33rd SLAC Summer Institute on Particle Physics (SSI 2005), 25 July - 5 August 2005Neutron Stars James M. Lattimer Dept. of Physics & Astronomy Stony Brook University Stony Brook, NY 11794-3800
[email protected] The structure, formation, and evolution of neutron stars are described. Neutron stars are laboratories for dense matter physics, since they contain the highest densities of cold matter in the universe. Combining laboratory experiments with observations of neutron stars will constrain the equation of state of relatively cold and dense matter. In addition, neutron stars are also laboratories for testing general relativity theory. 1. From Theory to Discovery Soon after the discovery of neutrons themselves by Chadwick in 1932, neutron stars were first proposed by Landau who suggested that in analogy to the support of white dwarfs by electron degeneracy pressure, neutron stars could be supported by neutron degeneracy pressure. Baade and Zwicky1 in 1934 firstsuggested neutron stars as being the remnants of supernovae. Tolman2 in 1939 produced a major study of theirtheoretical structure, employing the relativistic equations of stellar structure following from Einstein’s equations of general relativity. This work demonstrated, among other interesting ideas, that neutron stars could not be of arbitrarily large mass: general relativity introduces the concept of a neutron star maximum mass. Interestingly, the magnitude of the neutron star maximum mass is of the same order as the Chandrasekhar mass, which is the limiting mass for a white dwarf that exists in Newtonian gravity. Additional theoretical work followed, in which it was realized that neutron stars were likely to be rapidly rotating and to have intense magnetic fields. Pacini3 predicted that a rotating magnetized neutron starwould emit radio waves. But it was not until the 1967 Bell and Hewish4 discovery of radio pulsars that the existence of neutron stars was put on firm ground. Gold5 quickly made the connection between theextremely regular observed pulsations and a model of a highly magnetized, rapidly rotating and extremely compact configuration. The periods of pulsars range from seconds down to milliseconds. The rotation periods are obsered to be extremely stable, with P/P˙ ranging from thousands to millions of years. The periods are observed to be generally increasing. These three facts virtually guarantee that one is dealing with the rotation of a massive object. The short periods imply that L007 1 33rd SLAC Summer Institute on Particle Physics (SSI 2005), 25 July - 5 August 2005 the size of the pulsar is small. The distance cP is about 3000 km for P = 0.01 s. Unless relativistic beaming with γ ∼ 100 is occuring, a compact star is required. Vibrations of a star are ruled out because, generally, as vibrations decay their period decreases. Furthermore, the characteristic vibrational frequency is 2π(Gρ)−1/2, only a few milliseconds, and is too short. Finally, pulsing due to two orbiting objects is ruled out by the prediction of general relativity that a close compact binary will emit gravitational radiation, which again decreases its period as the orbit decays. A rotating rigid sphere with period P will begin to shed mass from its equator when the orbital period at the star’s surface equals P . Thus the limiting period is defined by 1/2 3/2 M R GM 2π = , Plim = 0.55 ms. R3 Plim M 10 km For a white dwarf, for example, the limiting rotational period would be of order 15 s. Only a neutron star can be more compact without forming a black hole. Although the details of the generating mechanism are vague, it is believed that the pulsations of a neutron star can be understood in terms of the energetics of a rotating dipole. If the magnetic and rotation axes are not perfectly aligned, then a beam of radiation emanating from the magnetic axes will pierce the surrounding space like a lighthouse beacon. The torque exerted by the beam will slowly spin down the pulsar. The magnetic moment magnitude of a sphere with a dipole field of strength B is |m| = BR3 /2. If the magnetic field is not aligned with the rotation axis, the magnetic field will change with time and therefore energy will be radiated. The emission rate is 2 B 2 R 6 Ω4 ¨ 2=− E˙ = − 3 |m| sin2 α, (1.1) 3c 6c3 where Ω = 2π/P and α is the misalignment angle. Ultimately, the source of the emitted energy is the rotational energy, 1 ˙ E˙ = IΩΩ (1.2) E = IΩ2 , 2 where I is the moment of inertia of the pulsar. If the original spin frequency is Ωi and the present frequency is Ω0 , one finds for the present age of the pulsar 2 −1/2 P Ω0 P 1− t= Ωi 2P˙ 2P˙ where we assumed Ωi >> Ω0 is the last step. We define T = 2P/P˙ as the characteristic age of the pulsar. In a few cases in which the age of the pulsar can be independently determined by direct observation (such as for the Crab pulsar) or inferred from kinematics, the characteristic age is only accurate to factors of 3 or so. Eliminating E˙ from Eq. (1.1) by using Eq. (1.2), one finds that the orthogonal component of the magnetic field is I G. (1.3) B sin α = 3 × 1019 P P˙ 45 10 g cm2 L007 2 33rd SLAC Summer Institute on Particle Physics (SSI 2005), 25 July - 5 August 2005 Figure 1.1: Period vs. period derivative for pulsars. Lines of constant characteristic age (red) and magnetic field strength (violet) are displayed. Generally, pulsars are born in the upper left part of the diagram, and evolve to longer periods and smaller period derivatives. It is not known on what timescale magnetic fields of pulsars decay. The lower right region represents a transition to pulsar death, when the emitted energy is too small to generate significant electron-positron pairs. The pulsars in the lower left corner are re-born, having been spun-up by accretion in binaries. The actual emission is believed to be due to curvature radiation from electrons trapped in the magnetic field. Since the potential drop is of order 1015−16 V, electrons will have energies of about 1012 eV, and the photons produced will have energies many times me c2 . Thus additional electron-positron pairs will be produced and the resulting pair-cascade shorts out the electric field propelling the charges into space. The high density of electrons makes maser activity possible, which ultimately results in radio waves. One major shortcoming of the magnetic dipole model concerns the braking index n=− ¨ ΩΩ , ˙2 Ω (1.4) ˙ ∝ −Ωn . The dipole model predicts this to which implies a power-law deceleration with Ω be 3, but observations of most pulsars yield a value significantly less than 3. L007 3 33rd SLAC Summer Institute on Particle Physics (SSI 2005), 25 July - 5 August 2005 Shortly after the first pulsar was discovered, a pulsar was found to exist at the center of the Crab nebula, the expanding gaseous remnant of a supernova known to have been visible in 1054. Fig. 1.2 shows a stone painting of the supernova with the Moon. Comparing the observed expansion of the ejecta with their angular distances from the center of the nebula confirms the time of the explosion. The Crab pulsar strengthened the connection between neutron stars and some types of supernovae, those that are collectively known as gravitational collapse supernovae. Figure 1.2: Anasazi rock painting showing SN 1054 and a crescent moon. In 1975, Hulse and Taylor6 discovered a pulsar which wasorbiting another star. Although this star has never been directly observed, its presence is deduced because of an additional periodicity imposed on the pulse stream. A straightforward application of Kepler’s Law The Doppler shift of the pulses, together with the orbital period, revealed that the combined masses of the two stars was about 2.8 M . The fact that no tidal distortions were apparently affecting the orbital motion, and the momentus discovery that the orbit was slowly shrinking, presumably due to the predicted emission of gravitational radiation, allowed the measurements of the individual masses in this system. The invisible companion’s mass turned out to be nearly equal to the pulsar’s mass, about 1.4 M . Although in principle, it could be a normal star or even a white dwarf, these possibilities are ruled out by its very invisibility. Finally, in 1987, the first supernova observed that year, in the close-by irregular galaxy known as the Large Magellanic Cloud, produced a brief neutrino pulse observed by at least two neutrino observatories, the IMB detector in the Morton Salt Mine near Cleveland, and the Kamiokande detector in the Japanese Alps. Models of supernovae predict that the aftermath of a gravitational collapse supernova will involve the formation of a proto-neutron star. Such a star is not only hot, with temperatures of nearly 1011 K, but is also lepton rich in comparison with a cold neutron star. The high concentration of leptons, which is the sum of the electron and neutrino concentrations, is mandated by the fact that neutrinos L007 4 33rd SLAC Summer Institute on Particle Physics (SSI 2005), 25 July - 5 August 2005 are trapped within the proto-neutron star at least for many seconds after its formation. Neutrinos are formed during gravitational collapse due to electron capture reactions induced by the increasing electron chemical potential. The duration of the neutrino burst, the average energy of the neutrinos, and the total number of neutrinos observed were all in accord with theoretical predictions7 . A massive amount of energy is released in a gravitational collapse supernova. Compared to the binding energy of the massive star that forms its progenitor, the binding energy of a neutron star is huge, about GM 2 /R ∼ 3 × 1053 ergs for M = 1.4 M and radius R = 15 km. The weak interaction cross section for neutrino scattering and absorption is σ0 ∼ 4 × 10−44 (Eν /MeV)2 cm2 , where Eν is the neutrino energy. We estimate the mean neutrino energy as the Fermi energy of degenerate neutrinos 1/3 1/3 1/3 ¯ c 6π 2ρN0 Yν 103 ρ14 Yν,0.04 MeV, (1.5) Eν = h where Yν,0.04 = Yν /0.04 is the concentration of neutrinos in dense proto-neutron star matter scaled to 0.04, and ρ14 is the density scaled to 1014 g cm−3 . The typical neutrino mean free path is then about λ 1 −5/3 −2/3 = 39.3 ρ14 Yν,0.04 cm. ρN0 σ0 (1.6) Compared to the core size of about 1/3 M1M 1/3 15M 16.8 km, R 8πρ ρ14 this is very small (M1M is the mass in units of solar masses). Diffusion of the neutrinos out of the core takes a time tdif f 1/3 3R2 = 7.2 ρ14 M1M Yν,0.04 s. cλ As the neutrinos diffuse out of the core, they both lose energy and encounter matter of lower density. Eventually, they reach a surface of last-scattering, or neutrinosphere, and escape. The condition on optical depth R dr 1, τ= R−∆R λ leads to an estimate for the depth ∆R √ under the surface where the neutrinosphere lies. One finds to order unity that ∆R λ0 R where λ0 is the central value of the mean free path. One can also see that the density at the neutrinosphere is approximately ∆R λ0 1/3 −1/3 −1/6 2ρ0 0.01 ρ14 Yν,0.04 M1M . (1.7) ρ (R − ∆R) 2ρ0 R R Although neutrinos at the neutrinosphere are not degenerate, using the degenerate approximation Eq. (1.5) allows us to estimate their mean escaping energy, which is 1/9 2/9 −1/18 Eν,esc 22 ρ14 Yν,0.04 M1M MeV. L007 5 (1.8) We will use units such that kB =1.1. Nevertheless. The extension of the equation of state to the highly interacting case will also be discussed. respectively.5 August 2005 Thus. the energies of the escaping neutrinos are very insensitive to assumptions about the central density and mass of the proto-neutron star (and by inference to the details of the neutrino opacity). muons and nucleons. Even if it were.1) The occupation index is the probability that a given momentum state will be occupied. it is: −1 E−µ . g = 1 for neutrinos). µ corresponds to the energy change when a particle is added to or subtracted from the system.2) f = 1 + exp T where µ is the chemical potential. using a phenomenological approach that can be considered as an extrapolation from the relatively well-understood matter found in atomic nuclei. in which the equation of state can be (mostly) adequately described by that of a perfect (non-interacting) gas. the equation of state of a non-interacting fermion gas serves as a useful framework. i. the equation of state would remain highly uncertain. 