the most straightforward definition of the spatial distance between two points in an expanding universe. If you use galaxies at d < 100 Mpc. you must also make allowances for the local inhomogeneity and anisotropy. Edwin Hubble was off by a factor of 7 when he estimated H0 ≈ 500 km s−1 Mpc−1 (see Figure 2. If we observe. instead. The second term is a correction due to the expansion of the universe during the time the light was traveling. Equation 7. we find −1 (7. is dp (t0 ) ≈ c(t0 − te ) + z= 1 −1 . is what the proper distance would be in a static universe – the lookback time times the speed of light. we knew the exact functional form of a(t). in the form of the Taylor expansion of equation 7.19) .15 gives us an approximate relation for the current proper distance to a galaxy with redshift z: dp (t0 ) ≈ c cH0 z 2 c 1 + q0 z − (1 + q0 /2)z 2 + = z 1− z .35): dp (t0 ) = c t0 te dt . they carry a stamp telling us the scale factor a(te ) at the time the light was emitted.18) t0 − te ≈ H0 z − (1 + q0 /2)z 2 . t0 − te . c(t0 − te ).14) cH0 (t0 − te )2 . The observed redshift z of a galaxy.130 CHAPTER 7. we may use the expansion 1 2 ≈ 1 − H0 (t − t0 ) + (1 + q0 /2)H0 (t − t0 )2 a(t) (7.14. t0 − te .17 to give the lookback time as a function of redshift. They don’t.16) in equation 7. we find that the proper distance to the galaxy is Using equation 7. (7. remember. a(t) (7. (7. (7. and hence could exactly compute dp (t0 ) for a galaxy of any redshift. 2 H0 2 H0 H0 2 Inverting equation 7.17) Substituting equation 7. Including the two lowest-order terms in the lookback time.15 would be extremely useful if the light from distant galaxies carried a stamp telling us the lookback time. MEASURING COSMOLOGICAL PARAMETERS current proper distance to that galaxy is (equation 5.15) 2 The first term in the above equation. we may write an approximate relation between redshift and lookback time: 2 z ≈ H0 (t0 − te ) + (1 + q0 /2)H0 (t0 − te )2 .4. a(te ) (7.13.18 into equation 7. If we have only partial knowledge of the scale factor.13) For the model universes examined in Chapters 5 and 6. the length of the astronomical unit. alternatively. you would need a tape measure which could be extended with infinite speed. LUMINOSITY DISTANCE 131 The linear Hubble relation dp ∝ z thus only holds true in the limit z 2/(1 + q0 ). If you tried to measure the distance to a galaxy with a tape measure. Since cosmology is ultimately based on observations. the star’s apparent position will shift by an angle θ. Neither of these alternatives is physically possible. the reflected radio waves are too faint to detect.1. we need some way of computing a distance from that galaxy’s observed properties. To measure the proper distance at time t0 . 3 . 7. is not a measurable property. the corrections due to relative motion during the time δt are minuscule. the distance would be continuously increasing as you extended the tape. Beyond ∼ 10 AU. if we want to find the distance to a galaxy. is now known to be 1 AU = 149. then the distance to the reflecting body is d = (c δt)/2. A favorite method for determining distances to other stars within our Galaxy is the method of trigonometric parallax.2 Luminosity distance Unfortunately. astronomers have found it useful to adopt and adapt the techniques used to measure shorter distances. The radar technique is useful only within the Solar System. Within the Solar System. When a star is observed from two points separated by a distance b. astronomers measure the distance to the Moon and planets by reflecting radar signals from them. If the baseline of observation is perpendicular to the line Since the relative speeds of objects within the Solar System are much smaller than c. If q0 > −1.61 km. the techniques used to measure relatively short distances. If δt is the time taken for a photon to complete the round-trip. then. for instance.7. 597. say) is less than would be predicted from the linear Hubble relation. for instance.2. dp (t0 ). In devising ways of computing the distance to galaxies. 870.3 The accuracy with which distances have been determined with this technique is impressive. Let’s examine. then the proper distance to a galaxy of moderate redshift (z ∼ 0. the current proper distance to a galaxy. you would need to stop the expansion of the universe at its current scale factor while you measured the distance at your leisure. if some class of astronomical object had luminosities which were the same throughout all of space-time. we can measure the redshift. If you know. z. The Hipparcos satellite. f . with an accuracy of ∼ 1 milliarcsecond.20) Measuring the distances to stars using the Earth’s orbit (b = 2 AU) as a baseline is a standard technique. However. One way of using measured properties to assign a distance is the standard candle method. The trigonometric parallaxes of galaxies at cosmological distances are too small to be measured with current technology. who used it to measure solar radiation. an accuracy of < 10 nanoarcseconds would be required. the accuracy with which the parallax distance can be determined is limited by the accuracy with which θ can be measured.” . we can measure its angular diameter. from the object. (7. A standard candle is an object whose luminosity L is known. The complete flux.132 CHAPTER 7. Let’s focus on the properties that we can measure for objects at cosmological distances. the flux over a limited range of wavelengths is measured. For instance. the luminosity of an object. We can measure the flux of light. If the light from the object has emission or absorption lines. then you can use its measured flux f to define a function called the luminosity distance: dL ≡ 4 L 4πf 1/2 . found the parallax distance for ∼ 105 stars. it was invented in 1881 by the astronomer Samuel Langley. they would act as excellent standard candles – if their unique luminosity L were known. is called the bolometric flux. Since the size of the Earth’s orbit is known with great accuracy from radar measurements. Langley devised the bolometer: / It’s really a kind of thermometer / Which measures the heat / From a polar bear’s feet / At a distance of half a kilometer. launched by the European Space Agency in 1989. in units of watts per square meter. given the difficulties of measuring the true bolometric flux. using the Earth’s orbit as a baseline. integrated over all wavelengths of light. the parallax distance will be b dπ = 1 pc 1 AU θ 1 arcsec −1 .4 ) More frequently. (7. (A bolometer is an extremely sensitive thermometer capable of detecting electromagnetic radiation over a wide range of wavelengths. If the object is an extended source rather than a point of light. δθ. to measure θ for a galaxy ∼ 100 Mpc away. by some means or other.21) As expressed more poetically in an anonymous limerick: “Oh. MEASURING COSMOLOGICAL PARAMETERS of sight to the star. θ. The photons which were emitted at time te are.2. and because it is what the proper distance to the standard candle would be if the universe were static and Euclidean. of known luminosity L. φ) at a time te (see Figure 7. though. θ. then the proper area . at comoving coordinate location (r. If space is flat (κ = 0). LUMINOSITY DISTANCE 133 Figure 7.1: An observer at the origin observes a standard candle.23) You are at the origin.25): ds2 = −c2 dt2 + a(t)2 [dr2 + Sκ (r)2 dΩ2 ] . At the present moment. Suppose.1). the propagation of light follows the inverse square law f = L/[4πd 2 ].22) (7. you see light that was emitted by a standard candle at comoving coordinate location (r. In a static Euclidean universe.7. The function dL is called a “distance” because its dimensionality is that of a distance. at the present moment. that you are in a universe described by a RobertsonWalker metric (equation 3. φ).   R0 sin(r/R0 )  (7. t = t0 . spread over a sphere of proper radius dp (t0 ) = r and proper surface area Ap (t0 ). with (κ = +1) (κ = 0) Sκ (r) =  r  R0 sinh(r/R0 ) (κ = −1) . (7. thanks to the expansion of the universe. MEASURING COSMOLOGICAL PARAMETERS of the sphere is given by the Euclidean relation Ap (t0 ) = 4πdp (t0 )2 = 4πr2 . More generally. by the time we detect the photons at time t0 . Ap (t0 ) > 4πr2 .26) Second. 1+z (7. the expansion of the universe causes the energy of each photon from the standard candle to decrease.27) 4πSκ (r)2 (1 + z)2 and the luminosity distance is dL = Sκ (r)(1 + z) . If two photons are emitted in the same direction separated by a time interval δte . If a photon starts with an energy Ee = hc/λe when the scale factor is a(te ). Ap (t0 ) = 4πSκ (r)2 . the relation between the observed flux f and the luminosity L of a distant light source is L f= . the expansion of the universe causes the observed flux of light from a standard candle of redshift z to be decreased by a factor of (1 + z)−2 . and the photons are spread over a smaller area than they would be in flat space. the time between photon detections will be greater. which would apply even in a static universe. (7. (7. the wavelength will have grown to λ0 = 1 λe = (1 + z)λe . In addition to these geometric effects. The net result is that in an expanding. and photons are spread over a larger area than they would be in flat space. First. a(te ) (7.134 CHAPTER 7. however.28) .25) and the energy will have fallen to E0 = Ee . the proper distance between them will initially be c(δte ). and we will detect them separated by a time interval δt0 = δte (1 + z). spatially curved universe. the proper distance between them will be stretched to c(δte )(1 + z). Ap (t0 ) < 4πr2 . when the scale factor is a(t0 ) = 1. When space is negatively curved. by the time we observe it.24) When space is positively curved. the . even if space is perfectly flat. When z 1. implying Sκ (r) ≈ r. matter-only universe. with a radius of curvature R0 which is larger than the current horizon distance dhor (t0 ). Objects with finite redshift are at proper distances smaller than the horizon distance. the dot-dash line for a flat.2.7. The available evidence indicates that our universe is nearly flat. LUMINOSITY DISTANCE 135 15 (H0/c) dL 10 5 0 Λ−only Benchmark matter−only 0 2 z 4 6 Figure 7. Thus. The bold solid line gives the result for the Benchmark Model.2: The luminosity distance of a standard candle with observed redshift z. and for two other flat universes. (7. and the dotted line for a flat.29) Thus. Figure 7. With our assumption that space is very close to being flat. one dominated by matter and one dominated by a cosmological constant.2 shows the luminosity distance dL as a function of redshift for the Benchmark Model. it is safe to make the approximation r R0 . the relation between the luminosity distance and the current proper distance becomes very simple: dL = r(1 + z) = dp (t0 )(1 + z) [κ = 0] . you will overestimate the actual ıve proper distance by a factor (1 + z). if you estimate the distance to a standard candle by using a na¨ inverse square law. lambda-only universe. and hence smaller than the radius of curvature. where z is the standard candle’s redshift. the luminosity distance is a good approximation to the current proper distance for objects with small redshifts. and hence is not expanding along with the universe as a whole. 7. by gravity or duct tape or some other influence. (7.33) This function of and δθ is called the angular-diameter distance. dp (t0 ) ≈ dL ≈ (7. you observed a standard yardstick. MEASURING COSMOLOGICAL PARAMETERS current proper distance may be approximated as dP (t0 ) ≈ 1 + q0 c z 1− z H0 2 . as shown in Figure 7. A standard yardstick is an object whose proper length is known. If δθ 1. it is convenient to chose as your yardstick an object which is tightly bound together. (7. you can compute a distance to the yardstick using the small-angle formula dA ≡ δθ .3 Angular-diameter distance The luminosity distance dL is not the only distance measure that can be computed using the observable properties of cosmological objects. and if you know the length of the yardstick. In most cases.32) In a universe described by the Robertson-Walker metric.30) In a universe which is nearly flat. The angular-diameter distance is equal to the proper distance to the yardstick if the universe is static and Euclidean. the luminosity distance may thus be approximated as dL ≈ c c 1 + q0 1 − q0 z (1 + z) ≈ z z 1− z 1+ H0 2 H0 2 c z.136 CHAPTER 7. Suppose that instead of a standard candle. H0 . and a redshift z for the light which the yardstick emits.3. .31) Note that in the limit z → 0. (7. Suppose a yardstick of constant proper length is aligned perpendicular to your line of sight. You measure an angular distance δθ between the ends of the yardstick. (7.7. θ1 . were (r. at comoving coordinate distance r. Suppose you are in a universe described by the Robertson-Walker metric given in equation (7. and thus find that Sκ (r)δθ . The distance ds between the two ends of the yardstick. if the universe is expanding or curved. ANGULAR-DIAMETER DISTANCE 137 Figure 7. the yardstick emitted the light which you observe at time t0 . can be found from the Robertson-Walker metric: ds = a(te )Sκ (r)δθ . for a standard yardstick whose length is known.3. At a time te .3: An observer at the origin observes a standard yardstick. φ).22). In general. measured at the time te when the light was emitted. the angulardiameter distance will not be equal to the current proper distance. θ2 . The comoving coordinates of the two ends of the yardstick. (7. Choose your comoving coordinate system so that you are at the origin. we can set ds = . at the time the light was emitted.35) = a(te )Sκ (r)δθ = 1+z . The yardstick is at a comoving coordinate distance r. As the light from the yardstick moves toward the origin.34) However. of known proper length . the angular size which you measure for the yardstick will be δθ = θ2 − θ1 . φ) and (r. Thus. though. it travels along geodesics with θ = constant and φ = constant. (7. we find that in the limit z → 0. equal to dp (te ). dp (t0 ) → dhor (t0 ) as z → ∞.28) shows that the relation between the angulardiameter distance and the luminosity distance is dA = dL . the state of affairs is very different in the limit z → ∞.) When z 1. Figure 7. with dhor (t0 ) dA (z → ∞) ≈ . the angular-diameter distance to highly redshifted objects approaches zero as z → ∞. MEASURING COSMOLOGICAL PARAMETERS Thus.39).4 is simply a replotting of of the lower panel in Figure 6. the angular-diameter distance dA to a standard yardstick is dA ≡ δθ = Sκ (r) .37) Thus. for these flat universes. and for two other spatially flat universes. However. 1+z (7. comparing equations (7. it is equal to the proper distance at the time the light from the object was emitted: dA = dp (t0 )/(1 + z) = dp (te ). The luminosity distance to highly redshifted objects.40) However. rather. the approximate value of dA is given by the expansion 3 + q0 c z . (7. one dominated by matter and one dominated by a cosmological constant. and (7.31).6.39) dA ≈ z 1− H0 2 Thus.30). diverges as z → ∞. therefore. (7.38) 1+z In a flat universe.41) z . dL [κ = 0] . (7. if the universe is spatially flat. Figure 7. the angular-diameter distance which you compute for the object will be smaller than the luminosity distance. in this case. if you compute the angular-diameter distance dA of a standard yardstick. Moreover. with dA (1 + z) = dp (t0 ) = dL (z → ∞) ≈ zdhor (t0 ) . it isn’t equal to the current proper distance dp (t0 ). (1 + z)2 (7. (Since dA is.4 shows the angular-diameter distance dA for the Benchmark Model.138 CHAPTER 7. (7. dA ≈ dL ≈ dp (t0 ) ≈ (c/H0 )z. In models with a finite horizon size. if you observe an object which is both a standard candle and a standard yardstick.36) Comparison with equation (7. 8 (H0/c) dA . and the dotted line for a flat.7. the dot-dash line for a flat.6 . The bold solid line gives the result for the Benchmark Model.2 0 0 2 Benchmark matter−only Λ−only z 4 6 Figure 7.3.4: The angular-diameter distance for a standard yardstick with observed redshift z. ANGULAR-DIAMETER DISTANCE 139 1 . lambda-only universe. . matter-only universe.4 . Moreover. will give you H0 . and the slope of the relation. For instance.41c/H0 = 1800 Mpc. Unfortunately for cosmologists. their angular size δθ would decrease with redshift out to z = zc .42) Both galaxies and clusters of galaxies are large enough to be useful standard candles. in the limit z → 0. δθ(min) = dA (max) = 1800 Mpc ≈ 0. Then. all of the same size . standard yardsticks can be used to determine H0 .39) to the data.) Correcting for these evolutionary trends is a difficult task. measure the redshift z and angular size δθ of each standard yardstick.140 CHAPTER 7. the angular-diameter distance dA has a maximum for standard yardsticks at some critical redshift zc . (For the Benchmark Model. redshifted yardsticks. tend to become larger with time. faint. A yardstick of physical size will have its angular size δθ minimized when it is at the critical redshift zc . you can estimate q0 by fitting equation (7.0 provide the best fit to the observed data. our Local Group will fall into the Virgo cluster. and a corresponding physical size . galaxies and clusters of galaxies do not have sharply defined edges.1 arcsec 1 kpc . In principle. zc = 1. (Eventually. but then would increase at larger redshifts. . the use of standard yardsticks to determine cosmological parameters has long been plagued with observational difficulties.6.) This means that if the universe were full of glow-in-the-dark yardsticks. to these objects is a somewhat tricky task. Clusters. To begin with. If you simply want a kinematic description. identify a population of standard yardsticks (objects whose physical size is known). For the Benchmark Model. a standard yardstick must have an angular size large enough to be resolved by your telescope. as galaxies fall into them. Compute the angular-diameter distance dA = /δθ for each standard yardstick. Galaxies tend to become larger with time as they undergo mergers with their neighbors. Given the difficulties involved in using standard yardsticks to determine cosmological parameters. Plot cz versus dA . galaxies and clusters of galaxies are not isolated. if you have measured the angular size δθ for standard candles at z ∼ zc . The sky would be full of big. more attention has been focused. MEASURING COSMOLOGICAL PARAMETERS In model universes other than the lambda-only model. you can see which values of Ωm. the shape of the cz versus dA plot can be used to determine further cosmological parameters. too. so assigning a particular angular size δθ. where dA (max) = 0. rigid yardsticks of fixed length. In practice.0 and ΩΛ. (7. If you are confident that the universe is dominated by matter and a cosmological constant. In addition. in recent years. 000 L . • Compute dL = (L/4πf )1/2 for each standard candle. as they are known. STANDARD CANDLES & H0 141 on the use of standard candles. lie in the range P = 1. The pulsational periods. Cepheids. • Measure the redshift z and flux f for each standard candle. therefore. their luminosity varies in response.5 One time-honored variety of standard candle is the class of Cepheid variable stars. partially due to the change in their surface area.5 → 60 days. Cepheids are pulsationally unstable. The recipe for finding the Hubble constant is a simple one: • Identify a population of standard candles with luminosity L.” the hardest step is the first one.4. It should also have a luminosity which is well determined. ¯ with mean luminosities lying in the range L = 400 → 40. A good standard candle is hard to find.4 Standard candles & the Hubble constant Using standard candles to determine the Hubble constant has a long and honorable history. is the saga of Edwin Hubble. Let’s first look. then focus on how they can be used to determine the acceleration of the universe. a standard candle should be bright enough to be detected at large redshifts. As they pulsate radially. In the 1929 paper which first demonstrated that dL ∝ z when z 1. this gives As with the apocryphal recipe for rabbit stew which begins “First catch your rabbit. A useful cautionary tale. it’s the method used by Hubble himself. • Measure the slope of the cz versus dL relation when z H0 . at how standard candles can be used to determine H0 . Hubble underestimated the luminosity distances to galaxies by a factor of ∼ 7 because he underestimated the luminosity of his standard candles by a factor of ∼ 49. • Plot cz versus dL . as reflected in the observed brightness variations of the star. 1. are highly luminous supergiant stars. and partially due to the changes in the surface temperature as the star pulsates. 7.7. in this regard. For cosmological purposes. 5 . 43) Thus. she was justified in assuming that the difference in mean flux for the Cepheids was due to differences in their mean luminosity. they would measure f and P for as many Cepheids as possible in each galaxy. For each Cepheid in the Small Magellanic Cloud (SMC). with stars having the longest period of variability also having the largest flux. Since the depth of the SMC. astronomers would not rely on a single Cepheid in ¯ each galaxy. Rather. you determine that ¯ fLMC = 230 . Suppose.44) Note that if you only know the relative fluxes of the two Cepheids. then Cepheids can act as a standard candle. averaged over one complete period. and found the mean flux f .000 L ) ten times farther away? The key to calibrating Cepheids was discovered by Henrietta Leavitt. she measured the period P by finding the time between maxima in the observed ¯ brightness. their mean luminosities range over two orders of magnitude. so you assume. front to back. you conclude that the luminosity distance to M31 is greater than that to the LMC6 by a factor dL (M31) = dL (LMC) ¯ fLMC ¯ fM31 1/2 = √ 230 = 15. ¯ She noted that there was a clear relation between P and f . for instance. you find a Cepheid star in the Large Magellanic Cloud (LMC) and another in M31. in all galaxies. Cepheids don’t seem sufficiently standardized to be standard candles. that they have ¯ the same mean luminosity L. Leavitt had discovered a period – luminosity relation for Cepheid variable stars. you will only know the relative distances of M31 and the In practice. (7. By careful measurement. not differences in their luminosity distance. and the inevitable error in measuring fluxes. ¯ fM31 (7. In the years prior to World War I. and not ¯ their luminosity L. a pair of relatively small satellite galaxies orbiting our own Galaxy. at Harvard College Observatory. They both have a pulsational period of 10 days.2 . given the intrinsic scatter in the period – luminosity relation. from the period – luminosity relation. then find the ratio of luminosity distances that would make the period – luminosity relations for the two galaxies coincide. If the same period – luminosity relation holds true for all Cepheids.142 CHAPTER 7. 6 . How can you tell whether you are looking at an intrinsically faint Cepheid (L ≈ 400 L ) or at an intrinsically bright Cepheid (L ≈ 40. is small compared to its distance from us. MEASURING COSMOLOGICAL PARAMETERS On the face of it. Leavitt was studying variable stars in the Large and Small Magellanic Clouds. 5. the parallax distance. only the very nearest Cepheids in our Galaxy have had their distances measured with even modest accuracy by the Hipparcos satellite. To fix an absolute distance to M31. One of the motivating reasons for building the Hubble Space Telescope in the first place was to use Cepheids to determine H0 . The nearest Cepheid is Polaris. If. and to other galaxies ¯ containing Cepheids. and the distances they yield. causing it to have a peculiar motion in that direction. Future space-based astrometric observatories. 7 . The net result of the Hubble Key Project to measure H0 is displayed in Figure 2. is given by van den Bergh (2000). you need to know the luminosity L for a Cepheid of a given period P . for instance. With the Hubble Space Telescope. There is a hidden difficulty involved in using Cepheid stars to determine H0 . the Local Group is gravitationally attracted toward the Virgo cluster. In fact. has yielded a distance dL (Virgo) = 300 dL (LMC) = 15 Mpc.4. scheduled for launch in 2009.7 π Unfortunately. will allow parallax distances to be measured with an accuracy greater than that provided by Hipparcos. 8 A good review of these methods. and the proper distance are identical. The next nearest is probably δ Cephei (the prototype after which all Cepheids are named). which is not expanding. that the recession velocity cz which we measure for the Virgo cluster Within our Galaxy. the fluxes and periods of Cepheids can be accurately measured out to luminosity distances of dL ∼ 20 Mpc. on this scale. as it turns out. you could measure the parallax distance dπ to ¯ a Cepheid within our own Galaxy. implying a distance to M31 of dL = 760 ± 50 kpc. Cepheids can take you out only to a distance dL ∼ 20 Mpc. The current consensus is that the Large Magellanic Cloud has a luminosity distance dL = 50 ± 3 kpc. Until the distances (and hence the luminosities) of nearby Cepheids are known with this great accuracy. the universe cannot be assumed to be homogeneous and isotropic. and use it to normalize the period – luminosity relation for Cepheids. at d π = 130±10 pc. Cepheids are rare stars. from dynamical models. at dπ = 300±50 pc. It is estimated. showing that the Cepheid data are best fit with a Hubble constant of H0 = 75 ± 8 km s−1 Mpc−1 .8 then using this distance to compute the mean luminosity of the LMC Cepheids. for instance. Observation of Cepheid stars in the Virgo cluster of galaxies. the luminosity distance. to the LMC. you could then compute its luminosity L = ¯ 4πd2 f .7. astronomers must still rely on alternate methods of normalizing the period – luminosity relation for Cepheids. such as the Space Interferometry Mission (SIM). The most usual method involves finding the distance to the Large Magellanic Cloud by secondary methods. STANDARD CANDLES & H0 143 LMC. 02. This attempt foundered on the lack of standardization among galaxies.9 For a standard candle to be seen at dL > 100 Mpc (to determine H0 with minimal effects from peculiar velocity) or at dL > 1000 Mpc (to determine q0 ). by contrast.45) At a redshift z = 0. the luminosity distance dL in the Benchmark Model (with q0 = −0. For instance. 9 . for instance. For any particular galaxy.144 CHAPTER 7.5 uses recession velocities which are corrected for this “Virgocentric flow”. To determine the value of q0 .2.55) is 5% larger than dL in an empty universe (with q0 = 0). as it “cannibalizes” smaller galaxies by merging with them. A galaxy in a rich cluster. we need to determine the luminosity distance to standard candles with dL > 100 Mpc. then you won’t have to go as deep into space to determine q0 accurately. as its stars exhaust their nuclear fuel and become dim stellar remnants. The plot of cz versus dL given in Figure 2. it’s difficult to tell which effect dominates. but any individual galaxy has a luminosity which evolves significantly with time. the luminosity distance at small redshift is dL ≈ c 1 − q0 z 1+ z . after an initial outburst of star formation. In terms of H0 and q0 . an isolated galaxy. can actually become more luminous with time. as it is called. H0 2 (7. Initial attempts to find a highly luminous standard candle focused on using entire galaxies as standard candles. MEASURING COSMOLOGICAL PARAMETERS is 250 km s−1 less than it would be if the universe were perfectly homogeneous. that you can determine luminosity distances with an accuracy much better than 5%. Not only do galaxies have a wide range of luminosities at the present moment. it must be very luminous. they aren’t particularly suitable for use as standard candles. we need to view standard candles for which the relation between dL and z deviates significantly from the linear relation which holds true at lower redshifts. or z > 0. optimistically. 7. Since the luminosity evolution of galaxies is imperfectly understood.5 Standard candles & the accelerating universe To determine the value of H0 without having to worry about Virgocentric flow and other peculiar velocities. If you think. will fade gradually with time. they are massive stars (M > 8 M ) whose cores collapse to form a black hole or neutron star when their nuclear fuel is exhausted. Type I supernovae contain no hydrogen absorption lines in their spectra. the white dwarf starts to collapse until its increased density triggers a runaway nuclear fusion reaction. type II supernovae contain strong hydrogen absorption lines. which are called type Ia and type Ib. unlike type II supernovae.000 times more luminous than even the brightest Cepheid. on the basis of their spectra. A supernova may be loosely defined as an exploding star. Gradually. blowing itself to smithereens. type Ia supernovae occur roughly once per century. type Ib and type II supernovae are driven by very similar mechanisms – their differences are superficial. The entire white dwarf becomes a fusion bomb. in the most literal sense. this is the maximum mass at which the electron degeneracy pressure can support a white dwarf against its own self-gravity. During the rapid collapse of the core. Within our Galaxy. Type Ib supernovae. Early in the history of supernova studies. The luminosity of an average type Ia supernova. Consider type Ia supernovae in the Virgo cluster. Although type Ia supernovae are not frequent occurrences locally. however. When the Chandrasekhar limit is exceeded. that is. type Ia supernovae do not leave a condensed stellar remnant behind. this means that type Ia supernovae can also be seen at z ∼ 1.4 M . They occur in close binary systems where one of the two stars in the system is a white dwarf. The transfer of mass from the companion star to the white dwarf eventually nudges the white dwarf over the Chandrasekhar limit of 1. For a few days. the standard candle of choice among cosmologists has been type Ia supernovae. on average. Although there’s only one type Ia . supernovae were divided into two classes. they are also reasonably standardized standard candles. Type I supernovae are actually two separate species. STANDARD CANDLES & ACCELERATION 145 In recent years. at peak brightness. is L = 4×109 L . Thus.5. the outer layers of the star are thrown off into space. Since moderately bright galaxies can be seen at z ∼ 1. it was realized that all type II supernovae are the same species of beast. they are extraordinary luminous. that’s 100. Not only are type Ia supernovae bright standard candles. are something completely different. are massive stars whose cores collapse after the hydrogen-rich outer layers of the star have been blown away in strong stellar winds. it is thought. and hence can be seen to large distances.7. Type Ia supernovae. a type Ia supernova in a moderately bright galaxy can outshine all the other stars in the galaxy combined. a stellar remnant which is supported against gravity by electron degeneracy pressure. when little was known about their underlying physics. by observing type Ia supernovae at z ∼ 0. have been conducting searches for supernovae in distant galaxies. the value of the Hubble constant which I have adopted in this text. has its roots in ancient Greece. Recently. according to their apparent brightness. and fifth magnitude. and have been found to have similar fluxes at maximum brightness. as indicated by the Cepheid results. the total luminosity of the Virgo cluster is a few hundred times that of our Galaxy. fourth. and intermediate stars were ranked as second. every year you can expect a few type Ia supernovae to go off in the Virgo cluster. Instead of all having L = 4 × 109 L . the supernova groups have been attempting to measure the acceleration (or deceleration) of the universe by observing type Ia supernovae at higher redshift. MEASURING COSMOLOGICAL PARAMETERS supernova per century in our own Galaxy. the faintest stars visible to the naked eye were of “sixth magnitude”. First. Thus. So far. To present the most recent supernova results to you. two research teams. divided the stars into six classes. In addition. type Ia supernovae sound like ideal standard candles. The Greek astronomer Hipparchus. their peak luminosities lie in the fairly broad range L ≈ 3 → 5 × 109 L . The magnitude system. Observation of supernovae in galaxies whose distances have been well determined by Cepheids reveal that type Ia supernovae do not have identical luminosities. The results of the different groups are in reasonable agreement with each other. however. very luminous and very standardized. in the second century BC. They have used the observed light curves and redshifts of type Ia supernovae to measure cosmological parameters. Several type Ia supernovae have been observed in the Virgo cluster in the recent past. third. Thus. it has also been noted that the peak luminosity of a type Ia supernova is tightly correlated with the shape of its light curve. the “Supernova Cosmology Project” and the “High-z Supernova Search Team”. supernovae with luminosities which rise and fall in a more leisurely manner are more luminous than average.146 CHAPTER 7. Type Ia supernovae with luminosities the shoot up rapidly and decline rapidly are less luminous than average at their peak.1. then the observed supernovae fluxes and redshifts are consistent with H0 = 70 ± 7 km s−1 Mpc−1 . I will have to introduce the “magnitude” system used by astronomers to express fluxes and luminosities. However. the rise and fall time of a type Ia supernova tells you its peak luminosity. Long after . There’s one complication. like much else in astronomy. just as the period of a Cepheid tells you its luminosity. The brightest stars were of “first magnitude”. If the distance to the Virgo cluster is pegged at dL = 15 Mpc. the value of H0 can be determined. 7.5. STANDARD CANDLES & ACCELERATION 147 the time of Hipparchus, it was realized that the response of the human eye is roughly logarithmic, and that stars of the first magnitude have fluxes (at visible wavelengths) about 100 times greater than stars of the sixth magnitude. On the basis of this realization, the magnitude system was placed on a more rigorous mathematical basis. Nowadays, the bolometric apparent magnitude of a light source is defined in terms of the source’s bolometric flux as m ≡ −2.5 log10 (f /fx ) , (7.46) where the reference luminosity is Lx = 78.7 L , since that is the luminosity of an object which produces a flux fx = 2.53 × 10−8 watt m−2 when viewed from a distance of 10 parsecs. The bolometric absolute magnitude of the Sun is thus M = 4.74. Although the system of apparent and absolute magnitudes seems strange to the uninitiated, the apparent magnitude is really nothing more than a logarithmic measure of the flux, and the absolute magnitude is a logarithmic measure of the luminosity. Given the definitions of apparent and absolute magnitude, the relation between an object’s apparent magnitude and its absolute magnitude can be written in the form dL M = m − 5 log10 , (7.48) 10 pc where dL is the luminosity distance to the light source. If the luminosity distance is given in units of megaparsecs, this relation becomes M = m − 5 log10 dL − 25 . 1 Mpc (7.49) where the reference flux fx is set at the value fx = 2.53 × 10−8 watt m−2 . Thanks to the negative sign in the definition, a small value of m corresponds to a large flux f . For instance, the flux of sunlight at the Earth’s location is f = 1367 watts m−2 ; the Sun thus has a bolometric apparent magnitude of m = −26.8. The choice of reference flux fx constitutes a tip of the hat to Hipparchus, since for stars visible to the naked eye it typically yields 0 < m < 6. The bolometric absolute magnitude of a light source is defined as the apparent magnitude that it would have if it were at a luminosity distance of dL = 10 pc. Thus, a light source with luminosity L has a bolometric absolute magnitude M ≡ −2.5 log10 (L/Lx ) , (7.47) 148 CHAPTER 7. MEASURING COSMOLOGICAL PARAMETERS Since astronomers frequently quote fluxes and luminosities in terms of apparent and absolute magnitudes, they find it convenient to quote luminosity distances in terms of the distance modulus to a light source. The distance modulus is defined as m − M , and is related to the luminosity distance by the relation dL m − M = 5 log10 + 25 . (7.50) 1 Mpc The distance modulus of the Large Magellanic Cloud, for instance, at dL = 0.050 Mpc, is m − M = 18.5. The distance modulus of the Virgo cluster, at dL = 15 Mpc, is m − M = 30.9. When z 1, the luminosity distance to a light source is 1 − q0 c z 1+ z . (7.51) dL ≈ H0 2 Substituting this relation into equation (7.50), we have an equation which gives the relation between distance modulus and redshift: m − M ≈ 43.17 − 5 log10 H0 70 km s−1 Mpc−1 + 5 log10 z + 1.086(1 − q0 )z . (7.52) For a population of standard candles with known luminosity L (and hence of known bolometric absolute magnitude M ), you measure the flux f (or equivalently, the bolometric apparent magnitude m) and the redshift z. In the limit z → 0, a plot of m − M versus log 10 z gives a straight line whose amplitude at a given value of z tells you the value of H0 . At slightly larger values of z, the deviation of the plot from a straight line tells you the value of q0 . At a given value of z, an accelerating universe (with q0 < 0) yields standard candles with a smaller flux than would a decelerating universe (with q0 > 0). The upper panel of Figure 7.5 shows the plot of distance modulus versus redshift for the combined supernova samples of the High-z Supernova Search Team (given by the filled circles) and the Supernova Cosmology Project (given by the open circles). The observational results are compared to the expected results for three model universes. One universe is flat, and contains nothing but matter (Ωm,0 = 1, q0 = 0.5). The second is negatively curved, and contains nothing but matter (Ωm,0 = 0.3, q0 = 0.15). The third is flat, and contains both matter and a cosmological constant (Ωm,0 = 0.3, ΩΛ,0 = 0.7, q0 = −0.55). The data are best fitted by the third of the models – which is, in fact, our Benchmark Model. The bottom panel of 7.5. STANDARD CANDLES & ACCELERATION 149 Figure 7.5: Distance modulus versus redshift for type Ia supernovae from the Supernova Cosmology Project (Perlmutter et al. 1999, ApJ, 517, 565) and the High-z Supernova Search Team (Riess et al. 1998, AJ, 116, 1009). The bottom panel shows the difference between the data and the predictions of a negatively curved Ωm,0 = 0.3 model (from Riess 2000, PASP, 112, 1284). 