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UNITEXT for PhysicsLorenzo Bianchini Selected Exercises in Particle and Nuclear Physics UNITEXT for Physics Series editors Michele Cini, Roma, Italy Attilio Ferrari, Torino, Italy Stefano Forte, Milano, Italy Guido Montagna, Pavia, Italy Oreste Nicrosini, Pavia, Italy Luca Peliti, Napoli, Italy Alberto Rotondi, Pavia, Italy Paolo Biscari, Milano, Italy Nicola Manini, Milano, Italy Morten Hjorth-Jensen, Oslo, Norway UNITEXT for Physics series, formerly UNITEXT Collana di Fisica e Astronomia, publishes textbooks and monographs in Physics and Astronomy, mainly in English language, characterized of a didactic style and comprehensiveness. The books published in UNITEXT for Physics series are addressed to graduate and advanced graduate students, but also to scientists and researchers as important resources for their education, knowledge and teaching. More information about this series at http://www.springer.com/series/13351 Lorenzo Bianchini Selected Exercises in Particle and Nuclear Physics 123 Lorenzo Bianchini Italian Institute for Nuclear Physics Pisa Italy ISSN 2198-7882 ISSN 2198-7890 (electronic) UNITEXT for Physics ISBN 978-3-319-70493-7 ISBN 978-3-319-70494-4 (eBook) https://doi.org/10.1007/978-3-319-70494-4 Library of Congress Control Number: 2017957685 © Springer International Publishing AG 2018 This work is subject to copyright. All rights are reserved by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. 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Printed on acid-free paper This Springer imprint is published by Springer Nature The registered company is Springer International Publishing AG The registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland To my wife Elisabetta, 1 Gv 4, 16 Preface The realm of particle physics is vast: multidisciplinary knowledge across several domains of physics and mathematics is required to understand the reactions that occur when particles collide and to master the functioning of the experiments built to study these reactions: classical and quantum mechanics, special relativity, electrodynamics, thermodynamics, chemistry, atomic and nuclear physics, quantum field theory, electronics, analysis, geometry, group theory, probability, informatics, among the others. Large-scale particle experiments, like those hosted in the main laboratories around the world, are perhaps the best example of how multidisciplinary this field can become: the successful operation of these complex structures relies on the synergetic work of hundreds of scientists and engineers; it is only the combination of their individual expertise that makes it possible to cover all the needs. Thanks to the maturity of this field (more than one hundred years old!), a huge collection of textbooks, topical schools, academic classes, and scientific literature is available, where both the theoretical and experimental foundations of particle physics can be elucidated to the desired level of detail. Yet, as for all the other domains of physics, particle physics should be more about solving problems rather than knowing concepts! The path towards a solid understanding of this discipline passes through the capability of solving exercises. This book collects a sample of about 240 solved problems about particle physics in general. About half of the exercises are drawn from the public exams that have been proposed by the Italian National Institute for Nuclear research (INFN) to select its scientific staff over the last decade. Additional material inspired by my personal experience as an undergraduate student at Scuola Normale Superiore di Pisa, researcher in the CMS experiment, and teaching assistant at ETH Zürich complements the selection. Throughout this book, the main emphasis is put on experimental problems, although some more theoretical ones are also included. Thus, this book is mostly addressed to experimentalists. The proposed exercises span several subjects in particle physics, although I must acknowledge that it has not been possible to be truly exhaustive. Several topics have been unfortunately, yet necessarily, discarded or only marginally mentioned. vii Per aspera ad astra: solving problems is the most difficult and painful task for students. from the more classical ones to the more recent technologies. Some of the exercises are used as prototypes for a class of problems. an experiment in particle physics usually starts by scattering particles: acceleration of particles in stable and repeatable beams is therefore another important topic. Whenever possible. one should better rely on computers rather than try the analytical approach. as to train one's capability to handle simple calculations using approximations or order-of-magnitude estimates. a personal cultural bias towards an LHC-centric vision of the field may have driven the focus towards the high-energy frontier at the expenses of other equally lively sectors of research. stored. experience teaches that there exist a few constants and formulas that are really worth keeping in mind! In other situations. what are the technologies at hand. Furthermore. rare decays. cosmology. it should provide a test bench for undergraduate students and young . examples of simple computer routines written in open-source pro- gramming languages are also proposed. References to the scientific literature and topical textbooks are then provided to help the reader go into the various subjects in greater detail. and data analysis enter naturally into the game. Particle detectors cannot be understood without first mastering the basics of the interaction between particles and matter. hadron spectroscopy. Indeed. but also one that unveils the true degree of comprehension of the subject. B physics. which need to be processed. Some exercises require a few lines of calculations. and what the future lines of research. Within each chapter. We hope that this book can serve as supporting material to back up existing and more complete textbooks on experimental and theoretical particle physics. one should always try to derive the symbolic solution analytically and carry out the numerical computation without the help of pocket calculators. so that each exercise can be propaedeutic to those that follow. dark matter. others one or more pages. like neutrino physics. Consistency of notation throughout the text is pursued to reduce at a minimum the confusion introduced by the abundance of acronyms and conventional symbols peculiar to this field. In this case. electronics. Much attention is devoted to the opera- tional principles of particle detectors. where the main topic of discussion is first introduced in an academic fashion. what is still unknown but important to study. Also. In the latter case. and a basic knowledge of both subjects is therefore required. a proper scientific maturity demands also an overall picture of the field: what is known.viii Preface In particular. beyond-standard model theories will not be much discussed here. the relevant concepts and the general-purpose formulas are derived once and recalled afterwards by pointing to the master exercise. Given that particle detectors typically provide electric outputs. Finally. the exercises are organised as much as possible according to a logical order. Several exercises go along this direction by discussing the state of the art on the field. and cleaned from the noise. which therefore represents another topic of foremost interest. Other exercises are instead chosen to introduce a particular topic. including ongoing or planned measurements and new experimental techniques. which is then explained in some more detail by dedicated mini-lectures. At the same time. The exercises are grouped by subject into five chapters. informatics. I also want to heartily thank the Institute for Particle Physics of ETH Zürich for granting me the time to work on this book and for the fantastic teaching experience I enjoyed amid its brilliant and lively students. Pisa.Preface ix researchers to validate their level of preparation and hopefully stimulate their curiosity on the field. I am much indebted to INFN for allowing me to profit from a large number of the exercises contained in this book. Italy Lorenzo Bianchini September 2017 . The richness and variety of topics covered in this immense reservoir of knowledge have been fundamental to shape this work. 344 References . . . . . . . . . . . . . . . . . 177 3. . . . . . . . . . . . . . . .3 Luminosity and Event Rates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .3 Cross Section . . . . . . . . . 107 2. . . . . 241 4. . . . . . . . 359 xi . . .3 Confidence Intervals . . . . . . . . 1 1. . . . . . . . . . . . . . . . . . . . . . . . . . . 105 2 Particle Detectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1 Lorentz Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .3 Functioning of Particle Detectors . . . . . . . . . . . . . . . . . . 267 References . . . . . . . . . . . . . . . . . 262 4. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1 Conservation Laws . . . . . . . 241 4. . 238 4 Statistics in Particle Physics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 177 3. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54 1. . . . . . . . . . . . . . . . . . . . . . .2 Accelerators .2 Center-of-Mass Dynamics and Particle Decays . . . . . . . . 293 5. . . . . . . . . . . . .1 Passage of Particles Through Matter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .4 Higgs Boson . . . . . . . . 151 References . . . . . . 138 2. . . . . . . . . . . . . . . . . . . . . . . . . . . . .1 Elements of Statistics . . . . . . 329 5. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .2 Particle Identification . . . . . . . . . . . . . . . . . . . . . . . 272 5 Subnuclear Physics . 355 Index . . . . . . . . . . . . . . 216 References . . . . . . . . . . . . . . . . . . . . . . . . . . 273 5. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .2 Electroweak and Strong Interactions . . . . 107 2. . . . . . . . . . . . . . . . . . . . . . . . .Contents 1 Kinematics . . . . . . . . . . . . . . . . .3 Flavour Physics . . . . . . 1 1. . . . . . . . . . . . . . 83 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175 3 Accelerators and Experimental Apparatuses . . . . . . . .1 Tracking of Charged Particles . . . . . . . . . . . . . . . . . . . . . . . . 273 5. . . . . . . . . . . . . . . . . . . 200 3. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .2 Error Propagation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Acronyms ATLAS A toroidal apparatus BNL Brookhaven National Laboratories BR Branching ratio CC Charged current CDF Collider Detector at Fermilab CERN Conseil européen pour la recherche nucléaire CI Confidence interval CKM Cabibbo–Kobayashi–Maskawa matrix CL Confidence level CMB Cosmic microwave background CMS Compact Muon Solenoid DIS Deep inelastic scattering DT Drift tube EM Electromagnetic ES Electron scattering EW Electroweak FCNC Flavour-changing neutral current FWHM Full width at half maximum GEM Gas electron multiplier LEP Large Electron–Positron collider LHC Large Hadron Collider LINAC Linear accelerator LNF Laboratory Nazionali di Frascati LO Leading order LST Limited streamer tubes MC Monte Carlo MIP Minimum ionising particle MWPC Multi-wire proportional chambers NC Neutral current N(N…)LO Next(-to-next-to-…) leading order xiii . xiv Acronyms PDF Parton density function PDG Particle Data Group PEP Positron Electron Project PMNS Pontecorvo–Maki–Nakagawa–Sakata matrix QCD Quantum chromodynamic QED Quantum electrodynamic R&D Research and development RF Radio frequency RMS Rootmeansquare RPC Resistive plate chambers SM Standard model STP Standard temperature and pressure TDC Time-to-digital converter TOF Time-of-flight TPC Time projection chamber UL Upper limit . equivalently. UNITEXT for Physics. The last part of the chapter is devoted to the concept of cross section. Large emphasis then is given to the transformation properties of velocities and angles.1). Rotations are determined by the three usual Euler angles. (1.Chapter 1 Kinematics Abstract The first chapter is dedicated to the kinematics of relativistic particles. .1) i=1 with g μν = diag(1.and three-body decays. −1. Rotations and boosts change both the four-momentum and the spin vector of a particle. https://doi. (1. p3 ) in endowed with the Minkowski norm defined by: 3 p2 ≡ pμ pν g μν = p20 − p2i . which plays a central role in particle physics.org/10. −1). while boosts are determined by the three components of the velocity v of the new reference frame R as measured by an observed at rest in the initial reference frame R. . The space of four-vectors p = (p0 .1 Lorentz Transformations The set of space-time transformations under which the laws of physics are postulated to be invariant form the so-called Poincaré group: they comprise four space-time translations.1) provides an example. By construction. Selected Exercises in Particle and Nuclear Physics. © Springer International Publishing AG 2018 1 L. (1. three spatial rotations. Bianchini. the Lorentz transformations preserve the Minkowski norm of Eq. The starting point is the introduction of the Lorentz group through its representations. or. it is possible to find a representation of each element in terms of square matrices acting on vector spaces: the four-dimensional space of four-vectors and the (2S + 1)-dimensional space of spin vectors for a particle of spin S. The centre-of-mass dynamics is studied in detail for two-to-two scattering and for two. . Space rotations and velocity transformations (the latter are often referred to as boosts) form the sub-group of Lorentz transformations. −1. and three velocity transformations. Since these transformations form a group. 1. Any function of four-vectors that has the same form in all reference frames related by a transformation of the Poincaré group is called an invariant: the squared norm of Eq. . by the dimensionless boost vector β = v/c.1007/978-3-319-70494-4_1 . and should not be confused with the velocity and gamma-factor of a particle as measured in a given reference frame: the suffix “p” in the latter thus reminds that these quantities are different from the boost parameters. The boost vector defines a privileged direction in space. (1. as opposed to the active transformation that changes the vector components in the same reference frame.17. (1. (1. This way of writing the transformation corresponds to the so-called passive transformation. and indeed the transformation distinguishes between the component parallel () and orthogonal (⊥) to β.5) would be the transformation of a normal vector under a spatial rotation. For most of the applications it is however enough to remember the matrix version of Eq. Given that γ 2 − (−βγ )2 = 1 and that γ ≥ 1. see Problem 1.3) Ep m In the four-momentum space. γp ≡ 1 − βp2 2 = (1. p). Were not for the imaginary “angle” α = iθ and the same-sign off- diagonal elements. the first of Eq. although the 1 explicit dependence is often omitted in the calculations. as done by ordinary rotations.4) are the parameters of a transformation. the suffix can be safely dropped to simplify the notation.2) The velocity and gamma-factor of a particle are then defined as |p| − 1 Ep βp ≡ . the four-momentum is subject to the mass-shell constraint: p2 − m2 = 0 ⇒ E = Ep ≡ |p|2 + m2 . The generic boost can be also written in a compact vectorial form. when no such ambiguity can arise. Eq. (1. (1. and not the Euclidean norm E 2 + p2 . In terms of these two components.4) p −βγ γ p with β = |β| and γ = (1 − β)− 2 .4) are the four-momenta components measured in the reference frame R that moves with velocity cβ with respect to R.5) p − sinh α cosh α p with α = atanh β.2 1 Kinematics The three-momentum of a particle can be embedded into a four-vector (E. The variables at the left-hand side of Eq. called four-momentum.4). . the generic boost transformation is given by: E γ −βγ E = . However. For a particle of mass m. Notice that γ is a function of β. A final word of caution: β and γ in Eq.4) can be equivalently written as: E cosh α − sinh α E = (1. p⊥ = p⊥ (1. each Lorentz transformation is represented by a 4 × 4 matrix which transforms p into a new four-momentum p . (1. The differences accounts for the fact that the transformation has to preserve the Minkowski norm E 2 − p2 . 507 × 1016 GeV−1 . s= (0. (1.52 × 1024 GeV−1 .1. one often needs to convert lengths.7) Finally. we can convert the second by simply using that: 3 × 108 c = 3 × 108 m s−1 . any dimensionful quantity is expressed in powers of the energy unit. since a theoretical calculation performed in natural unitswill yield the result as a power of GeV. .8) The inverse transformations into the MKS system laws are also very useful. cross sections.507 × 107 c eV−1 ⇒ m = 0. s] in natural units. In natural units.6 × 10−19 kg = 0. (1.52 × 1015 eV−1 ⇒ s = 1.56 × 1027 GeV. the decay rate of the φ meson 1/τ = 6. 2. it proves useful to remember the MKS value of the constant c: c = 197 MeV fm = 1. the electric charge e.6) To convert the metre. 1. c]. we can write: 9 × 1016 kg c2 = kg (3 × 108 m s−1 )2 = 9 × 1016 J = eV = 0. taken to be the eV or one of its multiple. (1.507 × 107 c eV−1 ) c s = 1. In order to convert the result back into the MKS system. Solution Remembering that m c2 is an energy. Discussion Natural units are of the greatest help in simplifying the calculations.97 × 10−7 m eV. and then to “silence” the result with respect to and c by setting them equal to unity.1 Express the MKS units [kg. 3. one needs to multiply the number obtained by a factor α cβ where α and β are chosen such that the overall dimension comes out right. In particular.56 × 1036 eV c−2 ⇒ kg = 0. m = 0.47 × 1021 s−1 .1 Lorentz Transformations 3 Problems Problem 1. m. m. The underlying idea is to trade off the usual MKS units [kg. s] by the three dimensionful units [eV. Using the results thus obtained. and time intervals into MKS: . write down the following quantities in natural units: 1.56 × 1036 eV. a cross section σ = 1 pb. 4). we have: e2 1 4π ≈ ⇒ e≈ = 0.26 × 10−8 GeV−2 .97 fm [cross section] GeV−2 = 0. Hence.4 1 Kinematics [length] GeV−1 = 1. where (E.52 × 1024 )−1 GeV = 4.47 × 1021 s−1 = 6.12) 4π c 137 137 measures d p ≡ dE d p and d p/Ep . (1. Then. (1. are invariant under a generic transformation of the Lorentz group.47 × 1021 (1. From the definition of the fine structure in Heaviside–Lorentz units.507 × 107 eV−1 )2 = 0. for every four- vector p and boost Λ it must hold: T T p Λ g (Λp) = pμ pν g μν = pμ pν g μν = pT g p ⇒ ΛT g Λ = g. (1.8) we have: Γφ = 6.13) The last equation implies (det Λ)2 = 1. (1.389 mbarn (1.1).303. . as it is the case for Eq.66 × 10 s Let’s now apply the results above to the three cases of interest. the proper Lorentz transformation are defined by the condition det Λ = +1.7) it follows that: 1 pb = 10−40 m2 = 10−40 (0.10) 2.2 Prove that the two form a four-vector and Ep = |p|2 + m2 . (1. (1. the Lorentz transformations preserves the Minkowski norm of Eq. (1.14) while under a space rotation. (1. (1. From Eq. As discussed in the introduction.9) −1 −24 [time] GeV = 0. d 3 p = d 3 p and dE = dE. In particular.11) 3. which proves that the measure is indeed invariant under the Lorentz group. Solution Let’s consider first the d 4 p measure. p) 4 3 3 Problem 1. 1.26 MeV. under a generic boost we have: d 4 p = | det Λ| d 4 p = d 4 p. such that the first of Maxwell’s equation becomes ∇ · E = ρ and [e] = kg1/2 m s. From Eq. 1 such must be the right-hand side. Alternatively. we can assume the x-axis to be aligned with the boost direction.1 Lorentz Transformations 5 Let’s now consider the second measure.15) Invariance under space rotation follows again from the fact that d 3 p = d 3 p and Ep = Ep . Without loss of generality.1. so that: d 3 p d γ (px − βdEp ) dpy dpz γ (dpx − βpx /Ep dpx ) dpy dpz d3p = = = Ep γ (Ep − βpx ) γ (Ep − βpx ) Ep (1. (1.16) ∂(p −m ) 2Ep ∂E +Ep where the Heaviside function Θ selects the positive root. we can write this measure in a way that if manifestly Lorentz- invariant. (1. Since the left-hand side of Eq.3 Prove that the scalar function: . Indeed: d3p d3p Θ(Ep ) δ(p2 − m2 ) d 4 p = 2 2 = .16) is manifestly Lorentz-invariant. Problem 1. I(p1 . p2 )2 = (E1 p2 − E2 p1 )2 − (p1 × p2 )2 = = E12 |p2 |2 + E22 |p1 |2 − 2E1 E2 p1 · p2 − |p1 |2 |p2 |2 + (p1 · p2 )2 = = E12 E22 − 2E1 E2 p1 · p2 + (p1 · p2 )2 −E12 p22 + E22 |p1 |2 − |p1 |2 |p2 |2 = (p1 p2 )2 = (p1 p2 )2 − (|p1 |2 + p21 )p22 + (|p2 |2 + p22 )|p1 |2 − |p1 |2 |p2 |2 = = (p1 p2 )2 − p21 p22 . . The invariance under boosts is less trivial and has to be proved explicitly. pi ) are a pair of four-vectors. is invariant under rotations and boosts. (1. The best way to do it is to find an equivalent expression for I that is manifestly Lorentz-invariant. it is convenient to take the square at both sides. Solution The invariance of the right-hand side of Eq. (1. To this purpose.17) under rotations follows from the fact that a rotation leaves unchanged both the 0th component of the four-vector and the relative angle between the three-vectors.17) where pi = (Ei . p2 ) = (E2 p1 − E1 p2 )2 − (p1 × p2 )2 (1.18) 1 It can be proved that the sign of the 0th component is also a Lorentz invariant. obtaining: I(p1 . (1. p2 ) = E1 p2 x − E2 p1 x = = |(cosh α E1 + sinh α p1 x )(sinh α E2 + cosh α p2 x )− − (cosh α E2 + sinh α p2 x )(sinh α E1 + cosh α p1 x )| = |(cosh2 α − sinh2 α)E1 p2 x − (cosh2 α − sinh2 α)E2 p2 x | = = |E1 p2 x − E2 p1 x | = I(p1 . we get: I(p1 . p2 ). The special case p1 ∝ p2 gives rise to a simpler expression. so must be the I(p1 . By using the matrix form of Eq. where the effect of a finite Lorentz transformation can be studied explicitly. (1.e. (1. where ρi is a space density ([ρ] = m−3 ) and ρi vi is a density flux ([ρv] = m−2 s−1 ). (1.5).17) becomes: . then Eq.: ji = (ρi .19) Discussion When the four-vectors are specialised to be the current densities2 of two beams. ρi vi ). i.6 1 Kinematics Since the left-hand side of Eq.18) is invariant under boosts. p2 ). which as to hold for any frame. (1. . although this is not a proper velocity (indeed. Problem 1.g. 2 2 I(j1 . (1.11.21) dτ dτ where p is the four-momentum of particle of mass m accelerated by external forces. The latter can be written in covariant notation as ∂μ jμ = 0.293).20) where β i are the particle velocities in units of c. hence jμ has to transform as a covariant vector since ∂μ is contravariant. The invariant I finds application in the general formula of the cross section. Solution We first express the four-momentum components in terms of the particle velocity β. while the time derivatives and the velocities are measured in a generic frame. j2 ) = ρ1 ρ2 β 1 − β 2 − β 1 × β 2 ≡ ρ1 ρ2 vrel (1. τ is the proper time of the particle.4 Prove the identity: dpμ dpμ 2 = −m2 γ 6 β̇ − (β × β̇)2 . and vrel is called relative velocity between the two particles. it can also exceed c in some reference frames). Eq. The interpretation of vrel in terms of particle velocities will be elucidated in Problem 1. namely: 2 The fact that j is a Lorentz-vector can be proved by noticing that the continuity equation ∂t ρ + divx ρv = 0 has to be invariant since it states the conservation of mass. see e. Problem 1. (1. where e is the electric charge of the particle [1]. mγ β). (1.24) dt 2 d(γ β) 2 2 = γ̇ β + γ β̇ = γ 2 γ 2 (β · β̇)β + β̇ = dt 2 2 2 = γ 6 β · β̇ β 2 + 2γ 4 β · β̇ + γ 2 β̇ = 2 2 2 = γ 4 (γ 2 − 1) β · β̇ β 2 + 2γ 4 β · β̇ + γ 2 β̇ = 2 2 2 = γ 6 β · β̇ + γ 4 β · β̇ + γ 2 β̇ (1.5) in an exponentiated form as to make the . so has to be the right-hand side. Indeed.25) Inserting these identities in Eq. (1.21) gives the Lieanard formula for the power P emitted by a charged particle accelerated by an external force. one can prove that P ∝ −e2 (dp/dτ )2 . (1. we obtain: dpμ dpμ 2 2 = −m2 γ 4 γ 2 β · β̇ + β̇ = dτ dτ 2 2 = −m2 γ 6 β 2 β̇ − (β × β̇)2 + β̇ (1 − β 2 ) = 2 = −m2 γ 6 β̇ − (β × β̇)2 . Discussion Equation (1. (1.22) Hence. Since the left-hand side is manifestly Lorentz- invariant.23). (1.5 Work out a heuristic representation of the boost generators K in dimension d = 2 starting from the finite boost transformation of Eq. p) = (mγ . we get: 2 dγ 2 2 = γ 3 β · β̇ = γ 6 β · β̇ (1.5). Solution Let’s denote by (χ1 . By applying the chain rule for the derivative of composite functions.1 Lorentz Transformations 7 p = (Ep . χ2 ) the four-vector components that transform non-trivially under boosts. (1.1. We then rewrite Eq.23) dτ dτ dτ dτ dt dt where we have made use of the relation dt = γ dτ .21).26) which proves the identity of Eq. we get: 2 2 2 2 dpμ dpμ dEp dp dγ d(γ β) = − =m γ 2 2 − . one can easily verify that J = σ /2 and K = −iσ /2 provide a representation of the generators. and considering the commutation rules in Eq. Since a generic element of a group can be always parametrised as exp{iK · α}. the so-called ( 21 . (1.27) 2 2 χ2 χ2 where |nα | = 1.27) yields the result: σ K = −i . For d = 2. The combinations J± = (J ± iK)/2 commute among themselves and satisfy individually the algebra of SU (2). since the commutation rules are all invariant under K → −K. [Ki . where K are the generators. we know that the σ /2 matrices provide a fundamental representation of the generators of SU (2).3 has important implication for the quantisation of spin-1/2 fields (Weil spinors). Kj ] = −i εijk Jk (1.28) 2 Discussion The six generators of the Lorentz group satisfy the Lie algebra: [Ji . 0) and (0.29). an immediate comparison with Eq.6 Prove that all RH (LH) spin-1/2 particles of mass m have helicity h = +1/2 (h = −1/2) in the limit |p| . The existence of two inequivalent representations of the boost vector in dimension d = 2. By defining the transformation parameter as α/2. Notice that K = +iσ /2 provides an equally valid representation. The latter can indeed exist in two chiralities. [Ji . 21 ). (1. Kj ] = i εijk Kk . which are related one-to-another by a parity operation. Jj ] = i εijk Jk . 2 01 2 10 i 0 0 −1 χ2 α α χ σ χ1 = cos i 1 − i sin i σ · nα 1 = ei(−i 2 )·α (1.29) where Ji are the generators of the space rotations and Ki of the boosts. (1. we could get directly the correct normalisation for the generators. depending under which representation of the Lorentz group they transform: right-handed (RH) and left-handed (LH) spinors. Given that J = J+ + J− and K = −i(J+ − J− ). Problem 1.8 1 Kinematics generators explicit: α α χ1 cosh − sinh χ1 = 2 2 = χ2 − sinh α2 cosh α χ2 2 α 10 α 01 0 −i 1 0 χ1 = cosh · − sinh · +0· +0· . 29).g. the second of Eq. and it also transforms as a vector under parity transformations. see e. 3K is a vector under rotations. since a parity operation must change the direction of the boost. (1. . m. 1 Lorentz Transformations 9 Discussion The helicity operator is defined as: S·p h= . i. sin 2δ ). for which the helicity is instead a conserved quantum number. (cos 2δ .L R. Notice that h is not a priori Lorentz-invariant. This is not the case for massless particles. where the particle three-momentum is p.31) exp i σ2 · θ + σ 2 · α LH The spin operator is represented by the matrix S = σ /2.30) |p| and it acts on the spin vector of the wave function.L + P−1/2 cos2 2δ (cosh α 2 ± cosh α2 )2 + sin2 2δ (cosh α 2 ∓ cosh α2 )2 α (cosh2 + sinh2 α2 ) cos δ ± 2 sinh α cosh α2 cos δ ± tanh α 2 2 α 2 = = cosh 2 + sinh2 α2 ± 2 sinh α 2 α cosh 2 cos δ 1 ± tanh α cos δ cos δ ± β = . (1. for a massive particle.e. (1. which is the space suitable to construct the spin states for spin-1/2 particles. since. Solution As discussed in Problem 1.L P+1/2 R. α) can be written as: i(J·θ −K·α) exp i σ2 · θ − σ · α RH Λ(θ . giving: ξ = Λ(α)ξ ∗ = cosh α2 1 − sinh α2 σ · (−ez ) ξ ∗ = cosh α2 1 + sinh α2 σz ξ ∗ η = Λ(α)η∗ = cosh α2 1 + sinh α2 σ · (−ez ) η∗ = cosh α2 1 − sinh α2 σz η∗ (1. Let’s also chose the basis of eigenvectors of σz . where the z-axis is assumed to be aligned with the boost direction.L P+1/2 − P−1/2 = 2 2 = R.33) 1 ± β cos δ . α) = e = 2 (1.1. The spinors ξ and η in the laboratory frame. the polarisation in the laboratory frame will be: cos2 2δ (cosh α ± cosh α2 )2 − sin2 2δ (cosh α ∓ cosh α2 )2 R. Let’s denote the spinor in the centre-of-mass frame by ξ ∗ for the RH fermion and by η∗ for the LH fermion. can be obtained by applying a boost with parameter α = −α ez .32) If we now take the spinor in the rest frame ξ ∗ and η∗ to a generic admixture of ± 21 eigenstates. The generic passive transformation Λ(θ.5 there are two inequivalent representation of the Lorentz group in dimension d = 2. it is always possible to find a reference frame where the momentum of the particle flips direction. In particular. if β → 1.40). see Eq.10 1 Kinematics The last equality makes use of Eq. (1.5) to relate the boost parameter to the velocity in the laboratory frame. as it is the case for |p| . It can be noticed that this expression is identical to the transformation law of the cosine of the polar angle of a massless four-vector under a boost β. (1. if one interprets δ as the polar angle in the centre-of-mass frame. Therefore. Suggested Readings The reader is addressed to a more complete textbook on quantum field theory. dm . A useful tool to investigate the behaviour of spin states under rotations is j provided by the so-called rotation matrix. m (θ ). since the latter transforms non trivially under rotations. Deter- mine the angular distribution of the scattered pions in the centre-of-mass frame. Chap. Ref. See e. The rotation matrix proves very useful when one wants to analyse transition probabilities between states related by a spatial rotation. (1. Problem 1. ultra-relativistic spin-1/2 particles of a given chirality will also have a net helicity. Invariance of the dynamics under rotations becomes an important selection rule when the particles involved carry spin. 9 of Ref. the polarisation in the laboratory frame is given by +β for RH fermions and by −β for LH. [2] or Chap. (1. [3]. m = dm .34) m Here. Discussion As discussed in the introduction.35) see e. independently from their spin state in the centre-of-mass frame: the helicity is positive for right-handed fermions and negative for left-handed.g. the vectors |j. m (1. −m . Assume the protons to be unpolarised. For an unpolarised state in the centre-of-mass (δ = π/2).g. m = (−1)m−m dm.7 The Δ(1232) resonance can be produced by scattering pions of appropriate energy against a proton target. As made evident by Eq. π . the elements of the rotation matrix are the linear coefficients of the rotated of the generic eigenstate by an angle θ around an axis orthogonal to the quantisation axis z.34). m = d−m. m. . m (θ )|j. the Lorentz group comprises the set of spatial rotations. It can be also shown that: j j j dm . [4]. m are the eigenstates of the angular momentum operators J2 and Jz . 3 of Ref. defined by: j exp −iθ J · ey |j. the polarisation tends to ±1 for 0 < δ < π and to 0 for δ = 0. +1/2 3 cos θ − 1 θ 2 3 cos θ + 1 θ 2 = cos + − sin = 2 2 2 2 1 θ θ 1 + 3 cos2 θ = 9 cos2 θ + 1 − 6 cos θ cos2 − sin2 = . the final state corresponds to the rotated by θ of either | 23 . Suggested Readings The reader is addressed to the original paper by E. At the resonance. [5] about the evidence of the Δ resonance and to the determination of its spin based on the distribution of the scattering angles.36) Notice that we have made use of Eq. − 21 .35) to simplify the calculations. p. 4 2 2 4 (1. Fermi et al. can take the values m = ±1/2.2 GeV. The eigenvalue of Ĵz . Prove that the light particle cannot be scattered exactly backwards.8 A massless spin-1/2 particle scatters elastically against a much heavier particle that can be assumed to be always at rest. The angular distribution in the centre-of-mass frame will therefore feature a dependence on the polar angle θ of the form ∼(1 + 3 cos2 θ ). The two particles exchange force through an helicity-conserving interaction which does not change the spin of the target particle. A compendium of formulas for the rotation matrices can be found in Table 43 of Ref.1 Lorentz Transformations 11 Solution When the centre-of-mass energy approaches 1. if any. called Δ. so the probability of decaying to an angle θ is given by the sum of the probabilities: 1 dΓ 1 1 dΓm=−1/2 1 1 dΓm=+1/2 = · + · = Γ d cos θ 2 Γ d cos θ 2 Γ d cos θ 3/2 2 3/2 2 m dm . Indeed. since only the proton can contribute to Jz through its spin. Let the angle between p and p be denoted by θ . . The Δ baryons are members of the j = 3/2 decuplet.1. Problem 1. the dominant state participating in the scattering can be therefore assumed to have j = 3/2. When the Δ resonance is produced. where z is the axis defined by the initial centre-of-mass momentum of the proton. The assumption of unpolarised protons implies that the two configurations must be equally likely. The two corresponds to orthogonal states. The decay amplitude is fully determined by angular momentum conservation. it decays into a new π p state with centre-of-mass momentum p . (1. [4]. the π p scattering becomes resonant due to a baryonic bound-state. −1/2 m dm . + 21 or | 23 . it is said to be chiral. It is convenient to choose the coordinate system so that β is aligned along the x-axis. ±1/2 = cos θ2 . r ) ⇔ ω(t − n · r) = ω (t − n · r ). Synchronising the two clocks so that at the time t = t = 0 the wave has a phase φ = 0 at the origin r = r = 0. Problem 1. 1 of Ref. since R. r) can be expressed in terms of (t .9 Determine the relativistic Doppler effect and the law of aberration of light for an observer moving with velocity v = βc with respect to the light source. since it only involves LH fermions.37) The coordinates (t. moves with velocity −β as seen from R .4). On the contrary. (1. In the frame R. 5. Solution A massless spin-1/2 particle can be in only one helicity state. the interaction between spin-1/2 particles and gauge bosons are helicity- conserving. it would imply a change of the overall angular momentum projection along the scattering axis by |Δm| = 1. When the theory is not symmetric for RH and LH particles. we can argument that the scattering amplitude at an angle θ should be proportional to d±1/2.2.12 1 Kinematics Discussion In the SM. Solution The invariance under Lorentz transformations requires the phase of a light wave to be the same for two reference frames R and R . r) = φ (t . moving one with respect to the other with constant velocity βc. the particle will have the same helicity before and after the scattering. while in R . the frequency of the wave is ω and its direction of propagation n. 1/2 Suggested Readings The reader is addressed to Chap. the same quantities are ω and n . r ) by using the transformation of Eq. depending on its chirality. We also write . with the only modification β → −β. QCD and the EM interaction treat RH and LH fermions on the same footing. In terms of the rotation matrices of Eq. If the interaction conserves the chirality. (1. If the projectile were to scatter exactly backwards. [2] for a deeper discussion on this subject. which indeed vanishes at θ = π . see Sect. since it distinguishes between the two chiralities. since they only involve combinations of vector V and axial A currents. we have: φ(t. If the target particle does not participate to the interaction through its spin. the frame where the light source is at rest. The SU (2)L interaction that under- lies the EWK force is chiral.34). it will also conserve its projection along the scattering axis. (1. which would violate angular momentum conservation. By equating the coefficients of the space-time coordinates in Eq. Solution Let the particle velocity in the reference frame R be denoted by v.39) by using the trigonometric identity cos θ = ±(1 + tan2 θ )− 2 . (1. one has: ⎧ ⎪ ⎪ dt = γ (dt − βdx) ⎪ ⎨dx = γ (−β dt + dx) (1. 1 Problem 1. we obtain the following system of equations: ⎧ ⎪ ⎨ω = ωγ (1 − β cos θ ) ≡ ωγ (1 − β · n) ω cos θ = ωγ (cos θ − β) (1.39) ω cos θ γ (cos θ − β) which describes the aberration of light.41) ⎪ ⎪ dy = dy ⎪ ⎩ dz = dz .38) ⎪ ⎩ ω sin θ = ω sin θ The first equation gives the relativistic Doppler effect. sin θ = .38). It is useful to write explicitly the two other trigonometric relations: cos θ − β sin θ cos θ = . r).4) applied to the four-vector (t. (1.10 Derive the transformation law for the velocity v of a massive particle under a generic Lorentz transformation.37). (1. From the Lorentz transformations Eq. Taking the ratio of the last two equations: ω sin θ sin θ = tan θ = . The reference frame R in which we want to calculate the particle velocity moves with velocity β as seen from R.1. It’s convenient to choose the coordinate system so that β is aligned along the x-axis. (1. (1.1 Lorentz Transformations 13 n = cos θ ex + sin θ ey .40) 1 − β cos θ γ (1 − β cos θ ) which can be derived from Eq. (1. or even directly from Eq. 42) dt γ (dt − β dx) γ (1 − βvx ) dz dz vz vz = = = dt γ (dt − β dx) γ (1 − βvx ) from which we can derive the transformation of polar angle: sin θ tan θ = . Then. so that θ < π2 and cos θ > 0. For sake of notation.39) for the case |v| = 1. To find such an angle. With this choice. This is not the case for β/β ∗ > 1. For later consistency. Eq. we redefine the variables using the same notation that will be adopted to study the kinematics in the centre-of-mass frame. Let’s now consider the two distinct cases for |v| < 1.43) γ (cos θ − β/|v|) which agrees with Eq. we get: . tan θ is always positive. namely: |v| < β and |v| ≥ β. because.44) γ (cos θ ∗ + β/β ∗ ) First.45) θ ∗ →π which means that a particle moving backwards in R ∗ will also appear moving backwards in R. (1.14 1 Kinematics Taking the ratios: dx −β dt + dx vx − β vx = = = dt dt − β dx 1 − βvx dy dy vy vy = = = (1. (1. (1. by Rolle’s theorem there must be an angle θmax ∗ giving the largest opening angle in R. we first use the relation cos x = ±(1 + tan2 x)− 2 to express cos θ as a function of θ ∗ . (1.43) becomes: sin θ ∗ tan θ = . in the same limit.46) 1 + tan2 θ γ 2 (cos θ ∗ + β/β ∗ ) + (1 − cos2 θ ∗ ) The choice of the “+” sign is motivated by the fact that for β > β ∗ . for β/β ∗ < 1. and then set its first 1 derivative to zero to find the maximum: 1 γ (cos θ ∗ + β/β ∗ ) cos θ = ± √ = + (1. Since tan θ = 0 for θ ∗ = 0. we define cos θmax ∗ =x and β/β ∗ = ξ . (1. we notice that. tan θ → 0+ . namely we replace |v| → β ∗ and β → −β. we have: lim tan θ = 0− . 6 (top).8 and three representative values of β ∗ . 1. Taking into account both the longitudinal and transverse components. (1.48) β At this centre-of-mass angle. dx θmax γ 2 (x + ξ )2 + 1 − x 2 0 = γ 2 (x + ξ )2 + 1 − x 2 − (x + ξ )[γ 2 (x + ξ ) − x] = 1 + xξ.9 (middle).11 Consider two particles with velocity vA and vB .50) γ (1 − v · β) β2 .8 for three different values of the particle velocity β ∗ : 0. (1.42) in a vectorial form.1 Relation between the cosine of the polar angle in two reference frames related by a boost β = 0. the opening angle in the laboratory frame is given by: ! 2 −1 β∗ β∗ β β∗ tan θmax = 1− γ − + ∗ = .1.1 shows a graph of cos θ as a function of cos θ ∗ for a boost β = 0. 0.47) from which: ∗ β∗ cos θmax =− . Problem 1. Solution It is convenient to first write Eq.1 Lorentz Transformations 15 Fig. (1.49) β β β γ β 2 − β ∗2 Figure 1. (1. Determine the velocity of particle B in the rest frame of A. and 1 (bottom) γ 2 (x + ξ )2 + 1 − x 2 − (x + ξ ) √2γ2 (x+ξ )−2x 2 d cos θ 2 γ (x+ξ )2 +1−x 2 0= ∗ = . (1. we have: 1 γ −1 v = v+ v·β −γ β . (1. therefore we also have |vB|A | = |vA|B |. we notice that: vrel |vB|A | = . the rest frame of either A or B. (1. (1.16 1 Kinematics To express the velocity vB|A of the particle B in the rest frame of A. (1.51) (1 − vA · vB )2 (1 − vA · vB ) The last equality is easy to prove since |vA ×vB |2 = |vA |2 |vB |2 −(vA ·vB )2 .53) are defined in a particular reference frame. ! (1 − |vA |2 )(1 − |vB |2 ) (vA − vB )2 − (vA × vB )2 |vB|A | = 1 − = . .12 Determine the minimum and the maximum opening angle between the two massless particles produced in the decay of a particle of mass m and momentum p.e. (1. we obtain: γA ρA|A ρB|A ρA|A ρB|A |vB|A | = (vA − vB )2 − (vA × vB )2 = γA (1 − vA · vB ) = ρA ρB (vA − vB )2 − (vA × vB )2 . This can be seen as another proof that the scalar variable I of Eq. so that Eq.52) 1 − vA · vB If we now multiply this expression by the density flux of B times the density of A in the rest frame of A.20). (1.20) is Lorentz-invariant. Problem 1. This expression is symmetric with respect to A ↔ B. γA (1 − vA · vB ) |vA |2 (γA − 1)2 2 − 2γA (γA − 1) v · v + γ 2 γA2 (1 − vA · vB )2 |vB|A |2 = |vB |2 + |vA |2 (v A · v B ) A B A |vA |4 |vA |2 γ −1 + 2(vA · vB )2 A 2 − 2γA vA · vB = |vA | 2 (γA − 1)2 2(γA − 1) = (vA · vB ) + − 2vA · vB γA (γA − 1) + γA + |vB |2 + |vA |2 γA2 = |vA |2 |vA |2 = γA2 (1 − vA · vB )2 − (1 − |vA |2 )(1 − |vB |2 ) . since all the quantities at the left-hand side of Eq. we need to perform a boost with parameter β = vA . (1.50) becomes: 1 γA − 1 vB|A = vB + v A · v B − γA v A .53) where we have used the fact that ρB|A is the transformed of the density ρB in the rest frame of A. Discussion From an immediate comparison with Eq. i. whereas γA ρA|A is the density of A in the laboratory frame. 4). It is convenient to express cos θ1. Then: cos φ = cos (θ1 + θ2 ) = cos θ1 cos θ2 − sin θ1 sin θ2 = β 2 − cos2 θ ∗ sin2 θ ∗ 2β 2 − 1 − β 2 cos2 θ ∗ = − (1 − β 2 ) = = 1 − β 2 cos2 θ ∗ 1 − β 2 cos2 θ ∗ 1 − β 2 cos2 θ ∗ 1 − β2 =1−2 .39).2 = γ (E ± β|p | cos θ ) = γ E (1 ± β cos θ ) ∗ ∗ ∗ ∗ ∗ p1.1.54) ⎪ ⎩ y ∗ ∗ p1. see Fig. so that: β + cos θ ∗ sin θ ∗ cos θ1 = .2 = γ (βE ± |p | cos θ ) = γ E (β ± cos θ ) x (1. with the modification β → −β: ⎧ ⎪ ∗ ∗ ∗ ∗ ∗ ⎨E1.56) 1 + β cos θ ∗ γ (1 + β cos θ ∗ ) β − cos θ ∗ sin θ ∗ cos θ2 = . Let θ ∗ be the polar angle of the photon moving along the y > 0 direction. The momentum components in the laboratory frame are therefore given by Eq. so that the other photon makes an angle of π −θ ∗ . (1.58) 1 − β 2 cos2 θ ∗ . or notice that cos θ1.2 .40) directly.55) p1. We choose the reference frame so that the x-axis is aligned along the direction of p and the decay takes place in the x–y plane.2 Representation of a R∗ R boost β from the laboratory frame R to the p∗ p1 centre-of-mass frame R ∗ θ∗ θ1 −βex +βex →R π − θ∗ θ2 → R∗ −p∗ p2 Solution In the centre-of-mass frame. the two massless particles have momenta ±p∗ and energy E ∗ = |p∗ | = m/2.57) 1 − β cos θ ∗ γ (1 − β cos θ ∗ ) Let’s define the opening angle between the two particles in the laboratory frame by φ. To this purpose. 1.2 = x p1.2 = y = . Seen from the laboratory frame.2 /E1. 1.2. sin θ2 = (1. (1.2 γ (β ± cos θ ∗ ) which agrees with Eq. we could either use Eq. (1.2 sin θ ∗ ⇒ tan θ1. (1.1 Lorentz Transformations 17 Fig.2 = |p | sin θ x p1.2 in terms of cos θ ∗ . (1. the centre-of-mass frame moves with velocity β = |p|/Ep . sin θ1 = (1. 58) in a monotonously decreasing function of cos2 θ ∗ ∈ [0. (1. so that: 1 − β2 cos(φmax ) = min (cos φ) = 1 − 2 = −1 θ∗ 1 − β 2 cos2 θ ∗ cos2 θ ∗ =1 ⇒ φmax = φ(θ ∗ = 0.59) 1−β 2 cos(φmin ) = max (cos φ) = 1 − 2 = 2β 2 − 1 θ∗ 1 − β 2 cos2 θ ∗ cos2 θ ∗ =0 π ⇒ φmin = φ(θ ∗ = ) = arccos(2β 2 − 1) = 2 arccos(β) (1.60) 2 An interesting case is when |p| . 1]. π ) = π (1.18 1 Kinematics The right-hand side of Eq. it is always possible for one of them to move backwards with respect to the direction of the mother particle. m. we obtain: φmin 2 2 2β 2 − 1 = cos φmin ≈ 1 − ⇒ φmin = 2 1 − β 2 = . For large boosts. π]. the backward-emitted photon gets increasingly red-shifted.38). so that β ≈ 1 and φmin is small. (1.61) 2 γ For large boosts. By Taylor- expanding the cosine around zero. or equivalently the first of Eq. for massless particles. the opening angle between the two photons is therefore contained in the range [2/γ . see the first of Eq. however. so that it eventually becomes of vanishing energy for γ . (1. Notice that. (1.54). (1. From this expression. 