2. The number and internal energy densities are given.4) ns = − 3 [f ln f + (1 − f ) ln (1 − f )] d3 p h L007 6 . It is quite remarkable that the observations of neutrinos from SN 1987A were in accord with all aspects of the above discussion: total emitted energy. When the particles are interacting. although the timescale for diffusion is proportional to the assumed central density. 2. (2. h h where g is the spin degeneracy (g = 2j + 1 for massive particles. thus T = 1 MeV corresponds to T = 1. This is in contrast to most other astrophysical objects. Equation of State of a Perfect Fermion Gas The energy of a non-interacting particle is related to its rest mass m and momentum p by the relativistic relation E 2 = m2 c4 + p2 c2 . For high density matter. g = 2 for electrons.e. (2. temperature and composition. 25 July . For fermions. E generally contains an effective mass and an interaction energy contribution. by g g 3 = 3 Ef d3 p (2. The Equation of State The equation of state refers to the equations describing how the pressure and other thermodynamic variables such as the free energy and entropy depend upon the quantities of density. where j is the spin of the particle. The entropy per baryon s can be expressed as g (2.16 × 1010 K.3) n = 3 f d p. such as is found in neutron stars.33rd SLAC Summer Institute on Particle Physics (SSI 2005). such as normal stars and white dwarfs. the composition is not well understood. timescale and average neutrino energy.. 5 August 2005 Figure 2. all energies are in MeV. and the thermodynamic relations P = .1: Occupation probabilities for various µ and T . 25 July .33rd SLAC Summer Institute on Particle Physics (SSI 2005). ∂ (/n) . n2 . 4) and (2.6) 3h ∂p Thermodynamics gives also that . P = 3 p (2.5)) are generally valid for interacting gases. for future reference. ∂n s = T sn + µn − (2.5) gives the pressure. also. (2. that g ∂E 3 f d p. Incidentally. the two expressions (Eqs. We also note. ∂P . . n= ∂µ . . T . ∂P . . ∂T . ns = . µ (2.7) Note that if we define degeneracy parameters φ = µ/T (useful in the relativistic case) and ψ = (µ − mc2 )/T (useful in the non-relativistic case) the following relations are valid: . . . ∂ . . ∂P . . ∂P . . ∂P . . P = − + n . (2. + T = ns + nφ. = ns + nψ.8) . . ∂n ∂T n ∂T . ∂T . non-relativistic (p << mc). extreme degeneracy and non-relativity (EDNR). A useful two-dimensional polynomial expansion has been developed by Eggleton. Fig. Ellis andLattimer9 . T φ ψ In general. and non-degeneracy and non-relativity (NDNR). extremely relativistic (p >> mc).2 shows thebehavior of various thermodynamic quantities in the density-temperature plane for an ideal gas. In many situations. L007 7 . 2. nondegenerate (φ → −∞). one or the other of the following limits may be realized: extremely degenerate (φ → +∞). Flannery and Faulkner8 and refined by Johns. In particular. it is useful to examine the cases of extreme degeneracy and extreme relativity (EDER). these equations are non-analytic except in limiting cases and full integrations are necessary. extreme cases of degeneracy and relativity. where nc = (g/2π 2)(mc/¯ The four regions where limiting approximations are valid. Contours of ψ (solid curves) and φ (dashed curves) run from lower left to upper right. These regions are separated by the curves ψ = 0 and the lines T = mc2 and n = 3nc . 2.2: Thermodynamics of a perfect fermion gas. 2 π g µ 3 +··· . 3 4 φ s =π 2 /φ + · · · L007 8 (2. The pressure is proportional to n4/3 .5 August 2005 Figure 2.1. n= 2 1+ 6π ¯hc φ 2 EDER nµ π =P = 1+ +··· .e. EDER and NDER. EDNR. thogonal solid lines show contours of log10 P/(nc mc2 ).1. the rest mass is negligible. Orh)3 . respectively. 25 July .. i. are indicated by NDNR. Extreme degeneracy and relativity In this case.33rd SLAC Summer Institute on Particle Physics (SSI 2005).9) . This situation is appropriate for trapped neutrinos.10) s =π 2 /2ψ + · · · .33rd SLAC Summer Institute on Particle Physics (SSI 2005).12) =P = + 1+2 . 3 5 2 ψ g n= 2 6π EDNR (2. 6π 2 ¯hc µ 2 4 gµ µ 3 7 πT πT (2. 3 2 T g πT n =n(+) − n(−) = 1+ . NDNR (2. s= 15 µ 6n (¯ hc)2 Note that the equation for the density can be inverted to yield µ = r − q/r. 8 ψ ¯h2 2 1 2nψT π 2 − nmc2 = P = 1+ +··· . 3/2 2 2mψT 1 π 1+ +··· .4. where µ = µ(+) is the chemical potential of the particles.1. 2.1. the chemical potential of antiparticles µ(−) = −µ(+) .13) q= 3 .5 August 2005 2. Non-degeneracy and non-relativity 3/2 mT n =g 2π¯h2 2 − nmc2 /3 = P =nT. 9 3π 2 n (¯ hc)3 . 25 July . L007 r= 2 q +r 2 1/2 1/3 +t . t= g (πT )2 . Relativistic including particle-antiparticle pairs In chemical equilibrium.2.3. eψ .5 MeV. (2. 2. 3 24π 2 ¯hc µ 15 µ 2 7 πT gT µ2 1+ . Extreme degeneracy and non-relativity The presure is proportional to n5/3 .1.11) s =5/2 − ψ. and also for electrons above 106 g cm−3 or T > 0. the so-called Thomas-Fermi approximation.33rd SLAC Summer Institute on Particle Physics (SSI 2005). np ) (2. The temperature contributions enter only from the Fermi statistics in the kinetic energy term.18) and ηt = (µt − Vt )/T is the degeneracy parameter. while Walecka-type models are examples of the latter. 1 + exp (u − η) (2. It is also clear that the uniform matter energy density can be written now as a sum of kinetic and potential contributions = HB = t ¯2 h τt + U (nn . np ) 2m∗t (nn . 2.2. We present only a brief descriptions of non-relativistic models10 . 2π 2 ¯h5 ¯h2 0 where Fi is the normal Fermi integral Fi (η) = 0 ∞ 1 2π 2¯h3 (2. 25 July . np ) and the auxiliary variables of kinetic energy densities (τn . Non-relativistic potential models The uniform matter energy density.1. np . ∂τt ∂nt 2m∗t (2.5 August 2005 2. Interacting Fermi Gas Most equations of state for neutron stars are based on one of two approaches – nonrelativistic potential models or relativistic field-theoretical models. np .17a) (2.14) which is a funcional of both nucleon densities (nn . The number and kinetic densities become 3 3 −1 ∞ ft (p. τp ) . τp). ¯h2 0 ∗ 5/2 3 5 −1 ∞ 2mt T 1 2 3 τt = 2π ¯h ft (p. the single particle energies of the nucleons are defined by Et (p) = ∂HB 2 ∂HB ¯h2 2 p + ≡ p + Vt .17b) (2. the bulk nuclear force is completely specified by the density dependance of m∗ and U .2. t = n. Therefore. Examples of the former include Skyrme forces. The occupation probabilities are ft (p.16) ui du.15) It is convenient to have thereby defined the effective nucleon masses m∗t and interaction potentials Vt . τn. p is the isospin index. L007 10 . T ) = [exp ([t (p) − µt ] /T ) + 1]−1 . T ) = HB (nn . which is dependent upon particle densities and temperature only. is given by a bulk Hamiltonian density HB (nn . (2. T ) p d p = F3/2 (ηt ) . Treating the single particle states as plane waves. T ) d3 p = nt = 2π ¯h 3/2 2m∗t T F1/2 (ηt ) .19) where U is the potential energy density. (2. the energy per baryon is finite at zero density and temperature unless B = −K/18. such as to very low and high densities.2. we will utilize this energy function with some success even under extreme extrapolations. For example. 6m∗t T 2π 3 ¯h2 0 (2. Eq. T.5 August 2005 From Eq.2) matter. 2 s ∂nt ∂nt (2. and this of course vanishes at zero density.24). ¯ 2 τt h nt ηt − 3m∗t + U.24).24) represents an three-dimensional expansion for the energy in the vicinity of n = n0 . Nevertheless.22) The free energy density F is F = − TS = t 2. K 225 MeV is the incompressibility parameter.7 for cold (T = 0). In other words. 25 July . symmetric (x = 1. are constrained by nuclear masses and K has been constrained by giant monopole resonances in nuclei. (2. this is an excellent approximation in the vicinity of the nuclear saturation density n ∼ n0 0.24) + Sv (1 − 2x)2 + a T2 . (2. B and Sv . (2. and to extremely neutron-rich matter.21) The pressure is P = (nt µt + T St ) − E = t t ¯ 2 τt h nt Vt + 3m∗t − U. in most applications we need the energy density rather than the specific energy. T = 0 and x = 1/2. The level density L007 11 . and is infinite at zero density and finite temperatures.2. in almost all situations its use remains justifiable. 1− (n.33rd SLAC Summer Institute on Particle Physics (SSI 2005). (2. Sv 30 MeV is the volume symmetry energy parameter.23) Schematic Hamiltonian density Models for the Hamiltonian density can be relatively complicated.15) it can be seen that Vt satisfies ¯ 2 ∂ (m∗s )−1 h ∂U Vt = τs + . x) = n B + 18 n0 n0 n Actually. to moderate temperatures.20) The entropy density of interacting Fermi-Dirac particles is ∞ 1 5¯h2 τt 3 ft ln ft + (1 − ft ) ln (1 − ft ) d p = St = − nt ηt . In Eq. B −16 MeV represents the binding energy of cold symmetric matter at the saturation density. However. Even though this model appear to break down in some circumstances . (2. Two of the parameters in Eq.16 fm−3 (corresponding to a mass density of about ρ14 = 2. and a (15 MeV)−1 is the nuclear level density parameter. but many of the concepts can be illustrated by using a parametrized energy density 2 n 2/3 n n K 0 (2. 33rd SLAC Summer Institute on Particle Physics (SSI 2005). From the energy density of uniform matter Eq. Plus. 25 July . the pressure of the two-phase mixture will no longer be negative. Because surface effects are generally reduced compared to volume effects by a factor A−1/3 . The latter is the correct behavior. This can be illustrated simply for the case of symmetric matter. and is negative in-between.5 August 2005 parameter is also constrained by laboratory measurements. The total free energy density of this system will satisfy F1 = uFI + (1 − u) FII . (2. they remain large even in the largest nuclei.27) Minimizaing F1 with respect to nI and u. (2. Ordinary matter cannot exist in such a condition. S =2an n These approximations have a number of believable aspects: the pressure vanishes both at zero density and at n0 . these constraints are not perfect because nuclei have finite size while Eq. chemical potentials and entropy density: n2 K n 2a n0 2/3 2 2 P = − 1 + Sv (1 − 2x) − n T . the a parameter) are fully equal to the volume contribution.3. =u + (1 − u) ∂nI ∂nI ∂nII 1 − u (2. in the density range from 0 to n0 is negative.26) (2. even though Eq. ∂u ∂nII 1 − u 1 − u L007 12 .28) ∂FII −nI nII ∂F1 =FI − FII + (1 − u) + = 0.24) is a degenerate expansion. both with the same pressure. valid for a non-degenerate gas. 2. We can illustrate that dividing such matter into two phases with different densities will result in a lower free energy than a single uniform phase at the same average density. one can determine the pressure. results in −u ∂FI ∂FII ∂F1 = 0. surface symmetry energies and incompressibility contributions are comparable to volume terms and difficult to uniquely determine.24) refers to infinite matter. n = unI + (1 − u) nII .25) n µ ˆ =4Sv (1 − 2x) .24). surface energies. µn =B + 18 n0 n0 n0 3 n (2. in this case 0. matter will separate into two phases.. it is believed that surface contributions to the nuclear specific heat (i. at zero temperature. n0 9 n0 3 n a n0 2/3 2 n n K n 1−3 + 2Sv 1 − 4x2 − 1− T . (2. Phase Coexistence We saw that the pressure. and the neutron and proton chemical potentials tend to negative infinity in the limit of low density. In a similar way. using the density constraint to eliminate nII . For example.e. Matter in the two phases in phase coexistence will have to be in chemical and pressure equilibrium. (2. Therefore surface and Coulomb effects have to be considered. n0 n 2/3 0 T. However. In practice. PI = PII . these equations have the solution nI = n0 and nII ≈ 0.) Since u = n/ns . At finite temperature.3 shows the phase coexistence region for the parametrized energy density L007 13 . F2 > F1 . The critical density and temperature are indicated by a filled black circle. Dashed lines connect the two densities satisfying equilibrium conditions. unlike system 2 for which P2 < 0. Parameters used were K = 225 MeV. one finds that F1 = nB. On the other hand.24) in the limit T → 0. uniform matter at the density n will have a free energy density 2 2 K K n n . 