150 CHAPTER 7. MEASURING COSMOLOGICAL PARAMETERS Figure 7.5 shows this result more clearly. It shows the difference between the data and the predictions of the negatively curved, matter-only model. The conclusion that the universe is accelerating derives from the observation that the supernovae seen at z ∼ 0.5 are, on average, about 0.25 magnitudes fainter than they would in a decelerating universe with Ωm,0 = 0.3 and no cosmological constant. The supernova data extend out to z ∼ 1; this is beyond the range where an expansion in terms of H0 and q0 is adequate to describe the scale factor a(t). Thus, the two supernova teams customarily describe their results in terms of a model universe which contains both matter and a cosmological constant. After choosing values of Ωm,0 and ΩΛ,0 , they compute the expected relation between m − M and z, and compare it to the observed data. The results of fitting these model universes are given in Figure 7.6. The ovals drawn on Figure 7.6 enclose those values of Ωm,0 and ΩΛ,0 which give the best fit to the supernova data. The results of the two teams (the solid ovals and dotted ovals) give very similar results. Three concentric ovals are shown for each team’s result; they correspond to 1σ, 2σ, and 3σ confidence intervals, with the inner oval representing the highest probability. The best fitting models lie along the line 0.8Ωm,0 − 0.6ΩΛ,0 ≈ −0.2. Note that decelerating universes (with q0 > 0) can be strongly excluded by the data, as can Big Crunch universes (labeled ‘Recollapses’ in Figure 7.6), and Big Bounce universes (labeled ‘No Big Bang’ in Figure 7.6). The supernova data are consistent with negative curvature (labeled ‘Open’ in Figure 7.6), positive curvature (labeled ‘Closed’ in Figure 7.6), or with a universe which is spatially flat. The results of the supernova teams made headlines when they were first announced; the discovery of the accelerating universe was named by Science magazine as the ‘Scientific Breakthrough of the Year’ for 1998. It is prudent to remember, however, that all the hoopla about the accelerating universe is based on the observation that type Ia supernova at z ∼ 0.5 and beyond have somewhat lower fluxes (by about 25%) than they would have in a decelerating universe. There are other reasons why their fluxes might be low. For instance, if type Ia supernovae were intrinsically less luminous at z ∼ 0.5 than at z ∼ 0, that could explain their low fluxes. (If a typical supernova at z ∼ 0.5 had L = 3 × 109 L rather than 4 × 109 L , that would explain their observed dimness, without the need to invoke a cosmological constant. Conversely, if the typical supernova at z ∼ 0.5 had L = 5 × 109 L rather than 4 × 109 L , that would require an even larger cosmological constant to explain their observed dimness.) However, the other properties of type Ia supernovae, such as their 7.5. STANDARD CANDLES & ACCELERATION 151 Figure 7.6: The values of Ωm,0 (horizontal axis) and ΩΛ,0 (vertical axis) which best fit the data shown in Figure 7.5. The solid ovals show the best-fitting values for the High-z Supernova Search Team data; the dotted ovals show the best-fitting values for the Supernova Cosmology Project data (from Riess 2000, PASP, 112, 1284). 152 CHAPTER 7. MEASURING COSMOLOGICAL PARAMETERS spectra, don’t seem to evolve with time, so why should their luminosity? Perhaps the fluxes of supernovae at z ∼ 0.5 are low because some of their light is scattered or absorbed by intervening dust. However, dust tends to scatter some wavelengths of light more than others. This would change the shape of the spectrum of distant type Ia supernovae, but no dependence of spectral shape on redshift is observed. In sum, the supernova results of Figure 7.6 provide persuasive (but, given the caveats, not absolutely compelling) evidence for an accelerating universe. We will see in future chapters how additional observational evidence interlocks with the supernova results to suggest that we live in a nearly flat accelerating universe with Ωm,0 ≈ 0.3 and ΩΛ,0 ≈ 0.7. Suggested reading [Full references are given in the “Annotated Bibliography” on page 286.] Liddle (1999), ch. 6: The relation among H0 , q0 , Ω0 , and Λ. Narlikar (2002), ch. 9, 10: Local observations (z < 0.1) and more distant observations (z > 0.1) of the universe, and what they tell us about cosmological parameters. Peacock (1999), ch. 5: A review of distance measures used in cosmology. Rich (2001), ch. 5.2: A summary of the supernova Ia results. Problems (7.1) Suppose that a polar bear’s foot has a luminosity of L = 10 watts. What is the bolometric absolute magnitude of the bear’s foot? What is the bolometric apparent magnitude of the foot at a luminosity distance of dL = 0.5 km? If a bolometer can detect the bear’s foot at a maximum luminosity distance of dL = 0.5 km, what is the maximum luminosity distance at which it could detect the Sun? What is the maximum luminosity distance at which it could detect a supernova with L = 4 × 109 L ? starting from the Robertson-Walker metric of equation (7. what was the variation timescale δte at the time the light was emitted? For the light from the quasar to vary on a timescale δte . STANDARD CANDLES & ACCELERATION 153 (7.6) You observe a quasar at a redshift z = 5.7) Derive the relation Ap (t0 ) = 4πSκ (r)2 . If the observed variation in flux is due to a variation in the intrinsic luminosity of the quasar. in units of the Hubble distance? (7.4343x in the limit |x| 1. and the standard candles are .4343 ln(1 + x) ≈ 0.16 m. What are the current proper distance dP (t0 ).22).5 km? In the Benchmark Model. standard candles of luminosity L are distributed homogeneously in space.) (7.5. What is Rmax for the observed quasar? What is the angular size of Rmax in the Benchmark Model? (7. as given in equation (7.0. what is the minimum possible angular size of the polar bear’s foot? (7. the bulk of the light must come from a region of physical size R ≤ Rmax = c(δte ).3) Suppose that you are in a spatially flat universe containing a single component with a unique equation-of-state parameter w. thus. what is Σ as a function of redshift? Would observing the surface brightness of this class of objects be a useful way of determining the value of the deceleration parameter q0 ? Why or why not? (7. (7.52) is correct in the limit of small z.8) A spatially flat universe contains a single component with equationof-state parameter w. (You will probably want to use the relation log10 (1 + x) ≈ 0. Σ ∝ f /(δθ)2 . and determine that the observed flux of light from the quasar varies on a timescale δt0 = 3 days. the luminosity distance dL and the angular-diameter distance dA as a function of z and w? At what redshift will dA have a maximum value? What will this maximum value be.5) The surface brightness Σ of an astronomical object is defined as its observed flux divided by its observed angular area.2) Suppose that a polar bear’s foot has a diameter of = 0.4) Verify that equation (7.7. The number density of the standard candles is n0 at t = t0 . What is the angular size δθ of the foot at an angular-diameter distance of dA = 0.24). For a class of objects which are both standard candles and standard yardsticks. In this universe. which have an infinite horizon distance. .53) when w = −1/3. the power per unit area per steradian of sky) from standard candles with redshifts in the range z → z + dz is dJ(z) = n0 L(c/H0 ) (1 + z)−(7+3w)/2 dz .54) What will be the total intensity J of all standard candles integrated over all redshifts? Explain why the night sky is of finite brightness even in universes with w ≤ −1/3. MEASURING COSMOLOGICAL PARAMETERS neither created nor destroyed. Show that the observed flux from a single standard candle at redshift z is f (z) = L(1 + 3w)2 1 1 − (1 + z)−(1+3w)/2 2 (1 + z)2 16π(c/H0 ) −2 (7.154 CHAPTER 7. 4π (7. What is the corresponding relation when w = −1/3? Show that the observed intensity (that is. in the previous chapter. and others. In order to determine Ωm.3 and ΩΛ. The supernova data are consistent with Ωm.7. over the years. However. The apparent magnitude (or flux) of type Ia supernovae as a function of redshift is consistent with a flat universe having Ωm. the matter content of the universe is not negligible today.0 .0 .0 ≈ 1. the density parameter in matter.4. and other familiar types of baryonic matter? What fraction of the density is made of dark matter? What constitutes the dark matter – cold stellar remnants. we will have to adopt alternate methods of estimating the matter content of the universe.Chapter 8 Dark Matter Cosmologists. they are also consistent with Ωm.0 = 0 if ΩΛ. First. one method of putting limits on Ωm. There are many reasons for this obsession. and find out what types of matter it contains. black holes. Ωm.0 is individually well-constrained by the supernova observations. Another reason for wanting to know the matter density of the universe is to find out what the universe is made of.0 ≈ 0.0 ≈ 0.0 nor ΩΛ. is important in determining the spatial curvature and expansion rate of the universe. have dedicated a large amount of time and effort to determining the matter density of the universe. and was the dominant component in the fairly recent past. We have already seen. Even if the cosmological constant is non-zero. What fraction of the density is made of stars. neither Ωm.0 more accurately. 155 .0 = 1 if ΩΛ. and in what quantities. exotic elementary particles. have driven astronomers to make a census of the universe.0 ≈ 0.7. or some other substance too dim for us to see? These questions. you must remember that the mass of a star includes all the fuel which it will require during its entire lifetime.2) . They have a wide range of masses and a wider range of B-band luminosities. at first glance. and ultraviolet range of the electromagnetic spectrum. let’s suppose that the mix of stars in the solar neighborhood is not abnormal. in addition to admitting blue light. That is.1) M/LB ≈ 4 M / L . DARK MATTER 8.B for the dimmest. and the remaining 82% is emitted at wavelengths too long to pass through the B filter. for instance. on average. help astronomers to detect them by broadcasting photons in all directions. the total luminosity density of stars within a few hundred megaparsecs of our Galaxy is j . or M/LB = 1 M / L . The Sun’s luminosity in the B band is L . however.B = 4. visible. the mass-to-light ratio of the stars works out to be Although a mass-to-light ratio of 170 tons per watt doesn’t seem. 1 = 1. As a first guess. least massive stars (the M stars). the mass-to-light ratio of the stars in a galaxy will depend on the mix of stars which it contains. Thus. the mass-to-light ratio ranges from M/LB ∼ 10−3 M / L . Within 1 kiloparsec of the Sun. most massive stars (the O stars in the classic OBAFGKM spectral sequence) to M/LB ∼ 103 M / L . If all stars were identical to the Sun. a B-band filter also lets through violet light. we need to know the mass-to-light ratio for the stars. we need to know how many kilograms of star. However.B for the brightest.B into a mass density ρ .7 × 1025 watts. it takes to produce one watt of starlight in the B band. 2 In the B band. (8.9 × 10−7 m to pass through.156 CHAPTER 8.B To convert a luminosity density j . powered by hydrogen fusion in their centers.0 × 10−7 m < λ < 4. (8. you install a B-band filter on your telescope. For main sequence stars. Suppose. 2 This is only 12% of the Sun’s total luminosity. For comparison. we could simply say that there is one solar mass of stars for each solar luminosity of output power. like a very high efficiency. stars are not uniform in their properties.1 The “B” in B-band stands for “blue”.B ≈ 170. your eyes detect photons in the wavelength range 4 × 10 −7 m < λ < 7 × 10−7 m.000 kg watt−1 .2 × 108 L . Such a filter allows only photons in the wavelength range 4.1 Visible matter Some types of matter. About 6% of the luminosity is emitted at ultraviolet wavelengths.B . Stars primarily emit light in the infrared. such as stars.B Mpc−3 . for instance.004 . as shown in Figure 8.0 = εc. and black holes) and brown dwarfs. there is a significant amount of gas between galaxies. ≈ 3 × 1013 M .004 is not a precisely determined one. In irregular galaxies such as the Magellanic Clouds.4×1011 M Mpc−3 . the stellar mass in the Coma cluster is small compared to the mass of the hot.0 = M/LB j . The Coma cluster contains thousands of galaxies. then the total mass of stars in the Coma cluster is MComa. The interstellar medium contains significant amounts of gas. the mass of interstellar gas is roughly equal to 10% of the mass of stars. (8. The density parameter in stars will be further increased if you include in the category of “stars” stellar remnants (such as white dwarfs. If the mass-to-light ratio of the stars in the Coma cluster is M/LB ≈ 4 M / L .3) Stars make up less than 1/2% of the density necessary to flatten the universe. Galaxies also contain baryonic matter which is not in the form of stars. then the mass density of stars in the universe is ρ .0 = ≈ 0.0 ≈ Ω . neutron stars. intracluster gas between the galaxies in the cluster.1. the number Ω .8.B ≈ 5 × 108 M Mpc−3 . In truth. ∼ 95% of the stellar luminosity comes from stars more luminous than the Sun. located 100 Mpc from our Galaxy. low-luminosity stars in galaxies.B . stellar remnants. in the direction of the constellation Coma Berenices. X-ray images.B . for instance. Because brown dwarfs and isolated cool stellar remnants are difficult to detect. the ratio of gas to stars is even higher. or brown dwarfs. Although 30 trillion solar masses represents a lot of stars. In our Galaxy and in M31. but ∼ 80% of the stellar mass comes from stars less luminous than the Sun. such as the one shown in Since the current critical density of the universe is equivalent to a mass density of ρc.0 ≈ 0. most of the light comes from the stars in the cluster’s galaxies.0 /c2 = 1. the current density parameter of stars is 5 × 108 M Mpc−3 ρ . Consider a rich cluster of galaxies such as the Coma cluster.4) ρc. In our Galaxy. (8.1. their number density is not well determined. their summed luminosity in the B band comes to LComa. largely because of the uncertainty in the number of low-mass. In addition. VISIBLE MATTER 157 If the mass-to-light ratio of the stars within a kiloparsec of us is not unusually high or low.4 × 1011 M Mpc−3 .B = 8 × 1012 L . At visible wavelengths.0 1. A brown dwarf is a self-gravitating ball of gas which is too low in mass to sustain nuclear fusion in its interior. equivalent to ∼ 1 Mpc at the distance of the Coma cluster.1: The Coma cluster as seen in visible light. DARK MATTER Figure 8. The scale is the same as that of the previous image. using NASA’s SkyView facility. courtesy Max-Planck-Institut f¨r extraterrestriche Physik. [From the ROSAT x-ray observatory.] .2: The Coma cluster as seen in x-ray light. produced at the Space Telescope Science Institute] Figure 8. The image shown is 35 arcminutes across. This figure and the u previous figure were produced by Raymond White.158 CHAPTER 8. [From the Digitized Sky Survey. which doesn’t absorb. The Sun. in fact.2. emitting x-rays with a typical energy of E ∼ kTgas ∼ 9 keV.2 Dark matter in galaxies The situation. is on such an orbit – it is R = 8. As it turns out. If the radius of the orbit is R and the orbital speed is v. helium. is even more extreme than stated in the previous section.gas ≈ 2 × 1014 M . most of the matter is not even baryonic. A classic method of detecting dark matter involves looking at the orbital speeds of stars in spiral galaxies such as our own Galaxy and M31. roughly six or seven times the mass in stars. lithium.0 = 0. (8. the efficiency with which fusion takes place in the early universe. 8. low-density gas.2. you are actually marveling at a minority of the baryonic matter in the universe.5 kpc from the Galactic center. fills the space between clusters. converting hydrogen into deuterium. with a typical temperature of T ≈ 1 × 108 K. within the disk. As we will see in Chapter 10.5) an order of magnitude larger than the density parameter for stars. The majority of the matter in the universe is nonbaryonic dark matter. then the star . Spiral galaxies contain flattened disks of stars. DARK MATTER IN GALAXIES 159 Figure 8. the best current limits on the baryon density of the universe come from the predictions of primordial nucleosynthesis. for instance.04 ± 0. When you stare up at the night sky and marvel at the glory of the stars. Most of the baryons are too cold to be readily visible (the infrared emitting brown dwarfs and cold stellar remnants) or too diffuse to be readily visible (the low density x-ray gas in clusters). and other elements. and has an orbital speed of v = 220 km s−1 . One way of detecting dark matter is to look for its gravitational influence on visible matter. Suppose that a star is on a circular orbit around the center of its galaxy. emit. Not only is most of the baryonic matter undetectable by our eyes. or scatter light of any wavelength. stars are on nearly circular orbits around the center of the galaxy. The total amount of x-ray emitting gas in the Coma cluster is estimated to be MComa.01 . depends on the density of protons and neutrons present. reveal that hot. Detailed studies of the amounts of deuterium and other elements present in primordial gas clouds indicate that the density parameter of baryonic matter must be Ωbary.8. In general. 4 99. Thus. R2 (8.9) R The surface brightness I of the disk of a spiral galaxy typically falls off exponentially with distance from the center: v= I(R) = I(0) exp(−R/Rs ) . Rs ≈ 6 kpc. 3 . If the acceleration is provided by the gravitational attraction of the galaxy. Rs ≈ 4 kpc.8) with the scale length Rs typically being a few kiloparsecs. is referred to as “Keplerian rotation”. true (the stars in the disk obviously have a flattened distribution).8% of the Solar System’s mass is contained within the Sun. v ∝ 1/ R.7): GM (R) v2 = . for M31. then a= a= GM (R) .3 The relation between v and M is found by setting equation (8. since it’s what Kepler found for orbits in the Solar System. R R2 or GM (R) .160 experiences an acceleration CHAPTER 8. Once you are a few scale lengths from the center of the spiral galaxy. strictly speaking. √ at of the mass in a galaxy. (8. This relation between orbital speed and orbital radius. or most. where the mass is strongly concentrated toward the center. a somewhat larger disk galaxy. (8. we won’t be seeing the disk perfectly face-on or edge-on. if stars contributed all.6) R directed toward the center of the galaxy.7) where M (R) is the mass contained within a sphere of radius R centered on the galactic center.4 The orbital speed v of stars within a spiral galaxy can be determined from observations. For our Galaxy.10) (8. but the flattening of the galaxy provides only a small correction to the equation for the gravitational acceleration. Consider a galaxy which has the shape of a thin circular disk.7) assumes that the mass distribution of the galaxy is spherically symmetric. DARK MATTER v2 . (8. the velocity would fall as v ∝ 1/ R√ large radii. Equation (8. the mass of stars inside R becomes essentially constant. This is not.6) equal to equation (8. and v(R) is the orbital speed at a distance R from the center of the disk. Since the redshift contains only the component of the stars’ orbital velocity which lies along the line of sight.3). we’ll see it at an inclination i.8. The disk we see in projection will be elliptical. = sin i 1 − b2 /a2 (8.11) For example. lines in light from the disk. This indicates that we are seeing M31 fairly close to edge-on. not circular. with an observed axis ratio b/a = 0. the galaxy M31 looks extremely elongated as seen from Earth. with an axis ratio b/a = cos i . resulting from the expansion of the universe.22.12) where vgal is the radial velocity of the galaxy as a whole. (8. with an inclination i = cos−1 (0. DARK MATTER IN GALAXIES 161 Figure 8. or emission.2. By measuring the redshift of the absorption. (8.13) . We can thus compute the orbital speed v(R) in terms of observable properties as v(R) = vr (R) − vgal vr (R) − vgal .22) = 77◦ .3: An observer sees a disk at an inclination angle i. we can find the radial velocity vr (R) = cz(R) along the apparent long axis of the galaxy. where i is the angle between our line of sight to the disk and a line perpendicular to the disk (see Figure 8. the radial velocity which we measure will be vr (R) = vgal + v(R) sin i . The mass of the dark halo provides the necessary . However. 327) at radio wavelengths (figure from van den Bergh.162 CHAPTER 8. in 1914. By measuring the redshift of this emission line. DARK MATTER Figure 8. The first astronomer to detect the rotation of M31 was Vesto Slipher. “extrapolation beyond that distance is a matter of taste. ApJ. we deduce the presence of a dark halo within which the visible stellar disk is embedded. 201. shown as the open circles in Figure 8. the solid dots with error bars show the results of Roberts and Whitehurst (1975. Since the orbital speed of the stars and gas at large radii (R > 3Rs ) is greater than it would be if stars and gas were the only matter present. 159. the orbital speed v at R > 3Rs = 18 kpc was not accurately measured until more than half a century later. M. Vera Rubin and Kent Ford looked at emission lines from regions of hot ionized gas in M31. In 1970. The open circles show the results of Rubin and Ford (1970.” At R > 4Rs . ApJ. there is still a small amount of atomic hydrogen in the disk of M31. Roberts and R. Whitehurst found that the orbital speed stayed at a nearly constant value of v(R) ≈ 230 km s−1 out to R ≈ 30 kpc ≈ 5Rs . 2000). as shown by the solid dots in Figure 8.4. 379) at visible wavelengths. given the difficulty of measuring the spectra at low surface brightness. Beyond R = 4Rs . as they wrote in their original paper. and were able to find the orbital speed v(R) out to a radius R = 24 kpc = 4Rs . Their results.4: The orbital speed v as a function of radius in M31. the visible light from M31 was too faint for Rubin and Ford to measure the redshift.4. give no sign of a Keplerian decrease in the orbital speed. which can be detected by means of its emission line at λ = 21 cm. B = 2. At R ≈ 20 kpc. typically. Rhalo > 20 kpc. The quantity Rhalo is the radius of the dark halo surrounding the luminous disk of our galaxy.2.3 × 1010 L .B . implying that the dark halo is an order of magnitude more massive than the stellar disk. thus.6 × 1010 M G v 220 km s−1 2 R 8.B . this would mean that our halo stretches nearly halfway to M31. For these hangers-on to remain gravitationally bound to our Galaxy.14). where the last detectable gas exists in the disk of our Galaxy. can be found from equation (8. if not all. taken as a whole. implying a total mass for our Galaxy of MGal ≈ 8 × 1011 M . the halo must extend as far as Rhalo ≈ 75 kpc.9): M (R) = v2R = 9. the mass of our Galaxy would be MGal ≈ 3 × 1012 M . . Some astronomers have speculated that the dark halo is actually four times larger in radius. this means that the mass-to-light ratio of our Galaxy. This mass-to-light ratio is ten times greater than that of the stars in our Galaxy. our own Galaxy has an orbital speed which actually seems to be rising slightly at R > 15 kpc. Thousands of spiral galaxies have had their orbital velocities v(R) measured. Since our Galaxy’s luminosity in the B band is estimated to be LGal. and a total mass-to-light ratio M/LB Gal ≈ 40 M/ L . including both the luminous disk and the dark halo.8.14) The values of v and R in the above equation are scaled to the Sun’s location in our Galaxy. With Rhalo ≈ 300 kpc. A rough estimate of the halo size can be made by looking at the velocities of the globular clusters and satellite galaxies (such as the Magellanic Clouds) which orbit our Galaxy. the mass of a spiral galaxy. and the total mass-to-light ratio would be M/LB Gal ≈ 150 M / L . (8. M31 is not a freak.B Rhalo 100 kpc . v is roughly constant at R > Rs . is M/LB Gal ≈ 50 M / L . If we approximate the orbital speed v as being constant with radius. DARK MATTER IN GALAXIES 163 gravitational “anchor” to keep the high-speed stars and gas from being flung out into intergalactic space. spiral galaxies have comparable dark halos.15) using v = 220 km s−1 in equation (8. For instance. (8.5 kpc .B . with Rhalo ≈ 300 kpc. The exact value of Rhalo is poorly known. instead of decreasing in a Keplerian fashion. the orbital speed shows no sign of a Keplerian decrease. most. and a velocity xi .0 ≈ 0. Thus. at a more mathematical level. Henri Poincar´ discussed the e possible existence within our Galaxy of “mati`re obscure” (translated as “dark matter” in e the standard edition of Poincar´’s works). DARK MATTER If our Galaxy is typical in having a dark halo 10 to 40 times more massive than its stellar component.04 → 0.5 To follow. in a Newtonian manner. Ωbary. a position ˙ xi .0 = (10 → 40)Ω . e 5 . .0002dhor . The acceleration of the Although Zwicky’s work popularized the phrase “dark matter”. given the uncertainty in the extent of their dark halos. and other clusters of galaxies.16 . the velocity dispersion within the Coma cluster is σComa ≈ 1000 km s−1 ≈ 0. the cluster must contain a large amount of “dunkle Materie”.3 Dark matter in clusters The first astronomer to make a compelling case for the existence of large quantities of dark matter was Fritz Zwicky.01. Clusters of galaxies are gravitationally bound objects. then the density parameter of galaxies (including their dark halos) must be Ωgal. with a mass mi (i = 1. the radius of the Coma cluster is RComa ≈ 3 Mpc ≈ 0.164 CHAPTER 8. . he noted that the dispersion in the radial velocity of the cluster’s galaxies was very large – around 1000 km s−1 .04 ± 0. Zwicky concluded.16) Although the total density of galaxies is poorly known. (8. .0 = 0. or (translated into English) “dark matter”.1). we can treat the dynamics of the Coma cluster. in 1908. For instance. let us suppose that a cluster of galaxies is comprised of N galaxies.003c. Because of these considerations. 2. The stars and gas visible within the galaxies simply did not provide enough gravitational attraction to hold the cluster together. he was not the first to use it in an astronomical context. In order to keep the galaxies in the Coma cluster from flying off into the surrounding voids. some part of the dark halos of galaxies is likely to be comprised of nonbaryonic dark matter. it is likely to be larger than the density of baryons. each of which can be approximated as a point mass. not expanding with the Hubble flow. N ). They are small compared to the horizon size. . The galaxies within a cluster are moving at non-relativistic speeds. in the 1930’s. 8. In studying the Coma cluster of galaxies (shown in Figure 8. Zwicky’s reasoning. (The factor of 1/2 in front of the double summation ensures that each pair of galaxies is only counted once in computing the potential energy. The gravitational potential energy of the system of N galaxies is W =− G 2 mi mj .) The potential energy of the cluster can also be written in the form W = −α GM 2 .17) Note that equation (8. the radius of a sphere centered on the cluster’s center of mass and containing a mass M/2. rh (8. |xj − xi |3 (8.j j=i This is the energy that would be required to pull the N galaxies away from each other so that they would all be at infinite distance from each other.17) assumes that the cluster is an isolated system.20) 2 i The kinetic energy K can also be written in the form 1 K = M v2 . |xj − xi | (8. 2 where v2 ≡ 1 M ˙ m i |x i | 2 (8.21) (8.18) i. is given by the Newtonian formula ¨ xi = G j=i 165 mj xj − x i . (8.22) i . The kinetic energy associated with the relative motion of the galaxies in the cluster is 1 ˙ K= m i |x i | 2 .8. it is found that α ≈ 0.19) where M = mi is the total mass of all the galaxies in the cluster. with the gravitational acceleration due to matter outside the cluster being negligibly small. DARK MATTER IN CLUSTERS ith galaxy in the cluster. then. For observed clusters of galaxies. α is a numerical factor of order unity which depends on the density profile of the cluster.4 gives a good fit to the potential energy. and rh is the half-mass radius of the cluster – that is.3. we can use equation (8. (8.29) .166 CHAPTER 8.17) to write ¨ mi (xi · xi ) = G mi mj i. |xj − xi | ¨ mj (xj · xj )  (8.26) and (8.i i=j j xj · (xi − xj ) . It is also useful to define the moment of inertia of the cluster as I≡ i mi |xi |2 .27) to find 1 ¨ mi (xi · xi ) = 2 = −  i i ¨ mi (xi · xi ) + j G 2 i.28) (it doesn’t matter whether we call the variable over which we’re summing i or j or k or “Fred”). we can combine equations (8.20).26) However.23) The moment of inertia I can be linked to the kinetic energy and the potential energy if we start by taking the second time derivative of I: ¨ I=2 i ¨ ˙ ˙ mi (xi · xi + xi · xi ) . |xj − xi |3 (8. (8. we can rewrite this as ¨ I=2 i ¨ mi (xi · xi ) + 4K .25) To introduce the potential energy W into the above relation. |xi − xj |3 (8.j j=i mi mj =W . (8. we could equally well switch around the i and j subscripts to find the equally valid equation ¨ mj (xj · xj ) = G mj mi j. DARK MATTER is the mean square velocity (weighted by galaxy mass) of all the galaxies in the cluster.j j=i i xi · (xj − xi ) .27) Since i ¨ mi (xi · xi ) = j ¨ mj (xj · xj ) (8.24) Using equation (8. or any other self-gravitating steady-state system: M= v 2 rh . used to estimate the mass of a cluster of galaxies. among other things. used to estimate the mass of a rotating spiral galaxy.8.3. (8. For instance. then dividing by the gravitational constant G.32). we find 1 α GM 2 M v2 = .14). Using equation (8. (This implies. Applying the virial theorem to a real cluster of galaxies. and equation (8.) If I = constant. 2 2 rh (8. is complicated by the fact that we have only partial information about the cluster. it applies perfectly well to a self-gravitating system of point masses.33) (8. we estimate the mass of a self-gravitating system by multiplying the square of a characteristic velocity by a characteristic radius. then the steady-state virial theorem is 0 = W + 2K . In either case. that the system is neither expanding nor contracting.34). . αG (8.31) This means we can use the virial theorem to estimate the mass of a cluster of galaxies. but as we have seen. The virial theorem is particularly useful when it is applied to a system in steady state. such as the Coma cluster. or K = −W/2 . the kinetic energy K is equal to −1/2 times the potential energy W .30) This relation is known as the virial theorem.32) That is. with a constant moment of inertia. the first term on the right hand side of equation (8.19) and (8. DARK MATTER IN CLUSTERS 167 Thus.21) in equation (8. It was actually first derived in the nineteenth century in the context of the kinetic theory of gases. and we may now write down the simple relation ¨ I = 2W + 4K .25) is simply 2W .34) Note the similarity between equation (8. and that you are using a coordinate system in which the center of mass of the cluster is at rest. (8. for a self-gravitating system in steady state. and thus do not know v 2 and rh exactly. 36) (8. we don’t know the distribution of dark matter in the cluster a priori . (8. but the velocity perpendicular to the line of sight is unknown.32 × 1012 m2 s−2 )(4.38) If we assume that the velocity dispersion is isotropic. we may estimate the mass of the Coma cluster to be MComa (2.4)(6.6 × 1022 m) v 2 rh ≈ = αG (0. (8.6 × 1022 m . the mean redshift of the cluster is found to be z = 0. DARK MATTER we can find the line-of-sight velocity of each galaxy from its redshift. if we assume that the mass-to-light ratio is constant with radius. then the three-dimensional mean square velocity v 2 will be equal to three times the one-dimensional 2 mean square velocity σr . then the sphere containing half the mass of the cluster will be the same as the sphere containing half the luminosity of the cluster. the total amount of dark matter is what we’re trying to find out.40) After all these assumptions and approximations. which can be translated into a radial velocity vr = c z = 6960 km s−1 and a distance dComa = (c/H0 ) z = 99 Mpc .5 Mpc ≈ 4.35) (8.7 × 10−11 m3 s−2 kg−1 ) ≈ 4 × 1045 kg ≈ 2 × 1015 M . in fact.37) The velocity dispersion of the cluster along the line of sight is found to be = 880 km s−1 . (8. σr = (vr − vr )2 1/2 (8. If we further assume that the cluster is intrinsically spherical.41) (8. After all. yielding v 2 = 3(880 km s−1 )2 = 2.168 CHAPTER 8. (8.39) Estimating the half-mass radius rh of the Coma cluster is even more perilridden than estimating the mean square velocity v 2 .32 × 1012 m2 s−2 .0232 . then the observed distribution of galaxies within the Coma cluster indicates a half-mass radius rh ≈ 1.42) . However. From measurements of the redshifts of hundreds of galaxies in the Coma cluster. 44) is the same equation which determines the internal structure of a star. Combined with the luminosity of the Coma cluster. where the inward force due to gravity is also exactly balanced by an outward force due to a pressure gradient. as a function of radius. The pressure of the gas is given by the perfect gas law.45) where T is the temperature of the gas.2. lost socks. and µ is its mass in units of the proton mass (mp ).B = 8 × 1012 L . and only ten percent consists of hot intracluster gas (MComa. (8. x-ray emitting intracluster gas.45): M (r) = 6 d ln ρ d ln T kT (r)r − − Gµmp d ln r d ln r .44) assumes. the gas in Coma isn’t perfectly spherical in shape.43) greater than the mass-to-light ratio of our Galaxy.3. and anything else.6 Of course. dr r2 (8.44) and (8. but it’s close enough to spherical to give a reasonable approximation to the mass. (8. DARK MATTER IN CLUSTERS 169 Thus.gas ≈ 2 × 1014 M ). is still in place. ρ is the density of the gas. If the hot intracluster gas is supported by its own pressure against gravitational infall. as equation (8.B .46) Equation (8. shown in Figure 8. including gas.B . and M is the total mass inside a sphere of radius r. if there were no dark matter to anchor the gas gravitationally. the hot gas would have expanded beyond the cluster on time scales much shorter than the Hubble time. P = ρkT . LComa.8. the total mass of the Coma cluster implies a mass-to-light ratio of M/LB Coma ≈ 250 M / L . ≈ 3 × 1013 M ). The mass of the cluster. less than two percent of the mass of the Coma cluster consists of stars (MComa. is found by combining equations (8. . The presence of a vast reservoir of dark matter in the Coma cluster is confirmed by the fact that the hot.44) where P is the pressure of the gas. The temperature and density of the hot gas in the Coma cluster can be used to make yet another estimate of the cluster’s mass. µmp (8. stars. dark matter. it must obey the equation of hydrostatic equilibrium: dP GM (r)ρ(r) =− . for instance. and chemical composition of the gas. assuming hydrostatic equilibrium. since any matter which is smoothly distributed in the intercluster voids will not be included in this number. when Leverrier and Adams deduced the existence of the dim planet Neptune by its effect on the orbit of Uranus. If the masses of all the clusters of galaxies are added together. as you’d expect if the chemical composition and ionization state of the gas is uniform throughout the cluster. DARK MATTER The above equation assumes that µ is constant with radius. temperature maps reveal relatively cool regions (at kT ≈ 5 keV) as well as hotter regions (at kT ≈ 12 keV). consistent with the mass estimate of the virial theorem. it is found that their density parameter is Ωclus. (8.2 . I have outlined the classical methods for detecting dark matter via its gravitational effects on luminous matter.7 We can detect dark matter around spiral galaxies because it affects the motions of stars and interstellar gas.0 ≈ 0.6 Mpc of the center. Typical mass-to-light ratios for clusters lie in the range M/LB = 200 → 300 M / L . The mass of the Coma cluster.B . 