1. we can compute the tangent of the opening angle φ: tan θ1 + tan θ2 tan(φ) = = 1 − tan θ1 tan θ2 β β1∗ +β2∗ √ γ β1∗ β1∗ 1 − cos2 θ ∗ = 2 . (1. giving an equation for tan θ1. (1. Problem 1. The denominator D(cos θ ∗ ) at the right-hand side of Eq.62) β1∗ −β2∗ −β 2 cos2 θ ∗ + β β1∗ β2∗ cos θ ∗ + ββ∗ β ∗ − 1 + β 2 1 2 ∗ ∗ where β1.2 are the velocities of two particles in the centre-of-mass frame.12.2 as in Eq.13 Determine the minimum and the maximum opening angle between two particles with mass m1 and m2 produced in the decay of a particle of mass m and momentum p. m2 > 0.67) is a second-degree polynomial with negative concavity. Let’s study its value for cos θ ∗ = ±1: . We can therefore start from Eq.2 = |p∗ |/E1.54) for the more general case m1 .44). (1. Solution This exercise is analogous to Problem 1. the denominator has to vanish at some point. while φmin is strictly larger than zero. (1. D(−1) < 0 if β ∗ < β1. (1. (1. ∗ ∗ • β2(1) > β.392. • β < min{β1∗ . and then it flips sign. One can find a numerical solution for φmax . or vanish twice.2 Hence. to θ ∗ = π/2. if β > max{β1∗ . Consequently. we have: 2 γ β β∗ tan(φmax )β1∗ =β2∗ = . (1.5. β2∗ }: the maximum opening angle φmax corresponds to atan (xmax ). as for Rolle’s theorem.3 provides an example of how to determine numerically argmax {tan φ} by using a computer program in Python. D(−1) > 0 if β ∗ < β2∗ and β ∗ > β1∗ ⎪ ⎩ ∗ D(+1) < 0. At this angle. starting from Eq.62). Since the concavity is positive.4 The case β1∗ = β2∗ = β ∗ allows to simplify further Eq.64) ⎪ ⎪D(+1) < 0.67). so that tan φ eventually approaches 0− as cos θ ∗ → ±1. since β ∗ < β for the other particle. one gets cos θmax 4 For ∗ = 0. We can easily see that such value does not correspond.2 ⎪ ⎨D(+1) > 0. Appendix 1. β1∗ ].1 Lorentz Transformations 19 β2 β β D(±1) = ± ∗ ∓ ∗ − 1. corresponding to a pair of collinear particles in the laboratory frame. (1. (1. which is in agreement with the numerical evaluation in Fig. corresponds to a backward-propagating particle in the laboratory frame. for example using Newton’s method to iteratively maximise tan φ. 1. In this case. using the input values β = 0. respectively. in general. if at least one among β1 and β2 is larger than β.3. Instead. D(−1) < 0 if β ∗ > β1∗ and β ∗ < β1∗ (1. example. we have: ⎧ ∗ ⎪ ⎪D(+1) > 0. β1∗ = 0.1.65) γ 2 β 2 − β ∗2 The minimum opening angle is φmin = 0. the solution with cos θ ∗ = −1 corresponds instead to a pair of collinear particles in the laboratory frame. D(−1) > 0 if β ∗ > β1. φmax = π like the previous scenario.63) β1∗ β2∗ β2 β1 These expressions are two parabola in β with roots [−β1∗ .62) is limited and has a global maximum for some value cos θmax . β2∗ }. β2∗ }. First one notices that tan(φ) becomes an even function of cos θ ∗ . β2∗ ] and [−β2∗ . ∗ tan φ in Eq. and it can be computed. . hence φmax = π . then D can be either always negative. β1(2) < β: the backward emission in the centre-of-mass frame of a particles with velocity larger than β. hence φmin = 0. We can summarise the various cases as follows: • β > max{β1∗ . again numerically. hence we the maximum has to occur at cos θ ∗ = 0. then D(cos θ ∗ ) > 0 for every angle θ ∗ . If instead both β1∗ and β2∗ exceed β.3 and β2∗ = 0. ∗ where cos θmax is the value that maximises the right-hand side of Eq. ∗ ∗ Conversely.62).8. we can use the approximate formula of Eq.20 1 Kinematics Fig.61) to express the minimum opening angle between the two photons as a function of the π 0 energy. (1. (1. 1. Problem 1.66) γ E 180◦ 0.2 ∗ > β.3 The trigonometric tangent of the opening angle φ in the laboratory frame between the decay products of a massive particle with velocity β = 0. The calorimeter is then used to detect π 0 ’s. In the rest frame of the mother particle. Bando n. What is the corresponding density Γ −1 dΓ /d cos θ in the laboratory frame? Find out an approximate formula valid for an isotropic decay in the limit |p| .8. Three different cases are ∗ = 1 > β. that is to say: 2 2 mπ 0 π 2 mπ 0 φmin = = > 5◦ · ⇒ E< ≈ 3.87 × 10−1 where we have used the PDG value mπ 0 = 135 MeV [4].14 An electromagnetic calorimeter is able to separate the showers induced by high-energy photons when the separation angle between the two photons is larger than 5◦ .3 shows tan φ as a function of cos θ ∗ . the angular distribution of the decay products is described by the probability density (Γ ∗ )−1 dΓ ∗ /d cos θ ∗ .2 centre-of-mass velocities of the two particles Figure 1.2 ∗ > β. ∗ are the where β1. assuming β = 0. where θ ∗ is the polar angle with respect to p. where θ ∗ is the polar angle in the centre-of-mass frame. The condition that all the π 0 ’s get reconstructed as two separate photons amounts to require that the minimum opening angle exceeds the resolution of the detector. considered: β1.2 = 1 > β. shown as a function of cos θ ∗ . What is the largest π 0 energy such that any π 0 decay can be reconstructed as a pair of distinct photons? Solution Given that a detector resolution of 5◦ is small.15 A particle of mass m and momentum p decays to a pair of massless particles.8 and for three ∗ representative cases: β1.2 β2∗ > β > β1∗ . β2∗ > β > β1∗ and β1. 18211/2016 Problem 1. and β1.1 GeV. m. 1.1 Lorentz Transformations 21 Solution If the decay distribution in the centre-of-mass frame is described by the density (Γ ∗ )−1 dΓ ∗ /d cos θ ∗ , in the laboratory frame one has: 1 dΓ 1 dΓ ∗ d cos θ ∗ = ∗ Γ d cos θ Γ d cos θ ∗ d cos θ . (1.67) The relation between cos θ ∗ and cos θ can be obtained from Eq. (1.56), giving: d cos θ ∗ |−(β cos θ − 1) − (β − cos θ )β| d cos θ = (1 − β cos θ )2 = 1 − β2 1 1 = = 2 , (1.68) (1 − β cos θ )2 γ (1 − β cos θ )2 where β = |p|/ |p|2 + m2 . Inserting this expression into the right-hand side of Eq. (1.67), we obtain: 1 dΓ 1 1 1 dΓ ∗ = . (1.69) Γ d cos θ γ 2 (1 − β cos θ )2 Γ ∗ d cos θ ∗ The Jacobian factor at the right-hand side of Eq. (1.68) implies that the angular distribution in the laboratory frame will be skewed towards the boost direction. The ratio between the Jacobian factor for backward- and forward-emitted photons is −4 (1 − β)2 /(1 + β)2 = (1 + β)γ : already for γ = 5, this ratio is 10−4 . Let’s now consider in more detail the case |p| m. In this limit, we can approximate: ! 1 1 1−β =1− 1− 2 ≈ . (1.70) γ 2γ 2 As already discussed, Eq. (1.68) implies a forward-peaked angular distribution, so that, for all practical purposes, we can assume θ 1 and consider only the first-order Taylor expansion of cos θ . With this approximation, Eq. (1.67) can be simplified to: 1 dΓ 1 1 1 dΓ ∗ 4γ 2 1 dΓ ≈ 2 2 Γ ∗ d cos θ ∗ ≈ 2 . (1.71) Γ d cos θ γ 1+γ θ2 2 Γ d cos θ ∗ 1 − β + β θ2 2 A case of special interest is for an isotropic angular distribution, which is the appropriate case for spin-0 particles or unpolarised beams. The angular distribution and the energy of the particle in the laboratory frame are then given respectively by: 22 1 Kinematics 1 dΓ 2γ 2 θ E∗ 2γ ≈ 2 , E(θ ) = ≈ E∗, (1.72) Γ dθ 1 + γ 2θ 2 γ (1 − β cos θ ) 1 + γ 2θ 2 where E ∗ is the energy of the massless particle in the centre-of-mass frame. The mean angle θ and RMS σθ can be computed from the first of Eq. (1.72) by assuming that the approximation is valid up to the maximum laboratory angle θ = π : " π 1 dΓ π 1 θ = dθ θ = +O (1.73) 0 Γ dθ 2γ γ " π # 2$ 1 dΓ 1 1 θ = dθ θ 2 = 2 log(γ 2 π + 1) − 1 + O (1.74) 0 Γ dθ γ γ2 # $ 1 π2 1 σθ = θ 2 − θ 2 = log(γ 2 π + 1) − 1 + +O (1.75) γ 4 γ2 The cumulative distribution F(θ ) can be obtained by integrating the differential distribution in Eq. (1.72). It can be used to determine the laboratory angle containing a given fraction α of the decay particles: 1 1 α F(θ ) = 1 − ⇒ θα = . (1.76) γ 2θ 2 +1 γ 1−α Problem 1.16 A particle of mass M and three-momentum along the z-axis decays into a pair of massless particles. Let the angular distribution of the decay products in the rest frame be described by the density (Γ ∗ )−1 dΓ ∗ /d cos θ ∗ . Show that the transverse momentum distribution Γ −1 dΓ /d|pT | develops an integrable singularity at |pT | = |p∗ |, where the trasverse momentum pT is defined as the projection of p onto the plane perpendicular to the z-axis, and p∗ is the centre-of-mass momentum of the decay particle. Consider now the transverse mass mT of the two daughter particles, defined as: mT2 = (ET 1 + ET 2 )2 − (pT 1 + pT 2 )2 = 2|pT 1 ||pT 2 |(1 − cos(Δφ)) (1.77) Show that the distribution Γ −1 dΓ /dmT features a singularity at mT = M. Solution The assumption that the momentum of the decaying particle purely longitudinal allows to relate the centre-of-mass kinematics to the transverse variables in laboratory frame through simple formulas, since: 1.1 Lorentz Transformations 23 |pT 1 | = |pT 2 | ≡ |pT | = |p∗ | 1 − cos2 θ ∗ 1 dΓ 1 dΓ ∗ d cos θ ∗ 1 dΓ ∗ |pT | = ∗ = = Γ d|pT | Γ d cos θ ∗ d|pT | Γ ∗ d cos θ ∗ |p∗ |2 1 − |pT |2 /|p∗ |2 1 dΓ ∗ 4|pT | = ∗ (1.78) Γ d cos θ ∗ M 2 1 − 4|pT |2 /M 2 where we have used the relation |p∗ | = M/2, see Eq. (1.90). Therefore, the transverse momentum distribution features a singularity at |pT | = |p∗ | = M/2, which is entirely due to the change of variables, hence the name of Jacobian peak. Notice that the singularity is integrable, since dΓ /d|pT | ∼ (1 − 4|pT |2 /M 2 )− 2 , which in the 1 neighbourhood of the singularity goes like ∼ε− 2 , where we have put ε = |p∗ |−|pT |. 1 In terms of the transverse mass of Eq. (1.77), we have: mT2 = 2|pT 1 ||pT 2 |(1 − cos(Δφ)) = 4|pT |2 ⎛ ⎞ 1 dΓ 1 dΓ d|pT | 1 dΓ ∗ ⎝ mT ⎠. = = (1.79) Γ dmT Γ d|pT | dmT Γ ∗ d cos θ ∗ M 2 1 − m2 /M 2 T Thus, the transverse mass features a Jacobian peak located at the mass of the decaying particle. Discussion The appearance of a Jacobian peak in both the transverse momentum and the transverse mass distribution of the decay products of a heavy resonance provides a powerful handle to distinguish such events from a non-resonant background. For example, the transverse mass has been extensively used at hadron colliders as the single most- efficient signature to identify the decay of W bosons into a charged lepton and a neutrino. While the charged lepton momentum can be fully reconstructed (if the lepton is either e or μ), the neutrino does not interact with the detector. An indirect evidence of its production is however provided by the momentum imbalance in the transverse plane, which, in the absence of other invisible particles, is just given by the neutrino transverse momentum. Thus, even in the presence of invisible particle, the transverse mass of Eq. (1.77) can be computed. Two remarks are due here. Firstly, in real experiments, the Jacobian peak is smeared by the finite detector resolution and by the natural width of the decaying particle, see e.g. Ref. [6] for the effect of the W boson width. Secondly, the expressions in Eqs. (1.78) and (1.79) have been derived under the assumption that the momentum of the decaying particle is purely longitudinal: if that is not true, i.e. if the particle has a momentum component orthogonal to the z-axis, the formula change. Differently from the transverse momentum, the transverse mass variable is Solution All transverse variables are invariant under longitudinal boosts. Problem 1.24 1 Kinematics less affected by a finite transverse momentum of the decaying particle.82) |pT | hence the transverse momentum changes already at the first order in β.81) If β is a small transverse vector. (1.83) This expression resembles closely the invariant mass squared m2 = 2p1 p2 for two transverse vectors. we can work out the transformation properties of pT : pT pT ≈ pT − Eβ ⇒ δ|pT | = −Eβ · . (1.4) in a vectorial form as: (γ − 1) p = (γ p · β̂)β̂ − β γ E β̂ + p − (p · β̂)β̂ = p + p·β −γ E β β2 (1. while for the transverse mass distribution Γ −1 dΓ /dmT such corrections start at O(β 2 ). This can be proved by noticing that . The transverse mass for two massless particles is defined as: mT2 = 2|pT 1 ||pT 2 | − 2pT 1 · pT 2 . (1. (1.78) for a particle of mass M decaying to massless particles. This is further elaborated in Problem 1. We then study their properties under transverse boosts. and the momentum p is almost longitudinal. receives O(β) corrections when the decaying particle velocity β has a transverse component. as one can easily verify.17 Show that the transverse momentum distribution Γ −1 dΓ /d|pT | of Eq.80) E = γ E − γ p · β (1.17. We first write Eq. but it has not the same properties under transverse Lorentz boosts. . since it coincides with the invariant mass squared (p1 + p2 )2 .84) is invariant under both longitudinal and transverse boosts. Since pz is invariant under transverse boosts. then it must hold: . |pT 1 |2 + p2z 1 |pT 2 |2 + p2z 2 − pT 1 · pT 2 − pz 1 pz 2 (1. . 85) E1 E2 . (1. δ (pT 1 · pT 2 ) = δ |pT 1 |2 + p2z 1 |pT 2 |2 + p2z 2 |pT 1 | δ|pT 1 | |pT 2 | δ|pT 2 | = E2 + E1 . The dashed histogram corresponds to a purely longitudinal velocity.4 Distribution of the transverse momentum |pT | (left) and transverse mass mT (right) obtained from a set of toy MC events where an unpolarised resonance of mass M decays into a pair of massless particles.5 GeV ×103 ×103 Events / 2. The response of the transverse momentum and transverse mass under boosts can be studied by using toy events generated with MC techniques.1 Lorentz Transformations 25 Events / 2.86) |pT 2 | |pT 1 | which vanishes because. in which the decay of an unpolarised resonance of mass M is simulated. Four different velocities β of the mother particle are assumed. Compute the minimum opening angle between the two particle directions after the scattering.85).8 β = 0. Hence.3. 1.1 T T β = 0. differing by the component on the transverse plane. this is clearly not the case for the transverse momentum.5 GeV 35 M = 80 GeV 25 M = 80 GeV β = 0. (1. An other interesting feature of the mT variable is that the location of the Jacobian peak location is not affected by boosts. The results are shown in Fig. in the centre-of-mass frame. A simple ROOT macro that performs the toy generation is illustrated in Appendix 1.5. while the other three distributions correspond to increasingly larger transverse boosts βT Using Eq.1 20 βT = 0.4 25 T T β = 0. the transverse mass does not change.1. see Problem 4. the mT variable is found to be significantly more stable against transverse boosts compared to |pT |. at first order in β. . whose peak value is significantly smeared by transverse boosts. initially at rest. Problem 1.18 A particle of mass m and momentum p scatters against an identical particle of mass m. 1. as one can readily verify by considering the O(β 2 ) term.82) and (1. E1 = E2 and |pT 1 | = |pT 2 |. (1.4 β = 0.0 30 βT = 0.8 T T 20 15 15 10 10 5 5 0 0 0 10 20 30 40 50 60 70 80 90 100 0 10 20 30 40 50 60 70 80 90 100 p (GeV) mT (GeV) T Fig.0 β = 0.4. This is however not true at second order. As expected. we get: pT 2 pT 1 δmT2 = (|pT 2 |E1 − |pT 1 |E2 ) β · − . To be more general. we can assume the two masses to be different.26 1 Kinematics Discussion This problem is a prototype for studying the kinematics of a fixed-target experiment. with particle (1) having momentum p and particle (2) at rest. The energy and momenta in the centre-of-mass frame are given by: √ . s = E1∗ + E2∗ = m12 + m22 + 2E1 m2 (1.87) . . (1.89) into the second of (1.87).88) The last equation. (E1∗ + E2∗ )(E1∗ − E2∗ ) = m12 − m22 . (1. (1. we get: 1 (s + m12 − m22 )2 (s + m22 − m12 )2 |p∗ |2 = − m12 + − m22 = 2 4s 4s 1 2 = 2s + 2(m12 − m22 )2 − 4 s m12 − 4 s m22 = 8s 1 2 = s + (m1 − m2 )2 (m1 + m2 )2 − 2 s m12 − 2 s m22 = 4s 1 = s + (m1 + m2 )2 s + (m1 − m2 )2 . in particular. √ √ s + m12 − m22 s (E1∗ − s + E1∗ ) = m12 − m22 ⇒ E1∗ = √ . (1. Inserting Eq. and symmetrising the expression for 1 ↔ 2 exchange. |p∗ | = E1∗ 2 − m12 = E2∗ 2 − m22 .89) 2 s An analogous result for E2∗ can be obtained by changing 1 ↔ 2.90) 4s from which: . implies: E1∗ 2 − m12 = E2∗ 2 − m22 . . 91) A special case is provided by m1 = m2 = m.92) 2 s 2 s Notice that.91) assumes a completely symmetric form under exchange of s ↔ m12 ↔ m22 : . after doing the algebra. (1. 2 s 2 s (1. E = . for which: √ √ ∗ s m2 ∗ s ∗ m2 |p | = 1−4 . the numerator of Eq. s − (m1 + m2 )2 s − (m1 − m2 )2 (s − m12 − m22 )2 − 4 m12 m22 |p∗ | = √ = √ . β = 1−4 . (1. 1.1 Lorentz Transformations 27 . . : |p∗ | = γ (|p| − βE1 ) = γ β m2 . particle (2). (1. the two particles must have equal and opposite momentum under this boost. (1.94) E2∗ which implies that β is also the velocity of particle (2) in the centre-of-mass frame. has to come to a rest under an inverse boost −β.96) E1 + m2 s The latter follows from: 1 E1 + m2 E1 + m2 γ = =. as one would have expected.e. i. that is to say: |p | 0 = γ (−|p | + βE2∗ ) ⇒ β= = β2∗ . m12 .95) from which we can derive the two relations: |p| E1 + m2 β= . which has momentum component along the boost direction −|p | and energy E2∗ . c) is known as the triangular function. The boost parameter to the centre-of-mass frame. β. s2 + m14 + m24 − 2s2 m12 − 2s2 m12 − 2m22 m12 λ(s. Besides. m22 ) |p∗ | = √ = √ (1. can be easily obtained by noticing that. γ = √ (1. b. by definition of centre-of-mass frame. in this frame.93) 2 s 2 s where λ(a. 95) and (1. = √ . we have E1∗ = |p∗ | + m2 = E2∗ ≡ E ∗ . since m1 = m2 = m. Also.96). 1.96) thus √ implies that the boost to the centre-of-mass is the velocity of a “particle” of mass s and total momentum p.43) to express the velocities in terms of the momenta and energies.99) γ (βE ∗ + |p∗ | cos θ ∗ ) γ (βE ∗ − |p∗ | cos θ ∗ ) . tan θ2 = . Furthermore.97) 1− β2 m12 + m22 + 2m2 E1 s Equation (1. one can relate the centre-of-mass momentum to the momentum in the laboratory frame as: |p| |p∗ | = γ β m2 = m2 √ . (1.5. (1. (1. (1. (1. With reference to Fig. Let’s define the opening angle by φ.98) s Solution We can use directly Eq. we then get: |p∗ | sin θ1∗ |p∗ | sin θ1∗ tan θ1 = . combining Eqs. 101) Finally.5 Elastic scattering p1 between two identical particles of mass m p1 θ1 m m θ2 p2 tan θ1 + tan θ2 tan φ = tan (θ1 + θ2 ) = = 1 − tan θ1 tan θ2 2βγ E ∗ |p∗ | sin θ ∗ = 2 2 ∗2 (1.19 Consider the decay of the KS0 meson into a pair of opposite-charge pions: calculate the energy and momenta of the charged pions in the KS0 rest frame. Solution Since the KS0 particle √ decays to a pair of particles of identical mass. 1.92) with s = mK to derive the energy E ∗ and momentum |p∗ | of the pions in the centre-of-mass frame: .96). hence the minimum of φ occurs for θ ∗ = π2 .100) γ (β E − |p∗ |2 ) + |p∗ |2 (γ 2 − 1) sin2 θ ∗ By using Eq. and the first of Eq. (1.28 1 Kinematics Fig. using a trigonometric identity.94). we can use Eq. we can further simplify the expression as: ! 1 (E ∗ − m)2 E∗ − m cos φmin = = = ∗ . Determine the energy distribution of the charged pions in the laboratory frame as a function of the beam momentum. Still using Eq. (1. the expression can be simplified to: √ 2γ 2 1 − β2 2 1 − |p∗ |2 /(m + E ∗ )2 2 2m2 + 2mE ∗ tan (φmin ) = 2 = = = .94). (1. we see that the first term at the denominator vanishes. 1 + tan2 φmin E ∗2 − 2 m E∗ 2 ∗ + m + 8mE + 8m 2 E + 3m (1. Consider now a monochromatic beam of KS0 decaying as before. (1. The right-hand side is a monotonously decreasing function of θ ∗ . γ −1 β2 |p∗ |2 /(m + E ∗ )2 E∗ − m (1.102) Problem 1. it’s decay into pions is isotropic in the rest frame. The laboratory energy Eπ of either of the two pions can be expressed in terms of the centre-of-mass polar angle θ ∗ via: 1 Eπ (cos θ ∗ ) = E ∗ γ (1 + ββ ∗ cos θ ∗ ) = E + |p|β ∗ cos θ ∗ . and β ∗ = |p∗ |/E ∗ ≈ 0. i. since it only depends on Eπ being a linear function of cos θ ∗ .15. see Problem 1. (1. (1. |p∗ | = 1−4 = 206 MeV. so that it’s distribution will be also uniformly distributed in the range [Eπ (−1). If the mother particle is relativistic. since E is not a linear function of |p|. (1.103) 2 2 mK where we have used the PDG values mK = 497 MeV and mπ = 139 MeV [4]. Furthermore. Eπ (1)]: 1 dΓ (|p|β ∗ )−1 if 21 (E − |p|β ∗ ) ≤ Eπ ≤ 1 (E + |p|β ∗ ) = 2 (1.104) 2 where E and |p| are the laboratory energy and momentum of the kaon.105) Γ dEπ 0 otherwise Discussion The fact that an isotropic distribution in the rest frame for a decay 1 → 2 gives rise to a rectangular distribution for the energy in the laboratory frame holds irrespectively of the mother energy and of the daughter mass. Eq.1 Lorentz Transformations 29 ! 2 mK mK mπ E∗ = = 248 MeV.1. whereas the kinetic energy T = E − m is. E .104) shows that the energy in the laboratory frame is a linear function of cos θ ∗ .e.829 is the centre-of-mass velocity of the pions. Notice that the momentum |p| is not uniformly distributed. Since the K 0 is a spin- 0 particle. mν mτ .7 GeV. and can decay via τ → π ντ . m. with mπ . τ leptons are abundantly produced from the decay of Z 0 or W bosons. the τ lepton has a mass of about 1. and the mass of the daughter particles is small compared to m.105) becomes: 1 dΓ 1 if 0 Eπ ≤ E ≈ Eπ (1. (1. so that Eτ 40 GeV . then Eq.106) Γ dEπ 0 otherwise For example. At colliders. Then. for example by averaging over the pion charge. the energy spectrum of the charged pion features an approximate rectangular distribution as in Eq. Although it is a spin-1/2 particle and the EWK dynamics is chiral. (1.8. one can usually consider unpolarised ensembles of τ leptons.106). so that the decay distribution in the τ rest frame can be still considered as isotropic. . mτ . see Problem 1. since γ − γ 2 − 1 < 1 and γ + γ 2 − 1 > 1 for any γ . It is also the only value featuring this property.19. (1. (1. where the latter is assumed massless. and let’s denote by θ ∗ the polar angle of a with respect to the direction of flight of B in the laboratory.89).108) Ea Furthermore. γ Ea∗ (1 + β)]. hence x is uniformly distributed in Iγ . into a pair of particles A and a. namely Ea∗ = (mB2 − mA2 )/2mA . Indeed. γ + γ 2 − 1 (1. the distribution of x will be given by: " " ∞ 1 dΓ g(λ) = dγ f (x|γ ) · g(γ ) = dγ . Solution Let’s consider the decay B → A a in the rest frame of B. it follows that x = 1 is contained in all intervals Iγ .107) where γ ≡ γB and β ≡ βB are the gamma-factor and velocity of B in the laboratory frame. it follows that Ea ∈ [γ Ea∗ (1 − β).30 1 Kinematics Another intriguing property of the energy distribution of Eq. Problem 1. Since Ea is a linear function of cos θ ∗ . (1.105) will be discussed in Problem 1. see Problem 1. the condition x ∈ Iγ can be obtained by solving the system: x > γ − γ2 − 1 1 1 ⇒ γ > x+ . Ea x ≡ ∗ ∈ Iγ ≡ γ − γ 2 − 1.20. one can always find a γ such that x ∈ / Iγ .110) Γ dx 1 2 ( x+ 1x ) 2 γ2 − 1 . It follows that: Ea = γ Ea∗ (1 + β cos θ ∗ ) (1.20 Show that for a two-body decay B → A a of an unpolarised particle B of mass mB . since for x = 1. for a fixed x. The centre-of-mass energy of a is given by Eq. If we assume that the boost factors of B are described by a distribution g(γ ). the energy spectrum of a in the laboratory frame has a global maximum at Ea∗ irrespectively of the momentum of B. (1.109) x < γ + γ2 − 1 2 x The assumption that B is unpolarised implies that Ea has a rectangular distribution. 1.110) is given by: ∂ 1 dΓ g 21 x + 1x 1 1 = − . (1.1 Lorentz Transformations 31 The derivative of Eq. γ + γ2 − 1 . In the latter case. if g(1) = 0.111) 2x 2 x Therefore: if g(1) = 0. An alternative way to convince oneself that x = 1 is indeed a maximum is to notice that this value is the only one that is contained by all intervals Iγ . 2 1 − = ∂x Γ dx 2 x2 2 21 x + 1x −1 sign(1 − x) 1 1 = g x+ .110). which plays here no role other than determining the centre-of-mass energy Ea∗ . The latter can be directly measured from the mode of the distribution of Ea from the relation: .112) Ea γ∗ γ∗ with γ ∗ = Ea∗ /ma . also for ma > 0. the derivative flips sign at x = 1. this condition is satisfied for any γ in the limit ma → 0. In both cases.113) γ∗ As expected. The same conclusion would hold. The condition 1 ∈ Iγ is then satisfied provided that: γ ∗2 − 1 γ− γ2 −1 <1 ⇔ γ < 2γ ∗2 − 1. under some more restrictive conditions. (1. (1. (1. (1. (1. hence it must have the highest probability density. hence this point represents a cusp in the distribution of x. Discussion This subtle property offers the possibility of measuring the mass of the parent particle mB regardless of both the kinematic of B and A. Eq.109) gets modified to: Ea γ ∗2 − 1 γ ∗2 − 1 x ≡ ∗ ∈ Iγ ≡ γ − γ 2 − 1 . x = 1 is a global maximum. then x = 1 is the unique maximum of Eq. since γ ∗ → +∞ and the inequality becomes γ < +∞. (1. Suggested Readings This problem is inspired by Ref.6 shows the simulated spectrum of log Ea . from which the notation and the mathematical proof have been also taken.114) Figure 1. mB = Ea∗ + mA2 − ma2 + (Ea∗ )2 . The idea of using the peak position of the energy spectrum . [8]. for the case where particle a is the b-jet produced in the decay of a top quark. 114) (taken 1 0 from Ref. and the velocities of the two daughter particles in the rest frame of the mother particle. that is: . (1.12.199 ± 0. In the rest frame of the mother particle.68 ± 0.25 GeV Uncertainty 2 Data-Fit using Eq.199 ± 0.151.5 GeV decays via K ∗ − → K − π 0 .01 ± 0.591 ± 0. βK ∗ . In particular. the K − momentum forms an angle θ ∗ = 55◦ with respect to p.554 and rπ ≡ mπ /mK ∗ = 0.2 4. it is convenient to write the formula in terms of the dimensionless ratios rπ ≡ mK /mK ∗ = 0.02 ± 0. βK∗ and βπ∗ .32 1 Kinematics Fig.15 GeV mt = 171. respectively.002.6 log(E) of the b-jets to measure the mass of the top quark.25 GeV corresponding to an 14 uncalibrated value of 12 mt = 171.91).115) |p| mK ∗ where we have used the PDG value mK ∗ = 892 MeV [4]. Problem 1. the opening angle φ depends on the centre-of-mass angle θ ∗ and on the velocity of the mother particle in the laboratory frame. The velocity and gamma factor of the K ∗− meson in the laboratory frame are respectively given by: 2 − 21 2 21 mK ∗ |p| β = 1− = 0. the largest and smallest? Solution We can use the results derived in Problem 1.5 GeV. 16 Epeak = 66. In order to make the numerical computation less error prone when using a calculator.4 4. What is the opening angle between the K − and π 0 ∗ ∗ momenta in the laboratory frame? Which are the centre-of-mass angles θmax and θmin for which the opening angle is. [7]) -1 -2 3. (1. (1. The Gaussian fit 20 yields a log Ea peak position 18 Uncalibrated Measurement of 4.002 Width=0.21 A K ∗ − meson with momentum |p| = 5. γ = 1+ = 6. The velocities of the daughter particles in the rest frame of the mother particle can be computed from Eqs.005 Simulation χ2/ndf=0. which decays via t → b W + .6 Fitted log Ea 8 TeV 26 distribution in a simulated 1/E dNbjets/dlog(E) Fit Results sample of t t̄ events with a 24 CMS Mean=4.987.25.8 4 4.89) and (1. has been first suggested by the same authors and first pursued at the CMS experiment [7].823 mass hypothesis of 22 172. 1. 1 Lorentz Transformations 33 .1. π = ∗ = = EK. |p∗ | mK2 ∗ − (mπ + mK )2 mK2 ∗ − (mπ − mK )2 ∗ βK.π mK2 ∗ ± mK2 ∓ mπ2 . (1. 4.0516) + atan(0.67) with respect to cos θ ∗ . the angular distribution of the muons in the centre-of-mass frame. 1 − (rπ + rK )2 1 − (rπ − rK )2 0. 5. Determine: 1.67) to derive φ as a function of θ ∗ . (1. the polar angle that contains 50% of the neutrinos. By using the numerical routine of Appendix 1. the decay products are isotropically distributed in the centre-of-mass frame. is a monotonously increasing function of θ ∗ .117) Since βK ∗ > min{βK∗ .π = = 6. (1.: 1 dΓ ∗ 1 = . (1.π φ = atan(0.254) = 17.906 π We can then use Eq. βπ∗ }.116) 1 ± rK2 ∓ rπ2 0.e. the beam energy threshold for which all muons move forwards.3. the fraction of neutrinos emitted with negative velocity.254 π βK. As seen in Problem 1. (1. for massless particles the polar angle in the laboratory frame.2◦ .119) Γ ∗ d cos θ ∗ dφ ∗ 4π 2.906 ⇒ φmax = 86◦ .22 A beam of K + mesons with energy E propagates along the z-axis. or equivalently use Eq.987 ∗ 0.25 · ± cos 55◦ + 0. we get: sin 55◦ 0. 3.56).118) Bando n. Consider the decay K + → μ+ νμ with massless neutrinos. whose tangent can be found by numerically maximising Eq. 13705/2010 Problem 1. which are the implications of the weak interaction for the muon helicity when the latter is emitted forwards or backwards along the z axis. (1.44) to calculate θK and θπ separately. i. θ . Following the latter approach.10. there must be a maximum opening angle φmax < π . we get a value: argmax {tan φ} = 0.502 K = = (1. Solution 1. Since the K + meson has spin-0.0516 K tan θK. and in particular Eq. (1. 2. Indeed: . since a monotonous mapping preserves the quantiles. For the neutrino to be emitted backwards.121) β cos π2 + 1 E 3. (1. (1. (1. which is trivially π/2 for an isotropic decay.122) Hence. the fraction αB of backward-emitted neutrinos as a function of E is given by: . Hence: m 2 cos π2 + β K θ50% = acos = acos β = acos 1 − . will also give the same quantile in the laboratory frame. Therefore.34 1 Kinematics d cos θ 1 − β2 = > 0.120) d cos θ ∗ (1 + β cos θ ∗ )2 where β is the K + velocity in the laboratory frame. one has the condition: pνz = γ (p∗z + βEν∗ ) = γ Eν∗ (cos θ ∗ + β) < 0 ⇒ cos θ ∗ < −β. the transformed to the laboratory frame of the polar angle θ ∗ giving the 50% quantile in the centre- of-mass frame. see Problem 1. 2 " −β " −β 1 − 1 − mEK 1 dΓ 1 αB = d cos θ ∗ = d cos θ ∗ = −1 Γ d cos θ ∗ 2 −1 2 (1. If the muon moves backwards along z in the laboratory frame. the neutrino moves backwards and hence hμ = −1/2). and then hμ can take both values of ±1/2. E > Eth ≡ 1+ = 1. The V − A structure of the charged-weak interaction implies that the neutrino is a pure left-handed particle. as to conserve the angular momentum in the z direction. then it must have hμ = −1/2 as to compensate for the forward-moving neutrino with hν = −1/2. If instead the muon moves forward. and since it is massless. In the latter case. from the case E > Eth . Since the K + is a spin-0 particle. m 2 mK2 − mμ2 mμ mK 2 K 1− > 2 . the neutrino can move either backwards or forwards. one needs: pμz = γ (p∗z + βEμ∗ ) = γ Eμ∗ (βμ∗ cos θ ∗ + β) > 0 ∀θ ∗ ⇔ β > βμ∗ . .124) (in which case. depending on the observed muon momentum. it is also in a helicity eigenstate with eigenvalue hν = −1/2. the muon and neutrino needs to be in opposite helicity eigenstate in the centre-of- mass frame. E mK + mμ2 2 mμ (1. we should consider separately the case where the K + energy is below the threshold Eth of Eq. For the muon to always move forward.124) 5.123) 4. (1.20 GeV.6. see Problem 1. We can transform it back to the C rest frame by applying a boost of magnitude βB . EB = . We can use Eq. (1. where a and b are massless particles.126) mB 2mB from which we get the result: .89) to express the energy of b.1 Lorentz Transformations 35 Problem 1.125) 2mC 2mC The same equation with the replacement C → B and B → A will also give the energy Ea∗ of particle a in the rest frame of B. since this configuration will also maximise the energy of a. followed by B → A a. (1. Hence: mab min = 0. where b has a fixed energy: since mab is an invariant. Assume now that the decay of B is isotropic in its rest frame: what is the expected distribution of mab in the laboratory frame? Solution max We study the problem in the rest frame of C. and C have non-zero masses mA .23 Consider the decay chain C → B b. mB .10. and mC .1. the results obtained in this particular frame will hold true for any other min frame. which can always happen if ma = mb = 0. Determine the lower and upper bounds on the invariant mass mab as a function of the mass of the three other particles. (1. which is the velocity of B in the rest frame of C: 2 2 ∗ mC + mB2 mB − mA2 mC2 − mB2 Ea = γB Ea (1 + βB ) = max 1+ 2 = 2mC mB 2mB mC + mB2 2 mC mB − mA2 = . The maximum invariant mass corresponds instead to a and b moving back-to-back. The minimum invariant mass mab corresponds to a and b moving collinear. respectively. whereas A. B. which is identical to the momentum of B since the former is assumed massless: mC2 − mB2 mC2 + mB2 Eb = |pB | = . 127) mB To study the distribution of mab . (1. (mC2 − mB2 )(mB2 − mA2 ) max mab =2 Eamax Eb = . as to obtain an expression which depends only on the latter. 1. we write cos θa as a function of Ea . (1.7. see Fig. We do so because we know that Ea is uniformly . Then. we first write it explicitly as a function of the kinematics of a and b: 2 mab = 2 Ea Eb (1 + cos θa ).128) where θa is the angle of a with respect to the direction of B in the rest frame of C. 129) Inserting this expression into Eq. A putative signal like the one discussed in this exercise would manifest itself as an edge in the mass distribution (dashed green line). Figure 1. it follows that mab has a triangular distribution: ⎧ √ ⎨ 2 2 1 dΓ mab if 0 ≤ mab ≤ mab max = max mab (1.g.126). 1. .89) and (1.130) A where the last equality has been obtained by means of Eqs. the edge location provides a constraint on the mass-scale of the new particles as shown by Eq. It is easy to verify that Eq. followed by B → A a in the rest frame of C distributed. which is uniformly distributed. βB γB Ea |pB |Ea (1. Eq.8 shows the spectrum of opposite-sign √ dilepton masses measured by the CMS experiment in proton-proton collisions at s = 8 TeV [9]. Since mab is a linear function of Ea . (1.131) Γ dmab ⎩0 otherwise Discussion Searching for end-points in the invariant mass spectrum of light particles provides an experimental technique to measure new heavy particles that decay to intermediate states.128) we get: |pB | Ea + EB Ea − mB Ea∗ 2Eb 2 = 2E E mab a b = Ea (EB + |pB |) − mB Ea∗ = |pB | Ea |pB | 2 − m2 ). = 2 mC Ea − (mB (1.105). Given that Ea∗ is constant.127).125). (1. for example the supersymmetric partners of the SM particles.128) can be recovered by using the expression at the right- 2 hand side of Eq. cos θa = = . (1. we get: γB Ea − Ea∗ EB Ea − mB Ea∗ Ea∗ = γB Ea (1 − βB cos θa ) .36 1 Kinematics pa a pb θa b C B A Fig. Furthermore. (1. if such is the angular distribution in the centre-of-mass frame. (1.7 Representation of the decay chain C → B b. see e. (1. the kinematics is fully specified by the polar angle θ ∗ with respect to v.4 fb-1 (8 TeV) distribution of same-flavour 200 opposite-sign lepton pairs 180 Data (e+ e− . Here. Solution We first consider the generic scattering in the centre-of-mass frame. 140 Events / 5 GeV DY The contribution from a 120 putative signal is shown as a Signal 100 green dashed-line histogram 80 SF central leptons featuring a peculiar 60 triangular shape: the end-point of the distribution 40 is related to the particle 20 masses.1 Lorentz Transformations 37 Fig.1. (1. Determine. and since the collision is elastic. what is the range of kinetic energy T in which the projectile particle can be found after the scattering.o. The velocity of the centre-of-mass in the laboratory frame. the velocity of the incoming particle before and after the collision are v∗ and v∗ .24 A non-relativistic particle of mass m and initial velocity v scatters elastically against a particle of mass A m.133) A+1 . [9]) Pull 0 -1 -2 -3 50 100 150 200 250 300 mll [GeV] Problem 1. vCM .m frame.8 Invariant mass CMS 19. (1.132) (A + 1) m A+1 In the c. we have: A |v∗ | = |v∗ | = |v − vCM | = |v|. Notice that no other known SM process produces 3 a similar invariant mass 2 1 shape (taken from Ref. is given by: m · r1 + A m · rA v rCM = ⇒ vCM = . as a function of A. assumed to be initially at rest. 1. where A is a positive coefficient. μ+ μ− ) measured by 160 Fit the CMS experiment √ in pp FS collisions at s = 8 TeV. T . the largest recoil energy transferred to the target particle. we obtain: v = v∗ + vCM . and A . 2 2 A 1 2A |v |2 = |v|2 + |v|2 + |v|2 cos θ ∗ . (1.134) implies that the maximum energy transfer. |v |2 = |v∗ |2 + |vCM |2 + 2|v∗ ||vCM | cos θ ∗ .38 1 Kinematics Expressing v back into the laboratory frame. is: 2 A−1 4A ΔTmax = T − Tmin = 1 − T= T. (1.134) (A + 1)2 A+1 Equation (1.e. A+1 A+1 A+1 A2 + 1 + 2A cos θ ∗ A−1 2 T = T ⇒ T ∈ T .135) A+1 (A + 1)2 Let’s now specify the relative energy exchange for the cases A 1. i. A = 1. 1: ⎧ ⎪ ⎨4/A if A . ΔTmax /T → 0 for A . 1 ΔTmax = 1 if A = 1 (1.136) T ⎪ ⎩ 4A if A 1 In particular. (1. like e. Indeed. the last limit implies that the maximum velocity of the target after the scattering is twice the velocity of the incoming particle. In particular. and is proportional to A for low values of the target mass.137) 2 2 Discussion The dependence of the relative energy transfer on the mass ratio between the projectile and the target has important implications on the possibility of slowing-down particles by elastic collisions. if A . This is easy to prove. neutrons produced in fission reactions. is maximal for A = 1. since we have: 1 1 (A m)|vA |2 = 4 A · m|v|2 ⇒ |vA | = 2 |v|.g. 1. H2 O is a good neutron moderator thanks to the presence of free protons in the molecule. so that it will take on average fewer binary collisions to achieve the desired moderation. For example. since the energy transfer ΔT can reach larger values. 1. elements with comparable mass are more effective in reducing the energy of the incoming particles. . (1.134). an elastic scattering implies only a very small energy loss per binary collision: massive elements are not efficient velocity moderators. On the contrary. as for from Eq. 0). let the four-momenta of the massless particle before and after the scattering be denoted by k = (E. 2. called Compton peak. m2 = m2 + 2kk + 2m(E − E ). Conservation of energy-momentum implies: k + P = k + P .8. Problem 1. k ). initially at rest. 2. Discussion The Compton scattering of a photon against atomic electrons falls into this class of problems. and the corresponding energy transfer is: 1 2 E/m 2k ΔTmax = E − E (θ = π ) = E 1 − =E ≡E . 2. (1. [10]. the initial particle is now. Suggested Readings For further details on this subject. P = k − k + P. which becomes P after the scattering.7 of Ref. cos θ = −1.1 Lorentz Transformations 39 Suggested Readings For further details on this subject.1 for more details. and let θ the angle between the two momenta. E E = . .139) with k ≡ E/m. k) and k = (E .1.24. 0 = −2EE (1 − cos θ ) + 2m(E − E ) = E [m + (1 − cos θ )E] − m E. the reader is addressed to Sect. The existence of a maximum energy transfer leads to a characteristic threshold in the energy distribution of the recoil electrons. Solution Differently from Problem 1. 1 + 2 E/m 1 + 2 E/m 1+ 2k (1. Determine the largest energy transfer from the massless particle to the target.138) 1 + mE (1 − cos θ ) The energy E is at a minimum for back-scattering. [10]. see Sect. It is most convenient to use the covariant formalism. For this purpose. the reader is addressed to Sect. The target initially has four-momentum P = (m. respectively.25 Consider the elastic scattering of a massless particle against a particle of mass m.1 of Ref. specialise the formula for the case of a non- relativistic particle. M .26 A particle of mass M and momentum p interacts elastically with a particle of mass m. Determine the maximum possible energy transfer involved in the scattering. initially at rest.40 1 Kinematics Problem 1. In particular. (2. valid in all regimes. can be obtained as a special case of a more general formula. the target energy for the case of back-scattering is given by: E + m s − M 2 + m2 |p| m E2 = γ (E2∗ ∗ + β|p |) = √ √ + |p| √ . In the laboratory frame. Here the projectile has a momentum p∗ and the target has opposite momentum −p∗ . (1. provides a natural energy scale for this kind of problems. The projectile energy in the laboratory frame is given by E = |p|2 + M 2 . (1. usually denoted by Wmax . s 2 s E+m s 2 E + m 2m + 2Em m|p| 2 m = √ √ + √ = (E + m)2 + |p|2 . whose derivation Problems 1.1). We can then calculate the corresponding energy in the laboratory frame by making a Lorentz transformation with boost parameter β. and (1. |p|. and in particular Eqs. Solution It’s convenient to study the kinematics of the scattering in the centre-of-mass frame. The maximum energy transfer to an atomic electron. see Eq.24 and 1. massless initial particle for the latter).89). as described by the so- called Bethe formula.18.96). We can use the results obtained in Problem 1.140) s .25 was based on specific assumptions (classical kinematics for the former. Discussion The results derived in this exercise will be useful when discussing the energy lost by a charged particle in the collision with the atomic electrons. (1. to express all quantities in terms of the energy- momentum in the laboratory frame. It is also interesting to see how the maximum energy transfer in the non-relativistic approximation and in the Compton scattering.98). M = 0. s s (E + m) s s m 2 m = E + 2mE + m + |p| + M − M = 2 2 2 2 s + 2|p|2 s s |p|2 =m 1+2 . The largest energy transfer to the target in the laboratory frame corresponds to its back-scattering in the centre-of-mass frame. provided that m > 0 (otherwise there would be no frame where the target is at rest). the other being the work needed to extract an electron from its orbital. and for the case of an initial massless particle. 1 Lorentz Transformations 41 Hence.1.141) s 1 + (m/M)2 + 2 (m/M) (βp γp )2 + 1 Let’s now specialise Eq. (1.141) to the following cases: 1. (1. the maximum energy transfer is given by: 2 m |p|2 2 m (βp γp )2 ΔTmax = = . M . γp = 10. while the target plays no role other than to specify the boost parameter.143) m2 + 2 m E 1 + 2 E/m which agrees with Eq. Four different cases are considered: γp ≈ 1. where v is the projectile velocity the laboratory frame. (1. (1. 2.9 shows the ratio ΔTmax /T between the maximal energy transfer and the kinetic energy of the projectile as a function of E/m.9 Ratio ΔTmax /T between the maximal energy transfer and the total energy of the projectile as a function of E/m. γp = 2.141) becomes: 2 m E2 2 E/m ΔTmax = =E . Indeed. γp = 10. In this approximation. E ≈ M and |p| ≈ M|v|. M=0 . (1. and the case of massless projectile. where m is the mass of the target. 1.139). It’s worth noticing that no assumption was made on the target being relativistic or not after the scattering.135) since A = m/M in the current notation. and the case M = 0. (1.135) was derived under the sole assumption that only the projectile is non-relativistic. |p| = E and Eq. Four different cases are considered: γp ≈ 1 (non-relativistic limit). Fig.142) (M + m)2 2 (m/M + 1)2 which agrees with Eq. γp = 2. (1. Hence: 2 m (M|v|)2 1 m/M ΔTmax ≈ = 2 M|v| 4 . Eq. |p|. In this case. (1. Figure 1. M = 0. Show that the mass M can be estimated from the following formula: 21 Ee + me M = |p| cos θe − 1 . an atomic electron is knocked out by the incoming particle. crosses a detector.144) Ee − me Solution Let the four-momentum of the unknown particle be P.42 1 Kinematics Suggested Readings The reader is addressed to Chap. |p|. 0. for which the polar angle θe with respect to p and the energy Ee are measured. 33 of Ref. the incoming particle will have four-momentum P . 0. 2 (1. After the scattering. Problem 1. At a certain depth inside the active material. 2. 0) . 0. With a convenient choice of the reference frame. we can write: P = ( |p|2 + M 2 . where the maximum energy transfer in a single collision is discussed. [10].2. [4]. 