25 July . a = (15 MeV)−1 . 2. (Note that one has to work in the limit T → 0 rather than use T = 0.33rd SLAC Summer Institute on Particle Physics (SSI 2005). Upper panel: The coexistence region in the densitytemperature plane is colored yellow. These result in the equilibrium conditions µI = µII . Fig.5 August 2005 Figure 2. and has a physically achievable pressure. where µ = ∂F/∂n. In this case PI = PII = 0 and µI = µII B. for the schematic energy density Eq. Approximately. phase coexistence is still possible. (2. Entropy per baryon contours are also displayed.3: Lower panel: Pressure isotherms for the schematic energy density for x = 1/2. System 1 is preferred. but for a lessened density range. F2 = n B + = F1 + n 1− 1− 18 n0 18 n0 Obviously. 29) and (2. PI = PI I. the phase coexistence model only demonstrates that matter divides into two phases. and the dashed lines connect the densities where the equilibrium conditions are satisfied.corrected the treatment of the surface energy. one can show that 1/3 1/3 1/2 1/2 5 12 5K 5Ka 5 Tc = . 12 12 32a 5 8 where sc is the entropy per baryon at the critical point. 12 and Ref. The density inside nuclei is close to the saturation density n0 .I = µp. Eq. where the three major finite-size effects concern surface. and further details can be sought therein. 13. To leading order.12 extended the treatment to finite temperature. and introduced the concept of using consistent nuclear interactions for matter both inside and outside the nuclei. (2.26) is modified: fLD (2. This critical temperature Tc . However. VN where the liquid droplet energy of a nucleus is fLD = fS + fC + fT . (2. Lattimer and Swesty13 extended the treatment to includenuclear shape variations and alternate nuclear interactions.24).24). sc = . the entire region with densities less than n0 consists of nuclei: the dense phase within nuclei is in equilibrium with the surrounding gas. respectively.33rd SLAC Summer Institute on Particle Physics (SSI 2005). the maximum temperature for coexistence is less than Tc and the pressure along isotherms steadily increases through the two-phase region rather than remaining constant.30) µn. where Ye is the ratio of protons to baryons (the number of electrons equals the number of protons for charge neutrality). an additional constraint corresponding to charge neutrality is enforced: (2. The lower part of the figure displays the pressure along isotherms.II . (2.5 August 2005 of Eq. While the resulting phase coexistence region remains similar to that found in Fig. (2. 25 July . 2. L007 14 .4. The treatment described here is based on Ref.29) nYe = unI xI + (1 − u) nII xII . = ∂n ∂n2 For the energy density of Eq. (2. is defined by the conditions ∂P ∂ 2P = 0. When matter is asymmetric. finite-size effects of nuclei can be described by a droplet model. 2. It is clear that a maximum temperature exists for which two-phase equilibrium is possible.31) F1 = u FI + + (1 − u) FII . Lattimer et al. (2.26) using Eqs.3. Minimizing the free energy density Eq. that is Ye < 1/2. and the accompanying critical density nc .II . This approach was considered for zero temperature nuclei in dense matter originally by Baym. Bethe and Pethick11 .27) results in the equilibrium coniditions µp. these effects can be considered separately. but does not indicate the sizes of the resulting nuclei.I = µn. Nuclear Droplet Model The results for phase coexistence illustrate that at zero temperature. Coulomb and translational contributions. Essentially. nc = n0 . To a very good approximation. 25 July . The background neutralizing electrons are uniformly VN = 4πRN distributed within Vc . fT = T ln nQ VN A3/2 2π¯h2 To simplify the following algebra.3.1. we have considered only the volume free energy of uniform nuclear matter. Finite temperature effects may be ignored. However.1. the uniform density nucleus is contained within a neutral sphere of volume Vc = VN /u. where 3 /3 is the nuclear volume. Pethick and Wilson14 showed. where RN is the nuclear radius.5 August 2005 2. as Ravenhall. Nuclear surface energy In principle. The minimization is a functional variation because the free energy must be optimized with respect to the density profile across the interface.2. For more discussion. non-degenerate and non-relativistic. in addition to their internal energies. and also see Fig. see Ref. However. 2 L007 15 (2. Hence. This introduces an additional gradient term to the total free energy density. In any case. 11.4. which to lowest order can be written 1 F = FI + F∇ = FI (n) + Q (n) (∇n)2 . (2. the density is rapidly varying. Nuclear Coulomb energy The Coulomb energy of a uniform density charged sphere is 3(Ze)2 /(5RN ). fT = 5/2 A0 A0 nQ A (2. Nuclear translational energy Nuclei themselves are (relatively) non-interacting.35) . Up until now.33) − T ≡ µT − T. The total Coulomb free energy of this configuration is 3 Z 2 e2 3 1/3 u 3 Z 2 e2 fc = D (u) .nuclear deformations and changes to rods or plates can be accomodated by a suitable modification of D. they have a translational free energy per nucleus of 3/2 u mb T nQ = .34) 0 where A0 60 is taken to be a constant. 13. But in the vicinity of the nuclear surface. the surface tension can be calculated by minimizing the total free energy involved in a semi-infinite interface. we choose to modify this result to VN nI VN nI unI (µT − T ) ≡ ln −1 . the surface tension is actually a surface thermodynamic potential density. As calculated in Ref.33rd SLAC Summer Institute on Particle Physics (SSI 2005).4. 2.4. one can consider the surface energy of a sphere to be the area multiplied by the surface tension. (2. in terms of thermodynamics. 2. at high densities this energy is modified by the close proximity of other nuclei (lattice effects).32) 1− u + ≡ 5 RN 2 2 5 RN The function D used here is for spherical nuclei. In the Wigner-Seitz approximation. 3. (2. First. Note that F − µn vanishes at large distances from the surface so that the integral is finite.39) n0 = 2Q FI − µn dn. (2. because the pressure at both 2 σ.38) into Eq. (2. The right-hand side of this expression represents the leptodermous expansion to a semi-infinite interface. (2.5 August 2005 where FI (n) is the uniform matter energy density and F∇ (n.39).24) in the case of symmetric matter and zero temperature. boundaries vanishes. 25 July . As x → +∞ this result is trivial. (2. We need to minimize the total free energy subject to the constraint of a fixed number of particles ∞ ∞ 3 2 ˜ (F − µn) d r 4πRN (F − µn) dx (2. consider symmetric matter so only one species of nucleon need be included. (2. consider the schematic free energy density Eq.33rd SLAC Summer Institute on Particle Physics (SSI 2005). 0 As an example.38) 2 We chose the constant of integration to ensure that the density gradient vanishes far from the interface. since both F and n vanish. where in the symmetric matter case n0 is the saturation density. it can be integrated to yield 1 2 Qn = FI − µn. (2.36) f= −∞ 0 where F is the free energy density and µ is a Lagrange parameter that turns out to be the chemical potential of the system. ∇n) contain gradient contributions. n0 1/2 n n n 4 1 3 QKn0 QKn30. (2. 0 Also note that since µA is the energy of A nucleons at the saturation density that the nuclear surface energy is just fS ≡ f˜m in. The surface energy becomes. although this is not necessary.36). although this step is not necessary to determine the surface tension. we assume for simplicity that Q is independent of density.36).40) 1− d = σ= 3 n n n 45 0 0 0 0 In this case there is no difference between the surface free energy per unit area and the surface tension. To proceed. As x → −∞ this requires that µ = F (n0 )/n0 . we find ∂FI − µ − Qn = 0. the surface radius is defined by ∞ 3 nd3 r = 4πn0RN /3 = A. (2. which is equivalent to minimizing the argument of the integral only.36) thus gives ∞ ∞ ∞ Q (F − µn) dx = (FI − µn) dx FI − µn + n dx = 2 σ= 2 −∞ −∞ −∞ (2.37) ∂n Derivatives with respect to density are indicated by ’s. Substituing Eq. We have µ = B. Minimizing Eq. or thermodynamic potential per unit area. using Eq. the minimized energy in Eq. Multiplying this equation by n . This equation can be further integrated to yield the density profile. The surface free energy is thus 4πRN L007 16 . In this model. 9 n0 (2.33rd SLAC Summer Institute on Particle Physics (SSI 2005).41) with respect to the neutron and proton density profiles. (2. ∂np ∂n ∂FI ∂FI −µ ˆ. (2. (2. Eq. one finds ∂FI − µn . (2. Qnp nn + Qpp np = ∂np Qnn nn + Qnp np = (2. In this case ∞ σ= (F − µn nn − µp np − P∞ ) dx.43) These can be written as ∂FI − µn + ∂nn ∂FI (Qnn − Qnp ) α = − µn − ∂nn (Qnn + Qnp ) n = ∂FI ∂FI − µp = 2 − µn − µp . n∞ = n0 .44) We introduced the asymmetry density as α = nn − np .41) −∞ where 1 2 2 Qnn (∇nn ) + 2Qnp ∇nn ∇np + Qpp (∇np ) . Thus we choose to write the surface tension as L007 17 .24). P∞ is the equilibrium pressure. For the case of the schematic energy density at zero temperature. n = 2 9 n0 Qnn − Qnp 2 2Sv α = (α∞ − α)2 . 2 n0 (2. (2.42) F = FI + 2 Also. (Qnn − Qnp ) α = n0 where α∞ = n∞ (1 − 2x∞ ).5 August 2005 For asymmetric matter.45) 4Sv (α − α∞ ) . Assuming P∞ = 0. + µp = 2 ∂np ∂α (2. P∞ will vanish unless the proton fraction in the dense phase is so small that the neutron chemical potential there becomes positive. There are two contraints. we choose x0 = np∞ /n∞ to be the proton fraction in the dense phase. Minimizing the integrand of Eq. 25 July . We introduced the equilibrium density n∞ in the dense phase. Each of Eq.47) It is apparent that the quantity σ is not a free energy but rather a thermodynamic potential density per unit area. one has n − n∞ K 3 (Qnn + Qnp ) n = −4 . defined by 9Sv n∞ =1− (2. for x∞ = 1/2. ∂nn ∂FI − µp . A thermodynamic potential is a perfect differential of the chemical potential and the temperature.36) can be integrated: 2 (Qnn + Qnp ) 2 K n∞ − n (n + 2n∞ − 2n0 ) . one each for proton number and neutron number.46) (1 − 2x∞ )2 . n0 K Of course. (2. 0= ∂RN 5 RN RN L007 18 .4. (2.II − xII µ ˆII (nII − nI ) 0= ∂u RN A0 4π 3 (nI xI eRN )2 D + uD .51a) + nI (xI eRN )2 D + 5 A0 8π ∂F1 xI nI (eRN )2 D .33rd SLAC Summer Institute on Particle Physics (SSI 2005).50) RN Using these two equations to relate nII and xII to other variables.51b) ˆII − µ 0= ∂xI 5 3u ∂F1 = (ˆ µII + µs ) .5 August 2005 σ(µs ) where µs is the chemical potential of the nucleons in the surface15 .49) 3 4π nI 2 (nI xI eRN ) + =u FI + (σ + µs νs ) + (µT − T ) + (1 − u) FII . (2. = unI µ ˆI + (2.51c) 0= ∂νs RN 3u ∂σ ∂FI = + νs .51e) µ ˆII νs − nI (xI − xII ) + + R 5 N 3 8π 3 ∂F1 (nI xI e)2 RN − 2 µ (2. we employed the relations of Eq. . The electron EOS does not affect the relative energies of the baryons. (2. µ ˆII = 4Sv (1 − 2xI ) n∞ /n0 . (2. RN 5 A0 In addition. we minimize F1 : 1 ∂F1 = µn. 2.12). if the surface tension is expressed as a quadratic expansion.48) The determination of the function σ(µs ) is discussed below. nYe = unI xI + (1 − u) nII xII + νs . For the calculations described below. except for influencing the composition of matter in beta equilibrium. so the perfect fermion EOS discussed earlier is appropriate for them. Thus. (2. one has σ = σ0 − σδ (1 − 2xI )2 .4.5. The number ofparticles associated with the surface per unit area is νs = −∂σ/∂µs from thermodynamics. we have the conservation equations 3u n + unI + (1 − u) nII . Liquid droplet equilibrium conditions The total nucleonic free energy density of matter containing nuclei in the droplet model can now be written: 2 4πRN 3 Z 2 e2 unI (σ + µs νs ) + D+ (µT − T ) F1 =uFI + (1 − u) FII + u VN 5 RN VN A0 (2. and P∞ = 0. 2. and the surface free energy density is then σ + µs νs . (2. 