7 .7 Mpc of the cluster center and (1 → 2)×1015 M within 3. In the Coma cluster. Starting from an x-ray spectrum. using the virial theorem applied to their galaxies or the equation of hydrostatic equilibrium applied to their gas. is computed to be (3 → 4)×1014 M within 0. We can detect dark matter in clusters of galaxies because it affects the motions of The roots of these methods can be traced back as far as the year 1846. averaging to kT ≈ 9 keV over the entire cluster. 8. so the Coma cluster is not unusual in the amount of dark matter which it contains. Other clusters of galaxies besides the Coma cluster have had their masses estimated. density.170 CHAPTER 8. The x-rays emitted from the hot intracluster gas are a combination of bremsstrahlung emission (caused by the acceleration of free electrons by protons and helium nuclei) and line emission from highly ionized iron and other heavy elements.4 Gravitational lensing So far. it is possible to fit models to the emission and thus compute the temperature.47) This provides a lower limit to the matter density of the universe. These objects have been collectively called MACHOs.4. neutron stars. it can act as a lens. a slightly strained acronym for MAssive Compact Halo Objects. the local curvature of space-time will cause the photon to be deflected by an angle α= 4GM . GRAVITATIONAL LENSING 171 Figure 8.8. c2 R (8.49) In 1919. an eclipse expedition photographed stars in the vicinity of the Sun. Comparison of the eclipse photographs with photographs of the same star field taken six months earlier revealed that the apparent positions of the stars were deflected by the amount which Einstein had predicted. after Einstein predicted a deflection of this magnitude. Some of the dark matter in the halo might consist of massive compact objects such as brown dwarfs. as well as in distant clusters of galaxies. galaxies and intracluster gas. start by considering the dark halo surrounding our Galaxy. white dwarfs. or a brown dwarf. as Einstein realized.5. light from a distant star which just grazes the Sun’s surface should be deflected through an angle α= 4G M = 1. acting as a gravitational lens. To see how gravitational lensing can be used to detect dark matter. If a photon passes such a compact massive object at an impact parameter b. Suppose a MACHO in the halo of our Galaxy passes . or a stellar remnant. For instance.7 arcsec .48) where M is the mass of the compact object. Thus. dark matter can bend and focus light. This result brought fame to Einstein and experimental support to the theory of general relativity. c2 b (8. but also the trajectory of photons. However. can deflect light.5: Deflection of light by a massive compact object. The effects of dark matter on photons have been used to search for dark matter within the halo of our own Galaxy. Since a star. dark matter will affect not only the trajectory of matter. and black holes. as shown in Figure 8. 5 (that is. if the MACHO is roughly halfway between the observer and the lensed star). then the image of the star will be distorted into two or more arcs instead of a single unbroken ring. the image produced is a perfect ring.6: Light from star in the Large Magellanic Cloud is deflected by a MACHO on its way to an observer in the disk of our Galaxy (seen edge-on in this figure). and xd (where 0 < x < 1) is the distance from the observer to the lensing MACHO. The angle θE is known as the Einstein radius.50) where M is the mass of the lensing MACHO. (8. it is possible. As the MACHO deflects the light from the distant star. Although the Einstein radius for an LMC star being lensed by a MACHO is too small to be resolved. Figure 8. directly between an observer in our Galaxy and a star in the Large Magellanic Cloud. If x ≈ 0. DARK MATTER Figure 8. with angular radius θE = 4GM 1 − x c2 d x 1/2 . in some cases.172 CHAPTER 8. with a MACHO which happens to be halfway between the observer and the star. (8. d is the distance from the observer to the lensed star. then θE ≈ 4 × 10−4 arcsec M 1M 1/2 d 50 kpc −1/2 . it produces an image of the star which is both distorted and amplified.6 shows such a situation. If the MACHO is exactly along the line of sight between the observer and the lensed star. to .51) If the MACHO does not lie perfectly along the line of sight to the star. with M ∼ 1014 M . It has been calculated that if the dark halo of our galaxy were entirely composed of MACHOs. Given the small size of the Einstein radius. the observed lensing events could be due. The long time scales of the observed lensing events. the probability of any particular star in the LMC being lensed at any moment is tiny. the search for MACHOs suggests that most of the matter in the dark halo of our galaxy is due to a smoothly distributed component. In any case. Gravitational lensing occurs at all mass scales. which would have the correct mass. suggesting that there is not a significant population of brown dwarfs (with M < 0. must be comparable to. then becomes dimmer as the angular distance increases again. To detect lensing by MACHOs.08 M ) in the dark halo of our Galaxy. Generally speaking. which have ∆t > 35 days.15 M . The typical time scale for a lensing event is the time it takes a MACHO to travel through an angular distance equal to θE as seen from Earth. GRAVITATIONAL LENSING 173 detect the amplification of the flux from the star. for instance.) Alternatively. for a MACHO halfway between here and the LMC. instead of being congealed into MACHOs of roughly stellar mass. The research groups which searched for MACHOs found a scarcity of short duration lensing events. the typical signature of a “lensing event” is a star which becomes brighter as the angular distance between star and MACHO decreases.52) where v is the relative transverse velocity of the MACHO and the lensed star as seen by the observer on Earth. to lensing objects within the LMC itself. Suppose. or smaller than. cold white dwarfs. Since the MACHOs in our dark halo and the stars in the LMC are in constant relative motion. The total number of lensing events which they detected suggest that as much as 20% of the halo mass could be in the form of MACHOs.4. this is d θE M ∆t = ≈ 90 days 2v 1M 1/2 v 200 km s−1 −1 . at a distance ∼ 500 Mpc from . as seen from Earth.8. the angular distance between the MACHO and the lensed star. (8. the Einstein radius. that a cluster of galaxies. (Perhaps the MACHOs are old. suggest typical MACHO masses of M > 0. then the probability of any given star in the LMC being lensed at any given time would still only be P ∼ 5 × 10−7 . more massive MACHOs produce larger Einstein rings and thus will amplify the lensed star for a longer time. various research groups took up the daunting task of monitoring millions of stars in the Large Magellanic Cloud to watch for changes in their flux. For the amplification to be significant. at least in part. 4 arcmin by 1. Couch [University of New South Wales] and NASA). they are background galaxies. which has a redshift z = 0.2 arcmin.7 are not oddly shaped galaxies within the cluster.5 arcmin 14 M 10 1/2 d 1000 Mpc −1/2 . The Einstein radius for this configuration will be M θE ≈ 0. equivalent to 0.54 Mpc by 0.18. Figure 8. our Galaxy.18. (8.53) The arc-shaped images into which the background galaxy is distorted by the lensing cluster can thus be resolved. The elongated arcs seen in Figure 8.27 Mpc at the distance of Abell 2218 (courtesy of W. and hence is at a proper distance d = 770 Mpc. displaying gravitationally lensed arcs. The mass of clusters can be estimated by the degree to which they lens background galaxies. DARK MATTER Figure 8. which are gravitationally lensed by the cluster mass.7: A Hubble Space Telescope picture of the rich cluster Abell 2218. . instead.174 CHAPTER 8. For instance. lenses a background galaxy at d ∼ 1000 Mpc.7 shows an image of the cluster Abell 2218. The masses calculated in this way are in general agreement with the masses found by applying the virial theorem to the motions of galaxies in the cluster or by applying the equation of hydrostatic equilibrium to the hot intracluster gas. at redshifts z > 0. The region shown is roughly 2. telling us that Ωm.5 What’s the matter? I have described how to detect dark matter by its gravitational effects.0 ≥ 0. the Cosmic Neutrino Background is a relic of the time when the universe was hot and dense enough to be opaque to neutrinos.8. rather than by the collapse of a massive star later on. the density of nonbaryonic matter is at least four times the density of the familiar baryonic matter of which people and planets and stars are made. Just as the Cosmic Microwave Background is a relic of the time when the universe was opaque to photons. As stated in section 5. On the other hand. equivalent to mBH ∼ 2 × 1035 kg. there should exist today a cosmic background of neutrinos. It is a sign of the vast ignorance concerning nonbaryonic dark matter that these two candidates for the role of dark matter differ in mass by 76 orders of magnitude.36 × 108 m−3 . with masses up to mBH ∼ 105 M . conjecture about the nature of the nonbaryonic dark matter has run rampant (some might even say it has run amok). equivalent to max ∼ 2 × 10−41 kg. One nonbaryonic particle which we know exists. Thus. some scientists have conjectured that the dark matter might be made of primordial black holes. The number density of each of the three flavors of neutrinos (νe .8 This is a rather high mass – it would take some 30 billion Earths to equal the mass of one primordial black hole (if they indeed exist). . To give a taste of the variety of speculation. A component of the universe which is totally invisible is an open invitation to speculation.1. is the neutrino. and ντ ) has been calculated to be 3/11 times the number density of CMB photons. yielding a total number density of neutrinos nν = 3(3/11)nγ = (9/11)(4. 8 (8. but I’ve been dodging the essential question: “What is it? ” Adding together the masses of clusters of galaxies gives a lower limit on the matter density of the universe.0 ≈ 0. However. WHAT’S THE MATTER? 175 8. a type of elementary particle with a rest energy of max c2 ∼ 10−5 eV.04.54) A primordial black hole is one which forms very early in the history of the universe.5. νµ . As you might expect. and which seems to have a non-zero mass. This is a rather low mass – it would take some 50 billion axions (if they indeed exist) to equal the mass of one electron.2. the density parameter of baryonic matter is only Ωbary. some scientists have proposed that the dark matter might be made of axions.11 × 108 m−3 ) = 3. Various supersymmetric models predict the existence of massive nonbaryonic particles such as photinos. another with 0. neutrinos make up all the nonbaryonic dark matter. ν If the masses of all three neutrinos turn out to be significantly less than mν c2 ∼ 4 eV. and neutrinos make up less than 0. this implies that a mean neutrino mass of mν c2 ≈ 0. then the density parameter in neutrinos is only Ων ∼ 10−3 . with the difference in the squares of the masses of the two neutrinos being ∆(m2 c4 ) ≈ 3 × 10−5 eV2 .26. another with mν c2 = 4. and so forth.36 × 108 m−3 (8. gravitinos.0 εc. Particle physicists have provided several possible candidates for the role of dark matter.0 eV. as mentioned in section 2. The minimum neutrino masses consistent ν with these results would have one flavor with mν c2 ∼ 0. Evidence indicates that neutrinos do have some mass. consider the extension of the Standard Model of particle physics known as supersymmetry. indicate that electron neutrinos oscillate into some other flavor of neutrino. about twenty million cosmic neutrinos are zipping through your body.0 . (8. Observations of muon neutrinos created in the ν Earth’s atmosphere indicate that muon neutrinos oscillate into tau neutrinos. and the third with mν c2 = 4.000004 eV would be in agreement with the deduced values of ∆(m2 c4 ).0004 eV. and the third with mν c2 are this small.005 eV. In order to provide all the nonbaryonic mass in the universe. sneutrinos.4. the average neutrino mass would have to be Ωdm. the masses of the three species would have to be very nearly identical.56) would be necessary to provide all the nonbaryonic dark matter in the universe. gluinos.05 eV. axinos. with ∆(m2 c4 ) ≈ 3 × 10−3 eV2 . If. one neutrino flavor with mν c2 = 4.0 ≈ 0. for instance. None of these “inos” have yet been detected in . For instance. If the neutrino masses mν c2 ∼ 0.5% of the nonbaryonic dark matter.55) mν c2 = nν Given a density parameter in nonbaryonic dark matter of Ωdm. on the other hand.005 eV. But how much? Enough to contribute significantly to the energy density of the universe? The observations of neutrinos from the Sun. “like photons through a pane of glass”.26(5200 MeV m−3 ) ≈ 4 eV 3.176 CHAPTER 8. DARK MATTER This means that at any moment. then the bulk of the nonbaryonic dark matter in the universe must be made of some particle other than neutrinos. are known generically as Weakly Interacting Massive Particles. no convincing detections have been made – but the search goes on. ch. Rich (2001). So far. which makes them intrinsically difficult to detect.9 Since WIMPs. experimenters have set up WIMP detectors to discover cosmic WIMPs. Suggested reading [Full references are given in the “Annotated Bibliography” on page 286.5. do interact with atomic nuclei on occasion. How far away would you expect the nearest MACHO to be? How The acronym “MACHO”. Peacock (1999). 2. The fact that supersymmetric particles such as photinos have not yet been seen in particle accelerator experiments means that they must be massive (if they exist).1) Suppose it were suggested that black holes of mass 10−8 M made up all the dark matter in the halo of our Galaxy.8. the hypothetical supersymmetric particles interact with other particles only through gravity and through the weak nuclear force. with mc2 > 10 GeV. like neutrinos. 12: Dark matter in the universe. ch. chapter 4 gives a good review of gravitational lensing. Also. encountered in the previous section. How far away would you expect the nearest such black hole to be? How frequently would you expect such a black hole to pass within 1 AU of the Sun? (An order-of-magnitude estimate is sufficient.) Suppose it were suggested that MACHOs of mass 10−3 M (about the mass of Jupiter) made up all the dark matter in the halo of our Galaxy. both baryonic and nonbaryonic.4: A discussion of the dark matter candidates. WHAT’S THE MATTER? 177 laboratories. Like neutrinos. ch.] Liddle (1999). Particles which interact via the weak nuclear force. was first coined as a humorous riposte to the acronym “WIMP”. 9 . but which are much more massive than the upper limit on the neutrino mass. Problems (8. or WIMPs. 8: A brief sketch of methods for detecting dark matter. 5 Mpc. What is the average time between collisions for a galaxy in the Coma cluster? Is this time greater than or less than the Hubble time? .7.2 km s−1 . What is the mean number density of galaxies within the half-mass radius? Suppose that the typical cross-section of a galaxy is Σ ≈ 10−3 Mpc2 .178 CHAPTER 8. R = 1. DARK MATTER frequently would such a MACHO pass within 1 AU of the Sun? (Again. would you expect it to significantly affect the dynamics of our Galaxy’s halo? Explain why or why not. I noted that galaxies in rich clusters are poor standard candles. (8.) Repeat your calculation for a white dwarf (M = 2. on average.2) The Draco galaxy is a dwarf galaxy within the Local Group.4) If the halo of our Galaxy is spherically symmetric. What is the mass of the Draco galaxy? What is its mass-to-light ratio? Describe the possible sources of error in your mass estimate of this galaxy.5) In the previous chapter.0 = 0. because they tend to grow brighter with time as they merge with other galaxies. (8. (8. The red giant stars in the Draco galaxy are bright enough to have their line-of-sight velocities measured.5 × 107 m) and for a neutron star (M = 3.0 × 1030 kg. Its luminosity is L = (1. The Coma cluster contains N ≈ 1000 galaxies within its half-mass radius of rh ≈ 1. Let’s estimate the galaxy merger rate in the Coma cluster to see whether it’s truly significant.0 × 1030 kg. what is the mass density ρ(r) within the halo? If the universe contains a cosmological constant with density parameter ΩΛ.5 ± 2.0 × 1024 kg. R = 1. The measured velocity dispersion of the red giant stars in the Draco galaxy is σr = 10.4 × 106 m). before it collides with another galaxy? The velocity dispersion of the Coma cluster is σ ≈ 880 km s−3 . R = 6. Through what angle α is the light ray bent by gravitational lensing? (Ignore the refractive effects of the Earth’s atmosphere. an order-of-magnitude estimate will suffice.8) × 105 L and half its total luminosity is contained within a sphere of radius rh = 120 ± 12 pc.2 × 104 m).) (8.8 ± 0.3) A light ray just grazes the surface of the Earth (M = 6. How far will a galaxy in the Coma cluster travel. 04. most of the light in the universe comes not from the hot balls of gas we call “stars”.0 = Ωbary. (9. unknown to Olbers.2) It is particularly enlightening to compare the energy density and number density of photons to those of baryons (that is. the number density of CMB photons in the universe is large: nγ.725 K rather than at a temperature of a few thousand degrees Kelvin. actually is uniformly bright – it’s just uniformly bright at a temperature of T 0 = 2.0 = αT0 = 0. protons and neutrons). However.0 ≈ 0.04(5200 MeV m−3 ) ≈ 210 MeV m−3 .3) .11 × 108 m−3 . 4 εγ. The current energy density of the Cosmic Microwave Background.1) is only 5 × 10−5 times the current critical density. since the energy per CMB photon is small (hfmean = 6. The night sky.34 × 10−4 eV).0 = 4. he would not have formulated Olbers’ Paradox.Chapter 9 The Cosmic Microwave Background If Heinrich Olbers had lived in intergalactic space and had eyes that operated at millimeter wavelengths (admittedly a very large “if”). (9. 179 (9. thousands of times longer than human eyes can detect. Given a current density parameter for baryons of Ωbary.0 ≈ 0. but from the Cosmic Microwave Background (CMB).0 εc. At wavelengths of a few millimeters.261 MeV m−3 . the current energy density of baryons is εbary. It was isotropic and constant with time. The number density of baryons.4) The ratio of baryons to photons in the universe (a number usually designated by the Greek letter η) is.2) and (9. Ebary 939 MeV (9. by a ratio of roughly two billion to one. THE COSMIC MICROWAVE BACKGROUND Thus. reflected from an orbiting communications satellite. Penzias and Wilson found a slightly stronger signal than they expected when they turned the antenna toward the sky. 1 . 9.11 × 108 m−3 (9.1 Observing the CMB Although CMB photons are as common as dirt. Wilson and Penzias were puzzled until they were put in touch with Robert Dicke and his research group at Princeton University. The horn-reflector radio antenna which they used had previously been utilized to receive microwave signals.5) Baryons are badly outnumbered by photons in the universe as a whole. At the time of their discovery. if it Actually.4). The excess signal remained.0 210 MeV m−3 ≈ ≈ 0. since dirt is made of baryons. They even shooed away a pair of pigeons that had roosted in the antenna and cleaned up they later called “the usual white dielectric” generated by pigeons. They did everything they could think of to reduce “noise” in their system. when you stop to think of it. Dicke had deduced that the universe. η= nbary.180 CHAPTER 9. Note.22 m−3 . Penzias and Wilson were radio astronomers working at Bell Laboratories.0 4. this is more than a trillion times the mean energy of a CMB photon. Turning from telecommunications to astronomy. therefore. much commoner than dirt.35 cm. though. of wavelength λ = 7.22 m−3 ≈ 5 × 10−10 . so it couldn’t be associated with an isolated celestial source. that the rest energy of a proton or neutron is Ebary ≈ 939 MeV. ≈ nγ. from equations (9. is much lower than the number density of photons: nbary.0 = εbary.0 0. the energy density in baryons today is about 800 times the energy density in CMB photons.1 Arno Penzias and Robert Wilson were surprised when they serendipitously discovered the Cosmic Microwave Background. Roll. and Wilkinson in a companion letter in this issue. Penzias and Wilson wrote a paper for The Astrophysical Journal in which they wrote. Measuring the spectrum of the CMB. Dicke and his group were in the process of building a microwave antenna when Penzias and Wilson told them that they had already detected the predicted microwave radiation.2 In fact.” The companion paper by Dicke and his collaborators points out that the radiation could be a relic of an early. . is not a simple task. Thus. and had fallen into obscurity during the intervening years. should now be filled with microwave radiation.9. then be absorbed a microsecond away from the Earth’s surface by a water molecule in the atmosphere. 1965). unfortunately. hot. is tiny compared to the energy required to break up an atomic nucleus (∼ 1 MeV) or even the energy required to ionize an atom (∼ 10 eV). The CMB can be measured at wavelengths shorter than 3 cm by observing from high-altitude balloons or from the South Pole. 2 . OBSERVING THE CMB 181 started in a hot dense state.5 K higher than expected. This excess temperature is. “Measurements of the effective zenith noise temperature of the 20-foot horn-reflector antenna . the existence of the cosmic background radiation had actually been predicted by the physicist George Gamow and his colleagues as early as 1948. 1964 – April. 3 The South Pole is nearly 3 kilometers above sea level. and confirming that it is indeed a blackbody. however. within the limits of our observations. isotropic. and opaque state of the universe. The best way to measure the spectrum of the CMB.35 cm because that was the wavelength of the signals that Bell Labs had been bouncing off orbiting satellites. Penzias and Wilson were observing at a wavelength 40 times longer than the wavelength (λ ≈ 2 mm) at which the CMB spectrum reaches its peak. even with modern technology. the mean photon energy is comparable to the energy of vibration or rotation for a small molecule such as H2 O. and free from seasonal variations (July.1. CMB photons can zip along for more than 13 billion years through the tenuous intergalactic medium. unpolarized. The current energy per CMB photon. The CMB spectrum To give credit where it’s due. is to go completely above the damp atmosphere of the Earth. However. ∼ 6 × 10−4 eV. at 4080 Mc/s have yielded a value about 3. Peebles. Thus. Gamow’s prediction wasn’t acted on at the time he made it. where the combination of cold temperatures and high altitude3 keeps the atmospheric humidity low. dense. Penzias and Wilson observed the CMB at a wavelength λ = 7. . Microwaves with wavelengths shorter than λ ∼ 3 cm are strongly absorbed by water molecules. A possible explanation for the observed excess noise temperature is the one given by Dicke. The second instrument. The uncertainties in the measurement are smaller than the thickness of the line (from Fixsen et al. How close is very close? FIRAS could have detected fluctuations in the spectrum as small as ∆ / ≈ 10−4 .7 mm.1. The Diffuse InfraRed Background Experiment (DIRBE) was designed to measure radiation at the wavelengths 0. was first measured accurately over a wide range of wavelengths by the COsmic Background Explorer (COBE) satellite.3 mm. Result number one: At any angular position (θ.1mm < λ < 10 mm. the spectrum of the Cosmic Microwave Background is very close to that of an ideal blackbody. The third instrument.24 mm. 1996.182 CHAPTER 9. was used to measure the spectrum of the CMB in the range 0. and 9. called the Far InfraRed Absolute Spectrophotometer (FIRAS). at these wavelengths. a wavelength band which includes the peak in the CMB spectrum.1: The spectrum of the Cosmic Microwave Background. 576). COBE actually contained three different instruments. THE COSMIC MICROWAVE BACKGROUND Figure 9. Three important results came from the analysis of the COBE data. was designed to make full-sky maps of the CMB at three different wavelengths: λ = 3. as measured by the FIRAS instrument on the COBE satellite. ApJ. launched in 1989. it was primarily detecting stars and dust within our own Galaxy.001 mm < λ < 0. into an orbit 900 km above the Earth’s surface. 473. 5.6 mm. φ) on the sky. No deviations were found at this level within the wavelength range investigated by . called the Differential Microwave Radiometer (DMR). as illustrated in Figure 9. and in the other half the spectrum is slightly redshifted to lower temperatures. shown in the lower panel of Figure 9. distorted by the map projection. are small in amplitude. the Virgo cluster is being accelerated toward the Hydra-Centaurus supercluster.725 K . The mean temperature. although each point on the sky has a blackbody spectrum. with a speed vLG = 630 ± 20 km s−1 = 0. The combination of these two accelerations. in one half of the sky the spectrum is slightly blueshifted to higher temperatures. be T (θ. This peculiar velocity for the Local Group is what you’d expect as the result of gravitational acceleration by the largest lumps of matter in the vicinity of the Local Group. φ) ≡ . φ). the nearest supercluster to us. the remaining temperature fluctuations. Let the temperature of the CMB. has launched the Local Group in the direction of Hydra. (9. and for the orbital motion of our Galaxy relative to the center of mass of the Local Group (v ∼ 80 km s−1 ). working over the age of the universe. it is found that the Local Group is moving in the general direction of the constellation Hydra. In addition.2. 4 . the nearest big cluster to us.9.2 represents the darker. caused by the net motion of the COBE satellite relative to a frame of reference in which the CMB is isotropic. is 1 T = T (θ.1. 5 The distorted “yin-yang” pattern in the upper panel of Figure 9.4 That is. The unique contribution of COBE was the precision with which it measured the temperature distortion. φ) sin θ dθ dφ = 2. φ) − T (θ. (9. using aircraft-borne and balloon-borne detectors.7) T T The dipole distortion of the CMB was first detected in 1977. cooler (yin?) hemisphere of the sky and the hotter. The Local Group is being accelerated toward the Virgo cluster. for the orbital motion of the Earth around the Sun (v ∼ 30 km s−1 ).2. averaging over all locations. OBSERVING THE CMB 183 FIRAS. brighter (yang?) hemisphere. at 0.5 This dipole distortion is a simple Doppler shift. for the orbital motion of the Sun around the Galactic center (v ∼ 220 km s−1 ).2% of the speed of light. at a given point on the sky. After correcting for the orbital motion of COBE around the Earth (v ∼ 8 km s−1 ). Result number three: After the dipole distortion of the CMB is subtracted away.6) 4π The dimensionless temperature fluctuation at a given point on the sky is δT T (θ.0021c. Result number two: The CMB has the dipole distortion in temperature shown in the top panel of Figure 9. (Courtesy of NASA Goddard Space Flight Center and the COBE Science Working Group.2: Top panel: The fluctuations in temperature in the CMB. The band across the middle is due to emission from the disk of our own Galaxy. THE COSMIC MICROWAVE BACKGROUND Figure 9.) . as measured by COBE. Bottom panel: The fluctuations in temperature remaining after subtraction of the dipole due to the satellite’s proper motion.184 CHAPTER 9. To avoid muddle. dense. if the surface of the Earth were smooth to 11 parts per million. 6 .) The fact that the temperature of the CMB varies by only 30 microKelvin across the sky represents a remarkably close approach to isotropy. 7 Cosmologists sometimes grumble that this should really be called the epoch of “combination” rather than the epoch of “recombination”.6 The observations that the CMB has a nearly perfect blackbody spectrum and that it is nearly isotropic (once the Doppler dipole is removed) provide strong support for the Hot Big Bang model of the universe. If the universe did not go through such a phase. I will distinguish among three closely related (but not identical) moments in the history of the universe. since this is the very first time when electrons and ions combined to form neutral atoms. A background of nearly isotropic blackbody radiation is natural if the universe was once hot. the highest mountains would be just seventy meters above the deepest ocean trenches. 9.9. you might define it as the instant in time at which the number density of ions is equal to the number density of neutral atoms. the epoch of photon decoupling is the time at which the rate at which photons scatter from electrons becomes smaller than the Hubble parameter (which tells us the rate at To make an analogy.1 × 10−5 .8) (This analysis excludes the regions of the sky contaminated by foreground emission from our own Galaxy.7 Second. and nearly homogeneous. First. Numerically. then any explanation of the Cosmic Microwave Background will have to be much more contrived. the root mean square temperature fluctuation was δT T 2 1/2 = 1. as it was in the Hot Big Bang scenario. RECOMBINATION AND DECOUPLING 185 From the maps of the sky made by the DMR instrument aboard COBE. opaque. (9. the epoch of recombination is the time at which the baryonic component of the universe goes from being ionized to being neutral. we’ll have to examine fairly carefully the process by which the baryonic matter goes from being an ionized plasma to a gas of neutral atoms.2.2 Recombination and decoupling To understand in more detail the origin of the Cosmic Microwave Background. it was found that after subtraction of the Doppler dipole. and the closely related process by which the universe goes from being opaque to being transparent. Surrounding every observer in the universe is a last scattering surface. This is not. before stars had a chance to pollute the universe with heavy elements. thus. THE COSMIC MICROWAVE BACKGROUND Figure 9.3: An observer is surrounded by a spherical last scattering surface. a strictly accurate assumption. from which the CMB photons have been streaming freely. The probability that a photon will scatter from an electron is small once the expansion rate of the universe is faster than the scattering rate. 8 . When photons decouple. illustrated in Figure 9. Even at the time of recombination. I concede. being continuously redshifted.8 However. All the significant physics of recombination can be studied in a simplified In the next chapter. the presence of helium is merely a complicating factor. The photons of the CMB travel straight to us from the last scattering surface. and the universe becomes transparent. they cease to interact with the electrons. there was a significant amount of helium present. I will assume that the baryonic component of the universe consisted entirely of hydrogen at the epoch of recombination. the epoch of last scattering is very close to the epoch of photon decoupling. which the universe expands). we will examine how and why this helium was formed in the early universe. To keep things from getting too complicated. the epoch of last scattering is the time at which a typical CMB photon underwent its last scattering from an electron.3. with no further scattering by electrons. Third.186 CHAPTER 9. the fractional ionization X will depend on the balance between photoionization and radiative recombination. as well. A photon with an energy hf > Q is capable of photoionizing a hydrogen atom: H + γ → p + e− . At early times.9) The value of X ranges from X = 1 when the baryonic content is fully ionized to X = 0 when it consists entirely of neutral atoms. the fractional ionization of the universe was very close to X = 1. electrons. the number density of free electrons must be equal to that of free protons: ne = np . almost immediately. and with a ratio of photons to baryons of nearly two billion.9. any hydrogen atoms that happened to form by radiative recombination were very short-lived. they were blasted apart into their component electron and proton by a high-energy photon.2.10) This reaction can run in the opposite direction.6 eV. (9. The degree to which the baryonic content of the universe is ionized can be expressed as the fractional ionization X.7kT ≈ 60 eV.) The temperature of the background radiation at this time was was T ≈ 3 × 105 K. and the average photon energy was hfmean ≈ 2. forming a bound hydrogen atom while a photon carries away the excess energy: p + e− → H + γ . RECOMBINATION AND DECOUPLING 187 universe containing no elements other than hydrogen. To maintain charge neutrality in this hydrogen-only universe. . or of a naked hydrogen nucleus. Q = 13. One useful consequence of assuming that hydrogen is the only element is that there is now a single relevant energy scale in the problem: the ionization energy of hydrogen. (In the Benchmark Model. With such a high energy per photon. np + n H nbary nbary (9. and photons. let’s choose the moment when a = 10−5 . defined as X≡ np ne np = = . corresponding to a redshift z = 105 . Let’s travel back in time to a period before the epoch of recombination. The hydrogen can take the form of a neutral atom (designated by the letter H). then. this scale factor was reached when the universe was seventy years old.11) In a universe containing protons. a proton and an electron can undergo radiative recombination. otherwise known as a proton (designated by the letter p). For concreteness. (9. such as baryons. a3 a3 (9. The cross-section for Thomson scattering is the Thomson cross-section of σe = 6.13) ne σ e Since photons travel with a speed c. The mean free path of a photon – that is. λ When the baryonic component of the universe is fully ionized. the rate at which a photon undergoes scattering interactions is c (9. THE COSMIC MICROWAVE BACKGROUND When the universe was fully ionized. and the main interaction mechanism was Thomson scattering: γ + e− → γ + e − . (9. As long as photons scatter frequently from electrons.4 × 10−21 s−1 = . about three times a week.65 × 10−29 m2 . as long as they remain in thermal equilibrium. (9. thanks to the electrons’ interactions with the protons).0 σe c 4. with the protons as well. ne = np = nbary .0 = 0. a3 (9.14) Γ = = n e σe c . all have the same temperature T . The photons remain coupled to the electrons as long as their scattering rate.188 CHAPTER 9. indirectly.4 × 10−6 s−1 . goes as 1/a3 . the photons remain in thermal equilibrium with the electrons (and. Currently. photons would scatter from electrons at a rate Γ = 4.0 . the mean distance it travels before scattering from an electron – is 1 λ= .16) nbary.22 m−3 . then the electrons are being diluted by expansion .15) This means. that at a = 10−5 .12) The scattering interaction is accompanied by a transfer of energy and momentum between the photon and electron. the number density of baryons is nbary. the free electron density was ne = nbary = and the scattering rate for photons was Γ= nbary. and protons. is larger than H. so when the early universe was fully ionized. the rate at which the universe expands. The photons. The number density of conserved particles. When the photon scattering rate Γ drops below H. this is equivalent to saying that their mean free path λ is shorter than the Hubble distance c/H. for instance. photons interacted primarily with electrons. Γ. electrons. 1 × 10−20 s−1 = . Rather.3. Taking into account the transition from a radiation-dominated to a matter-dominated universe. as electrons combine with protons to form hydrogen atoms.1 × 10−10 s−1 . take place? It’s easy to do a quick and dirty approximation of the recombination temperature. then photons would have remained coupled to the electrons and protons until a relatively recent time. the CMB photons are too low in energy to keep the hydrogen ionized. we can compute that if hydrogen had remained fully ionized. and the resulting change in the expansion rate. THE PHYSICS OF RECOMBINATION 189 more rapidly than the photons can interact with them. caused by the abrupt plummeting of free electron density during the epoch of recombination. (9. If hydrogen remained ionized (and note the qualifying if ).4 × 10−6 s−1 at the same scale factor.18) This means. the photons were well coupled to the electrons and protons. that at a = 10−5 . Thus. The photons then decouple from the electrons and the universe becomes transparent. one could argue. it is a relatively sudden process.0 H2 = 4 . H= a2 a2 1/2 (9.0 2. During the early stages of the universe (a < arm ≈ 3 × 10−4 ) the universe was radiation dominated. and the consequent photon decoupling.023. must take place when the mean .9. 