0) k = (me . the PDG review dedicated to the passage of particles through matter. and the four-momenta of the electron before and after the scattering be k and k .2 of Ref.27 A charged particle of unknown mass M and momentum |p| much larger than the electron mass me . See also Sect. . (1.145) k = (Ee , Ee2 − me2 cos θe , Ee2 − me2 sin θe , 0) P = P + k − k Squaring the last of Eq. (1.145), we get rid of the unknown kinematics, obtaining: P 2 = P2 + 2Pk − 2Pk + k 2 + k 2 − 2kk , M 2 = M 2 + 2(me − Ee ) |p|2 + M 2 + 2|p| Ee2 − me2 cos θe + 2me (me − Ee ), ! Ee + me 0 = |p| + M − |p| 2 2 cos θe + me , Ee − me ! ! M2 Ee + me me 0= 1+ 2 − cos θe + |p| Ee − me |p| 21 21 Ee + me Te + 2me M ≈ |p| cos2 θe − 1 = |p| cos2 θe − 1 , (1.146) Ee − me Te 1.1 Lorentz Transformations 43 where the factor me /|p| has been neglected, and we have introduced the kinetic energy Te = Ee − me . Discussion This formula was used by Leprince-Ringuet and collaborators to estimate the mass of a new long-lived particles discovered in cosmic rays. A magnetised nuclear emulsion had been exposed to cosmic rays on the top of the french Alps. Among the others, it recorded an event that could be interpreted as a charged particle, identified as such by the bubble density of the track impressed on the emulsion, transferring a sizable fraction of its momentum to a single atomic electron, a so-called δ-ray. The latter was identified as a spiraling track originating from the kink of the primary track, and its momentum, like the momentum of the unknown particle, could be measured from the radius of curvature of the track. With these measurements at hand, the mass of the unknown could be measured with enough accuracy to establish the discovery of a new particle, later identified as the charged kaon. Suggested Readings The reader is addressed to the paper by Leprince-Ringuet and Crussard [11], reporting the first evidence for a particle with a mass of about one-thousand times the electron mass contained in cosmic rays. The rather precise estimation of the particle mass was based on the kinematics of a fully-reconstructed event. This experiment is also discussed in Ref. [12]. Problem 1.28 A high-energy positron beam on a fixed-target experiment can produce the reaction e+ e− → f f¯ , where f and f¯ have the same mass M. Show that M can be estimated using the formula: 21 1 θ1 − θ2 2 |p| M= 2me |p| 1 − 1− θ1 θ2 , (1.147) 2 θ1 + θ2 2me where |p| is the beam momentum, assumed to be much larger than me and M, and θi are the polar angles of f and f¯ with respect to the beam direction. Solution Let’s denote the momenta of the outgoing particles by p1 and p2 . Momentum conservation along the beam and its orthogonal axis implies: |p1 | sin θ1 = |p2 | sin θ2 (1.148) |p1 | cos θ1 + |p2 | cos θ2 = |p| 44 1 Kinematics We now work in the assumption θi 1, which is indeed justified if |p| me , M because of the large boost of the centre-of-mass frame, see Problem 1.15. With this approximation: |p1 |θ1 = |p2 |θ2 θ12 θ22 (1.149) |p1 |(1 − 2 ) + |p2 |(1 − 2 ) = |p| Inserting the first equation into the second: θ2 θ1 θ2 θ22 θ1 + θ2 |p2 | − +1− = |p2 | + O(θi ) = |p|, 2 θ1 2 2 θ1 |p1 | = θ1θ+θ 2 |p| ⇒ 2 (1.150) |p2 | = θ1θ+θ 1 2 |p| Let’s now denote the four-momentum of the positron, electron, and of two outgoing particles by k, P, p1 , p2 , respectively. We can treat the electron/positron as massless. Energy-momentum conservation implies: p1 + p2 = k + P, 2M 2 + 2p1 p2 = 2me |p|, M 2 + ( |p1 |2 + M 2 )( |p2 |2 + M 2 ) − |p1 ||p2 | cos(θ1 + θ2 ) = me |p|, M2 M2 (θ1 + θ2 )2 M + |p1 | + 2 |p2 | + − |p1 ||p2 |(1 − ) = me |p|, 2|p1 | 2|p2 | 2 1 |p2 | |p1 | (θ1 + θ2 )2 M2 1 + + = me |p| − |p2 ||p2 | , 2 |p1 | |p2 | 2 1 θ1 θ2 |p| M2 1 + + = me |p| 1 − θ1 θ2 , 2 θ2 θ1 2me (θ1 + θ2 ) 2 |p| M2 = me |p| 1 − θ1 θ2 , 2θ1 θ2 2me −1 θ1 − θ2 2 |p| 2M 1 − 2 = me |p| 1 − θ1 θ2 , θ1 + θ2 2me 21 1 θ1 − θ2 2 |p| M= 2me |p| 1 − 1− θ1 θ2 . (1.151) 2 θ1 + θ2 2me Discussion This formula was used by the NA7 Collaboration in a fixed-target experiment aiming at measuring the pion form factor in the time-like region in the range 0.1 < q2 /GeV2 < 0.18, where q2 = 2me |p| [13]. These values of q2 correspond to beam energies ranging from 100 up to 175 GeV. The experimental apparatus con- 1.1 Lorentz Transformations 45 sisted in a liquid hydrogen target followed by a planes of MWPC, see Problem 2.52, giving an angular resolution of 0.02 mrad for particles emerging from the target within 7 mrad from the beam direction. A magnetic spectrometer, complemented by electromagnetic calorimeters, allowed to separate electrons/positrons from muons and pions. The latter two were further separated by performing an angular analysis of the two polar angles θ1 and θ2 , since the two angles are correlated by the mass of the produced particle. The comparison between the cross sections for e+ e− → μ+ μ− and e+ e− → π + π − allowed to extract the pion form factor Fπ (q2 ) in the q2 region corresponding to the available beam energies. Suggested Readings The reader is addressed to the NA7 paper [13], where the angular analysis used to disentangle between pions and muons is discussed in detail. Problem 1.29 The rapidity of a particle of momentum p and mass m is defined as: 1 E + pz y = ln . (1.152) 2 E − pz 1. How does y transform under a boost β = βez ? 2. Write down the phase-space measure d 3 p/Ep in terms of new set of variables (|pT |, φ, y), where |pT |2 = p2x + p2y and φ is the azimuthal angle around the z-axis. 3. Show that, in the limit |p| m, the rapidity reduces to the purely geometrical quantity, η, called pseudorapidity. Express η in terms of the polar angle θ . Solution We first derive a set of equations that will prove useful in the following. The rapidity y can be also expressed as: p 1 E + pz pz ey − e−y z y = ln , = y = tanh y, y = atanh . (1.153) 2 E − pz E e + e−y E By introducing the transverse mass mT ≡ |pT |2 + m2 , we also get: 1 E + pz 1 (E + pz )2 E + pz y= ln = ln = ln , 2 E − pz 2 E 2 − p2z mT 154) If the particle momentum p is aligned with the z-axis.153) gives: . Eq. mT ey − pz = mT2 + p2z . (1. pz = mT sinh y. −2pz ey = mT (1 − e2y ). (1. 155) 2 1 − βp from which: βp = tanh y. Δφ. We want to find the Jacobian J = |∂(px .160) holds under integration over the azimuthal angle. py . 3. y)| of the transformation: ⎧ ⎪ ⎨px = |pT | cos φ px = |pT | sin φ (1. 2 E − pz 1+β 2 1−β (1. Let’s now consider the case E .156) 1. the rapidity expressed in the new reference frame becomes: 1 E + pz 1 γ (E − βpz ) + γ (−βE + pz ) y = ln = ln = 2 E − pz 2 γ (E − βpz ) − γ (−βE + pz ) 1 E + pz 1−β 1 1+β = ln = y − ln = y − atanh β. γp = cosh y βp γp = sinh y. pz )/∂(|pT |. as are differences between azimuthal angles.157) so that the difference Δy between the rapidity of two particles is an invariant under longitudinal boosts. we first notice that dpx dpy = |pT |d|pT | dφ. 2. φ. By using the fact that ∂Ep /∂pi = pi /Ep . so that all we are left to do is to compute ∂pz /∂y.160) Ep 2 The last equality in Eq.46 1 Kinematics 1 1 + βp y= ln = atanh βp . (1.159) hence: dp 1 = |pT |d|pT | dφ dy = d|pT |2 dφ dy = π d|pT |2 dy (1. After a boost β = βez . (1. we get: ∂y 1 Ep − p z (pz /Ep + 1)(Ep − pz ) − (Ep + pz )(pz /Ep − 1) 1 = = .158) ⎪ ⎩ pz = m + |pT | sinh y 2 2 To this purpose. ∂pz 2 Ep + pz (Ep − pz )2 Ep (1. (1. . The momentum |p| can be Taylor-expanded around E to give: m2 1 E + E cos θ − 2E 2 cos θ + . . 2 1 − cos θ 2 sin2 θ2 2E 2 . . 1 1 + cos θ 1 cos2 θ2 y = ln ≈ ln = ln = 2 E − E cos θ + m2 cos θ + . m. . we also get two more useful trigonometric relations: 1 1 + cos θ 1 η= ln = atanh (cos θ ). (1.164) dy γ (1 − β)2 e2y + 4e−2y which goes to zero for y .162) Discussion Rapidity is a useful variable in hadron colliders thanks to its transformation properties: a partonic differential cross section dσ/dy.163) γ e (1 − β)2 + e−2y (1 + β)2 − 2(1 + β 2 ) + 4 where we have used Eq. we have: dσ dσ d cos θ 1 1 dσ d tanh y = = · = dy d cos θ dy γ 2 (1 − β cos θ )2 d cos θ ∗ dy 1 1 1 = = 2γ 2 (1 − β tanh y)2 cosh2 y 2 1 = 2 2y .161). Differential cross sections at hadron colliders are often forward-peaked due to the large longitudinal momentum of the initial-state partons. (1. When expressed in terms of the rapidity.15.161). (1. see Problem 1. computed in any reference frame (e. (1. differential cross sections become more smooth for large boost factors γ . cos θ = tanh y. (1. and assuming y ≈ η.157). provided one replaces y by the linearly transformed value as for Eq.1. has the same form in any other frame. the centre-of-mass frame of the parton-parton scattering). This expression simplifies in the limit β → 1 to: dσ 2 1 ≈ 2 . sin θ = . Indeed.161) 2 From Eq. 2 1 − cos θ cosh y (1. for an isotropic process in the centre-of-mass frame.1 Lorentz Transformations 47 θ = − ln tan ≡η (1.g. 11. see e. Another reason of interest for y (or η) at hadron colliders is due to the fact that a variety of soft processes.g. Figure 1. by Fig. turns out to be almost uniformly distributed in η. the differential cross section in y is a smooth and broad function. like the productions of particles in minimum bias hadron- hadron collisions. whereas the differential cross section in cos θ is squeezed around cos θ = 1. 1. superimposed on the same axis for illustration purposes.10 shows dσ/d cos θ and dσ/dy. for a scattering process characterised by a large boost factor γ . This variable then becomes the relevant metric when designing a detector . A comparison between the two shows that. 1 and has a maximum at y = ln(2γ ). 11 Distributions of the pseudorapidity density of charged hadrons in the region |η| < 2 in inelastic pp collisions at 13 TeV measured in data (markers) and predicted by two of LHC event generators (curves). superimposed on the same axis for illustration. 2 Problem 1.30 Write down the invariant mass m12 of two particles with four- momentum p1 and p2 in terms of their transverse momenta pT . giving: . The boost factor of the centre-of-mass is taken to be γ =5 Fig. transverse mass mT . [14] for occupancy and radiation hardness. and rapidity y with respect to the same axis.10 Comparison between the differential cross section dσ/d cos θ and dσ/dy for a scattering process which is isotropic in the centre-of-mass frame. This plot has been taken from Ref.48 1 Kinematics Fig. 1. (1. 1. Find out an approximate formula for m12 valid in the case that the mass of the two particles is small compared to their energy and the momenta are almost collinear. Solution We can replace the canonical variables by the transverse variables by means of Eq.154). (1.165) In the limit |pi | .1 Lorentz Transformations 49 2 m12 = m12 + m22 + 2(E1 E2 − p1 · p2 ) = m12 + m22 + + 2 mT 1 (ey1 − sinh y1 ) mT 2 (ey2 − sinh y2 ) − mT 1 mT 2 sinh y1 sinh y2 − pT 1 · pT 1 ey2 − e−y2 y1 ey1 − e−y1 y2 = m12 + m22 + 2 mT 1 mT 2 ey1 ey2 − e − e − pT 1 · pT 2 2 2 = m12 + m22 + 2 mT 1 mT 2 cosh(y1 − y2 ) − pT 1 · pT 2 .1. and we have introduced the Euclidean metric R12 in the (η.. Determine the largest rapidity of the mother particle for which the experiment is sensitive to the mother particle. 2. Suggested Readings For an application of the R metric in jet-clustering algorithms. mi and θ12 1. where we have exploited the fact that yi → ηi in this limit... assuming two definitions of acceptance: 1.. whose geometrical coverage is however limited to the pseudorapidity region |η| ≤ ηacc . see Problem 1. Jet-clustering algorithm often rely on Rij to quantify the distance between two particles such that they can be associated with the same jet. . φ) space.167) Discussion The Euclidean metric Rij in the (η. ⎣2pT 1 pT 2 ⎝cosh(Δη) ⎠⎦ (1. 1− Δφ2 +.166) 2 2 1+ Δη2 +. The interaction point is surrounded by a cylindrical detector around the z-axis. The level of isolation of a particle is usually defined by energy collected within a cone of radius R centered around the particle direction. (1.31 A particle of mass M and momentum p parallel to the z-axis is produced at the interaction point of an accelerator and then decays to a pair of identical particles of mass m.29. the reader is addressed to Ref. defined by: R12 = (η1 − η2 )2 + (φ1 − φ2 )2 . we have: ⎡ ⎛ ⎞⎤ 21 ⎢ ⎜ ⎟⎥ m12 ≈ ⎢ ⎜ − cos(Δφ)⎟⎥ ≡ √pT 1 pT 2 R12 . both daughter particles are within acceptance. φ) space is widely used at hadron colliders to define the geometric “distance” between two particles. Problem 1. [15]. at least one of the daughter particles is within acceptance. (1.g.92) with s = M. and the rapidity is y = atanh β. When β > β ∗ .47). The velocity β ∗ √ of the daughter particles in the centre-of-mass frame in given by Eq.155). (1. regardless of the definition of acceptance. θ2 } ≤ θmax . see e. (1. If β < β ∗ . the condition that none of the daughter particles fall within the acceptance translates to θmax < θacc . (1. while the velocity of the mother particle in the laboratory frame is β. The condition that at least one of the daughter particles is within the detector acceptance amounts to require θacc ≤ max{θ1 . it is always possible to find a polar angle θ ∗ in the centre-of-mass frame such one of the two particles emerges at an angle larger than θacc ≡ that 2 tan−1 e−ηacc in the laboratory frame. the polar angle θ is bounded by a maximum angle θmax satisfying Eq. see Eq.168) By using a trigonometric identity and Eq. Since the other particle is emitted at an angle π − θ ∗ in the centre-of-mass frame. Problem 1. or: ∗ 1 √ β θmax tan θmax γ β 2 −β ∗2 − ln tan = ln = ln . it’s enough to consider the case pz > 0 and get an upper bound on y: symmetry will then imply that the same bound. in absolute value. will hold for |y|.49). (1.50 1 Kinematics Solution Since the problem has cylindrical symmetry. it will emerge at an even larger polar angle in the laboratory frame: the experiment therefore is sensitive to a non-zero fraction of decays up to y = atanh β ∗ .10. then θ1 ≤ θ (θ ∗ = π2 ) ≡ θ⊥ . then θ2 < θ⊥ . we notice that θ ∗ = π/2 plays a special role. π/2]. (1. θ2 } ≤ θ⊥ . • if θ ∗ ∈ (π/2. then θ1 > θ⊥ . one needs: θacc ≤ min{θ1 . in order to have both daughter particles within the acceptance. (1. θmax ∗ ]. To obtain the maximum rapidity y such that both particles have θ > θacc .169) γβ − γ γ ∗ β 2 − β ∗2 sinh y − γ ∗ tanh2 y − β ∗2 cosh y where Eq. (1.170) The condition that none of the daughter particles fall within the acceptance translates to θ⊥ < θacc . Hence. π ). ∗ ∗ • if θ ∈ (θmax . since: • if θ ∗ ∈ [0. or: . = 2 1 + tan2 θmax − 1 ∗2 1 + γ12 β 2β−β ∗2 − 1 γ ∗β ∗ γ ∗β ∗ ln = ln > ηacc .156) has been used to express γ and βγ in terms of the rapidity y. but θ2 < θ⊥ . 12 Maximum detector pseudorapidity ηacc as a function of the particle rapidity ymax such that the detector is sensitive to y > ymax . 1.1. for two definitions of acceptance: requiring that both daughter particles have η < ηacc (solid curve) or requiring that at least one has η < ηacc (dashed curve) θ⊥ tan θ⊥ β ∗ /γβ − ln tan = ln = ln .1 Lorentz Transformations 51 Fig. and the difference between the time of flight and kinetic energy of the two particles. Determine which fraction of undecayed pions and muons arrives to the detector. or in a n-tuple. (1. Indeed. Problem 1.171) sinh2 y + β ∗2 − sinh y Figure 1.12 shows a graph of the functions at the left-hand side of Eqs.169) and (1. For smaller centre-of-mass velocities. and (3) saves the values (η1 . the problem could have been studied by a MC program that (1) generates particle decays in the centre-of-mass for random values of θ ∗ and for discrete values of β ∗ . The beams are detected by a detector located d = 21 m downstream of the primary target. η2 ) in a binned 2D-histogram. = 2 1 + tan2 θ⊥ − 1 ∗2 1 + γβ2 β 2 − 1 β∗ = ln > ηacc . this exercise offers an example of how problems in particle physics are often better tackled by using Monte Carlo methods. For a sufficiently large number of toy events. Discussion Although we have arrived at an analytical result. the maximum rapidity becomes smaller than ηacc . The case β ∗ = 1 (massless particles) corresponds to y < ηacc : the maximum particle rapidity for which both daughter particles are within the acceptance coincides with the pseudorapidity range of the detector. we can then get the acceptance map for the tested values of (β.32 A beam made of pions and muons of momentum |p| = 170 MeV are produced from a short proton pulse against a fixed target.171) for a few values of β ∗ . (2) applies the Lorentz boost to the laboratory frame for some values of β. η2 ). (1. . β ∗ ) by studying the joint distribution of (η1 . it is more convenient to quote the length cτ in place of τ .: P(t ∗ | τ ) = exp −t ∗ /τ .5 ns μ t= = 1+ = · (1.8 m for pions. In some cases. which is given by: ! 2 d d m 21 m 1. defined as the inverse of the decay probability per unit of proper time. the time measured in the rest frame of the particle.e.6 × 102 m for muons and 7. while states with smaller lifetimes are more frequently referred to as resonance. like the laboratory frame where particles have |p| > 0. and masses mμ = 106 MeV and mπ − = 139 MeV [4]. While τ is a constant. and the natural width Γ = /τ .18 = 82.29 = 90. Finally we can compute the difference between the TOF for the two beams. the probability P(t ∗ | τ ) that the particle survive up to time t ∗ . which. given that at time t ∗ = 0 it has not decayed.98 μ exp − = 139 MeV 21 m (1.175) βc c |p| 3 × 108 m/s 1. a time interval is however not a Lorentz invariant. ∼10−18 s).g.4 ns π . and one often wants to express a survival probability.f. i. (1.52 1 Kinematics Discussion Unstable particles are associated with a mean life-time τ .d. Solution In the rest frame of an unstable particle. is provided by the cumulative of the exponential p.6×102 m = 0.11 π where we used the values cτ = 6.172) In our case. we need to compute the probability that the beam particles have survived over the time t that takes them to travel the whole length d in the laboratory frame. (1. is reported instead. It is customary to define particle any quantum state with τ above some conventional lower bound (e. for the proper time obeys an exponential law with time constant τ . in a different reference frame. This time interval corresponds to a proper time: t 1 d md m t∗ = = = = d.174) |p| τ exp − 170 MeV 7.173) γ γ βc m βγ c |p| It is easy to see that a consistent use of cτ together with masses in units of GeV/c2 and momenta in units of GeV/c gives the correct numerical result: 106 MeV 21 m m d exp − 170 MeV 6.8 m = 0. These particles are produced in the decay reaction: μ+ → e+ νe ν̄μ .8 m = 1100 ± 300 m π + L cτ = 0. thus allowing for a sizable fraction of the mesons to be decayed.7 m = 150 ± 40 m K+ The beam purity can be affected by the presence of ν̄μ and νe . K + → μ+ νμ .180) The mean muon energy in the two main decays can be estimated from Eq. Estimate the length of the pipe and at least one process that can reduce the beam purity. We want the beam to be as pure as possible in νμ .3 MeV μ T = E −m = m⎣ 1+ − 1⎦ = m 139 MeV · 0.89 = 94.1.58 = 80. (1. (1. Both mesons decay with lifetimes of order 10−8 s. 18211/2016 Problem 1.1 MeV.176) from which |ΔT | = 14.178) |p| cτ If we want this probability to be small. For a fixed beam momentum.139 GeV (1. (1. A magnetic selector filters positively charged particles of momentum |p| = 20±5 GeV. The kinetic energy is given by: ⎡! ⎤ 2 |p| 106 MeV · 0.179) m 20±5 0.177) + where the last decay (Kμ3 ) is suppressed with respect to the direct decay by a factor of 20. The main decay reactions giving rise to a νμ are π + → μ+ νμ . thus giving a beam profile and probability of survival up to distance d: m Prob [x ≥ d] = exp − d (1.497 GeV GeV · 3. which then decay inside a pipe filled with helium.33 A neutrino beam is produced by a 120 GeV proton accelerator. Bando n. K + → π 0 μ+ νμ .2 ns.6 MeV π (1. Solution The selected beam consists of π + and K + . the probability of decay per unit length is constant. the pipe length L needs to be larger than the decay length |p| 20±5 GeV · 7.109) giving .1 Lorentz Transformations 53 from which |Δt| = 8. Electron neutrinos can be also produced directly by the helicity-suppressed decays π + → e+ νe and K + → e+ νe . although with a probability about 10−4 ÷ 10−5 smaller than for the corresponding muonic decays. . (1.e. the decay reaction K + → π 0 e+ νe (BR = 5%) represents the largest source of background. quantum states of definite three-momentum pi .00 electrons t= = 1− = · (1. respectively. pass through two scintillators separated by a distance d = 30 m. The relativistic- invariant phase-space measure for a system of n particles of prescribed energy and momentum is defined by 1 n dpi dΦn (P. In practice one is usually concerned with the measurement or the prediction of transition probabilities per unit volume.K mμ 16 GeV π + Eμ ≈ γπ.100 µs and 0. 1. .2 Center-of-Mass Dynamics and Particle Decays The position and three-momenta of the particles (ri . the TOF is simply given by: d d m 2 − 21 30 m 1.179) is therefore: Eμ cτμ 1% π+ ≈ (1. and between free-particle states.2% K + where we have used cτμ = 660 m. .54 1 Kinematics 2 # $ mπ.K 10 GeV K + The probability of muon decay over a distance of order L as given by Eq. . both of energy E = 2 GeV. What is the time of flight between the two scintillators for the two particles? Solution Since the velocity is constant.181) 2 mπ. 18211/2016 Problem 1.184) i=1 (2π )3 2Epi . Therefore.13 protons giving approximately 0. For kaons. the measure in the phase-space (ri . pi ) define a set of canonical phase-space variables.K 1+ ≈ (1.183) βc c E 3 × 10 m/s 8 1.113 µs for electrons and protons. Bando n.34 A proton and an electron.182) mμ L 0. p1 . pi ) comes out quite naturally in calculations. pn ) = (2π )4 δ 4 (P − pi ) . (1. i. p1 . Given a particle decay. they cannot be “visited” by the system. . (1. Due to the factor of (2E)−1 at the denominator of Eq. (1. i. . .187) Γ f Problems Problem 1.e. since the allowed states only live in a hyper-surface such that pi − P = 0: all other states have null measure. p22 = 0. the differential width transforms like the inverse of a time. . .184). .2 Center-of-Mass Dynamics and Particle Decays 55 The factor of 2Ep at the denominator comes from the conventional free-particle normalisation: # $ p|p = 2Ep (2π )3 δ 3 (p − p ). pn ) (1. the probability of falling into a given channel |f is called branching ratio (BR): " Γi→f Γ = dΓi→f . (1. the solid angle of particle (1) in the laboratory frame. Feynman’s rules help organ- ising the perturbative expansion of iM and calculate the various terms up to the desired perturbative order. The total width Γ is obtained by integrating the differential width in the rest mass of the particle over the full phase-space and over all decay channels. the solid angle of particle (1) in the centre-of-mass frame. . . The volume factor d 3 ri is here omitted because it usually cancels and is not much relevant in practice.186) 2E where: • Mfi is the relativistic matrix element of the interaction Hamiltonian between the initial and final state. 2. .1. see Problem 1. 3. BR(i → f ) = (1.2. . the Mandelstam invariant t = (pc − pa )2 for a scattering a + b → c + d. p1 . pn )|2 dΦn (P. The differential decay width of a particle with total four-momentum P in the initial state |i into a channel |f is given by the formula: 1 dΓi→f (P.185) which makes dΦn Lorentz-invariant. and in particular for the two cases p21 = p22 and p21 = 0. The presence of a δ 4 function in Eq. p1 . pn ) = |Mfi (P.186). . (1. When the theory is perturbative. .184) implies that not all of the 3n variables are independent. • dΦn is the phase-space measure of Eq. .35 Express the relativistic two-body phase-space measure dΦ2 as a function of: 1. (1. .89). p2 ) = (2π )−2 δ( s − Ep1 . d p2 = δ s − Ep∗ .56 1 Kinematics Solution At the price of adding some burden to the notation. one has to integrate out the remaining ones and perform suitable changes of variables. |p1 | = |p2 | = |p∗ |. m2 . m2 " 3 1 √ . In the centre-of-mass frame. m2 )δ(p1 + p2 ) 2Ep1 .184) becomes: √ dp1 dp2 dΦ2 (P. p1 . mi = |pi |2 + mi2 . m1 − E−p∗ . . (1. m1 E−p∗ . In order to reduce the phase-space measure into a measure on a subset of phase-space variables. m1 2Ep2 . so that Eq. m2 d 3 p1 = (2π )2 4Ep∗ . 1. we will write Ep2i . and use the notation Ei∗ for the centre-of-mass energy of the i-th particle as in Eq. m1 − Ep2 . . . d|p∗ | = ∗ dΩ ∗ = √ dΩ ∗ . 1 √ ∗ |2 + m2 − |p∗ |2 + m2 |p∗ |2 d|p∗ | dΩ ∗ = δ s − |p 16π 2 E1∗ E2∗ 1 2 " 1 |p∗ |2 1 |p∗ | . (1. .188) ∗ ∗ 16π E1 E2 |p ∗| + |p ∗| 2 ∗ 16π 2 s E 1 E2 Two special cases are worth being considered: ⎧. (1.189) ⎩ 1− m 2 dΩ for m1 = 0. p1 . m1 EP−p1 . This second parametrisation is useful for treating e. m2 d 3 p1 = (2π )2 4Ep1 . .184) we get: dp1 dp2 dΦ2 (P. m2 )δ(P − p1 − p2 ) 2Ep1 . where one is interested in one particle only (the projectile). fixed-target scatterings. m1 2Ep2 .92). m1 − Ep2 . m2 . p2 ) = (2π )−2 δ(E − Ep1 . ⎨ 1 − 4 m2 dΩ ∗ = β ∗ dΩ ∗ for m = m = m dΦ2 = s 32π∗2 32π 2 1 2 (1. P). (1. see Eq. 2. m2 = m s 32π 2 where β ∗ is the velocity of both particles in the centre-of-mass frame. m2 " 1 . Let the total four-momentum be (E. d 3 p2 = δ E − Ep. . m1 − EP−p1 .g. Starting from Eq. and wants to integrate- out the degrees of freedom of the struck target. . . 16π E1 E2 |p1 | + |p1 |−|P| cos θ1 2 16π 2 E − E1 |P| cos θ1 1 E E 2 |p1 | (1. 1 = 2 δ E − |p1 |2 + m12 − |P − p1 |2 + m22 |p1 |2 d|p1 | dΩ1 16π E1 E2 " 1 |p1 |2 1 |p1 | .190) . d|p1 | = dΩ1 = dΩ1 . . m22 = s + m12 − 2p1 P.1. This is best achieved by using the equation of conservation of four-momentum. . and squaring it in order to get rid of the struck target kinematics: p2 = P − p1 .2 Center-of-Mass Dynamics and Particle Decays 57 It remains to express |p1 | as a function of cos θ1 . E E . 2 |P|2 |P| 0 = |p1 |2 1 − 2 cos2 θ1 − 2|p1 | cos θ1 E1∗ − |p∗ |2 . s + m12 − m22 E |p1 |2 + m12 = |p1 ||P| cos θ1 + . (1. Indicating the centre-of- mass momentum before (after) the scattering by p∗ab (p∗cd ).193) dE (2π ) 3 dE1 2E1 (2π ) 3 (2π )3 3.89) and (1. and the factor of (2m1 )−1 that comes from the relativistic normalisation of the wave function of Eq.191) 1 − |P| 2 E2 cos2 θ1 where E1∗ and |p∗ | are given by Eqs. It is interesting to study Eq.192) 2 16π m2 2m2 (2π ) 2m1 (2π )3 3 Modulo the numerical factor within parentheses. (1. we have: 1 t = (pc − pa ) =2 ma2 + mb2 − 2|p∗cd ||p∗ab | ∗ − cos θ . (1. In this case: 1 |p1 | (2π )4 1 m1 |p1 | dΦ2 (P. |P| ∗ |p∗ |2 + |P| 2 E E 1 cos θ1 + E2 m12 cos2 θ1 |p1 | = . respectively. (1. (1. p1 . To fix the ideas. this expression coincides with the relativistic version of the density of states dNNR /dE for a non-relativistic particle of momentum p. In the centre-of-mass frame. p2 ) ≈ dΩ1 = dΩ1 .185).194) .91).190) in the non-relativistic limit |p1 | E ≈ m2 . so that E1 ≈ m1 . the momentum magnitude before and after the scattering is in general different if there is some inelasticity. βa∗ βc∗ dt = 2|p∗cd ||p∗ab | d cos θ ∗ . (1. which accounts for the phase- space of the struck particle and the (2π )4 factor in front of the delta function. and dE1 /d|p1 | ≈ 2E1 /|p1 | while the target is a heavy nucleus at rest of mass m2 . (1. we can imagine that the projectile is a classical particle. since: dNNR |p1 |2 dΩ1 d|p1 | |p1 |3 m1 |p1 | = = dΩ1 = dΩ1 . pj+1 . pn ) 2π (1. (1. . p1 . . n 1 n dpi (2π )4 δ(P − pi ) = (2π )3 2Epi i=1 i=1 ⎡ ⎤ j 4q n 1 n d dpi = ⎣(2π ) δ(q − 4 pi ) ⎦ × (2π )4 δ(P − q − pi ) = (2π )4 (2π )3 2Epi i=1 i=j+1 i=1 n = dΦj (q. . . . will be introduced between square brackets.58 1 Kinematics We then use Eq.188) and get rid of cos θ ∗ in favour of t. which. . Suggested Readings The PDG review on kinematics. 47 of Ref. . pj ) dΦn−j+1 (P. obtaining: 1 |p∗cd | dt 1 dt dΦ2 = √ dφ ∗ = √ dφ ∗ . p1 .196) In order to prove it. (1. . . Beware that the definition of dΦn may differ by a factor of (2π )4 due to the normalisation of the four-momentum conserving δ function. we introduce twice the identity into Eq. (1. . . . q. Chap. . . pn ) = dΦj (q. .184) in a suitable form. . Problem 1. so this formula remains valid (modulo φ ∗ → φ) in any reference frame where pa and pa are collinear. offers a complete summary of the most important results for two-body phase-space. p1 .195) 16π 2 s 2|p∗ab ||p∗cd | 16π 2 2|p∗ab | s Notice that φ ∗ in the formula above stands for the azimuthal angle around p∗a in the ab rest frame. Discussion The phase-space measure dΦn for an arbitrary particle multiplicity n can be constructed with the recursive formula: dq2 dΦn (P. . for sake of clarity.36 Explicate the three-body phase-space measure dΦ3 in the centre-of- mass frame by using the known expression of dΦ2 . . [4]. pj ) × (2π )4 δ(P − q − pi )× i=j+1 1 d4q 2 n dpi × δ(q − (p1 + . + pj )2 ) dq2 = (2π ) 2Epi (2π ) 3 4 i=j+1 . p1 . This can be done by means of Eq. p1 . (1. .1. after integrating over all angles. m 2Ep∗ . p3 ) = (2π )−5 δ( s − Ep∗1 . q. . p2 . . and m1 → m12 . m3 1 1 2 2 3 " . .184) and specialise the four-momenta of the three particles in their centre-of-mass frame: √ dΦ3 (P. (1. m2 → m3 for the latter. m 2Ep∗∗ . (1. with the replacements s → m12 for the former.188) and obtain: 1 |p3 | 1 |p∗1 | ∗ dm12 2 dΦ3 (P.197) 2π where we have made use of Eq. . p3 ) = √ dΩ 3 · dΩ 1 · = 16π 2 s 16π 2 m12 2π 1 1 = √ |p∗ ||p3 | dm12 dΩ1∗ dΩ3 (1. (1. . m3 )δ(p∗1 + p∗2 + p∗3 )× d 3 p∗1 d 3 p∗2 d 3 p∗3 × 2Ep∗∗ .196) to the two-body phase-space (1. pj+1 .37 Determine the three-body phase-space measure dΦ3 as a function of the centre-of-mass energies of two of the particles. p1 .198) (2π )5 8 s 1 To conclude. pj )(2π )4 δ(P − q − pi ) = (2π )3 2Epi (2π )3 2Eq 2π i=j+1 i=j+1 dq2 = dΦj (q.2 Center-of-Mass Dynamics and Particle Decays 59 n 1 n dpi d3q dq2 = dΦj (q. p1 . pn ) (1.16) to transform δ(q2 − μ2 ) d 4 q into d 3 q/2Eq . . . one needs to express |p∗1 | and |p3 | as a√function of m12 . m1 − Ep∗2 . Problem 1. .91). m2 − Ep∗3 . pj ) dΦj (P. . . Solution We can apply Eq. p2 . Solution Let’s start from Eq. . . . 3 ∗ √ ∗ 2 ∗ 2 ∗ ∗ 2 . . d p3 = δ s − |p1 | + m1 − |p2 | + m2 − |p1 + p2 | + m3 × 2 2 2 (2π )−5 ∗ 2 × |p | d|p∗1 | dΩ1∗ |p∗2 |2 d|p∗2 | dΩ2∗ 8E1∗ E2∗ E3∗ 1 " . . . . . . √ . dΩ1∗ dφ2∗ = δ s − |p∗1 |2 + m12 − |p∗2 |2 + m22 − |p∗1 + p∗2 |2 + m32 × 2(2π )−3 |p∗1 |2 |p∗2 |2 × d|p∗1 | d|p∗2 | d cos θ12 ∗ 8E1∗ E2∗ E3∗ . one notices that: √ (mi + mj )2 ≤ mij2 ≤ ( s − mk )2 . and where we have used the fact that dEp /d|p| = |p|/Ep = βp . d cos θ12 ∗ ∗ = d|p∗1 | d|p∗2 | 8E1∗ E2∗ E3∗ |p1 ||p2 | E∗ 4(2π )3 E1∗ E2∗ 3 dE1∗ dE2∗ = . respectively. larger or smaller. m23 2 ) space is. in general. Without loss of generality. (1.60 1 Kinematics " ∗ = 2(2π )−3 |p∗1 |2 |p∗2 |2 d|p∗1 | d|p∗2 | 1 |p∗1 ||p∗2 | . while the energy of particle (3) can be obtained by applying a boost from the three-body rest frame to the rest frame of (12): 2 + m2 − m2 m12 (12) 2 1 E2 = (1. In this frame. (1.201) 16(2π )3 s 16(2π )3 s 16(2π )3 s Equation (1. First of all.200) Hence. ( s −m3 )2 ] × [(m2 + m3 )2 . (1. The border of the domain in the (m12 2 . and particle k be at rest in the three-body centre-of-mass frame. we can consider the first of Eq. the phase-space volume dΦ3 expressed in terms of the mij2 becomes: 2 2 2 2 2 2 dm12 dm23 dm13 dm23 dm13 dm12 dΦ3 = = = . Finally.203) 2m12 (12) E3 = γp∗ E3∗ − βp∗ (−|p∗3 |) = (12) (12) √ s + m122 − m2 s − m2 + m2 3 12 3 |p∗3 |2 2 s = √ √ + 2 − m2 = 2 s m12 2 s s + m12 3 . At a given value of m12 . the energy of particles (2) is given by Eq. (1. (1. all other cases giving values of m23 that are. respectively. Equation √ (1. a non-trivial curve. .89). whose sides are also tangent to the domain boundary. . (1. one notices that P = p1 + p2 + p3 implies that: √ √ mij2 = (P − pk )2 = s − 2Ek∗ s + mk2 ⇒ dmij2 = −2 s dEk∗ for i = j = k.201) shows that the three-body phase-space is uniform over the invariant masses of any two-pairs of particles.201).199) 4(2π )3 where θ12 is the angle between p∗2 and p∗1 .202) implies that the√domain is contained within the rectangle [(m1 +m2 )2 . ( s − m1 )2 ].202) where the boundaries correspond to particles (ij) being at rest in their centre-of-mass frame. the range of m23 can be determined by requiring the momentum of particle (3) in the rest frame of (12) to be either parallel or antiparallel to the momentum of particle (2). (1.204) 4s m12 2m12 The last result could have been obtained way more easily by noticing that.1.2 Center-of-Mass Dynamics and Particle Decays 61 ⎛ ⎞ 2 − m2 s + m12 s − m 2 + m2 s − (m12 − m3 )2 s − (m12 + m3 )2 = √ 3 ⎝ 12 √ 3 + ⎠= √ 2 − m2 ) 2 s m12 2 s 2 s(s + m12 3 1 2 2 2 2 + m2 ) = s − m12 − m3 .87) gives m12 2 + m32 + 2E3(12) m12 = s. = 2s − 2s(m12 3 (1. Hence: 2 s − m12 + m22 − m32 2 m23 (m12 ) − 2m12 ⎛. the kinematics looks like the one of a fixed-target experiment. for which Eq. in the rest frame of (12). . Discussion If the matrix element squared is uniform over mij2 . m23 2 ). Non-uniformities of the amplitude squared over either of the mij2 variables. this is the case if the decay can be mediated by an intermediate resonance: indeed. Such a distribution for a three-body decay is called Dalitz plot.205) where the two signs correspond to the lower and upper bound. if a narrow resonance of mass m0 and width Γ0 is present in e. the differential decay probabilities dΓ /dm122 dm232 is uniform in the domain of (m12 2 .206) (m12 2 − m02 )2 + m02 Γ02 m0 Γ0 and the corresponding Dalitz plot will feature a cluster of experimental points around 2 the line m12 = m02 . then: 1 π |M |2 ∝ ≈ δ(m12 2 − m02 ). . ⎞2 (m12 2 + m22 − m12 )2 + 4m22 m12 2 (s − m12 2 − m32 )2 + 4m32 m12 2 −⎝ ± ⎠ . respectively. would lead to a non-uniform distribution of the experimental points. For example. (1. the maximum value of the three-momentum is taken by the particle with the largest mass. Problem 1.38 Prove that in the centre-of-mass of a three-body decay. 2m12 2m12 (1. the (12)-channel.g. 210) Therefore. mk ≥ mi . it is more convenient to write Eq. (1. |p∗k | is a decreasing function of mij . Experimentally.207) 4s where mij is the invariant mass of the (ij) pair and s is the mass squared of the mother particle.186) using the non-relativistic normalisation . i.39 Consider the β-decay A → A e− ν̄e . so that the maximum value corresponds to the minimum value of mij .208) is always negative since (mij − mk )2 ≤ (mij + mk )2 .e. Solution Let’s denote the three momenta of the decay particles by pp . Therefore: 2 2 s − mi + mj + mk s − mi + mj − mk |p∗k |2max = . (1. the derivative of |p∗k |2 with respect to m12 is: ∂|p∗k |2 ∝ −mij s − (mij − mk )2 − mij s − (mij + mk )2 − ∂mij − mk s − (mij − mk )2 + mk s − (mij + mk )2 .91): 2 2 s − mij + mk s − mij − mk |p∗k |2 = . where A and A are two heavy nuclei and mA − mA − me = Q is large compared to me . Assume that the non- relativistic matrix element squared |MNR |2 for this transition is approximately constant. (1. (1. Hence. the particle that can take the maximum centre-of-mass momentum is also the one with the largest mass. Problem 1. (1.209) 4s and for any pair of indices i. (mi + mj ). (1. 2(mi − mk ) ≤ 0. this value corresponds to the end-point of the |p∗i | distribution over several decays. Modulo some positive constants. Show that the decay width is proportional to Q5 . Since the non-relativistic matrix element is a pure constant. pe and pν .208) The last row of Eq.62 1 Kinematics Solution The centre-of-mass momentum of particle k is given by Eq. k we have: |p∗k |max ≥ |p∗i |max ⇔ s − (mi + mj − mk )2 ≥ s − (mj + mk − mi )2 mi + mj − mk ≤ mj + mk − mi . (1. the differential decay width becomes: d 3 pA d 3 pe d 3 pν dΓ = |MNR |2 (2π )4 δ 3 (pA + pe + pν )δ(mA − EA − Ee − Eν ) = (2π )3 (2π )3 (2π )3 .2 Center-of-Mass Dynamics and Particle Decays 63 # $ p|p = (2π )3 δ 3 (p − p ).211) With this convention. (1.1. . one can notice that the two measures are different. which is a linear function of the electron energy with an end-point related to the Q-value of the reaction.214) shows that the total decay width is proportional to the fifth power of the Q-value. (1. (1. Discussion One may wonder whether this result could have been obtained by plugging in the phase-space measure of Eq. Equation (1. the integration is trivial. (1. Since the intagrand grows like |pe |2 .212).212) can be also written as: ! 1 dΓ = Q − Te .212) (2π )3 Q Let’s see in more details the approximations that went into Eq. a property know as Sargent rule. which is indeed a good approximation since Temax = Q.211). it was assumed that the electron energy is much smaller than the proton mass. (1.199). (1.213) |pe | d|pe | The left-hand side of Eq. we can integrate over the electron momenta. (1. The other assumption is that the neutrino mass is negligible. |MNR |2 2 − |p |2 + m2 − |p | × = δ mA − |pe + pν |2 + mA e e ν (2π ) 5 × |pe |2 d|pe | dΩe |pν |2 d|pν | dΩν = ⎛ ⎞2 4|MNR |2 ⎜ ⎟ = |pe |2 ⎝mA − mA − me −Te ⎠ d|pe | (1. since . Firstly. If we then make the approximation me = 0.212).213) is the so-called Kurie plot. this approximation allows to neglect the recoil energy taken by the nucleus. Indeed. which is perfectly fine here. giving: " |M |2 |MNR |2 Q 2 2 2 NR 5 1 1 1 Γ ≈ d|pe | |pe | Q − 2Q|pe | + |pe | = Q − + = 2π 3 0 2π 3 3 2 5 |MNR |2 Q5 = . corrected for the non-relativistic normalisation (1.214) 60π 3 Equation (1. Coming back to Eq. the integral is dominated by the high-energy part of the electron spectrum. (1.64 1 Kinematics Eq. if the third particle is much heavier than the maximum energy available to the two lighter particles. Indeed. (1. looking back at Eq.212) is quadratic in |p|. if m3 . and the integration of the last delta function is not valid anymore.199). ∗ one can see that the integration over the polar angle between the two momenta θ12 lied on the assumption that there exists only one such angle so that the energy conservation is satisfied for a given value of |p∗1 | and |p∗2 |. However.199) is linear in the two energies (after introducing the non-relativistic normalisation). the equation ∗ of conservation of energy becomes nearly independent of θ12 . so that it can be considered at rest for what concerns the energy balance. (1. while Eq. Therefore. since E1 + E2 = m − E3 ≈ m − m3 = Q: the phase-space measure is therefore given by the product of the two particle phase- spaces. Determine the photon energy Eγ .2 .3 of Ref. we have: . initially at rest in the laboratory frame. 7. which are just the number of states inside a spheric layer of radius |p|. Problem 1. and the matrix element is constant.1 ÷ 1 MeV. In the assumption ε/M 1.40 A heavy nucleus of mass M ∗ . whereas nuclear masses are at least three orders of magnitude larger. [16] for another derivation of the Kurie plot of β-decays. Discuss a possible technique to suppress the energy shift due to the nuclear recoil. m1. then the two light particles momenta are uncorrelated in direction and fully anti-correlated in modulus. which is generally the case since the nuclear transitions produce photons with energy of order 0. Solution Let’s denote the photon energy in the laboratory by Eγ and the energy gap between the two nuclear levels by ε. such that M ∗ = M + ε. Suggested Readings The reader is addressed to Sect. decays to the ground state of mass M by emitting a photon. 215) 2M Eγ2 ε2 ε Eγ = ε − ≈ε− =ε 1− . However. the photon energy is smaller than the energy gap by a fraction ε/2M 1 due to the nuclear recoil. Eγ2 M + ε = Eγ + M 2 + Eγ2 ≈ Eγ + M + . Discussion If the width of the excited level is much smaller than the recoil energy taken by the nucleus. the emitted photon won’t be anymore at the resonance. when the . 2M 2M 2M Hence. (1. Let the vertices of the triangle be located at √ √ 2 3 1 3 1 r1 = 0. r3 = + . is equal to the distances of the point from the sides. − 13 ) (+ 33 . − 13 ) Fig. The sides of the triangle are then defined by the three straight line equations: y (0. y). |pi |2 /2mQ. 1. This behaviour is called Mössbauer effect and is vastly employed in spectroscopy. y).13. + .1. The phase-space for this decay can be represented inside an equilateral triangle centred around the origin of a cartesian coordinate system (x.2 Center-of-Mass Dynamics and Particle Decays 65 nucleus is bound inside a crystal. Solution Consider an equilateral triangle of unit height centred around the origin of a cartesian coordinate system (x. such that the kinetic energy of each particle in units of the Q-value. 1.− . is equal to the distances of the point from the sides. the crystal lattice behaves as a collective object. 23 ) 2 1 x 3 √ √ (− 33 . r2 = − . |pi |2 /2m Q. represented as an equilateral triangle centred around the origin of a cartesian coordinate system (x. resulting in a much larger effective mass M and thus no recoil energy will be taken away: the emitted photon is then capable of inducing the inverse reaction γ + M → M ∗ with an enhanced cross section.13 The phase-space for a three-body decay into non-relativistic particles of energy εi and same mass m. Only the shaded circle is however compatible with the three-momentum conservation .− (1. Problem 1. assumed to be non-relativistic in the rest frame of the decaying particle.216) 3 3 3 3 3 as shown in Fig. Show that the allowed kinematic configurations live inside a circle of equation x 2 + y2 − 19 ≤ 0. y). such that the kinetic energy of each particle in units of the Q-value.41 Consider the decay of a particle of mass M into three particles of identical mass m. (1. However. the phase-space points live inside an equilateral triangle such that the distances of its internal points from the sides are given by εi .221) By inserting any pair of equations in (1. Let’s denote the adimensional quantity |pi |2 /2m Q ≡ εi .219) 2m 2m 2m Therefore.218) ⎪ ⎪ 2 ⎩ε = y + 1 3 0 3 which shows that an equilateral triangle of unit height is the geometric locus of points for which the sum of the distances from the three sides is equal to unity. being more restrictive than Eq. (1. In particular: 2 1 − 2ε1 − 2ε2 1 cos θ12 = 2 √ ≤1 ⇒ ε12 + ε22 + ε1 ε2 − ε1 − ε2 + ≤ 0. so that: |p1 |2 |p2 |2 |p3 |2 + + = M − 3m = Q ⇔ ε1 + ε2 + ε3 = 1 (1. (1. y). 2. say the first and the last.219).218). ⎪ ε2 = = 23 x0 − 21 y0 + 13 (1. so that Eq. This condition.222) 2 2 3 3 4 .219) can be also written as: √ 2ε1 + 2ε2 + 2 ε1 ε2 cos θ12 = 1. we can now express the inequality above in terms of the coordinates (x.66 1 Kinematics ⎧ √ ⎪ ⎨+√3x + y − 2 3 =0 − 3x + y − 2 =0 (1.217) ⎪ ⎩ 3 y + 13 = 0 The distances of a given point (x0 . y0 ) inside the triangle from the three sides are therefore given by: ⎧ √ √ ⎪ ⎪ ε = |+ 3x0 +y0 − 23 | = − 23 x0 − 21 y0 + 1 ⎪ ⎨ 1 2 3 √ √ |− 3x0 +y0 + 23 | ⇒ ε1 + ε2 + ε3 = 1. namely: √ 2 √ 3 1 1 1 2 3 1 1 1 − x− y+ + y+ + − x− y+ y+ − 2 2 3 3 2 2 3 3 √ 3 1 1 1 1 − − x− y+ − y+ + ≤0 (1. 2 ε1 ε2 4 (1. i = 1.220) where θ12 is the angle between two of the particles. 3. conservation of momentum implies that p1 + p2 + p3 = 0. places additional constraints on the allowed phase-space points. namely K + → π + π + π − . .223) 9 which describes a circle of radius 1/3 centred around the origin and tangent to the three sides of the triangle in their middle points.42 The thrust is an event-shape variable which. (1. Solution Let’s denote the three-momenta of the involved particles by pi . Let n̂ be the direction that maximises T . where xi = 2|pi |/ s and s is the centre-of-mass energy. By definition. in a three-body decay the phase-space points are evenly distributed in the (E1 . one can find there interesting considerations on how different matrix elements would affect the distribution of events in the (x. The representation of decay events in this plane has been very popular in the early years of particle physics as a tool to infer the properties of the interaction responsible for the decay. The (x. x2 . so that the (x. A famous example is provided by the study of the decays of a long-lived particle. in the centre-of-mass frame of a n-particle event is defined as i |pi · n| T (p1 . (1. 3 of Ref. . In particular. y) plane. into three pions.13. y) points will be still evenly distributed.1. y) variables introduced in this exercise transform a tiny square of area dE1 dE2 into a parallelogram uniformly in E1 and E2 . originally called τ and later-on identified as the charged kaon. pn ) = max . [12]. This is shown by the solid circle in Fig. E2 ) plane. where E1 and E2 are the energies of any two particles. 