25 July . Electron energy Electrons form a nearly ideal and uniform gas.4.51d) ∂µs RN ∂µs 3 nI ∂F1 = FI − FII + (σ + µs νs ) + µT + µn.51f ) = u − 2 (σ + µs νs ) + ˆII νs .I − xI µ ˆI − µn.II + xII µ ˆII + µ ˆ II (xI − xII ) 0= u ∂nI µT 8π . for a given charge ratio. modified by surface and Coulomb effects. A0 (2. Note that nuclei persist to high temperatures even for very neutron-rich matter. althought the mass fractions of nuclei in the most extreme case are only a few percent even at the lowest temperatures. Such extremely neutron-rich nuclei are not observed in the laboratory. lastly. the inner and outer cores. due to Baym.1 −0. Internal Composition of Neutron Stars A schematic view of the insides of a neutron star. the crust. The electron contributions to all thermodynamics have been included.4 and Fig. 1+ 2 RN D 1/3 15σ . µn = µn.52d) (2. modified by translation and Coulomb effects. pressure equality. which states that the optimum nuclear size.52a) 3σ .35 and Ye = 0. R N n I xI −1 ∂σ ∂σ ∂µs νs = − =− . Also. but the atmosphere plays an important role in shaping the emergent photon spectrum. primarily contains nuclei. 3.52e) These have simple physical interpretations: neutron and proton chemical equalities. but rare-isotope accelerators hope to create some of them. L007 19 . respectively.5 August 2005 These can be more compactly written as µn. is shown in Fig. is set when the surface energy equals twice the Coulomb energy. extending approximately 1 to 2 km below the surface. the Nuclear Virial Theorem. and range from 56 Fe for matter with densities less than about 106 g cm−3 to nuclei with A ∼ 200 but x ∼ (0. courtesy of D. Page. The atmosphere and envelope contain a negligible amount of mass.I + µT . 3. A neutron star can be considered as having five major regions. and. for Ye = 0. The crust. ∂µs ∂xI ∂xI 3 σ uD PII = PI + .02. 2.5. We see that this theorem is correct only if the surface “energy” actually refers to the surface thermodynamic potential. The dominant nuclei in the crust vary with density.52c) (2. 2. Bethe and Pethick. The overall pressure and chemical potentials of matter with nuclei can be simply stated: P = PII + Pe . (2. 25 July . the envelope and the atmosphere. µ ˆ=µ ˆII .2) near the core-crust interface at n ≈ n0 /3.1.33rd SLAC Summer Institute on Particle Physics (SSI 2005).53) Some results of the liquid droplet EOS are displayed in Fig.II = µn. RN = 8πn2I x2I e2 D µ ˆII = − µs = µ ˆI + (2. photon contributions are also included.52b) (2. and the envelope crucially influences the transport and release of thermal energy from the star’s surface. the definition of the nucleon surface density.II . 4: Liquid droplet EOS. to 2-D cylindrical nuclei (spaghetti). the nuclear lattice turns “inside-out“ and forms a lattice of voids.1n0 . there could be a continuous change of the dimensionality of matter from 3-D nuclei (meatballs). If so. beginning at about 0. Ye = 0. Lower right: and µ ˆ − µe . Upper right: Contours of µn 2 ν contours.35 Upperleft: Pressure and entropy contours. cm−3 L007 20 . 12. which is eventually squeezed out at densities near n0 .5 August 2005 Figure 2.33rd SLAC Summer Institute on Particle Physics (SSI 2005). At the core-crust interface. more of the matter resides in the neutron fluid than in nuclei. parameters from Ref. Z and Ns = 4πRN s Contours of mass fractons of heavy nuclei (XH ) and α particles (Xα). as described in Ref. 13. 25 July . Lower left: A. at densities above the so-called “neutron drip” density 4 × 1011 g where the neutron chemical potential (the energy required to remove a neutron from the filled sea of degenerate fermions) is zero. neutrons leak out of nuclei. It could well be that at somewhat lower densities. Within the crust. nuclei are so closely packed that they are virtually touching. At the highest densities in the crust. 5 August 2005 Figure 2. to 2-D cylindrical voids (ziti). could play a major role in pulsar glitch phenomena.4. Second.1 before an eventual transition to uniform nucleonic matter (sauce). This is important because.5: Same as Fig.02. but for Ye = 0. first. electrons and muons. 25 July .1 MeV the neutron fluid in the crust forms a 1 S0 superfluid. being loosely coupled to the crust. to 1-D slabs of nuclei interlaid with planar voids (lasagna). The neutrons could form a 3 P2 superfluid and the protons a 1 S0 superconducter within the outer core. it would form a reservoir of angular momentum that. it would alter the specific heat and the neutrino emissivities of the crust. The core comprises nearly all the mass of the star. 3. thereby affecting how neutron stars cool. It seems likely that for temperatures smaller than about 0. In the inner core exotic particles such as strangeness-bearing hyperons and/or Bose condensates (pions or kaons) may become abundant. This series of transitions is thus known as the “nuclear pasta”. At least in its outer portion. it consists of a soup of nucleons.33rd SLAC Summer Institute on Particle Physics (SSI 2005). It is even possible that a transition to a mixed phase of L007 21 . to 3-D voids (ravioli. 2. or Swiss cheese in Fig. to spherical nuclei at low densities. Recent years have seen intense activity in delineating the phase structure of dense cold quark matter. a crystalline phase.1: The internal composition of a neutron star. In quark phases with finite gaps.5 August 2005 Figure 3. The densities at which these phases occur are still somewhat uncertain. from uniform matter at high densities. Novel states of matter uncovered so far include color-superconducting phases with and without condensed mesons. L007 22 . 25 July . The top band illustrates the geometric transitions that might occur. hadronic and deconfined quark matter develops. Superfluid aspects of the crust and core are shown in insets.33rd SLAC Summer Institute on Particle Physics (SSI 2005). and a gapless superconducting phase. initial estimates indicate gaps of several tens of MeV or more in contrast to gaps of a few MeV in baryonic phases. a color-flavor-locked (CFL) phase. even if strange quark matter is not the ultimate ground state of matter. Examples in the former case include a two-flavor superconducting (2SC) phase. i.02. will vary more. Crustal Neutron Star Matter: n < n0 It is instructive to examine some consequences of the equilibrium conditions derived above as they pertain to the composition of matter in the crust of a neutron star. the liquid droplet model predicts 1/3 3 4nI ¯ c 3π 2 nxI . The proton fraction inside nuclei.02n0 with a value of about 0. and assuming that the abundance of outside nucleons is very small. Left panel: Ye contours.3.e. 3. first of all.2. that the density inside nuclei will remain close to n0 irrespective of the overall matter density n. it is apparent that the nuclear mass number steadily increases with density until nuclei dissolve around 0.2: Beta equilibrium matter for the SKM∗ interaction. 3. (2.33rd SLAC Summer Institute on Particle Physics (SSI 2005). 25 July . which results in 1/3 µ ˆ = µe = h ¯ c 3π 2 nYe . since neutrinos can freely escape on the timescales of interest. using the schematic interaction Eq.5 August 2005 3. Right panel: nuclear XH and α particle Xα abundances. Note that at zero temperature. A cold neutron star is in beta equilibrium. Details of the nuclear composition are shown in Fig. The pressure condition. where the density n < n0 . (This value of density marks the minimum value of Ye . ensures. Also. Ye has a minimum in the vicinity of 0. In such matter..02 fm−3 . In beta equilibrium. however. 4Sv (1 − 2xI ) + xI D 2 = h 5 3π A L007 23 (3.) Figure 3. the beta-equilibrium properties of the matter are shown in Fig.2) .1.24). the abundance of heavy nuclei abruptly decreases above 10−4 fm−3 .1) where the electrons are assumed to be degenerate and relativisitic. the optimum composition is obtained by finding the optimum value of Ye by free energy minimization. (3. For the SKM* interaction. and the relatively large value of the incompressibility parameter K. xI must decrease.5 August 2005 Figure 3. (3.4) xd 2 2Sv The linear symmetry energy overestimates xd compared to the models shown in Fig. as the density n increases. (2. σ − σ (1 − 2x ) 0 I δ 2nI x2I e2 D (3. (3. The “neutron drip point” is the density where Ye xd .25). 25 July .33rd SLAC Summer Institute on Particle Physics (SSI 2005). neutrons flood out of nuclei. Around the density n ≈ 10−4 n0 . the neutron chemical potential becomes positive.41. Above this density.2 (recall that Ye xI ). 3. A= 5π 2 .3: Beta equilibrium matter for the SKM∗ interaction. As a result. Left panel: A and Z contours. one has 1 B 1+ 0.2. Right panel: presure and entropy contours. as observed in Fig. From Eq. L007 24 . A increases.3) According to Eq. 3.2). In Fig. First.varies in a power law fashion with mass and the constant K in the polytrope law: R ∝ K n/(e−n) M (1−n)/(3−n) .2.16 fm−3 . It is easily seen that the pressure of matter L007 25 . Nevertheless.4. 16. 16. a lot of freedom still exists. Figure 3. dimensional analysis shows that the stellar radius of an object with a constant polytropic index. or to proton fractions below 0.44 M .4.33rd SLAC Summer Institute on Particle Physics (SSI 2005). we have two constraints. extrapolating these interactions to twice nuclear density and beyond. A survey of many often-used nuclear forces is described in Ref. namely causality and the existence of neutron stars of at least 1. From the Newtonian equation of hydrostatic equilibrium. The figure notes a number of important points. and for normal EOS’s. is dangerous. P = Kρn+1 . 3.5) Second. Ultra-High Density Matter Despite the existence of many models of nuclear matter that successfully predict several aspects of nuclear structure. 25 July . n0 0. we compare the pressure-density relations of beta-equilibrium neutron star matter for some of these models. (3.4: Pressure-density relation for EOS’s described in Ref. at nuclear saturation density. the effective polytropic index n = ∂P/∂ρ − 1 of most normal nuclear EOS’s is about 1. the uncertainty in the pressure is about a factor of 6.5 August 2005 3. From a practical standpoint. Third.5). this might be somewhat too steep. (2. Einstein’s equations for this metric are: 1 λ (r) 8πρ (r) = 2 1 − e−λ + e−λ . matter in our universe. when n = n0 . General Relativistic Structure Equations We confine attention to spherically symmetric configurations. for example. and is thus a limit to the ultimate energy density of cold. Neutron Star Structure Newtonian hydrostatic equilibrium. The metric for the static case can generally be written ds2 = eλ(r) dr 2 + r 2 dθ 2 + sin2 θdφ2 − eν(r) dt2 .24). Obviously. the strange-quark matter EOS’s. one might anticipate that the radius of a star constructed with a weaker symmetry energy would be smaller than one constructed with a stronger symmetry dependence. This will lead to significant differences in the structure of pure strange quark matter stars. As derived in any text on GR. Probably the most important defect is the inability to predict the existence of the maximum mass. The schematic energy density in Eq. but in addition. 4. In practice. it should be noted that many normal EOS’s display significant softening (flattenting) of the pressure-density relation in the range 2 − 4n0 . 8πp (r) = − 2 1 − e−λ + e−λ r r p (r) + ρ (r) p (r) = − ν (r) . and leads to about a 50% uncertainty in the predicted neutron star radius. (4.5 August 2005 in the vicinity of n0 is completely determine by the symmetry parameter Sv and its density dependence. (2. Suprisingly. breaks down for neutron stars.33rd SLAC Summer Institute on Particle Physics (SSI 2005). who demonstrated that the escape velocity GM/R could eventually exceed the speed of light. 