9. Once the photons are decoupled from the electrons and protons. and the Friedmann equation was Ωr. for instance. the decoupling of photons is not a gradual process. the baryonic portion of the universe is no longer compelled to have the same temperature as the Cosmic Microwave Background. then decoupling would have taken place at a scale factor a ≈ 0.3 The physics of recombination When does recombination. at such a low temperature. the Hubble parameter was H0 Ωr. Recombination. caused by the continuous lowering of free electron density as the universe expands. Since this is much smaller than the scattering rate Γ = 4. corresponding to a redshift z ≈ 42 and a CMB temperature of T ≈ 120 K. However.17) 2 H0 a Thus. the Hubble parameter was H = 2. the number density nx of particles with mass mx is given by the MaxwellBoltzmann equation nx = g x mx kT 2π¯ 2 h 3/2 exp − mx c2 kT . we expect the recombination temperature to depend on the baryonto-photon ratio η as well as on the ionization energy Q.7kT . For instance. (9. requires a smattering of statistical mechanics.7(8.19) Alas. one in 3 million will have E > 20kT .7) trailing off to high energies.21) as long as the particles are non-relativistic. Since the mean CMB photon energy is ∼ 2.6 eV Q ∼ ∼ 60. and be photoionized. . the average CMB photon is no longer able to photoionize hydrogen. about one photon in 500 will have E > 10kT . this reaction will be in statistical equilibrium.6 × 10−5 eV K−1 ) (9. THE COSMIC MICROWAVE BACKGROUND energy per photon of the Cosmic Microwave Background falls below the ionization energy of hydrogen. and one in 30 billion will have E > 30kT . It doesn’t take into account the fact that CMB photons are not of uniform energy – a blackbody spectrum has an exponential tail (see Figure 2. When a reaction is in statistical equilibrium at a temperature T . The vast swarms of photons that surround every newly formed hydrogen atom greatly increase the probability that the atom will collide with a photon from the high-energy tail of the blackbody spectrum. In equation (9. When this happens. this crude approximation is a little too crude to be useful. (9. as a function of η and T .6 eV.000 K . this line of argument would indicate a recombination temperature of Trec ∼ 13.7k 2. Q = 13. gx is the statistical weight of particle x.190 CHAPTER 9. Although extremely high energy photons make up only a tiny fraction of the CMB photons. Although the mean photon energy is 2. the total number of CMB photons is enormous – nearly 2 billion photons for every baryon. with kT mx c2 . 2. electrons. An exact calculation of the fractional ionization X.20) While the photons are still coupled to the baryonic component. with the photoionization rate (going from left to right) balancing the radiative recombination rate (going from right to left).7kT . Let’s start with the reaction that determines the value of X in the early universe: H+γ p + e− .21). Thus. THE PHYSICS OF RECOMBINATION 191 protons. we may write η= 9 np . This yields me kT 1−X = np X 2π¯ 2 h −3/2 exp Q kT . we can make the substitution nH = 1−X np . and e− . nH = np ne me kT 2π¯ 2 h −3/2 exp Q kT . . p. (9. and neutrons (and their anti-particles as well) all have a statistical weight gx = 2. we can make the substitution ne = np . In most cosmological contexts. Our next job is to convert the Saha equation into a relation among X.9). we can set mH /mp = 1. the binding energy Q = 13.26) If you are a true thermodynamic maven. you will have noted that equation (9. µ. are gp = ge = 2. From the definition of X (equation 9. Second. we can construct an equation which relates the number densities of these particles: nH gH = np ne gp g e mH mp me 3/2 kT 2π¯ 2 h −3/2 exp [mp + me − mH ]c2 kT .3. T . The resulting equation. (9. since the mass of an electron is small compared to that of a photon. Thus. we recall that η ≡ nbary /nγ .25) To eliminate np from the above equation.24) and from the requirement of charge neutrality. and η.23) is called the Saha equation. while the statistical weight of a hydrogen atom is gH = 4.9. Xnγ (9.21) omits the chemical potential term.6 eV is given by the formula (mp + me − mH )c2 = Q.22) can be simplified further. which appears in the most general form of the MaxwellBoltzmann equation. In a universe where hydrogen is the only element. the chemical potential is small enough to be safely neglected. the factor gH /(gp ge ) can be set equal to one. as it turns out.9 From the Maxwell-Boltzmann equation for H. corresponding to their two possible spin states. (9. and a fraction X of the hydrogen is in the form of naked protons. The statistical weights of the proton and electron. X (9. First.22) Equation (9. slightly higher than the melting point of tungsten.30) exp Q kT . THE COSMIC MICROWAVE BACKGROUND Since the photons have a blackbody spectrum. (9.31) If we define the moment of recombination as the exact instant when X = 1/2. kTrec = 0.26) and (9. (9. .243 kT hc ¯ 3 .404 π2 kT hc ¯ 3 = 0. however. was trec = 240. as shown in Figure 9. In degrees Kelvin. for which nγ = 2. when the age of the universe.84η kT me c2 3/2 (9. it proceeded fairly rapidly.192 CHAPTER 9.4. in the Benchmark Model. whose positive root is √ −1 + 1 + 4S .323 eV corresponds to a temperature Trec = 3740 K.243Xη hc ¯ 3 . (9. 42 (9.28) Substituting equation (9.25). then (assuming η = 5.84η 2 X me c2 3/2 exp Q kT .9 at a redshift z = 1475 10 Not that there was any tungsten around back then to be melted. η) = 3. we finally find the desired equation for X in terms of T and η: 1−X kT = 3.27) to find kT np = 0.27) we can combine equations (9.5 × 10−10 ) the recombination temperature is kTrec = 0. The fractional ionization goes from X = 0. Recombination was not an instantaneous process. (9.29) This is a quadratic equation in X. the exact value of η doesn’t strongly affect the value of Trec .323 eV = Q .10 The temperature of the universe had a value T = Trec = 3740 K at a redshift zrec = 1370. X= 2S where S(T.000 yr.32) Because of the exponential dependence of S upon the temperature.28) back into equation (9. 0 σe c . A baryon-to-photon ratio of η = 5.0 (1 + z)3 . the numerical value of the scattering rate is Γ(z) = 4.0 = 0.9 to X = 0. 2 H0 a (9.9. so the Hubble parameter is given by the relation Ωm. the universe is matter-dominated. the numerical value of the Hubble parameter during the epoch of recombination is H(z) = 1.5 × 10−10 is assumed.8 X . (9.04.0 H2 = 3 = Ωm.24 × 10−18 s−1 (1 + z)3/2 .35) Using Ωm.1 is ∆t ≈ 70.34) While recombination is taking place. (9. Since the number density of free electrons drops rapidly during the epoch of recombination.3.6 .4 × 10−21 s−1 X(z)(1 + z)3 .4 . In the Benchmark Model.0 = 0.36) . THE PHYSICS OF RECOMBINATION 193 1 .000 yr. the time of photon decoupling comes soon after the time of recombination. when the hydrogen is partially ionized.3.4: The fractional ionization X as a function of redshift during the epoch of recombination.1 at a redshift z = 1255.33) Using Ωbary. to X = 0. the time that elapses from X = 0. The rate of photon scattering. (9.2 0 1200 1300 1400 1500 1600 z Figure 9. is Γ(z) = ne (z)σe c = X(z)(1 + z)3 nbary. THE COSMIC MICROWAVE BACKGROUND The redshift of photon decoupling is found by setting Γ = H.000 yr in the Benchmark Model. H(a) a (9. the photoionization reaction is no longer in equilibrium. The time t for which τ = 1 is the time of last scattering.4). or (combining equations 9. if we detect a CMB photon at time t 0 .38) t The dimensionless number τ is the optical depth.39) using the fact that H = a/a.34 and 9.0 . we find that τ (a) = 1 a Γ(a) da = a ˙ 1 a Γ(a) da .36). when the age of the universe was tdec ≈ 350. when Γ starts to drop below H. zdec ≈ 1100. and the decoupling of photons is therefore delayed. corresponding to a temperature Tdec ≈ 3000 K. where Γ(t) is the scattering rate at time t. let’s content ourselves by saying.40) . the exact redshift of photon decoupling is somewhat smaller than this value. at redshifts smaller than ∼ 1200. in round numbers. The p + e− is in equilibrium. (9.38) from t to a.37) Using the value of X(z) given by the Saha equation (shown in Figure 9. X(zrec )2/3 (9. and represents the time which has elapsed since a typical CMB photon last scattered from a free electron. the fractional ionization X is larger than would be predicted by the Saha equation. During a brief time interval t → t + dt. the redshift of photon decoupling is found to be zdec = 1130. the expected number of scatterings it has undergone since an earlier time t is t0 τ (t) = Γ(t)dt . we can find the optical depth as ˙ a function of redshift by making the substitution 1 + z = 1/a: τ (z) = z 0 Γ(z) dz = 0. (9. Alternatively. the probability that a photon undergoes a scattering is dP = Γ(t)dt. As a consequence. When we examine the CMB with our microwave antennas. the photons we collect have been traveling straight toward us since the last time they scattered from a free electron.194 CHAPTER 9. 1 + zrec = 43.0035 H(z) 1 + z z 0 X(z)(1 + z)1/2 dz . Without going into the details of the non-equilibrium physics. Saha equation assumes that the reaction H + γ However. If we change the variable of integration in equation (9. In truth. Thus. 1. and last scattering took place when the universe was matter-dominated. Since the photons traveled about at the speed of light. kicking the electrons By the time the universe becomes Λ dominated. THE PHYSICS OF RECOMBINATION Table 9. 11 .35 Here. and are dependent on the cosmological model you choose.000 of its present age. emphasizing the fact that recombination.) When we look at the CMB. If you scoop up two photons from the CMB. For purposes of comparison. electrons.11 As it turns out.24 0. To sufficient accuracy for our purposes. Not all the CMB photons underwent their last scattering simultaneously. and protons coupled together. consisting of photons. so the Saha equation doesn’t action H + γ strictly apply. there existed a single photonbaryon fluid.047 0. Thus. we are getting an intriguing glimpse of the universe as it was when it was only 1/40. the universe doesn’t choreograph its microphysics that well.34) and (9. one may have undergone its last scattering at z = 1200. The epoch of photon decoupling marked an important change in the state of the universe. at z = 1000. I have made use of equations (9. we can see a little way into the “electron fog” which hides the early universe from our direct view. photon decoupling.35 0. the “last scattering surface” is really more of a “last scattering layer”. the free electron density has fallen to negligibly small levels. so using the Hubble parameter for a matter-dominated universe is a justifiable approximation in computing τ . the table also contains the time of radiation-matter equality. we can state that the redshift of last scattering was comparable to the redshift of photon decoupling: zls ≈ zdec ≈ 1100. while the other may have scattered more recently.9.36). the last scattering of a typical CMB photon occurs after the photoionization rep + e− falls out of equilibrium.3. Before photon decoupling. just as we can see a little way into a fog bank here on Earth. The relevant times of various events around the time of recombination are shown in Table 9. Note that all these times are approximate.1: Events in the early universe event redshift temperature (K) time radiation-matter equality 3570 9730 recombination 1370 3740 photon decoupling 1100 3000 last scattering 1100 3000 195 (megayears) 0. (I have chosen the Benchmark Model in calculating these numbers. all have provided maps of δT /T down to scales of δθ ∼ 10 arcminutes. however. a good dA = . stars. For instance. we see a Doppler shift in the radiation of the CMB. Thus. and the other dense knots of baryonic matter which make the universe so rich and strange today. Thus.2. Since the last scattering surface is at a redshift zls = 1100 1. shown in the top panel of Figure 9.41) δθ where dA is the angular-diameter distance to the last scattering surface. The angular size of the temperature fluctuations reflects in part the physical size of the density and velocity fluctuations at z ≈ 1100.196 CHAPTER 9. tell us that the universe was not perfectly homogeneous at the time of last scattering (at z ≈ 1100). one of photons and the other of neutral hydrogen. MAXIMA (a balloon-borne experiment). and BOOMERANG (a balloon-borne experiment launched from Antarctica). The angular size δθ of a temperature fluctuation in the CMB is related to a physical size on the last scattering surface by the relation . THE COSMIC MICROWAVE BACKGROUND before them as they went. 9. planets. instead of being kicked to and fro by the photons. they tended to smooth out any density fluctuations in the photon-baryon fluid smaller than the horizon.2. Because we are gravitationally accelerated towards the nearest large lumps of matter. The COBE DMR experiment had limited angular resolution. we are looking backward in time to an important epoch in the history of the universe – the epoch when the baryons. cosmologists. (9. shown in the bottom panel of Figure 9. More recent experiments have provided higher angular resolution. the photon-baryon fluid became a pair of gases. and thus form galaxies. were free to collapse gravitationally. free from the harassment of photons. results from the fact that the universe is not perfectly homogeneous today (at z = 0). the hydrogen gas was free to collapse under its own self-gravity (and the added gravitational attraction of the dark matter). DASI (an experiment located at the South Pole).4 Temperature fluctuations The dipole distortion of the Cosmic Microwave Background. The distortions on a smaller angular scale. when we look at the Cosmic Microwave Background. and was only able to detect temperature fluctuations larger than δθ ≈ 7◦ . Although the two gases coexisted spatially. they were no longer coupled together. After photon decoupling. φ) . the fluctuations at the time of last scattering were not gravitationally bound objects. or (1 + zls ) ≈ 40 Mpc today. shows δT /T as measured by COBE at low resolution over the entire sky. 1100 (9.9. Figure 9.000 Mpc ≈ 13 Mpc . The most important statistical property of δT /T is the correlation function C(θ).42) In the Benchmark Model. so the angular-diameter distance to the surface of last scattering is dA ≈ 14.22 Mpc δθ 1◦ (9. Consider the density fluctuations δT /T observed by a particular experiment.44) at the time of last scattering.04 Mpc at the time of last scattering. Thus. for instance. the fluctuations that gave rise to the fluctuations seen by COBE (with δθ > 7◦ ) had a proper size > 1. Since δT /T is defined on the surface of a sphere – the celestial sphere. the current horizon distance is dhor (t0 ) ≈ 14. (9. much larger than the biggest superclusters. the fluctuations seen by COBE correspond to physical scales of (1 + zls ) > 1700 Mpc today.45) T l=0 m=−l where Ylm (θ. but their statistical properties.4. DASI.41): dA ≈ dhor (t0 ) . fluctuations on the last scattering surface with an observed angular size δθ had a proper size = dA (δθ) = 13 Mpc δθ 1 rad = 0. φ) are the usual spherical harmonic functions. The higher-resolution experiments such as MAXIMA. and as measured by BOOMERANG at higher resolution over part of the sky.43) Thus. and BOOMERANG see fluctuations corresponding to scales as small as ≈ 0. What concerns cosmologists is not the exact pattern of hot spots and cold spots on the sky. TEMPERATURE FLUCTUATIONS approximation to dA is given by equation (7. Thus.6 Mpc. However. zls 197 (9. Consider two points on the last .5. they were expanding along with the universal Hubble expansion.000 Mpc. φ) = alm Ylm (θ. in this case – it is useful to expand it in spherical harmonics: ∞ l δT (θ. about the size of today’s superclusters. The white spots are regions 300 microKelvin warmer than average. the black spots are regions 300 microKelvin cooler than average. after the dipole and galaxy foreground are subtracted. Lower right: the temperature fluctuations measured by the BOOMERANG experiment over a region ∼ 20◦ across. THE COSMIC MICROWAVE BACKGROUND Figure 9. (Courtesy of the BOOMERANG collaboration.) .5: Upper left: the temperature fluctuations measured by the COBE DMR instrument.198 CHAPTER 9. 45). l=0 (9. The limited angular resolution of available observations is what makes the spherical harmonic expansion of δT /T . For a given experiment. Generally speaking. so useful.9. the multipole l is interchangeable. they are in the directions n and ˆ n .47) where Pl are the usual Legendre polynomials: P0 (x) = 1 P1 (x) = x 1 P2 (x) = (3x2 − 1) 2 (9. multiply together the values of δT /T at the two points. (9. for all practical purposes. Relative to an observer. shown in equation (9. since they tell us about the fluctuations present at the time of last scattering.4.46) C(θ) = T T n·ˆ =cos θ ˆn If cosmologists knew the precise value of C(θ) for all angles from θ = 0 to θ = 180◦ . Unfortunately. then average the product over all points separated by the angle θ: δT δT (ˆ ) (ˆ ) n n . It is the moments with l ≥ 2 which are of the most interest to cosmologists. The l = 0 (monopole) term of the correlation function vanishes if you’ve defined the mean temperature correctly. TEMPERATURE FLUCTUATIONS 199 scattering surface. a term Cl is a measure of temperature fluctuations on the angular scale θ ∼ 180◦ /l. the value of Cl will be nonzero for angular scales larger than the resolution of the experiment and smaller than the patch of sky examined. . and are separated by an angle θ given by the relation cos θ = n · n . they would have a complete statistical description of the temperature fluctuations over all angular scales. To ˆ ˆ ˆ find the correlation function C(θ). a measured correlation function C(θ) can be broken down into its multipole moments Cl . The l = 1 (dipole) term results primarily from the Doppler shift due to our motion through space. Using the expansion of δT /T in spherical harmonics. with the angular scale θ. the correlation function can be written in the form 1 C(θ) = 4π ∞ (2l + 1)Cl Pl (cos θ) . In this way. Thus.48) and so forth. the CMB measurements which tell us about C(θ) contain information over only a limited range of angular scales. 6. THE COSMIC MICROWAVE BACKGROUND Figure 9. we will examine. Rev. 2002. and how we can extract cosmological information from the temperature anisotropy of the Cosmic Microwave Background.6. ∆T . corresponding to an angular size of ∼ 1◦ . Figure 9. very briefly. Note that the temperature fluctuation has a peak at l ∼ 200. In the next section. In presenting the results of CMB observations. D). & Zaldarriaga. Tegmark.49) 2π since this function tells us the contribution per logarithmic interval in l to the total temperature fluctuation δT of the Cosmic Microwave Background. as shown in Figure 9.200 CHAPTER 9. The detailed shape of the ∆T versus l curve. is a plot of ∆T as a function of the logarithm of l. the physics behind the temperature fluctuations. it is customary to plot the function 1/2 l(l + 1) Cl ∆T ≡ T (9. contains a wealth of information about the universe at the time of photon decoupling. Phys. .6: The anisotropy of the CMB temperature. expressed as a function of the multipole l (from Wang. which combines data from a large number of CMB experiments. 4 : 1.5 What causes the fluctuations? At the time of last scattering. since εγ ∝ a−4 ∝ (1 + z)4 . at the time of last scattering. was εγ (zls ) = Ωγ.51) It is no coincidence that the peak in the ∆T versus l curve (Figure 9.50) where I have used equation (9. ≈ dA 13 Mpc (9.8 × 1012 MeV m−3 . 1. is the Hubble distance.0 (1 + zls )3 (9.52) since the energy density of matter is ∝ a−3 ∝ (1 + z)3 . A patch of the last scattering surface with this physical size will have an angular size. The density of nonbaryonic dark matter at the time of last scattering was εdm (zls ) = Ωdm. and baryons having energy densities in roughly the ratio 6. c/H(zls ) ≈ 3. These temperature fluctuations arise from the gravitational effect of primordial density fluctuations in the distribution of nonbaryonic dark matter. (9.54) The density of photons at the time of last scattering. (9.55) Thus.4 : 1. Plugging in the appropriate numbers.0 c. a particularly interesting length scale.8 × 1011 MeV m−3 . we find that εdm (zls ) ≈ (0.015 rad ≈ 1◦ . The origin of temperature fluctuations with θ > θH (l < 180) is different from those with θ < θH (l > 180). as seen from Earth. εdm > εγ > εbary .5.26)(5200 MeV m−3 )(1101)3 ≈ 1.0 c.0 (1 + zls )4 ≈ 3.24 × 10−18 s−1 (1101)3/2 (9.9. (9. The density of baryonic matter at the time of last scattering was εbary (zls ) = Ωbary.6 × 1021 m ≈ 0.6) occurs at an angular scale θ ∼ θH .2 Mpc ≈ 0.2 Mpc .8 × 1011 MeV m−3 .0 (1 + zls )3 ≈ 2. WHAT CAUSES THE FLUCTUATIONS? 201 9. zls ≈ 1100.36) to compute the Hubble parameter at the redshift of last scattering.53) equivalent to a mass density of ∼ 3 × 10−18 kg m−3 . with dark matter. of θH = c/H(zls ) 0. photons. Consider first the large-scale fluctuations – those with angular size θ > θH . cosmologically speaking.0 × 108 m s−1 ≈ 6. .0 εc. c2 (9. the spatially varying component of the energy density. Consider the fate of a CMB photon which happens to be at a local minimum of the potential at the time of last scattering. In the Newtonian approximation. the fluctuations in its density would necessarily have given rise to fluctuations in the gravitational potential. it loses energy. A detailed general relativistic calculation. (9.57) Unless the distribution of dark matter were perfectly smooth at the time of last scattering. δε. The cool (redshifted) spots on the COBE temperature map correspond to minima in δΦ at the time of last scattering. tells us that δT 1 δΦ = . the origin of the temperature fluctuations in the CMB is complicated by the behavior of the photons and baryons. and thus is blueshifted. THE COSMIC MICROWAVE BACKGROUND The nonbaryonic dark matter dominated the energy density . Conversely. first performed by Sachs and Wolfe in 1967.202 CHAPTER 9. (9. Then we could write the energy density of the dark matter as ¯ ε(r) = ε + δε(r) . (Minima in the gravitational potential are known colloquially as “potential wells”. the hot (blueshifted) spots correspond to maxima in δΦ. and consequently is redshifted. of the universe at the time of last scattering. a photon which happens to be at a potential maximum when the universe became transparent gains energy as it falls down the “potential hill”. and hence the gravitational potential. On smaller scales (θ < θH ). Suppose that the density of the nonbaryonic dark matter at the time of last scattering was not perfectly homogeneous.56) where ε is the spatially averaged energy density of the nonbaryonic dark ¯ matter. and δε is the local deviation from the mean.58) T 3 c2 Thus.) In climbing out of the potential well. the temperature fluctuations on large angular scales (θ > θH ≈ 1◦ ) give us a map of the potential fluctuations δΦ present at the time of last scattering. but varied as a function of position. The creation of temperature fluctuations by variations in the gravitational potential is known as the Sachs-Wolfe effect. . The link between δε and δΦ is Poisson’s equation: 2 (δΦ) = 4πG δε . gives rise to a spatially varying gravitational potential δΦ. in tribute to the work of Sachs and Wolfe. the highest peak in the ∆T curve (at l ∼ 200 or θ ∼ 1◦ in Figure 9. whose energy density is only about a third that of the dark matter. the liberated photons will be slightly cooler than average. The inward and outward oscillations of the photonbaryon fluid are called acoustic oscillations. electrons. if the photon-baryon fluid within a potential well is at maximum expansion at the time of decoupling. and hence have angular sizes of Note that if the size of the well is larger than c/H(zls ). the photon-baryon fluid. will not have time to fall to the center by the time of last scattering tls ∼ 1/H(zls ). The cycle of compression and expansion which is set up continues until the time of photon decoupling. its pressure starts to rise. will be hotter than average. it will fall to the center of the well. however. since they represent a type of standing sound wave in the photon-baryon fluid. These potential wells had proper sizes ∼ c/H(zls ) at the time of last scattering. and why the temperature fluctuations on these large scales are dictated purely by the distribution of dark matter. depending on whether the photon-baryon fluid was moving away from our location or toward it at the time of photon decoupling.5. and the liberated photons.12 As the photon-baryon fluid is compressed by gravity.6) represents the potential wells within which the photon-baryon fluid had just reached maximum compression at the time of last scattering. which travels at a speed < c. If the photonbaryon fluid finds itself in a potential well of the dark matter. WHAT CAUSES THE FLUCTUATIONS? 203 Consider the situation immediately prior to photon decoupling. Eventually. If the photon-baryon fluid is in the process of expanding or contracting at the time of decoupling. its density will be higher than average. Thus. The equation-of-state parameter wpb of the photon-baryon fluid is intermediate between the value w = 1/3 expected for a gas containing only photons and the value w = 0 expected for a gas containing only cold baryons and electrons. Generally speaking. 12 . and protons together make a single photon-baryon fluid.9. Conversely. the pressure is sufficient to cause the fluid to expand outward. since T ∝ ε1/4 . If the photon-baryon fluid within a potential well is at maximum compression at the time of photon decoupling. As the expansion continues. This is why the motions of the photons and baryons are irrelevant on scales θ > θH . the pressure drops until gravity causes the photon-baryon fluid to fall inward again. The photons. rather than under its own self-gravity. however. the Doppler effect will cause the liberated photons to be cooler or hotter than average. Computing the exact shape of the ∆T versus l curve expected in a particular model universe is a rather complicated chore. the photon-baryon fluid moves primarily under the gravitational influence of the dark matter. and the computed density of matter in clusters (as discussed in Chapter 8). . with Ωm. As we’ll see in the next chapter.7 shows the values of Ωm. or l > 180.3 and ΩΛ.0 ≈ 0. this baryon density is consistent with that found from the entirely different arguments of primordial nucleosynthesis.7. with the currently available data. A spatially flat universe. or Ω0 = 1.0 = 0. The location and amplitude of the highest peak in Figure 9. As it turns out. In a negatively curved universe (κ = −1). the peak would be seen at an angle θ > 1◦ . Note that the shaded areas which show the best fit to the CMB data are roughly parallel to the Ω = 1 line (representing κ = 0 universes) and roughly perpendicular to the region permitted by the type Ia supernova results.04 ± 0. THE COSMIC MICROWAVE BACKGROUND ∼ θH as seen by us at the present day. The oval curve marked “68%” in Figure 9. The observed position of the peak is consistent with κ = 0. the supernova results. The amplitude of the peak is dependent on the sound speed of the photon-baryon fluid prior to photon decoupling.0 and ΩΛ.02 .7 represents the region consistent with both the CMB results and the supernova results. the angular size θ of an object of known proper size at a known redshift is smaller than it is in a positively curved universe (κ = +1). yields Ωbary. the amplitude of the peak is a useful diagnostic of the baryon density of the universe. Since the sound speed of the photon-baryon fluid is √ cs = wpb c . If the universe were negatively curved. Figure 9. (9. the angle θ ≈ 1◦ at which it’s located is dependent on the spatial curvature of the universe. The angular size corresponding to the highest peak in the ∆T versus l curve gives useful information about the density parameter Ω and the curvature κ. This happy concurrence is the basis for the “Benchmark Model” which I have been using in this book. Detailed analysis of the ∆T curve.0 ≈ 0.0 which are permitted by the present CMB data.204 CHAPTER 9. agrees with the CMB results. (9.59) and the equation-of-state parameter wpb is in turn dependent on the baryonto-photon ratio. if the universe were positively curved. the peak in ∆T would be seen at an angle θ < 1◦ .60) assuming a Hubble constant of H0 = 70 km s−1 Mpc−1 .6 is a very useful cosmological diagnostic. or l < 180. 2001.5. The light gray area near the Ω = 1 line shows the bestfitting values of Ωm. 561.0 and ΩΛ.9. for the combined CMB and SN Ia data (from Stompor et al.0 permitted by the MAXIMA and COBE DMR data. ApJ. at the designated confidence levels. 95%. L7). The SN Ia data (as seen earlier in Figure 7.7: The values of Ωm.0 . WHAT CAUSES THE FLUCTUATIONS? 205 Figure 9. . The curves marked 68%. and 98% give the best fits.6) are also shown.0 and ΩΛ. ch. THE COSMIC MICROWAVE BACKGROUND Suggested reading [Full references are given in the “Annotated Bibliography” on page 286.] Coles (1999). The ionization energy of helium (that is. How much does this change in η affect the computed value of the recombination temperature Trec . Liddle (1999). At what temperature would the fractional ionization of the helium be X = 1/2? Assume that η = 5. what is the numerical value of f? (9.2) Suppose the temperature T of a blackbody distribution is such that kT Q. Rich (2001).3) Imagine that at the time of recombination.5 × 10−10 and that the number density of He++ is negligibly small. affects the recombination temperature in the early universe. η.6 eV.9: A discussion of the anisotropies in the CMB. (The relevant statistical weight factor for the ionization of helium is gHe /(ge gHe+ ) = 1/4. the baryonic portion of the universe consisted entirely of 4 He (that is. 9: A simplified overview of photon decoupling and the origin of the CMB. in the range 3000 K < T < 4500 K. Part I. ch. 7.) .1) The purpose of this problem is to determine how the uncertainty in the value of the baryon-to-photon ratio. Problems (9. What fraction f of the blackbody photons are energetic enough to ionize hydrogen? If T = Trec = 3740 K. helium with two protons and two neutrons in its nucleus). then assuming η = 8 × 10−10 . first make the plot assuming η = 4 × 10−10 .206 CHAPTER 9. ch.6 eV is the ionization energy of hydrogen. the energy required to convert neutral He to He+ ) is QHe = 24. where Q = 13. and their use as diagnostics of cosmological parameters. if we define Trec as the temperature at which X = 1/2? (9. 5: An insightful overview of the origin of the temperature anisotropies in the CMB. Plot the fractional ionization X as a function of temperature. 5. To what redshift z∗ does this value of t∗ correspond? . and that the baryonic component of the universe is pure hydrogen.61) equal one? For simplicity. τ= t0 t∗ Γ(t)dt = t0 t∗ ne (t)σe cdt . at some time between trec and t0 . the intergalactic medium must have been reionized. WHAT CAUSES THE FLUCTUATIONS? 207 (9. Thus.9.4) What is the proper distance dp to the surface of last scattering? What is the luminosity distance dL to the surface of last scattering? Assume that the Benchmark Model is correct. (9. For what value of t∗ does the optical depth of reionized material. The fact that we can see small fluctuations in the CMB places limits on how early the reionization took place. and that the redshift of the last scattering surface is zls = 1100.5) We know from observations that the intergalactic medium is currently ionized. Assume that the baryonic component of the universe instantaneously became completely reionized at some time t∗ . (9. assume that the universe is spatially flat and matter-dominated. 35 Myr). or the surface of the Sun. From a theoretical viewpoint.1) or equivalently kT (t) ≈ 1 MeV 208 t 1s −1/2 . when radiation is strongly dominant over matter. is given by the convenient relation t T (t) ≈ 10 K 1s 10 −1/2 . many properties of the early universe should be quite simple.8 × 10−4 . the expansion of the universe has the simple power-law form a(t) ∝ t1/2 . (10. at scale factors a arm ≈ 2. However. our curiosity is piqued. although some information is carried by cosmic rays and by neutrinos. and we wish to find out what conditions are like in the opaque regions so tantalizingly hidden from our direct view. the opacity of the early universe prevents us from directly seeing what the universe was like at t < tls . most of what we know about the universe beyond our Solar System has come in the form of photons.000 yr. or times t trm ≈ 47.Chapter 10 Nucleosynthesis & the Early Universe The Cosmic Microwave Background tells us a great deal about the state of the universe at the time of last scattering (tls ≈ 0. The temperature of the blackbody photons in the early universe. which decreases as T ∝ a−1 as the universe expands. For instance. Photons are the “messenger boys” of astronomy. Looking at the last scattering surface is like looking at the surface of a cloud. (10.2) . designed to accelerate protons to an energy of 1 TeV. hydrogen (H) nuclei all have Z = 1. we sample over 31 orders of magnitude in particle energy. atomic physicists. since it provided particles of very high energy without running up an enormous electricity bill or having Congress threaten to cut off its funding. when considering the ionization of atoms. in Batavia. The ionization energies of other elements (that is. the early universe is referred to as “the poor man’s particle accelerator”. The ionization energy of hydrogen is Q = 13. by studying the universe as it expands. lithium (Li) nuclei have Z = 3. typically deal with energies of ∼ 10 eV. and so on. Thus. as we have already noted.1.3) The Fermi National Accelerator Laboratory. in round numbers. (10.7kT (t) ≈ 3 MeV t 1s −1/2 209 . Illinois. the entire universe was a Tevatron. through the complete . the most interesting energy scale is the ionization energy of an atom. Thus. For instance. The total number of nucleons within an atomic nucleus is called the mass number. The proton number Z of a nucleus determines the atomic element to which that nucleus belongs. or a thousand times the rest energy of a proton.1 Nuclear physics and cosmology As the universe has expanded and cooled. Well. For instance. where Z ≥ 1 and N ≥ 0. Within this wide energy range. NUCLEAR PHYSICS AND COSMOLOGY Thus the mean energy per photon was Emean (t) ≈ 2. some energies are of more interest than others to physicists. and is given by the formula A = Z + N . when the universe was one picosecond old (t = 10−12 s). that’s 1012 eV. the mean energy per photon has dropped from Emean (tP ) ∼ EP ∼ 1028 eV at the Planck time to Emean (t0 ) ≈ 6 × 10−4 eV at the present day. to physicists studying recombination and photoionization. An atomic nucleus contains Z protons and N neutrons. beryllium (Be) nuclei have Z = 4. is justly proud of its “Tevatron”. but with the much higher energy processes of fission and fusion (splitting or merging atomic nuclei). 10. Thus.10.6 eV. the energy required to remove the most loosely bound electron in the neutral atom) are roughly comparable. Protons and neutrons are collectively called nucleons. Nuclear physicists are concerned not with ionization and recombination (removing or adding electrons to an atom). helium (He) nuclei have Z = 2. so protons and neutrons must have fused to form deuterium when the temperature dropped sufficiently far below the binding energy of deuterium (BD = 2. we may also write p in place of 1 H when considering nuclear reactions. or deuterium. NUCLEOSYNTHESIS & THE EARLY UNIVERSE periodic table. and is symbolized as 4 He. compared to other atomic nuclei.8 MeV. compared to other light nuclei (that is. Figure 10. Thus. so studies of nuclear fusion and fission deal with an energy scale of ∼ 8 MeV (a typical binding energy per nucleon). when a neutron and a proton are bound together to form a deuterium nucleus.30 MeV. Note that 4 He. just as studies of ionization and recombination deal with an energy scale of ∼ 10 eV (a typical ionization energy).22 MeV is released: p+n D + 2.07 MeV. (Since an ordinary hydrogen nucleus is nothing more than a proton. D.) Ordinary helium contains two protons and two neutrons.4) The deuterium nucleus is not very tightly bound.1 plots the binding energy per nucleon (B/A) for atomic nuclei with different mass numbers. The most tightly bound nuclei are those of 56 Fe and 62 Ni. which both have B/A ≈ 8. Although all atoms of a given element have the same number of protons in their nuclei. (Since deuterium is very frequently mentioned in the context of nuclear fusion. it is the energy released when a nucleus is fused together from individual protons and neutrons. The epoch of recombination must have been preceded by an epoch of nuclear fusion. is symbolized as 1 H. Thus. just as electrons and protons combined to form neutral hydrogen atoms when the temperature dropped sufficiently far below the ionization energy of hydrogen (Q = 13. A particular isotope of an element is designated by prefixing the mass number A to the symbol for that element. contains one proton and one neutron. nuclei with A ≤ 10). (10.22 MeV . Nuclei less massive than iron or nickel can release energy by fusion – merging into heavier nuclei. is relatively tightly bound.6 eV). nuclei more massive than iron or nickel can release energy by fission – splitting into lighter nuclei. with one proton and no neutrons. commonly called . For instance. The binding energy B of a nucleus is the energy required to pull it apart into its component protons and neutrons. and a binding energy per nucleon of B/A = 7. and is symbolized as 2 H. Moreover. For instance.210 CHAPTER 10. it has its own special symbol. Equivalently. with a total binding energy of B = 28. an energy of BD = 2. different isotopes of an element can have different numbers of neutrons.) Heavy hydrogen.22 MeV). a standard hydrogen nucleus. NUCLEAR PHYSICS AND COSMOLOGY 211 8 B/A (MeV) 6 4 2 D 0 1 4 He 3 H 3 He 10 A 100 Figure 10.10.1: The binding energy per nucleon (B/A) as a function of the number of nucleons (protons and neutrons) in an atomic nucleus.1. . Note the absence of nuclei at A = 5 and A = 8. corresponding to a time tnuc ≈ 300 s. From an energy viewpoint.24. we find that they are primarily 4 He. When we look at astronomical objects of different sorts. three-fourths of the baryonic component (by mass) is still in the form of unbound protons. Moreover.28. the Sun is made of recycled interstellar gas.6 × 105 (3740 K) ≈ 6 × 108 K. Yp is the mass density of 4 He divided by the mass density of all the baryonic matter. the entire saga of Big Bang Nucleosynthesis takes place when the universe is only a few minutes old. For instance. but it certainly gives the right order of magnitude. the helium fraction before nucleosynthesis begins in stars) is usually expressed as the dimensionless number Yp ≡ ρ(4 He) . baryonic objects such as stars and gas clouds are all at least 24% helium. The First Three Minutes. then proceeds to form heavier nuclei by successive acts of fusion. As indicated in the title of Steven Weinberg’s classic book. deuterium synthesis occurred at a temperature Tnuc ≈ 1. Nucleosynthesis in the early universe starts by the fusion of neutrons and protons to form deuterium. This estimate. we find a minimum value of Y = 0.5) That is. as a rough estimate. a relatively lightweight nucleus compared to 56 Fe and 62 Ni. ρbary (10. the synthesis of deuterium to occur at a temperature 1.6×105 times higher than the recombination temperature Trec = 3740 K. we would expect. your helium fraction is 24%. is that it was shockingly inefficient. when we look for nuclei heavier than 1 H. That is.6 × 105 . gives a temperature slightly too low. as we’ll see when we do the detailed calculations. The primordial helium fraction of the universe (that is. The Sun’s atmosphere has a helium fraction (by mass) of Y = 0. the preferred universe would be one in which the baryonic matter consisted of an iron-nickel alloy. 