1. Prove that T = max{x1 .2 Center-of-Mass Dynamics and Particle Decays 67 After some straightforward algebra. Suggested Readings More details on the so-called θ -τ puzzle for the three-charged pion decay of charged kaons. a large number of cancellations occurs and one gets: 1 x 2 + y2 − ≤ 0. we must have: .224) i |pi | n with |n| = 1. x3 }. .37. Problem 1. . can be found in Chap. such that pi = 0. Consider a three-particle decay and assume the √ three particles √ to be massless. Discussion As discussed in Problem 1. under some suitable assumptions on the matrix- element. α13 between three vectors p1 . or p3 . (1. it follows that . Since n̂ · δ n̂ = 0. the vector inside the square brackets is proportional to one of the particle momentum.227) ⎪ ⎪ ⎩ 2e√1 ·p3 − s ≤ x3 if p1 · p2 > 0 Similarly. then the other two opening angles must be larger than π/2. one has: (|p1 | + e1 · p2 + ||p1 | + e1 · p2 |) T (e1 ) = √ = s ⎧ 2|p1 | ⎪ ⎪ + √s = x1 if p1 · p2 < 0.226) where ei are the directions of the three momenta. p3 · p2 < 0 we notice that the angles α12 .228) ⎪ ⎩ ≤ x3 if p1 · p2 > 0 ⎧ ⎪ ⎨≤ x1 if p3 · p1 < 0. T (e2 ). p2 · p3 < 0 (1.225) s where θi = ±1 are signs such that the individual scalar products are all positive. (1. In this latter case. p1 · p3 < 0 ⎨ = − 2e√1 ·ps 2 ≤ x2 if p1 · p2 < 0. so that: T = max{T (e1 ). therefore. p2 . that i pi = 0 satisfy α12 + α23 + α13 = 2π and αij < π . for n̂ = e1 . or p2 . p1 · p3 > 0 (1. α23 . we find: ⎧ ⎪ ⎨≤ x1 if p1 · p2 < 0.229) ⎪ ⎩ = x3 if p3 · p1 < 0. p3 such Then. T (e3 )}. p3 · p2 > 0 T (e3 ) ≤ x2 if p3 · p1 > 0 (1. For example. if αij < π/2.68 1 Kinematics θ1 p1 + θ2 p2 − θ3 (p1 + p2 ) · δ n̂ 0 = δT (n̂) = √ . it also follows that n̂ has to be parallel to either p1 . For any choice of the signs θ1 . p2 · p3 > 0 T (e2 ) = x2 if p1 · p2 < 0. the vector that “recoils” against the two vectors with opening angle smaller than π/2 has also the largest modulus among the three. (1. |pj |}.230) i. If we now consider all possible cases. |pk | = |pi |2 + |pj |2 + 2 pj · pj ≥ max{|pi |.e. we obtain: . . Show that the S is not. . x2 . . . the Kinoshita–Lee–Nauenberg theorem ensures that inclusive-enough observables are IR-safe. 3. Consider the three event-shape variables: i |pi · n| T = max (1.236) In+1 (p1 . Problem 1. [17]. (1 − λ)pn ) = I(p1 . . x . T (e ) ≤ x 1 2 2 2 1 3 3 2 if p1 · p3 > 0 (1. . λpn . . For QCD.2 Center-of-Mass Dynamics and Particle Decays 69 ⎧ ⎪ ⎪T (e1 ) ≤ x3 .235) 2 n i pi 1. .234) i |pi | 3 (pi × n)2 S = min i 2 . T (e ) = x > x . . . .232) Suggested Readings More informations on the thrust can be found in Ref.5 of Ref. . which is a function of the four-momenta pi of an arbitrary number of partons. . . x3 }.43 Event-shape variables are widely used to describe the structure of the hadronic events and to test perturbative chromo-dynamics (pQCD). λk eigenvalues of Θ = (1. This is equivalent to requiring that the observable does not distinguish between configurations related to each other by a soft gluon emission or by the collinear splitting of a parton. . . pn ).233) i |pi | n pi pTi i |pi | C = 3(λ1 λ2 + λ1 λ3 + λ2 λ3 ).231) ⎪ ⎪T (e ) = x > x . x2 if p1 · p2 > 0 ⎪ ⎨T (e ) ≤ x . T (e ) ≤ x . T (e ) ≤ x if p2 · p3 > 0 ⎪ ⎩ 1 1 2 3 2 1 3 1 T (e1 ) = x1 . is said to be infrared-safe (IR-safe) if: In+1 (p1 . Show that T and the C are infrared-safe observables. T (e3 ) = x3 if pi · pj < 0 ∀i. j For all the above cases (which exhaust all the possibilities). Discussion An observable In (p1 . T (e2 ) ≤ x3 . T (e3 ) = x3 > x1 . (1.g. T (e2 ) = x2 . . pn ) for any particle n. pn ) (1.1. it always holds that: T = max{x1 . (1. Sect. 2. [18]. . see e. 0) = In (p1 . x . . . pn . . . . . it suffices to verify that the Θ tensor satisfies the conditions (1. . . . .241) i pi .236) for an arbitrary set of four-momenta: n i|pi · n| + |0 · n| Tn+1 (p1 . . . pn . . . . . Indeed: 1 0 = det(Θ − λ1) = λ3 − Tr{Θ}λ2 + (Tr{Θ})2 − Tr{Θ 2 } − det Θ (1. (1 − λ)pn ) = max i n−1 = i=1 |pi | + λ|pn | + (1 − λ)|pn+1 | n = Tn (p1 . pn . .237) The variable C is proportional to the second invariant of the symmetric tensor Θ.236): n pi pTi i |p| + 0 Θn+1 (p1 . . . . To study the IR properties of C. pn ) (1. . .238) 2 (λ1 λ2 +λ2 λ3 +λ1 λ3 )= C3 where λi are the eigenvalues of Θ. 0) = n = Θn (p1 . . λpn . . .239) Finally. pn ) i=1 |pi | + 0 n n−1 |pi · n| + λ|pn · n| + (1 − λ)|pn · n| Tn+1 (p1 . pn ) i=1 |pi | + 0 n−1 pi pTi λ2 p pT (1−λ)2 p pT i |pi | + λ|pnn |n + (1−λ)|pnn |n Θn+1 (p1 . . λpn . (1. .70 1 Kinematics Solution In order to prove that T is IR-safe. . . we need to verify the conditions (1. . . . pn ) (1. . We now introduce an auxiliary tensor W defined as: T i pi pi W = 2 = W T.240) 2 n pi 2 n i pi where p⊥ is the momentum component orthogonal to the direction n. (1. (1 − λ)pn ) = n = i=1 |pi | + λ|pn | + (1 − λ)|pn | = Θn (p1 . . 0) = max n = Tn (p1 . we notice that S can be also written as: 2 3 i (pi × n)2 3 p S = min 2 = min i ⊥2 i . . (1. λpn . . it holds: 2 p2 i p n W n = 2 = 1 − i ⊥2 i . [17. (1. pn ). . pn ) (1. (1. . . .244) 2 To study the IR properties of S. .1. T i (1.247) π n i |pi | called sphericity. . differently from the sphericity. . 0) = ni = Wn (p1 . . .242) i pi i pi Hence. pn . (1 − λ)pn ) = i n i=1 |pi | + λ |pn | + (1 − λ) |pn | 2 2 2 2 2 = Wn (p1 . . . We conclude by mentioning that. Another way to see that S is not collinear-safe is to consider the case where n is aligned with one of the particle momenta p: in this case. 18] for more details on IR properties of observables relevant for experimental tests of pQCD.2 Center-of-Mass Dynamics and Particle Decays 71 For any vector n. the event-shape variable: 2 4 i |pi × n| S = min . a collinear splitting of that particle would not change the numerator (n × p = 0). . .243) This implies that n is the eigenvector of the W tensor corresponding to the largest of its three eigenvalues λ1 ≥ λ2 ≥ λ3 : 3 S= (1 − λ1 ) .236): n pi pTi + 0 Wn+1 (p1 . but it would change the denominator due to the quadratic dependence on the momenta.246) The last inequality is a consequence of the non-linear dependence of W on the particle momenta. .245) i=1 |pi | + 0 2 n−1 pi pTi + λ2 pn pTn + (1 − λ)2 pn pTn Wn+1 (p1 . (1. Suggested Readings See Refs. . we need to test whether W satisfies the conditions (1. is instead IR-safe. . minimising S(n) is equivalent to maximise the quadratic form nT W n subject to the constraint nT n = 1: 0 = ∇n nT W n − λ (1 − nT n) ⇔ W n = λn. 42. (1. q̄. Show that the allowed values for the xi parameters are located inside a triangle in the plane (xq . It is also interesting to notice that for three partons: 2 3 2 3 (pi + pj )2 s − 2pk P min = min = min {1 − xk } = 1 − max {xk } = 1 − T . then the event contains exactly two jets. g and P = i pi . i=j s k s k k (1. also known ad JADE distance. The algorithm will then cluster the two “closest” partons. The replacement of parton-level observables with jet-based ones allows to recover a well-posedness necessary to carry out the perturbative calculations. Metrics based on the Rij distance of Eq. while the Lorentz- invariant metric of the exercise. where three-jet configurations can arise. This is easy to prove since: 2pi pj ( pi )2 y12 + y13 + y23 = = i = 1. 2pi P/s with pi = q. xq̄ ). see Problem 1. Solution If y > 13 . the concept of jet is introduced. The three xi parameters are not independent since they satisfy the relation: . These divergences can be consistently cate- gorised and re-absorbed for IR-safe observables. An event is said to contain three jets if mini=j yij > y for some value of the jet parameter y. We then focus on the more interesting case 0 ≤ y ≤ 13 . A problem which often occurs in theories of massless particles like QCD and QED is connected with the divergences associated with soft or collinear splitting of the massless particles. leaving two resolved jets. Consider now the metric yij ≡ 2pi pj /s. A jet is defined by a jet-finding algorithm that determines how to cluster together an ensemble of particles in an iterative procedure based on a definition of distance between particles. (1.248) i=j s s hence at least one of the three measures has to be smaller than 1/3.44 Consider the parton-level reaction e+ e− → q q̄ g. A jet algorithm can be studied in an experiment by using as input the experimental signature of the theoretical particles. are more popular at lepton colliders. if one wishes to compare its predictions to the experiment. Observables related to a jet-finding algorithm should be predictable within a given theory.249) where T is the thrust parameter introduced in Problem 1.167) are quite popular at hadron colliders.43.47. where all final- Define the adimensional parameters xi = state partons are assumed to be massless. Discussion In this problem. Determine the geometric locus of three-jet events in the (xq . see Problem 1. xq̄ ) plane as a function of y.72 1 Kinematics Problem 1. Figure 1.252) k=1. Since all Fig.q̄. (1.14 The phase-space xq̄ representation of a qq̄g event in terms of the qg g→0 Lorentz-invariants variables xi = 2pi P/s computed in the 1−y centre-of-mass frame.253) ⎪ ⎩ xq + xq̄ ≥ 1 + y. xq̄ ) plane. Notice that for y > 0.14 shows the triangle for a generic value of y. which identifies a rectangular triangle in the (xq .250). (1.1. xq̄ ) plane.3 The three-jet events are therefore associated with points located inside a triangle similar to Eq. In the (xq . since xq + xq̄ → 2 implies xg → 0 as for Eq.251) ⎪ ⎩ xq + xq̄ ≥ 1.e.2.g > y. The top-side (left) of this triangle corresponds to xq̄ = 1 (xq = 1). (1. the sides of the larger triangle are excluded from the three-jet parameter space. For a three-jet event one has instead: min {1 − xk } > y ⇒ 1 − xq. The 3 − jet 2 domain corresponding to 3 3-jet events according to the q̄ g JADE algorithm with parameter y is the one 2y delimited by the light lines 2y 2 xq 1−y 3 . (1. i. 1. The top-right vertex of the triangle corresponds instead to a soft gluon emission.250) s Each event can be parametrised by two energy fractions xq and xq̄ .2 Center-of-Mass Dynamics and Particle Decays 73 2(q + q̄ + g)P xq + xq̄ + xg = = 2. but with shorted sides defined by: ⎧ ⎪ ⎨2y ≤ xq ≤ 1 − y 2y ≤ xq̄ ≤ 1 − y (1. the allowed points lie in the geometric locus defined by: ⎧ ⎪ ⎨xq ≤ 1 xq̄ ≤ 1 (1.251). a collinear emission of the gluon by the quark (antiquark). pΛ = (mΛ . pp = (mp .89). they can be written as: pK = (EK . In the laboratory frame. pπ = (Eπ . Solution Let’s denote the four-momenta of the involved particles by pK .255) 2(mΛ − mp ) Discussion Equation (1. Indeed. (1. this particular configuration corresponds to a Λ emitted forward (θ ∗ = 0) . mπ2 = mK2 + mp2 + mΛ 2 + 2EK mp − 2EK mΛ − 2mp mΛ . (1. we get rid of it by squaring the π 0 four-momentum: pπ = pK + pπ − pΛ . [18]. 2EK (mΛ − mp ) = mK2 + mp2 + mΛ 2 − 2mp mΛ − mπ2 . 0). the three-jet phase-space is free from IR divergences. Bando n.74 1 Kinematics the divergence related to a soft or collinear splitting are located on the top and right side of the triangle. (1.45 The reaction K − p → π 0 Λ0 is studied in a fixed-target experiment. pK ). pπ and pΛ . 0). pp . Notice that not all scatterings at energy EK produce Λ’s at rest in the laboratory. Since nothing is known about the π 0 kinematics after the scattering. (mΛ − mp )2 + mK2 − mπ2 EK = . with the proton at√rest. which is analogous to a two-body decay where only a centre-of-mass energy s = mΛ − mp is available for the two mesons. pK ). Determine the beam energy for which the Λ0 baryon can be produced at rest in the laboratory frame.255) √ has the same form of the centre-of-mass energy in a two-body decay with s = mΛ − mp . Suggested Readings A detailed calculation of the LO three-jet cross section with the JADE algorithm can be found in Ref.254) Conservation of energy and momentum implies that pK + pp = pπ + pΛ . Indeed. we could have guessed this result by noticing that this scattering reaction is kinematically identical to the decay Λ0 → p K − π 0 . see Eq. 13705/2010 Problem 1. 2 Center-of-Mass Dynamics and Particle Decays 75 in the centre-of-mass frame: for any other angle θ ∗ . 3. giving s ≈ 20 GeV.96). the velocity of the centre-of-mass in the laboratory frame. the centre-of-mass energy of a binary pp collision. The velocity of the centre-of-mass frame is the velocity of a “particle” of four- momentum p1 + p2 . Let’s denote the four-momenta of the colliding protons by pi = (Ei . a different kinematics in the laboratory will arise.256) E1 E2 The proton mass mp = 0. 4. the energy that a proton beam must have in order to generate the same centre-of- mass energy in a fixed-target collision.95 GeV. p p 1− (1. 2. hence: . Solution 1. the angle in the laboratory of a relativistic particle produced at 90◦ in the centre- of-mass. 2.1. so that a first-order √ expansion of the square roots is accurate√enough for most purposes. The centre-of-mass energy is given by the square-root of the Mandelstam variable: s = (p1 + p2 )2 = 2mp2 + 2(E1 E2 + |p1 ||p2 |) = ⎛ ! ⎞ 2 ! 2 m m = 2mp2 + 2E1 E2 ⎝1 + 1 − ⎠. Bando n. pi ). 1N/R3/SUB/2005 Problem 1. Determine: 1. see Eq.938 GeV is small compared to either of the two energies. (1. A complete calculation yields s = 19.46 A proton beam of energy E1 = 20 GeV collides head-on against an other beam of protons of energy E2 = 5 GeV. . the boost parameter to the laboratory frame is given by −β. (1.9◦ . A relativistic particle with centre-of-mass velocity β ∗ ≈ 1 and polar angle θ ∗ = π/2 with respect to the beam direction. (1.07 GeV β= = = 0.32) = 52.44) and (1. γ (cos(π/2) + β) sβ γ |p2 | − |p1 | 15. then.257) to get: √ √ sin(π/2) s s 19. 2 E1 − mp2 − E22 − mp2 15. (1.257) E1 + E2 25 GeV 3. If we chose the direction of the x axis as aligned with the most energetic proton.95 GeV tan θ = =√ = = = 1.44).07 GeV θ = atan(1. We can therefore use Eqs. will emerge at an angle θ in the laboratory frame given by Eq.258) . (1.32.603. For a proton collision against a fixed-target at the same centre-of-mass energy.76 1 Kinematics 4. the beam momentum Efix must satisfy: . Bando n. say particle a.938 GeV which is more than one order of magnitude larger than the highest proton energy for head-on collisions. 2mp (mp + Efix ) = s. and γ → e+ e− in the neighbourhood of an electron. . .938 GeV)2 Efix = = = 211 GeV.47 Can a photon decay to an electron-positron pair in vacuum? Can a particle radiate a photon in vacuum? Consider a photon conversion in the following two cases: γ → e+ e− in the electromagnetic field of a heavy nucleus of mass mN . Indeed. 13153/2009 Problem 1.. . .259) 2mp 2 · 0. . determine the threshold energy of the photon such that the reaction can take place. In both cases. (1. with b at rest in the laboratory frame. . . . . it then follows that (pa + pb + pc + . √ 2mp2 + 2mp Efix = s.95 GeV)2 − (0. its minimum value corresponds to the minimum possible value of s. . . initially at rest. .)2 . (1. For both cases.260) If instead both particles are massless. the sum of two such four-vectors is always time-like.261) The right-hand side of this inequality is the tightest lower bound and corresponds to a configuration where all particles are at rest in their centre-of-mass frame. one can make a boost to its rest frame and compute explicitly the invariant: (pa + pb )2 = p2a + p2b + 2pa pb = ma2 + mb2 + 2ma Eb(a) ≥ ≥ ma2 + mb2 + 2ma mb = (ma + mb )2 > 0. which is given by the left-hand side . 1N/R3/SUB/2005. if at least one of the particles is massive. s − 2mp2 (19.)2 ≥ (mc + md + . also at rest. Bando n. ma2 + mb2 + 2Ea mb = (pc + pd + . By iteratively clustering pairs of four-momenta. then in any frame (pa + pb )2 = 2Ea Eb (1 − cos θab ) ≥ 0. we then have: s = (pa + pb )2 = (pc + pd + . (1.)2 ≥ (ma + mb + mc + .262) Since the laboratory energy of Ea is a linear function of s. (1.)2 .)2 . Discussion The four-momentum p of a particle is either time-like (p2 > 0) or light-like (p2 = 0). For a reaction a + b → c + d + . . where pγ . (1. (1.48 A photon converts to γ → e+ e− in the electromagnetic field of a heavy nucleus of mass mN .)2 − ma2 − mb2 Eathr = . p2e− = p2 e− + pγ + 2pγ pe− . Notice that the only possibility for Eq. (1.1.263) 2mb Solution Let’s consider the reaction γ → e+ e− . whereas photon conversion in the presence of a spectator particle is kinematically allowed. The threshold condition for a fixed-target scattering corresponds to a beam energy such that: (mc + md + . p2γ = 2me2 + 2pe+ pe− .261). one gets the condition: pe− = pe− + pγ . (1. Like in the previous case.2 Center-of-Mass Dynamics and Particle Decays 77 of Eq. photon conversion in vacuum is not allowed. As seen from Eq. (1. Conservation of energy and momentum implies: pγ = pe+ + pe− .264) to be still fulfilled.264). .265) In the rest frame of the final electron. The reaction can occur if the photon energy is above the threshold for e+ e− production. hence the last equality in Eq. pe+ . 0 = me2 + pe+ pe− . the Lorentz-invariant product pe+ pe− is Ee + me > 0. Let’s consider the reaction γ X → e+ e− X. namely: ⎧ ⎨ (2me +me )2 −me2 = 4 me electron Eγthr = (2me +m2me (1. Notice that this is not the case if me were zero: in this latter case. Problem 1. Estimate an order of magnitude for the opening angle between the two leptons in the laboratory frame. or the emission of an infinitely soft electron or positron. and pe− are the four-momenta of the tree particles.265) to be satisfied is again through a collinear splitting in the limit me → 0. photon conversion in the electromagnetic field of a heavy nucleus requires a factor of two smaller threshold energy compared to the conversion in the electron field. 2 0 = pγ pe− (1. pγ pe− = Eγ me > 0. (1. Let’s now consider the reaction e− → e− γ .264) cannot be satisfied.264) In the rest frame of the electron. where X can be either an electron or a heavy nucleus at rest in the laboratory frame. or through an infinitely soft photon emission.266) N ) −mN 2 2 ⎩ 2mN = 2N mNe + O mNe m 4m m ≈ 2 me nucleus Thus. . (1. Calculate the minimum momentum transfer to the nucleus |q| when the photon has an initial energy Eγ = 1 GeV. would allow Eq. either a collinear splitting with pe− parallel to pe+ . In this way: pγ → pX = (E. In order to transmute into a pair of massive electrons. Exchanging three-momentum with a negligible energy loss is indeed possible if mN . It can be heuristically explained like follows. the photon needs to “acquire” a mass. which gives p2γ = Eγ2 − Eγ2 = 0. A photon has four-momentum pγ = (Eγ . This can be achieved by transfering a quantity q of its momentum to the nearby nucleus with no energy loss. Eγ ex ). E − |q|) and p2X is now positive. after the authors who first computed its theoretical cross section. provided that the total four-momentum is conserved.78 1 Kinematics Discussion The process by which a photon converts in matter is known as Bethe–Hadler. Eγ . Let’s define: . Solution Conservation of energy-momentum implies that pγ + pN = pe+ + pe− + pN . + − q = pγ − pX = pN − pN = ( |q|2 + mN2 − mN .269) where θ is the angle of q with respect to the photon direction. q) (1. we see that Eq. (1. From Descartes’ rule of sign. (1. Terms suppressed by powers of mN have been neglected.q . (1. then the exchanged momentum |q| is also much smaller than mN . Explicitly: .269) admits a valid solution only if cos θ > 0.268) 2mN From the momentum conservation equation pX = pγ − q.267) If the energy of the photon is much smaller than the nuclear mass mN . so that: |q|2 q≈ . it follows: |q|4 Eγ + 1 + |q|2 − 2Eγ |q| cos θ + mX2 ≈ 4mN2 mN |q|2 − 2Eγ |q| cos θ + mX2 = 0. pX = pe + pe . which is given by mXmin = 2me . so that the minimum momentum transfer corresponds to a nucleus recoil parallel to the incoming photon momentum and to the minimum possible invariant mass of the e+ e− pair. (1. see Eq.270) It can be easily verified that ∂|q|± /∂mX2 ≶ 0 and ∂|q|± /∂ cos θ ≷ 0. Hence: . |q|± = Eγ cos θ ± Eγ2 cos2 θ − mX2 .261). (1. (1.m frame is given by Eq. (1.1. If mX = 2me . hence φ = 0. The largest opening angle for a two-body decay as a function of the velocity β of the mother particle (X in our case) and on the velocity β ∗ of the daughter particles in their c.92): any boost from the reference frame to the laboratory frame will maintain the two momenta parallel to each other. (1.2 Center-of-Mass Dynamics and Particle Decays 79 ! 4m2 2me2 |q|min = Eγ 1− 1 − 2e ≈ ≈ 0.65): .o. the velocity of the two particles in their rest frame is zero as for Eq.271) Eγ Eγ We now estimate the typical size of the opening angle φ between the electron and positron.5 KeV. 49 The strong reaction p p → X K + K − is studied in laboratory. 2γ β β ∗ 2|pX | mX2 − 4me2 tan(φmax ) = = . Determine the minimal energy in the laboratory frame necessary to produce X in a fixed target experiment. Equation (1. Since |q|min Eγ . 13153/2009 Problem 1. Suggested Readings Reference [19] provides a complete review of the photo-production mechanism of lepton-antilepton pairs. and thus to determine the decay plane of lepton pair. the momentum |pX | of the e+ e− pair is |pX | ≈ Eγ . Let the mass of X be twice the proton mass. and baryon number of X. a virtual photon is exchanged between the nucleus and a virtual electron.273) then gives |φmax − φmin | me 2 φ∼ ≈ f − 1 = O(mrad). at leading order. one can expect the differential cross section to feature the typical ∼1/|q|4 behaviour. [20].272) γ 2 β 2 − β ∗2 |pX |2 − (mX2 − 4me2 ) Since pair-production is an electromagnetic process where. (1. where X denotes an unknown particle. since the orientation of the decay plane is correlated with the polarisation of the photon. opens the possibility to use the Bethe–Hadler process as a polarimeter of the incoming photon. Determine the values of the electric charge. so that the bulk of the conversions will have |q| |q|min . or equivalently mX = f · 2me . Bando n. for small transfered momenta. (1. with f of order one.273) 2 Eγ The possibility to measure the opening angle. strangeness. The idea of using the double photon-conversion for the measurement of the CP properties of the Higgs boson in the H → γ γ channel has been investigated by the phenomenological work of Ref. . The threshold energy is therefore given by Eq. Consider now the case that the target consists of a heavy material. Since EF mp . so that the target proton is actually a bounded nucleon. in terms of kinetic energy.263): 2 (2mp + 2mK )2 − 2mp2 mK mK Epthr = = mp 1+4 +2 = 3. (1. being characterised by an energy kB T ≈ 25 meV (T /300 K). and BX = 2. it follows that QX = +2.262) gets modified to: . Assume a Fermi energy EF = 30 MeV. The threshold condition of Eq. and the baryon number (B). is totally negligible. 5 for more details.80 1 Kinematics Solution Since the strong interaction conserves the electric charge (Q). when using a proton beam on a liquid-hydrogen target.275) 2mp or. (1.274) where we have used the PDG values mp = 0. The thermal motion.263): (4mp )2 − 2mp2 Epthr = = 7mp = 6. Tpthr = 6mp = 5. Problem 1. The most favourable kinematical configuration corresponds to a nucleon of momentum |pF | moving against the incoming proton. By how much does the threshold energy get reduced by the nuclear motion? Solution Let’s first consider the case where the target consists of liquid hydrogen. (1.938 GeV and mK = 0. SX = 0. the threshold energy for a fixed-target collision is given by Eq. 2mp mp mp (1.50 Determine the threshold energy for the antiproton production p p → p p p p̄. See Chap. If the target proton is bounded inside a nucleus. its momentum can be as large as the Fermi momentum |pF |.56 GeV.6 GeV. so that the target proton can be safely assumed at rest.494 GeV [4]. the strangeness (S). (1. the bounded nucleon can be treated as classical.43 GeV. If mX = 2mp . |pF |2 2mp2 + 2Epthr mp + + 2 (Epthr )2 − mp2 |pF | = (4mp )2 . (1.276) 2mp Introducing the adimensional parameter ε = |pF |/mp = 2EF /mp ≈ 0.23, we can simplify the above equation by neglecting terms of O(ε2 ): 1.2 Center-of-Mass Dynamics and Particle Decays 81 ε2 (Epthr )2 − 14mp1+ + (7mp )2 + ε2 mp2 = 0 2 ε2 Epthr = 7mp 1 + − (7mp )2 (1 + ε2 ) − (7mp )2 − ε2 mp2 = 2 ε2 EF = 7mp 1 + − 48 mp2 ε2 = 7mp 1 + − 96 mp EF = 5.24 GeV, 2 mp (1.277) corresponding to a proton kinetic energy Tpthr = 4.3 GeV, i.e. about 1.3 GeV less than that necessary with a hydrogen target. Suggested Readings This reaction was first produced in laboratory by a E. Segrè et al. at the AGS accelerator at Berkeley. The reader is encouraged to study the discovery paper [21]. A guided discussion on the experimental set up can be also found in Ref. [12]. Bando n. 18211/2016 Problem 1.51 Given the Boltzman constant kB = 8.6 × 10−5 eV K−1 , estimate the typical wavelength of the cosmic background radiation. Which part of the EM spectrum does it belong to? Solution The cosmic microwave background (CMB) features a black-body spectrum with temperature T ≈ 2.7 K. The energy density per unit of frequency is given by Planck’s law dE 8π h ν3 = 3 . (1.278) dν c exp (hν/kB T ) − 1 The peak frequency mode has an energy of about 3kB T , so that: hc hc 3kB T = hν̄ = , λ̄ = = λ̄ 3kB T 6.6 × 10−34 J s · 3 × 108 m s−1 = ≈ 1.7 mm, (1.279) 3 · 8.6 × 10−5 K−1 · 1.6 × 10−19 J · 2.7 K corresponding to a frequency ν̄ = 160 GHz. The typical radiation spectrum is therefore located in the microwave domain. Bando n. 13705/2010 82 1 Kinematics Problem 1.52 A proton of the cosmic radiation can produce π 0 by interacting with the cosmic microwave background (CMB) at 3 K. You can assume that the CMB photons are uniformly distributed in space with an energy density ργ = 0.38 eV/cm3 , and have an average energy Eγ = 0.7 meV. What is the proton energy threshold? At the energy threshold, what fraction of the proton energy gets lost, and what is the minimum angle between the two photons in the π 0 decay? Let us assume that the photo-absorption cross section above threshold is σγ p = 200 μbarn. Provide an estimate of the proton mean free path. Discussion The existence of a threshold energy for the inelastic reaction p γCMB → p π 0 , where γCMB is a photon of the cosmic microwave radiation, implies a suppression of the cosmic ray spectrum above that threshold, known as the GZK cut-off. Solution The lowest proton energy threshold corresponds to a CMB photon of energy Eγ moving against the proton. For this configuration, one has: mp2 + 2mp mπ + mπ2 = mp2 + 2Epthr Eγ + 2|pp |Eγ . (1.280) Since Epthr /mp ∼ mπ 0 /Eγ 1, it can be safely assumed that |pp | ≈ Ep . Therefore: 2mp + mπ 0 (2 · 0.938 + 0.135) GeV Epthr = mπ 0 = 0.135 GeV = 4Eγ 4 · 7 × 10−13 GeV = 0.9 × 1018 eV. (1.281) At the threshold, both the proton and the neutral pion are produced at rest in their centre-of-mass frame. The gamma factor of the centre-of-mass is given by γ = (Epthr + Eγ )/(mp + mπ 0 ) ≈ Epthr /(mp + mπ 0 ). The relative energy loss suffered by the proton after the photon absorbtion is therefore given by: mp Epthr − Ep mπ 0 Ep = γ mp = E thr ⇒ = ≈ 0.12. (1.282) mp + mπ 0 p Epthr mp + mπ 0 The minimum open angle between the two photons from the π 0 decay is given by Eq. (1.61): 2 2(mp + mπ 0 ) φmin = = ≈ 2 × 10−19 rad. (1.283) γ Epthr Given a process with cross section σ , the interaction length λ is defined as the inverse of the probability of interaction per unit length and per incoming particle, so that dNpγ /dΦp dt is the number of interactions per incoming proton. By definition of cross section. nb . dt d 2 A dx dt dNγ p 1 1 = 2 ργ σγ p ⇒ λ= = −3 = (dΦp dt) dx 2 ργ σγ p 2 0. The latter is defined as the number of particles a crossing per unit time and unit area the normal surface at the position of the volume d 3 r: Ja = na |va |. and write: dNX /dt d 3 r · d 3 r dNX /dt d 3 r σ = = (1. (1. = c ρp d 2 A dx · 2 ργ σγ p = dΦp dx · 2 ργ σγ p . The quantity nb na |va | at the denominator can be written in terms of the density and velocity in the generic reference frame by means of the . 1.20). is defined as the number of reactions X measured per unit time.1. (1.285) can be generalised to an arbitrary kinematics of the involved particles.2 Center-of-Mass Dynamics and Particle Decays 83 see Eq. 6 (1. and per unit of incoming flux density: 1 dNX σ = . the relative velocity is vrel = 2c. Firstly. The dimension of σ is therefore [σ ] = cm2 . Since both the proton and the CMB photon are moving in opposite direction at speed c. we introduce the particle density of the target. respectively.7×10 eV = 5 × 10 24 cm = 1. the number of interactions per unit time and unit volume is given by: dNγ p dNγ p = ρp ργ vrel · σγ p . where a is a moving projectile and b is at rest. see Eq. dΦp in the intermediate calculations is the flux of incoming proton entering the infinitesimal volume d 2 A dx.291). The definition (1.286) where na and |va | are the particle density ([n] = m−3 ) and velocity.285) Ja dt dNb where dNb is the number of scattering centers in an infinitesimal volume d 3 r irradiated by the flux density Ja . per unit scattering center. so that the distinction between projectile and target can be ultimately ignored.3 Cross Section The cross section σ of a scattering process a b → X. (1.38 eVcm −3 · 2 × 10−28 cm2 0.5 × 10 parsec. (1.284) Here.287) nb · d 3 r na |va | nb na |va | where we have introduced at the numerator the number of reactions X observed per unit time and unit volume. Dividing both sides of Eq. (1. pn )|2 dΦn (p1 + p2 . p2 ) (1.2.291) nb σ is called interaction length of the process under consideration. so that: dNX /dt d 3 r dNX /dt d 3 r σ = = .285) and introduce the useful concept of interaction length. see Sect. The differential cross section for a scattering p1 p2 → p3 . . pn ) 4 I(p1 . since dNX is a pure number and d 4 x ≡ dt d 3 r is invariant under transformations of the Poincaré group. p2 . with Φa being the total flux across the surface A ([Φ] = s−1 ) irradiating a target of thickness δx and uniform particle density nb . . (1. p3 . 3. For different beam structure.3. Since λ−1 is a probability of interaction per incoming particle and per unit length. .288) na nb (va − vb )2 − (va × vb )2 /c2 na nb vrel see Problem 1. it gives the luminosity L of a fixed target experiment. (1. (1.3: the cross section is therefore a Lorentz-invariant. p3 . From Eq. in the presence of multiple exclusive processes of interaction between a and b. Under a generic boost the numerator of Eq. (1. (1.288) as the coefficient of proportionality between the total event rate and the cross section. (1. .84 1 Kinematics relative velocity of Eq. and once multiplied by the flux. .292) . as discussed in Problem 1.285) we have: dNX Φa = · (nb · A · δx) · σ = Φa (nb δx) ·σ. We now come back to the early definition of Eq. . . . . the luminosity can be still defined from Eq. Let’s consider a uniform flux density Ja = Φa /A. . (1. see Problem 1. pn ) = |M (p1 .289) dt A L The quantity nb δx is the surface density of the target. . . .288) is invariant. The same holds for the combination na nb vrel . and its inverse 1 λ≡ . and the total interaction length is therefore given by the inverse of the sum of all inverse interaction lengths. Notice that [L ] = cm−2 s−1 . (1. .11. the probability of interaction per incoming particle is obtained: 1 dNX = probability of interaction per particle = (nb σ ) · δx.290) by the incoming flux Φa . (1.20).290) Φa dt so that nb σ is the probability of interaction per particle and per unit length. the probability of any such interaction is given by the sum of all probabilities. pn can be calculated theoretically using the formula: 1 dσ (p1 . .e.295) 4 where the identity has been proved in Eq. see Eq. (1. Solution The relativistic invariant I at the denominator of Eq.296) 4 |p∗i | s 16π 2 s dΩ ∗ 64π 2 s |p∗i | .. (1. For a free particle.293) by r1 .211). When the incoming particles are unpolarised and spin is not observed. the factor 4 I = (2E1 ) (2E2 ) vrel is just but the product of the two beam densities times their relative velocity as in Eq. (1.293) can be also written as: (s − m12 − m22 )2 + 4m12 m22 I 2 = (p1 p2 )2 + m12 m22 = = s |p∗ |2 .91). (1..184).293) (2S1 + 1)(2S2 + 1) which has often the virtue of greatly simplifying the calculations. see Eq. (1. (1. p2 ) is the invariant discussed in Problem 1. (1. (1. and: • I(p1 . eigenstates of the linear momentum and spin. Indeed. i.294) dΩ ∗ 64π 2 s |p∗i | where |p∗i | and |p∗f | are the centre-of-mass momenta before and after the scattering. and accounts for the incoming particle flux. For example. • dΦn is the relativistic phase-space measure of Eq. (1. The choice of wave function normalisation is convention as long as it is used consistently.1. (1.3 Cross Section 85 where the incoming and outgoing particles are asymptotic states of the free Hamil- tonian. (1.293) to simplify the notation. one can replace the matrix element squared in Eq. the space density is given by the squared norm of the wave function |ψp (r)|2 = 2Ep . although one should always remember that the cross section depends in general on the spin vectors ri of the scattered particles.53 Prove that the differential cross section for a 2 → 2 scattering in the centre-of-mass frame can be written as: ∗ dσ 1 |pf | = |M |2 .288). • M is the relativistic scattering amplitude.188). another popular normalisation is the non- relativistic normalisation of Eq. (1.186). we get: 1 2 1 |p∗f | dσ 1 |pf | ∗ dσ = √ |M | √ dΩ ∗ ⇒ = |M |2 .3. Problems Problem 1.rn |M |2 |M | ≡ . By making use of the two- body phase-space measure expressed in terms of the centre-of-mass solid angle.. Spin indices are omitted in Eq. (1. As an example. Discussion Consider the 2 → 2 scattering a b → c d.54 What is the π + p cross section at the peak of the Δ(1232) resonance? . this technique was employed by Steinberger et al. [22]. denoted by (2). In particular. if the two reactions are studied at the same centre-of-mass energy. Suppose that both this reaction.293) together with Eq.296) implies dσ(1) /dΩ ∗ (2Sc + 1)(2Sd + 1) |p∗cd | 2 = . and if the scattering particles are unpolarised and the measurement is inclusive with respect to the spin of the final-state particles. An introduction to the same topic can be found in Chap.298) dσ(2) /dΩ ∗ (2Sa + 1)(2Sb + 1) |p∗ab | where one has to further assumed that the interaction is invariant under time-reversal so that the spin-averaged matrix element squared for (1) and (2) are identical at the same centre-of-mass angle cos θ ∗ . (1. Suggested Readings The measurement of the pion spin from detailed balance arguments is documented in Ref. (1. 2 of Ref. (1. and its time-reversed c d → a b. (1. which could be both obtained in fixed-target collisions. denoted by (1). [22] to measure the spin of the charged pion profiting from the reaction π + d → p p and its inverse p p → π + d. [12].86 1 Kinematics If the scattering amplitude is normalised using a non-relativistic normalisation. can be performed in the laboratory under controlled conditions. then Eq. then the above expression becomes: ∗ ∗ ∗ ∗ ∗ |p∗ | 1 |pf | 2 = 1 (2E1 ) (2E2 ) (2E3 ) (2E4 ) f |M |2 = ∗ |M | 64π 2 (E1 + E2 )(E3 + E4 ) |p∗i | ∗ ∗ ∗ ∗ NR 64π 2 s |pi | 1 1 |p∗f | 1 |p∗f |2 = ∗ |MNR |2 = |MNR |2 4π 2 1∗ + 1∗ 1 + 1 |p | 4π 2 (|v | + |v |) (|v | + |v |) 1 2 3 4 E E E∗ E∗ i 1 2 3 4 1 |p∗f |2 = |MNR |2 . the spin of the fourth can be measured by comparing the event rates of (1) and (2). Problem 1.297) 4π 2 vi vf rel rel where the relative velocity has been used. If the spin of three of the four particles is known. and Sπ = 0.300). we have made use of Eq.11 of Ref. (1. and p∗ is the centre-of-mass momentum given by Eq. Suggested Readings A concise summary of the Breit–Wigner scattering can be found in the PDG review dedicated to kinematics.9) to convert the result into SI units. respectively. is described by the Breit–Wigner formula: 4π 2J + 1 Γ 2 /4 σf (E ∗ ) = BRi BRf . (1.299) |p∗ |2 (2S1 + 1)(2S2 + 1) (E ∗ − E0 )2 + Γ 2 /4 where BRi. (1. Bando n. 18211/2016 Problem 1. (1. (1. [4].g.91).f are the branching ratio of the resonance into the initial and final state particles. 47 of Ref.3 Cross Section 87 Discussion When the centre-of-mass energy E ∗ approaches the mass E0 of a resonance of spin J and width Γ . the cross section at the peak is independent of the width Γ and depends only on the resonance mass E0 : 2 8π 32π mΔ σΔ = ∗ = 2 2 = |p | 2 mΔ − (mp + mπ )2 mΔ − (mp − mπ )2 = 485 GeV−2 = 188 mbarn.55 How was it possible to measure the number of SM neutrino families at LEP by studying the Z 0 resonance? Discussion In the neighbourhood of the Z 0 mass. Fig. see e. the scattering cross section between two unpolarised particles with spin S1 and S2 . the cross section for e+ e− → Z 0 → hadrons for unpolarised electron-positron beams features an energy dependence described by a relativistic Breit–Wigner: . Chap. Solution According to the Breit–Wigner formula of Eq. This result is in good agreement with the experimental data. Sp = 1/2.299). detected in the channel f .300) where we have used the values BRπ + p = 1. (1.1. J = 3/2. In the last row of Eq. 2. [16]. This can be best implemented as fit to the lineshape√of the hadronic cross section σhad (s) of Eq. Part of these assumptions can be relieved by using additional observables measured at the Z 0 peak.301) measured at different value of s. In Eq.302) Radiative corrections due to soft and collinear photon radiation from the incoming leptons distort the lineshape by inducing an asymmetric tail and by reducing the peak cross section to a value: peak 12π Γe Γhad σhad = (1 − δrad ) ≡ σhad 0 (1 − δrad ).301).30 and Nν < 4 at 98% CL. The ratio between the hadronic and .e. (1.301) mZ ΓZ (s − mZ ) + s2 ΓZ2 /mZ2 2 2 2 2 where ΓZ is the width at the Z 0 mass.299) when s ≈ mZ2 . The result published by the ALEPH Collaboration in 1989 gave a result Nν = 3.27 ± 0.305) The lineshape-based analysis relies on model assumptions for the partial widths and on the lineshape itself. charged under the SM group. and Γe .304) where Nν is the number of neutrino families that couple to the Z 0 and δτ is a phase- space corrections that accounts for the large τ mass. (1. By taking the SM values for δrad . (1. the only unknown parameters in the fit are mZ and Nν .301) reduces to Eq. (1. i. (1. (s − mZ )2 + s2 ΓZ /mZ ( s − mZ )2 ( s + mZ )2 + mZ ΓZ ( s − mZ )2 + ΓZ2 /4 (1.88 1 Kinematics 12π Γe Γhad s ΓZ2 σhad (s) = . It should be noticed that the sensitivity to Γν arises from both the width of the Breit–Wigner and from the cross section value at the peak. Indeed: s ΓZ2 mZ2 ΓZ2 ΓZ2 /4 2 2 2 ≈ √ √ 2 2 = √ . The first method is based on subtracting the leptonic and hadronic width from the total visible width ΓZ . which can be therefore simultaneously extracted from the fit. [4].303) mZ2 ΓZ2 In the SM. Γν . (1. leptonic. Solution There are three main methods to measure at LEP the number of light and active neutrinos. (1. the Z 0 boson width is given by the sum of the hadronic. (1. Notice that the kinematic part of Eq. the Breit– Wigner width grows with s. see Ref. Γhad . and neutrino partial widths: ΓZ = Γhad + Nν Γν + (3 + δτ )Γ . modulo the mass- related corrections for τ leptons. the value of Rinv thus obtained can be related to the number of active light neutrinos as: Nν ΓνSM Γ 0 Rinv = ⇒ Nν = 0 Rinv (1.1. where the final-state radiation (FSR) photon is required to trigger the event. By assuming universality of the charged lepton coupling to the Z 0 boson. one can therefore express the ratio 0 Rinv ≡ Γinv /Γ . Write down the LO cross section for p p → X → γ γ in terms of the resonance parameters and of the proton PDF. compatible with the method of Eq. (1. σhad0 . Suggested Readings The measurement of the number of light neutrino families from the Z 0 lineshape is documented in Ref. as a function of mZ . Solution At LO in perturbation theory. [24]. is an observable that could be measured to large accuracy at LEP. Two partonic channels contribute to the resonance production: q q̄ → X and g g → X.984 ± 0. The details on the combined LEP measurement of Nν can be found in Ref.56 A neutral narrow resonance X of mass M. Rμ0 . followed by the disintegration of X into a pair of photons. The third method is based on a direct measurement of the invisible width Γinv 0 from the cross section of the process e+ e− → ν ν̄ γ . By using the hadronic cross section at the peak. and Rτ0 can be assumed identical. natural width √ Γ . σhad 0 . mZ ΓZ 2 2 mZ (Rinv + 3 + δτ + R0 )2 0 1 12π R0 2 0 Rinv = − R0 − (3 + δτ ).05. (1.008 [23]. the production amplitude is described by the s-channel diagrams q q̄ → X and g g → X. See also the dedicated PDG review on this subject [4] for more details.92 ± 0. and R0 as: 12π Γe Γhad 12π R0 σhad 0 = = 2 . the ratios Re0 . For values of the partonic centre-of-mass energy ŝ in the neighbourhood of .306) mZ2 σhad 0 0 If one further assumes that the total width arises from neutrinos.3 Cross Section 89 leptonic decay widths.307) ΓSM Γν SM The combined result from the four LEP experiments is Nν = 2. The combined LEP measurement gave a result of Nν = 2. [23]. Problem 1. R0 = Γhad /Γ . is produced in proton-proton collisions at a centre-of-mass energy s and detected through its decay to a pair of photons. although plagued by larger uncertainties. and spin J. encoded in a correction factor Rτ0 /R0 ≡ 1 + δτ .307). characterised by an energy scale ŝ.g.309) (ŝ − M 2 )2 + M2Γ 2 MΓ The factor of π at the numerator ensures the proper normalisation. the partonic cross section is described by the relativistic Breit–Wigner similar to Eq. resulting in a dilution factor ρi and ρj . see e. For unpolarised initial-states. which is parametrised as in Eq. the cross section can be factorised into a short-distance component. and antiquark PDF’s: " ij fi (x1 .90 1 Kinematics M 2 . (1. Specialising Eq. Under this narrow-width assumption.301). 1 + δij 16π (2J + 1) Γγ γ Γij M2Γ 2 with σ̂ = 1 + δij (1. ŝ MΓ s ρi ρj τ ≡ Ms x 2 x ij =Cij (τ ) (2J + 1) M2 ≡ Γγ γ Cij Γij (1. All of the short-distance physics is encoded in the partonic cross section σ̂ . (1. one has to average over the possible configurations of quantum numbers as described by the appropriate density matrix. ŝ) σ = dx1 dx2 σ̂ (ŝ = x1 x2 s). [18].310) MΓ s s ij The long-distance physics in fully encoded into the parton luminosity factor Cij (τ ). (1. we have: 2 41 (2J + 1) Γgg × π8 τ dxx g(x) g τx gg →X σ = Γγ γ × 4 Γqq̄ × 4π9 τ dxx q(x) q̄ τx + q̄(x) q τx 2 1 MΓ s q q̄ → X (1. The symmetry fractor (1 + δij ) accounts for the undistinguishability of the initial- state particles. quark. ŝ) fj . (1.310) to the case of interest. An heuristic motivation for this extra factor will be provided later. Notice that the energy-dependence of the width has been suppressed because we can work under the assumption that Γ M.301). In high-energy hadron-hadron interactions. fi (x.309) over ŝ. we can further approximate the Breit–Wigner by a delta function: 1 π ≈ δ(ŝ − M 2 ). (1. so that ŝ ∼ M 2 . as one can readily verify by integrating both sides of Eq. The total cross section thus become: " 16π 2 (2J + 1) ij fi (x1 . Ref. ŝ) is the PDF of parton i evaluated at the momentum fraction x and at a factorisation scale ŝ. ŝ) fj (x2 . ŝ) fj (x2 . and a long-distance component that accounts for the parton distribution inside the protons. ŝ) σ = (1 + δij )Γγ γ Γij dx1 dx2 δ(x1 x2 s − M 2 ) MΓ ρi ρj 1 + δij (2J + 1) 16π 2 " 1 dx τ = Γγ γ Γij fi (x. described by the gluon.308) M 2 ρi ρj Γ 2 (ŝ − M ) + M Γ 2 2 2 2 Here.311) . (1. (1.r4 ρ 2 4|p | s (s − M 2 ) 2 + Γ 2M2 a propagator " 1 1 2 1 2M 1 = 2M dΦ 2 |Mb | |Ma |2 Φ2 2M 2 ρa2 2M r . at a resonance mass M = 750 GeV and at a centre-of-mass energy For s = 13 TeV. the parton luminosities using the MSTW2008NLO PDF set are reported in Table 1. The two-body phase-space for massless particles is given by Eq.293) using the spin-average of Eq.313) (2J + 1) Cij Γ Discussion We now give an heuristic motivation for the symmetry factor (1 + δij ) appearing in the Breit–Wigner of Eq. Consider for example the case of a spin-0 particle produced in the s-channel by massless particles a. (1. we have used the fact that the sum over the gluon PDF’s is symmetric under i ↔ j. using the MSTW2008NLO PDF set Cd d̄ Cuū Css̄ Ccc̄ Cbb̄ Cgg 627 1054 83 36 15 2137 To derive this result.189). The cross section can be computed directly from Eq.314) M 2 ρa2 (s − M 2 )2 + Γ 2 M 2 √ where we have approximated |p∗ | = M/2 and s = M.308).3 Cross Section 91 √ value of the parton luminosity factors2 C(τ ) 2evaluated at τ = M /s with Table 1. by measuring σ .1.r2 r3 . (1. one can constrain the product of the branching ratios into photons and into the channel ij in the combination: σs M BR(X → γ γ ) · BR(X → i j) = . and Γ . M.r 3 4 1 2 Γb Γa 2 (Φ2 )−1 BRa BRb M2Γ 2 = .r (s − M 2 )2 + Γ 2 M 2 Φ2 2M r . and decaying to a pair of massless particles b.312) 1 2·2 1 NC2 = 1 4·32 = 1 36 q q̄ √ example. and factorisation scale μ = M . In one further assumes that one production channel dominates over the others (for example. (1. If the particles are distinguishable.293): 1 " 1 1 σ = ∗ √ dΦ2 |Ma |2 |Mb |2 = r1 . then. and that the density factors are 1 2·2 1 NA2 = 1 4·82 = 1 256 gg ρ= (1. (1.1. one .1 Numerical 2 M = 750 GeV and s = 13 TeV. due to the largeness of Cij (τ ) or of the partial width Γij ). 7 × 10−11 keV.187).316) dΩ sin θ dθ where dΩ = dφ d cos θ is the solid angle parametrised by the polar and azimuthal angles with respect to p. [28] for further details.8). and b = b(θ. i. Ref. |p|) is the so-called impact parameter.4 keV photon (t1/2 = 68 ns). and the usual factor of 16π at the numerator.92 1 Kinematics has to integrate over the full solid angle. see Eq. (1.g.58 In classical mechanics. Suggested Readings A putative new high-mass resonance decaying to a pair of photons at a mass of about 750 GeV was reported by both the ATLAS and CMS Collaborations in December 2016 [25.299) is assumed. the full-width-at-half- maximum is given by ΓFWHM = Γ = 0. Discussion The natural width of spectral lines is generally very small. in natural units. 26]. Problem 1. however. [27]. A lot of theoretical speculation was stimulated by the early observation.57 An excited state of 57 Fe decays by emitting a 14. Solution The spectral width Γ and decay time τ are related by the relation Γ = /τ . (1. Γ = τ −1 . The relative width ΓFWHM /E is therefore of order 10−12 .315) t1/2 If a Breit–Wigner distribution as in Eq. dominated by prompt diphoton production. which was then attributed to a mere statistical fluctuation of the background. 