2 L007 26 . This is actually true in practice. static. the pressure at zero temperature is Sv n0 for neutron-rich matter. the kinetic energy density of the fermionic baryons has a symmetry energy contribution that scales as n2/3 . this limit carries over into general relativity (GR). then the pressure of pure neutron matter at n0 is pn0 Sv . Using. This behavior has consequences for limiting the value of the neutron star maximum mass. which have a finite density at zero pressure.2) . Compactness limits were described over 200 years ago by Laplace.25). Fourth. The fact that the pressure is uncertain is a direct statement about our lack of knowledge of the symmetry energy’s density dependence. the schematic expansion Eq. the pressure is given in Eq. 25 July . GR also predicts that the measurement of any neutron star mass leads to a limit on the maximum density inside any neutron star. have a completely different low-density behavior than normal matter EOS’s. 4. which is adequate for most stars. In general. (2.1. GR predicts a number of further constraints on compactness. r r ν (r) 1 (4. Besides the additional limit imposed on the mass. if the symmetry energy scales as np .1) The functions λ(r) and ν(r) are referred to as metric functions. as we will observe in the next chapter. For one thing. According to the scaling in Eq. (3.24) utilized a linearly-varying symmetry energy per particle. The nucleon number is R 2 λ/2 4πr e N= 0 R n (r) dr = 0 2m (r) 4πr n (r) 1 − r 2 −1/2 dr. The constant of integration for the number density can be established from conditions at the surface of the star. also known as the Tolman-Oppenheimer-Volkov (TOV) equation in GR: ν (r) m (r) + 4πr 3p (r) −p (r) = = . we have ρ = n(m+e). (4. exactly the limit deduced by Laplace. which terminates at the radius R. From the above. ρ+p 2 dP dn = dρ .6) Another quantity of interest is the total number of nucleons in the star. L007 27 (4.3) e = 1 − 2m (r) /r. r≥R (4. if there is uniform entropy per nucleon.5 August 2005 Derivatives with respect to the radius are denoted by . ρ (r) + p (r) 2 r (r − 2m (r)) r≤R (4.475 km. p = n2 de/dn. (4. Outside the distribution of mass. eν = e−λ = 1 − .4) Near the origin. and Einstein’s equations give 2M m (r) = m (R) ≡ M. The first of Eq. the enclosed gravitational mass. (4.33rd SLAC Summer Institute on Particle Physics (SSI 2005). This is not just M/mb (mb being the nucleon mass) since in GR the binding energy represents a decrease of the gravitational mass.95 km for 1 M .5) r the so-called Schwarzschild exterior solution. ρ = ρo and e = eo when P = 0. N . there is vacuum with p(r) = ρ(r) = 0. so that d (log n) = 1 dρ dρ =− dν. If e is the internal energy per nucleon.7) and the total binding energy is BE = N mb − M. From thermodynamics. the first law gives ρ 1 0=d + pd n n where n is the number density. We employ units in which G = c = 1. one has ρ (r) = p (r) = m(r) = 0. so that 1 M is equivalent to 1. 0 The second and third of Einstein’s equations form the equation of hydrostatic equilibrium. h where h = (ρ + p)/n is the enthalpy per nucleon or the chemical potential. If n = no . which is 2. 25 July .2) can be exactly integrated. one finds r −λ m (r) = 4π ρr 2 dr . The black hole limit is seen to be R = 2M .8) . where the pressure vanishes (it is not necessary that the energy density or the number density also vanish there). (4. one finds ρo − mno = no eo and mn (r) = (ρ (r) + p (r)) e(ν(r)−ν(R))/2 − no eo . Defining the constant of integration so obtained as m(r). 16. R4 d¯ ¯ ω 2 r=R ω dj = .9) which is shown in Fig. (4. (4. R It is convenient to define j = exp[−(λ + ν)/2]. Key to EOS’s is in Ref. I =− 3 r=0 Ω 6Ω dr R L007 28 (4. The moment of inertia of a star in the limit of a small rotation rate Ω is obtained from the expression 8π R 4 ω ¯ r (ρ + P ) e(λ−ν)/2 dr.1: Binding energy per unit mass of neutron star models.11) with the surface boundary condition R ω ¯ =Ω− 3 d¯ ω dr R 2I =Ω 1− 3 . Lattimer and Prakash16 gave an approximate relation between BE andM/R as BE/M 0. 25 July .9).12) (4.13) . Then.5 August 2005 Figure 4. The yellow shaded band indicates approximation of Eq.5β) . The thicker curves with larger text symbols represent various analytic solutions.10) 3 0 Ω where the metric function ω ¯ is a solution of ω 4 −(λ+ν)/2 d¯ + 4r 3 ω ¯ de−(λ+ν)/2 = 0 d r e dr (4. 4. I= (4.1 along with representative EOS’s and analytical solutions.33rd SLAC Summer Institute on Particle Physics (SSI 2005).6β/ (1 − 0. 237 ± 0. 4 dω d ξ j + 4ξ 3 ωdj = 0. 4.14) R M R M This is shown in Fig. (4.2 together with representative EOS’s and analytic solutions. dξ where ξ = r/R.33rd SLAC Summer Institute on Particle Physics (SSI 2005).11). Eq. 2 MR β dξ 1 6ω1 + 2 (dω/dξ)1 where we use the values ω1 and (dω/dξ)1 obtained at ξ = 1. to ξ = 1. 25 July . 16. In practice. (4. Lattimer and Schutz17 found an approximation valid for normalEOS’s that don’t display severe softening just above n0 : 4 M km M km I (0.2: Moments of inertia of neutron star models. yields I 1 1 dω = . The thicker curves with larger text symbols represent various analytic solutions.14). from the origin where the initial values ω(0) = ω ¯ (0)/Ω = 1 and dω(0)/dξ = 0. The shaded gray band indicates the approximation Eq.008) M R2 1 + 4.2 + 90 .5 August 2005 Figure 4. Inset shows the behavior for small M/R. Then application of the surface bondary condition. one integrates the dimensionless second order equation found from Eq. (4.12). (4. L007 29 . Key to EOS’s is in Ref. Fig. In addition. Given the relation P (ρ) where ρ is the mass-energy density. The dashed line is a limit derived from Vela pulsar glitches. with a value of approximately 0. Black curves are for normal nucleonic EOS’s. Orange curves are contours of radiation radii R∞ = R/ 1 − 2GM/R. and causality represent limits to physically realistic structures (see text).09 M . Note the dramatic differences among normal EOS’s. Mass-Radius Diagram for Neutron Stars Figure 4. and also the difference between normal and strange quark matter EOS’s. The red region labeled rotation shows a limit derived from the most rapidly rotating pulsar. but this is no displayed since Rmin is of order 200 km. The presence of a maximum mass for each EOS is apparent. the TOV equations can be integrated. The lines denoted GR.33rd SLAC Summer Institute on Particle Physics (SSI 2005).5 August 2005 4.2. This behavior is related to the approximate n = 1 polytrope behavior observed previously. It is interesting to note that many normal nucleonic EOS’s have the property that in the mass range near 1 M the radius is relatively independent of the mass.3: Mass-Radius diagram. while z = 0. 16.35 is the redshift of candidate spectral lines on a neutron star. 25 July . The notation for the EOS’s is detailed in Ref. there is a minimum mass as well.3 shows the mass as a function of radius for selected EOS’s. while green curves (SQM1 and SQM3) are for pure strange quark matter stars. P < ∞. L007 30 . 4. Here. However. there are only 3 that satisfy the criteria that the pressure and energy density vanish on the boundary R.15) n = constant. at R. Clearly. These are discussed below. β < 4/9 or else the denominator has a zero and the central pressure will become infinite. the expansion becomes BE e −1 3β 9β 2 e 1+ + +··· . 4π 3 e−λ = 1 − 2β (r/R)2 . (4. but not energy density. Analytic Solutions to Einstein’s Equations It turns out there are hundreds of analytic solutions to Einstein’s equations.87β − 0. together with two of the infinite number of known solutions that have vanishing pressure. cs = ∂p/∂ρ is infinite.3.17) The moment of inertia can be approimated as −1 . Uniform density model Among the simplest analytic solutions is the so-called Schwarzschild interior solution for a constant density fluid. 25 July . In this case.16) − 1 − 2β − 1 M 4β 5 14 2β In the case that e/m is finite.33rd SLAC Summer Institute on Particle Physics (SSI 2005). 3 2 3 1 2 ν 1 − 2β − 1 − 2β (r/R) .3. The binding energy for the incompressible fluid is analytic (taking e = 0): √ 3 sin−1 2β 3β 9β 2 BE √ = + +··· (4. IInc /M R2 (2/5) 1 − 0. ρr . 4πR2 √ 2 3 1 − 2β − 1 − 2β (r/R) m (r) = ρ = n (m + e) = constant.3β 2 L007 31 (4.18) .5 August 2005 4. e = 2 2 √ 1 − 2β (r/R)2 − 1 − 2β 3β p (r) = . ρ(r) = constant.1. and that the speed of sound. β ≡ M/R. and that the pressure and energy density decrease monotonically with increasing radius. This solution is technically unphysical for the reasons that the energy density does not vanish on the surface. 4. It can be shown that this limit to β holds for any star. − + M m m 5 14 (4. 20) Here. energy and number densities. nc mn c2 = 72βp∗ (1 − 2β)3/2 . p∗ p (4.2.19) ρ = 12 p∗ p − 5p and found eν = (1 − 2β) (1 − β − u) (1 − β + u)−1 .47β 2 L007 32 (4. and binding energy are π . cs = 6 .c < 1. 25 July . He assumed an equation of state √ (4. Buchdahl18 discovered an extension of the Newtonian n = 1polytrope into GR that has an analytic solution. This solution is limited to values of β < 1/6 for cs. u = β Ar r = r (1 − β + u)−1 (1 − 2β) . R = (1 − β) 288p∗ (1 − 2β) (4. and r is.3. central pressure. with u. the radius. β β 2 3β 3 BE = (1 − 1. IBuch /M R2 2/3 − 4/π 2 1 − 1. Buchdahl’s solution In 1967. eλ = (1 − 2β) (1 − β + u) (1 − β − u)−1 1 − β + β cos Ar 8πp = A2 u2 (1 − 2β) (1 − β + u)−2 . −2 . For this solution. 8πρ = 2A2 u (1 − 2β) (1 − β − 3u/2) (1 − β + u)−2 .22) pc = 36p∗ β 2 .5 August 2005 4.81β + 0.33rd SLAC Summer Institute on Particle Physics (SSI 2005).21) . A2 = 288πp∗ (1 − 2β) −1 (4. −1 3/2 p p∗ √ 2 1 −5 mn = 12 pp∗ 1 − 3 . M 2 2 4 The moment of inertia can be approximated as −1 .23) . p∗ is a parameter.5β) (1 − 2β)−1/2 (1 − β)−1 − 1 ≈ + + + ···. ρc = 72p∗ β (1 − 5β/2) . a radial-like variable −1 sin Ar . Tolman2 discovered that the simple density function ρ = ρc [1 − (r/R)2] has an analytic solution. w = log x − 5/6 + e / (3β) . Tolman VII solution In 1939. e−λ = 1 − βx (5 − 3x) .c are . φc = φ (x = 0) . −λ w1 = w (x = 1) . .4.24) In the above. x = (r/R)2. It is known as the Tolman VII solution: eν = (1 − 5β/3) cos2 φ. (ρ + P ) cos φ 1 β −λ n= P = 3βe tan φ − (5 − 3x) . 25 July . The central values of P/ρ and the square of the sound speed c2s.5 August 2005 4. (4. −1 φ1 = φ (x = 1) = tan β/ [3 (1 − 2β)].33rd SLAC Summer Institute on Particle Physics (SSI 2005). 2 4πR 2 m cos φ1 φ = (w1 − w) /2 + φ1 . c 3 P . 2c2s. . . = ρ . or else Pc becomes infinite. M 21 18018 306306 (4.2698.c < 1 if β < 0. but in expansion 11β 7187β 2 68371β 3 BE ≈ + + +··· .c 15 β c2s. or β < 0. There is no analytic result for the binding energy. For causality cs.27) .1β − 0. tan φc + 3 (4.3862.26) A fit to the moment of inertia is −1 .6β −2 L007 33 (4.c = tan φc β .25) This solution is limited to φc < π/2. IT V II /M R2 (2/7) 1 − 1. 25 July .33rd SLAC Summer Institute on Particle Physics (SSI 2005). and the causality limit is β = 0.223.4.21β 2 . ρc 3 1 2 tan2 g (0) − tan2 f (0) . Nariai IV solution In 1950. e = 1 − 3β R c cos f (r ) 2 2 r r 3β 3β 1− 1− . 1 − 3β − sin f r 2 − R 2 R √ 3β c2 √ ρ r = 4πR2 1 − 2β e2 2 3β r 3 − cos2 f r .32β − 0. This solution is quite similar to Tolman VII.1. 2e2 + (1 − e2 ) (7e2 − 3) (5e2 − 3) 2 2 The pressure-density ratio and sound speed at the center are Pc 1 √ = 2 cot f (0) tan g (0) − 2 .29) L007 34 . 3 sin f r cos f r − 4 R 2 r 3 e tan f (r ) 3β r 3β (1 − 2β) 1 − tan f r . The leading order term in the binding energy is identical to Tolman VII. e = (1 − 2β) 2 . and the moment of inertia expansion is −1 I/M R2 (2/7) 1 − 1. It is known as the Nariai IV solution. Nariai19 discovered yet another analytic solution.c = 3 The central pressure and sound speed become infinite when cos g(0) = 0 or when β = 0. r= c cos f (r ) √ cos f (r ) c2 tan g r p r = 3β 2 cos f r 4πR2 e2 (4. f r = cos−1 e + 4 R 2 R r e 1 − 2β. c2s. g r = cos−1 c + . (4.4126. and is expressed in terms of a parametric variable r : 2 2 2 r e2 cos g (r ) −λ ν tan f r . m r = 2 R c cos f (r ) 4 R The quantities e and c are √ 2 + β + 2 1 − 2β e = cos f R = 4 + β/3 2e2 c2 = cos2 g R = −1 .28) 2 2 r r 3 tan f r 3β tan f r .5 August 2005 4. (4. L007 35 .5 August 2005 4. Tolman IV variant Lake20 discovered a variant of a solution Tolmandiscovered in 19392 and since known as the Tolman IV solution: 2/3 2 − 2β (2 − 5β + βx)2 ν −λ . the central and the surface sound speed are both c2s = 3/10.3624.1β 3 .‘ (4.31) 1/3 5 (2 − 5β) (2 − 2β)2/3 This solution has non-vanishing energy density at the surface where the pressure vanishes. P = 4πR2 2 − 5β + βx 2 − 5β + 3βx 5/3 (2 − 5β + 3βx) (2 − 5β + 3βx) 2 + (2 − 5β) − 5β 2 x2 .33) A good approximation to the moment of inertia for this solution is −1 I/M R2 = (2/5) 1 − 0. and tends to zero for β = 2/5. cs (R) = c when β = 0. and cs (0) = c when β ≥ 0.surf = 2 − 2β 5 (2 − 4β)3 2 − 2β + (2 − 5β)2 − 5β 2 . 