1 .1 Condensed objects which have undergone chemical or physical fractionation can be much lower in helium than this value. One thing we can say about Big Bang Nucleosynthesis. However. Currently. or 1 H. That is.212 CHAPTER 10. we do not live in such a universe. NUCLEOSYNTHESIS & THE EARLY UNIVERSE the epoch of Big Bang Nucleosynthesis (BBN). after taking a look at the present-day universe. which was contaminated by helium formed in earlier generations of stars. Since the binding energy of deuterium is larger than the ionization energy of hydrogen by a factor BD /Q = 1. Obviously. via the interactions n + νe p + e− (10. However. neutrons and protons were in equilibrium with each other.9) and n + e+ p + νe . their number density is given by The ‘half-life’. once the universe was several hours old. At that time. a neutron which is bound into a stable atomic nucleus is preserved against decay. so there were positrons as well as electrons present at t = 0. That is. it contained essentially no free neutrons.9) and (10. the temperature was T ≈ 3 × 1010 K.29 MeV . 2 .2.1 s. is related to the decay time by the relation t1/2 = τn ln 2 = 617 s.2 Since the energy Qn released by the decay of a neutron into a proton is greater than the rest energy of an electron (me c2 = 0.7) (10. because they’ve been tied up in deuterium.10). Let’s consider the state of the universe when its age is t = 0. a fraction f = exp(−t/τ n ) will remain. the time it takes for half the neutrons to decay. With a decay time of only fifteen minutes.1 s.51 MeV). if you start out with a population of free neutrons. helium. and other atoms. NEUTRONS AND PROTONS 213 10. and the mean energy per photon was Emean ≈ 10 MeV. ¯ (10.10) As long as neutrons and protons are kept in equilibrium by the reactions shown in equations (10. the remainder of the energy is carried away by the kinetic energy of the electron and the energy of the electron anti-neutrino. the existence of a free neutron is as fleeting as fame. created by pair production: γ+γ e− + e + . This energy is much greater than the rest energy of a electron or positron. A free neutron is unstable.1 s. decaying via the reaction n → p + e − + νe . ¯ (10. after a time t.6) The decay time for a free neutron is τn = 890 s. (10. There are still neutrons around today.10.2 Neutrons and protons The basic building blocks for nucleosynthesis are neutrons and protons.8) At t = 0. The rest energy of a neutron is greater than that of a proton by a factor Qn = mn c2 − mp c2 = 1. the cross-sections are small. The cross-sections for weak interactions have the temperature dependence σw ∝ T 2 . corresponding to T 1. there would be only one neutron for every million protons.11) and (10. The number density of neutrons is then nn = g n mn kT 2π¯ 2 h 3/2 exp − mn c2 kT (10.12) Since the statistical weights of protons and neutrons are equal. from equations (10. (10. First. at the temperatures we are considering. neutrons and protons do not remain in equilibrium for nearly that long. is nn = np mn mp 3/2 exp − (mn − mp )c2 kT .10). Second. At temperatures kT Qn = 10 1.214 CHAPTER 10. NUCLEOSYNTHESIS & THE EARLY UNIVERSE the Maxwell-Boltzmann equation. However. as the temperature starts to drop below 1.9) and (10. (10. protons begin to be strongly favored.5 × 10 K and t 1 s. involve the interaction of a baryon with a neutrino (or anti-neutrino). the neutron-to-proton ratio.29 MeV. The interactions which mediate between neutrons and protons in the early universe. If the neutrons and protons remained in equilibrium.11) and the number density of protons is np = g p mp kT 2π¯ 2 h 3/2 mp c2 exp − kT . there will be no great loss in accuracy if we set this factor equal to one.14) = exp − np kT illustrated as the solid line in Figure 10. the equilibrium neutron-to-proton ratio has the particularly simple form nn Qn . then by the time the universe was six minutes old. A typical cross-section for .29 MeV. the difference in rest energy of the neutron and proton is (mn − mp )c2 = Qn = 1.5 × 1010 K. However. (10. with gp = gn = 2.12). shown in equations (10. Neutrinos interact with baryons via the weak nuclear force. as discussed in section 9. Thus.3. (mn /mp )3/2 = 1.13) The above equation can be simplified. and the neutron-to-proton ratio plummets exponentially.2. the number of neutrons is nearly equal to the number of protons.002. the temperature falls at the rate T ∝ a(t)−1 ∝ t−1/2 .01 nn/np . the interaction of a neutrino with any other particle via the weak nuclear force is 2 kT −47 2 . the dotted line gives the value after freezeout. An exact calculation of the temperature Tfreeze at .0001 10 10 −5 −6 100 10 9 T (10 K) 1 Figure 10. NEUTRONS AND PROTONS 215 1 .10.2. and thus the cross-sections for weak interactions diminish at the rate σw ∝ t−1 . the neutrinos decouple from the neutrons and protons. and the ratio of neutrons to protons is “frozen” (at least until the neutrons start to decay. (10. The number density of neutrinos falls at the rate nν ∝ a(t)−3 ∝ t−3/2 .2: Neutron-to-proton ratio in the early universe.15) σw ∼ 10 m 1 MeV (Compare this to the Thomson cross-section for the interaction of electrons via the electromagnetic force: σe = 6.001 freezeout . When Γ ≈ H.) In the radiationdominated universe.16) Meanwhile.1 . (10.65 × 10−29 m2 . and hence the rate Γ with which neutrons and protons interact with neutrinos via the weak force falls rapidly: Γ = nν cσw ∝ t−5/2 . at times t ∼ τn ). the Hubble parameter is only decreasing at the rate H ∝ t−1 . The solid line assumes equilibrium. of order σw . for two reasons. the “freezeout temperature” turns out to be kTfreeze = 0. since protons are all positively charged. First. since it is a weak interaction. once the temperature drops below Tfreeze . they must surmount the Coulomb barrier between them in order to fuse. The universe reaches this temperature when its age is tfreeze ∼ 1 s. fusion in the Sun is a very slow process. It is the scarcity of neutrons relative to protons that explains why Big Bang Nucleosynthesis was so incomplete.17) At times tfreeze < t τn . the neutron-proton fusion reaction is p+n D+γ . If you pick out any particular . A neutron will fuse with a proton much more readily than a proton will fuse with another proton. NUCLEOSYNTHESIS & THE EARLY UNIVERSE which Γ = H requires a knowledge of the exact cross-section of the proton and neutron for weak interactions. a positron must be emitted (to conserve charge). and thus has a minuscule cross-section. this is a strong interaction (one involving the strong nuclear force). Using the best available laboratory information. It’s possible. Second. for instance.8 MeV. of course.29 MeV 0. When two protons fuse to form a deuterium nucleus. However.216 CHAPTER 10. It’s happening in the Sun.8 MeV ≈ 0. even as you read this sentence. this means that an electron neutrino must also be emitted (to conserve electron quantum number). the cross-section for proton-proton fusion. is frozen at the value Qn nn = exp − np kTfreeze ≈ exp − 1. to coax two protons into fusing with each other. The rate of proton-proton fusion is much slower than the rate of neutronproton fusion. is minuscule compared to the crosssection for neutron-proton fusion. or Tfreeze = 9 × 109 K. The proton-proton fusion reaction can be written as p + p → D + e + + νe . leaving three-fourths of the baryons in the form of unfused protons. The neutron-to-proton ratio.2 (10. (10.18) The involvement of a neutrino in this reaction tells us that it involves the weak nuclear force. By contrast. The cross-section for interactions involving the strong nuclear force are much larger than for those involving the weak nuclear force. there was one neutron for every five protons in the universe. (10.19) No neutrinos are involved. The remaining 8 protons. if f ≡ nn /np . we can consider a representative group of 2 neutrons and 10 protons. The mass fraction of 4 He will then be 4 1 Ymax = = . will remain unfused. then the maximum possible value of Yp is Ymax = 2f /(1 + f ). it has only one chance in ten billion of undergoing fusion during the next year. by strong contrast. though. this is because of the scarcity of neutrons relative to protons. In this approximation. Thus. and even those high-speed protons have only a tiny probability of quantum tunneling through the Coulomb barrier of another proton and fusing with it. ¯ (10. (10. that BBN proceeds until every free neutron is bonded into an atomic nucleus. with 0 ≤ f ≤ 1. I should also note that the rate of neutronneutron fusion in the early universe is negligibly small compared to the rate of neutron-proton fusion.2. the presence of a neutrino (an electron anti-neutrino. For the sake of completeness. the temperature drops as T ∝ t−1/2 and the density of baryons drops as nbary ∝ t−3/2 . To compute the maximum possible value of Yp .10. but mainly it is because of the small cross-section for neutronneutron fusion. although their is no Coulomb barrier between neutrons. is a stable environment. as a lowest order approximation. we can compute the maximum possible value of Yp . n + n → D + e − + νe . In part. with the leftover protons remaining solitary. Given the alacrity of neutron-proton fusion when compared to the leisurely rate of proton-proton and neutron-neutron fusion. In the early universe. in this case) tells us this is an interaction involving the weak nuclear force. though. The core of the Sun. The 2 neutrons can fuse with 2 of the protons to form a single 4 He nucleus. Given a neutron-to-proton ratio of nn /np = 1/5. it’s in hydrostatic equilibrium. and its temperature and density change only slowly with time. The reaction governing neutron-neutron fusion is Again. the fraction of the baryon mass in the form of 4 He. we can state. NEUTRONS AND PROTONS 217 proton in the Sun’s core. After less than an hour.20) . Big Bang Nucleosynthesis is a race against time.21) 12 3 More generally. suppose that every neutron present after the proton-neutron freezeout is incorporated into a 4 He nucleus. Only exceptionally fast protons have any chance of undergoing fusion. the temperature and density have dropped too low for fusion to occur. the neutron-neutron fusion rate is tiny. and thus reduce the amount of 4 He produced. (10. (10. they will not contribute to Yp .22 MeV . the energy released (and carried away by a gamma ray) is the binding energy of a deuterium nucleus: BD = (mn + mp − mD )c2 = 2. there are various factors which act to reduce the actual value of Yp below its theoretical maximum. Big Bang Nucleosynthesis takes place through a series of two-body reactions. as well as the abundances of other isotopes. are still strongly coupled to the protons and neutrons. if some neutrons escape fusion altogether. The neutron-toproton ratio is nn /np = 0.22) should have a haunting familiarity if you’ve just read Chapter 9. The neutrinos. that we didn’t really understand the process of proton-neutron freezeout. Next. are now decoupled from the rest of the universe. However. 10. Fortunately.22) When a proton and a neutron fuse. that too will reduce Yp . the fact that the observed value of Yp is less than Ymax is not worrisome. building heavier nuclei step by step. The photons. a photon with energy ≥ BD can photodissociate a deuterium nucleus into its component proton and neutron.23) Conversely. just after proton-neutron freezeout is complete. First. much of the statistical mechanics we will need is just a rehash of what we used when studying recombination. if nucleosynthesis didn’t take place immediately after freezeout at t ≈ 1 s. that would be a cause for worry. which ceased to interact with electrons about the same time they stopped interacting with neutrons and protons. however. or end in nuclei lighter than 4 He (such as D or 3 He). . The essential first step in BBN is the fusion of a proton and a neutron to form a deuterium nucleus: p+n D+γ . Finally. it might mean. In order to compute Yp accurately. then the spontaneous decay of neutrons would inevitably lower the neutron-to-proton ratio.3 Deuterium synthesis Let’s move on to the next stage of Big Bang Nucleosynthesis. NUCLEOSYNTHESIS & THE EARLY UNIVERSE If the observed value of Yp = 0. if nucleosynthesis goes on long enough to produce nuclei heavier than 4 He. The reaction shown in equation (10.218 CHAPTER 10.24 were greater than the predicted Ymax . it will be necessary to consider the process of nuclear fusion in more detail. for example. The time is t ≈ 2 s.2. A major difference between nucleosynthesis and recombination. To acceptable accuracy. The statistical weight factor of the deuterium nucleus is gD = 3. Around the time of recombination. with a gamma-ray photon being emitted. we can make the substitution [mp +mn −mD ]c2 = BD . Q = 13. the relative numbers of free protons. (10.24) p + e− A comparison of equation (10. Around the time of deuterium synthesis. The binding energy of deuterium. for instance.22) with equation (10. and that ionized hydrogen is favored in the limit kT → ∞. free electrons.10. 3 . is 160. (10.22): gD nD = np nn gp gn mD mp mn 3/2 kT 2π¯ 2 h −3/2 exp [mp + mn − mD ]c2 kT . free neutrons. you can release much more energy by fusing atomic nuclei than by simply shuffling electrons around. is between the energy scales involved. and deuterium nuclei is given by an equation directly analogous to equation (9. many of the equations used to analyze recombination can be re-used to analyze deuterium nucleosynthesis.3 Despite the difference in energy scales. −3/2 Q nH me kT exp = .6 eV. we may write mp = mn = mD /2. with an ultraviolet photon being emitted. and neutral hydrogen atoms is given by the Saha equation. two particles become bound together to form a composite object. a proton and neutron are bonded together by the strong nuclear force to form a deuterium nucleus.000 times greater than the ionization energy of hydrogen.200. the relative numbers of free protons. of course. a proton and electron are bonded together by the electromagnetic force to form a neutral hydrogen atom. In the case of photoionization.25) 2 np ne kT 2π¯ h which tells us that neutral hydrogen is favored in the limit kT → 0.23). In the case of nucleosynthesis.26) From equation (10. DEUTERIUM SYNTHESIS 219 it has the same structural form as the reaction governing the recombination of hydrogen: H +γ .3. BD = 2. These substitutions yield the nucleosynthetic As the makers of bombs have long known.24) shows that in each case. in comparison to gp = gn = 2 for a proton or neutron. (10. with the excess energy carried away by a photon.000 eV. (10. Before the start of deuterium synthesis. Thus.220 CHAPTER 10. To define a precise temperature Tnuc at which the nucleosynthesis of deuterium takes place.5η mn c2 3/2 exp BD kTnuc .8η 0.30) This function is plotted in Figure 10. mn kT nD =6 np nn π¯ 2 h −3/2 exp BD kT .27) holds true.28).5 × 10−10 . Just as recombination is a process which takes a finite length of time.31) .3. 5 out of 6 baryons (or ∼ 83%) were in the form of unbound protons. that is. and that free protons and neutrons are favored in the limit kT → ∞. As long as equation (10. Even today. if we don’t want to be fanatical about accuracy.28) We can write the deuterium-to-neutron ratio as a function of T and the baryon-to-photon ratio η if we make some simplifying assumptions.29) exp BD kT . we find that the deuteriumto-proton ratio is a relatively simple function of temperature: nD kT ≈ 6. (10. we know that ∼ 75% of all the baryons in the universe are in the form of unbound protons. NUCLEOSYNTHESIS & THE EARLY UNIVERSE equivalent of the Saha equation. the temperature at which half the free neutrons have been fused into deuterium nuclei. though. The temperature Tnuc of deuterium nucleosynthesis can be found by solving the equation kTnuc 1 ≈ 6. so is nucleosynthesis. (10. to define Tnuc as the temperature at which nD /nn = 1. the deuterium-to-neutron ratio can be written as nD mn kT = 6np nn π¯ 2 h −3/2 exp BD kT .5η nn mn c2 3/2  . (10.27) which tells us that deuterium is favored in the limit kT → 0.8ηnγ = 0. we can write np ≈ 0. (10.243 hc ¯  3  Substituting equation (10. assuming a baryon-to-photon ratio of η = 5.29) into equation (10. we need to define what we mean by “the nucleosynthesis of deuterium”.8nbary kT = 0. It is useful. 10. .0001 .7 9 T (10 K) .1 .01 .6 Figure 10.8 .3.9 . DEUTERIUM SYNTHESIS 221 10000 1000 100 10 nD/nn 1 .27) is assumed to hold true. The nucleosynthetic equivalent of the Saha equation (equation 10.001 .3: The deuterium-to-neutron ratio during the epoch of deuterium synthesis. Note that the time delay until the start of nucleosynthesis. Once a significant amount of deuterium forms. a deuterium nucleus can fuse with a proton to form 3 He: D+p 3 He + γ .6 × 108 K. emitting an electron and an electron anti-neutrino in the process. (10.36) . τn = 890 s. D+D 3 H+p .30).066 MeV.15 . it is more likely that the interaction of two deuterium nuclei will end in the formation of a tritium nucleus (with the emission of a proton). and η = 5.2 (10. also known as “tritium”: D+n 3 H+γ . (10. BD = 2. Deuterium nuclei can also fuse with each other to form 4 He: D+D 4 He + γ . is not negligible compared to the decay time of the neutron. corresponding to Tnuc ≈ 7. there are many possible nuclear reactions available. the decay time of tritium is approximately 18 years.32) This in turn lowers the maximum possible 4 He mass fraction from Ymax ≈ 0. The temperature drops to this value when the age of the universe is tnuc ≈ 200 s. NUCLEOSYNTHESIS & THE EARLY UNIVERSE With mn c2 = 939. tritium can be regarded as effectively stable. neutron decay has slightly decreased the neutron-to-proton ratio from nn /np = 1/5 to nn exp(−200/890) 0.33 to Ymax ≈ 0. it can fuse with a neutron to form 3 H.35) However.5 × 10−10 . np 5 + [1 − exp(−200/890)] 5. By the time nucleosynthesis actually gets underway.4 Beyond deuterium The deuterium-to-neutron ratio nD /nn does not remain indefinitely at the equilibrium value given by equation (10.33) Alternatively.34) Tritium is unstable.8 ≈ ≈ ≈ 0. 10.22 MeV. However. tnuc ≈ 200 s.6 MeV. during the brief time that Big Bang Nucleosynthesis lasts. it spontaneously decays to 3 He.222 CHAPTER 10. For instance.27. (10. (10. the temperature of deuterium synthesis is kTnuc ≈ 0. it won’t work. D. For instance. (10. they are all involve the strong nuclear force. they are converted to 4 He by reactions such as H+p 3 He + n 3 H+D 3 He + D 3 4 4 He + γ He + γ 4 He + n 4 He + p . 5 He and 5 Li are not stable nuclei. however.4.38) involves neutrinos.37) There is never a large amount of 3 H or 3 He present during the time of nucleosynthesis. .33) through (10. (10. By contrast.41) The synthesis of nuclei with A > 7 is hindered by the absence of stable nuclei with A = 8. (10.10.1. there are no stable nuclei with A = 5. Thus.42) the 8 Be nucleus falls back apart into a pair of 4 He nuclei with a decay time of only τ = 3 × 10−16 s. 4 He is exceptionally tightly bound. 4 He is resistant to fusion with protons and neutrons. For such a light nucleus. (10. Soon after they are formed. BEYOND DEUTERIUM or the formation of a 3 He nucleus (with the emission of a neutron). if 8 Be is made by the reaction 4 He + 4 He → 8 Be . Small amounts of 6 Li and 7 Li. the orderly march of nucleosynthesis to heavier and heavier nuclei reaches a roadblock. once nucleosynthesis begins. Thus. and 3 He are all efficiently converted to 4 He. If you try to fuse a proton or neutron to 4 He. 3 H. (10.40) and 4 He + 3 H 7 In addition.39) (10. Once 4 He is reached. and have large cross-sections and fast reaction rates.38) None of the post-deuterium reactions outlined in equations (10. D+D 3 223 He + n . are made by reactions such as 4 He + D 6 Li + γ Li + γ . as illustrated in Figure 10. small amounts of 7 Be are made by reactions such as 4 He + 3 He 7 Be + γ . the two stable isotopes of lithium. The precise yields of the different isotopes involved in BBN are customarily calculated using a fairly complex computer code. T85. which follows the mass fraction of different isotopes as the universe expands and cools. By the time the temperature has dropped to T ∼ 4 × 108 K. A baryon-to-photon ratio of η = 5. at T 109 K.4: Mass fraction of nuclei as a function of time during the epoch of nucleosynthesis. Big Bang Nucleosynthesis is essentially over. Fusion of heavier nuclei occurs much less rapidly. The complexity is necessary because of the large number of possible reactions which can occur once deuterium has been formed. NUCLEOSYNTHESIS & THE EARLY UNIVERSE Figure 10. fusion up to the tightly bound 4 He nucleus proceeds very rapidly. all of which have temperature-dependent cross-sections. at t ∼ 10 min.224 CHAPTER 10. 3 He. 12). The results of a typical BBN code. and 4 He are formed. the point is eventually reached where significant amounts of 3 H. almost all the baryonic matter is in the form of free protons and neutrons. however. Nearly all the baryons are in the form of free protons or 4 He nuclei. As the deuterium density climbs upward. (2000) Physica Scripta.1 × 10−10 is assumed (from Tytler et al.4. Initially. is shown in Figure 10. The bottom line is that once deuterium begins to be formed. . there’s a good deal of bookkeeping involved. Thus. the primordial helium fraction. BEYOND DEUTERIUM 225 The small residue of free neutrons decays into protons. Conversely. we know immediately that the baryon-to-photon ratio can’t be as small as η ∼ 10−12 . the direct production of 7 Li by the fusion of 4 He and 3 H is a decreasing function of η. Within the range of η plotted in Figure 10.) Very small amounts of 6 Li.5. (The 7 Be is later converted to 7 Li by electron capture: 7 Be + e− → 7 Li + νe . as shown in Figure 10. that is. getting an earlier start means that nucleosynthesis is more efficient at producing 4 He. BBN would be extremely inefficient. and 7 Li depend on various physical parameters. 3 He.) The yields of D. The dependence of the 7 Li density on η is more complicated. If it were. This is because the deuterium abundance is strongly dependent on η in the range of interest.5. Pinning down the value of η more accurately requires making accurate observations of the primordial densities of the light elements.01). determining η with a fair degree of accuracy would . we know that the baryon-to-photon ratio can’t be as large as η ∼ 10−7 . Yp . and Yp would be near its maximum permissible value of Ymax ≈ 0. while the indirect production of 7 Li by 7 Be electron capture is an increasing function of η. determining the deuterium abundance with only modest accuracy will enable us to determine η fairly well. the densities before nucleosynthesis in stars started to alter the chemical composition of the universe. a tribute to the incomplete nature of Big Bang Nucleosynthesis. Small amounts of D. By contrast. Note that larger values of η produce larger values for Yp (the 4 He mass fraction) and smaller values for the deuterium density. A high baryon-to-photon ratio increases the temperature Tnuc at which deuterium synthesis occurs. it is most useful to determine the primordial abundance of deuterium. as explained above. 7 Li. and we would expect only tiny amounts of helium to be produced (Yp < 0. In determining the value of η. 4 He. Since BBN is a race against the clock as the density and temperature of the universe drop. Thus. has only a weak dependence on η for the range of interest.5. Broadly speaking. and hence gives an earlier start to Big Bang Nucleosynthesis. nucleosynthesis would have taken place very early (before neutrons had a chance to decay).10. The net result is a minimum in the predicted density of 7 Li at η ≈ 3 × 10−10 . 3 H. they depend on the baryon-to-photon ratio η. If it were. leaving less D and 3 He as “leftovers”. the universe would be essentially deuterium-free.33. and 3 He are left over. A plot of the mass fraction of various elements produced by Big Bang Nucleosynthesis is shown in Figure 10.4. 6 Li. Most importantly. and 7 Be are made. Thus. (The 3 H later decays to 3 He. and Turner (2001) ApJ. and 7 Li expressed as a fraction of the H number density. . and the number densities of D. L1).226 CHAPTER 10. The width of each line represents the 95% confidence interval in the density (from Burles. 552. D+3 He. Nollett. NUCLEOSYNTHESIS & THE EARLY UNIVERSE Figure 10.5: The mass fraction of 4 He. If an intergalactic gas cloud contains no detectable stars. Currently.4 Deuterium abundances are customarily given as the ratio of the number of deuterium atoms to the number of hydrogen atoms (D/H). the greater mass of the nucleus causes a small isotopic shift in the electron’s energy levels.0 ± 0. we will see a strong absorption line at λH (1 + zcl ).6 × 10−5 .000 ordinary hydrogen atoms. When an interstellar gas cloud collapses gravitationally to form a star. Thus. the transition of the atom’s electron from the ground state (n = 1) to the next higher energy level (n = 2). that is. the most promising way to find the primordial value of D/H is to look at the spectra of distant quasars. In an ordinary hydrogen atom (1 H).4) × 10 −5 . λD = 121.10. there’s one deuterium atom for every 60. we just want to use the quasar as a flashlight to illuminate the intergalactic gas clouds which lie between it and us.6 × 10−5 . In a deuterium atom.4. 4 . Neutral hydrogen atoms within these intergalactic clouds will absorb photons whose energy corresponds to the Lyman-α transition.57 nm. that is. and has very low levels of elements heavier than lithium. BEYOND DEUTERIUM 227 require measuring Yp with fanatic precision. the first fusion reactions that occur involve the fusion of deuterium into helium. the Lyman-α transition in deuterium corresponds to a slightly shorter wavelength. astronomers find an average value D/H ≈ 1.54 nm. A major problem in determining the primordial deuterium abundance is that deuterium is very easily destroyed in stars. due to absorption from ordinary hydrogen. due to absorption from deuterium. This There are no mechanisms known that will create deuterium in significant amounts after Big Bang Nucleosynthesis is complete. We don’t care about the deuterium within the quasar itself. When we look at light from a quasar which has passed through an intergalactic cloud at redshift zcl . As a consequence. the Lyman-α transition corresponds to a wavelength λH = 121. the Sun and the interstellar medium are contaminated with material that has been cycled through stellar interiors. in the local interstellar medium within our Galaxy. Thus. the abundance of deuterium in the universe tends to decrease with time. we expect the primordial deuterium-to-hydrogen value to have been (D/H)p > 1. For instance. and a much weaker absorption line at λD (1 + zcl ). instead. Detailed studies of the strength of the absorption lines in the spectra of different quasars give results which are consistent with the ratio (D/H) = (3. and hasn’t been driven downward by the effects of fusion within stars. we can hope that its D/H value is close to the primordial value. However. A mean photon energy of 2. Remember.7kTnuc ≈ 0.0 is not the uncertainty in the baryon density. The mass density of baryons at the time of BBN was ρbary (tnuc ) = Ωbary.0 = εbary.0 ρc.47) 7 × 108 2.43) (10.007 kg m−3 .0 = 210 ± 20 MeV m−3 . that the energy density at the time of BBN was almost entirely in the form of radiation. you are bombarded with photons of about that energy .02 m−3 .5) × 10−10 . but the uncertainty in the critical density (which in turn results from the fact that H0 is not particularly well determined).0 210 ± 20 MeV m−3 = 0. = εc. though.44) (10.5 ± 0.0 = (mp c2 )nbary. (10. or about twice the density of water.46) This energy density corresponds to a mass density of εnuc /c2 ≈ 2000 kg m−3 .0 5200 ± 1000 MeV m−3 translates into a baryon-to-photon ratio of η = (5. it’s equal to the density of the Earth’s stratosphere. (10.725 3 ≈ (0.04 ± 0. A density of several grams per cubic meter is not outlandishly high. The value of η can be converted into a value for the current baryon density by the relation nbary. εbary.2 × 10−27 kg m−3 ) ≈ 0. to acceptable accuracy.0 Tnuc T0 3 (10.04)(9.45) Note that the largest source of uncertainty in the value of Ωbary.228 CHAPTER 10. NUCLEOSYNTHESIS & THE EARLY UNIVERSE Since most of the baryons are protons. by everyday standards. by everyday standards. we can write.0 = ηnγ. The current density parameter in baryons is thus Ωbary.0 = 0.01 .5 Baryon – antibaryon asymmetry The results of Big Bang Nucleosynthesis tell us what the universe was like when it was relatively hot (Tnuc ≈ 7 × 108 K) and dense: 4 εnuc ≈ αTnuc ≈ 1033 MeV m−3 . 10.2 MeV is not outlandishly high.23 ± 0. n ¯ which in turn was tiny compared to the number density of photons. is a remarkably small number. the number density of antibaryons (¯ and p) was tiny compared to the number density of baryons. The laws of physics demand the presence of antiprotons (¯). has its origin in the physics of the very early universe. or a “down” quark and an “anti-down” quark. The baryon-to-photon ratio. for instance. such that δq ≡ 5 nq − n q ¯ nq + n q ¯ 1. though.5 × 10−10 . containing one “antin 5 up” quark and two “anti-down” quarks apiece. an “up” quark and an “anti-up” ¯ quark. it is found that the universe has an extremely large excess of protons and neutrons over antiprotons and antineutrons (and hence an excess of quarks over antiquarks).10. . It’s also worthy of remark that the universe seems to have a strong preference for baryons over antibaryons. however. BARYON – ANTIBARYON ASYMMETRY 229 when you have your teeth X-rayed at the dentist.5. but formed a sea of free quarks (sometimes referred to by the oddly culinary name of “quark soup”). the universe seems to have a strong preference for photons over baryons. suppose there were a very tiny asymmetry between quarks and antiquarks. When the temperature of the early universe was greater than kT ≈ 150 MeV. Some of the initial conditions for Big Bang Nucleosynthesis. “anti-up” quarks. η ≈ 5. when the quark soup was piping hot.48) where q and q could represent. as well as antineutrons (¯ ). and photons were nearly equal to each other. nor is “anti-down” equivalent to “up”. quarks and antiquarks were constantly being created by pair production and destroyed by mutual annihilation: γ+γ q+q . “down” quarks. the quarks which it contained were not confined within baryons and other particles. containing two “anti-up” quarks and one p “anti-down” quark apiece. The physics of Big Bang Nucleosynthesis is well understood.49) Note that an “anti-up” quark is not the same as a “down” quark. During this period of quark pair production. ¯ (10. are rather puzzling. as they are today. During the first few microseconds of the universe. However. At the time of Big Bang Nucleosynthesis. (10. In practice. “anti-down” quarks. the numbers of “up” quarks. This imbalance. nantibary nbary nγ . .1) Suppose the neutron decay time were τn = 89 s instead of τn = 890 s. Suggested reading [Full references are given in the “Annotated Bibliography” on page 286. however. with all other physical parameters unchanged. Weinberg (1993) The First Three Minutes: Gives some of the historical background to the development of Big Bang Nucleosynthesis. there would be a residue of quarks with number density nq ∼ δq . as far back as the year 1967. the product of the annihilations. However. 5. however. quark-antiquark pairs would no longer be produced.000. if there were 1. the precise mechanism by which the quarks first developed their few-parts-per-billion advantage over antiquarks still remains to be found.000 quarks for every 999. Estimate Ymax . the very strong asymmetry between baryons and antibaryons today and the large number of photons per baryon are both products of a tiny asymmetry between quarks and antiquarks in the early universe.997 antiquarks in the early universe. because of the small excess of quarks over antiquarks. 6: Puts nucleosynthesis in the larger context of particle physics and thermodynamics.999. would be surrounded by roughly 2 billion photons. three lucky quarks in a billion would be left over after the others encountered antiquarks and were annihilated. The existing antiquarks would then annihilate with the quarks. (10. Thus.50) nγ Thus.000. The exact origin of the quark-antiquark asymmetry in the early universe is still not exactly known. the resulting baryon-to-photon ratio would be η ∼ 5 × 10−10 . Problems (10. The physicist Andrei Sakharov. was the first to outline the necessary physical conditions for producing a small asymmetry.] Bernstein (1995).230 CHAPTER 10. 4. NUCLEOSYNTHESIS & THE EARLY UNIVERSE As the universe expanded and the quark soup cooled. After the three quarks were bound together into a baryon at kT ≈ 150 MeV. ch. The leftover quarks. BARYON – ANTIBARYON ASYMMETRY 231 the maximum possible mass fraction in 4 He. (Unfortunately. assuming that the fusion of H into 4 He is the only significant energy source? If the baryonic mass of our Galaxy is M ≈ 1011 M .) Let’s see if you can reproduce Gamow’s line of argument. the maximum possible mass fraction in 4 He. will this increase or decrease T0 for fixed values of Tnuc and t0 ? Explain your answer.4 × 1010 L . (10.4) The total luminosity of the stars in our Galaxy is L ≈ 1. Gamow knew that nucleosynthesis must have taken place at a temperature Tnuc ≈ 109 K. by what amount has the helium fraction Y of our Galaxy been increased over its primordial value Y4 = 0. How much energy has our Galaxy emitted in the form of starlight during that time? Most stars are powered by the fusion of H into 4 He.24? (10.5. (10.10.2. Assume that the universe is flat and contains only radiation.3) A fascinating bit of cosmological history is that of George Gamow’s prediction of the Cosmic Microwave Background in 1948. (10. Suppose that the luminosity of our Galaxy has been constant for the past 10 Gyr. by the time the CMB was actually discovered in the 1960’s. with the release of 28. assuming that all available neutrons are incorporated into 4 He nuclei. with all other physical parameters unchanged.5) In section 10. his prediction was premature.29 MeV. his prediction had fallen into obscurity.129 MeV instead of Bn = 1. Estimate Ymax .4 MeV for every helium nucleus formed. what was the energy density ε at the time of nucleosynthesis? What was the Hubble parameter H at the time of nucleosynthesis? What was the time tnuc at which nucleosynthesis took place? What is the current temperature T0 of the radiation filling the universe today? If the universe switched from being radiation-dominated to being matter-dominated at a redshift zrm > 0. assuming that all available neutrons are incorporated into 4 He nuclei. With these assumptions. and that the age of the universe is currently t0 ≈ 10 Gyr. How many helium nuclei have been created within stars in our Galaxy over the course of the past 10 Gyr. it is asserted that the maximum possible value of the .2) Suppose the difference in rest energy of the neutron and proton were Bn = (mn − mp )c2 = 0. as pointed out in Chapter 5. What is the approximate interaction cross-section σw for one of these cosmic neutrinos? Suppose you had a large lump of 56 Fe (with density ρ = 7900 kg m−3 ). regardless of the type of particle the neutrino interacts with.) . Prove that this assertion is true. would a cosmic neutrino travel through the iron before interacting with a proton. NUCLEOSYNTHESIS & THE EARLY UNIVERSE primordial helium fraction is Ymax = 2f . (10. neutron.51) where f = nn /np ≤ 1 is the neutron-to-proton ratio at the time of nucleosynthesis.6) The typical energy of a neutrino in the Cosmic Neutrino Background. is Eν ∼ kTν ∼ 5 × 10−4 eV.232 CHAPTER 10. neutrons. 