18211/2016 Problem 1. see e. the differential cross section dσ/dΩ of a particle of mass m and momentum p scattered by a central potential is given by the formula: dσ b db = . giving: ln 2 Γ = τ −1 = = 6. (1. the energy resolution is dominated by other effects. Determine the FWHM of the energy distribution of the emitted photon. whereas identical particles give a factor of two smaller phase-space. or.308). hence the symmetry factor (1 + δij ) of Eq.e. By using Eq. (1. (1. giving Φ2 = 1/8π . Ref. disproved the excess. we can express t1/2 in keV−1 . In hot media.7 × 10−12 keV. Bando n. and thus too small to be measured in spectroscopy. like collisional and Doppler broadening.g. the . (1. The analysis of additional data. see e. To this purpose. Use this formula to derive the Rutherford cross section for a particle of charge e scattered by a heavy nucleus with charge Z e. |p|) for the Coulomb potential. and m and |p| gets replaced by the reduced mass μ = m M/(M + m) and centre-of- mass momentum. Eq. with the x-axis parallel to the initial momentum.316) gives the correct result. 1. the momentum after the scattering is p = |p|(cos θ. Let’s imagine that the source of the potential is located at the origin of the reference frame. sin θ ).15. hence it’s possible to define a function b = b(θ. φ). The angular momentum for a particle with impact parameter b is given by . (1.317) still holds provided that the polar angle θ is measured in the centre-of-mass frame. |p|) θ dσ = 2πb db distance from the polar axis at time t = −∞ of those particles that get scattered at an angle θ at time t = +∞.3 Cross Section 93 Fig. Discussion It is easy to verify that Eq. Solution We need to compute b(θ. Those particles that at time t = +∞ are scattered at a polar angle θ have momentum p = |p| eθ . see Fig.1. dσ = d 2 b = b db dφ is the infinitesimal cross section for scattering at the polar angle centred around (θ. (1.317) dφ dφ d cos θ sin θ dθ If the classical source is replaced by a pointlike source of finite mass M. Since the potential is central. |p|). hence: dσ dσ b db = b db ⇒ = (1. In the orbital plane. By definition. angular momentum with respect to the origin is conserved: all particles with impact parameter b will end up at the same polar angle θ . respectively. 1.15 Cartoon showing p the classical scattering of a particle against a central potential b p b(θ. it is convenient to exploit the time-invariance of L = r × p. coincides with the (non-relativistic) quantum-mechanical expectation from the exchange of a virtual photon.320) π 4π |p|b θ 4π |p|b from which we get the relation: m Z e2 1 + cos θ m Z e2 b(θ.323) dΩ 4T sin2 2 and the conversion into MKS units can be done by either remembering that α → (1/137) · 197 MeV fm. Rutherford’s formula becomes: 2 dσ Zα = θ . In natural units. (1. (1. the amplitude squared for the scattering .319) 4π |r| |q|2 so that dpy Z e2 = −∂z V (r) = sin θ (t). (1. The above result. which has been obtained according to the laws of classical mechanics.322) 4 4π |p|2 sin4 θ2 4T sin2 θ2 where T = |p|2 /2m is the kinetic energy of the particle and α = e2 /(4π c). See Problem 1.318) dt |p|b The Coulomb potential V (r) and its Fourier transform Ṽ (q) for a point-like source of charge Z e at the origin are given by Ze Ze V (r) = . (1. we thus obtain: m Z e2 m Z e2 dσ 1 1 1 = − 2θ θ = dΩ sin θ 4π |p|2 tan θ2 4π |p|2 tan 2 2 cos2 2 2 2 1 m Z e2 1 Z α( c) = = (1. Ṽ (q) = .9). |p|) = = . or by using the conversion GeV−2 → mbarn of Eq.59 for its relativistic generalisation. (1.94 1 Kinematics dθ m|r|2 |L| = −m|r|2 = |p|b ⇒ dt = − dθ. which reproduces the well-known non-relativistic Rutherford’s formula.317).321) 4π |p|2 sin θ 4π |p|2 tan θ2 Using Eq. (1. dt 4π |r|2 Z e2 m|r|2 m Z e2 dpy = sin θ − dθ = − sin θ dθ 4π |r|2 |p|b 4π |p|b " θ " m Z e2 π m Z e2 |p| sin θ = dpy = dθ sin θ = (1 + cos θ ) . Indeed. (1. (1. where q is the four-momentum exchange. the latter being much heavier than the former so that the scattering occurs without any energy transfer.323).326) dΩ 2|p| sin2 2 2 2 where p is the electron momentum.1. (1. (1. Hence.58. Since that the reaction is elastic.16. see Problem 1. M being the mass of the source generating the potential.324) can be written as: 2 dσ 4 Z 2 e4 |p∗ |2 Zα = θ = θ . • Show that the formula reduces to the Rutherford cross section in the classical limit |p| m. • What is the value of the total cross section σ ? • Show that in the ultra-relativistic limit the formula reduces to: dσ α 2 cos2 θ2 ≈ . Indeed. which is not present in Rutherford’s formula? Discussion The well-known Rutherford’s formula applies to the non-relativistic scatettering between two charged particles.2 (1.324) dΩ 4π 2 vrel For the Rutherford scattering.327) dΩ 4E 2 sin4 θ2 where E is the total relativistic energy. the very same formula can be obtained by using the classical picture of a point-like charged particle interacting with the static electric field of the target. |q| = 2|p| sin θ/2. a moving projectile and a steady target. vrel = |v|. (1.3 Cross Section 95 is |MNR |2 = |Ṽ (q)|2 = Z 2 e4 /|q|4 . . and Eq. 1. see Fig. and the centre-of-mass frame coincides with the laboratory frame if m M. Eq. (1.297) gives: dσ 1 |p∗ |2 Z 2 e4 = ∗2 |q|4 .325) dΩ (4π )2 16|v|2 |p∗ |4 sin4 2 4T sin2 2 which agrees with Eq. (1. the velocity and centre-of-mass momenta are identical before and after the scattering.59 The differential cross section of a spin-1/2 particle of mass m scattering against a point-like heavy particle is described by the Mott formula: 2 dσ α θ = θ m + |p| cos 2 2 . Problem 1. What is the reason behind the cos2 θ/2 dependence of the cross section. T ≡ E − m. (1. ε].329) 0 sin 2 0 θ 0 θ3 This is a consequence of the electromagnetic interaction range being infinite. The total cross section is obtained by integrating the differential cross section over the full solid angle. (1. (1. For θ ∈ [0.328) dΩ 2(mv)2 sin2 2 4T sin2 2 where T is the kinetic energy.326) reduces to: 2 2 dσ α α ≈ θ m = 2 θ . the integral goes like: " ε " ε " ε 1 θ dθ dθ sin θ 4 θ ∼ dθ 4 = = +∞. the formula of Eq. 1. In the ultrarelativistic limit |p| ≈ E .96 1 Kinematics Fig.16 Cartoon of the p scattering of an electron off a much heavier particle of mass M e− p θ e− M Solution In the limit |p| m. ε 1. m. The ratio between the . A particle detector.8. A velocity selector filters ions of the same initial velocity |v| = 0. counts the number of scattered ions. thus changing the Jz component of the state along the scattering axis. collide against a fixed gold target. since this configuration would result in a spin-flipped state. Problem 1. an initial relativistic fermion of a given chirality will also have a fixed helicity h as discussed in Problems 1. The two beams provide the same integrated flux. located at a polar angle θ with respect to the beam direction.6 and 1. one composed of 42 He+ ions and the other of A3 Li+ ions of unknown mass number A.60 Two ion beams.330) dΩ 2E sin2 2 4E 2 sin4 θ2 The cos θ/2 term at the numerator arises from the spin-1/2 nature of the electron: since the electromagnetic interaction does not change the chirality.1 c. A left-handed electron with h = −1/2 cannot be scattered exactly backward. (1. the Mott formula reduced to: 2 dσ α cos θ2 α 2 cos2 θ ≈ θ = 2 . A scintillating detector is placed at an angle θ = 30◦ from the beam axis. molar mass A = 27 g/mol) of thickness of d = 1 cm.93 ± 0. we can use Rutherford’s formula (1.323). (1. The most probable value for the mass number of the lithium ions is therefore A = 7.332) A 2 R from which we estimate A = (6. The active surface of the detector has a cross section of 1 cm × 1 cm as seen from the target. Under the conditions assumed by the problem. The beam flux at the target is Φ = 109 s−1 .290): ρ NA 2.e. and L = 1 m away from the target.1. (1. i.1 GeV collides against a fixed target of aluminium (density ρ = 2. velocity.333) Using the Rutherford’s cross section formula (1. we find δA 1 δR = .2 GeV−2 /std = dΩ 4 16 T 2 sin 2 16 · 0.1.323).61 A beam of alpha particles of energy T = 0.95 · 10−26 cm2 /std.11). (1. (1. which is compatible with the closest integer to the 1σ level. we get: dσ 2 Zα2 ZAl α2 22 · 132 · (1/137)2 (θ ) = θ = = 50. or charge of the two beams. What is the most probable mass number A of the lithium ions? Solution Given that the target nuclei are much heavier than any of the two ions.12 · sin4 ( 30·π/180 2 ) (1. A= R · 4 = 6. (1.335) . and that the ion velocity is small enough to treat the scattering particles as classical.7 g cm−3 · 6.334) = 1. Solution The instantaneous luminosity L of the experimental set-up is given by Eq.3 Cross Section 97 number of countings for the two beams is R = NHe /NLi = 3. which corresponds indeed to a stable isotope.02 × 1023 mol−1 L =Φ d = 109 s−1 · · 1 cm A 27 g mol−1 ≈ 6 × 1031 cm−2 s−1 . Problem 1.331) NLi mHe |vHe |2 ZLi 4 By propagating the error on the ratio.93.0 ± 0. Estimate the counting rate measured by the detector.7 gcm−3 .: 2 2 NHe mLi |vLi |2 ZHe A √ R= = = . deviations between the two event counts can only arise from differences in the mass. (1. any δ-ray with energy above 103 MeV can be assumed to originate from a pion scattering.62 A beam made of π + and K + with momentum |p| = 5 GeV is directed towards a bubble chamber filled with liquid hydrogen. (1. Estimate the probability of such events if the chamber has a total length L = 1 m.339) 2 2|p∗ |2 Inserting this last expression into Eq.336) dΩ Using the instantaneous luminosity of Eq.324). see Eq. (1. we have 2 me |p|2 1.9). we get: dσ 1 |p∗ |2 Z 2 e4 π 4π Z 2 α 2 = ∗2 = ∗2 q4 . according to Eq. (1. (1. (1.05 GeV π + Tmax = = (1.26. The integrated cross section in a small solid angle ΔΩ = (1 cm × 1 cm)/1 m2 = 10−4 std is therefore given by: dσ σ ≈ (θ ) · ΔΩ = 1. Determine the minimum kinetic energy T of the δ-ray such that the incoming particle can be unambiguously identified as a pion.340) dq2 4π 2 vrel q4 |p∗ |2 vrel where vrel is the relative velocity in the centre-of-mass frame frame. (1. we integrate the differential cross section. Indeed. The latter can be expressed in terms of the velocity v of the beam particles in the laboratory frame .338) me2 + M 2 + 2me |p|2 + M 2 103 MeV K + Therefore.98 1 Kinematics The last equality in Eq. π To compute the probability of a δ-ray emission with energy Tmax K ≤ T ≤ Tmax .95 × 10−30 cm2 . (1. This is best done by expressing the differential cross section as a function of the four-momentum transfer q2 . see Problem 1.141). Solution The discrimination between pions and kaons is possible because the maximum energy transfer at fixed momentum |p| decreases with the mass of the incoming particle. giving: θ∗ dφdq2 q2 = −|q∗ |2 = −4|p∗ |2 sin2 ⇒ dΩ ∗ = .333). Events where an energetic δ- ray is produced are selected and the energy of the emitted electron is measured with negligible uncertainty.334) comes from the conversion GeV−2 → cm2 .337) dt Problem 1. we therefore expect an average rate of dN = L σ ≈ 120 Hz. (1. 291): L ρ NA p≈ =L· σ = λ A 0.8 × 10−27 cm2 · = 6.06 g cm−3 · 6.340) into a cross section differential in the kinetic energy of the recoiling electron: dσ 2π Z 2 α 2 1 = ∗2 . (1.343) 2me Since q0 = T . (1.511 MeV (1 + 0.12. (1. we can easily transform Eq. me2 = me2 + 2me q0 + q2 .89).341) where βe∗ and βπ∗ are the velocities in the centre-of-mass frame. Tmax ] is therefore given by: " Tπ max dσ 2π Z 2 α 2 1 1 σ = = ∗2 K − π = K Tmax dT vrel me Tmax Tmax 2 1 )2 (197)2 MeV2 10−26 cm2 1 + mπ 2 /|p|2 2π · ( 137 MeV−1 MeV−1 = − 0.02 × 1023 mol−1 102 cm · 1. if we denote the four-momentum of the initial (final) electron by k (k ). (1. The latter can be computed from Eqs.05 × 103 = 1. thus justifying the use . q0 = .1. (1.8 × 10−27 cm2 .52): ∗ vrel = |v|(1 + βe∗ βπ∗ ). (1. Since the pion mass is much larger than the electron mass.3 Cross Section 99 through Eq. Notice that this probability comes out to be small.5 × 10−3 1 g mol−1 (1.06 g cm−3 for the mass density of liquid hydrogen. βπ∗ ≈ ≈ 2 = 0.91) to (1. it follows that −q2 k = k + q.346) where we have used the value ρ = 0.345) The probability p of such an emission across a length L = 1 m can be estimated by introducing the interaction length of Eq. (1. (1.344) dT vrel me T 2 π The total cross section for T in the range [Tmax K .342) s − mπ2 s + mπ2 mπ + me E Furthermore. we can approximate: s − mπ2 s − mπ2 me E βe∗ ≈ ≈ 1. (1.12)2 103 1. sets an intrinsic limitation to the position accuracy achievable by such detectors. see e.3).344).17 taken from Ref. 13705/2010 Problem 1. Taken from Ref. 2. The reader is recommended to study in detail that chapter to find more information on the subject.17 Probability of single collisions in which released electrons have an energy E or larger (left scale) and practical range of electrons in Ar/C H4 (P10) at NTP (dot-dashed curve. (1. By using the same beam. The neutrons are detected by a counter located at the end of the cylinder. 11. The number of countings for the two set-up is given by Eq. 5 × 106 countings are measured when the chamber is empty.14). [16]. Chap.100 1 Kinematics Fig. [4] of Eq.6 × 106 when the chamber is filled with H2 . Suggested Readings This problem is inspired from a similar exercise that can found in Ref. [4]. see Problem 2. 1. whose range scales like T 2 (see Problem 2.g. see Chap. right scale). Solution Let’s denote the detection efficiency of the detector by ε. Bando n.346) to estimate the probability of interaction (otherwise we should have taken the cumulative of the exponential distribution. (1.290). Discussion According to Eq. Estimate the neutron-proton cross section and its statistical uncertainty. The emission of energetic electrons. Fig. This property is relevant in gaseous detectors which detect the passage of a particle by the ionisation trail left behind by the passage of a charged particle. namely: . and 4. the probability of emitting an electron of energy equal or larger than T follows an approximate T −1 law. 1.63 A neutron beam passes through a chamber of length d = 1 m which can be either empty or filled with hydrogen at 20 ◦ C and 760 mmHg. (1. (1. Estimate: • the energy of the pion beam (assumed monochromatic) that has generated the neutrino beam. the unknown efficiency ε cancels. giving: N1 N2 − N1 1 = 1 − 2 n d σ.1. The factor of two in front of the gas density accounts for the fact that each hydrogen molecule contains two protons. from which: N P NA 760 · 133 Pa · 6. 1N/R3/SUB/2005 Problem 1. see Sect. By standard error propagation.350) σ N1 N2 − N1 hence: σ = (16 ± 1) barn.314 J mol−1 K−1 · 293 K (1.3 Cross Section 101 N1 = N0 ε (1 − 2 n d σ ) (1. V RT 8.02 × 1023 mol−1 n= = = = 2. 4. (1. The molecular density n can be obtained from the law of ideal gases P V = N R T . • the order-of-magnitude for the neutrino-nucleon cross section. • the ratio between the cross section on protons and on electrons.348) Taking the ratio between the two countings. σ = = N2 N1 2nd 5. see Problem 4. and σ is the total neutron-proton cross section.11. . N12 N1 N1 N1 N2 ! √ δσ N2 N1 + N2 = = 0.0 × 106 − 4.5 × 1019 cm−3 · 102 cm The event counts are uncorrelated and large enough so that the Gaussian statistics applies.8 × 10−2 . Bando n.347) N2 = N0 ε where N0 is the total number of neutrons entering the chamber. it follows that: 2 2 N2 2 1 N2 1 1 (δσ ) ∝ 2 (δN1 ) + 2 (δN2 ) = 2 + .349) 5.0 × 106 2 · 2. • the divergence of the neutrino beam at the far-end detector located at a distance d = 100 km downstream of the beampipe.2.64 A neutrino beam of mean energy Eν = 20 GeV is produced from the decay of charged pions.6 × 106 1 = = 16 barn. with N = N/NA . n is the molecular density of the gas when the chamber is filled.5 × 1019 cm−3 . • the mean free path of the neutrinos in a detector with the density of water. However. we know that the energy of the neutrino in the laboratory frame is uniformly distributed in the range Eν∗ (γ ± γ 2 − 1). mπ 2mπ 1 − (mμ /mπ )2 (1.109). where Eν∗ = (mπ2 − mμ2 )/2mπ is the centre-of-mass energy of the neutrino in the most probable decay π + → μ+ νμ . the polar angle in the laboratory frame can range up to π radians. for large boosts.351) For a massless neutrino.19 and (1. the probability of decaying at an angle θ .102 1 Kinematics Solution From Problem 1. and γ = Eπ /mπ . The mean energy is therefore: Eπ mπ2 − mμ2 2 Eν Eν = γ Eν∗ = ⇒ Eπ = = 96 GeV. Since γ ≈ 670 . 1/γ is negligible. we can make use of Eqs.82). (1. The charged-current (CC) neutrino cross section on an isoscalar target in the deep inelastic scattering (DIS) regime.56 × 10−38 cm2 Q+ . (1. σθ ≈ 4. Problem 5. (1. and M indicates the average nucleon mass. θ90% ≈ 4. From DIS experiments. we get a beam spread of about 450 m along the transverse coordinate. See e.5.291).g. Thus. Taking M = 0. one measures Q̄ Q ≈ 0.938 GeV. (1. we get: (1. (1. see e. and using Eq. σν ū = 4π 2 4π 2 4 1 G 2F Q̄ ⇒ σνN = σνp + σνn = M Eν Q + .353) 2 π 3 4 4 where Q = dx x q(x) (Q̄ = dx x q̄(x)) is the average momentum carried by quarks (antiquarks) inside the proton.938 GeV · Eν Q + π 3 Eν Q̄ ≈ 1.5 × 10−3 .3 × 10−3 . (1. For anti-neutrinos. appropriate for this value of the neutrino energy. [4] for more informations. we get: .76) to obtain: θ ≈ 2. giving: G 2F s G 2F s (1 + cos θ )2 σνd = . can be computed by using the effective Fermi Lagrangian of Eq.9 × 10−3 . (2.20.9) to convert the result into SI. 1.73) and (1. Ref. one needs to swap Q ↔ Q̄.166 GeV−2 )2 Q̄ σνN = · 0.352) Taking the 90% quantile as an estimator of the beam divergence of the far-end detector located at d = 100 km.354) GeV 3 The mean free-path for neutrinos in a medium of density ρ = 1 g cm−3 is given by Eq.g. Appendix 1 We report here a simple computer program in Python which implements Newton’s method for finding the roots of a real-valued function f .1. [4]. When this happens. Δxi /xi−1 < ε.56 × 10−38 cm2 · 20 · 0. The loop stops when the desired accuracy is attained. or the maximum number of iterations is exceeded.e. β1∗ . 49 of Ref.355) The same effective theory predicts the total cross section of the CC interaction νμ e− → μ− νe to be as the first of Eq. the roots of f are searched for by iteratively incrementing the variable x as: f (xi−1 ) xi−1 → xi = xi−1 − f (xi−1 ) starting from an initial value x0 . (1. one can try tuning the starting value x0 until the convergence is attained. A critical point of such a method applied to the case of interest arises from the fact that x ∈ [−1.02 × 1023 mol−1 ≈ 2 × 10 m. (1. 8 of Ref. specialised here to the case where f is the first derivative of tan(φ) in Eq. i.67) with respect to x ≡ cos θ ∗ .353): G 2F s 2 G 2F σνe 2 me σνe = = me Eν ⇒ = ≈ 2 × 10−3 . After initialising the program with the values of β. . and β2∗ . while the intermediate values xi may occasionally fall outside of this range. (1.356) π π σνN M Q + Q̄/3 Suggested Readings For more details on the interaction of high-energy neutrino with matter. A compendium of useful formulas can be also found in Sect.3 Cross Section 103 A 18 g mol−1 λ= = σ ρ NA 1. 14 (1. 1]. [12].5 · 1 g cm−3 · 6. the reader is addressed to Chap. c = b0*(b1-b2)/(b1*b2) self.5) result = newton.d = (b0*b0/(b1*b2)-1+b0*b0) self. b=self.1. it_max=10. b0.py newton = Newton(b0=0. This routine profits from a number of built-in functions available in ROOT that implement a good deal of four-vectors algebra.0. val0 = a/math. b2=0.2) return val # The second derivative of tan(phi) [=f’ in Newton’s method] def f_second(self.a. res=0.a = b0/math. .pow(1-x*x.iterate(x_start=0. . val = a*(b*x*x*x-(d+2*b)*x-c)/math.b. n_it) # Run from command line: $ python newton. x_start=0.a.01): x_min = x_start n_it = 0 for it in xrange(it_max): n_it += 1 if abs(x_min)>1: self.sqrt(1-x*x)/math.pow(b*x*x*x + c*x +d. result[0] Appendix 2 The computer program below illustrates the generation of toy MC events where an unpolarised resonance of mass M decays into a pair of massless particles. b1.0.f_second(x_min) delta = -f0/f1 x_min += delta if(abs(delta)<res): self. x): # need to define a=self.01) print"Result:".. 0=OK. it_max=10.b = -b0*b0 self..sqrt(1-b0*b0)*(b1 + b2)/(b1*b2) self..b. x): # need to define a=self. .104 1 Kinematics import math class Newton: def __init__(self. b2): self.8.err = -1 #-1=NOT CONVERGED. 1=ERROR # The first derivative of tan(phi) [=f in Newton’s method] def f_prime(self.err = 1 break f0 = self. b1=0.(d+2*b))*(1-x*x)*(b*x*x + c*x + d) val2 = (b*x*x*x-(d+2*b)*x-c)*(-5*b*x*x*x-3*c*x*x-(d-2*b)*x+2*c) return val0*(val1-val2) # Do up˜to it_max iterations starting from x_start until accuracy<res def iterate(self.err = 0 break return (x_min.. res=0.3.5)/math.pow(b*x*x+c*x+d.3) val1 = (3*b*x*x .f_prime(x_min) f1 = self. b=self. 1. 0.936 6. M.3 Cross Section 105 import ROOT import math # the random number generator ran = ROOT.D.cos(phi)) v2_cm = -v1_cm # the 4-vectors in the centre-of-mass frame p1_cm = ROOT. mT.pi) # the 3-momentum in the centre-of-mass frame v1_cm = ROOT.. Phys.ch/record/2053086 8.Pz() mT = math.sqrt(mT2) print "Result:". Rev. pT_1 References 1. https://doi. H. 100001 (2016) 5.1738 7.pi. E_cm) p2_cm = ROOT.Pz()) pT_2 = math. https://cds. +math.1].V.SetPy(E_cm*math.1) phi = ran. Advanced Book Program (Westview Press.TVector3() v1_cm. Phys. Classical Electrodynamics.85. 936 (1952). 1995) 3. K. phi in [-pi. Boulder. https://doi.sqrt(p1_cm.Py()-2*p1_cm.sin(phi)) v1_cm. beta = boost vector\index{Boost} def toys(M=80. Jackson.cern.pi] cos = ran. CMS Collaboration. Schroeder.Boost(beta) # compute mT and pT pT_1 = math.sqrt(1-cos*cos)*math.Py()*p2_cm. https://doi. New York... C. D 88(5). Chin. Rev. Pisa.sqrt(p2_cm. CMS-PAS-TOP-15-002 (2015). D. 85. 1999) 2. C 40. Di Giacomo.. 057701 .Pz()*p2_cm.Uniform(-1. 1738 (1983). ntoys = number of MC\index{Monte Carlo} events.Py() + p1_cm.sqrt(1-cos*cos)*math. Patrignani et al.SetPz(E_cm*math..1)): # this is the same for all events E_cm = M/2.TRandom3() # M = resonance\index{Resonance} mass.1103/PhysRev. 1992) 4.Pz()) mT2 = 2*pT_1*pT_2-2*p1_cm. J..TLorentzVector(v1_cm.TLorentzVector(v2_cm. Phys.1103/PhysRevLett. 50.Pz()*p2_cm. An Introduction to Quantum Field Theory. 0.Py()*p1_cm.SetPx(E_cm*cos) v1_cm.org/10. ntoys=1000. Lett. for ntoy in xrange(ntoys): # cos(theta*) is uniform in [-1. Agashe et al.L.Uniform(-math.Boost(beta) p2_cm.Py()*p2_cm.1103/PhysRevD. J.88.TVector3(0.org/10. Particle data group. Andreson et al. E_cm) # apply a boost to the lab frame p1_cm.50.Pz()*p1_cm. beta=ROOT. 057701 (2013). Phys. Lezioni di Fisica teorica (Edizioni ETS. 3rd edn. Smith et al.org/10. (Wiley. Peskin.Py() + p2_cm. A. Rev.. M. Y.R. Ellis. I.-S. 2nd edn. Goldhaber. Phys.1016/j. R. https://doi. Cambridge. 04. (Cambridge University Press. Cambridge. 03. 519–529 (1989). B 768.1007/ JHEP09(2016)001 27.org/10.org/10. CMS Collaboration.H.org/10. 02. B. Francsechini et al. Introduction to Hig Energy Physics. https://doi. O. CMS Collaboration. the OPAL Col- laboration.010 28. W.. Techniques for Nuclear and Particle Physics Experiments. L. 117. (Springer. https://doi. High Energy Phys. https://doi.100. Eur.10.org/10. Durbin et al.1007/ JHEP04(2014)084 21. https://doi.org/10. 637686 (2010).1016/0370- 2693(84)91938-5 14. Cahn. Phys. 646 (1951). 4th edn. 57–80 (2017). High Energy Phys. NA7 Collaboration.1016/j. Radium 7.1007/ JHEP04(2015)124 10. 46. Crussard. Knowles. 1 (2016). G. B 751. Phys. Phys.physrep. Rept.815 20.R.1103/PhysRev.46. the L3 Collaboration.physletb. 09. 618–620 (1944).. CMS Collaboration.G. Phys. 2005. https://doi. Phys. High Energy Phys. Chamberlain et al. Oxford.org/10. The Experimental Foundations of Particle Physics. 816 (1974). 2000) 17.org/10.83. the SLD electroweak. 2002) 19. Phys.2017. Webber. https://doi.947 22.1140/epjc/s10052-010-1314- 6 16. G.. J. Lett.117. Tsai. 051802 (2016). 2nd edn.1016/0370- 2693(89)90704-1 25. Perkins.646 23.physletb.org/10. 947 (1955).1103/ PhysRevLett. 257–454 (2006).051802 26. 65–69 (1946) 12. Quantum Chromodynamics: High Energy Exper- iments and Theory (Oxford University Press. Leprince-Ringuet.. Phys. Rev. the LEP Electroweak Working Group. F. J.1016/j. Lett. QCD and Collider Physics (Cambridge Monographs. 2015. 04. 124 (2015). 1993) 11. J. the SLD Collaboration. 454–458 (1984). Lett. the DELPHI Collaboration. Cambridge.1103/RevModPhys. Lett. Lett. The ALEPH Collaboration. https://doi. J. Phys. Rend. Rev. https://doi. Stirling. High Energy Phys. R. W. Berlin.1103/PhysRev. 100. https://doi.org/10. https://doi.org/10. D. Mod. https://doi. heavy flavour groups. J. 144 (2016).004 15.K. 1989) 13. 219. (Cam- bridge University Press. 084 (2014). J. 83.1007/ JHEP(2016)144 . Schmelling.org/10. B 138. The ALEPH Collaboration. Dissertori. Leo.org/10. https://doi. R.006 24.12. Salam. J. CMS Collaboration. R.org/10. 143–163 (2015). 1996) 18. B 231. C 67.J. Phys.106 1 Kinematics 9. G. https://doi. Rev. Phys.org/10. Bishara et al. ATLAS Collaboration. 427. Compt. Rev. 1) dx A m e c2 β 2 I2 Z © Springer International Publishing AG 2018 107 L. 2. multiple scattering. The third section is dedicated to the functioning of particle detectors. Depending on the particle type. bremsstrahlung. charged-current neutrino interactions.1 Passage of Particles Through Matter The kinematics of a particle moving through matter is affected by the interaction with the medium.Chapter 2 Particle Detectors Abstract The subject of the second chapter is the interaction of particles with matter. Particular emphasis is given to the concept of energy loss and range in matter. Compton scattering. In the 10−1 βγ 103 regime. the rate of energy loss per unit of traversed length. or linear stopping power. When the interaction is elastic. (2. on its energy. The first section discusses the mechanism by which various types of particles interact with different media.org/10. which can be traced back to one or multiple incoherent collisions with the scattering centres. and on the properties of the medium. This is the case of the energy loss by electron collision. and can also give rise to new forms of radiation or leave behind excited states. d E/d x. UNITEXT for Physics. https://doi. This is for example the case of photon conversion. Selected Exercises in Particle and Nuclear Physics. The formula describing the average rate of energy loss. Bianchini. Energy Loss by Collision Moderately relativistic charged particles lose energy mostly by the interaction with the electromagnetic field of atoms (electron collision). The second section focuses on the experimental techniques for particle identification. is called Bethe formula and is given by: dE Z α 2 (c)2 z 2 2 m e c2 γ 2 β 2 Wmax C(I. the particle transfers to the medium part of its energy or momentum at each collision.1007/978-3-319-70494-4_2 . β) − = 2π NA ρ ln − β 2 − δ(β) − 2 . one mechanism usually dominates over the others. or to coherent effects that involve the medium as a whole. neutron capture. depends almost exclusively on the particle velocity β and on the properties of the medium. Inelastic reactions absorb or transmute the particle into something else. in units of MeV g−1 cm2 rather than in MeV cm−1 . Z mass density (g cm−3 ). when dealing with particles of sufficiently large initial energy. and neglecting both shell and density corrections. atomic weight (g mol−1 ). and gamma-factor of the incident particle Wmax maximum energy transfer in a binary collision (see Problem 1. Using the approximation m e /M 1 in Eq.2.1). The functional form features a fast rise as β approaches 0 due to the β −2 factor. The units of d E/d x deserve a few more words. or γ between about 3 and 4. β. and then rises logarithmically with γ .e. (1. A. velocity. Since Z /A is quite uniform across different materials.108 2 Particle Detectors with: NA . It is quite common to express the energy loss as a mass stopping power. (2. δ. the energy loss per unit of surphace density is less dependent on the medium. and atomic number of the material z.9 eV. (2.02 × 1023 mol−1 ) and fine structure constant (α ≈ 1/137) ρ.5 g−1 mol. and given that the term within square brackets is slowly varying with γ between 10 and 15.3) dx g cm−3 β 2 Figure 2. C density and shell corrections factors. given by the approximate formula I ≈ 16 · Z 0. (2. Z NA ρ/A. we can arrive an approximate formula: dE Z z2 2 m e c2 γ 2 β 2 − ≈ 0. As shown by Eq. Particle sitting on the minimum and on the plateau of their d E/d x curve are characterised by a rather uniform and close-to-minal energy loss.307 MeV mol−1 cm2 ρ ln − β 2 .e. This is motivated by the fact that the energy loss by collision is proportional to the density of scattering centers. see Problem 2. it approaches a global minimum at around β ≈ 0.2) for Z /A = 0. i. i. for a fixed medium the energy loss by collision depends only on the particle velocity β and on its charge z.5 g−1 mol and for two extreme values of the ionisation potential I . one can often use an average value: 2 dE ρ z − ≈ 1. α Avogadro number (6.5 ÷ 2. (2.1 shows the function at the right-hand side of Eq.2) dx A β2 I Considering that for most of the elements and their compounds Z /A ≈ 0.97. the reader is addressed to more advanced textbooks on the topic.0 MeV/cm.141) for the maximum energy transfer.26 for its derivation) I mean excitation potential of the material. γ electric charge in units of e.94 ÷ 0. and for this reason they are said to be minimum ionising particles (MIP). For their parametrisation. The num- . which are however relevant for large γ . The relative difference between Eqs. it turns out that the stopping power for heavy ions and electrons and positrons with the same velocity β are rather similar.8) for a few illustrative values of γ is reported in Table 2.3) to be d Ne 1 dE ≈ . however.1 The approximate Bethe formula of Eq. to account for the smaller mass and for the identity of the incident electron with the electrons that it ionizes. and add a number of extra β-dependent terms inside the square brackets of Eq. [1] or Sect.4 of Ref.1) holds.4) dx I dx For example. indeed they are consistent with each other to within about 15% up to γ factors of about 100.2.4 of Ref.2) as a function of the γ factor of the incident particle and for two extreme values of the mean excitation potential I .1). a unit-charge MIP produces about 105 electrons/cm.1) and (2. (2.g. (2. (2. (2. the mass stopping power is in the ballpark of 2 MeV g−1 cm2 ber of electrons extracted from their orbitals per unit length by the interaction with a MIP can be crudely estimated from Eq. Sect. [2]. for a typical ionisation potential I = 20 eV and a water-like mass density. For electrons and positrons moving inside matter. For values of γ above the argmin of the function.5) 2 I2 γ2 γ2 8 γ Numerically. In particular. 33. a formula similar to Eq. The global minimum at around γ = 3 ÷ 4 and the logarithmic growth are evident from the curves. (2.1. 2. see e. one needs to replace Wmax by m e c2 (γ − 2)/2 and 2 m e c2 by m e c2 in the argument of the logarithm. giving: dE 0. . 2. (2. after which energy loss by radiation prevails anyway.1 Passage of Particles Through Matter 109 Fig.307 MeV mol−1 cm2 Z z2 − = ρ × dx 2 A β2 m e c2 γ 2 β 2 (m e c2 (γ − 1)) 1 2γ − 1 1 γ −1 2 × ln + − ln 2 + − δ(β) (2. A few modifications have to be introduced. or a scintillation photon. One strength of these materials relies on their small band gap energy. Because of such property. for which the fluorescence originates from the band structure of the crystal (possibly activated by the introduction of suitable inpurities).g.995 9. Semiconductor materials are also largely employed in experiments. these materials are called scintillators. For example. .2 shows the mean energy loss necessary to produce one signal carrier.3% (17. As shown in the table. plastic. or from electron-ion or ion-ion recombination. Through appropriate doping and polarisation of the semiconductor.3% (17.140 11. a few eV infact. an electron-hole pair.g.417 10.968 8.110 2 Particle Detectors Table 2. Scintillators can be classified into two families: organic.866 8. for which the scintillation mechanism relies on the fluorescence of organic molecules (e.1 0. The total light output per unit length is approximately proportional to the stopping power. to be finally collected for signal generation. depending on the excitation mechanism. a property which can also allow one to measure the total particle energy for fully absorbed particles.01 0.2%) 100 0.5%) 4 0. and the emitted radiation is called scintillation light. Table 2. Some of the most popular scintillators materials in HEP are actually characterised by relatively low light yield. liquid noble-gas detectors. and inorganic. etc. the largest signal yields are provided by semiconductors. [1]. this is the working principle of gaseous detectors like proportional chambers.1%) 1.3%) 2 0. A key property of the scintillation mechanism is that the medium is transparent to its own light over distances large enough that the photons can be efficiently collected. a number of electron-hole pairs are produced by the electrons being excited from the valence to the conductive band.999 12. RPC.7% (16.4%) 10 0.48% (12. these electron-hole pairs can drift across the medium without significant recombination. that subsequently decay by emitting photons of characteristic wavelength (fluorescence).06% (12. 7 of Ref. A broader overview on the field can be found in e. Chap. Other materials have the property of converting a fraction of the energy lost by a moving charged particle in the form of molecular or electronic excitation of long- lived states.10% (11.1 Relative difference between the energy loss by collision d E/d x for ions and electrons at the same velocity and for a value of the mean excitation potential I = 20 (800) eV. The density correction δ is neglected d E ion /d x−d E e /d x γ β d E ion /d x 1. yielding a large number of signal carriers. organic crystals). which can be either a ion-electron pair.5%) The ionisation charge and the residual atomic excitation produced by the passage of a charged particle can be detected through various methods and thus yield a measurement of the particle position or energy. followed by the best scintillators and by ionisation in noble gases. drift and streamer tubes. When a charged particle moves inside a semiconductor. which in the standard theory is given by: 2 Es 1 4π Θs2 = . [2]. However. The quantity that char- acterises multiple scattering through small angles is the mean square angle per unit length Θs2 . The values are taken from Ref. the mean energy is defined as the inverse of the light yield (LY) in [γ /MeV]. with E s = m e c2 = 21 MeV. there is also some finite probability that the scattering occurs at large angles. Their collective effect it to randomise the direction of the incoming particle with no significant energy loss. (2.62 for how to estimate such a probability).58). they should be considered more as an order-of-magnitude estimate. Considering their projections onto a plane.d. see Problem 2.2 Mean energy loss necessary to produce one signal carrier. The probability of multiple scattering through small angles is large because of the sin−4 θ/2 dependence of the Rutherford cross section (see Problem 1. or δ-ray (see Problem 1. the reader is addressed to the technical literature Material Excitation Mean excitation energy ε [eV] Ge (77 K) Electron-hole 3. For more precise values.0 Si Electron-hole 3.1 Passage of Particles Through Matter 111 Table 2. γ 300 PbWO Scintill. listed in increasing order. a quantity that is commonly used to quantify the brightness of the scintillator. More informations on the subject can be found in Ref. γ 5000 Multiple Scattering Multiple scattering (MS) through small angles refers to the ensemble of incoherent elastic collisions against the nuclear fields that charged particles undergo when crossing a piece of material. with subsequent emission of a knocked-out electron. [1.2. γ 100 BGO Scintill. 2] and.6 Cs I (Tl) Scintill.f: . the displacement y and angle θ y are described by the joint p. γ 22 Xe Electron-ion 22 Isobutane Ionisation 23 Ar Electron-ion 26 CO2 Ionisation 33 LISO (Ce) Scintill. The effect of MS inside a medium of length L and radiation length X 0 is to randomise the position and direction of a charged particle at the exit of the medium. γ 35 He Electron-ion 41 Plastic Scintill.34. since they usually depend on the ambiental conditions.6) βc|p| X0 α Notice that the quantity E s is the same that enters the definition of the Molier radius for the lateral width of an electromagnetic shower. γ 12 Na I (Tl) Scintill. For scintillators. θ y | L) = exp − 2 θ y − 2 + 2 (2.112 2 Particle Detectors √ 2 3 1 4 3 y θy 3 y2 p(y.7) π Θs2 L 2 Θs L L From Eq. one can easily compute the standard deviation of θ y and y.7). (2. and their correlation: . Θ 2L 14.8) 2 βc|p| X0 .8 MeV L θ y2 = z s =z (2. 54 MeV L y = θ y √ = z 2 2 L 3 βc|p| X0 √ y θ y 3 = . (2.8) to the well-known formula: . (2. L 8.9) θ y y 2 2 2 A more accurate treatment of MS modifies the first of Eq. (2.12) dx X0 where X 0 . 0. In units of mass per unit area. An approximate parametrisation for the critical energy for electrons and positrons is provided by the formula 800 MeV Ec = . the radiation length is provided by the approximate expression: (m e c2 )2 A X0 = 4 Z (Z + 1) NA α 3 ( c)2 ln(183Z −1/3 ) − f (Z ) 716 A g cm−2 A ≈ √ ≈ 180 2 g cm−2 .0136 GeV L L θ y = z 2 1 + 0.038 ln .11) (Z + 1. the energy loss per unit length is approximately proportional to the energy itself: dE E − = . (2. is approximately independent of E.10) βc|p| X0 X0 See Ref. [2] for further informations.13) Z (Z + 1) ln(287 Z ) Z .2) In the bremsstrahlung-dominated regime. (2. (2. Energy Loss by Bremstrahlung For energies above a material-dependent threshold known as critical energy (E c ). energy loss by radiation in the electromagnetic field of the atoms (bremsstrahlung) prevails. called radiation length. (2.2.8 BGO 8.14) dx c where θc is the angle of the shock-wave direction with respect to the particle direction. see e. 44 42 H2 O 36 36 Si 21.0 1.5 2.8 1.g. the latter through Z incoherent scatterings of strength e (hence the term ∼Z ). photodetectors have a limited range of sensitivity which depends of the quantum efficiency of the photocathode. The energy loss per unit length is given by: dE α − = z2 dω ω sin2 θc (ω).3 Radiation lengths for some materials that can be commonly found in particle physics experiments. Notice that the radiation length X 0 is proportional to the mass squared of the charged ion (m e in Eq. which satisfies the relation: 1 cos θc = .9 9. The next-to-lightest charged particle is the muon with a mass nearly 200 times larger than m e .4 NaI 9. (2.6 Fe 13. The energy loss by radiation is the dominant mechanism of energy degradation for ultra-relativistic charged particles. we can integrate Eq. From Ref. [1] Material X 0 [g/cm2 ] X 0 [cm] Air 36 300 × 102 Scintill. The last of Eq. The threshold at which energy loss by radiation starts to be comparable to energy loss by collision is therefore much higher. the particle emits energy in the form of Cherenkov radiation of wavelength λ = 2π c/ω. [1]. listed by decreasing order or X 0 [cm].15) βn Although the Cherenkov spectrum is continuous. More informations on f (Z ) can be found in dedicated textbook. Ref.13) over the relevant spectrum to yield: .1 Pb 6.1 Passage of Particles Through Matter 113 Table 2. Notice that both the nucleus and the atomic electrons contribute to this O(α 3 ) process: the former through a charge Z e (hence the term ∼Z 2 ).56 where A is the mass number in units of g mol−1 . (2. Energy Loss by Coherent Radiation: Cherenkov and Transition Radiation If β > 1/n(ω). The radiation length for a few representative materials commonly found in particle physics experiments are reported in Table 2.3. In order to estimate the number of photons to which the detector will be sensitive.13) is a further approximation that helps remembering the order-of-magnitude of X 0 and its dependence on the atomic and mass number. (2.4 0. (2. n(ω) being the refraction index of the medium at the frequency ω.13)). 114 2 Particle Detectors d Nγ 2π α z 2 d Nγ λ2 − λ1 = sin2 θc (λ) ⇒ ≈ 2π αz 2 sin2 θc (2.e. (2.) output. which is of order 100 cm−1 for realistic photodetectors sensitive in the visible-UV range: practical counters in experiments feature values of the quality factor ranging between 30 and 180 cm−1 [2]. C O2 .e. In the first case. When coupled to a photodetector. (2. which is appropriate if n is slowing varying. (2.7). a simple model based on a collection of damped electronic oscillators with resonant frequency ωk and damping constant νk would give [3]: fk n(ω) = 1 + 2π re c2 N . with a photodetector sensitive in the range 300 to 500 nm. or silica aerogel. f (0.16) d x dλ λ2 dx λ1 λ2 Nγ 1. see Eq. like He. L (λ̂/400 nm) λ̂ (2.20) k ωk2 − ω2− iνk ω . which trigger the passage of a particle with velocity above the Cherenkov threshold. this gives Nγ 500 photons/cm for a particle with z = 1. which is approximately proportional to the initial particle energy. k) is the forward scattering amplitude and k is the wave-number vector.17) √ where λ̂ = λ1 λ2 and the mean value of sin2 θc is used. For example. see Problem 2. the employment of threshold Cherenkov detectors for particle identification becomes problematic since the index of refraction needs to approach one.16 for z = 1 can be written as: Np. Indeed. k) n =1+ N. For highly energetic particles with β ≈ 1. radiators with very low density. the refraction index for a homogeneous medium depends on the density according to the relation: 2π f (0. = L N0 sin2 θc (2. to be compared with the about 105 electrons/cm electrons released by a MIP from collision loss. Equation 2. the geometric and quantum efficiency of the photocathode further reduce the photo-electrons (p. Detectors based on the detection of Cherenkov radiation can be used for measuring the total energy of the crossing particle as well as for particle identification. For example. which are instead designed to exploit the angle of emission of individual Cherenkov photons. are commonly used.18) where N0 is the so-called Cherenkov detector quality factor.15 × 103 2 Δλ ≈ z sin2 θc .19) |k|2 where N is the density of scattering centres. For the second purpose. one should distinguish between threshold detectors. To this purpose.3. and imaging detectors. one exploits the proportionality between the collected light yield and the range of the particle. For example.4 Refraction index and γ threshold for various radiators commonly used in Cherenkov detectors. [2] for the index of refraction of some popular radiators. if a threshold Cherenkov is used for particle identification in a beam of fixed momentum p.7 × 10−4 43 CO2 (NTP) 4.1 ÷ 8. the light output becomes small as for Eq.4 and 6. though.2. In the β → 1 regime. say β2 . the refraction index can be set to the inverse velocity of the slowest particle.33 1.46 ÷ 0.21) β12 β1 2 |p| + m 21 which decreases like the square of the beam momentum.13 2. and then: β22 1 − m 21 /|p|2 − 1 + m 22 /|p|2 m 2 − m 21 sin2 θc = 1 − = = 22 . (2.1 Passage of Particles Through Matter 115 Table 2.3 × 10−5 123 Air (NTP) 2.5 H2 O 0. When a relativistic charged particle crosses the boundary between vacuum and a medium.2 Silica aerogel 0.3 × 10−4 34 C5 H12 (NTP) 1.75 1. The total energy radiated depends linearly on the γ factor of the particle according to the formula: . The values refer to wavelengths in the visible domain Material n−1 γ He (NTP) 3.16).52 Glass 0.1 of Ref. a coherent radiation is emitted in the forward region θ ∼ 1/γ .7 × 10−3 17.007 ÷ 0.37 See Tables 2.22 ÷ 1. (2. Rayleigh and Compton scattering.07 z 2 eV γ .1 γ < ω/ωp < γ . and pair-production.22) 3 g cm−3 A √ where ωp = 4π n e /m e is the plasma frequency of the medium [3]. one mechanism at the time usually dominates over the others. so that more energetic particles give rise to a harder spectrum. ω p ρ Z I = α z2 γ = 0. as it is usually done in the so-called transition radiation detectors (TRD). the total yield for particles with large γ . The latter find applications as tracking devices with built-in particle-identification capability. Depending on the material and on the photon energy. Interaction of Photons with Matter Photons interact with matter by three mechanisms: photoelectric effect. like GeV-electrons can be enhanced by interleaving several layers of medium. The . [2]. In terms of emitted photons. (2. the spectrum is concentrated in the region 0. More informations on the subject can be found in Ref. Although the energy emitted per each crossing is rather small. 