1/3 ρc 3 (2 − 5β) (2 − 2β)2/3 c2s.2.30) 4πR2 2 − 5β + 3βx 2 − 5β + 3βx 2/3 2 − 2β β 1 2 − (2 − 5β + 5βx) .34) The central pressure. energy density and sound speed become infinite for β ≥ 2/5. 25 July . (4. The surface sound speed is c2s. On the other hand.4. The ratio of the surface to central energy densities is ρsurf (2 − 5β)2/3 2 = (3 − 5β) . e = 1 − 2βx . In the limit of small β. (4. ρc 3 (2 − 2β)5/3 (4. The sound speed at large densities in strange quark matter tends to c2s = 1/3 because of asymptotic freedom.32) which is unity for β → 0.c = 1 1 + 1 .58β − 1.33rd SLAC Summer Institute on Particle Physics (SSI 2005).3978. which makes this an interesting solution to compare with strange quark matter stars. e = 4 (1 − 2β) 2 − 5β + 3βx 2/3 3 2 − 2β 2 m =βRx . 2 − 5β + 3βx 2/3 6 − 15β + 5βx 1 2 − 2β ρ= β . c2s = 2/3 5 (2 − 5β + βx)3 (2 − 2β) The central values of P/ρ and c2s are 1 2 Pc = −1 . 4 (1 − 2β) ρc √ For the Tolman VII solution at the causal limit. r = x/ ρo.38) Since the most compact configuration is achieved at the maximum mass.4 ρs /ρc M . One also finds for this equation of state that Rmax = 18. M= 8πρc Finally.36) results in a neutron star maximum mass that is practically independent of the equation of state for ρ < ρo .22 pointed out. For the Buchdahl solution at the causal limit. which lead to 33/2 β 1/2 sin f (0) cos f (0) β M= < 3. which lead to πβ 3 (1 − 5β/2) M = (1 − β) < 2.33rd SLAC Summer Institute on Particle Physics (SSI 2005).2 ρs /ρo M .27 and p/ρ = 2/( 75β) 0. = 4πdx2dx.7 · 1014 g cm−3 is the nuclear saturation density.5 August 2005 4.35) Rhoades and Ruffini21 showed that thecausally limiting equation of state p = po + ρ − ρo ρ > ρo (4.5 ρs /ρo km. β = 1/6 and p/ρ = β/(2 − 5β).44. Here ρs = 2.9 ρs /ρc M . Neutron Star Maximum Mass and Compactness Limit The TOV equation can be scaled by introducing dimensionless variables: √ √ p = qρo .228 and p/ρ 0. This result wasreinforced by Glendenning23 . Some justification for the Rhoades-Ruffini result appears from the analytic solutions of Einstein’s equations.5. for the Nariai IV solution at the causal limit.. (4. as Lattimer et al.37) Mmax = 4.14 ρs /ρc M . (q + d) z + 4πdx3 dz dq =− . 25 July . this represents the limiting value of β for causality.33 .246. dx x (x − 2z) dx (4. β 0. βmax 0. ρ = dρo . cos f (R ) 4πρc L007 36 . β 0. who performed aparametrized variational calculation to find the most compace possible stars as a function of mass. m = z/ ρo . which lead to 15β 3 < 4. and is (4. 6. We assume K = −GM/R. In the so-called Roche model. one finds h (r) − GM/r − (1/2) Ω2 r 2 = K = −GM/re − (1/2) Ω2 re2 . In more realistic models. the value obtained for a non-rotating configuration. Maximal Rotation Rates for Neutron Stars The absolute maximum rotation rate is set by the “mass-shedding” limit. Using it anyway.33rd SLAC Summer Institute on Particle Physics (SSI 2005). and the equation of hydrostatic equilibrium is (1/ρ) ∇P = ∇h = −∇ΦG − ∇Φc . ρc /¯ ρ 54. such as ρ = ρc [1 − (r/R)2]. when the rotational velocity at the equatorial radius (R) equals the Keplerian orbital velocity Ω = GM/R3 . The potential Φ ≡ ΦG + Φc is maximized at the . For an n = 3 polytrope. where re is the equatorial radius and h(re ) = 0.5 August 2005 4. (4. so this would be a good approximation. the actual limit on the period is larger because rotation induces an increase in the equatorial radius. Integrating this from the surface to an interior point along the equator. 25 July . for which ρc /¯ ρ = 5/2. as described in Shapiro and Teukolsky24 one treats the rotating star as beinghighly centrally compressed. for which ρc /¯ gravitational potential near the surface is ΦG = −GM/r and the centrifugal potential is Φc = −(1/2)Ω2 r 2 sin2 θ. However. the trope.39) R M for a rigid sphere. this approximation is not as good. and an n = 1 polyρ = π 2 /3.55 ms (4. or 1/2 3/2 10 km M rigid Pmin = 0.40) where h = dP/ρ is the enthalpy per unit mass. . point where ∂Φ/∂r . 44) R M L007 37 . re has rc the largest possible value when re = rc = 3R/2. Ω = 3 = rc 3 R3 The revised minimum period then becomes 1/2 3/2 10 km M Roche ms.03 ms.43) Pmin 0. (4.03 Rmax M An even more useful form describes the maximum rotation rate that a non-rotating object of mass M and radius R can be spun: 1/2 3/2 10 km M arbitrary Pmin 0. or 3 GM GM 2 2 . (4.41) (4.0 R M (4. including the increase in maximum mass for a rotating fluid. Pmin = 1.82 ± 0. can be accurately expressed in terms of its non-rotating maximum mass and the radius at that maximum mass as: 3/2 1/2 10 km Mmax EOS ms. = 0.96 ± 0. Thus. or where rc3 = gM/Ω3 and Φ = −(3/2)GM/rc .42) Calculations including general relativity show that the minimum spin period for an equation of state. 46) 2 M This result is illustrated in Fig. could have a central density larger than this value.3. (In GR. β = 5/7 for Tolman VII for which ρ = ρc [(1 − (r/R)2].Inc 13. the larger the measured mass. much larger values. 4.5 × 10 g cm−3 . β = 3/2 for an n = 3 polytrope. we can write T = α(2/3)3GM 2 /R and |W | = βGM 2 /R. 4. Each larger mass star that is measured will lower this bound. By calculating the structures of a large number of neutron stars.2 M sets an upper bound of about 8n0 . α = 0. β = 3/5 for an incompressible fluid. (4. for these four cases. It is interesting to compare the rotational kinetic energy T = IΩ2 /2 with the gravitational potential energy W at the mass-shedding limit.Inc = 4πM 2 3G M A precisely measured neutron star mass would thus imply a value for the central density of the star.33rd SLAC Summer Institute on Particle Physics (SSI 2005). and if another star was more massive. which is perhaps dangerously close to predicted L007 38 . It is this limit that is plotted in Fig. A lower mass star cannot have a higher central density than that star. The figure illustrates that a measured mass of about 2. an incompressible ellipsoid becomes secularly (dynamically) unstable at T /|W | = 0.7. static environment in our universe.0377. the uniform density EOS is not realistic: it violates causality and the density at the surface ρsurf ace = 0. We therefore find that T /|W | is 0. Furthermore. respectively. interestingly enough.0593.1375(0.0516. Lattimer and Prakash26 found no EOS has agreater ρc for given Mmax than that predicted by the Tolman VII solution: 2 5 15 M ρc. the smaller the central density of that star. It is moderately violated for strange quarkmatter stars. No other star. Maximum Density Inside Neutron Stars If the uniform density model was a good model for a neutron star. α = 1/7. and can be used in the manner described above: the largest precisely measured neutron star mass determines an upper bound to the density of matter in a cold. 4. 0.2738).5 August 2005 a result found to be valid for normal EOS’s25 .4. (4. I is the moment of inertia about the rotation axis: 8π R 4 I= r ρdr 3 0 for Newtonian stars. 641 Hz from PSR B1937+21. 25 July . For comparison.45) ρc.V II = ρc. using the highest observed spin period of a pulsar.) Using Ω2 = (2/3)3GM/R3 . 0. We have α = 1/5.45). However. a similar relation deduced from the Tolman VII analytic solution apparently bounds the relation between central density and maximum mass.8 × 10 g cm−3 .00745 and 0.0988. α = 1/3 − 2/π 2. β = 3/4 for an n = 1 polytrope. it would have to have a smaller central density according to Eq. one must take into account frame-dragging as well as volume and redshift corrections. (4. the causality limit would imply a central density 2 3 2 3 c 15 M 5. at the mass-shedding limit. no matter what its mass. But. 4. astrophsysical measurements of neutron star masses may be able to rule out the existence of deconfined quark matter. Polytropic relations Eq.5.47) where P (n) is the total pressure inclusive of leptonic contributions evaluated at the density n. NR are non-relativistic potential models. The Exotica points include self-bound strange quark matter stars. (4.5) implied the value of the radius is connected to the constant K in the EOS.4: The central energy density and mass of maximum mass configurations. 4.theoretical models. a Bose condensate. n).5 August 2005 Figure 4. indeed. Neutron Star Radii As previously noted. The correlation has the form: R (M. there is a strong correlation between the radius of stars with masses 1 − 1. Symbols reflect the nature of the EOS’s selected from Ref. For comparison.5 M and the pressure in the vicinity of 1 − 2n0 . and M is the stellar gravitational mass. The dashed line for 2. due to the occurence of hyperons. 25 July . values for the density at which nucleonic matter gives way to deconfined quark matter. The constant C(M. 16.2 M serves to guide the eye. or quark matter. Lattimer and Prakash16 found that.25 .33rd SLAC Summer Institute on Particle Physics (SSI 2005). This correlation is shown in Fig. In other words. in units of km fm3/4 L007 39 . (3. at least in cold matter. and Exotica refers to NR or R models in which strong softening occurs. the central density – maximum mass relations for the Tolman VII and uniform density (incompressible) models are shown. many EOS’s feature the property that in the vicinity of 1 M their radii are relatively independent of the mass. or to the value of the pressure at a characteristic density.8. n) [P (n)]0. R are field. n) C (M. 11 for the 1. n0 . MeV−1/4 . The only analytic solution that explicitly relates the radius.84±0.5ns and 2ns cases. for the densities n = ns . (4.21. the baryon density and stellar radius are given in Eqs. for EOS’s listed in Ref.5: Empirical relation between pressure.47) can thus be found: −1 . This is a general relativistic effect. (4.03 and 5. however.16 ± 0.5n0 and 2n0 . Note.5 August 2005 Figure 4.15 and 5. The different symbols show values of RP −1/4 evaluated at three fiducial densities. in units of MeV fm−3 . 16. is 9.5ns and 2ns .04 for the 1 M case. 1. 1.11±0. respectively. and R.22).20) and Eq. that this exponent is not 1/2 as the n = 1 Newtonian polytrope predicts.57±0. (4. the lower panel is for 1. mass and pressure is that due to Buchdahl. The exponent in Eq. 7.4 M case.53 ± 0.82±0. The correlation is seen to be somewhat tighter for the baryon density n = 1.4 M stars. and 9. 6.33rd SLAC Summer Institute on Particle Physics (SSI 2005).07. In terms of the parameters p∗ and β ≡ GM/Rc2 . The upper panel shows results for 1 M (gravitational mass) stars. as we now demonstrate by using an analytic solution to Einstein’s equations. in km. 25 July . . d ln R . 48) 1+ = 12 1 − . (1 − β) (1 − 2β) 5 P 1 P . (4. If M and R are about 1. the value characteristic of an n = 1 Newtonian polytrope. one has P → 0 and d ln R/d ln P |n. respectively.22) gives L007 40 .M → 1/2. Finite values of β and P reduce the exponent. for example.4 M and 15 km.M 6 p∗ 6 p∗ (1 − 3β + 3β 2 ) In the limit β → 0.14 and Eq. d ln P n. β 0. (4. . x) = −16 + K 18 1− n n0 2 + K 27 1− n n0 3 + Esym (n) (1 − 2x)2 . As shown in Fig. in large part. The pressure depends primarily upon Sv .2 and Eq. so the equilibrium proton fraction at n0 is x0 (3π 2 n0 )−1 (4Sv /¯hc)3 0. so that the energy per particle is E (n.24). Eq. L007 41 . (4. in the schematic expansion Eq. (2. 5. . The symmetry energy parameter Sv ≡ Esym (n0 ).49) Here. x0 ) = n0 (1 − 2x0 ) n0 Sv (1 − 2x0 ) + Sv x0 n20 Sv . a white dwarf or a main-sequence star.28. Sv ≡ (∂Esym/∂n)ns . in which the masses are 1. we introduce an additional term.3867 ± 0.5n0 ) = 2. respectively. the skewness. Matter in neutron stars is in beta equilibrium. (4.5 August 2005 p∗ = π/(288R2) ≈ 4. determined by the symmetry properties of the EOS. approximately the energy difference at a given density between symmetric and pure neutron matter. K and K are the incompressibility and skewness parameters. But the compact nature of several binary pulsars permits detection of relativistic effects.e. which constrains the inclination angle and permits measurement of each mass in the binary.51) and it is noted that the contributions from K and K largely cancel.0002 M .4414 ± 0. such as Shapiro delay or orbit shrinkage due to gravitational radiation reaction. these include pulsars orbiting another neutron star. Leptonic contributions Ee = (3/4)¯ hcx(3π 2 nx4 )1/3 must be added.1. Evaluating the pressure for n = 1.5ns . respectively.04. i.5n0. For the present discussion. P (1.85 · 10−5 km−2 (in geometrized units). The textbook case is the binary pulsar PSR 1913+16.1. This correlation is significant because the pressure of degenerate neutron-star matter near the nuclear saturation density ns is. from which the total mass of the binary can be deduced. 