1+f (10. on average. or electron? (Assume that the cross-section for interaction is simply σw . What is the number density of protons. and electrons within the lump of iron? How far. The abundances of light elements such as deuterium and helium tell us about the universe at the time of Big Bang Nucleosynthesis (znuc ≈ 3 × 108 . So far.35 Myr). and supernovae at relatively small redshift (z < 6) tell us about the universe at times t > 1 Gyr. the observation that primordial gas clouds are roughly one-fourth helium by mass. called the flatness problem.” The horizon problem can be summarized by the statement.” Just as every triumphant general must be reminded of his flawed. and ΩΛ. tells us that we have a fair understanding of what was happening at the time of neutron-proton freezeout (zfreeze ≈ 4 × 109 . the horizon problem. “The universe is nearly flat today. imperfect nature. The flatness problem can be summarized by the statement. lest he become insufferably arrogant. The properties of the Cosmic Microwave Background tell us about the universe at the time of last scattering (zls ≈ 1100. “The universe is nearly isotropic and 233 . whispering in his ear. and was even flatter in the past. “Remember. Whenever a conquering general made a triumphal progress into ancient Rome. has three underlying problems.0 ) remaining to be worked out. In fact. with only minor details (such as the exact determination of H0 . rather than being all helium or all hydrogen. tls ≈ 0. quasars.Chapter 11 Inflation & the Very Early Universe The observed properties of galaxies. The standard Hot Big Bang scenario. Ωm. in which the early universe was radiation-dominated. so every triumphant theory should be carefully inspected for flaws and imperfections.0 . tfreeze ≈ 1 s). you are mortal. my description of the Hot Big Bang scenario has been a triumphal progress. tnuc ≈ 3 min). and the monopole problem. a slave stood behind him. 1 The flatness problem Let’s start by examining the flatness problem. invoke coincidence by saying that the initial conditions for the universe just happened to produce Ω0 ≈ 1 today. 11.0 H(t)2 = 4 + 3 . the value Ω0 = 10−6 or Ω0 = 106 without violating any laws of physics.29.” To see why these simple statements pose a problem to the standard Hot Big Bang scenario.0 Ωm.4) When the universe was dominated by radiation and matter.2 aren’t that close to one. (11. H(t)2 a(t)2 |1 − Ω0 | ≤ 0. the Hubble parameter was given by equation (6.1) and (11.2) R0 H 0 The results of the type Ia supernova observations and the measurements of the CMB anisotropy are consistent with the value (see Figure 9.3) (11. After all.2). “The universe is apparently free of magnetic monopoles.234 CHAPTER 11. By combining equations (11. 2 R0 a(t)2 H(t)2 (11.8 or Ω0 = 1.1) (Here. 2 H0 a a (11.” The monopole problem can be summarized by the statement.7) Why should the value of Ω0 be so close to one today? It could have had. the density parameter and curvature are linked by the equation κc2 1 − Ω0 = − 2 2 . of course.5) . We could. we find the equation that gives the density parameter as a function of time: 1 − Ω(t) = 2 H0 (1 − Ω0 ) . (11.35): Ωr. it is necessary to go a little deeper into the physics of the expanding universe. at times t tmΛ ≈ 10 Gyr. and was even more so in the past. when you extrapolate the value of Ω(t) backward into the past. for instance. The spatial curvature of the universe is related to the density parameter Ω by the Friedmann equation: 1 − Ω(t) = − κc2 . Ω0 = 0.2 . INFLATION & THE VERY EARLY UNIVERSE homogeneous today. I am using the Friedmann equation in the form given by equation 4. However. the closeness of Ω to unity becomes more and more difficult to dismiss as a coincidence.) At the present moment. If. aP ≈ 2 × 10−32 . (11.0 = 0. at anuc ≈ 3.0 = 8. (11. At the time of radiation-matter equality. If we push our extrapolation as far back as we dare. |1 − Ω|r ∝ a2 ∝ t .9) If we extrapolate backward to the time of Big Bang Nucleosynthesis.4 × 10−5 . we find that the density parameter ΩP was extraordinarily close to one: |1 − ΩP | ≤ 1 × 10−60 .11) The number 10−60 is. with Ωr.8) Suppose.7 ± 0.1.1.11. of course. the density of the universe was equal to the critical density to an accuracy of one part in 30 trillion. very tiny. 1 − Ω(t) = Ωr.1. as the available evidence indicates. (11.3 ± 0.2. for instance. therefore. |1 − Ω|m ∝ a ∝ t2/3 . in order to change the Sun’s mass by one part in 1060 . During the radiationdominated phase. Ωm. the universe would have collapsed in a Big Crunch or expanded to a low-density Big Bore . the density parameter evolved at the rate (1 − Ω0 )a2 . the density parameter falls within the limits |1−Ω0 | ≤ 0. THE FLATNESS PROBLEM Thus.0 235 (11. to the Planck time at tP ≈ 5 × 10−44 s.0 + aΩm. (11. the density parameter Ωrm was equal to one with an accuracy |1 − Ωrm | ≤ 2 × 10−4 .0 = 0. (11. the deviation of Ω from one at the time of nucleosynthesis had been one part in 30 thousand instead of one part in 30 trillion. that the universe is described by a model close to the Benchmark Model.7) During the later matter-dominated phase. Our very existence depends on the fanatically close balance between the actual density and the critical density in the early universe. the deviation of the density parameter Ωnuc from one was only |1 − Ωnuc | ≤ 3 × 10−14 . To make an analogy. At the present. you would have to add or subtract two electrons. and ΩΛ.10) At the time that deuterium was forming.6 × 10−8 .6) During the period when the universe was dominated by radiation and matter. the deviation of Ω from one was constantly growing. 236 CHAPTER 11. INFLATION & THE VERY EARLY UNIVERSE after only a few years. In that case, galaxies, stars, planets, and cosmologists would not have had time to form. You might try to dismiss the extreme flatness of the the early universe as a coincidence, by saying, “ΩP might have had any value, but it just happened to be 1 ± 10−60 .” However, a coincidence at the level of one part in 1060 is extremely far-fetched. It would be far more satisfactory if we could find a physical mechanism for flattening the universe early in its history, instead of relying on extremely contrived initial conditions at the Planck time. 11.2 The horizon problem The “flatness problem”, the remarkable closeness of Ω to one in the early universe, is puzzling. It is accompanied, however, by the “horizon problem”, which is, if anything, even more puzzling. The “horizon problem” is simply the statement that the universe is nearly homogeneous and isotropic on very large scales. Why should we regard this as a problem? So far, we’ve treated the homogeneity and isotropy of the universe as a blessing rather than a curse. It’s the homogeneity and isotropy of the universe, after all, which permit us to describe its curvature by the relatively simple Robertson-Walker metric, and its expansion by the relatively simple Friedmann equation. If the universe were inhomogeneous and anisotropic on large scales, it would be much more difficult to describe mathematically. The universe, however, is under no obligation to make things simple for cosmologists. To see why the large scale homogeneity and isotropy of the universe is so unexpected in the standard Hot Big Bang scenario, consider two antipodal points on the last scattering surface, as illustrated in Figure 11.1. The current proper distance to the last scattering surface is dp (t0 ) = c t0 tls dt . a(t) (11.12) Since the last scattering of the CMB photons occurred a long time ago (tls t0 ), the current proper distance to the last scattering surface is only slightly smaller than the current horizon distance. For the Benchmark Model, the current proper distance to the last scattering surface is dp (t0 ) = 0.98dhor (t0 ). Thus, two antipodal points on the last scattering surface, separated by 180◦ as seen by an observer on Earth, are currently separated by a proper distance of 1.96dhor (t0 ). Since the two points are farther 11.2. THE HORIZON PROBLEM 237 Figure 11.1: In the standard Hot Big Bang scenario, the current proper distance to the last scattering surface is 98% of the current horizon distance. apart than the horizon distance, they are causally disconnected. That is, they haven’t had time to send messages to each other, and in particular, haven’t had time to come into thermal equilibrium with each other. Nevertheless, the two points have the same temperature (once the dipole distortion is subtracted) to within one part in 105 . Why? How can two points that haven’t had time to swap information be so nearly identical in their properties? The near-isotropy of the Cosmic Microwave Background is still more remarkable when it is recalled that the temperature fluctuations in the CMB result from the density and velocity fluctuations that existed at the time of last scattering. In the standard Hot Big Bang scenario, the universe was matter-dominated at the time of last scattering, so the horizon distance at that time can be approximated by the value dhor (tls ) = 2 c H(tls ) (11.13) appropriate to a flat, matter-only universe (see section 5.4). Since the Hubble distance at the time of last scattering was c/H(tls ) ≈ 0.2 Mpc, the horizon distance at that time was only dhor (tls ) ≈ 0.4 Mpc. Thus, points more than 238 CHAPTER 11. INFLATION & THE VERY EARLY UNIVERSE 0.4 megaparsecs apart at the time of last scattering were not in causal contact in the standard Hot Big Bang scenario. The angular-diameter distance to the last scattering surface is dA ≈ 13 Mpc, as computed in section 9.4. Thus, points on the last scattering surface that were separated by a horizon distance will have an angular separation equal to θhor = dhor (tls ) 0.4 Mpc ≈ ≈ 0.03 rad ≈ 2◦ dA 13 Mpc (11.14) as seen from the Earth today. Therefore, points on the last scattering surface separated by an angle as small as ∼ 2◦ were out of contact with each other at the time the temperature fluctuations were stamped upon the CMB. Nevertheless, we find that δT /T is as small as 10−5 on scales θ > 2◦ (corresponding to l < 100 in Figure 9.6). Why should regions which were out of causal contact with each other at tls have been so nearly homogeneous in their properties? Invoking coincidence (“The different patches just happened to have the same temperature”) requires a great stretch of the imagination. The surface of last scattering can be divided into some 20,000 patches, each two degrees across. In the standard Hot Big Bang scenario, the center of each of these patches was out of touch with the other patches at the time of last scattering. Now, if you invite two people to a potluck dinner, and they both bring potato salad, you can dismiss that as coincidence, even if they had 105 different dishes to chose from. However, if you invite 20,000 people to a potluck dinner, and they all bring potato salad, it starts to dawn on you that they must have been in contact with each other: “Psst...let’s all bring potato salad. Pass it on.” Similarly, it starts to dawn on you that the different patches of the last scattering surface, in order to be so nearly equal in temperature, must have been in contact with each other: “Psst...let’s all be at T = 2.725 K when the universe is 13.5 gigayears old. Pass it on.” 11.3 The monopole problem The monopole problem – that is, the apparent lack of magnetic monopoles in the universe – is not a purely cosmological problem, but one that results from combining the Hot Big Bang scenario with the particle physics concept of a Grand Unified Theory. In particle physics, a Grand Unified Theory, or GUT, is a field theory which attempts to unify the electromagnetic force, the weak 11.3. THE MONOPOLE PROBLEM 239 nuclear force, and the strong nuclear force. Unification of forces has been a goal of scientists since the 1870s, when James Clerk Maxwell demonstrated that electricity and magnetism are both manifestations of a single underlying electromagnetic field. Currently, it is customary to speak of the four fundamental forces of nature: gravitational, electromagnetic, weak, and strong. In the view of many physicists, though, four forces are three too many; they’ve spent much time and effort to show that two or more of the “fundamental forces” are actually different aspects of a single underlying force. About a century after Maxwell, Steven Weinberg, Abdus Salam, and Sheldon Glashow successfully devised an electroweak theory. They demonstrated that at particle energies greater than Eew ∼ 1 TeV, the electromagnetic force and the weak force unite to form a single “electroweak” force. The electroweak energy of Eew ∼ 1 TeV corresponds to a temperature Tew ∼ Eew /k ∼ 1016 K; the universe had this temperature when its age was tew ∼ 10−12 s. Thus, when the universe was less than a picosecond old, there were only three fundamental forces: the gravitational, strong, and electroweak force. When the predictions of the electroweak energy were confirmed experimentally, Weinberg, Salam, and Glashow toted home their Nobel Prizes, and physicists braced themselves for the next step: unifying the electroweak force with the strong force. By extrapolating the known properties of the strong and electroweak forces to higher particle energies, physicists estimate that at an energy EGUT of roughly 1012 → 1013 TeV, the strong and electroweak forces should be unified as a single Grand Unified Force. If the GUT energy is EGUT ∼ 1012 TeV, this corresponds to a temperature TGUT ∼ 1028 K and an age for the universe of tGUT ∼ 10−36 s. The GUT energy is about four orders of magnitude smaller than the Planck energy, EP ∼ 1016 TeV. Physicists are searching for a Theory of Everything (TOE) which describes how the Grand Unified Force and the force of gravity ultimately unite to form a single unified force at the Planck scale. The different unification energy scales, and the corresponding temperatures and times in the early universe, are shown in Figure 11.2. One of the predictions of Grand Unified Theories is that the universe underwent a phase transition as the temperature dropped below the GUT temperature. Generally speaking, phase transitions are associated with a spontaneous loss of symmetry as the temperature of a system is lowered. Take, as an example, the phase transition known as “freezing water”. At temperatures T > 273 K, water is liquid. Individual water molecules are 240 CHAPTER 11. INFLATION & THE VERY EARLY UNIVERSE gravity ?? TOE GUT ew E (TeV) T (K) t (sec) 10 32 10 −43 10 16 strong weak 10 28 10 −36 10 12 electromagnetic 1 16 10 −12 10 Figure 11.2: The energy, temperature, and time scales at which the different force unifications occur. randomly oriented, and the liquid water thus has rotational symmetry about any point; in other words, it is isotropic. However, when the temperature drops below T = 273 K, the water undergoes a phase transition, from liquid to solid, and the rotational symmetry of the water is lost. The water molecules are locked into a crystalline structure, and the ice no longer has rotational symmetry about an arbitrary point. In other words, the ice crystal is anisotropic, with preferred directions corresponding to the crystal’s axes of symmetry.1 In a broadly similar vein, there is a loss of symmetry when the universe undergoes the GUT phase transition at tGUT ∼ 10−36 s. At T > TGUT , there was a symmetry between the strong and electroweak forces. At T < TGUT , the symmetry is spontaneously lost; the strong and electroweak forces begin to behave quite differently from each other. In general, phase transitions associated with a loss of symmetry give rise to flaws known as topological defects. To see how topological defects form, consider a large tub of water which is cooled below T = 273 K. Usually, the freezing of the water will start at two or more widely separated nucleation sites. The crystal which forms about any given nucleation site is very regular, with well-defined axes of symmetry. However, the axes of symmeSuppose, for instance, that the water freezes in the familiar six-pointed form of a snowflake. It is now only symmetric with respect to rotations of 60◦ (or integral multiples of that angle) about the snowflake’s center. 1 11.3. THE MONOPOLE PROBLEM 241 try of two adjacent ice crystals will be misaligned. At the boundary of two adjacent crystals, there will be a two-dimensional topological defect, called a domain wall, where the axes of symmetry fail to line up. Other types of phase transitions give rise to one-dimensional, or line-like, topological defects (in a cosmological context, these linear defects are known as cosmic strings). Still other types of phase transitions give rise to zero-dimensional, or point-like, topological defects. Grand Unified Theories predict that the GUT phase transition creates point-like topological defects which act as magnetic monopoles. That is, they act as the isolated north pole or south pole of a magnet. The rest energy of the magnetic monopoles created in the GUT phase transition is predicted to be mM c2 ∼ EGUT ∼ 1012 TeV. This corresponds to a mass of over a nanogram (comparable to that of a bacterium), which is a lot of mass for a single particle to be carrying around. At the time of the GUT phase transition, points further apart than the horizon size will be out of causal contact with each other. Thus, we expect roughly one topological defect per horizon volume, due to the mismatch of fields which are not causally linked. The number density of magnetic monopoles, at the time of their creation, would be nM (tGUT ) ∼ 1 (2ctGUT )3 ∼ 1082 m−3 , (11.15) and their energy density would be εM (tGUT ) ∼ (mM c2 )nM ∼ 1094 TeV m−3 . (11.16) This is a large energy density, but it is smaller by ten orders of magnitude than the energy density of radiation at the time of the GUT phase transition: 4 εγ (tGUT ) ≈ αTGUT ∼ 10104 TeV m−3 . (11.17) Thus, the magnetic monopoles wouldn’t have kept the universe from being radiation-dominated at the time of the GUT phase transition. However, the magnetic monopoles, being so massive, would soon have become highly nonrelativistic, with energy density εM ∝ a−3 . The energy density in radiation, though, was falling off at the rate εγ ∝ a−4 . Thus, the magnetic monopoles would have dominated the energy density of the universe when the scale factor had grown by a factor ∼ 1010 ; that is, when the temperature had fallen to T ∼ 10−10 TGUT ∼ 1018 K, and the age of the universe was only t ∼ 10−16 s. (11.” There is not yet a single. When the physicist Alan Guth first proposed the idea of inflation in 1981. Thus. he introduced it as a way of resolving the flatness problem. a 3c (11. there is no strong evidence that they exist at all.4 The inflation solution What is inflation? In a cosmological context. the Friedmann equation was a ˙ a 2 = Λi . 3 (11. an epoch when a > 0. “Where have the monopoles gone?” by saying. you can always answer the question. the horizon problem. and thus had an acceleration equation that could be written in the form Λi a ¨ = >0. The monopole problem can thus be rephrased as the question. However.19) a 3 In an inflationary phase when the energy density was dominated by a cosmological constant. and in some variants on the GUT theme. “Where have all the magnetic monopoles gone?” Now. inflation can most generally be defined as the hypothesis that there was a period. The usual implementation of inflation states that the universe was temporarily dominated by a positive cosmological constant Λi (with w = −1). Every north magnetic pole which we can find is paired with a south magnetic pole. and vice versa. “There were never any monopoles to begin with. definitive Grand Unified Theory.18) tells us that a > 0 when P < −ε/3.242 CHAPTER 11. that is. 11. INFLATION & THE VERY EARLY UNIVERSE Obviously. In fact. The acceleration equation. when the expansion was accelerating outward. the universe is not dominated by magnetic monopoles today. ¨ a ¨ 4πG = − 2 (ε + 3P ) . the flatness and horizon problems are not so readily dismissed. magnetic monopoles are not produced. inflation would have taken place if ¨ the universe were temporarily dominated by a component with equation-ofstate parameter w < −1/3. and the monopole problem with a single cosmological mechanism. early in the history of our universe. There are no isolated north poles or isolated south poles.20) . then N was large. and monopole problems. when the exponential growth was switched off instantaneously. and the scale factor grew exponentially with time: a(t) ∝ eHi t . suppose the exponential growth was switched on instantaneously at a time ti . a(ti ) where N . was N ≡ Hi (tf − ti ) .22) Thus. tf − ti . with the value Hi = (Λi /3)1/2 .4. the number of e-foldings of inflation. For concreteness. suppose that the universe had a period of exponential expansion sometime in the midst of its early. (11. with a Hubble parameter Hi ≈ t−1 ≈ 1036 s−1 . was long compared to the Hubble time during inflation. and lasted for GUT N ∼ 100 Hubble times.25) a(ti ) Note that the cosmological constant Λi present at the time of inflation was very large compared to the cosmological constant which seems to be present . we can write the scale factor as t < ti a(t) =  ai eHi (t−ti ) ti < t < t f  ai eHi (tf −ti ) (t/tf )1/2 t > tf   ai (t/ti )1/2  (11. the growth in scale factor during inflation was a(tf ) ∼ e100 ∼ 1043 . Hi−1 . and lasted until some later time tf . and the growth in scale factor during inflation was enormous. (11.21) Too see how a period of exponential growth can resolve the flatness. let’s take one possible model for inflation. For simplicity. the scale factor increased by a factor a(tf ) = eN . ti ≈ tGUT ≈ 10−36 s. This model states that exponential inflation started around the GUT time. In this particular model. (11. radiation-dominated phase. In this simple case. horizon. and the universe reverted to its former state of radiation-dominated expansion. THE INFLATION SOLUTION 243 The Hubble constant Hi during the inflationary phase was thus constant.23) If the duration of inflation.24) (11.11. between the time ti when the exponential inflation began and the time tf when the inflation stopped. 29) and the difference between Ω and one decreases exponentially with time.31) After a hundred e-foldings of inflation.0 ≈ 0. with |1 − Ω(ti )| ∼ 1 . 8πG 8πG i (11. (11. the difference between Ω and one decreases with time. can be written in the form |1 − Ω(t)| = c2 2 R0 a(t)2 H(t)2 (11. we find |1 − Ω(tf )| = e−2N |1 − Ω(ti )| .26) over 107 orders of magnitude larger. the evidence is consistent with an energy density in Λ of εΛ. INFLATION & THE VERY EARLY UNIVERSE today. If the universe is expanding exponentially during the inflationary epoch. Currently. (11. If the universe is dominated by a single component with equation-of-state parameter w = −1. To produce exponential expansion with a Hubble parameter Hi ≈ 1036 s−1 .28) Thus. the universe was actually fairly strongly curved. How does inflation resolve the flatness problem? Equation (11. the deviation of Ω from one would be |1 − Ω(tf )| ∼ e−2N ∼ e−200 ∼ 10−87 .65) ε Λi = c2 3c2 2 Λi = H ∼ 10105 TeV m−3 . (11. (11.004 TeV m−3 . The current .27) for any universe which is not perfectly flat. which gives Ω as a function of time.7εc. H ∝ t−1 . and |1 − Ω(t)| ∝ t2(1+3w)/(3+3w) . then a ∝ t2/(3+3w) .1).30) Suppose that prior to inflation. (11. If we compare the density parameter at the beginning of exponential inflation (t = ti ) with the density parameter at the end of inflation (t = tf = ti + N/Hi ). then |1 − Ω(t)| ∝ e−2Hi t .0 ≈ 0. the cosmological constant during inflation would have had an energy density (see equation 4. if w < −1/3.244 CHAPTER 11.32) Even if the universe at ti wasn’t particularly close to being flat. a hundred efoldings of inflation would flatten it like the proverbial pancake. .4. the horizon size is boosted exponentially from submicroscopic scales to nearly a parsec.34) The horizon size at the end of inflation was dhor (tf ) = ai eN c ti 0 dt + ai (t/ti )1/2 tf ti dt ai exp[Hi (t − ti )] . with a Hubble parameter Hi ≈ 1036 s−1 .37) During the brief period of ∼ 10−34 s that inflation lasts in this model.11. How does inflation resolve the horizon problem? At any time t. (11.38) (11. and lasted for n ≈ 100 e-foldings. the universe was radiation-dominated. is large. the horizon distance dhor (t) is given by the relation dhor (t) = a(t)c t 0 dt . then the horizon size immediately before inflation was dhor (ti ) = 2cti ≈ 6 × 10−28 m . ai (t/ti )1/2 (11.35) If N . (11. if inflation took place around the GUT time. the number of e-foldings of inflation. the horizon size reverts to growing at a sedate linear rate. At the end of the inflationary epoch. The horizon size immediately after inflation was dhor (tf ) ≈ eN 3cti ≈ 2 × 1016 m ≈ 0.36) An epoch of exponential inflation causes the horizon size to grow exponentially. then the horizon size at the end of inflation was dhor (tf ) = eN c(2ti + Hi−1 ) . Thus. it’s possible that N may have been very much greater than 60. THE INFLATION SOLUTION 245 limits on the density parameter. the horizon time at the beginning of inflation was dhor (ti ) = ai c ti 0 dt = 2cti . a(t) (11. However.2. since the observational data are entirely consistent with |1 − Ω0 | 1.8 pc . |1−Ω0 | ≤ 0.33) Prior to the inflationary period. (11. imply that N > 60. If inflation started at ti ≈ 10−36 s. (11. was decreasing exponentially (nM ∝ e−3Hi t ). folks. INFLATION & THE VERY EARLY UNIVERSE The net result of inflation is to increase the horizon length in the postinflationary universe by a factor ∼ eN over what it would have been without inflation. consider the entire universe directly visible to us today. if they were neither created nor destroyed.41) Note that this distance is 16 orders of magnitude smaller than the horizon size immediately prior to inflation (dhor (ti ) ∼ 6×10−28 m. Currently. the proper distance to the surface of last scattering is dp (t0 ) ≈ 1. For instance.4 × 104 Mpc .9 m . obviously gargantuan enough for the entire last scattering surface to be in causal contact. the horizon size at last scattering would have been ∼ 1043 Mpc. the region bounded by the surface of last scattering. immediately after inflation. that is.4 Mpc. To look at the resolution of the horizon problem from a slightly different viewpoint.40) Immediately after inflation. . the currently visible universe was enclosed within a sphere of proper radius dp (ti ) = e−N dp (tf ) ∼ 3 × 10−44 m . that corresponds to a scale factor af ∼ 2 × 10−27 .246 CHAPTER 11. if inflation started around the GUT time. Given a hundred e-foldings of inflation in the early universe. If there were N = 100 e-foldings of inflation. the number density of monopoles. in the absence of inflation. the horizon size at the time of last scattering was dhor (tls ) ≈ 0. you ain’t heard nothin’ yet. During a period when the universe was expanding exponentially (a ∝ eHi t ). (11. How does inflation resolve the monopole problem? If magnetic monopoles were created before or during inflation. But. in this model. the portion of the universe which we can see today had plenty of time to achieve thermal uniformity before inflation began. we found that. For instance. all the mass-energy destined to become the hundreds of billions of galaxies we see today was enclosed within a sphere only six feet across. however.) Thus. Thus. the portion of the universe currently visible to us was crammed into a sphere of proper radius dp (tf ) = af dp (t0 ) ∼ 3 × 10−23 Mpc ∼ 0. then immediately prior to the inflationary epoch. (11.39) If inflation ended at tf ∼ 10−34 s. to quote Al Jolson. then the number density of monopoles was diluted to any undetectably low level. However. Suppose the universe contains a scalar field φ(r. the number density would have been nM (tf ) = e−300 nM (tGUT ) ≈ 5 × 10−49 m−3 ≈ 15 pc−3 .8 km s−1 . a scalar field can have an associated potential energy V (φ). ensuring its homogeneity over large scales.11. t) whose value can vary as a function of position and time. THE PHYSICS OF INFLATION 247 when the number density of magnetic monopoles was nM (tGUT ) ≈ 1082 m−3 . t) the inflaton field. and its potential V has units of energy density.5. however. 2 . (Some early implementations of inflation associated the scalar field φ with the Higgs field which mediates interactions between particles at energies higher than the GUT energy. and driving down the number density of magnetic monopoles which it contains.) Generally speaking. in this case. 11. The probability of finding even a single monopole within the last scattering surface would be astronomically small. at least in broad outline. we will have to examine. The associated potential energy. why doesn’t it reduce the number density of photons to undetectably low levels? If inflation is so efficient at flattening the global curvature of the universe. after the additional expansion from a(tf ) ≈ 2 × 10−27 to a0 = 1. At present. the physics behind inflation.2 If φ has units of energy. is the gravitational potential V = gφ. because the Cosmic Microwave Background isn’t perfectly isotropic. I have not yet answered many crucial questions about the inflationary epoch. to keep the discussion general. What triggers inflation at t = ti . As a simple illustrative example. where g = 9. by flattening it. let’s just call the field φ(r. To answer these questions. and (just as important) what turns it off at t = tf ? If inflation reduces the number density of monopoles to undetectably low levels. would then be nM (t0 ) ≈ 1 × 10−61 Mpc−3 . I will restrict myself to speaking in general terms about one plausible mechanism for bringing about an inflationary epoch. The number density today. suppose that the scalar field φ is the elevation above sea level at a given point on the Earth’s surface. there is not a consensus among cosmologists about the exact mechanism driving inflation. why doesn’t it also flatten out the local curvature due to fluctuations in the energy density? We know that the universe wasn’t perfectly homogeneous after inflation. then after 100 e-foldings of inflation.5 The physics of inflation Inflation explains some otherwise puzzling aspects of our universe. Substituting from equations (11. (11. we find ˙ the equation which governs the rate of change of φ: dV ¨ ˙ φ + 3H(t)φ = −¯ c3 h . the fluid equation for the energy density of the inflaton field. The pressure of the inflaton field is 1 1 ˙2 Pφ = φ − V (φ) . ¯ (11. Thus.45) Thus. ˙ (11. ˙ Under what circumstances are the conditions for inflation (small φ and ˙ start with large V ) met in the early universe? To determine the value of φ. dφ (11.248 CHAPTER 11. dφ (11. INFLATION & THE VERY EARLY UNIVERSE then the energy density of the inflaton field is εφ = 1 1 ˙2 φ + V (φ) 2 hc3 ¯ (11. with ˙ φ2 hc3 V (φ) .48) . an inflaton field can drive exponential inflation if there is a temporary ˙ period when its rate of change φ is small (satisfying equation 11. 3H φ. and its potential V (φ) is large enough to dominate the energy density of the universe. with εφ ≈ −Pφ ≈ V (φ) .47) mimics the equation of motion for a particle which is being accelerated by a force proportional to −dV /dφ and being impeded by a frictional force proportional to the particle’s speed. so the inflaton field ¨ can reach “terminal velocity” (with φ = 0) when dV ˙ 3H φ = −¯ c3 h .43).44).42) and (11.43) 2 hc3 ¯ If the inflaton field changes only very slowly as a function of time.46) where H(t) = a/a. (11. which slows the transition of the inflaton field to a value which will minimize the potential V . εφ + 3H(t)(εφ + Pφ ) = 0 .42) in a region of space where φ is homogeneous. the expansion ˙ of the universe provides a “Hubble friction” term. Just as a skydiver reaches terminal velocity when the downward force of gravity is balanced by the upward force of air resistance.44) then the inflaton field acts like a cosmological constant.47) Note that equation (11. where the potential is V (φ) ≈ V0 . If the slope of the inflaton’s potential is sufficiently shallow. and if the amplitude of the potential is sufficiently large to dominate the energy density of the universe. As a concrete example of a potential V (φ) which can give rise to inflation. translates into dV dφ 2 9H 2 V . consider the potential shown in Figure 11. hc3 ¯ (11. this means that the Hubble parameter is H= 8πGεφ 3c2 1/2 = 8πGV 3c2 1/2 .50) If the universe is undergoing exponential inflation driven by the potential energy of the inflaton field. hc5 ¯ (11. The global minimum in the potential occurs when the value of the inflaton field is φ = φ0 . satisfying equation (11.50) becomes dV dφ which can also be written as EP dV V dφ 2 2 24πGV 2 .5. If 2 V02 dV (11.3.49) φ=− 3H dφ If the inflaton field has reached this terminal velocity.51) and equation (11. the inflaton field contributes an energy density εφ ≈ V0 ≈ constant to the universe. however. that the inflaton field starts at φ ≈ 0. (11. (11.54) 2 dφ EP on the “plateau” where V ≈ V0 .53) where EP is the Planck energy. . then while φ is slowly rolling toward φ0 .52) 1. then the requirement ˙ that φ2 hc3 V .11. Suppose. THE PHYSICS OF INFLATION or 3 249 hc dV ¯ ˙ . necessary if the inflaton field is to play the role of a cosmo¯ logical constant. then the inflaton field is capable of giving rise to exponential expansion. (11.53). 250 CHAPTER 11. INFLATION & THE VERY EARLY UNIVERSE V0 V 0 0 φ0 φ Figure 11.3: A potential which can give rise to an inflationary epoch. The global minimum in V (or “true vacuum”) is at φ = φ0 . If the scalar field starts at φ = 0, it is in a “false vacuum” state. 11.5. THE PHYSICS OF INFLATION 251 When an inflaton field has a potential similar to that of Figure 11.3, it is referred to as being in a metastable false vacuum state when it is near the maximum at φ = 0. Such a state is not truly stable; if the inflaton field is nudged from φ = 0 to φ = +dφ, it will continue to slowly roll toward the true vacuum state at φ = φ0 and V = 0. However, if the plateau is sufficiently broad as well as sufficiently shallow, it can take many Hubble times for the inflaton field to roll down to the true vacuum state. Whether the inflaton field is dynamically significant during its transition from the false vacuum to the true vacuum depends on the value of V0 . As long as εφ ≈ V0 is tiny compared to the energy density of radiation, εr ∼ αT 4 , the contribution of the inflaton field to the Friedmann equation can be ignored. Exponential inflation, driven by the energy density of the inflaton field, will begin at a temperature Ti ≈ or kTi ≈ (¯ 3 c3 V0 )1/4 ≈ 2 × 1012 TeV h This corresponds to a time ti ≈ c2 GV0 1/2 V0 α 1/4 ≈ 2 × 1028 K V0 10105 TeV m−3 V0 105 TeV m−3 10 1/4 (11.55) 1/4 . (11.56) ≈ 3 × 10 −36 V0 s 105 TeV m−3 10 −1/2 . (11.57) While the inflaton field is slowly rolling toward the true vacuum state, it produces exponential expansion, with a Hubble parameter Hi ≈ 8πGV0 3c2 1/2 ≈ t−1 . i (11.58) The exponential expansion ends as the inflaton field reaches the true vacuum at φ = φ0 . The duration of inflation thus depends on the exact shape of the potential V (φ). The number of e-foldings of inflation, for the potential shown in Figure 11.3, should be N ∼ Hi φ0 ∼ ˙ φ EP dV V0 dφ −1 φ0 EP . (11.59) Large values of φ0 and V0 (that is, a broad, high plateau) and small values of dV /dφ (that is, a shallowly sloped plateau) lead to more e-foldings of inflation. 