2 k 2 σComp = Z σKN ≈ Z r (2. thus leaving the atom in an excited state.48 for the kinematics of this reaction. with β = 4 ÷ 5. and E K ≈ 100 keV for lead (measured value 88 keV). [9] for a compendium of measured values.25. Above the K -threshold. the cross section changes with the atomic number of the medium.25) 3 e (1 + 2 k)2 with 8π /3re2 = 0.23) where hν is the photon energy and Be is the electron binding energy. the so-called K -shell. see e.116 2 Particle Detectors photoelectric effect consists in the absorption of the photon by an atom. Conversely. after which pair-production becomes dominant.25) gives the Thomas cross section for Z free electrons. As the atomic electrons are bound in discrete states. but modifies its energy and direction. (2. For k ≡ E/m e 2.665 barn. or Compton scattering). The K -shell threshold for high-Z elements can be crudely estimated by using the energy levels formula for the hydrogen atom: 1 2 2 E(n) = − α Z m e c2 . the photoelectric and Compton scattering cross sections become of comparable size. (2. or Rayleigh scattering) or kick-out the electron (incoherent. At lower energies. (2. The scattering can either leave the atom in the ground state (coherent. For energies above the innermost level. photon scattering against the atomic electrons does not destroy the photon. while the k-dependent term reduces the cross section for increasing photon energies. At low energy. photoelectric effect prevails. The differential cross section in the recoil energy of th electron T can be obtained from the Klein–Nishina formula.g. In the Compton scattering.1 keV). See Ref. Depending on the photon energy. the cross section steeply falls with energy like ∼E 7/2 . the total cross sections is approximately given by the Klein– Nishina formula for Z incoherent scattering centers [9]: 8π 2 1 + 2 k + 1. Pair-production is the conversion of a photon into e+ e− in the electromagnetic field of the atom. giving . it is roughly proportional to Z β .g. one expects E K ≈ 10 keV for metals like iron (measured value 7. Problem 1. The low-energy limit of Eq. the photoelectric cross section as a function of the photon energy features a number of thresholds corresponding to the opening of new atomic level. For MeV photons. the L and M atomic levels give rise to as many new thresholds. a fraction of the photon energy is transferred to the outgoing electron. see e.24) 2 n2 From this approximations. The latter changes mildly with energy for photon energies up to the pair-production threshold. The cross section at the K -threshold is of the order of 103 barn in lead and about 106 barn in iron. with the subsequent expulsion of an electron of energy E e = hν − Be . Problem 1. n . Neutrons The interaction between neutrons and matter depends strongly on the neutron energy. When the photon energy exceeds the e+ e− threshold. or faster in hydrogenised materials. Fast neutrons. Notice that the appearance of macroscopic properties of the medium in the cross section. (2. slowly thermalise by elastic scattering in high-Z materials. the interaction cross section varies logarithmically with the photon energy. with the production of primary hadrons (e. since they exactly cancel the same quantities inside X 0 .1 Passage of Particles Through Matter 117 T 2 T dσComp 3 σTh T 2 = 2 + E + E − (2. the differential cross section rises steeply with T up to the kinetic bound. neutrons initiate a hadronic cascade. Due to the second term within square brackets in Eq. See Ref.25.g. i.2 Z 2 mbarn.24. pions) sharing a fair fraction of the initial neutron energy. 1− 2k see also Problem 1. see Problem 1. and we have used the last formula in Eq.2.26).27) where x is the photon energy fraction transferred to the electron/positron. we get dσpair A 4 7 A = 1 − x (1 − x) ⇒ σpair = ≈ 7. dx X 0 NA 3 9 X 0 NA (2. [2] for more details. Using Tsai’s formula [4]. The latter is conveniently introduced to show that the interaction length for e+ e− production is indeed related to the radiation length by λpair = (9/7) X 0 .13) to approximate X 0 . from a few hundreds of keV to a few tens of MeV. For energies in excess of about 100 MeV. Inelastic scattering.26) dT 8 m e c2 k 2 k 1− E 2 T 2 1 − T E E k 2k with 0 ≤ T ≤ E. giving rise to a characteristic peak in the electron spectrum known as Compton peak.e. (2. like A(n. like the mass number and the Avogadro number. are fictitious. For energies below about 10 MeV. and then becomes almost independent of energy. pair-production in the nuclear and electronic fields dominates. )B. A(n. 2n . . undergo preferentially nuclear reactions. A(n.e. )B.e. 13153/2009 Problem 2. Problems Bando n. can also occur in the presence of nuclear resonances. nuclear spallation A(n. around 25 meV. i.1 eV to about 100 keV. p)B. γ )B. Epithermal neutrons.1 Give a qualitative description of how the energy loss by ionisation of a charged particle of mass m depends on the particle momentum. like radiative neutron capture A(n. α)B. and thermal neutrons. and nuclear fission. from about 0. i. 1). (2. namely: dE − = z 2 f (β) = z 2 f . it is a function of the particle velocity and charge only. To good approximation.118 2 Particle Detectors Solution The energy loss by ionisation d E/d x of a particle of mass m and charge z e is described by the Bethe formula of Eq. (|p|) (2.28) dx see also Problem 2.4. At a given value of m. the function f . (|p|) features the following qualitative behaviour: ⎧ ⎪ ⎪a |p|−2 ln |p| |p| (m/m e ) I ⎪ ⎨b |p|−2 + c |p| m f . 2. In this respect. (|p|) ∼ (2. compared to its piecewise approximation in four momentum ranges . it assumes that the electrons are at rest compared to the moving particle.2 Motivate the presence of the density and shell correction terms to the Bethe formula.u. At this point.) as a function of |p|. until the momentum reaches a few times the mass value.2 The Bethe formula d E/d x in arbitrary units (a. Bando n.29) ⎪ ⎪ c |p| ≈ 3m ÷ 4m ⎪ ⎩ c + d ln |p| |p| m In words: it first decreases as |p|−2 ln |p| at small momenta. it plateaus and increases only logarithmically with |p|. 2. which is nearly unaffected by each Fig. 2. see Fig. 5N/R3/TEC/2005 Problem 2. Discussion The Bethe formula describes the energy loss of a charged particle due to the elastic collisions with the atomic electrons. (2. [2] for further details on this topic.1 Passage of Particles Through Matter 119 individual binary collision. as shown in Eq.31) dx β .1). see e. for sufficiently large initial energy. (2. 2. The range of a particle is the average distance it travels before losing all of its kinetic energy and thus come to a stop. In the continuous slowing-down approximation (CSDA). Suggested Readings The reader is addressed to Sect.2 of Ref. Problem 2. Solution The Bethe formula turns out to be accurate in the high. A non-relativistic version of the Bethe-Bloch equation can be indeed obtained by considering the total momentum transfer that an infinitely massive moving charge has on a free electron initially at rest and located at an impact parameter b with respect to the direction of flight. which decreases the effective volume available for electron collision. β) has to be included. and is more relevant at high-energy. [1]. The former accounts for the polarisation of the medium by the electric field of the incident particle.2 of Ref.g.and low-velocity regimes only if the density δ and shell C corrections are added. [2] for a parametrisation of δ.2.g. one can neglect the dependence of this term on the velocity β and use an approximate version of the type: dE C z2 =− 2 . then the assumption that the electrons are at rest breaks down and a correction C(I. However. As such.3 Derive an approximate expression for the range R of a charged particle of mass m and initial energy E that loses energy by collision with the atomic electrons. which is of order α. 2. (2. See e. Conversely.1). How does R depends on the initial kinetic energy in the ultra-relativistic limit E m and in the classical limit? Solution The energy loss by collision is given by the Bethe formula of Eq. 2.30) E d E/d x The analytical integration of the Bethe formula is an hard task to due to the logarithmic term. it tends to reduce the energy loss. it can be obtained by integrating the inverse linear stopping power over the full range of kinetic energy. as we have seen in the introduction Sect.e. i. Sect. (2. [1] and Chap. 33 of Ref. Ref. if the particle velocity is comparable with the electron velocity.1. The net effect is obtained by considering an ensemble of such electrons up to a maximum value of b such that the momentum transfer is above the mean ionisation energy necessary to strip the electron from its orbital.: m 1 R(E) = dE . iron. and 5000 Pb lead.0 20. Fig.0 Muon momentum (GeV/c) 0.33) underestimates the true range by a fair amount. carbon.31) in place of the full Bethe formula.0 2. C z2 E m C z2 E C z2 γ (2.0 10.7 MeV ρ/(g cm−3 ) cm−1 .0 2.1). as a function of βγ (from Ref. [2]).33) can be directly compared to Fig. This is a consequence of having neglected the logarithmic term in the stopping power.02 0. 2.3). A good numerical agreement is found with the approximate formula of Eq.5 1.3.5 1. 2.1 0.02 0.0 10. [2] 2000 R/M (g cm−2 GeV−1) H2 liquid 1000 He gas 500 200 100 50 20 10 5 2 1 0.3 Range of heavy 50000 charged particles in liquid 20000 C (bubble chamber) hydrogen.0 2 5 10. Eq.1 0.0 10. For smaller values of β.32) Hence.2 0. From Ref.0 Pion momentum (GeV/c) 0. (2. (2.33) m C z2 γ C z2 1 + (βγ )2 The second of Eq.5 1. (2.0 2 5 100. (2.0 Proton momentum (GeV/c) .0 5. (2.33) up to βγ 1.1 2 5 1. (2.1 0.0 2.120 2 Particle Detectors with C ≈ 1.2 0.0 βγ = p/Mc 0.0 5. the range is given by: E E β2 1 m2 R(E) = dE = dE 1 − 2 = m C z2 C z2 m E 1 1 1 1 (E − m)2 m (γ − 1)2 = (E − m) + m 2 − = = . see Eq.05 0.0 5. (2. which shows the range of a heavy ion in different materials as obtained from a full integration of Eq. 10000 Fe helium gas.05 0.2 0. Using Eq. we find that R/m is a function of γ = E/m: 2 R 1 (γ − 1) 1 2 1 + (βγ )2 − 1 = = .0 50. the range in air predicted by Eq.5.37) z 12 d x m2 .1). the range for an ultra-relativistic particle is proportional to its energy: mγ E Rγ 1 ≈ = .35) is about 5.32). At this point. one should remember that for β 0.34) C z2 C z2 This result comes out intuitive if one considers that a particle with γ 1 moves at the speed of light. the range is instead a quadratic function of the kinetic energy T : (E − m)2 T2 RNR ≈ = . (2. moving through the same medium.2. so that R ∼ E/(d E/d x|MIP ) ∼ E. (2. which. (2.1 Passage of Particles Through Matter 121 According to Eq. whereas the CSDA range from a full integration of the Bethe function gives about 5 cm [5]. What is the relation between the range R1 and R2 of the two particles under the same conditions? Solution The energy loss by collision is given by the Bethe formula of Eq.36) dx mi so that: d E2 |p| z 22 2 m 1 |p| (|p|) = z 22 f = z f dx m2 z 12 1 m2 m1 z 22 d E 1 m1 = |p| . can be written as: d Ei |p| (|p|) = z i2 f . For a non-relativistic particle. the stopping power steepens due to the β −2 dependence and the residual energy gets degraded in a short path. Suggested Readings A good starting point to learn more about the concept of range is Chap.4 Determine the relation between the stopping power d E/d x for two particles of masses m 1 and m 2 .32 /(2 · 10−3 · 22 · 4 × 103 ) ≈ 1 cm. the energy loss is approximately constant until the velocity drops below c. for an α particle emitted in the decay of 210 Po with T = 5. (2. (2. (2. (2.31) is not valid anymore and the resulting range is underestimated. and same momentum |p|. as a function of the particle momentum. (2. 2 of Ref.3 MeV. For example.35) C z2 m C z2 m However. Problem 2. [1]. the approximation of Eq. electric charges z 1 e and z 2 e. 122 2 Particle Detectors Owing to such scaling law. and by a scaling on the vertical axis by the ratio of the squared charges. (2.38) T d E i /d x T z i2 f E mi so that: 0 0 1 z2 0 1 z2 m 1 R2 (T ) = dE = 1 d E = 1 2 m d E.3. the stopping power d E(|p|)/d x as a function of |p| for different particle types are all related by a uniform scaling of the orizontal axis equal the mass ratio. Let’s now consider the range as defined in Problem 2. For a given kinetic energy T . the range is given by: 0 0 1 1 Ri (T ) = dE = dE . . 37). Suggested Readings For an overview on the TPC. .7 MeV cm−1 . or of its kinetic energy. or of its velocity. and β is the particle velocity. See also Ref.40) dx β where C ≈ 1. [1] for the scaling law of stopping powers and ranges. z is the ion charge in units of e. provides a tool to identify the particle type thanks to the scaling law of Eq. • What kind of interaction between the ion and the material is responsible for this energy loss? • Explain how the mass of a charged particle can be determined from the simultaneous measurement of d E/d x and of the momentum |p|. (2. (2. the energy loss per unit length of traversed material can be approximated by the formula dE C z2 =− 2 . Problem 2. the reader is encouraged to consult the PDG review on detectors for accelerators [2]. (2.39) z2 m1 m2 Discussion The simultaneous measurement of the stopping power d E/d x and of the particle momentum. The canonical example of a detector that allows for a simultaneous measurement of these quantities is the time projection chamber (TPC).5 The range R of a particle is the distance over which the particle loses all of its kinetic energy. E 2 m E 2 z2 m 1 m T T z 22 f m 2 z2 T z 12 f m 21 m 1 1 2 z 12 f mE1 z2 m2 m1 = 12 R1 T . For a heavy ion. 40) and using the fact that T = |p|2 /2m for a classical particle: 0 1 T β2 T 2T. Solution As discussed in Sect. 2.2. We can obtain the same result starting from Eq.35) obtained from the limit γ → 1 in Eq.32).1. the simultaneous measurement of the two quantities allows to measure m for different ansatz on z. A comparison of the mass values thus obtained with the spectrum of known particles allows one to identify the particle type. In order to estimate the range of a proton in water. we can use Eq. heavy ions moving in matter lose energy due to elastic collision with the atomic electrons. Since d E/d x ∼ z 2 f (β) and |p| = mβγ .1 Passage of Particles Through Matter 123 • Estimate the range R in water of a proton with T = 60 MeV. (2. (2. (2. 1 T2 R(T ) = dE = dE = dT . Since E = E(β). By tuning the initial particle energy T to attain a certain range R.41) 10 MeV · 12 · 1.9. Due to the dominant 1/β 2 behaviour at velocities below about 0. Solution The energy loss of a charged ion in matter is described by the Bethe formula (2. the energy deposition per unit length becomes increasingly more intense as the particle velocity decreases.7 MeV cm−1 to be compared with a CSDA value of 3. Bando n. the so-called Bragg peak. giving a solution d E(x)/d x. (2. the Bethe formula can be solved as an ordinary differential equation (ODE) in β. by using the approximation (2.1 cm from a full integration of the Bethe formula [5]. 13153/2009 Problem 2. yielding: d 1 z2 C dβ z2 C z2 C m β2 =− .1 cm. = = T d E/d x 0 z2 C 0 m p c2 z 2 C m p c2 z 2 C (60)2 MeV2 = 3 = 2.3) and assuming the ion to be non-relativistic.1). mβ =− 2 . Indeed. The latter features a peak at x ≈ R.6 Discuss the characteristics of the Bragg peak and its main applications. β 3 dβ = − d x. the ODE can be easily solved analytically. dx 2 β2 dx β m . the Bethe formula predicts that most of T will be infact dissipated near the end of the trajectory. β (x) = β0 1 − 2 2 x = β0 1 − .42) m m β04 R . 4 z2 C 4 z2 C x β 4 − β04 = − x. 2 (2. 43) dx β0 1 − x/R The energy ΔE deposited in the interval [λR. (2.44) dx λR The value of λ such that a fraction α of the initial energy is lost in the interval [λR.35) to define the range R of the particle. (2. A beam of particles of energy 10 GeV impinges perpendicularly to the scintillator: . reducing significantly the stopping power. when βγ 0. Bando n. R] is therefore given by λ = 1 − α 2 . and 25% in the trailing 6% of the path. R] can be easily computed from Eq. 50% of the kinetic energy T is lost in the last quarter of the particle path.4. and the resulting stopping power gets largely overestimated. such that all the scintillation light can be assumed to be detected.124 2 Particle Detectors Fig.43) has been obtained under the assumption that d E/d x ∼ β −2 . (2. Overall. 2. Furthermore. the shell corrections are relevant. For example. (2. The Bragg peak finds one major application in medical physics as a tool for curing solid tumors: the intense energy deposition in the neighbourhood of the beam range allows to burn selected tissue depths with reduced damage to the upstream tissue.43). the Bragg curve is much less peaked than predicted by Eq. (2. (2.42) we therefore get: dE z2 C 1 (x) = − 2 √ . From Eq.7 A 2 cm-thick plastic scintillator is coupled to a photomultiplier with gain G = 106 and detection threshold Q th = 1 pC.4 Sketch of a typical −dE/dx Bragg curve for protons or heavy ions moving in a dense medium x where we have used Eq. see Fig. 2. This is a poor approximation for βγ 1. and infact the maximum occurs before the full range is attained.1. and the Bethe formula ultimately breaks down. (2. A caveat: Eq. 13705/2010 Problem 2.43) to give: R dE √ ΔE(λ) = dx = T 1 − λ. e.1 Passage of Particles Through Matter 125 • Estimate the charge collected at the anode. The enhancement factor. is the called gain (G) of the photomultiplier.e. undergo a multiplicative enhancement between the photocathode and the anode. a photomultiplier is only sensitive to a fraction εC of the total light output. Because of the geometry of the medium and of the QM-nature of the photoelectric effect. A scintillator is always coupled with a photomultiplier that transforms the scintillating photons in photoelectrons (p. if the beam is made of muons. For this reason. characterised by an electronic noise Ne . After coupling the amplification stage to the read-out electronics.e. The enhanced charge is finally read-out at the anode by a chain of amplifiers which transforms it into voltage or currents. the relative energy resolution from a scintillator that produces n γ Poisson-distributed photons for a particle of energy E.1. can be parametrised as [2]: .e. the primary p.e. estimate the minimum scattering angle on protons such that the neutron can be detected.). the total output charge per initial p.2.e. such p. so that they can initiate a chain reaction that brings to the fan exponential charge multiplication. By themselves.. This can be for example achieved by accelerating them with intense electric fields. above the electronic noise. i. A key point in all this procedure is that the proportionality between the initial number of p. 2. do not usually represent an amount of charge large enough to generate a significant signal. • If the beam is made of neutrons. and the final signal amplitude is preserved.e. of which only a fraction εQ is actually converted into p.e. Discussion Scintillators have been briefly discussed in Sect. i. Solution A 10 GeV muon loses energy mostly by collision with the atomic electrons as discussed in Sect. 2 σ (E) fN Ne = + . the energy loss is given by: dE − ≈ 2.46) dx While crossing a thickness d = 2 cm.2. (2. the total energy lost by the muon is ΔE = |d E/d x| · d ≈ 4 MeV. 2.45) E n γ εQ εC Q n γ εQ εC where f N is the called excess noise factor and arised from the amplification process. the mass density is approximately ρ ≈ 1 g cm−3 . it behaves as a MIP. (2. Assuming ε = 100 eV. With this value. (2.3).1. The role of the gain factor in reducing the signal uncertainty is made clear by Eq. εC = 1. The mean excitation energy for a plastic scintillator can be found in Table 2. For a plastic scintillator. and its mean energy loss per unit length is provided by Eq.0 MeV g−1 cm2 · 1 g cm−3 = 2 MeV cm−1 . and εQ = 1. In particular. (2.45). we expect to collect an average charge at the anode of about: . (2.126 2 Particle Detectors ΔE 4 MeV Q=G· · e = 106 · · 1. more than three orders of magnitude larger than the threshold charge Q th . Indeed. If the beam is made of neutrons. It is easy to show that for very small recoil energy. their detection proceeds through the measurement of the recoil energy of protons and other nuclei that interact with the beam particles.62 keV. The threshold energy such that a recoil proton gives rise to a detectable signal is determined by the condition: Tth 10−12 C · 102 eV G· · e = Q th .e.4 nC.47) ε 100 eV i. (2.48) ε 106 · 1.6 × 10−19 C = 6. ⇒ Tth = = 0.6 × 10−19 C which is small compared to the proton mass and to the beam momentum. if we indicated the four-momenta of the initial (final) neutron and proton by p and k ( p . momentum has to be exchanged perpendicularly. and k . and the angle that the recoiling proton forms with the beam momentum as θ p . ). then: p. = p + k − k . . m 2n = m 2n + 2m 2p − 2E n m p − 2(E n E . p − |pn ||p. p | cos θ p ) − 2E . E .p m p . p (E n + m p ) − m p (E n + m p ) T p (E n + m p ) cos θ p = = ≈ |pn ||p. p | |pn ||p. p | . if |p. Tp En + m p ≈ . (2. and given that the factor within square brackets is of order one. and conservation of momentum implies that the momentum received by the extra neutron is also is also a vector perpendicular to the beam direction. the resulting angle turns out to be pretty much π/2.49) 2m p |pn | Since T m p for our case.p | m p . and the scattering angle of the neutron is therefore given by: √ |p. the neutron momentum magnitude after the scattering is almost unchanged. Since Tth E n . if the gas consists in a argon-isobuthan mixture 60%–40%? Which additional factors acting on the statistics of the produced electrons determine the standard deviation of the signal? . (2. 1N/R3/SUB/2005 Problem 2.938 · 0.50) |pn | |pn | 10 Bando n. on average.p | 2m p Tth 2 · 0.62 × 10−6 θn ≈ = = = 1.8 A MIP generates. What is the typical value of n.1 × 10−4 rad. n electron-ion pairs per cm in a gaseous detector at standard pressure. it gets lost for later multiplication. Another typical problem with gaseous detectors is the formation of avalanches in random points of the chamber created by energetic photons emitted by the accelerated electrons. Table 6. [7]. the Bethe formula predicts an energy loss per mass surphace of about 2. 2.52) In a gaseous ionisation detector. (2. This undesired effect limits the operation rate and resolution of the detector.5 MeV g−1 cm2 .4 · 23) eV 24.1 of Ref. Finally. an amount of primary ionisation electrons like in Eq. or if it gets trapped by the gas molecules to give rise to an ion. The gain (see Problem 2.5 × 104 cm3 mol−1 = 1.2.4 · 58) g mol−1 34 g mol−1 ρ= = −1 −1 = = R T /P 8.8 eV (2. [5]. the reader is addressed to Ref. see e. These effects can be limited by adding appropriate amounts of organic quenchers. The density of the gas at STP conditions can be calculated from the law od ideal gases: A (0. can limit this effect. see e. see also Fig.g.6 · 18 + 0. Taking a weighted average of the two components.4 × 10−3 g cm−3 n= = = 140 cm−1 .05 ÷ 0.314 J mol K · 298 K/10 Pa 5 2.52) is not sufficient to produce a detectable signal.51) The mean excitation energy for the two molecules can be read from Table.7).3). Indeed.1. 7].2 of Ref. it introduces an additional fluctuation in the number of signal carriers.6 · 26 + 0. which have a small ionisation potential I ≈ 12 eV. [8]. (2. If an electron-ion pair recombines before the formation of the avalanche. the primary electrons need to be accelerated by an intense electric fields until they trigger the formation of an avalanche. If the gas is made of argon and isobutane. [1] or Ref. Bando n. which for typical gases is in the range 0. Suitable amounts of electronegative gases. Table 6. we get: |d E/d x| 2.g. [1. a MIP releases an amount of energy per unit length described by Eq. (0. like isobutane. (2. Suggested Readings An introduction to the physics of electronic avalanches in gas can be found in Refs. 2. depends on the choice of the gas.5 MeV g−1 cm2 · 1. one should remember that the resolution of a gaseous ionisation √ detector that absorbs all of the particle kinetic energy scales better than 1/ n by the so-called Fano factor.1 Passage of Particles Through Matter 127 Solution On average. Noble gases are usually chosen because of their large gain factors. Since the charge-multiplication process is intrinsically random. For a more comprehensive review of gaseous detectors.4 × 10−3 g cm−3 .2. like freon. 5N/R3/TEC/2005 .20. and hence the final statistics of signal carriers. The number of electron-hole pairs produced by the passage of such a particle across a thickness d = 100 µm of silicon is therefore given by: |d E/d x| · d 3. (2. energy loss by collision with the atomic electrons prevails. γ ≈ 4. we can explicitly compute the right-hand side of Eq. Ref. The energy loss per unit length is given by the Bethe formula of Eq. while it goes like T −1 at smaller energies. Bando n.9 MeV cm−1 [5].53) which agrees well with the more accurate prediction of 3.2) for a pure silicon medium. (2.307 MeV mol cm ·2 ln −1 = dx 28.1 g mol−1 16 · 140. (2.1. • How does the energy loss by ionisation and by radiation depend on the material? • How do they depend on the electron energy? • The critical energy is defined as the energy at which the two energy losses are equal: which between a muon and an electron has the smallest critical energy? Solution The energy loss of relativistic electrons and positrons is discussed in Sect.2). The material enters mostly through its electron density n e = NA ρ Z /A and the average ionisation potential I .511 MeV · 42 − = 0. see e.9 How many electrons does a charged particle produce on average when crossing 100 µm of silicon? Solution Let us assume that the charged particle have z = 1 and that they behave like a MIP. (2.128 2 Particle Detectors Problem 2.7 MeV cm−1 . giving: dE −1 2. For energies below the critical energy E c . Assuming the particle to be in the neighborhood of the global minimum. i.2.e. For a MIP. [2]. the dependence of d E/d x on the particle energy is mainly through the logarithmic term ∼ ln γ . 1N/R3/SUB/2005 Problem 2. 2. . A residual dependence on the atomic number Z comes from the shell and density effects.6 eV where we have used the mean excitation energy for silicon as in Table 2.33 g cm3 · 14 2 · 0.10 A relativistic electron loses energy by both ionisation and by radiation when moving inside matter. For electrons with energy in excess of a few MeV. the rate of energy loss by collision is almost independent of the electron energy.9 eV = 3. The stopping power is proportional to n e and depends logarithmically on I .g.7 MeV cm−1 · 10−2 cm n eh = = ≈ 104 .54) ε 3. 13153/2009 Problem 2. (2. see Eq.55) (Z + 1. Fig. Since the energy loss by radiation is inversely proportional to m 2 . In particular.g.5 Electron and muon critical energy for the chemical elements. Bando n. since it is roughly given by the position of the intersection point between two curves in the (d E/d x. is expected to be about 4×104 times larger than for electrons. (2. 2. as shown by Eq.13).03) 0. [2] Energy loss by radiation prevails above the critical energy. Define the critical energy and explain how it depends on the atomic number Z of the material. 2. while the other (energy loss by radiation) is a straight line of slope proportional to m −2 .24 (Z + 1.1 Passage of Particles Through Matter 129 400 4000 7980 GeV 200 2000 (Z + 2.12). where m is the mass of the incident particle.13).879 100 1000 Eμc (GeV) 710 MeV ________ Ec (MeV) 610 MeV ________ Z + 0. An exact scaling does not hold however.838 400 20 Solids Gases Gases 200 Solids 10 H He Li Be B C N O Ne Fe Sn H He Li Be B C N O Ne Fe Sn 100 5 1 2 5 10 20 50 100 1 2 5 10 20 50 100 Z Z Fig. The critical energy E c is defined as the energy at which the two rates of energy loss become identical. Furthermore. it is proportional to the combination Z 2 ρ/A. From Ref.2. see e.5 taken from Ref. The material enters through the atomic density n = NA ρ A and through the atomic number Z . (2. 2. Solution Energy loss by collision and radiation are discussed in Sect.11 An electron moving in a material loses energy by a variety of mechanisms. the critical energy for muons.2) . see Eq. it is proportional to the energy itself. it follows that the critical energy must be approximately go as ∼m 2 . [2]. (2. and the critical energy E μc is a factor of about 3 smaller than the naive scaling E μc ≈ (m μ /m e )2 E c . one of which is roughly constant (energy loss by collision). while the energy loss by ionisation is independent of m for sufficiently high energies. E) plane.47) 0.1. An approximate formula for E c is given by 800 MeV Ec = .92 700 5700 GeV 50 Z + 1. E μc . According to this picture. 130 2 Particle Detectors see e. Ref. the critical energy decreases with Z . The rate of energy loss per unit length is therefore given by dE E =− .13 How much energy does an electron with initial energy of 1 GeV lose by crossing a material with thickness equal to one radiation length? Solution An energy of 1 GeV is above the critical energy E c of Eq.13): 716 A X0 ≈ √ g cm−2 . In particular.5. 2. the radiation length scales as ∼A Z −2 . it goes like Z −1 for Z 1. The electron energy as a function of the traversed length is then obtained by integrating Eq. therefore the electron loses energy mostly by radiation. (2. (2. (2.11). of which one is flat versus energy and goes like ∼Z (energy loss by collision).56) Z (Z + 1) ln(287 Z) where A is the mass number in units of g mol−1 and Z is the atomic number. Hence. [1. 18211/2016 Problem 2. Bando n.368 E 0 . (2. This can be understood by the following argument: the critical energy is roughly given by the position of the intersection point between two curves in the (d E/d x.g.57) dx X0 where X 0 is the radiation length measured.58) e . 2]. see Fig.57) to give: E0 E(x) = E 0 e−x/ X 0 ⇒ E(X 0 ) = = 0. Solution An approximate version of X 0 has been derived in Eq.12 Provide an approximate formula for the radiation length X 0 in terms of the atomic and mass numbers of the material. (2. while the other has a positive slope and goes approximately like ∼Z 2 at large values of Z (energy loss by radiation). 5N/R3/TEC/2005 Problem 2. hence the intersection point should scale as ∼Z −1 . (2. for sufficiently large values of Z . Bando n. Hence. E) plane. the two become of similar size at energies of about 10 (500) KeV. the probability of interaction per unit length is given by the interaction length of Eq. Solution Electrons lose energy mostly by radiation at high energy. Compton scattering (incoherent photon-electron scattering) becomes significant for energies above the K -threshold and below a few times 2 m e . =− ⇒ I (x) = I0 e−x/λ (2. 2. Elastic scattering can instead deflect the electron from its original trajectory and remove it from the beam. Ref.632 E 0 . and then by elastic collison with atomic electrons at lower energies. 13153/2009 Problem 2. after which pair-production dominates.2.1 Passage of Particles Through Matter 131 The energy lost in the medium is therefore ΔE = (1 − 1/e) E 0 = 0.1. the beam particles will traverse the full thickness and emerge with an energy distribution centred around a smaller value. Bando n.59) λ I λ The intensity varies exponentially with the traversed length. The cross section for pair-production is inversely proportional to the radiation length X 0 . Bando n. The transition between the photoelectric and Compton- dominated regime depends on the medium (see below). . see e. If the beam is monochromatic and the thickness d exceeds the electron range in the material.15 In which energy interval does Compton scattering dominate in the interaction of photons with matter? What kind of interaction prevails at lower and higher energies? How does it depend on the absorber? Solution The interaction of photons with matter is discussed in Sect. Let’s assume that the reaction which removes electrons from the beam is characterised by a cross section σ and let’s denote the density of scattering centres by n. The photoelectric cross section for energies in the MeV region is goes as ∼Z β with β = 4 ÷ 5. (1. [2]. The absorber type enters mostly through the atomic number Z . 13153/2009 Problem 2.g. The Compton cross section is instead proportional to the number of electrons per atomi. At low energy. the intensity at a distance x + d x is given by: dx dI dx I (x + d x) = I (x) − I (x) . the photoelectric effect (photon absorption with electron emission) is the main interaction mechanism.291).14 Determine the law by which a beam of electrons of intensity I0 gets attenuated while crossing a layer of material of thickness d. If the beam has an intensity I (x) at a depth x. 1N/R3/SUB/2005. hence it goes as ∼Z . hence it is roughly proportional to ∼Z 2 for large atomic numbers. For carbon (lead). Bando n. By definition. namely λ = 1/(nσ ). [6]. we can refer to Fig. Solution To solve this exercise. 1. To validate the extrapolation.e. A large amount of tabulated data can be found in Ref. [6. 2.132 2 Particle Detectors Suggested Readings Photon interaction in matter is discussed in a large number of textbooks. Bando n.6. 10 MeV photons on C. [2]. together with the known Z -dependence of the cross sections. .16 How does the photoelectric cross section vary as a function of the photon energy? How does it depend on the atomic number Z ? Solution The interaction of photons with matter is discussed in Sect. A 1 MeV photon is just below the pair-production threshold. 100 keV photons on Fe. Bando n. 10 MeV photons on Pb. 5. ≈ 10−3 barn.17 Determine which process dominates in the photon-matter interaction for the following reactions: 1. For energies above the innermost level (K -shell). [6] we find σComp ≈ 3 barn and σp. the cross section steeply falls with energy as ∼E −7/2 and it grows with the atomic number as ∼Z β with β = 4 ÷ 5. Aluminium has atomic number Z = 13. 2. and then use these values. Suggested Readings See Problem 2. 5N/R3/TEC/2005 Problem 2. [1] and to the PDG review [2]. Indeed.7 of Ref. the reader is addressed to Sect. 2. therefore Compton scattering has to be by far dominant at such an energy. 9]. taken from Ref. At low energy. Aluminimum must be in-between. to read the cross section values for carbon and lead. 4.15 and references therein. we can use the values tabulated in Ref. For a primer. 100 keV photons on H2 . The photoelectric cross section as a function of the photon energy features a number of edges corresponding to the opening of new atomic levels. 2. the photoelectric effect (photon absorption with electron emission) prevails.1. in order to extrapolate to other materials. 1 MeV photons on Al. from Ref. 18211/2016 Problem 2. The energy at which Compton and pair-production become similar is about 500 keV in lead and about 10 keV in carbon. 3. ≈ 10−6 barn.experimental σtot Cross section (barns/atom) σp. Since Compton scattering dominates the photon-matter interaction at this energy for carbon. [6] we find σComp ≈ 0. A 100 keV photon on hydrogen cannot undergo pair-production. 33.2. since the photoelectric cross section decreases as a function of Z much faster compared to the Compton cross section.e. showing the contributions of different processes.15 of Ref. Indeed.e. From Fig.experimental σtot 1 Mb σp. Cross section (barns/atom) σRayleigh 1 kb κ nuc σg. from Ref. A 100 keV photon on iron cannot undergo pair-production. 1 kb σRayleigh 1b κ nuc σCompton κe 10 mb (b) Lead (Z = 82) .1 Passage of Particles Through Matter 133 (a) Carbon ( Z = 6) 1 Mb .5 barn and σp. [2]. [2] 2.e. 3.r.6 Photon total cross sections as a function of energy in carbon and lead. 2. 1b σCompton κe 10 mb 10 eV 1 keV 1 MeV 1 GeV 100 GeV Photon Energy Fig. the photoelectric (Compton) cross section in lead at that energy is around .d. it will be a fortiori dominant in hydrogen. Taken from Ref. [2] we see that the critical energy for oxygen is about 900 GeV. see e. we can use the approximate formula of Eq. The stopping power for a MIP muon in water is about 2. so the same conclusions hold. Since γ = E/m = 3. the latter ranges from ≈26 at γ = 3. By which process can it be detected? Estimate the depth at which the muon arrives before decaying. As before. the logarithmic term is non-negligible. [5]. 5. so that. from Ref. hence still larger than the initial muon energy. Using the value I = 80 eV [5]. Therefore: 400 GeV R≈ = 1. Indeed from Ref. The dominant energy loss mechanism is therefore by electron collision as described by the Bethe formula of Eq. (2. although the two are still comparable.g. Indeed from Ref. 18211/2016 Problem 2. Indeed. the muon is an unstable particle with life-time .216 km from Table II-28 of Ref [10].e.24 of Ref. at very large energies. and the critical energy for muons is about 1 TeV [5]. the constant term can be approximated as 2.0 × 20/12 ≈ 3.134 2 Particle Detectors 103 barn (10 barn). 33. [6] we find σComp ≈ 4 barn and σpair ≈ 12 barn.3 MeV g−1 cm2 · 1 g cm−3 This result is in good agreement with the more accurate estimate of 1. by assuming a ∼Z 5 (∼Z ) scaling. However.8×103 down to ≈12 at γ = 4 (MIP). Referring to Fig. (2. but Compton scattering is sizable at that energy. This time. the naive scaling is only approximate. 500] nm.8 × 10−1 barn. as discussed in Problem 2.33) emits Cherenkov radiation at a rate of about 200 γ /cm in the wavelength range [300. we should expect roughly the same cross sections. it’s the former to be larger because of the ∼Z 2 scaling compared to just a ∼Z scaling of Compton scattering. see Eq.15 of Ref. (2. one sees that Compton cross section on carbon is larger than pair-production. hence far above the initial muon energy of 400 GeV.3 barn and σpair ≈ 0.3 MeV g−1 cm2 . However. Solution A muon of energy E = 400 GeV moving in water (n = 1. ≈ 20 barn. 4.8 × 103 1.60) 3.18 A muon with energy of 400 GeV penetrates vertically into the sea. From Fig. the critical energy for muons is expected to be in excess of 3 TeV.10. where C is a constant that sets the plateau level of the Bethe formula. 33.16). one should expect the pair-production and Compton cross sections to be of the same order. [6] we find σComp ≈ 0.34) to predict the range R in water (ρ = 1 g cm−3 ) to be R ≈ E/C. A 10 MeV photon can undergo pair-production. The critical energy for electrons in water is about 80 MeV. Bando n.2 km. [2]. From the ∼m 2 scaling of the critical energy with the particle mass.1). (2. Taking an intermediate value of 20. [6] we find σComp ≈ 12 barn and σp. Ref.0 MeV g−1 cm2 [5]. However. 19 An underground experiment located at a depth d = 1 km from the top of the mountain measures the momentum of cosmic muons arriving vertically from above. Eq.62) dx where a and b are slowly varying functions of energy for E 1 TeV. which can be detected through their emission of Cherenkov light by an array of PMT’s located deep into the ice. Although the muon momentum progressively changes as the muon penetrates deeper into the sea.61) 2. Estimate the muon energy at the top of the mountain if the muon momentum at the detector is |p| = 1. For example. ln = −b x.0 TeV. its mean path before decaying would be βcτ γ ≈ 2 km.94. including e+ e− pair production. The range calculation of Eq. Problem 2.63) . (2. is sensitive to the CC interaction of very-high energy neutrinos. At a velocity β = 0. Discussion The exploitation of large sea volumes as Cherenkov radiators allows one to study cosmic radiation of very high energy. and photonu- clear interaction. (2. Assuming constant values for a and b. E 0 = eb x E + E μc − E μc .60) will hold only if the muon does not decay before coming to a stop. so much larger than the residual path before stopping completely. and the residual energy is dissipated after traversing a length of about 106 MeV (3 − 1)2 = 70 cm R. (2. the IceCube neutrino observatory at the South Pole.1 Passage of Particles Through Matter 135 τ = 2. (2. bremsstrahlung.0 MeV g−1 cm2 · 1 g cm−3 3 Were the muon to conserve γ = 3. Solution Energetic muons lose energy by electron collision and by various forms of electromagnetic radiation. a+bE 1 + E 0 /E μc (2. or γ ≈ 3.62) can be solved exactly to yield the solution E 0 = E 0 (E. namely: − ddEx = a + b E E(0) = E 0 dE 1 + E/E μc − = d x. The overall stopping power can be parametrised as dE − = a(E) + b(E) E. the muon is at the minimum of the stopping power curve.2.2 × 10−6 s. This is indeed the case with high probability. x). time dilatation makes such that the muon decay probability over a fixed length in the Earth frame is significant only for small velocities. we get E μc = 0. and the muon spectrum goes approximately as E μ−2. For E min = 1 GeV.136 2 Particle Detectors where.0 TeV. E μc ≡ a/b is the energy at which energy loss by ionisation equals the energy loss by radiation. where E th is the detector threshold energy.69 − 0. see Problem 2. The neutrino-induced flux can be thus estimated to be: . Owing to the continuous slowing down and subsequent decay. Solution At a depth d larger than a few km w.9 × 10−6 g−1 cm2 from Table 29. [2]. 33..4 and Sect. a νμ of similar energy is produced. (2.19. Equation (2. namely muon production from charged-current interaction of muon neutrinos with the rock. 29. the spectrum settles to a constant value. the muon flux becomes so weak that another source of underground muons takes over.7 .e. the offset R − r is about 200 m.69 TeV and: E 0 = exp 3. by definition. In this energy regime.2 of Ref. The latter is almost independent on the depth.64) Suggested Readings For more details on cosmic muons and their interaction with matter. only if the muon interacts with the rock within a distance r = R − E th /(d E μ /d x) from the underground level d (we make the approximation E μ ≈ E ν ).9 × 10−6 g−1 cm2 · 2. For example. At some point. At this point. E ν + d E ν ]: they will contribute to the measured muon flux of energy E μ ≥ E th .20 The vertical flux of cosmic muons with E μ > 1 GeV at the sea level is about 70 m−2 s−1 sr−1 .e. the reader is addressed to Sect.19. see Problem 1.65 g cm−3 · 105 cm · 1. since for every muon. and by assuming the standard rock density ρ = 2.65) implies an exponential suppression of the flux at large depths. (1.19. Explain this behaviour and provide a rought estimate of the muon flux deep underground.6 of Ref.e. see Eq.7 MeV g−1 cm2 and b = 3. only muons with energies of order of E μc or larger can make their way through the underground soil.291). let’s consider the infinitesimal flux of neutrinos with energy in the range [E ν . the range scales logarithmically with the muon energy at the sea level E 0 : E0 R(E 0 ) ≈ b−1 ln 1 + . Problem 2.65 g cm−3 . [2]. the muon spectrum underground reduces with depth untill a depth of about 10 km w. Using the values a = 2. The neutrino spectrum can be assumed to be similar to the muon spectrum. The probability for such interaction is r/λ 1. where λ is the interaction length and depends on the neutrino energy. (2.65) E μc where a and b are the constants introduced in Problem 2.69 TeV = = 4. however. = 105 g cm−2 ) is attained. (1 km w. 9 MeV g−1 cm2 [5] and the cross section of Eq (1.65).23 × 10 ≈ 3−α GeV2 ≈ 10−9 m−2 s−1 sr−1 . From Ref.g. (2.6 × 10 cm = 1 GeV d Eν 1.66) E min d Eν λ(E ν ) The maximum energy E max can be assumed to be of order of E μc . (2. Although the muon spectrum at E μ 10 GeV decreases slower than E μ−2. This order-of-magnitude estimate is in a decent agreement with the measured spectrum. see e.2.68) The result depends only mildly on the choice of E max .67) d Eν with Φ0 = 70 m−2 s−1 sr−1 and α = 2. By using d E μ /d x = 1. (2. 2.354) for the neutrino-nucleon scattering (with Q ≈ 1).7 taken from Ref.9 MeV g−1 cm2 · ρ A GeV α−1 E 3−α − (1 GeV)3−α −12 α − 1 (1 GeV) μc = Φ0 · 0.7 . Fig. [2].7.1 Passage of Particles Through Matter 137 E max dΦν0 R(E ν ) − 200 m Φμdeep ≈ d Eν (2.7 Vertical muon intensity versus depth. and neglecting for simplicity the offset of 200 m. see Eq. [2] 1 2 5 10 1 10 100 . and thus it will contribute by one power less to the muon flux. since for larger energies the range becomes only mildly dependent on the muon energy. Fig. 2. for an order-of-magnitude estimate we can assume for simplicity: dΦν0 = (α − 1)(1 GeV)α−1 Φ0 E ν−α . we have: E μc deep dΦν0 Eν ρ · NA −38 2 Eν Φμ ∼ d Eν · 1. 2 Particle Identification Particle identification (PID) is a common problem in particle physics experiments. Relativistic particles can emit Cherenkov lights when moving inside a refractive medium. • Cherenkov-light detection. A non-exhaustive list of techniques for PID includes: • Measurement of the range. [11]. about 50 times faster. a variety of methods can be deployed in experiments. Even if the particle range is not fully contained within the active volume of a detector. or momentum. the presence of backgrounds and imperfections in the detector makes PID a statistical test rather than a deterministic decision: the probability of correctly identifying a given particle (efficiency) has always to be weighted against the probability of wrongly identifying a background event (fake-rate). Each particle loses energy by interaction with matter at a different rate.21). can be found in Ref. with typical interaction lengths of tens of centimetres for condensed materials.e. The angle of emission and the number of emitted photons depend on the particle velocity β as for Eq. see Eq. The stopping power can be measured from the energy deposited within the detector. (2. i. proportional chambers. which are sensitive to the light emitted by charged particles while . Time Projection Chambers. A simultaneous measurement of the particle momentum and of the Cherenkov light can be thus used to determine the particle mass. (2. nuclear emulsions. 