5. At a fiducial density n = 1. A sufficiently well-observed system can have masses determined to impressive accuracy. Experimental constraints to the compression modulus K. this is equivalent in geometrized units to n = 2.50) due the small value of x0 . 25 July . and generalize the symmetry energy. The skewness parameter K has been estimated to lie in the range 1780–2380 MeV.2. µe = µn − µp = −∂E/∂x. Ordinarily. (4.02 × 10−4 km−2 . . (4. The pressure at n0 is P (n0 . . and Esym is the symmetry energy function. The equilibrium pressure at moderately larger densities similarly is insensitive to K and K . 5.0002 and 1.25n0 K/18 − K /216 + n0 (1 − 2x)2 (∂Esym/∂n)1.5n0 . Observations of Neutron Stars Masses The most accurately measured neutron star masses are from timing observations of the radio binary pulsars. observations of pulsars in binaries yield orbital sizes and periods from Doppler phenomenon.33rd SLAC Summer Institute on Particle Physics (SSI 2005). (4. or n/p∗ 4.20) then implies P/p∗ 0. most importantly from analyses of giant monopole resonances give K ∼ = 220 MeV.48) yields d ln R/d ln P 0. 4 M is about 4σ from the optimum value. which show a broader mass range than binary pulsars having neutron star companions. leading to a restricted range of neutron star masses. Letters in parentheses refer to references cited in Ref. including the fascinating case27 of PSR J0751+1807 in which the estimated mass with1σ error bars is 2. It has been suggested that a rather narrow set of evolutionary circumstances conspire to form double neutron star binaries. In addition. For this neutron star.33rd SLAC Summer Institute on Particle Physics (SSI 2005). 26.4 M value. silver and blue regions) and in x-ray accreting binaries (green). a mass of 1.2 ± 0.1: Measured and estimated masses of neutron stars in radio binary pulsars (gold. the mean observed value L007 42 . 25 July . One particularly significant development is mass determinations in binaries with white dwarf companions. Evidence is accumulating that a few of the white dwarf binaries may contain neutron stars larger than the canonical 1.2 M .5 August 2005 Figure 5. This restriction is relaxed for other neutron star binaries. Strongly magnetized hydrogen is relatively simple. d is the distance. However. Continued observations guarantee that these errors will be reduced. is a quantity that can be estimated if F∞ . radius and temperature. the observed absence of narrow spectral features.13(R∞/km)M−1 .1. The system of Vela X-1 is noteworthy because its lower mass limit (1.7M ) is at least mildly constrained by geometry28 . This optical excess is a natural consequence of the neutron star atmosphere. especially those in which substantial softening begins around 2 to 3ns . although pulsars 12 are thought to have intense magnetic fields > ∼ 10 G. In many cases a heavy-element atmosphere appears to fit the global spectral distributions from x-ray to optical energies while also yielding neutron star radii in a plausible range. 25 July . M and R could be determined. However.e.e. The redshift is z = (1 − 2GM/Rc2 )−1 − 1. This could be extremely significant. Such emissions are difficult to utilize in terms of constraining the star’s global aspects. they are predominately x-ray sources.5 August 2005 of the white dwarf-neutron star binaries exceeds that of the double neutron star binaries by 0. These stars are observed to have optical fluxes factors of 3 to 5 times greater than a naive blackbody extrapolation from the x-ray range would imply.. the 1σ errors of all but one of these systems extends into the range below 1.3. F∞ and L∞ refer to the temperature.25 M . they would obey 4 F∞ = L∞ /4πd2 = σT∞ (R∞ /d)2 . but if redshift information is available.1) where. T∞ and d are known. Masses can also be estimated for another handful of binaries which contain an accreting neutron star emitting x-rays. flux and luminosity redshifted relative to their values at the neutron star surface. Neutron star atmosphere models are mostly limitedto non-magnetized atmospheres. Bose condensates. R∞ = R(1 + z). the socalled radiation radius.45 M . Radii and Redshifts Most known neutron stars are observed as pulsars and have photon emissions from radio to x-ray wavelengths dominated by non-thermal emissions believed to be generated in a neutron star’s magnetosphere. Some of these systems are characterized by relatively large masses. F∞ = F/(1 + z)2 . but magnetized heavy element atmospheres are still in a state of infancy. 4. or quarks) generally reduce the maximum mass appreciably.2. i. R∞ is a function of both M and R. 5.6 to 1. i. A useful constraint is provided by a few cases in which the neutron star is sufficiently close for detection of optical radiation. Mass estimates of selectedx-ray binaries are also shown in Fig. A value of R∞ requires both that R < R∞ and M < (3 3G)−1 c2 R∞ 0. Contours √ of R∞ are displayed in Fig. However. (5. such as mass. but the estimated errors are also large. predicted by L007 43 . If their total photon fluxes were that of a blackbody.. since exotica (hyperons. (For example.33rd SLAC Summer Institute on Particle Physics (SSI 2005). approximately a dozen neutron stars with ages up to a million years old have been identified with significant thermal emissions. A serious problem in determining R∞ and T∞ is that the star’s atmosphere rearranges the spectral distribution of emitted radiation. and results in an inferred R∞ substantially greater than that deduced from the x-ray blackbody. Stars of these ages are are expected to have surface temperatures in the range of 3×105 K to 106 K. and T∞ .) As a result. 5. T∞ = T /(1 + z). Raising the limit for the neutron star maximum mass could eliminate entire families of EOS’s. they are not blackbodies29 . These estimates suggest relatively large neutron star radii with R∞ > 14 km begin favored. is puzzling.5 August 2005 Figure 5. 25 July . 10 km star is disfavored. As a consequence. Distances to pulsars can be estimated by their dispersion measures. although errors are still large. the canonical 1. while in a plausible range. The estimation of radii from isolated neutron stars is also hampered by uncertainties of source distance. are not sufficiently precise at present to usefully restrict properties of dense matter. 5. Walter and Lattimer30 and Braje andRomani31 separately deduced values of the neutron starmass and radius shown in Fig. but in a three cases (Geminga.4 M . In the case of RX J185635-3754. values of R∞ determined from thermally-emitting neutron stars. 30 and Ref.2. RX J185635-3754 and PSR B0656+14) parallax distances have been obtained.2: Estimated mass and radius for RX J185635-3754 from Ref.33rd SLAC Summer Institute on Particle Physics (SSI 2005). 31 (dark blue region). heavy-element atmosphere models. L007 44 . The explanation could lie with broadening or elimination of spectral features caused by intense magnetic fields or high pressures. the distances to these sources will likely become relatively well known in the near future. Heavy elements could be present only if accretion was ongoing. which would decrease radii.4 M .5 SQM3 APR 1 MS1 PCL2 0. In addition. respectively. 25 July . From Ref. small heavy element contamination cannot be ruled out. These systems contain rejuvenated 10 billion year-old neutron stars heated by recent episodes of mass accretion from their companions.2. Measurements from the neutron star X7 in the Globular Cluster 47 Tucanae yield R∞ 17. 32.5 August 2005 5.33rd SLAC Summer Institute on Particle Physics (SSI 2005).3)32 . Interestingly. Estimates from Quiescent X-ray Bursters in Globular Clusters In this context.this implies a radius range fom 12 to 17 km.5 10 12 14 16 18 Figure 5. 5.3: Estimated ranges of masses and radii of the neutron stars X7 in 47 Tucanae. since this material obscures 50% or more of the x-ray flux and by the distance to the globular cluster. for hydrogen atmosphere models. although a pure H atmosphere was assumed. 3 2.5 ± 2. reducing a source of error that plagues interpretations of isloated neutron stars. Confidence contours are 1-.5 2 FPS 1. Accuracy is limited by systematic uncertainties in the intervening interstellar hydrogen column density. L007 45 . but the stability of the observed X-ray flux implies that it is not. For 1.and 3-sigma.5 km (see Fig. and accretion is known to quench magnetic fields. these stars may have the simplest of all possible atmospheres: non-magnetic hydrogen. the recent discovery of thermal radiation from quiescent x-ray bursters in globular clusters is particularly interesting. 2.1. Since the accreted material was dominated by hydrogen. 25 < Pt /(MeV fm−3 ) < 0. the occasional disruption of the otherwise regular pulses. their magnitudes and stochastic behavior suggests they are global phenomena 37 . 5.30with a small dependence upon assumed source distance. Both are spinning.6β 2 .2) The dependence on nt is weak. this leads to the limit shown in Fig. For given values of ∆I/I and M .3.2) is obtained employing the largest Pt value in this range. The resulting light curves can be modelled taking into account light-bending which constrains the star’s compactness. something breaks and the spin rates are brought closer in alignment. I 3 GM 2 1 + (2Pt /nt β 2 ) (1 + 5β − 14β 2 ) (5.33rd SLAC Summer Institute on Particle Physics (SSI 2005). (5. For the source XTE J1814-338.3 M ). Estimates from Active X-ray Bursters Neutron stars that are actively accreting material from companion stars produce xray bursts. Since this star is a member of an eclipsing binary. amounts to at least 1. Their spins and spin-down rate provide estimates of magnetic field strengths and ages. While the origin of glitches is unknown. The angular momentum observed to be transferred between the components.65 among published EOS’s. the smallest R compatible with Eq.5 < R < 15 km (corresponding to 1. These estimates are based on a geometric technique and are argued to be insensitive to spectral modelling.5 < M < 2. L007 46 . This inference received additional credibility by themeasurement 36 of a 45 Hz neutron starspin frequency. which could fix R.2. The superfluid is weakly coupled with the normal matter and its rotation rate is not diminished.5 August 2005 5.38. Pt and nt depend on the symmetry energy’s density dependence as well as on the incompressibility. If this also corresonds to the fraction of the moment of inertia residing in the neutron star crust. The fastest pulsars provide constraints on neutron star radii. Shaposhnikov & Titarchuk34 found 0.35 35 . there is a range 0. This limit has a relatively weak dependence on −1/4 the EOS (R ∝ Pt ).4 M star.and He-like Fe lines and imply z 0. For the source 4U 1820-30.33 foundz < 0. in the case of the Vela pulsar. 25 July .2. But when the difference in spin rates becomes too large. These techniques could potentially be extended to other x-ray bursters. however.67β − . 4. it applies only to the Vela pulsar. but the normal crust isdecelerated by the pulsar’s magnetic dipole radiaion. Hence. and it depends upon the crustal superluid coupling hypothesis. Two lines observed in x-ray burst spectra of EXO 0748-676 have been suggested to be H. For a 1. This low spin rate is consistent with the observed equivalent widths of these lines if they are due to Fe and further implies 9. limits can be set on the star’s mass and radius. at the base of the crust.20 < z < 0. A potentially rich source of data are pulsar glitches. Bhattacharyya et al. an independent mass measurement might yet be possible. The leading glitchmodel involves angular momentum transfer in the crust from the superfluid to the normal component 38 . Pulsar Glitches Pulsars provide several sources of information concerning neutron star properties. this fraction is37 28π Pt R4 ∆I 1 − 1.3. In terms of the density and pressure nt and Pt .4% of the star’s total 37 . Since the orbital angular momentum dominates the spin angular momenta. |L| P (MA + MB ) (5. (5. 