252 CHAPTER 11. INFLATION & THE VERY EARLY UNIVERSE After rolling off the plateau in Figure 11.3, the inflaton field φ oscillates about the minimum at φ0 . The amplitude of these oscillations is damped by ˙ the “Hubble friction” term proportional to H φ in equation (11.47). If the inflaton field is coupled to any of the other fields in the universe, however, the oscillations in φ are damped more rapidly, with the energy of the inflaton field being carried away by photons or other relativistic particles. These photons reheat the universe after the precipitous drop in temperature caused by inflation. The energy lost by the inflaton field after its phase transition from the false vacuum to the true vacuum can be thought of as the latent heat of that transition. When water freezes, to use a low-energy analogy, it loses an energy of 3 × 108 J m−3 , which goes to heat its surroundings.3 Similarly, the transition from false to true vacuum releases an energy V0 which goes to reheat the universe. If the scale factor increases by a factor a(tf ) = eN a(ti ) (11.60) during inflation, then the temperature will drop by a factor e−N . If inflation starts around the GUT time, and lasts for N = 100 e-foldings, then the temperature drops from a toasty T (ti ) ∼ TGUT ∼ 1028 K to a chilly T (tf ) ∼ e−100 TGUT ∼ 10−15 K. At a temperature of 10−15 K, you’d expect to find a single photon in a box 25 AU on a side, as compared to the 411 million photons packed into every cubic meter of space today. Not only is inflation very effective at driving down the number density of magnetic monopoles, it is also effective at driving down the number density of every other type of particle, including photons. The chilly post-inflationary period didn’t last, though. As the energy density associated with the inflaton field was converted to relativistic particles such as photons, the temperature of the universe was restored to its pre-inflationary value Ti . Inflation successfully explains the flatness, homogeneity, and isotropy of the universe. It ensures that we live in a universe with a negligibly low density of magnetic monopoles, while the inclusion of reheating ensures that we don’t live in a universe with a negligibly low density of photons. In some This is why orange growers spray their trees with water when a hard freeze threatens. The energy released by water as it freezes keeps the delicate leaves warm. (The thin layer of ice also cuts down on convective and radiative heat loss, but the release of latent heat is the largest effect.) 3 11.5. THE PHYSICS OF INFLATION 253 ways, though, inflation seems to be too successful. It makes the universe homogeneous and isotropic all right, but it makes it too homogeneous and isotropic. One hundred e-foldings of inflation not only flattens the global curvature of the universe, it also flattens the local curvature due to fluctuations in the energy density. If energy fluctuations prior to inflation were δε/¯ ∼ 1, ε a na¨ calculation predicts that density fluctuations immediately after 100 ıve e-foldings of inflation would be δε ∼ e−100 ∼ 10−43 . ε ¯ (11.61) This is a very close approach to homogeneity. Even allowing for the growth in amplitude of density fluctuations prior to the time of last scattering, this would leave the Cosmic Microwave Background much smoother than is actually observed. Remember, however, the saga of how a submicroscopic patch of the universe (d ∼ 3 × 10−44 m) was inflated to macroscopic size (d ∼ 1 m), before growing to the size of the currently visible universe. Inflation excels in taking submicroscopic scales and blowing them up to macroscopic scales. On submicroscopic scales, the vacuum, whether true or false, is full of constantly changing quantum fluctuations, with virtual particles popping into and out of existence. On quantum scales, the universe is intrinsically inhomogeneous. Inflation takes the submicroscopic quantum fluctuations in the inflaton field and expands them to macroscopic scales. The energy fluctuations that result are the origin, in the inflationary scenario, of the inhomogeneities in the current universe. We can replace the old proverb, “Great oaks from tiny acorns grow,” with the yet more amazing proverb, “Great superclusters from tiny quantum fluctuations grow.” Suggested reading [Full references are given in the “Annotated Bibliography” on page 286.] Islam (2002), ch. 9: A general overview of inflation, avoiding technical concepts of particle physics. Liddle (1999), ch. 11: A brief, clear discussion of how inflation solves the horizon, flatness, and monopole problems. 254 CHAPTER 11. INFLATION & THE VERY EARLY UNIVERSE Liddle & Lyth (2000): A thorough, recent review of all aspects of inflationary cosmology. Problems (11.1) What upper limit is placed on Ω(tP ) by the requirement that the universe not end in a Big Crunch between the Planck time, tP ≈ 5 × 10−44 s, and the start of the inflationary epoch at ti ? Compute the maximum permissible value of Ω(tP ), first assuming ti ≈ 10−36 s, then assuming ti ≈ 10−26 s. (Hint: prior to inflation, the Friedmann equation will be dominated by the radiation term and the curvature term.) (11.2) Current observational limits on the density of magnetic monopoles tell us that their density parameter is currently ΩM,0 < 10−6 . If monopoles formed at the GUT time, with one monopole per horizon of mass mM = mGUT , how many e-foldings of inflation would be required to drive the current density of monopoles below the bound ΩM,0 < 10−6 ? Assume that inflation took place immediately after the formation of monopoles. (11.3) It has been speculated that the present-day acceleration of the universe is due to the existence of a false vacuum which will eventually decay. Suppose that the energy density of the false vacuum is εΛ = 0.7εc,0 = 3600 MeV m−3 , and that the current energy density of matter is εm,0 = 0.3εc,0 = 1600 MeV m−3 . What will be the value of the Hubble parameter once the false vacuum becomes strongly dominant? Suppose that the false vacuum is fated to decay instantaneously to radiation at a time tf = 50t0 . (Assume, for simplicity, that the radiation takes the form of blackbody photons.) To what temperature will the universe be reheated at t = tf ? What will the energy density of matter be at t = tf ? At what time will the universe again be dominated by matter? Chapter 12 The Formation of Structure The universe can be approximated as being homogeneous and isotropic only if we smooth it with a filter ∼ 100 Mpc across. On smaller scales, the universe contains density fluctuations ranging from subatomic quantum fluctuations up to the large superclusters and voids, ∼ 50 Mpc across, which characterize the distribution of galaxies in space. If we relax the strict assumption of homogeneity and isotropy which underlies the Robertson-Walker metric and the Friedmann equation, we can ask (and, to some extent, answer) the question, “How do density fluctuations in the universe evolve with time?” The formation of relatively small objects, such as planets, stars, or even galaxies, involves some fairly complicated physics. Consider a galaxy, for instance. As mentioned in Chapter 8, the luminous portions of galaxies are typically much smaller than the dark halos in which they are embedded. In the usual scenario for galaxy formation, this is because the baryonic component of a galaxy radiates away energy, in the form of photons, and slides to the bottom of the potential well defined by the dark matter. The baryonic gas then fragments to form stars, in a nonlinear magnetohydrodynamical process. In this chapter, however, I will be focusing on the formation of structures larger than galaxies – clusters, superclusters, and voids. Cosmologists use the term “large scale structure of the universe” to refer to all structures bigger than individual galaxies. A map of the large scale structure of the universe, as traced by the positions of galaxies, can be made by measuring the redshifts of a sample of galaxies and using the Hubble relation, d = (c/H0 )z, to compute their distances from our own Galaxy. For instance, Figure 12.1 shows a redshift map from the 2dF Galaxy Redshift Survey. By plotting redshift as 255 1: A redshift map of ∼ 105 galaxies. from right ascension α ≈ 10 h to α ≈ 15 h. THE FORMATION OF STRUCTURE Figure 12. (Image courtesy of the 2dF Galaxy Redshift Survey team. and ∼ 8◦ wide.256 CHAPTER 12. in a strip ∼ 75◦ long.) . from declination δ ≈ −5◦ to δ ≈ 3◦ . A supercluster will contain one or more clusters embedded within it. narrow strip of the sky. as traced by the distribution of galaxies. as discussed in section 8. a function of angular position for galaxies in a long. There are “voids” such as the Appalachian void.3. the galaxies obviously do not have a random Poisson distribution. Superclusters are objects which are just in the process of collapsing under their own self-gravity. In a slice such as that of Figure 12.1. the distribution of luminous objects shown in Figure 12. and hence obeys the virial theorem. a “slice of the universe” can be mapped.3. which reaches to z ≈ 0.2: The northeastern United States and southeastern Canada at night. the influences which determine the exact . such as the Boswash supercluster stretching from Boston to Washington.1 are superclusters and voids. a cluster is a fully collapsed object which has come to equilibrium (more or less).257 Figure 12.2. Consider. When gazing at the large scale structure of the universe. the underdense voids are roughly spherical in shape. as seen by a satellite from the Defense Meteorological Satellite Program (DMSP). In comparison to the flattened or elongated superclusters. or dp (t0 ) ≈ 1300 Mpc. as an analogy. Superclusters are typically flattened (roughly planar) or elongated (roughly linear) structures. There are “superclusters” of cities. The distribution of illuminated cities on the Earth’s surface is obviously not random. However. The most prominent structures in Figure 12. Being able to describe the distribution of galaxies in space doesn’t automatically lead to an understanding of the origin of large scale structure. cosmologists are likely to call it “bubbly” or “spongy” or “frothy” or “foamy”. but whosoever hath not. draw matter to themselves from the surrounding underdense regions. the complicated politics that went into determining the location of Washington. THE FORMATION OF STRUCTURE location of cities are often far removed from fundamental physics. (12.258 CHAPTER 12. in addition. In less biblical language. DC. for instance. We know. t) − ε(t) . The basic mechanism for growing large structures. 1 .2) The value of δ is thus negative in underdense regions and positive in overdense regions. 12. is gravitational instability. The minimum possible value of δ is δ = −1. ε(t) ¯ (12. t) ≡ ¯ ε(r. the density of the universe had slight inhomogeneities. if their density is sufficiently great. When the universe is matter-dominated. for instance. that such density fluctuations occurred at the time of last scattering. the overdense regions expand less rapidly than the universe as a whole. It is useful to define a dimensionless density fluctuation δ(r.1 Gravitational instability To put our study of gravitational instability on a more quantitative basis.1) ε(t) = ¯ V V To ensure that we have found the true average. t) is a function of position as well as time. the rich get richer and the poor get poorer.1 Fortunately. the spatially averaged energy density is 1 ε(r. they will collapse and become gravitationally bound objects such as clusters. consider some component of the universe whose energy density ε(r. from him shall be taken away even that he hath” (Matthew 13:12). the distribution of galaxies in space is more closely tied to fundamental physics than is the distribution of cities on the Earth. The effect of gravity on density fluctuations is sometimes referred to as the Matthew Effect: “For whosoever hath. Consider. corresponding to ε = 0. the volume V over which we are averaging must be large compared to the size of the biggest structure in the universe. such as voids and superclusters. and he shall have more abundance. to him shall be given. At a given time t. Suppose that at some time in the past. since they left their stamp on the Cosmic Microwave Background. t)d3 r . The dense clusters will. due to the excess mass. the air in a closed room is approximately static and homogeneous. there are conditions under which we can consider some region of the universe to be approximately static and homogeneous. GRAVITATIONAL INSTABILITY 259 R(t) ρ = ρ(1+δ) Figure 12. we add a small amount of mass within a sphere of radius R. will be G(∆M ) G ¨ R=− =− 2 2 R R 4π 3 R ρδ ¯ 3 . To get a feel for how small density contrasts grow with time. there is no upper limit on δ.3. we can successfully use linear perturbation theory. represent a region of space where the baryon density has δ ≈ 2 × 1030 . matteronly universe.3: A sphere of radius R(t) expanding or contracting under the influence of the density fluctuation δ(t). with |δ| 1. with δ ¯ 1. Start with a static. This is the awkward fact that inspired Einstein to introduce the cosmological constant. However. (At this point.3) . The study of how large scale structure evolves with time requires knowing how a small fluctuation in density. matter-only universe with uniform mass density ρ. we stumble over the ¯ inconvenient fact that there’s no such thing as a static. homogeneous. In the limit that the amplitude of the fluctuations remains small. it is stabilized by a pressure gradient with a scale length which is much greater than the height of the ceiling. homogeneous. grows in amplitude under the influence of gravity. consider a particularly simple case. as seen in Figure 12. If the density excess δ is uniform within the sphere. In principle. You.1. (12. then the gravitational acceleration at the sphere’s surface. for instance.) In a region of the universe which is approximately static and homogeneous. so that the density within the sphere is ρ(1+δ).12. For instance. This problem is most tractable when |δ| remains very much smaller than one. (12.9) in the limit that δ 1.4) and (12. 3 3 Thus. M= 4π ρ[1 + δ(t)]R(t)3 . we find a tidy equation which tells us how the small overdensity δ evolves as the sphere collapses: ¨ δ = 4πG¯δ .4) contains two unknowns.5) remains constant during the collapse.7) 1.8) Taking the second time derivative yields 1 ¨ 1 ¨ ¨ R ≈ − R0 δ ≈ − R δ .10).260 or CHAPTER 12. R(t) and δ(t). (12. R 3 (12. If we want to find an explicit solution for δ(t). we may make the approximation 1 R(t) ≈ R0 [1 − δ(t)] . mass conservation tells us that ¨ 1¨ R ≈− δ R 3 (12.4) ¨ Thus.11) The most general solution of equation (12.12) . ρ (12.11) has the form δ(t) = A1 et/tdyn + A2 e−t/tdyn . Conservation of mass tells us that the mass of the sphere. 3 (12. a mass excess (δ > 0) will cause the sphere to collapse inward (R < 0). where R0 ≡ When δ 3M 4π ρ ¯ 1/3 (12.10) (12. THE FORMATION OF STRUCTURE ¨ R 4πG¯ ρ =− δ(t) . ¯ 3 (12. Thus. we need a second equation involving R(t) and δ(t). we can write another relation between R(t) and δ(t) which must hold true during the collapse: R(t) = R0 [1 + δ(t)]−1/3 . Combining equations (12. Equation (12.6) = constant . 3 However.12.12) depend on the initial conditions of the ˙ sphere. THE JEANS LENGTH where the dynamical time for collapse is tdyn 1 = = (4πG¯)1/2 ρ c2 4πG¯ ε 1/2 261 . but will only approach zero in the limit that the temperature approaches absolute zero. gravity tends to make small density fluctuations in a static. For instance. (12. if the overdense sphere starts at rest. After a few dynamical times. is pressure. What keeps small density perturbations in the air from undergoing a runaway collapse over the course of a few days? The answer. Thus. A non-relativistic gas. has an equation-of-state parameter w≈ kT . however. and not on R. pressureless medium grow exponentially with time.2. a pressure gradient will build up which tends to counter the effects of gravity.13) Note that the dynamical time depends only on ρ. 12.12) is significant. µc2 (12. ¯ 2 yielding a dynamical time for collapse of tdyn ≈ 9 hours. then A1 = A2 = δ(0)/2. 2 . When a sphere of gas is compressed by its own gravity. the pressure of a ideal gas will never totally vanish. with δ = 0 at t = 0. The con¯ stants A1 and A2 in equation (12.2 The Jeans length The exponential growth of density perturbations is slightly alarming. at first glance. For instance. only the exponentially growing term of equation (12. as shown in section 4. the density of the air around you is ρ ≈ 1 kg m−3 .14) where T is the temperature of the gas and µ is the mean mass per gas particle. 3 A star is the prime example of a dense sphere of gas in which the inward force of gravity is balanced by the outward force provided by a pressure gradient. Thus.3. the state in which gravity is exactly balanced by a pressure gradient. Consider an overdense sphere Slightly longer if you are using this book for recreational reading as you climb Mount Everest. cannot always be attained. hydrostatic equilibrium. of course. (12. 4 What is sound. Overdense regions smaller than the Jeans length merely oscillate in density. given by the relation λJ ∼ cs tdyn ∼ cs c2 G¯ ε 1/2 . after all. Any change in pressure travels at the sound speed. The steepening of the pressure gradient.17) For hydrostatic equilibrium to be attained. (12. In a medium with equation-of-state parameter w > 0.262 CHAPTER 12. doesn’t occur instantaneously.15) tdyn ∼ (G¯)1/2 ρ G¯ ε If the pressure is nonzero. If pressure were not present. but a traveling change in pressure? .19) The length scale λJ is known as the Jeans length. it would collapse on a timescale 1/2 c2 1 ∼ . they constitute stable sound waves.16) where cs is the local sound speed. Overdense regions larger than the Jeans length collapse under their own gravity. the attempted collapse will be countered by a steepening of the pressure gradient within the perturbation. THE FORMATION OF STRUCTURE with initial radius R. we find that for a density perturbation to be stabilized by pressure against collapse. the pressure gradient must build up before the overdense region collapses. (12. however. the time it takes for the pressure gradient to build up in a region of radius R will be tpre ∼ R .4 Thus.15) with equation (12.16). implying tpre < tdyn .18) Comparing equation (12. who was among the first to study gravitational instability in a cosmological context. the sound speed is dP cs = c dε 1/2 = √ wc . it must be smaller than some reference size λJ . (12. cs (12. after the astrophysicist James Jeans. 21) Comparison of equation (12. the “radiation” component of the universe. With w = 1/3.13) with equation (12. for instance. the sound speed in a gas of photons or other relativistic particles is √ cs = c/ 3 ≈ 0. To consider the behavior of density fluctuations on cosmological scales. the Jeans length for that component will be 2 1/2 √ c (12. including all the appropriate factors of π. H (12. THE JEANS LENGTH 263 A more precise derivation of the Jeans length. You don’t have to worry about density fluctuations in the air undergoing a catastrophic collapse. H −1 = 3c2 8πG¯ ε 1/2 . yields the result λJ = c s πc2 G¯ ε 1/2 = 2πcs tdyn . (12. (12.23) If we focus on one particular component of√ universe. with equation-ofthe state parameter w and sound speed cs = wc.2.20) The Jeans length of the Earth’s atmosphere. (12. (12. 3 H Consider. far longer than the scale height of the Earth’s atmosphere.24) λJ = 2π w .22tdyn .22) The Jeans length in an expanding flat universe will then be λJ = 2πcs tdyn = 2π 3 2 1/2 cs .58c . consider a spatially flat universe in which the mean density is ε. The characteristic time contains density fluctuations with amplitude |δ| for expansion of such a universe is the Hubble time.25) . where the sound speed is a third of a kilometer per second and the dynamical time is nine hours. but which ¯ 1.12.21) reveals that the Hubble time is comparable in magnitude to the dynamical time tdyn for the collapse of an overdense region: H −1 = 3 2 1/2 tdyn ≈ 1. for instance. is λJ ∼ 105 km. the Jeans length of the photon-baryon fluid was roughly the same as the Jeans length of a pure photon gas: λJ (before) ≈ 3c/H(zdec ) ≈ 0. THE FORMATION OF STRUCTURE The Jeans length for radiation in an expanding universe is then √ c 2π 2 c ≈ 3. 4π 3 λ . is defined as the mass of baryons contained within a sphere of radius λJ . The gravitational collapse of the baryonic component of the universe is complicated by the fact that it was coupled to photons until a redshift zdec ≈ zls ≈ 1100.264 CHAPTER 12. Since the photons were still dominant over the baryons at the time of decoupling.0 .28) (12. Just before decoupling. (12.0 × 10−19 kg m−3 4π (1. electrons.2 Mpc. corresponding to a mass density ρbary ≈ 5.4εbary .9 × 1022 m)3 3 ≈ 1. it must have a non-relativistic compo√ nent. (12. with εγ > εbary . the photons. In order for a universe to have gravitationally collapsed structures much smaller than the Hubble distance. and baryons were all coupled together to form a single photon-baryon fluid. with w 1.27) The baryonic Jeans mass.5. the baryonic Jeans mass was MJ ≡ ρbary MJ (before) ≈ 5.6 Mpc ≈ 1. (12.3 × 1049 kg ≈ 7 × 1018 M .9 × 1022 m . and will not collapse under their own gravity.29) . MJ . the Hubble distance at the time of last scattering (effectively equal to the time of decoupling) was shown to be c/H(zdec ) ≈ 0. The energy density of baryons at decoupling was εbary ≈ 2. and the energy density of photons was εγ ≈ 3.8 × 1011 MeV m−3 ≈ 1. if we regard the baryons as a minor contaminant. Although a universe containing nothing but radiation can have density perturbations smaller than λJ ∼ 3c/H. In section 9. we can regard the baryons (with only mild exaggeration) as being a dynamically insignificant contaminant in the photon-baryon fluid.0 × 10−19 kg m−3 . 3 J Immediately before decoupling. they will be stable sound waves.26) λJ = 3 H H Density fluctuations in the radiative component will be pressure-supported if they are smaller than three times the Hubble distance.8 × 1011 MeV m−3 . Prior to decoupling. and is very much smaller than the baryonic mass of our own Galaxy. by contrast. from supercluster scales down the the size of the . (12. Now consider what happens to the baryonic Jeans mass immediately after decoupling. plummeting from MJ (before) ≈ 7 × 1018 M to MJ (after) = F 3 MJ (before) ≈ 1 × 105 M .5 × 10−5 cs (baryon) ≈ ≈ 2. Perturbations in the baryon density.26 eV 1140 × 106 eV 1/2 c ≈ 1. the photons and baryons form two separate gases. cs (photon) 0.33) Decoupling causes the baryonic Jeans mass to decrease by a factor F 3 ≈ 1. their associated Jeans length decreased by a factor F = 1. which is ∼ 1011 M .58c .5 × 10−5 c . Thus.26 eV. Thus. and the mean rest energy of the atoms in the baryonic gas was mc2 = 1. THE JEANS LENGTH 265 This is approximately 3×104 times greater than the estimated baryonic mass of the Coma cluster. (12. and represents a mass greater than the baryonic mass of even the largest supercluster seen today. (12.6 × 10−5 .34) This is comparable to the baryonic mass of the smallest dwarf galaxies known. instead of a single photon-baryon fluid.31) At the time of decoupling.30) The sound speed in the baryonic gas. taking into account the helium mass fraction of Yp = 0. the sound speed of the baryonic gas immediately after decoupling was cs (baryon) ≈ 0.12.2. Once the photons are decoupled. The sound speed in the photon gas is √ cs (photon) = c/ 3 ≈ 0.8 × 10−14 . (12. once the baryons were decoupled from the photons.22mp c2 ≈ 1140 MeV.32) only 5 kilometers per second. is cs (baryon) = kT mc2 1/2 c. The abrupt decrease of the baryonic Jeans mass at the time of decoupling marks an important epoch in the history of structure formation.58 (12. the thermal energy per particle was kTdec ≈ 0.24. in the absence of self-gravity. The global expansion of the universe. For the dark matter. ε −1 2 1/2 H ∼ (c /G¯) . assumed that the universe was static as well as pressureless. would have become non-relativistic long before the time of radiation-matter equality at zrm ≈ 3570. The analysis of section 12. this loss of pressure support happens abruptly at the time of decoupling. the timescale for the growth of a density perturbation by selfgravity. which yielded δ ∝ exp(t/tdyn ).3 Instability in an expanding universe Thus.1.35) a 5 × 10−4 eV . causes ε overdense regions to become more dense with time. tdyn ∼ (c2 /G¯)1/2 . dropped still further with time as the universe expanded and the baryonic component cooled. When considering the Cosmic Neutrino Background. they must both be taken into account when computing Density perturbations smaller than the Hubble distance can grow in amplitude only when they are no longer pressure-supported. couldn’t grow in amplitude until the time of photon decoupling. causes overdense regions to become less dense with time. the growth of density perturbations in the baryonic component was off and running. we found (see equation 5.266 CHAPTER 12.35 Myr. Once the pressure (and hence the Jeans length) of some component becomes negligibly small. the WIMPs. the loss of pressure support occurs more gradually. when the Jeans length for baryons drops suddenly by a factor F ∼ 3 × 10−5 . (12. is comparable to the timescale for expansion. For the baryonic matter. does this imply that the amplitude of density fluctuations is free to grow exponentially with time? Not necessarily. which has a temperature comparable to the Cosmic Microwave Background. if they have a mass mW c2 2 eV. for instance. already small by cosmological standards at the time of decoupling. in the absence of global expansion. 12. when the universe had reached the ripe old age of tdec ≈ 0. if the universe contains a Cosmic WIMP Background comparable in temperature to the Cosmic Neutrino Background.18) that neutrinos of mass mν became non-relativistic at a redshift 1 mν c2 1+z = ≈ . Because the timescales for these two competing processes are similar. The baryonic Jeans mass. In an expanding Friedmann universe. After decoupling. as the thermal energy of the dark matter particles drops below their rest energy. Self-gravity. THE FORMATION OF STRUCTURE smallest dwarf galaxies. 41) yields ¨ ˙ R a 1¨ 2 a ˙ ¨ = − δ− δ. ρ 3 3 3 (12. so that the density within the sphere is ρ(t) = ρ(t)[1 + δ(t)] .40) (12. we are implicitly assuming that the radius R is small compared to the Hubble distance and large compared to the Jeans length. (In performing a Newtonian analysis of this problem. Taking two time derivatives of equation (12. INSTABILITY IN AN EXPANDING UNIVERSE 267 the time evolution of a density perturbation. or removed. As the universe expands.37) The equation of motion for a point at the surface of the sphere can then be written in the form ¨ 4π 4π R = − G¯ − ρ G¯δ . let’s do a Newtonian analysis of the problem.36) with |δ| 1. M= 4π ρ(t)[1 + δ(t)]R(t)3 .42) . its radius will grow slightly less rapidly than the scale factor a(t). ¯ (12.38) R 3 3 Mass conservation tells us that the mass inside the sphere. Suppose you are in a universe filled with pressureless matter which has mass density ρ(t). if the sphere is slightly overdense.41) That is. To get a feel how small density perturbations in an expanding universe evolve with time. If the sphere is slightly underdense. the density decreases at the ¯ −3 ¯ rate ρ(t) ∝ a(t) . it will grow slightly more rapidly than the scale factor. similar in spirit to the Newtonian derivation of the Friedmann equation given in Chapter 4. ¯ 3 (12. Thus.12. (12. ρ (12.39) remains constant as the sphere expands. a small amount of matter is added. Within a spherical region of radius R. R(t) ∝ ρ(t)−1/3 [1 + δ(t)]−1/3 .3. since ρ ∝ a−3 . R a 3 3a (12.) The total gravitational acceleration at the surface of the sphere will be GM G ¨ R=− 2 =− 2 R R 4π 4π 4π 3 ρR = − G¯R − ρ G(¯δ)R . ¯ or. ¯ R(t) ∝ a(t)[1 + δ(t)]−1/3 . equa˙ tion (12. c (12.44) from equation (12. we find the equation which governs the growth of small perturbations: 4π 1¨ 2 a ˙ ˙ δ = − G¯δ . By subtracting equation (12. In multiple-component universes.44).11): ¨ δ = 4πG¯δ .268 when |δ| CHAPTER 12. ρ (12. εm − ε m ¯ δ= . That is. is sometimes called the “Hubble friction” term. such as radiation (w = 1/3) or a cosmological constant (w = −1). Combining equations (12.47) ˙ The additional term.44) which is the correct acceleration equation for a homogeneous.42). with H = 0.46) reduces to the result which we have already found in equation (12.43) reduces to a ¨ 4π = − G¯ . ρ a 3 3a 3 3 (12. In a static universe.49) εm ¯ . (12. − δ− ρ 3 3a 3 or ¨ ˙ δ + 2H δ = 4πG¯δ .43) to leave only the terms linear in the perturbation δ.46) remembering that H ≡ a/a. THE FORMATION OF STRUCTURE 1. found in an expanding universe.45) (12.48) This form of the equation can be applied to a universe which contains components with non-negligible pressure. however.43) When δ = 0. isotropic universe containing nothing but pressureless matter (compare to equation 4. it should be remembered that δ represents the fluctuation in the density of matter alone. A fully relativistic calculation for the growth of density perturbations yields the more general result ¨ ˙ 4πG ¯ δ + 2H δ = 2 εm δ . equation (12. ∝ H δ.38) and (12. we find a 1¨ 2 a ˙ ¨ ˙ 4π 4π − δ− δ = − G¯ − ρ G¯δ . ρ (12. ρ a 3 (12. it acts to slow the growth of density perturbations in an expanding universe. 12. might be only a small fraction of ¯ ε(t). Take. which has a solution of the form δ(t) ≈ C1 + C2 e−2HΛ t . ¯ It is only when matter dominates the energy density that fluctuations in the matter density can grow at a significant rate.51) takes the form 2 ¨ 4 ˙ δ + δ − 2δ = 0 .56) . fluctuations in the matter density reach a constant fractional amplitude. (12. Rewritten in terms of the density parameter ¯ for matter.53) (12. and equation (12. t which has a solution of the form δ(t) ≈ B1 + B2 ln t . In a flat. Ωm = 1. the average total density. while the average matter density plummets at the rate εm ∝ e−3HΛ t . the density parameter for matter will again be negligibly small. In the far future. εm ¯ 8πG¯m ε Ωm = = 2 2 . for instance. and equation (12.52) During the radiation-dominated epoch.55) (12. (12.54) In a lambda-dominated phase.3.51) During epochs when the universe is not dominated by matter. Ωm 1 and H = 1/(2t). the average matter density. meaning that equation (12.51) takes the form ¨ 1˙ δ+ δ≈0.50) εc 3c H equation (12.51) will take the form ¨ ˙ δ + 2HΛ δ ≈ 0 . H = 2/(3t). density perturbations in the matter do not grow rapidly in amplitude. the Hubble parameter will have the constant value H = HΛ . 3t 3t (12. 2 (12. if the universe is indeed dominated by a cosmological constant.48) takes the form ¨ ˙ 3 δ + 2H δ − Ωm H 2 δ = 0 . therefore. the early radiation-dominated phase of the universe. density fluctuations in the dark matter grew only at a logarithmic rate. matter-dominated universe. During this epoch. INSTABILITY IN AN EXPANDING UNIVERSE 269 where εm (t). (12. In these simulations. THE FORMATION OF STRUCTURE If we guess that the solution to the above equation has the power-law form Dtn . The less concentrated nonbaryonic matter forms the dark halo within which the stellar component of the galaxy is embedded. as in the real universe. (12. its evolution can no longer be treated with a simple linear perturbation approach. the general solution for the time evolution of density perturbations in a spatially flat. When the growing mode is the only survivor. it breaks away from the Hubble flow and collapses. (12.57) 4 2 n(n − 1) + n − = 0 . ∝ t2/3 .60) as long as |δ| 1. . If baryonic matter were the only type of non-relativistic matter in the universe. The centrally concentrated baryons can then proceed to form stars. The decaying mode. plugging this guess into the equation yields n(n − 1)Dtn−2 + or 2 4 nDtn−1 − 2 Dtn = 0 . eventually becomes negligibly small compared to the growing mode. matter-only universe is δ(t) ≈ D1 t2/3 + D2 t−1 . matter-only universe grow at the rate δ ∝ t2/3 ∝ a(t) ∝ 1 1+z (12. the density perturbations in a flat. in which the matter filling the universe is modeled as a distribution of point masses interacting via Newtonian gravity.270 CHAPTER 12. When an overdense region attains an overdensity δ ∼ 1. If the baryonic matter within the structure is able to cool efficiently (by bremsstrahlung or some other process) it will radiate away energy and fall to the center. the overdense region attains virial equilibrium as a gravitationally bound structure. After one or two oscillations in radius. Studies of the nonlinear evolution of structure are commonly made using numerical computer simulations.58) 3 3 The two possible solutions for this quadratic equation are n = −1 and n = 2/3. Thus. when a region reaches an overdensity δ ∼ 1. ∝ t−1 .59) The values of D1 and D2 are determined by the initial conditions for δ(t). then density perturbations could have started to grow at zdec ≈ 1100. 3t 3t (12. becoming the visible portions of galaxies that we see today. The situation is schematically illustrated in Figure 12.12. The density perturbations in the dark matter started to grow effectively at zrd ≈ 3570. the mominant form of non-relativistic matter is dark matter.4. as they would in a universe without dark matter. . At the time of decoupling. and they could have grown in amplitude only by a factor ∼ 1100 by the present day.3. Having nonbaryonic dark matter allows the universe to get a “head start” on structure formation. perturbations in the matter density can start growing at zrd ≈ 3570 rather than zdec ≈ 1100. the baryons fell into the preexisting gravitational wells of the dark matter.4: A highly schematic drawing of how density fluctuations in different components of the universe evolve with time. INSTABILITY IN AN EXPANDING UNIVERSE 271 Figure 12. However. 61) Under these circumstances. I assumed that the perturbation was spherically symmetric. as shown in section 3. 5 .4 The power spectrum When deriving equation (12. 5 Let us consider the properties of the early density fluctuations at some time ti when they were still very low in amplitude (|δ| 1). The bubbly structure shown in redshift maps of galaxies. at some time ti . it is useful to set up a comoving coordinate system. but before At the risk of carrying the arboreal analogy too far. we can deduce the properties of the early. the proper distance of any point from the origin can be written in the form dp (ti ) = a(ti )r . (12. THE FORMATION OF STRUCTURE 12. By looking at the current large scale structure (the “tree”). which determines the growth rate of density perturbations in a Newtonian universe.1. as shown in Figures 9. great pine trees from tiny pinenuts grow. then r will remain very nearly constant as long as |δ| 1.272 CHAPTER 12.62) where the comoving distance r is what the proper distance would be at the present day (a = 1) if the expansion continued to be perfectly isotropic. low-amplitude.5. has grown from the density perturbations which were present when the universe became matter dominated. (The exact value of ti doesn’t matter. I should mention that the temperature fluctuations of the Cosmic Microwave Background. offer us a look at the “sapling”.48).4. Thus. In fact.2 and 9. cosmologists typically consider δ(r). the expansion of the universe is still nearly isotropic. such as Figure 12. Choose some point as the origin. If we label each bit of matter in the universe with its comoving coordinate position r = (r.25): ds2 = −c2 dt2 + a(t)2 [dr2 + Sκ (r)2 dΩ2 ] . as long as it’s a time after the density perturbations are in place.46). since the density perturbations in the real universe are not spherically symmetric. density fluctuations (the “nut”). (12. when considering the regime where density fluctuations are small. Great oaks from tiny acorns grow – but then. This is fortunate. both apply to low-amplitude perturbations of any shape. equation (12. and the geometry of the universe is still well described by the Robertson-Walker metric (equation 3. θ. φ). the density fluctuation at a comoving location r. equation (12.46) and its relativistically correct brother. In a universe described by the Robertson-Walker metric. As long as the density fluctuations are small in amplitude. conversely.66) as long as the proper wavelength.4. Each Fourier component is a complex number.64) V In performing the Fourier transform. THE POWER SPECTRUM 273 they reach an amplitude |δ| ∼ 1. which can be written in the form δk = |δk |eiφk . of comoving volume V . A similar decomposition of δ(r) is also useful. Similarly. you can reconstruct the density field δ(r). a useful expansion of δ is in terms of Fourier components. if you know all the Fourier components. a(t)2π/k. the density fluctuation field δ(r) can be expressed as δ(r) = V (2π)3 δk e−ik·r d3 k . is large compared to the Jeans . (12.) When discussing the temperature fluctuations of the Cosmic Microwave Background. (12. Switching to a different value of ti .63) where the individual Fourier components δk are found by performing the integral 1 δk = δ(r)eik·r d3 r . If you have complete. under these restrictions. (12. Since δ is defined in three-dimensional space (rather than on the surface of a sphere).65) When |δk | 1. I pointed out that cosmologists weren’t interested in the exact pattern of hot and cold spots on the last scattering surface. and not its shape. When studying the temperature fluctuations of the CMB. then each Fourier component obeys equation (12. back in Chapter 9. 2 (12. uncensored knowledge of δ(r). cosmologists are not interested in the exact locations of the density maxima and minima in the early universe. Within a large comoving box. you can compute all the Fourier components δk . but rather in the statistical properties of the field δT /T (θ. θ) in spherical harmonics.51). each with comoving wavenumber k and comoving wavelength λ = 2π/k. it is useful to expand δT /T (φ. but rather in the statistical properties of the field δ(r).12. simply changes the amplitude of δ(r). φ). you are breaking up the function δ(r) into an infinite number of sine waves. 