29. Hadronic particles interact strongly with the nuclei. provides a handle to distinguish between different particles. Therefore. As a general rule. including a review of theoretical calculation. which are often equipped with a redundance of detectors as to be able to identify the particle type besides measuring their kinematics. 2. [2] for more details on the muon flux underground. • Transition-light detection. Cherenkov detectors as particle identifiers become inefficient.15) and Eq. For example. while an electron of the same energy loses energy by radiation at a rate of about 550 MeV cm−1 . (2. Depending on the particle type and on its energy. For high-energy particles. The reader is addressed to Sect. a 10 GeV muon loses energy by collision at a MIP rate of about 11 MeV cm−1 . An alternative to using the β-dependence of Cherenkov detectors is provided by the use of transition radiation detectors.138 2 Particle Detectors Suggested Readings More details on cosmic ray fluxes. the capability of muons to penetrate massive detectors exceeds by far larger that of other particles. solid-state detectors are examples of detectors which can measure the energy loss across the particle trajectory. • Measurement of the stopping power. so that the measured range can be used to differentiate between different particle types.21).4 of Ref. the simultaneous measurement of the stopping power d E/d x and of the particle energy. this allows to infere the particle life-time (see Problem 1. the kinematics of the decay products can be used to reconstruct the decay process. (2. allows to determine the particle mass. see Problem 2. and 1. In scattering experiments where the kinematics of the initial and final state can be measured. Problems Bando n. from hadrons. particles of a given momentum. from which the mass of the mother particle can be inferred.32). a combination of tracking and energy measurements in segmented calorimeters is sometimes used for PID: a calorimeter consisting of an electromagnetic (ECAL) and an hadronic (HCAL) section with independent read-out offers the possibility to separate electrons. indicating the range of applicability and their complementarity.16. In high-energy experiments. 1. like pions and electrons. 13153/2009 Problem 2. Solution At small velocities. However. At larger energies. the stopping power (by collision) becomes only logarithmically sensitive to the particle velocity. The attempt to reconstruct and identify each and every particle in a HEP event is called particle flow and was pioneered at LEP [12]. Since the intensity of the emitted radiation is proportional to the γ -factor of the particle as for Eq. and hence the particle type. or of the stopping power d E/d x. • Kinematics. see e.27. see e. the simultaneous measurement of the particle momentum |p| and of its time-of-flight over a known distance. whose intensity is proportional to the γ -factor of the particle. can be efficiently separated by measuring their transition light. A simultaneous measurement of the particle momentum and of the TOF over known distances.21 Mention two methods of identification for charged particles.2 Particle Identification 139 crossing the separation surface between vacuum and a dielectric material. which are stopped in ECAL. while for Cherenkov detectors this is due to the fact that the sensitivity to mass differences is suppressed by |p|−2 .20. four-momentum conservation can be used to infere the mass of the particles involved in the scattering.g. Once combined with momentum information.37.62. or of the Cherenkov light emission. one can instead exploit the emission of transition radiation. the TOF can be measured from the distance traveled by the particle before decaying. 1. and 1.22). which interact in both. all these methods become inefficient due to the saturation of the particle velocity to β → 1. represent canonical techniques for PID.g. Problems 1.2. For unstable particles. • Measurement of the time-of-flight. For unstable particles that decay in reconstructable vertices. High-energy electrons can be discriminated from other charged particles thanks to their larger emission of bremstrahlung radiation. at higher energies. but very different mass. so that the TOF over a baseline distance L saturates to L/c for all particles.28. . 1.25. Problem 1.23. α-particles. detected by using liquid. or inorganic scintillators. or gaseous ionisation detectors. 6 Li (n. whose molecules contain hydrogen. Bando n. devices that degrade the initial hadron energy by initiating a hadronic cascade and measure the visible energy deposited by the cascade particles. p) t. Bando n. α) 7 Li(∗) . The active material is conveniently loaded with suitable nuclei like 3 He.7 of Ref.e. The kinetic energy of the emitted particles (protons. the reader is encouraged to read about PID within the particle flow algorithm as implemented in the CMS event reconstruction [14]. these materials can also offer n/γ discrimination by pulse-shape analysis. tritium. 6 Li. see Problem 2. like Li I (Eu). B F3 . α). p) scatterings. (2. 13153/2009 Problem 2. For thermal neutrons. Besides the already quoted ALEPH publication [12]. t) 4 He. Given the different fluorescent response of organic compounds to particles of different ionisation power.g.22 Discuss a few techniques for neutron detection as a function of the neutron energy. [13] are also indicated for a first overview on the subject. Introductory textbooks like Ref. The detection of fast neutrons relies on the detection of the recoil proton in (n. one usually relies on the nuclear reactions (n. 10 B (n. Suggested Readings Chapter 7. glass. which have large cross sections for the reactions: 3 He (n. γ ) and (n. Li ions) peakes at values determined by the Q-value of the reactions. Solution Neutrons with energies in excess of a few GeV are best measured by hadronic calorimeters. like 3 He. and 10 B. 1N/R3/SUB/2005 . thus allowing to separate the neutron signals from other backgrounds.69) respectively. which is usually proportional to the incoming neutron energy.35 for more details. which can be e. [1] describes the pulse-shape technique with scintillators and provides an introduction to various experimental techniques for neutron detection. most notably by photon interactions. This is best achieved by using plastic or liquid organic scintillators. i.140 2 Particle Detectors Suggested Readings The PDG review of particle detectors at colliders provides a comprehensive and up-to-date overview of detectors for PID. Cherenkov detectors are often integrated with spectrometers or other detectors that can measure the momentum of the particle.0006 no π emit ⎪ ⎨ n > 1.49. one can use two threshold Cherenkov detectors operated in series.0195 all π emit (2. one could set counter A at a value of n A = 1.12] GeV π <β< ⇒ |p| ∈ (2. respectively. while n − 1 ≈ 7 × 10−3 can .23 In order to separate K + and π + in a momentum window between 700 MeV and 4 GeV. then the ambiguity is lifted. if one considers pions and kaons with velocities in the range [1/n A .0195. there remains an ambiguity for the case where only counter A records a signal. while the vertical arrows mark the upper and lower momenta at which pions fail to generate a signal in B and kaons generate a signal in A. Neglecting possible inefficiencies of the detectors near the threshold.5. The dashed lines indicate the indexes chosen for counters A and B.4 to identify possible candidates. we see that a value of n − 1 ≈ 2 × 10−2 can be obtained for example by using aerogels. and propose a suitable radiator.0] GeV K so that a simultaneous measurement of the particle momentum and of the Cherenkov counters can discriminate between the two particles.2 Particle Identification 141 Problem 2. For example.0076. we can refer to Table 2.70.71) nA nB [2.2. 1/n B ]. 1. a scintillator located along the beam direction). Discussion Although not mentioned explicitly. so that a signal in that counter would imply that the particle is not a kaon (π -tag). so that no signal there would imply that the particle is a kaon (K -tag).8 shows the critical index 1/β for the two particle types as a function of |p|. the corresponding momenta span two non-intersecting ranges: 1 1 [0. Concerning the choice of radiator medium. namely: ⎧ ⎪ ⎪ n < 1.70) ⎪ ⎪ n < 1. With this scheme one has three possibilities. Cherenkov detectors can be employed to select particles of a given type from a composite beam of given momentum. 4. Indeed.22 all K emit With two counters at hand. summarised in Table 2. The third row (all counters with no-signal) represents a useful event only if the experiment is equipped with an independent trigger (e. determine which values of the refraction index can be chosen.0076 no K emit ⎪ ⎩ n > 1. and counter B at a value n B = 1. Figure 2.g. However. In particular. Solution The momentum acceptance of the experiment provides four threshold velocities and as many refraction indexes. If one further assumes that the particle momentum can be measured. respectively be obtained by using e. where n is the refraction index. the identification of three particles of different mass but same momentum |p|.8 The critical index 1/β for the two particle types as a function of |p|. Bando n. By definition. thus producing no signal in any of the two counters. 2. β = 1/n. vacuum has n = 1.142 2 Particle Detectors Table 2. while the vertical arrows mark the upper and lower momenta at which pions fail to generate a signal in B and kaons generate a signal in A. i. so that: • particle (1) is below threshold in both counters (β1 ≤ 1/n A . How can the three particles be identified? Solution The Cherenkov threshold is the velocity β that equals the group velocity of light in the medium. 1/n B ).g.4 for a few representative materials.24 Explain how the Cherenkov threshold depends on the refraction index of the medium.e.5 Possible outcomes of a single-particle event using two threshold Cherenkov detectors in series with NA > n B A B Particle 1 1 π 1 0 π or K 0 0 K 0 1 Not possible Fig. . and n > 1 for any other medium. can be achieved by setting the refraction index of the two counters at n A = 1/β1 and n B = 1/β2 . see Table 2. such that the three particles have velocities β1 < β2 < β3 . Given two threshold Cherenkov detectors A and B operated in series. The dashed lines indicate the indexes chosen for counters A and B. pentane (C5 H12 ) or perfluoropentane (C5 F12 ) of appropriate temperature and pressure. Three particles of different mass but same momentum |p| cross a system of two Cherenkov detectors arranged in series. 13153/2009 Problem 2. 25 A Cherenkov imaging detector measures the angle θ of Cherenkov photons with a resolution σθ = 2 mrad. thus producing a signal in both counters. The reader is addressed to this reference for more information on the subject. but below threshold in counter B (β2 ≤ 1/n B ). What is the largest beam momentum |p| such that kaons and pions can be discriminated to better than 3σ by the angular measurement only.2 Particle Identification 143 • particle (2) is above threshold in counter A (β2 > 1/n A ).5 of Ref.474) or fluorocarbon gas (n = 1.72) 2σθ n 2 − 1|p|2 Hence. .00172 −1 (2. we get: Δθ dθ 1 β 1 β 2 dm 2 ≈ = d cos θ = d = σθ σθ σθ sin θ σθ β 2 n 2 − 1 β 2 σθ β 2 n 2 − 1 |p|2 |m 2K − m 2π | ≈ √ . [2].2. the largest momentum for which the statistical separation is in excess of Nσ = 3σ is provided by: ⎧ |m 2K − m 2π | ⎨√ 0. Problem 2. An analysis of the signal output in the two counters can thus reveal which of the three particles has crossed the detector.474 GeV √ = 18 GeV fluorocarbon Nσ · 2σθ n −1 2 3·2·2×10−3 · 1. By approximating finite differences by their differentials. 34. (2.73) Suggested Readings This problem is inspired by Sect. if the Cherenkov radiator consists of fused silica (n = 1. thus producing a signal in only one counter.4742 −1 |p| < 21 = √ ⎩√ 0.2 GeV silica 3·2·2×10−3 · 1. 1/n B ) in both detectors. This configuration also maximises the light yield when the particle is above threshold.474 GeV √ = 4.0017)? Solution Let the Cherenkov angle be denoted by θ . A separation to better than 3σ amounts to require Δθ /σθ ≥ 3. • particle (3) is above threshold (β3 > 1/n A . 26 Tellurium dioxide (Te O2 ) crystals (n = 2. The Q-value of the reaction. Show that the simultaneous measurement of Cherenkov photons and calorimetric energy would allow to separate α particles from signal events. a neutrinoless double-β decay (0νββ) does not produce neutrinos A in the final state.53 MeV. behave like background events by releasing their energy in the calorimeter. typical of radioactive decays. where a nucleus ZA X decays to − Z +2 Y + 2ν + 2e .717 in Te O2 is above the velocity of α’s. Alpha particles of a few MeV energy. 600] nm produced by a signal event in a few centimetres long crystals. is entirely taken by the two electrons: their energy sum is therefore a line around Q smeared by the detector resolution. A major background to this process is represented by α-decays of radioactive contaminants. [15]. Estimate the mean number of Cherenkov photons with wavelengths in the range [350.39. but below the velocity of at least one of the electrons. The experimental signature is provided by an energy deposit around 2. Solution In order to prove that the electrons radiate Cherenkov light while the α particles do not. it suffices to verify that the threshold velocity β = 1/n = 0. one has . This also implies that the electron energies are fully anticorrelated. ρ = 6 g cm−3 ) have 52 Te →54 Xe been used to search for the putative neutrinoless double-beta decay 130 130 in bolometric calorimeters. The theoretical energy distributions for this decay can be found in Ref. Assuming Tα = 2.4. Discussion Differently from an ordinary double-β decay (2νββ).53 MeV.144 2 Particle Detectors Problem 2. see Problem 1. 2Tα 2 · 2. (2.53 MeV βα ≈ = = 0.037 < β.73 GeV while for a 0νββ decay: .74) mα 3. 2 . though. the range is a linear function of energy as for Eq.32). To circumvent the lack of tabulated data and the mild dependence on the kinematics. (2.511 MeV max βe > 1− = 1− = 0. (2. thanks to the anti- correlation between the two energies. and one should rather use the full calculation.32) is expected to underestimates the true range for small values of γ . we consider a particular decay configuration. (2. 2 me 0. so the linearity is lost.75) Q/2 + m e 1. for γ 1. and is larger when the energy sharing is more asymmetric. Eq. the total range and the number of Cherenkov photons are independent of the energy sharing between the two electrons. In the case of interest.77 MeV To good approximation. Furthermore.958 > β. the average kinetic energy is comparable to m e . a numerical investigation shows that the total range is constant to within 15% over the allowed electron spectrum. However. . Indeed. which contains Iodine.32). we get R1 ≈ 0. (2. which averages the range over the proper energy spectrum.76) Taking the mean.64 g cm−2 .5 MeV. and Ti O2 . while pions do not.16) with sin2 θ ≈ 1 − 1/n 2 . [17].2 Particle Identification 145 namely T1 = 1. R2 ≈ 0. At T = 1 MeV. The light output in the wavelength window [350.69. (2. [16] gives: R1 (Na I) = 0. or 0. Problem 2. giving a ratio R2 /R1 = 1.55 g cm−2 .27 A threshold Cherenkov detector is used to separate muons from pions in a beam with momentum |p| = 150 MeV. is given by: . a Tellurium neighbour in the periodic table. from which the problem is largely inspired.175 cm. Hence.15 × 103 cm−1 1 600 − 350 Nγ ≈ (0. [17]. We then approximate the stopping power by averaging the tabulated values for two similar materials: Na I.10 cm. but we can use the scaling predicted by Eq. (2.0 MeV and T2 = Q − T1 ≈ 1. What values of the refraction index n can be used? Solution The condition for which muons emit Cherenkov light.69 g cm−2 . There are no values tabulated for T = 1.42 600 · 350 = 46 + 79 = 125. R1 (Ti O2 ) = 0.5 MeV.175) cm √ 1− √ 600 · 350/400 2. which is also a metal dioxide.2.77) which agrees with the more accurate expectation of Ref. giving: 1. Ref. 600] nm can be estimated by using Eq. (2.10 + 0. Suggested Readings The idea of exploiting Cherenkov radiation in bolometric detectors has been first proposed in Ref. . 1N/R3/SUB/2005 Problem 2. mμ 2 mπ 2 1/βμ < n < 1/βπ ⇔ +1<n < + 1. Can each particle be identified? With which precision can the pion fraction be determined? . The instrumentation has a time resolution σt = 0. |p| |p| giving the result: 1.2 ns.37.22 < n < 1.28 An experiment needs to distinguish pions from kaons of momentum |p| = 2 GeV by measuring the time flight on a L = 2 m baseline. Bando n. 146 2 Particle Detectors Solution The time-of-flight (TOF) for pions and kaons in the beam is given by: . or by the electron-scattering (ES) . (2. Hence.79) The classical theory of estimators predicts that the asymptotic variance of the ML estimator is given by 2 1 ∂ ln f (x.139 2 = 6. if we decided to tag a particle as a K if the TOF is in excess of 6. σt )] i=1 i=1 (2.e. i.81) N Problem 2.68 ns π t= = 1+ 2 = 2 (2. ∂επ ε̂π N N with L = f (X i . detected a neutrino burst that was attributed to a supernova event occurred at a distance d = 5. the water Cherenkov detector Kamiokande-II in the Kamioka mine (Japan). For example. (2.19 ns ≈ σt . The energy and arrival time at the detector could be measured for those (anti)neutrinos that interacted via the charged-current (CC) scattering ν̄ p → n e+ .1. the standard deviation on the pion fraction will be given by: 1 σε̂π ≈ √ .80) N I (ε̂π ) ∂ 2 επ see Sect.03 (0. with I (ε̂π ) = E − .78) βc c |p| 3 × 108 m s−2 ⎩ 1 − 0. ⎧ L L m2 2m ⎨ 1 − 0. for a fake-rate of about 50%.29 In 1987. επ ) Var ε̂π = . particle-by-particle identification is affected by a large statistical uncertainty.67 ns. the Type-II error is large for any given efficiency to identify the correct particle type.5 × 104 kpc from the Earth. επ ) = [επ N (X i | tπ . Assuming N independent and gaussian distributed measurements X = {X i }. 4. the pion (or kaon) fraction of the beam can be estimated with large accuracy for a sufficiently large number of measurements. Even though an event-by-event classification is not very accurate. The result is a number of O(1): for example.80).87 − 1σt = 6.3) one gets I = 1. the selection efficiency would be 84%.494 2 = 6. the maximum-likelihood (ML) estimator of the pion fraction ε̄π is given by the solution of the equation: ∂ L(X.3.87 ns K 2 Since Δt = 0. επ ) 0= . The information can be computed numerically using a simple program for different values of επ . see Appendix 2. σt ) + (1 − επ ) N (X i | t K .1 (0. for επ = 0. provided E ν̄ 2 MeV. an electron antinutrino interacts mostly 16 16 through the CC reaction ν̄ p → n e+ . The single-PMT time resolution was 13 ns. the Kamiokande-II experiment consisted of a cylindric water tank containing over 2000 t of water instrumented with uniformly distributed PMT’s covering about 20% of the total surface. production vertex. • The Kamiokande experiment could not distinguish electrons from positrons by using the sole Cherenkov light. An electron neutrino with energy of about 10 MeV interacts mostly through ES on the atomic electrons. The main background to ∼10 MeV electrons and positrons is represented by cosmic muons. • Determine a lower bound to the νe lifetime. 2. Solution The separation between electrons and positrons is possible on a statistical basis. within the fiducial volume of the detector. The event trigger. During a time interval Δt = 12 s. direction.2. and by γ /n radiation from the cavern walls. one notices that the velocity of . The PMT’s were sensitive to the Cherenkov light in the range 300÷500 nm. thus allowing an efficient light collection all across the fiducial volume. 2.9 Scatter plot of energy and time for the twelve supernova candidate events recorded by Kamionkande in 1987 (from Ref.2 Particle Identification 147 Fig. a total of 12 events were registered. estimate an upper bound to the electron neutrino mass m νe . Conversely. the CC scattering for antineutrino energies E ν̄ ≈ 10 MeV is isotropic in the laboratory frame. the light attenuation length exceeds 50 m. while the relative energy resolution was estimated from simulation to be about 20%. Indeed. [18]) reaction νe e− → νe e− . β-decays of unstable isotopes polluting the water. Discussion As of 1987. The CC interaction 6 O → 7 F is instead suppressed by the large mass difference with the transmutation 16 16 B(6 O) − B(7 F) ≈ 16 MeV. First.9. This can be proved as follows. At these wavelengths. The time vs energy diagram of the signal events is reported in the Fig. and energy of the particles were reconstructed by using the charge and time stamp of all PMT with a signal above the noise. • Using the data reported in the plot. How was it then possible to separate νe from ν̄e ? • Explain how the antineutrino energy E ν̄ could be measured from the positron energy E e+ . 85) βν c c 1 − (m ν /E ν )2 c 2 Eν Two neutrinos of energies E 1 and E 2 . is an indication that neutrinos have a mass. (2. i. (2. since the centre-of-mass velocity is now β = E ν̄ /(E ν̄ + m e ) ≈ 1. has to be large enough so that the neutrinos can make it to Earth.e.148 2 Particle Detectors the centre-of-mass frame is β = E ν̄ /(E ν̄ + m p ) ≈ 10−2 . and the other around t = 2 s.83) ! " 1. From Problem 1. In this case.84) c γν c (E ν /m ν ) c (E ν /m ν ) Eν where we have used the relation 1 pc ≈ 3.3 c · y mν τν = = = 1. as described by the Fermi Lagrangian: GF LF = √ cos θC n̄γμ (1 − αγ5 ) p ν̄γ μ (1 − γ5 )e . which is larger than the expected duration of a supernova burst (a few seconds).: d 5. (2. (2.8 × 105 y. will arrive at destination with a time separation: d 1 1 t1 − t2 = m 2ν 2 − 2 . we can see that the cross section is roughly isotropic in the laboratory frame. since otherwise they would have arrived all in one shot.5 × 104 pc 5. the dynamics is governed by the exchange of a virtual W boson. if the neutron recoil is neglected compared to the nucleon mass.15. τν . it becomes proportional to ( pe+ p p )( pν̄ pn ) ≈ E e+ E ν̄ m p m n . hence the ampitude squared itself is constant.5 × 104 · 3. the arrival time at the detector is: d d 1 d 1 mν 2 t= = ≈ 1+ . The presence of two populations of events. which gives rise to a very forward-peaked differential cross section in the laboratory frame. that the pattern of . see Problem 1. (2. though. (2. one located within the first second. E e+ is also a constant.3 MeV The neutrino lifetime. energy conservation implies E ν̄ + m p = E e+ + m n ⇒ E ν̄ ≈ E e+ + (m n − m p ) .85). we observe the presence of a few neutrino events separated by about 10 s from the the first burst events. emitted at the same time t = 0. This is not the case for the ES. For the antineutrino scattering. By taking α = −1.82) 2 The amplitude squared can be obtained with the usual Casimir’s tricks.86) 2c E1 E2 From the recorded data. In the latter. so that the centre-of-mass is almost at rest in the laboratory frame.53 and the considerations above. indicates.3 c · y. If the neutrino burst starts at the time t = 0. which are not distributed according to Eq. 87) d E2 − E2 2 1 Suggested Readings This problem is inspired by the Kamiokande publication of Ref.g. νμ X → μ− Y . In both cases. Explain this behaviour. A conservative upper limit on m ν can thus be obtained by considering. [18]. Bando n. for a smaller fraction of the events. 18211/2016 Problem 2.2 Particle Identification 149 neutrino emission from the supernova has some non-trivial time dependence. A fraction of the events features a few metres long track starting from the interaction point.64. Yet. those events that feature the largest energy difference |ΔE| among the first and last arrived events. t1 ) = (35 MeV. while. the neutrinos can interact via either the charged current. and with some broad spectrum of energies.86). some of the neutrinos must have been created simultaneously.5 s). νμ X → νμ X . so that any difference in arrival time has to be attributed to the non-zero neutrino mass. we take e. i.e. (2. From the plot. the latter having a cross section suppressed by a factor of m e /m N .5 s) and (E 2 . 12. one cannot assume a perfectly synchronous burst. (2. Inverting Eq. all tracks are contained within a small volume. t2 ) = (10 MeV. 1.2. or the neutral current interaction.: (E 1 .30 A νμ beam with an energy of 30 GeV enters a detector containing liquid Ar. respectively. we have: 2c (t1 − t2 ) E 1 E 2 mν ≈ 20 eV. Solution As already discussed in Problem 1. neutrinos can undergo interactions with both the nuclei and and the atomic electrons. is highly penetrating in the Ar medium and can be therefore identified as a long track.88) σCC ν 2 27 see e. the only detectable signal is provided by the recoil of the struck nucleus.31. the interaction is inelastic and results in a number of hadronic particles which. . being much heavier than the muon and less energetic. have smaller range. in the occurrence of a NC interaction. Ref. [19].g. . (2. When a neutrino of energy E ν = 30 GeV interacts via CC. Conversely. The EWK theory predicts the ratio between neutral and charged current cross section in terms of the Weinberg angle θW to be: σNC 1 20 4 = − sin2 θW + sin θW ≈ 0. thus appearing as a set of short tracks emerging from the interaction point. Since DIS prevails in this energy regime. which being a MIP. it produces a muon of similar energy. Bando n. 10−12 s. (2.89) where z is the particle charge in units of the proton charge e.32 Describe which methods could be used to measure lifetimes of order 109 years. see Problem 3.20 |z| GeV. and 10−22 s.00 ± 0.00)2 m 2 = (T + m)2 − |p|2 . Let NC be the number of countings after background-subtraction. Solution Lifetimes of order 109 years are typical of radioactive decays. The kinetic energy of the particle is measured to be T = (2. non-exotic particles have integer charges. which does not match any known particle within the experimental uncertainty. The particle momentum |p| is instead given by the formula: |p| = 0. The radius of curvature of the track is R = 7. For |z| = 2. Solution The charge sign is fixed by the direction of curvature.03) GeV. Determine which type of particle is most probably being measured. Under the assumption τ Δt. the lifetime can be measured from the relation: .00) · 0.31 A charged particle is moving inside a uniform magnetic field of intensity B = 1.03 GeV.150 2 Particle Detectors Suggested Readings The reader is addressed to Chap.20 · z)2 − (2.3 |z| (B/T) (R/m) GeV = 2. [19] for more information on neutrino interactions in matter.20 · z/2.25 m with negligible error.09) GeV.00 (2. 2T 2 · 2. Such lifetimes can be measured by counting the number of decays in a sample and in a given time interval Δt.90) gives m = (210 ± 30) MeV. For |z| = 1. We can therefore try different ansatz values of |z| and compare the result with the known spectrum of particles. Problem 2. 12 of Ref.73 GeV at the 1σ level. Eq.0 T.84 ± 0. m= = GeV. which is compatible with the mass of the α particle m α = 3. (2.91) 2 2 2 Δm = ∂T 2 2 Stable. The particle mass m is therefore given by: |p|2 − T 2 (2. one has m = (3.90) The uncertainty on m can be obtained by propagating the uncertainty on T : ∂m ΔT = 1 + |p| /T ΔT = 1 + (2. (2. 13153/2009 Problem 2.3. The selected problems want to discuss the main technologies and introduce general concepts. 1N/R3/SUB/2005 Problem 2. In modern experiments.2. (2. dead time. Can we conclude that the energy resolution for an electron of energy E = 50 GeV is 1%? . 2. Lifetimes of 10−22 s are characteristics of strongly decaying particles. which interacts with the particle. with intrinsic spatial resolutions of a few tens of microns or better. and finally shape the signal according to some logic suitable for later processing in the experiment or for persistent data storage. (1. efficiency. or the Δ baryon.92) A NC where V is the volume of the sample being observed. Detectors are usually composed of an active volume. are ideal candidates to build vertex detectors with sufficient resolution to resolve such decays. Such lifetimes are therefore indirectly estimated from the decay width Γ of Eqs. or τ leptons.33 In an √ electromagnetic calorimeter. detectors are commonly operated by computers. when the particles are produced at relativistic energies. hosting the electronics required to generate an electric signal.186). Problems Bando n. the energy and direction of the incoming particle. the stochastic contribution to the resolution is 0. No attemp is made here to give a comprehensive overview on this subject. Depending on the detector type and on the form of radiation it is sensitive to.3 Functioning of Particle Detectors Particle detectors record the passage of particles. and sometimes even identify the type of particle. The field of particle detection is vast and finds application that range from pure research to industry. provide signal amplification to improve the signal-over-noise ratio. like D and B mesons. as measured from the invariant mass distributions of the decay products. Silicon detectors. detectors can be used to measure the position and time of arrival of a given particle at the detector location. and a readout component. see Problem 2.43. like resolution. Lifetimes of order 10−12 s are characteristics of weakly decaying particles. like the ρ and ω mesons.2 Particle Identification 151 V ρ NA Δt τ= . or from the production cross section. the decay vertexes of such particles are of order 300 µm. The distance of flight is far too small to be measurable by any position-measuring device.07/ E. which supervise their correct functioning and take care of data acquisition. Since c = 3 × 102 µm/ps. dependence of the response with the particle impact point. which uses instead a sampling liquid-Ar calorimeter.g. etc. In most applications. after a proper calibration is performed. gas sampling.27%. semiconductive. and ionisation. The importance of the noise term a depends on the signal collection type: scintillation and Cherenkov calorimeters coupled to high-gain PMT suffers the least from the electronic noise. In general. As a general rule. noise. Cherenkov light emission. and a local constant term of 0. depending on whether the active medium is composed of the same material. This problem is somehow relieved by the ATLAS setup. the goal constant term needs to be kept below 0. (2. When averaged over the full detector acceptance.3%/ E/GeV. or interleaved with layers of inactive absorbers which degrade the energy of the incoming particle.g. As an example.93) E E E where the symbol ⊕ indicates sum in quadrature. only a fraction of the total initial energy is converted into a visible signal: the proportionality between the measured signal and the total energy allows to measure the latter. but achieve similar physics performances. The three contributions are called stochastic. overall. and on other geometrical properties of the detector (e.152 2 Particle Detectors Discussion Electromagnetic calorimeters are detectors that measure the kinetic energy of charged particles by exploiting one or more interaction mechanisms between charged particles and matter. and constant term. the constant term ends up to be the limiting factor to the ultimate energy resolution. which is challenging since the whole detector is composed of about hundred thousand crystals that need to be inter-calibrated.). while for sampling calorimeters the stochastic term is in the range 5 ÷ 20%. the relative energy resolution can be parametrised in terms of these three contributions as: σ (E) a b = √ ⊕ ⊕ c. where particles with energies of hundreds of GeV need to be measured. including fluorescence. since a preamplifier is the first element in the readout chain. Electromagnetic calorimeters can be broadly classified into two categories: homogeneous and sampling. The CMS detector makes use of a homogeneous scintillation calorimeter based on PbWO √ 4 crystals. at the price of increasing the stochastic term. homogeneous √ calorimeters shine for their small stochastic term of order 1% in units of 1/ E/GeV. electronic noise). on the signal generation and electronics (efficiency of the photodetector. For this contribution to be subleading in the GeV range. a noise term of 0. uniformity. For use in high-energy experiments.g.5%. in the same units. The total energy resolution depends on the choice of active material. which determines the statistics of signal carriers per unit of deposited energy (e. respectively. the electromagnetic calorimeters employed by the CMS and ATLAS experiments at the LHC are built with different technologies. while the noise is usually larger for calorimeters that collect the signal in the form of charge (e. the parameter b needs to be kept at the 100 MeV level per channel.19/(E/GeV). the statistics of scintillation photons). and noble-gas calorimeters). A test beam on a prototype . A test beam on a small prototype yielded a stochastic term of 3. (2.1 ⇒ 2 1% (2.95) dt Γ (a) where t = x/ X 0 and a and b are constants that depend on the material. We can estimate an upper limit to the noise and constant terms such that they do not contribute individually to the total relative resolution by more than a certain fraction f .94) 1% ⎩1 c 2 0. An electron of few GeV energy loses energy mostly by radiation. while the signal generated in a block of lead glass of the same size is only 103 p. Simplifying the shower development as a series of 1 → 2 branches (e± → e± γ and γ → e+ e− ) . a noise term of 0.5% 2 1% We can therefore conclude that the energy resolution for an electron of energy E = 50 GeV is about 1% provided that the noise and constant term are below about 200 MeV and 0. the energy resolution of an electromagnetic calorimeter depends √ with E = 50 GeV and a calorimeter on the energy as in Eq.3 Functioning of Particle Detectors 153 √ yielded a stochastic term of 10%/ E/GeV./GeV.1.3%./GeV. c 0.25/(E/GeV). The latter has to be added in quadrature to the constant and noise term to obtain the total relative energy resolution. With this choice: ⎧ σ (E)/E − 1% ⎨ 1 b/50 GeV 2 0.34 A relativistic electron releases energy in a block of BGO.e. 1N/R3/SUB/2005 Problem 2. that we can conventionally set to e. The emitted bremsstrahlung photons undergo pair-production. respectively. the stochastic term is 7%/ 50 ≈ 1%. (2.1.93).g. f = 0. generating a signal of about 106 p. Bando n. For an electron with a = 7%. Solution As discussed above. How can such a difference be explained? Discussion Both BGO and lead glass feature a radiation length X 0 of about 1 cm and a critical energy of about 10 MeV [5].5%.e. [20]. with subsequent photon emission. The resulting electromagnetic shower is characterised by an energy profile dE (b t)a−1 e−b t = E0 b . and a local constant term of 0. More informations on the state-of-the-art in calorimetry can be found in the PDG review [2] and references therein.1. Suggested Readings A succint but complete review of calorimetry in particle physics can be found in Ref. b 220 MeV < 0.2. 97). and a total charged track length as in Eq. . Solution BGO. The difference between the two materials can be therefore ascribed to the different mechanism by which photoelectrons are produced in the two materials.21).2 and is about 300 eV/γ . (2. Eq.e. an acronym for (Bi2 O3 )2 (Ge O2 )3 .e. which makes it a good Cherenkov radiator. so that the secondary e+ e− pairs from γ conversion are preferably soft. is a scintillating crystal.8) and is transparent to visible wavelengths. /E ≈ 103 /GeV [20]. (2.97) Ec A more refined treatment of shower development.98) 10 MeV where the factor of 2/3 accounts for the fact that only electrons and positrons produce Cherenkov light. tmax = ln(E/E c )/ ln 2. [1]. It has a large refraction index (n ≈ 1. electrons and positrons excite the fluorescent levels of the crystal. it follows that the total track length L(t) from electrons. and ln E/E c X0 L = 2 ln 2 X 0 = E. and photons. Assuming a quality factor N0 of about 90 cm−1 .96) The maximum number of radiation lengths tmax is determined by the condition that the electron/positron energy falls below the critical energy E c .3 cm = 90 cm−1 · 0. which is in the ballpark of the value reported by the problem (the ultimate p.g. after traversing t radiation lengths is given by L(t) = 2t X 0 .e. i. (2. This estimate does not account for the fact that the simple shower model is not well representative of the energy distribution within the shower: the bremsstrahlung cross section dσ/dν for emitting one photon with frequency ν is approximately proportional to ν −1 . will still predict the total track length L to be proportional to the initial energy. The mean excitation energy per photon is reported in Table 2. see Eq. (2.154 2 Particle Detectors with equal energy sharing and separated by a distance X 0 . A more accurate estimation would yield a smaller value Np. an upper limit to the number of p. per GeV can be estimated as: Np.e. characterised by an average excitation energy ε. Lead glass (Pb O) is an amorphous material and does not scintillate. Np. so that the total photon output Nγ is still proportional to the initial energy E.68) of Ref. positrons. or 3 × 106 γ /GeV. with implications on the total Cherenkov light yield. statistics depends on the PMT collection and quantum efficiency). so that the energy per constituent at a depth t is E/2t . (2. (2. Along their path. see e.69 · (2/3) = 5 × 103 /GeV. L X0 = ≈ 90 cm−1 sin2 θc · (2/3) = E L E Ec 1.e.e. The latter can come delayed up to 1 µs with respect to the primary interaction. the fraction of energy drained away in the form of π 0 → γ γ photons increases with energy. 13153/2009 Problem 2. electrons and positrons. involves the production of energetic secondary hadrons through strong interactions with typical interaction lengths of about 35 A1/3 g cm−2 .35 Measuring the energy of hadronic particles through calorimetric methods is a fundamental ingredient in HEP experiments.. The ratio between the response to an hadron h and to an electromagnetic particle. like iron. etc.3 Functioning of Particle Detectors 155 Suggested Readings The reader is addressed to Ref. photons.2. on average 30% of the initial energy is transformed into “invisible” energy. Since the number of high-energy interactions that produce pions increases with energy. The low energy spectrum of the hadronic cascade is dominated by neutrons. and overall account for about 30% of the total cascade energy. The hadronic shower in usually initiated inside the so-called radiator. followed by the degradation of their energy by nuclear reactions that produce nuclear excitation.99) ηe where Fh = 1 − Fe is the hadronic energy fraction. lead. Let ηe (ηh ) be the efficiency of detecting the energy contained in the electromagnetic (hadronic) component. Discussion The physics of hadronic cascades is by far more involved compared to the development of electromagnetic showers due to the richness of interactions that hadronic particles undergo when crossing matter. fission. is therefore: e −1 h E vis ηh ≡ = 1 + 1 − Fh (E). (2. Both the hadronic and electromagnetic component of the cascade contribute to the energy measurement in the active material. The total energy measured by the interaction of a high-energy hadron with initial energy E is therefore given by: ηh h E vis = [ηe Fπ 0 (E) + ηh Fh (E)] E = ηe 1 + 1 − Fh (E) E. [20] for a primer on calorimetry for particle physics. The interaction of a high-energy hadron with a typical calorimetric material. whereas the energy measurement is performed in the active material that samples the cascade. Photons are produced by two main mechanisms: from π 0 → γ γ and from nuclear de-excitations and (n. like an electron. resulting in particles with characteristic nuclear energy (100 keV÷a few MeV). (2. which depends on the initial hadron energy [20]. γ ) reactions.100) e E vis π ηe . although with different efficiencies. the latter produced by the interaction of photons with matter. evaporation. When a hadron produces a shower. spallation. or copper. Bando n. Indicate which mechanisms are responsible for the production of invisible energy and discuss at least one method to recover it. 156 2 Particle Detectors Since ηh = ηe in general. i.35). the response of the scintillator to fast neutrons is only marginally affected. in a sampling calorimeter made of high-Z material like brass. • the energy resolution is worse than for an electromagnetic calorimeter due to the stochastic fluctuations on Fh . uranium. By tuning the e/π ratio to unity. and because of the dependence of Fh with energy. Solution The origin of invisible energy in hadronic cascades can be tracked down to the production of delayed photons. Ge detectors place among the most precise detectors for γ spectroscopy below a few MeV. or lead.2.36 Which processes among pair-production. with quantitative description of compensation in real detectors. This problem can be greatly miti- gated by tuning the ratio ηh /ηe to unity. Bando n. the efficiency of detecting a photon which is entirely absorbed . Compton scattering.e. the fraction of active material. interleaved with a plastic organic scintillator. Eq.4 as a result of the lower efficiency of detecting the hadronic component. the energy resolution can be grearly enhanced.100) implies that • the energy response of a hadronic calorimeter is in general non-linear. which is again drained away in the form of low-range nuclear decays. and to the production of nuclear binding energy. are non-negligible in the interaction of γ emitted by a 60 Co source with a Ge detector? Which process has necessarily to happen in order to measure the total photon energy? Discussion Thanks to the large Z value and the small excitation energy. in a homogeneous calorimeter. the response to the electromagnetic cascade gets reduced proportionally to the sampling fraction. it is possible to compensate for it in a statistical sense by decreasing the sensitivity of the detector to the electromagnetic component. an important property of the detector is the photo- peak efficiency. since a recoil proton with T ∼ 1 MeV has a range of a few tens of microns. On the contrary. (2. Although such energy is not measurable.e. i. • the energy response is not gaussian.e. see Eq. For example. hence it will always interact in the active material regardless of its thickness. e/π ≈ 1. When dealing with γ radiation. (2. i. For example. 18211/2016 Problem 2. and photoelectric effect. Suggested Readings The review article [20] gives a concise but clear discussion of the phenomenology of hadron cascades. soft neutrons that undergo nuclear reactions giving low-range particles. The latter can be tuned by varying the thickness of the scintillator layers. see Table 2. by compensating the calorimeter for the intrinsically different response to the hadronic component. namely: 2k Tmax = E γ = 0. while Compton scattering should be the dominant interaction mechanism. After annihilation with an atomic electron.1 MeV. and ranges below 2 mm. hence the photoelectric effect is expected to be small for the 60 Co photons. the emitted e± have an energy of about (E γ − 2m e )/2 ≈ 75 and 150 keV and ranges of about 25 and 85 µm.139).e.101) so there is a finite probability that the photon undergoes one Compton scattering only before leaving the active volume. . 10. Fig.g.3 g cm−3 · 6 × 1023 mol−1 λComp = (n σComp )−1 = · 6 barn ≈ 4 cm.33 MeV. the 2m e rest energy of the e+ e− pair restores the full energy measurement if the two photons from positronium annihilation interact with the active material (the interaction length for 0. [1]. ≈ 5 × 10−2 barn.102) 1+ 2k where k = E γ /m e .32 MeV.1 MeV is about 1. The only reactions that guarantee a full energy measurement are therefore the photoelectric effect (probability ≈1%). If the photon undergoes Compton scattering. hence there is a non-negligible chance that the recoil electron escapes the active volume. if the latter is a a few mm thick.2%). (2. The maximum electron recoil energy is given by Eq. from Ref. the photo-peak efficiency is 1%. 