1/2 A1P N 6 (1 − e2 ) M a2 MA PA (5.5 August 2005 5. To lowest post-Newtonian order.4) the amplitude of the P is the binary orbital period. the spin SA and orbital angular momenta evolve according to 39 . L007 47 .4. L S L · S L × S 2π 2π ˙ A A A SO ˙A = S L .3) = . the precession of the pulsar spins produces a compensating change in the orientation of the orbital plane.7) where φA is the angle between the line of sight to pulsar A and the projection of SA on the orbital plane. Then. IA . If θA is the angle between SA and L. respectively. L −3 3 PpA |L| PpA |L| |L| where a and e are the orbital semimajor axis and the eccentricity. an effect also known as geodetic precession39−40 . 25 July . The spin-orbit effects usually simplify because one star (hereafter called A) spins much faster than the other star B.5) where PA is A’s spin period. and the precession of the pulsar spins about the direction of the total angular momentum of the system. MA + MB c c M A a 2 PA (5. all observable spin-orbit effects are proportional to the moment of inertia of pulsar A. Since the total angular momentum remains fixed. the geodetic precession amplitude is very small while the associated spin precession amplitudes are substantial.33rd SLAC Summer Institute on Particle Physics (SSI 2005). 1/2 |L| a2 MA MB (1 − e2 ) PA (5.6) The ratio of the periastron advances due to spin-orbit coupling and to first-order post Newtonian contributions is 40 ApA IA (4MA + 3MB ) P = (2 cos θA + cot i sin θA sin φA ) . 2 1 − e2 1/2 M a 2πM A B = . and the are precesion period PpA and |L| PpA = 2aP (MA +MB )c2 (1−e2 ) GMB (4MA +3MB ) . Spin-orbit coupling produces two relativistic effects that could be measured: a small extra advance of the periastron of the orbit beyond the standard post-Newtonian advance. This will cause a periodic departure from the expected time-of-arrival of pulses from pulsar A of amplitude (for e 0) δta = a MB a IA P δi cos i = sin θA cos i . change in the orbital inclination i due to A’s precession is δi = P |SA | IA (MA + MB ) sin θA sin θA . Moments of Inertia From Relativistic Binaries It might be possible to measure the entire moment of inertia of a neutron star using pulsar timing in relativistic binaries. Coupled with 0737’s shorter precession period.4: The moment of inertia scaled by M 3/2 . 16. L007 48 . However. The curve Crab illustrates a limit for the Crab pulsar41 . since M is precisely known. these same attributes render the spin-orbit contribution to the periastron advance about 6 times larger than for the other two systems. The dashed curve labelled Crab is a lower limit for the Crab pulsar. EOS labels are described in Ref. i 90◦ . For a 1. The shaded band illustrates a ±10% error on a hypothet−1/2 ical I/M 3/2 measurement of 50 km2 M .5 August 2005 Figure 5. The importance of a measurement of I to within ±10% is illustrated in Fig. Those families of models lying close to the measured values would have their parameters limited correspondingly.4. which has remained undetectable in other systems. It is noteworthy that the net timing delays due to inclination shifts due to precession are more than an order of magnitude larger for PSR 1913+16 and PSR 1534+12 than for PSR 0737-3039. a 10% uncertainty in K would result in a radius estimate with about 6 to 7% uncertainty.14) should then allow a measurement of R. It is expected that IA can be determined to about 10% after a few years observations17 . Using Eq. the error bar is for M = 1. (4.4 M star. 5. due to their smaller inclinations. this produces a factor 24 in observability for this effect.34 M . and a small misalignment angle θA that make the inclination timing delays extremely small. PSR 0737-3039 has a nearly edge-on orbit.33rd SLAC Summer Institute on Particle Physics (SSI 2005). In particular. 25 July . It is clear that relatively few equationsof state would survive these constraints. having lost energy to neutrinos and from nuclear dissociation of the material it has plowed through (stage (I) in Fig. which within seconds accelerates outwards. The escape of neutrinos from the interior occurs on a diffusion time of seconds. expelling the massive stellar mantle. The shock wave propagates only about 100 to 200 km before it stalls. In the core. which triggers the formation of a shock wave at the core’s outer edge.5 August 2005 Figure 6. 6. Newly-born neutron stars or proto-neutron stars are rich in leptons. The radius R and central temperatures Tc for the neutron star are indicated as it evolves in time t. but it is probable that neutrinos play a crucialrole.1). Evolution of Neutron Stars Neutron Star Birth Neutron stars are created in the aftermath of the gravitational collapse of the core of a massive star (> 8 M ) at the end of its life. Apparently. 6. eventually resuscitate the shock.8) shows that neutrinos escape with Eν. The loss of neutrinos and lepton number neutronizes the matter. The proto-neutron star left behind rapidly shrinks due to pressure losses from neutrino emission in its periphery (stage II). 25 July . Core collapse halts when the star’s interior density exceeds n0 . assisted perhaps by rotation. The energy L007 49 . 6. which triggers a Type II supernova explosion. The detailed explosion mechanism of Type II supernovae are not understood 42 . Neutrinos observed from SN 1987A in the Large Magellanic Cloud confirmed this time scale and the overall energy release of 3 × 1053 ergs43−44 .1. Roman numerals indicate various stages described in the text. convection and magnetic fields. mostly e− and νe . the µνe ∼ 300 MeV but Eq. (1. neutrinos from the core.1: The main stages of evolution of a neutron star.33rd SLAC Summer Institute on Particle Physics (SSI 2005).esc ∼ 20 MeV. the net electron neutrino abundance Yν . baryon density nB . A second mode of black hole creation ispossible47 since the maximum massis enhanced relative to a cold star by extra L007 50 . in some cases. the steady emission of neutrinos begins to cool the interior45 . reaching ∼ 50 MeV.33rd SLAC Summer Institute on Particle Physics (SSI 2005). however. unless the star’s mass has exceeded the theoretical maximum mass. Taken from Ref. and its cooling rate accelerates. 6. more than doubling the core temperature (stage III). The star becomes transparent to neutrinos (stage IV). After 10 to 20 s. First.5 August 2005 Figure 6. proto-neutron stars accrete mass which has fallen through the shock. The proto-neutron star. It would then collapse and its neutrino signal would abruptly cease46 . T . the simulation uses the EOS GM3 for nucleons-only matter. Labels indicate time in seconds. the condition λ > R is achieved in about 50 s. 25 July . This could occur in two different ways. This accretion terminates when the shock lifts off. and the net electron concentration Ye inside a proto-neutron star. The abscissa shows the enclosed baryonic mass. 45.2: Temporal and radial variation of entropy per baryon s. left behind warms the stellar interior (Fig. collapsing instead into a black hole. might not survive its early evolution. electron neutrino chemical potential µνe . Because the cross section σ ∝ λ−1 scaleslike the square of the mean neutrino energy.2). The modified Urca rate is reduced by a factor of (T /µn )2 ∼ 10−4 to 10−5 compared to the direct Urca rate. The overall time that a neutron star will remain visible to terrestrial observers is not yet known. Ye 0. and neutron star cooling is correspondingly slower. The star continuously emits photons. However. 6. in which thermally excited particles alternately undergo beta and inverse-beta decays. dominantly in x-rays. the thermal evolution time of the crust. longer than in the first case. Sv 30 MeV. (3. the direct Urca process might still occur above some density threshold48 . heat transported by electron conduction into the interior. Either there is presently no neutron star. where it is radiated away by neutrinos. expected from a rotating magnetized neutron star.2) where.neutrino cooling must occur by the modified Urca process n + (n.048 n Sv (6. The energy loss from photons is swamped by neutrino emission from the interior until the star becomes about 3 × 105 y old (stage VI).3) in which an additional nucleon (n. or Ye ≥ 1/9.1).1) suggests that the proton fraction Ye satisfies 3 n0 Esym (n) (1 − 2Ye ) . p) → p + (n. Because Ye generally increases with density. p) → n + (n. yet there is no evidence that a neutron star now exists in this supernova’s remnant. 1/3 1/3 1/3 The condition nn ≤ np + ne with np = ne = nYe requires that the proton to neutron ratio exceeds 1/8. collapse to a black hole would occur on a diffusion time of 10 to 20 s. The most efficient Urca process is the direct Urca process involving nucleons: n → p + e− + ν¯e . and thermal energy is thus continuously lost.2. 25 July . The star’s mass might be between the proto-neutron star maximum mass and the cold maximum mass. (6. Eq.33rd SLAC Summer Institute on Particle Physics (SSI 2005). p) + e− + ν¯e . Within 10 to 100 y. p) participates in order to conserve momentum. p → n + e+ + νe . The remnant’s observed luminosity is fully accounted for by radioactivity in the ejected meaning that any contribution from magnetic dipole radiation. but there are two possibilities: the standard and enhanced cooling scenarios. if the direct process is not possible. or its spin rate or magnetic field are substantially smaller than those of typical pulsars. protons and electrons. p + (n. The standard cooling L007 51 . Neutron Star Cooling The interior of a proto-neutron star loses energy at a rapid rate by neutrino emission. The dominant neutrino cooling reactions are of a general type. However. which is far above the value found in neutron star matter in the vicinity of n0 . A delayed collapse scenario could account for these observations. known as Urca processes. in a mixture of neutrons.1) This process is only permitted if energy and momentum can be simultaneously conserved.5 August 2005 leptons and thermal energy. Perhaps such a scenario could explain the enigma of SN 1987A. (6. creates an isothermal structure (stage (V) in Fig. p) + e+ + νe . 6. If so. with an effective temperature Tef f that tracks the interior temperature but about 100 times smaller. The 10 s duration of the neutrino signal confirmed the existence of a proto-neutron star. is very small. typically. Each reaction produces a neutrino or antineutrino. such as hyperons. Superfluidity quenches cooling from the direct Urca process. pions. There are two additional issues affecting cooling trajectories of neutron stars: superfluidity and envelope composition. Once again. It also plays an essential role in determining the threshold densities of other particles. The question of whether or not the direct Urca process occurs in neutron stars is of fundamental importance. Green boxes indicate sources from which both thermal optical and X-ray emissions have been observed. 25 July . scenario assumes that direct Urca processes cannot occur. enhanced cooling cannot occur. (6.4 M .3: Observational estimates (boxes) of neutron star temperatures and ages together with theoretical cooling simulations (curves) for M = 1.5 August 2005 Figure 6. and its inherent uncertainty means that it is presently unknown if direct Urca processes can occur in neutron stars. described in Ref. kaons or quarks. However. the quantity Sv (n) plays a crucial role for neutron stars. The density dependence of the symmetry energy function Esym determines the values of Ye and the threshold density at which the nucleonic direct Urca process occurs (see Eq. 51. 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