3 ¨ ˙ δk + 2H δk − Ωm H 2 δk = 0 . such as galaxies and clusters of galaxies. the growth of the longer wavelength perturbations is still described by equation (12. then no information is lost. if we average the power spectrum over all angles. as long as a(t)2π/k > c/H(t)). have already collapsed. If δ(r) is a Gaussian field. power-law form: P (k) ∝ k n . (12.68) where the standard deviation σ can be computed from the power spectrum: σ= V (2π)3 P (k)d3 k = V 2π 2 ∞ 0 P (k)k 2 dk . then δ(r) is called a Gaussian field. then all its statistical properties are summed up in the power spectrum P (k). that we can use linear perturbation theory to study the growth of very large scale structure even after smaller structures. homogeneous Gaussian field.66). then the value of δ at a randomly selected point is drawn from the Gaussian probability distribution p(δ) = √ δ2 1 exp − 2 2σ 2πσ . THE FORMATION OF STRUCTURE length and small compared to the Hubble distance c/H.67) where the average is taken over all possible orientations of the wavenumber k. helpfully enough.70) When a sine wave perturbation has a wavelength large compared to the Hubble distance. 6 (12. Even after fluctuations with a short proper wavelength have reached |δk | ∼ 1 and collapsed. its crests are not causally connected to its troughs.69) The study of Gaussian density fields is of particular interest to cosmologists because most inflationary scenarios predict that the density fluctuations created by inflation (see section 11. (12. (If δ(r) is isotropic. statistically speaking. .) When the phases φk of the different Fourier components are uncorrelated with each other. As long as the crests remain out of touch with the troughs (that is. This means. The mean square amplitude of the Fourier components defines the power spectrum: P (k) = |δk |2 .5) will be an isotropic. (12. the expected power spectrum for the inflationary fluctuations has a scale-invariant.6 The phase φk remains constant as long as the amplitude |δk | remains small. If a Gaussian field is homogeneous and isotropic. In addition. the amplitude of a perturbation grows at the rate δ(t) ∝ a(t).274 CHAPTER 12. The mean square density fluctuation of the mass inside each sphere is a function of the power spectrum and of the size of the sphere: M− M M 2 ∝ k 3 P (k) . The Harrison-Zel’dovich spectrum. (12.71) However. Since the average mass within a sphere is M ∝ N . 3 c (12. if the power spectrum has the form P (k) ∝ k n . with n = 1. and others will be slightly overdense. so that they formed a Poisson distribution. THE POWER SPECTRUM 275 with the favored value of the power-law index being n = 1.72) where the comoving wavenumber associated with the sphere is k = 2π/L. where N is the expected number of point masses within the sphere. or n = 0.4. What would a universe with P (k) ∝ k n look like? Imagine going through such a universe and marking out randomly located spheres of comoving radius L. thus will produce more power on small length scales than a Poisson distribution of points. (Note that if you scattered point masses randomly throughout the universe. is often referred to as a Harrison-Zel’dovich spectrum.12. the root mean square mass fluctuation within spheres of comoving radius L will be δM ≡ M M− M M 2 1/2 ∝ L−(3+n)/2 . Thus. the Harrison-Zel’dovich spectrum. (12. Thus.0 L 2 . some spheres will be slightly underdense. For n < −3. a Poisson point distribution has δM/M ∝ M −1/2 . which would be Bad News for our assumption of homogeneity on large scales. The preferred power spectrum.73) This can also be expressed in the form δM/M ∝ M −(3+n)/6 . the mass fluctuations diverge on large scales. is the only power law which prevents the divergence of the potential fluctuations on both large and small scales.) Note that the potential fluctuations associated with mass fluctuations on a length scale L will have an amplitude δΦ ∼ GδM/L ∝ δM/M 1/3 ∝ M (1−n)/6 . P (k) ∝ k. you would expect mass fluctuations of amplitude δM/M ∝ N −1/2 . . The mean mass of each sphere (considering only the non-relativistic matter which it contains) will be M = 4π 3 εm. with n = 1. the actual mass of each sphere will vary. If they exist.75) much larger than the mass of even a fairly large galaxy such as our own (Mgal ≈ 1012 M ).276 CHAPTER 12. THE FORMATION OF STRUCTURE 12.0 = ρc. a weakly interacting particle which remains relativistic until the universe becomes matter-dominated will act as hot dark . the expected power spectrum for density perturbations has the form P (k) ∝ k n . axions would act as cold dark matter. Thus. the shape of the power spectrum will be modified between the end of inflation at tf and the time of radiation-matter equality at trm ≈ 4. The shape of the power spectrum at trm . Axions are a type of elementary particle first proposed by particle physicists for non-cosmological purposes.0 ρc.0 3/2 (12. Cold dark matter consists of particles which were non-relativistic at the time they decoupled from the other components of the universe. More specifically.4 × 1017 M . by contrast. and which remained relativistic until the mass contained within a Hubble volume (a sphere of proper radius c/H) was large compared to the mass of a galaxy. Hot dark matter. however.5 Hot versus cold Immediately after inflation. If any primordial black holes had formed in the early universe.0 so the mass within a Hubble volume at that time was √ 3/2 4π c3 Ωm.026 Mpc . their peculiar velocities would have been much smaller than c at the time they formed. thus primordial black holes would also act as cold dark matter. H(trm ) 2H0 Ω2 m. However. 3 2 3 3 3 H(trm ) arm 3 H0 Ωm. with very low thermal velocities.0 2π c3 Ωr.8ctrm ≈ 0. depends on the properties of the dark matter. For instance.74) (12. when density perturbations start to grow significantly in amplitude. they would have formed out of equilibrium in the early universe. with an index n = 1 being predicted by most inflationary models. it depends on whether the dark matter is predominantly cold dark matter or hot dark matter. WIMPs would have had thermal velocities much smaller than c at the time they decoupled. Thus. and hence qualify as cold dark matter.0 ≈ 1. as well.7 × 104 yr. consists of particles which were relativistic at the time they decoupled from the other components of the universe. the Hubble distance at the time of radiation-matter equality was c Ωr.0 c =√ ≈ 1. In the Benchmark Model. The label “hot”. This happens at a temperature mh c2 mh c2 . HOT VERSUS COLD 277 matter. For instance. there will be no density fluctuations on scales smaller than the physical scale λmin mh c2 ∼ cth ∼ 20 kpc 2 eV −2 . (12. The initially relativistic particles cool as the universe expands. this corresponds to a cosmic time (equation 10. .76) ∼ 8000 K Th ∼ 3k 2 eV In the radiation-dominated universe.77) 2 eV Prior to the time th . when the temperature of the Cosmic Neutrino Background is only two degrees above absolute zero. a neutrino with mass mν c2 < 2 eV doesn’t become non-relativistic until after radiation-matter equality. neutrinos decoupled at t ∼ 1 s. consider what would happen in a universe filled with weakly interacting particles which are relativistic at the time they decouple. when the universe had a temperature kT ∼ 1 MeV. the hot dark matter particles move freely in random directions with a speed close to that of light. simply means that the neutrinos were hot enough to be relativistic back in the radiation-dominated era. the net effect of free streaming in the hot dark matter is to wipe out any density fluctuations whose wavelength is smaller than ∼ cth .5. (12. in this case. (12. until their thermal velocities drop well below c when 3kT ∼ mh c2 . Moreover.79) It may seem odd to refer to neutrinos as “hot” dark matter. a neutrino with mass mν c2 1 MeV was hot enough to be relativistic at the time it decoupled.725 K 2 eV −1 .78) corresponding to a comoving length scale Lmin = 7 mh c2 Th λmin ∼ λmin ∼ 60 Mpc a(th ) 2.2) −2 mh c2 th ∼ 2 × 1012 s .1. Thus. as discussed in section 5. Thus. called free streaming. (12. This motion. and hence qualifies as hot dark matter.7 To see how the existence of hot dark matter modifies the spectrum of density perturbations. When the hot dark matter particles become non-relativistic.12. acts to wipe out any density fluctuations present in the hot dark matter. However. (This scenario. Smaller structures. the opposite seems to be true in our universe.278 CHAPTER 12. the first structures to collapse are the size of superclusters. δM/M ∝ (k 3 P )1/2 . when the proper wavelength of a density perturbation is large compared to the Hubble distance.3. Superclusters are just collapsing today. The lower panel of Figure 12. for which free streaming has been negligible. when the universe is radiationdominated. most of the dark matter in the universe must be cold dark matter. given the absence of free streaming.80) assuming Ωm. such as neutrinos. have a maximum amplitude at a mass scale M ∼ 1016 M . Thus. once the hot dark matter has cooled enough to become non-relativistic.0 = 0. they decouple from the radiation at a time td ∼ 1 s. Note that for wavenumbers k 2π/Lmin . The upper panel of Figure 12. If the dark matter is contributed by neutrinos with rest energy of a few electron volts. (12. in which the largest observable structures form first. THE FORMATION OF STRUCTURE The total amount of matter within a sphere of comoving radius Lmin is Mmin 4π 3 Lmin Ωm. such as clusters and galaxies then form by fragmentation of the superclusters. This implies that in a universe filled with hot dark matter. when the universe was less than a gigayear old. then the free streaming will wipe out all density fluctuations smaller than superclusters. the free streaming of the hot dark matter results in a severe loss of power for wavenumbers k 2π/Lmin . when the . and that galaxies would be relatively young. regardless of whether the universe is radiation-dominated or matter-dominated. is quite different from the evolution of the power spectrum for hot dark matter. is called the top-down scenario. as long as their proper wavelength a(t)2π/k is small compared to the Hubble distance c/H(t). Remember.0 ρc.) If most of the dark matter in the universe were hot dark matter. In fact.0 ∼ 5 × 1016 M = 3 mh c2 2 eV −3 . If the cold dark matter consists of WIMPs.5 shows the power spectrum of density fluctuations in hot dark matter. density fluctuations δk in the dark matter do not grow appreciably in amplitude. the power spectrum of hot dark matter (shown as the dotted line) is indistinguishable from the original P ∝ k spectrum (shown as the dashed line). then we would expect the oldest structures in the universe to be superclusters.5 shows that the root mean square mass fluctuations in hot dark matter. The evolution of the power spectrum of cold dark matter. However. its amplitude will be able to increase. while galaxies have been around since at least z ∼ 6. 01 . δM/M ∝ (k 3 P )1/2 .001 10 10 10 −5 −6 −7 P∝k CDM . The line types are the same as in the upper panel.5: Upper panel – The power spectrum at the time of radiationmatter equality for cold dark matter (solid line) and for hot dark matter (dotted line).0001 10 −5 12 14 16 log10M 18 20 Figure 12.01 P .1 .1 1/2 (k P) 3 .001 . The normalization of the power spectrum is arbitrary.0001 HDM . HOT VERSUS COLD 279 1 . Lower panel – The root mean square mass fluctuations.1 −1 k [Mpc ] 1 10 10 1 .12.01 .001 .5. . are shown as a function of M ∝ k −3 (masses are in units of M ). The initial power spectrum produced by inflation (dashed line) is assumed to have the form P (k) ∝ k. 5). At that time. with galaxies forming first.026 Mpc. density fluctuations with a wavenumber k < krm ≈ 0.5 are largest in amplitude for the smallest mass scales. the first objects to form are the smallest. when matter dominates. the Hubble distance is c/H ∼ 2ctd ∼ 6×108 m. By contrast. This implies that in a universe filled with cold dark matter. is c/H ≈ 1. as long as their wavelength remains longer than the Hubble distance. the Hubble distance.5 shows. corresponding to a comoving wavenumber 2πarm krm ≈ ≈ 0.07 Mpc−1 . The top panel of Figure 12. the power spectrum for cold dark matter retains the original P (k) ∝ k form which it had immediately after inflation (see the upper panel of Figure 12. corresponding to a comoving wavenumber kd ∼ 2πad ∼ 105 Mpc−1 .74).8ctrm Thus. with the suppression being greatest for the largest wavenumbers (corresponding to shorter wavelengths. the power spectrum for cold dark matter is suppressed in amplitude. THE FORMATION OF STRUCTURE scale factor is ad ∼ 3 × 10−10 . when the scale factor has grown to arm ≈ 2.8 × 10−4 . The root mean square mass fluctuations in the cold dark matter. then superclusters.7 × 104 yr. the power spectrum for cold dark matter at the time of radiation-matter equality. and for wavenumbers k < krm ≈ 0. cold dark matter density perturbations with a wavenumber kd > k > krm will be able to grow in amplitude only until their physical wavelength a(t)/(2πk) ∝ t1/2 is smaller than the Hubble distance c/H(t) ∝ t.82) 1. Note the broad maximum in the power spectrum at k ∼ krm ≈ 0. At the time of radiation-matter equality.07 Mpc−1 . shown in the bottom panel of Figure 12. Thus. for wavenumbers k > krm .8ctrm ≈ 0. and will be able to grow freely in amplitude. Density fluctuations with k > kd will remain frozen in amplitude until matter starts to dominate the universe at trm ≈ 4.280 CHAPTER 12.07 Mpc−1 . density fluctuations with a wavenumber k < kd will have a wavelength greater than the Hubble distance at the time of WIMP decoupling. as the solid line. as given in equation (12. which come within the horizon at an earlier time). At the time of WIMP decoupling. then clusters.81) Thus. This scenario.07 Mpc−1 will grow steadily in amplitude during the entire radiation-dominated era. and density perturbations smaller than the Hubble distance are free to grow again. (12. 2ctd (12. their amplitude will be frozen until the time trm . called . 12.3. but most like it cold – the majority of the dark matter in the universe must be cold dark matter. What is the functional form of δ(t) in such a universe? . and thus that the universe contains at least some hot dark matter. there is strong evidence that neutrinos do have some mass. free streaming of the hot dark matter particles would make the universe too smooth on small scales.] Liddle & Lyth (2000): The origin of density perturbations during the inflationary era.5. and how it ties into cosmology. Longair (1998): For those who want to know more about galaxy formation. Cosmologists studying the large scale structure of the universe can adjust the assumed power spectrum of the dark matter. is consistent with the observed ages of galaxies and superclusters.1) Consider a spatially flat.0 ≤ 0. Assuming that the dark matter is purely cold dark matter gives a much better fit. for instance. However. in figure 12. 7: The origin and evolution of density fluctuations. For Ωm. Suggested reading [Full references are given in the “Annotated Bibliography” on page 286. Problems (12.) Comparison of the assumed power spectrum to the observed large scale structure (as seen.04. by mixing together hot and cold matter.1) reveals that ∼ 13% or less of the matter in the universe consists of hot dark matter. If there were more hot dark matter than this amount. HOT VERSUS COLD 281 the bottom-up scenario. matter-dominated universe (Ω = Ωm = 1) which is contracting with time. Rich (2001). (It’s a bit like adjusting the temperature of your bath by tweaking the hot and cold water knobs. Assuming that the dark matter consists of nothing but hot dark matter gives a poor fit to the observed large scale structure of the universe. Some like it hot.0 = 0. and their growth thereafter. ch. this implies ΩHDM. 3. and the pressure is P = Pγ = εγ /3. galaxies have a root mean square velocity of v 2 1/2 ≈ 1520 km s−1 relative to the center of mass of the cluster. the half-mass radius of the Coma cluster is rh ≈ 1.5) Within the Coma cluster. What is tmin for our own Galaxy? What is the maximum possible redshift at which you would expect to see galaxies comparable in v and Rhalo to our own Galaxy? (Assume the Benchmark Model is correct.3) A volume containing a photon-baryon fluid is adiabatically expanded or compressed. Using arguments similar to those of the previous problem.) What is the value of ρ for our Galaxy. compute the minimum time tmin required for the Coma cluster to form by gravitational collapse.74). If a dynamically insignificant amount of matter (Ωm 1) is present in such a universe.27). What is dP/dε for the photon-baryon fluid? What is the sound speed.) (12. how do density fluctuations in the matter evolve with time? That is.5 Mpc. as discussed in section 8. What is the average density ρ of the matter in the ¯ galaxy.2.6) Derive equation (12.2) Consider an empty. what is the functional form of δ(t)? (12. expanding universe.2. how large of an error did I make in my estimate of λJ (before) by ignoring the effect of the baryons on the sound speed of the photon-baryon fluid? (12. as described in section 5. such as a galaxy. giving the Hubble distance at the time of radiation-matter equality. at a distance Rhalo from the galaxy’s center.4) Suppose that the stars in a disk galaxy have a constant orbital speed v out to the edge of its spherical dark halo. The energy density of the fluid is ε = εγ + εbary . forms by gravitational collapse of an initially small density perturbation. cs ? In equation (12. What was the Hubble distance at the time of matter-lambda equality.282 CHAPTER 12. negatively curved. Show that tmin ≈ Rhalo /v for a ρ disk galaxy. the minimum time √ for collapse is tmin ≈ tdyn ≈ 1/ G¯. THE FORMATION OF STRUCTURE (12. including its dark halo? (Hint: go back to section 8. in the Benchmark Model? How much matter was contained within a Hubble volume at the time of matter-lambda equality? . (12. assuming v = 220 km s−1 and Rhalo = ¯ 100 kpc? If a bound structure. 5. what mass must they have in order to have become non-relativistic at t ∼ tWDM ? . at what time tWDM was the mass contained within a Hubble volume equal to Mgal = 1012 M ? If the warm dark matter particles have a temperature equal to that of the cosmic neutrino background. HOT VERSUS COLD 283 (12. In the Benchmark Model.12.7) Warm dark matter is defined as matter which became non-relativistic when the amount of matter within a Hubble volume had a mass comparable to that of a galaxy. many questions about the cosmos remain unanswered. However. Here are a few of the questions that currently nag at cosmologists: • What are the precise values of cosmological parameters such as H 0 . dense state. Only 30% of the energy density is contributed by matter (and only 4% is contributed by the familiar baryonic matter of which you and I are made). and ΩΛ. the growing weight of evidence pointed toward the Hot Big Bang model. As I write this epilogue. • What is the dark matter? It can’t be made entirely of baryons. is it provided by a false 284 . but gradually cooled as it expanded. • What is the “dark energy”? Is it vacuum energy which plays the role of a cosmological constant.Epilogue A book dealing with an active field like cosmology can’t really have a neat. or is it some other component of the universe with −1 < w < −1/3? If it is vacuum energy. Our understanding of the universe is still growing and evolving. on a sunny spring day in the year 2002. Most of the dark matter must be in the form of some exotic stuff which has not yet been detected in laboratories. Ωm. the available evidence is explained by a Benchmark Model which is spatially flat and which has an expansion which is currently accelerating. cosmological evidence was gathered at an increasing rate. At the end of the twentieth century and the beginning of the twenty-first.0 ? Much effort has been invested in determining these parameters. refining our knowledge of the universe. in which the universe started in a hot. It can’t be made entirely of neutrinos. tidy ending. q0 . It seems that 70% of the energy density of the universe is contributed by a cosmological constant (or other form of “dark energy” with negative pressure). but they are still not pinned down precisely. During the twentieth century.0 . however. a list of unanswered questions in cosmology might have included “How high is the dome which makes up the sky?” and “What’s the dome made of?” Severely erroneous models of the universe obviously give rise to irrelevant questions. Indeed. unsettling possibility exists that future observations will render the now-promising Benchmark Model obsolete. It was expanding yesterday because it was expanding the day before yesterday. What determined this set of initial conditions? In other words. that learning about cosmology from this book has encouraged you to become a cosmologist yourself. driving a temporary inflationary stage. patient reader. However.. The exciting. horizon.. or are we finally seeing the true vacuum energy? • What drove inflation during the early universe? Our knowledge of the particle physics behind inflation is still sadly incomplete. I hope. and to join the scientists who are laboring to make my book a quaint. out-of-date relic from a time when the universe was poorly understood. you find that the universe was expanding then with a Hubble parameter H ∼ 1/tP .285 vacuum. . some cosmologists pose the questions. The universe is expanding today because it was expanding yesterday. and monopole problems?” • Why is the universe expanding? At one level. this question is easily answered. are those which we are still too ignorant to pose correctly. in ancient Egypt. “What put the Bang in the Big Bang?” The most interesting questions. For instance. when you extrapolate back to the Planck time. “Did inflation take place at all during the early universe? Is there another way to resolve the flatness. Introduction to Cosmology (Englewood Cliffs.) A broad overview of cosmology. Freeman & Co. Weinberg’s revision has brought the original 1977 version more nearly up-to-date. it doesn’t skimp on the physics. 1993. H. S. Darkness at Night: A Riddle of the Universe (Cambridge: Harvard University Press) A comprehensive discussion of Olbers’ Paradox and its place in the history of cosmology. J. Although aimed at a popular audience (with all mathematical formulas banished to an appendix). Cosmology and Controversy (Princeton: Princeton University Press) A well-reseached history of the Big Bang vs. Popular • Harrison. Steady State debate.Annotated Bibliography Works described as ‘Popular’ contain little or no math. J. • Weinberg. A fascinating book if you are at all interested in the sociology of science. appropriate for study at a graduate level. 1987. Those described as ‘Intermediate’ are at roughly the same level as this book. NJ: Prentice Hall) Has a slightly greater emphasis on particle physics than 286 . H. 1996. The First Three Minutes (revised edition) (Perseus Books) A classic of popular science. • Kragh. R. E. Intermediate • Bernstein. The Big Bang (third edition) (New York: W. 1995. 2001. Those described as ‘Advanced’ have a higher level of mathematical and physical sophistication. • Silk. this book also contains longer essays on cosmological topics of current interest. from “absorption line” to “Zel’dovich-Sunyaev effect”. Malcolm S. ed. • Liddle. Introduction to Cosmology (third edition) (Cambridge: Cambridge University Press) Particularly useful for its section on alternative (non-Friedmann) cosmologies. a long. J. 1999. 2000. • Coles. N. • Harrison. • Cox. • Longair. An Introduction to Mathematical Cosmology (second edition) (Cambridge: Cambridge University Press) A book which emphasizes (as its name implies) the mathematical rather than the observational aspects of cosmology. and discussing its philosophical and religious implications.287 most cosmology texts. An Introduction to Modern Cosmology (Chichester: John Wiley & Sons) A clear and concise introductory work. 1998. Cosmological Physics (Cambridge: Cambridge University Press) A large. 2000. J. 1999. E. A. detailed discussion of galaxy formation and clustering. Allen’s Astrophysical Quantities (fourth edition) (New York: Springer-Verlag) A standard reference book of astronomically relevant data. J. among other useful things. Cosmology: The Science of the Universe (second edition) (Cambridge: Cambridge University Press) A wide-ranging book. well-stuffed grabbag of cosmological topics. Galaxy Formation (Berlin: Springer-Verlag) A well-written introduction to galaxy formation. 2002. 2002. P. J. The Routledge Critical Dictionary of the New Cosmology (New York: Routledge) In addition to a dictionary of cosmology-related terms. approached from a cosmological perspective. A. V. Fundamentals of Cosmology (Berlin: Springer-Verlag) Aimed primarily at physicists. • Peacock. 1999. Contains. • Islam. placing the science of cosmology in its historical context.. A. N. provides a self-contained introduction . from the Euler vectors of the Nazca plate to the intensity of the extragalactic gamma-ray background. 2001. • Narlikar. • Rich. The Early Universe (Redwood City. of cosmological topics. . P. 1996. E. and its dark matter content. Cosmology (third edition) (Oxford: Oxford University Press) Has a slightly greater emphasis on astronomical observations than most cosmology texts.. Advanced • Kolb. and how it gives rise to structure in the universe. Cosmological Inflation and LargeScale Structure (Cambridge: Cambridge University Press) Gives an in-depth treatment of inflation. and Lyth. The Annual Review of Astronomy and Astrophysics regularly provides reviews. D. M. • Peebles. W. 2000. CA: Addison-Wesley) A text which helped to define the field currently known as “particle astrophysics”. Keeping Up to Date Popular astronomy magazines such as Sky and Telescope and Astronomy provide non-technical news updates on advances in cosmology.. H. includes more in-depth articles on cosmological topics. Principles of Physical Cosmology (Princeton: Princeton University Press) A classic comprehensive book. • Liddle. on a more technical level. 2000. A. The Galaxies of the Local Group (Cambridge: Cambridge University Press) Cosmology begins at home: contains information about the distance scale within the Local Group. telling you as much as you need to know for cosmological purposes. R. 1990. • Rowan-Robinson. S. and Turner.288 ANNOTATED BIBLIOGRAPHY to general relativity. M. E. • van den Bergh. J. Scientific American. 1993. from time to time. 112 empty universe. 35. 23 energy density. 174 accelerating universe. 180. 264 after decoupling. 113 evidence against. 118–123. 212 Big Bounce. 131 axions. 94 radiation only. 116. 229 in early universe. 235. 164. 68. 230 smallness of. 228 bear polar. 278–280 Coma cluster. 167–170 . 180. 21 BOOMERANG. 106. 281 power spectrum. 179. 220. 148. 197 bottom-up scenario. 104. Hermann. 225. 196. 84. 111. 223 Big Bang. 140 related to luminosity distance. 111 matter only. 87 flat universe matter + lambda. 142 cold dark matter. 20–22 Big Bang Nucleosynthesis. 268 with cosmological constant. 276. 106 blackbody radiation. 276 baryon-antibaryon asymmetry. 23 289 baryons. 134 Euclidean. 15 bolometric flux (f ). 6. 105. 265 before decoupling. 228 number density. 95 single-component. 141 period-luminosity relation.Index Abell 2218 (cluster of galaxies). 73 acoustic oscillations. 92 angular-diameter distance (dA ). 150 Big Chill. 242. 150 time of (tcrunch ). 139. 229 baryonic Jeans mass (MJ ). 203 age of universe (t0 ) Benchmark Model. 281 Cepheid variable stars. 264 baryonic matter. 132. 75. 230 baryon-to-photon ratio (η). 136. 204. 102–103. 2. 106. 157. 85. 109. 236 beryllium (Be). 113 evidence against. 132. 204. 113 Big Crunch. 138. 195. 278. 67. 25 blueshift. 209. 190. 47 Benchmark Model. 138 astronomical unit (AU). 128. 133 Bondi. 150 acceleration equation. 152–153 teddy. 196 maximum value. 183. 84 galaxies (Ωgal ). 185 and baryon density. 84 stars (Ω ). 14 perfect. 197. 157 deuterium (D). 65 baryons (Ωbary ). 84. 83 number density. 204 and flatness of universe. 64 current value. 27. 168. 21 INDEX cosmological proper time. 127 Benchmark Model. see cosmological principle Cosmic Background Explorer (COBE). 28. 179 number density. 201–204 Cosmic Neutrino Background. 64 curvature negative. 199. 30 dipole distortion. 164 primordial black holes. 269 matter-dominated era. 170 cosmological constant (ΩΛ ). 179 temperature fluctuations. 212. 183 discovery of. 197 instruments aboard. see cosmic time critical density (εc ). 179 blackbody spectrum. 129 sign convention. 168 mass of. 127 density fluctuations lambda-dominated era. 269 density parameter (Ω). 182 Cosmic Microwave Background. 68. 175 cosmic time (t). 180–181 energy density.290 distance to. 46 cosmological constant (Λ). 46 dark energy. 274 radiation-dominated era. 175 neutrinos. 210 deuterium abundance. 170 comoving coordinates. 83 non-detection. 204 correlation function. 162 dark matter. 200 origin of. 159. 28. 196 amplitude. 71– 75. 159. 196 de Sitter universe. 227 determination of. 43. 228 clusters of galaxies (Ωclus ). 182. 28. 175 DASI. 40 curvature constant (κ). 243 energy density. 182 compared to starlight. 270 power spectrum. 164 matter (Ωm ). 216. 175 nonbaryonic. 73 cosmological principle. 41 positive. 227 deuterium synthesis. 164 axions. 98 deceleration parameter (q0 ). 28. 71 dark halo. 83. 218 . 155 radiation (Ωr ). 28. 46 Copernican principle. 73 pressure. 82–83 cooling of. 175 energy density. 244–245 fluid equation. 57. 73 Friedmann. 80 radiation. 73. 3 empty universe. George. 113 instability. 74 radius of curvature. 55 gigayear (Gyr). Robert. 87 static.INDEX compared to recombination. 58 fundamental force. Enrico. 50 false vacuum. 222 deuterium-to-neutron ratio. 24 first law of thermodynamics. 163 Galilei. 160 luminosity of. 174 Einstein’s static universe. 79. Kent. 29. 163 mass-to-light ratio. 74. 33 strong nuclear. 79 Ford. 10 distance modulus (m − M ). 67. 38 and teddy bears. 33 non-Euclidean. 7. 32–38. 3 . Alexander. 171 cosmological constant. 180 Digges. 94 electron. 251 291 driving inflation. Galileo. 216 Friedmann equation. 274 geodesic. 23 electron volt (eV). 74. 71 Einstein-de Sitter universe. 239 Grand Unified Force. 69. 34 and photons. 179 CMB. Thomas. 251 Fermat’s principle. 66 equation of state. 86 energy density (ε). 32. 66 flatness problem. 86 expanding. 92 matter. 66 superluminal speed. 74 Einstein. 234 during inflation. 68. 32 electroweak. 2. 162 Fourier transform. 32 weak nuclear. 11. 219 temperature of. 79 equivalence principle. 35 expansion adiabatic. 220 Dicke. 277 freezeout. 58. 81 entropy. 68. 239 gravitational. 61 Newtonian. 148 Einstein radius (θE ). 233–236 resolved by inflation. 192 free streaming. 38 geometry Euclidean. 60 with cosmological constant. 187. 239 electromagnetic. Albert. 242 general relativistic. 179 flat universe single-component. 181 Gaussian field. 38 Fermi. 80 baryons. 273 fractional ionization (X). 172. 215 Galaxy (Milky Way). 34 Gamow. 32. 3 mass of. 57 additive. 21 INDEX Hubble constant (H0 ). 143 homogeneity. 252. 234. 217–218 High-z Supernova Search Team. 63. 10. 6. 276–277 power spectrum. 13 horizon distance (dhor ). 248. 127 Hubble distance (c/H). 233 hot dark matter. 15 consequence of expansion. 225 primordial. 275 helium (He). Sheldon. David. 20. 19. 262 radiation. 245–246 Hot Big Bang. 186 . 15. 245 at last scattering. 209 hydrostatic equilibrium. 240 gravitational instability. 258–261 gravitational lens. 11–13 Jeans length (λJ ). Samuel. 233. Alan. 169. 38 Guth. 63 −1 Hubble time (H0 ). 11. 201 Hubble friction. 264 Jeans. 171 cluster of galaxies. 129. 236–238 and isotropy of CMB. 18. 143 large scale structure. 17. 209. 21 Grand Unified Theory (GUT). 120 flat universe matter only. 242–253 and density perturbations. 160 kinetic energy. 237 Benchmark Model. 244–245 and horizon problem. 238 resolved by inflation. 50 at time of last scattering. 187 isotropy. 238 phase transition. 95 single-component. 94 radiation only. 233 and production of CMB. 174 MACHO. 20. 93. 223 helium fraction (Y ). 30. Einstein. see cosmological constant Langley. Edwin. 262 Keplerian rotation. 212. 15. 142. 146 Hilbert. 268 Hubble parameter (H). 239 Gold. 261 inflation. 18 Hubble. 141 Hydra-Centaurus supercluster. 185 problems with. 132 Large Magellanic Cloud. 92 Hubble’s Law. 183 hydrogen (H). Newton vs. 165 gravity.292 Glashow. 245–246 and monopole problem. 171 gravitational potential energy. 242 Harrison-Zel’dovich spectrum. 75 relation to age of universe. 247. Fred. 143. 253 and flatness problem. James. 93 horizon problem. 252 ionization energy (Q). Thomas. 255 last scattering. 42 Hipparcos satellite. 246–247 inflaton field. 165 lambda. 20. 278 Hoyle. 172 distance to. 258 MAXIMA. 175 flavors. 105 293 positive curvature. James Clerk. 120 Benchmark Model. 135 related to angular-diameter distance. 104–105 ultimate fate of. 277 neutron. 43 homogeneous & isotropic. 25 oscillation. 143 MACHO. 112 Benchmark Model. 132. 214.INDEX last scattering surface. 238– 242 resolved by inflation. 190. 45 Robertson-Walker. 118 Matthew Effect. 214 mean free path. 213 decay. 20. 209 . 108 matter-lambda equality. 147 mass gravitational (mg ). 2. 210. 215 after freezeout. 233. 43 curved space. 24. 136 flat universe. 163 stars. 116 lookback time. 241 number density. 170 Galaxy. 23 lithium (Li). 87 moment of inertia. 195 Leavitt. 239 Maxwell-Boltzmann equation. 55 loitering universe. 111. Henrietta. Nikolai. 156 cluster of galaxies. 156 luminosity distance (dL ). 34 mass-to-light ratio. 46 Milne universe. 24 massive. 11. 205 Maxwell. 188 megaparsec (Mpc). 216 Newton. 176 neutrinos. 171–173 magnetic monopole. 156 matter-dominated universe negative curvature. 142 Lemaˆ universe. 33 inertial (mi ). 210. 146 absolute. 32–38 nuclear binding energy (B). 238 thickness of. 147 apparent. 213. 166 monopole problem. 43. 2 metric. 196. 210 nucleons. 223. 246–247 neutrino. 138 M31 (Andromeda galaxy). 44 flat space. 44 Minkowski. 234. 113 evidence against. 186 angular-diameter distance to. 241 magnitude. 241 energy density. 209. 211 of deuterium. 196. 122 luminosity density. 218 of helium. 222 neutron-to-proton ratio. 23. 197. 225 Lobachevski. 25. 23. 15. 160–162 distance to. Isaac. 84. 113 ıtre lepton. 275 Poisson’s equation. 4 Planck temperature (TP ). 97 radiation-matter equality. Heinrich. 203. 46 Rubin. 276 proper distance Robertson-Walker metric. 94 radiation only. 50–52 related to time. 69 perturbation theory. 279 scale invariant. 93 time of observation. 89 flat universe lambda only. 61 Penzias. 45 Robertson-Walker. Howard. 187 radius of curvature (R). 46 recombination. 164 Poisson distribution. Henri. 194. 193 photon decoupling time (tdec ). 195 photon-baryon fluid. 3 Planck mass (MP ). 179 Olbers. 49 proper distance (dp ). 95 single-component. 130 proton. 86 redshift survey. 264 Planck energy (EP ). Arno. 4 Planck time (tP ). 117 Benchmark Model. 192. 259 phase transition. 91 at time of observation. 187 photon. 4. 213 Pythagorean theorem. 72. 68 primordial black holes. 202 with cosmological constant. 40 quantum gravity. 218 photoionization. 28. 239 photodissociation. 189. 25. 274 pressure (P ). 51 Olbers’ Paradox. 25 photon decoupling. 194 parallax distance. 23. Vera. 255 reheating. 185. 57. 195. 20–21. 57 INDEX additive. 25. 219 recombination temperature crude approximation. 274. 195 redshift (z). 15 related to scale factor. 80 negative. 252 Robertson. 45 in Minkowski space. 132 parsec (pc). 7 optical depth (τ ). 162 . see horizon distance peculiar motion. Edgar Allen. 7–11. 180 perfect gas law. 73 power spectrum. 239 loss of symmetry. 98 matter only. 43. 190 refined calculation. 96 Poe. 192 recombination time (trec ). 2 particle horizon distance. 45 Robertson-Walker metric. 118 radiative recombination. 10 e Poincar´. 47 at time of emission.294 null geodesic. 85. 185 compared to deuterium synthesis. 4 Planck length ( P ). 190 statistical weight (g). 220 Sakharov. 144 virial theorem. 241 domain wall. 257 Walker. 239 Wheeler. 106 Taylor expansion. 108 positive curvature matter only. 239 Thomson cross-section (σe ). 265 Space Interferometry Mission (SIM). 38 Wilson. 95 single-component. 136 angular resolution. Planck energy density. 241 magnetic monopole. 39 tritium (3 He). 201 top-down scenario. 257 295 Supernova Cosmology Project. 27.INDEX Sachs-Wolfe effect. 219 nucleosynthetic analog. 117 radiation only. 143 spherical harmonics. 140 statistical equilibrium. Allan. 197 standard candle. 76 vs. 128 Slipher. Steven. 167 visible universe. 177. 215 Thomson scattering. 278 topological defect. 190 Steady State. John. 188 time of last scattering (tls ). 93 void. 3. 194. 276 year (yr). 222 Turner. 162 Small Magellanic Cloud. 21–22 supercluster. 2 sound speed (cs ). Arthur. Michael. 70. Fritz. 262 baryonic gas. 164 . 265 photon gas. 2. 240 cosmic string. 146 vacuum energy density. 13. 40. 143. 142 solar luminosity (L ). 180 WIMP. 212. 98 matter + lambda. 202 Saha equation. 266. 128 scale factor (a). 183 distance to. 46 empty universe. 143 Virgocentric flow. 132 determining H0 . 195 Hubble distance. Andrei. 109. 191. 176 Theory of Everything (TOE). 91 negative curvature matter only. 15. 94 radiation + matter. Robert. 71 type Ia supernova. 241 triangle in curved space. 156 solar mass (M ). 3 Zwicky. 230 Salam. 76 Virgo cluster. 28. Abdus. 141 standard yardstick. 42 in flat space. 13. 111 matter only. 188. 145 luminosity. 145. 87 flat universe lambda only. 45 Weinberg. 146 supersymmetry. 126. Vesto. 75. 239 Sandage.
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