72 g mol−1 (2. 6 keV. (1. 18211/2016 Problem 2.20 of Ref. and are therefore very likely to be fully contained in the active volume.96. hence just above the pair-production threshold E th = 2m e ≈ 1. respectively.2. [6]. and pair- production (probability ≈ 0. the range in Ge for an electron of kinetic energy 1.4 cm). σComp ≈ 6 barn.02 MeV. The interaction length for photons in Ge is given by: −1 5. the 60 Co isotope produces two monochromatic photon lines of energy 1.3 Functioning of Particle Detectors 157 by photoelectric effect.25 MeV. Bando n. In the latter case. The K -shell for Ge is located at 11 keV [9].2 mm [5]. like in practical Ge detectors. 1.16 and 1. For example.17 and 1. we find σp. Solution In its β-decay chain. and σpair ≈ 10−2 barn for E γ = 1. Indeed. and 15 keV. only the energy deposited by the recoil electron can be measured by the detector. The same holds for the photoelectrons. with full electron confinement. For Ge detectors and photons of order 1 MeV energy. see e.5 MeV photons is about 2.37 Estimate the contribution to the energy resolution (FWHM) due to the stochastic fluctuations in silicon calorimeters generated by photons of energy 2 keV. which have energies E γ − B ≈ 1. The Fano factor for ionisation detectors has been discussed in Problem 2. However. feature a Fano factor of about 0. Indeed. so that one can still assume that the whole photon energy is absorbed with little energy partitioning. This is seldom the case.103) When dealing with energy resolution with particle detectors. the resulting photo-electron will be rather soft as for Eq.f equals half of its value at the mean position μ. If a particle produces on average N = E/ε signal carriers through independent random interactions √ characterised by probability p. with F < 1. The mean excitation . 4 of Ref. σ ) = N (0. if the detector cannot but absorb all of the particle energy by converting it into detectable signal carriers. However. the ionised atom is in an excited state. the stochastic fluctuation in√this number is N from Poisson statistics. which undergo photoelectric effect from L-shells with the emission of secondary photoelectrons. We can therefore assume that the photon interacts with one atom by emitting an electron of a few keV energy. the multiplicity of the latter is ideally fixed to N and there would be no stochastic fluctuations at all. the photoelectric effect dominates the interaction of photons with silicon. i. or to the emission of short-range Auger electrons [5]. the secondary particles will release energy in the active medium. with F ≈ 0. see e. since there is in general a partioning of the energy transferred by the particle to the active material into more channels.12. Solution At energies below 15 keV. the photoelectron will most likely originate from a K -shell emission.158 2 Particle Detectors Discussion If the measured energy E is distributed according to a Gaussian law with mean μ = 0 and standard deviation σ . Since the K -edge in silicon is at 1839 eV [9]. σ ) ⇒ x± = ± 2 ln 2 σ ≈ ±1. At E = 2 keV.104) E E Semiconductors that absorb the full particle energy into eh-pairs. [6]. which will bring to the emission of either K -α and K -β photons. μ. i. Ref.: 1 √ N (x± .: σ Fε = (2. More informations can be found in Chap. In any case.e.35 σ (2.177 σ 2 σFWHM = (x+ − x− ) = 2.d.12 for silicon. an important concept is the so-called Fano factor (F). the FWHM resolution is defined as the interval such that the p. μ.g.e. and the relative energy resolution is 1/ N . [7]. This reduces the standard deviation √ the number of electron-hole pairs Neh from the Poisson of √ expectation of 1/ Neh to F/Neh .23). in some circumstances it is observed that the relative energy variance is smaller than the Poisson expectation by an empirical factor F. The photo-produced electron loses energy by collision loss and creates additional electon-hole pairs along its track. some of which may not produce signal carriers. (2.8. 106). [2].6 eV ⎨ σFWHM = 2. .2.106) E E · εQ εC 661 keV · 0. is used to measure the 137 Cs line: estimate the energy resolution by listing the contributing factors.2 of Ref. An other contribution may come from the dependence of the response with the photon impact point and from an imperfect shower containment. read-out by a phototube.2.0% E = 6 keV (2. see e. and on the overall efficiency of the photocathode. (2.2 for a typical PMT. The 137 Cs isotope produces a monochromatic X-ray emission with energy E = 661 keV.3% E = 15 keV Suggested Readings Reference [21] discusses in more detail the use of silicon detectors for γ spectroscopy.6 eV.45). A broader discussion on the phenomenology of photoelectric absorption in matter can be found in Ref.3 Functioning of Particle Detectors 159 energy is ε = 3. G 1). (2.35 = 3. see Table 2. Table 34. the relative energy resolution (FWHM) can be estimated to be: σFWHM ε 22 eV = 2. since there is no evidence for its presence in scintillators. where ε is the mean excitation energy.35 = 2.0%.37. which depends on the mean number of photons n γ = E/ε.2. 1N/R3/SUB/2005 Problem 2. see Table 2. [7]. 5N/R3/TEC/2005 Problem 2.4% E = 2 keV Fε 0. Assuming εQ εC = 0. No Fano factor has been accounted for in Eq. with examples of measured spectra from nuclear candles. (2.g.105) E E ⎪ ⎩ 1.38 A piece of Na I (Tl) scintillator. The electronic noise plays also an important role. We can therefore estimate the FWHM of the measured signal to be: ⎧ ⎪3. Bando n.35 = 2. Solution The energy resolution for a coupled scintillator-phototube detector is described by Eq.12 · 3. Bando n.2 see Problem 2. and negligible noise from the electronics and amplification statistics ( f N = 1. The main contribution to the energy resolution comes from the statistics of photoelectrons.39 Estimate the energy resolution at 140 keV of a photo-detector equipped with Na I (Tl) crystals.35 = 2. 2. see Table 2. Calculate the resolution (FWHM) for a 4 MeV particle assuming a total light collection efficiency εC εQ =1. the mean signal charge Q collected at the electrodes of the p-n junction and its standard deviation are given respectively by: E FE Q = e.41 Calculate the energy resolution for photons of energy E measured by a solid state detector with ionisation energy ε. Bando n.160 2 Particle Detectors Solution We can refer to Problem 2. the relative energy resolution (FWHM) can be estimated to be: σFWHM ε 10−4 MeV = 2. (2.35 = 2. 18211/2016 Problem 2.35 = 2. (2.2 for a typical PMT and negligible noise from the electronics and amplification statistics. Solution If the photon energy is intirely absorbed by the detector.109) ε ε . Assuming εQ εC = 1 and negligible noise from the electronics and amplification statistics. (2.45).38 for determining the energy resolution of a similar setup.35 = 1. σQ = e. Solution The energy resolution of the detector is described by Eq. (2.35 = 6.40 A scintillator emits 104 γ /MeV. leakage current Id .108) E E 4 MeV See Problem 2. and integration time of the associated electronics equal to TS . 13153/2009 Problem 2.37 for the definition of FWHM. Bando n. See Problem 2. Assuming εQ εC = 0. the relative energy resolution (FWHM) can be estimated to be: σFWHM ε 22 eV = 2.2%.107) E E · εQ εC 140 keV · 0.6%.37 for the definition of FWHM.2 where ε = 22 eV is the mean excitation energy for Na I (Tl). [2]. the leakage current contributes to the noise via an equivalent squared-charge: Q 2n = 2 e Id TS .3 Functioning of Particle Detectors 161 where F ≈ 0.8 of Ref.g. (2.2. 34. Since the noise from the leakage current and the statistical fluctuation in the number of signal carriers are uncorrelated.110) see e. In the integration time TS taken by the electronics to shape the signal. the relative energy resolution is given by the sum in quadrature: . Sect.12 is the Fano factor in silicon. 52. Electrons can instead reach much higher velocities compared to ions. (2. For ions moving in a gas. while it is almost independent of the electric field. Drift tubes are built according to this concept. and the time of formation of the electric signal at the anode. 2. σ Q2 + Q 2n σE ε 2 Id TS ε = = F+ (2.8 of Ref.112 implies a proportionality between the distance from the anode of the primary ionisation position and the drift time. Eq.112) where μ is called mobility and depends on the pressure P and temperature T of the gas. The latter is defined as the time interval between a fast trigger. [2]. 1N/R3/SUB/2005 Problem 2. Drift tubes (DT) are gaseous ionisation detectors that measure the time taken by the primary ionisation electrons to drift from their point of formation up to the anode. and the mobility μ depends on E in such a way that a saturation of the velocity at values of order 50 µm/ns is reached for E ∼ 1 kV/cm at STP. The position resolution is determined by the sum in quadrature . the reader is addressed to Sect. What does it imply for a multiwire chamber with wire spacing s = 2 mm. Bando n. Typical position resolutions achievable with DT are 100 µm over few drift lengths d of a few cm.111) E Q E e E Suggested Readings For a concise overview of low-noise front-end electronics for particle detectors. the drift velocity v is roughly proportional to the electric field intensity: v = μ E.42 The drift velocity of electrons in some gas mixture is v = 5 cm/µs. and what for a drfit chamber read-out by a TDC with 500 MHz clock? Discussion Multiwire chambers have been briefly discussed in Problem 2. By making the electric field as uniform as possible in the drift region. that provides the start time to the clock. 34. The position uncertainty induced by the TDC clock is therefore given by: σ y.113) 12 12 √ The factor 12 accounts for the fact that the particles arrive at the detector uniformly distributed across y. The time resolution is therefore given by σ yMW /2 580 µm/2 σtMW = = ≈ 5. Figure 2.10 provides a qualitative description of the position resolution as a function of the drift length d.114). The total resolution Diffusion is broken up into three main contributions: statistics of 50 the primary ionisation.162 2 Particle Detectors Fig. has a spatial resolution along the coordinate y orthogonal to the wires: s 2 mm σ yMW = √ = √ ≈ 580 µm. Solution A MWPC with wire spacing s = 2 mm.58 ns. (2. 2. noise Electronics from the electronics. and charge diffusion Primary statistics 0 d (mm) 0 10 20 of three dominant contributions: the statistics of primary ionisation (relevant at small d).116) .115) 12 12 √ Again.DTclock = σtDT · v = 0. the factor of 1/2 at the numerators comes from the fact that a primary ionisation generated outside of the ±s/2 range from a given wire will be detected by one of the two neighbouring wires. (2.8 ns (2. and electron diffusion (proportional to d). (2. the TDC clock period f −1 sets a minimum time resolution f −1 2 ns σtDT = √ = √ = 0.√the electronic noise (independent from d).58 ns · 5 cm/µs ≈ 29 µm. one has to divide by 12 since the actual arrival time at the anode is uniformly distributed across the time interval f −1 between subsequent clocks. For a DT readout by a time-to-digital converter (TDC).114) v 5 cm/µs In Eq. (2.10 Typical position σ (μm) resolutions in a drift chamber 100 as a function of the drift length d. with SNR = # ≡ . provides an estimator of the impact position with typical resolution of about d σxana ∼ . one surface of the wafer is grounded and the strips are implanted on the opposite side and connected to the bias voltage via DC or AC coupling. as discussed above. This can be easily proved by using the standard propagation of error for uncorrelated measurements. (4.#the one with # the largest signal yield. (2. 1N/R3/SUB/2005 Problem 2.6 eV. with p-n junctions shaped in the form of long and thin parallel strips separated by a distance (pitch) ranging between 20 and 200 µm. [22]. for a typical wafer thickness. i Qi j 2 j # j j j i Qi j Qj # d j Qj S σx̄ = . In a possible setup. Given that the average excitation energy is ε = 3. What is its spatial resolution? Discussion A silicon microstrip is a solid-state detector consisting of a wafer of doped silicon.33 MeV/cm ≈ 3. which is anyway smaller than the one from a typical MWPC. Typical position resolutions of a conventional DT is about 100 µm. the reader is addressed to Ref. The charge division method consists in an analog measurement of the signal from the strips close to the one which recorded the hit. for example.2.43 A depleted microstrip silicon detector has a strip pitch of 50 µm and operates without charge division.117) SNR where d is the strip pitch and SNR is the signal-over-noise ratio.118) SNR δQ 2 N j j . Bando n. The signal carriers drift under the effect of the bias voltage and the induced charge is measured by the front-end electronics. (2. where i runs over the strips and xi (Q i ) are the strip positions (measured signal). a total of 3 × 104 eh-pairs are produced on average across a 300 µm-thick junction. see Eq.66 · 2. 2.73): # # ∂ x̄ 2 x j − x̄ 2 2 j δQ j i Q i xi δ Q2 = x̄ = # ⇒ σx̄2 = ∂Q # δ Q 2 = d 2 2 . The overall position resolution depends however on other factors. it gets completely depleted by a bias voltage of order 100 V. A MIP loses 1.87 MeV/cm in silicon [5].10.e. Suggested Readings For a comprehensive review of DT. The centre-of-mass of the strip charges x̄ = i Q i xi / i Q i . of a high-resistivity n-type with typical thickness of about 300 µm. The junction may be realised by p + -type silicon and.3 Functioning of Particle Detectors 163 This term contributes to the electronic noise shown in Fig. i. 44 A silicon detector is made of a pixels with dimension 100 µm × 200 µm. [1]. thanks to their superior spatial resolutions in two dimensions. a lower bound to the spatial resolution (FWHM) in the two directions is given by: . we can estimate the spatial resolution (FWHM) as: d σFWHM x = 2. 10. see Prob- lem 2. 18211/2016 Problem 2. Solution If the detector is operated in digital readout.119) 12 Additional sources of uncertainty affecting the collection of charge carriers. δ-rays. which are independently read-out. Solution In the absence of charge division. see Problem 2. Bando n. like thermal diffusion. if the detector has digital readout? Discussion Pixel detectors are semiconductive detectors where the active volume is segmented in small picture elements (pixels). multiple-scattering.35 √ = 34 µm. Suggested Readings For a first introduction to microstrip detectors. Planar pixel detectors are commonly employed in HEP experiments as vertex detectors.43. (2. if the strips can be read in digital mode only. (2. What is the smallest spatial resolution in the two dimensions. Since the particle flux can be asumed to be uniformly distributed across the microstrip detectors. and their close-to- ideal efficiency to detect the passage of ionising particles. which allows for a small occupancy even at the closest distance to the interaction point. the position resolution is given by the strip pitch: d σxdig = √ .120) 12 √ where the factor of 1/ 12 comes from the assumed flux uniformity.6 of Ref.164 2 Particle Detectors Conversely.42. the reader is addressed to Sect. should be also considered for realistic detectors. the spatial resolution of a microstrip detector is primarily determined by the pitch size d. 18211/2016 Problem 2.42.3 Functioning of Particle Detectors 165 σFWHM x = 2. Bando n.2. is approximately given by: . [21].121) σFWHM = 2.35 √12 y = 136 µm √ where the factor of 1/ 12 comes from the assumed flux uniformity.45 Why is a diode used as radiation detector usually operated with an inverse bias? Solution A p-n junction operated at inverse bias give rise to an active region depleted from mobile charge where an intense electric field can sweep out free charges liberated by a ionising particle. The thickness of the depletion zone for the case of a silicon p-n junction realised by a p + -doped material put into contact with a lightly doped n region. Suggested Readings For a comprehesive introduction to pixel detectors in HEP experiments.35 √d12 x = 68 µm y d (2. see Prob- lem 2. the reader is addressed to Ref. the thickness of the depletion region would change from 70 to 700 µm. For example.46 Consider a D 0 meson produced with an energy of 20 GeV. 20 of Ref. The importance of applying an inverse bias to the junction as to enlarge the active volume is made clear by Eq. for typical values ρn = 2 × 104 Ω cm. Determine the spatial resolution necessary to measure the production and decay vertex position. ρ V + V n 0 bias W = 0. if a reverse bias Vbias = 100 V is applied. Suggested Readings An introduction to the physics of semiconductors for particle detectors can be found in Chap.5 µm.122) Ω cm V where ρn is the resistivity of the n-type region. and Vbias is the bias voltage.g.122). 13705/2010 Problem 2. (2. Bando n. and indicate which detectors are best suited for an efficiency exceeding 90%. [2]. (2. V0 ∼ 1 V is the barrier voltage. see e. . [1]. Ref. 2. (2.4 cm and a MS mean angle of about 2 × 10−4 rad at |p| = 10 GeV. either pixel. (2. see Ref. if we want to reconstruct at least 90% of the D 0 decays from their decay vertex. see Problem 3. Discussion For E = 20 GeV. the vertex resolution must be smaller than about 140 µm. (1. Assuming a MS-dominated regime.11. Modulo resolution effects. The probability of surviving up to a distance d ot more from its production vertex is given by Eq. 25]. Conversely. which . with equally likely values of sip .126) where r1 is the distance of the innermost silicon layer from the interaction point and θ12 is given by Eq.9.8). see Fig. one has r1 = 4. Therefore.124) where we have used the fact that γ ≈ 11 is large. A more realistic simulation.174): mc d 1 d P [x ≥ d] = exp − = exp − (2. This property can be exploited to define tagging algorithms for displaced vertexes and experimental methods to measure their efficiency in data [24.105 · 123 · 11) µm ≈ 140 µm. A good channel to reconstruct its decay is D 0 → K ± π ∓ . In this regime. the impact point resolution is given by: σip ≈ r1 θ12 .or mictrostrip-based. the quantity j sip = sign ip j · (SV − PV) (2. the detector rsolution smears the impact point of tracks emerging from the PV around zero.41 ps. assuming the design of the CMS pixel detector. and is related to the resolution on the position of the secondary vertex (SV). giving σip ≈ 10 µm.9) c τ γ = (0. (2.123) |p| cτ γ − 1 cτ 2 Requiring this probability to be at least 90% is equivalent to impose that the flight distance should be in excess of: d90% = (− ln 0. multiple scattering usually dominates the tracking resolution when using silicon detectors with pixel/pitch size 100 µm. the decay products have energy of about 10 GeV each.166 2 Particle Detectors Solution The D 0 meson decays via the electroweak interaction with a lifetime of about 0. The impact point resolution (σip ) is the uncertainty on the position of closest approach of the track extrapolation to the primary vertex point (PV). [26]. This can be easily achieved by silicon-based vertex detectors. corresponding to cτ ≈ 120 µm [2]. For example.125) should be positive for tracks emerging from the same secondary vertex. at room temperature. Suggested Readings For an overview of tracking and vertexing performances at the LHC. 18211/2016 Problem 2.127) where E gap is the energy gap.7 of Ref.11 Cartoon pπ illustrating the two-body decay of a D 0 meson. Bando n.41). gives about a factor of 2 larger resolution. For example. the reader is recommended to read the review article [26].48 Which property of a SiPM makes it a preferable solution cimpared to a conventional PMT for an integrated imaging PET-MRI system? .3 Functioning of Particle Detectors 167 Fig. 34. which still satisfies the constraint of Eq.124). Although the same effect exists in silicon. see e. 18211/2016 Problem 2. corresponding to a factor of 3 × 103 larger leakage current for the latter. the energy gap in the latter is larger than in germanium. The pK distance of closest approach of the extrapolation of the SV daughter particles d ipK trajectories to the primary vertex (PV) is called impact PV parameter ipπ includes measurement uncertainty and MS in the beam pipe. 2 kB T Id (T1 ) T1 2 kB T2 T1 (2.1 eV for silicon and 0.g.47 Explain why Ge sensors need to be cooled. (2. Bando n. while Si sensors do not. The bias current depends exponentially on the temperature T : 2 E gap Id (T2 ) T2 E gap 1 1 Id (T ) ∝ T exp − 2 ⇒ = exp − − . Sect. [2]. Solution Germanium detectors are commonly operated at liquid nitrogen temperature (T = 77 K) to reduce the leakage current Id due to thermal excitation.7 eV for germanium. one has E gap = 1.2. see Problem 2. 2. and hence the electronic noise and power consumption. In the first case. Suggested Readings For an introduction to SiPM’s. drift chamber.168 2 Particle Detectors Discussion Silicon photomultipliers (SiPM).5 mm2 to 5 × 5 mm2 .1 of Ref. which suffers from the presence of magnetic fields. When a eh-pair is created in the depleted region. Bando n. Table 7. for example by altering the gain. see e. See Problem 3.5 × 0.e. [1]. with decay times of a few ns. Solution SiPM’s represent a convenient alternative to PMT’s for applications in environment with intense magnetic field. Solution Plastic scintillators are generally faster than inorganic scintillators. if we assume that the first event occurred at time t0 . since the amplification stage in a SiPM does not require the photoelectrons to be accelerated along the dynode of conventional PMT’s. the new event does not change the detector state at all. also known as Pixelized Photon Detectors (PPD) are photodetectors composed by an array of pixel-size photodiodes with typical size ranging from 25 × 25 µm2 to 100 × 100 µm2 . Which one would you chose for a time measurement with resolution of a few hundred ps? Discussion The dead time τ is the time required by a detector to process one event and be ready to accept a new event. regardless of what happens meanwhile. the reader is addressed to the dedicated PDG review [2] and references therein. 1N/R3/SUB/2005 Problem 2. plastic scintillator. two types of dead-time exist: extendable or not-extendable. The proportionality with the total input photons is restored by summing the binary cell outputs from the full array. typically from 0. i. the arrival of a new event at a time t1 < t0 + τ shifts the time at which the detector is ready to accept and process a new event to at least t2 = t1 + τ . like in positron emission tomography-magnetic resonance imaging (PET-MRI) applications. Depending whether the detector is sensitive or not to a new event while processing the previous one. In the second case. the intense electric field triggers the formation of an avalanche.38 for more details. with a bias voltage in excess of the break-down voltage. and operated in Geiger mode. Fast photodetectors can also have risetimes . but proportionality between the number of photons impinging on a given cell and the collected charge is lost.49 Order the following detectors by decreasing dead time: silicon. and the subsequent event can be accepted and processed at any time t1 ≥ t0 + τ . The high bias voltage provides large gains per incident photon and per pixel. packed over a small area.g. 18211/2016 Problem 2.3 Functioning of Particle Detectors 169 below 1 ns. [2]. i. the dead time is mostly due to the time taken by the primary ionisation electrons (ions) to drift to the anode (cathode). 5.2. can be found in Sect. Solution For what concerns the pileup of multiple events. 2. (3.e.1 of Ref. Ref. A time measurement with a few hundreds ps time resolution is best accommodated with plastic scintillators coupled to fast photo multipliers. [2]). 34. Bando n. see Problem 2. with a fast sampling frequency of the readout electronics as to allow for the full pulse shape reconstruction. A coupled scintillator-photodetector system is ready to accept and process a new event after the fluorescent excitation from the previous event have decayed to the ground level. we can therefore invert the equation and express the true rate ν as a function of the measured rate m and of the dead time τ . we can use the same line of thought used to relate the true and measured rate in a non-paralyzable system. see e. The readout electronics further increases the processing time to at least 50 ns. [1]. During this time. Estimate an upper bound to the processing time of the analog pre-amplifier and shaper. Table 34. but the time needed to collect the full charge released in the depleted zone can take a few tens of nanoseconds (10 ns for electrons and 25 ns for holes in a 300 µm thich detector. (2.50 The mean counting rate on single electrode for a given detector is 150 kHz.186) with ε = 1. [2] summarises the typical resolutions and dead times of common charged particle detectors. a new event would cause pile-up and confusion on the time measurement. which can take about 10 ns for fast scintillators. like microchannel plate (MCP) or gas electron multipliers (GEM). see Sect.128) 1−mτ Requiring that the pile-up is less than δ = 3% amounts to require that the ratio between the measured rate and the true rate is larger than 1 − δ. or equivalently: .38. In a drift chamber.42.g. Suggested Readings More details on the dead time of particle detectors. A silicon strip or pixel detector has time resolutions of a few ns. see Problems. including techniques for measuring it in the laboratory. Referring to Eq.7 of Ref. if the pile-up probability has to be maintained below 3%. For a typical electron velocity of 5 cm/µs. the time needed to drift over 1 cm is about 200 ns.7 of Ref.49 and 3. m ν= . 10% C H4 ) at STP can be estimated from Diethorn formula: V0 ln 2 V0 ln M = ln ≈ 7. wire radius r . see e. R = 5 cm. c0 E(d) = . (2. By using the cylindrical symmetry of this configuration.130) d where c0 is a constant that depends on the boundary conditions.e. . and anodic tension V0 .g. Table 6. For example. this fully determines the potential to be: V0 V (d) = ln(d/R). the gas multiplication factor M for a cylindrical chamber filled with P-10 gas (90% Ar.5 × 103 . What is the value of the electric field at a distance d ≤ R from the anode? Solution Let’s assume that the anode is connected to a potential V0 > 0 and that the cathode is grounded.3 ⇒ M = 1. (2. V0 = 2 kV.131) ln(r/R) from which we get the result: ∂V V0 1 E(d) = − = .132) ∂d ln(R/r ) d Discussion The d −1 scaling of the electric field makes the cylindrical tube suitable for charge multiplication. Since E = −∇V . 18211/2016 Problem 2. it is easy to prove that the electric field must be radial.133) where p is the gas pressure and K and ΔV are gas-specific parameters. As an eample. (2. i. Together with the boundary conditions at the two electrodes. [7] for a few examples. i. the field intensity E(r ) must scale as d −1 . which is enough to trigger the formation of an avalanche with its resulting charge multiplication. the electric potential V (d) must be proportional to ln d.170 2 Particle Detectors m δ 0. (2.e. the electric field at a distance of 100 µm from the wire is about 20 kV/cm.1 of Ref.51 A proportional cylindrical tube has inner radius R.03 =1−mτ >1−δ ⇒ τ< = = 200 ns. assuming typical values r = 20 µm. E = E(d) er .5 × 105 Hz Bando n.129) ν m 1. ln(R/r ) ΔV p r ln(R/r ) K (2. The wire acquires a charge with uniform linear density. By virtue of Gauss law. A gaseous detector consists in a pair of electrodes kept at different electrostatic potentials and separated by a gaseous medium.2. A plane capacitor with small inter-plane distance is an other option. and even RPC.e. only the y coordinate can be measured with good resolution.6 of Ref. like in the form of parallel strips. the x resolution is instead determined by the granularity of the cathode. and couple the anode directly to the readout electronics.52 The cathode readout can be used in wire detectors. limited streamer tubes (LST). The cathode confines the electric field and shields the detector from the outside.53. The anode is usually shaped in a way as to produce intense electric fields nearby its surface. The usual way of opertaing a gaseous detector is to ground the cathode and read the anode in AC-coupling. 1N/R3/SUB/2005 Problem 2. The passage of a ionising particle induces the formation of an electron avalanche in the neighbourhood of the anode. offers the possibility of measuring also the x coordinate. LST. [1]. Solution Let’s consider the case where the anode consists in a set of parallel wire with small inter-distance. i. which is e. If the cathode is segmented along y. What does it mean? What are the main advantages of this setup? Discussion Multiwire proportional chambers (MWPC). one can set the cathode at a negative bias voltage.3 Functioning of Particle Detectors 171 Suggested Readings An introduction to the physics of electronic avalanches in gas can be found in Ref. then a cathode readout. 6. The positive ions drift towards the cathode inducing a signal (a time-dependent voltage pulse) between the two electrodes. and resistive plate chambers (RPC) are all examples of gaseous ionisation detectors that measure the ionisation charge left behind by particles interacting with the gas. [8]. i. A metallic wire kept at a positive voltage bias is the solution at the basis of the MWPC. [1. separating the bias voltage from the readout electronics by means of a capacitor. A more advanced and complete reference on the subject is provided by . For a more comprehensive review of gaseous detectors. If the detector is operated in anode readout. If the readout is digital-only. the reader is addressed to Ref. If the cathode readout is analogic. a centre-of-gravity method allows to measure the x position with high precision (indeed. time projection chambers (TPC).e. used in RPC detectors. Suggested Readings An introduction to gaseous detectors and to their readout can be found in Sect. 7]. like multiwire chambers. TPC. and LST technology. and let y be the orthogonal coordinate. see Problem 2. Bando n. a measurement of the pulse induced at the cathode. stretched along the coordinate x. Alternatively. only limited by the noise of the electronics).g. TPC. 172 2 Particle Detectors Fig. 2.12 AC and DC V0 couplings for a generic detector R R C readout readout detector detector V0 Ref. [8]. A stimulating discussion on this subject can be also found in the Nobel lecture by G. Charpak (1992), the inventor of the MWPC. Bando n. 18211/2016 Problem 2.53 Does a radiation detector AC-coupled to its electronics have a larger noise compared to a DC-coupled detector with the same electronics? Discussion The readout electrode of a charge-sensitive detector, like a microstrip silicon detector, an RPC, a MWPC, etc., can be either set to a large bias voltage or be grounded. In the former case, the front-end electronics, which usually starts with a pre-amplifier, needs to be decoupled from the bias voltage by a capacitance (AC-coupling). In the latter case, the electrode can be directly accessed by the pre-amplifier (DC-coupling), see Fig. 2.12. Solution AC-coupling offers the advantage of having the opposite electrode (e.g. the cathode, for wire detectors) grounded, resulting in a convenient configuration to insulate the detector. However, it provides an extra decoupling capacitance in input to the readout chain, thus increasing the electronic noise compared to a DC-coupling. Indeed, for a capacitive sensor, the charge-equivalent noise Q n can be parametrised as: en2 Fv Q 2n = i n2 Fi TS + + Fv f A f C 2, (2.134) TS where C is the sum of all capacitances shunting the input, i n2 and en2 are the quadratic current and voltage noise densities, TS is the characteristic shaping time, and Fi,v,v f are devise-specific constants [2]. 2.3 Functioning of Particle Detectors 173 Suggested Readings For an introductory discussion on the readout of silicon detectors, the reader is addressed to Sect. 10.9 of Ref. [1]. More details on low-noise electronics for capacitive detectors can be found in the dedicated PDG review [2]. Bando n. 1N/R3/SUB/2005 Problem 2.54 A discriminator is operated with a threshold Vth = 0.4 V and receives in input signals that have a constant rise-time equal to TS = 10 ns, but an amplitude variation between Vmin = 0.5 V and Vmax = 1 V. Estimate the lower bound on the time resolution due to the variable amplitude. Which technique would you use to reduce such an effect? Discussion A discriminator is a device that produces a digital signal when an analogical input pulse overcomes a predefined threshold. A discriminator in combination with a TDC device can be used for timing measurements of signals. When the input signals differ in amplitude and/or rise-times, the time measurement performed by a discriminator with fixed threshold is affected by event-by-event fluctuations on the pulse shape, giving rise to the so-called time walk. A number of time-pickoff methods can be deployed to mitigate the walk effect. A common method is based on the constant fraction triggering (CFT), which consists in analysing the zero-crossing of a signal obtained by a linear combination of the pulse V , delayed by a fixed time τd , with −k V , where k is an attenuation coefficient. The triggering time tR is defined as the time at which: V (tR − τD ) − k V (tR ) = 0. (2.135) Since Eq. (2.135) is homogeneous in V , signals with the same time-shape, but different amplitude, will give the same triggering time tr , see Fig. 2.13. This method is however affected by a residual walk effect if the pulse shape differ from one event to another. In this case, one can try to reduce the delay time τD as to trigger on the rising edge of the signals, where event-by-event changes are smaller, a technique known as amplitude and risetime compensation (ARC). Solution The time derivative of the signal is distributed in the range dV Vmin Vmax V ∈ , = [0.05, 0.1] . (2.136) dt TS TS ns Signals with time derivatives at the edge of the interval of Eq. (2.136) will trigger the disciminator at times: 174 2 Particle Detectors Fig. 2.13 Application of the CFT technique to a pair of Gaussian-like signals with different amplitude Vth 0.4 V tmin = = = 4 ns Vmax /TS 0.1 V/ns Vth 0.4 V tmax = = = 8 ns (2.137) Vmin /TS 0.05 V/ns The time walk Δt is therefore given by Δt = tmax − tmin = 4 ns. (2.138) A technique to eliminate the time walk is for example the constant fraction triggering, which is appropriate for this case since the signals feature the same rise-time. Suggested Readings Discriminators are briefly discussed in Sect. 14.0 of Ref. [1], while a few time- pickoff methods are described in Sect. 17.2 of the same reference. The reader is also addressed to Ref. [23] for application of discriminators in experiments. Appendix 1 The computer program below illustrates the numerical evaluation of the information Iεπ from Problem 2.28. The algorithm approximates the Rieman integral by the finite sum of rectangles: xi+1 − xi d x f (x) ≈ f · Δx i 2 2.3 Functioning of Particle Detectors 175 The integral to be approximated is given by: +∞ 2 ∂ 2 f (x, επ ) ∂ ln f (x, επ ) Iεπ =E − ≡ dt f (t, επ ) − , (2.139) ∂ 2 επ −∞ ∂ 2 επ with: ∂ ln f (x, επ ) N (t; tπ , σt ) − N (t; t K , σt ) = (2.140) ∂επ f (t, επ ) ∂ 2 ln f (x, επ ) [N (t; tπ , σt ) − N (t; t K , σt )]2 =− (2.141) ∂ 2 επ f (t, επ )2 import math # gaussian function def gaus(x, m, s): return 1./math.sqrt(2*math.pi)/s * math.exp(-math.pow(x - m,2)/2/s/s) m_pi = 6.68 $ TOF\index{Time of flight@Time-of-flight} for pi m_k = 6.87 # TOF\index{Time of flight@Time-of-flight} for K sigma = 0.2 # std of TOF\index{Time of flight@Time-of-flight} measurement def integrate(x_l=6.0, x_h=7.5, step=0.01, f_pi=0.5): integ = 0.0 n_step = int((x_h-x_l)/step) for s in xrange( n_step ): t = x_l + (s+0.5)*step g_pi = gaus(t, m_pi, sigma) g_k = gaus(t, m_k, sigma) val = math.pow(g_pi - g_k, 2)/(f_pi*g_pi + (1. - f_pi) * g_k ) integ += val*step return integ ############################## for f_pi in [0.1, 0.3]: res = integrate(x_l=5., x_h=10, step=0.001, f_pi=f_pi) print f_pi"==>", res References 1. W.R. Leo, Techniques for Nuclear and Particle Physics Experiments, 2nd edn. (Springer, Berlin, 1993) 2. C. Patrignani et al., (Particle Data Group). Chin. Phys. C 40, 100001 (2016) 3. J.D. Jackson, Classical Electrodynamics, 3rd edn. (Wiley, New Jersey, 1999) 4. Yung-Su Tsai, Rev. Mod. Phys 46, 815 (1974) 5. http://pdg.lbl.gov/2016/AtomicNuclearProperties 176 2 Particle Detectors 6. National Institute of Standards and Technology. XCOM: Photon Cross Sections Database (2012), http://nist.gov/pml/data/xcom/index.cfm 7. G.F. Knoll, Radiation Detection and Measurement, 4th edn. (Wiley, New Jersey, 2010) 8. F. Sauli, Gaseous Radiation Detectors : Fundamentals and Applications (Cambridge Univer- sity, Cambridge, 2014) 9. A. Thompson et al., X-ray Data Booklet, LBNL/PUB-490 Rev.3 (2009) 10. D.E. Groom et al., Atomic Data Nucl. Data Tab. 78, 183 (2001), https://doi.org/10.1006/adnd. 2001.0861 11. A.A. Kochanov et al., Astropart. Phys. 30, 219233 (2008), https://doi.org/10.1016/j. astropartphys.2008.09.008 12. ALEPH Collaboration Nucl, Instr. Methods A 360, 481 (1995) 13. M. Peskin, D.V. Schroeder, An introduction to Quantum Field Theory (Advanced Book Pro- gram, Westview Press, 1995) 14. CMS Collaboration, Accepted by JINST, CMS-PRF-14-001, CERN-EP-2017-110, arXiv:1706.04965 (2017) 15. H. Primakoff, S. Rosen, Rep. Prog. Phys. 22, 121–166 (1959) 16. Stopping-power and range tables for electrons, ICRU Report No. 37 (1984) 17. T. Tabarelli de Fatis, Eur. Phys. J. C 65, 359 (2010), https://doi.org/10.1140/epjc/s10052-009- 1207-8 18. K.S. Hirata et al., Phys. Rev. D 38, 448–457 (1988), https://doi.org/10.1103/PhysRevD.38.448 19. R. Cahn, G. Goldhaber, The Experimental Foundations of Particle Physics, 2nd edn. (Cam- bridge University, Cambridge, 1989) 20. C.W. Fabjan, F. Gianotti, Rev. Mod. Phys. 75, 1243–1286 (2003), https://doi.org/10.1103/ RevModPhys.75.1243 21. L. Rossi et al., Pixel Detectors (Spinger, Berlin, 2006) 22. W. Blum, W. Riegler, L. Rolandi, Particle Detection with Drift Chambers (Springer, Berlin, 2008) 23. R. Frühwirth et al., Data Analysis Techniques for High-Energy Physics, 2nd edn. (Cambridge Press, Cambridge, 2000) 24. ATLAS Collaboration, JINST 11 P04008 (2016), https://doi.org/10.1088/1748-0221/11/04/ P04008 25. C.M.S. Collaboration, JINST 8, P04013 (2013), https://doi.org/10.1088/1748-0221/8/04/ P04013 26. L. Rolandi, F. Ragusa, New J. Phys. 9, 336 (2007), https://doi.org/10.1088/1367-2630/9/9/336 Chapter 3 Accelerators and Experimental Apparatuses Abstract The subject of the third chapter is the motion of charged particles induced by electromagnetic fields. The first section focuses on the kinematics of charged particles inside static fields and its application for particle tracking. The second section is dedicated to the physics of accelerators. The last section is devoted to the concept of luminosity and event rates at colliders. 3.1 Tracking of Charged Particles Tracking a charged particle is the general problem of determining the particle trajectory by interpolating a collection of position measurements. In some circumstances, one can assume that the particle trajectory gets sampled by the detector with negligible impact on the particle kinematics: the equation of motion of the particle is therefore the same as if there were no detector at all. While this is in general the case for high-energy and minimum ionising particles and for sufficiently thin detectors, this assumption breaks down when the particles have a high probability of interacting inside the detector. In general, the interaction of the particle with the detector has to be accounted for, and a variety of techniques, either analytical or Monte Carlo-based, exist for the purpose of estimating the kinematics of the particle cleared from detector effects. Two main cases should be considered when dealing with the tracking of charged particles. In the first situation, the particle momentum is assumed to be known by other means, or perhaps is not relevant at all, and one is rather interested on the track direction and/or position at an arbitrary location in space, given a set of measurements. In the second situation, tracking is performed in a static magnetic field at the purpose of extracting the particle momentum. Linear Tracking Let us assume that the tracking system consists of N + 1 measuring stations located at the positions x0 , . . . , x N , equally spaced across the spectrometer length L (also known as lever arm), and characterised by the same spatial resolution σ . In the absence of a magnetic field, the particle trajectory is a straight line. The measured points y = (y1 , . . . , y N ) can be therefore fitted to a linear function: © Springer International Publishing AG 2018 177 L. Bianchini, Selected Exercises in Particle and Nuclear Physics, UNITEXT for Physics, https://doi.org/10.1007/978-3-319-70494-4_3 178 3 Accelerators and Experimental Apparatuses Fig. 3.1 Examples of linear L tracking using N + 1 equally spaced measuring stations over a total lever arm L x0 x1 ... xc ... xN −1 xN y(x) = a + b x, (3.1) see Fig. 3.1. Since the vector y of measurements depends linearly on the unknown parameters a and b, the theory of χ 2 estimators can be applied analytically to yield the estimators of the intercept (â) and slope of the line (b̂). The variance of the estimators are given by [1]: σ2 12 N σ2 Var b̂ = , Var â = + x 2 Var b̂ , meas. L 2 (N + 1)(N + 2) meas. N +1 c meas. (3.2) where xc = (x N + x0 )/2 is the coordinate of the middle point of the spectrometer. Furthermore, if xc = 0, the two estimators are uncorrelated. Given N + 1 measuring stations and a total lever arm L, the configuration leading to the smallest possible value of Var b̂ corresponds to half stations clustered at the front and half at the rear of the spectrometer, giving opt. σ2 2 Var b̂ = 2 (3.3) meas. L (N + 1) The effect of multiple scattering through small angles (MS), on the slope, see Sect. 2.1, is given by the second of Eq. (2.8), namely: y2 θ y2 Var b̂ = = , (3.4) MS L2 3 where θ y2 can be computed according to Eq. (2.7) taking into account the full material budget in the the N + 1 stations. Tracking in Magnetic Field In the presence of a static magnetic field, the trajectory of a charged particle is no longer a straight line. In the simplest case, the magnetic field can be assumed uniform inside the spectrometer, B(r) = B0 . The trajectory is then given by a helix, The bending angle θ = L/R is assumed to be small. However. the lever arm.2. For large transverse momenta. the magnetic field. b̂. The momentum component onto this plane is usually denoted as transverse momentum pT ≡ |p| cos λ. (3. so that Eq. the measurement vector y depends linearly on the unknown parameters. the magnetic field intensity B. it follows that: θ θ2 L2 s = R 1 − cos ≈R = . the radius R is usually much larger than the lever arm L. (3. and is usually denoted by s. (3.7) 2 8 8R Regardless of the details of its design. For small uncertainties δs one has: δpT δR δs δy 8 δy R 8 δy = = ∼ = 2 = pT pT meas. etc.1 Tracking of Charged Particles 179 see Problem 3. such that the bending power of Eq.43) is small.e. namely the lever arm of the spectrometer L. a simplified treatment of tracking in magnetic field for energetic particles. the trajectory is an arc of length L and radius R given by Eq.3.e. the material budget.e. and ĉ. the position resolution of the measuring stations. and the material budget. i.3 z B L 2 . The momentum resolution δpT / pT achievable by tracking a charged particle in magnetic field depends on the layout of the spectrometer. (3. i. (3.5) can be approximated to first order to give: y(x) = a + b x + c x 2 . let’s consider the trajectory of a charged particle on the plane transverse to the magnetic field.45). from which the particle momentum can be inferred.2. allows to derive the dependence of δpT / pT upon the main dimensionful quantities involved in the problem. The uncertainty on the sagitta will be of the order of the single-point uncertainty δy. the spectrometer will sample the track in a number of points. (3. (3. R s s L q B L2 8δy = pT . the number and location of the measuring layers. Again. δs ∼ δy. To this purpose.5) q |B| where λ is the angle between the particle momentum and the plane orthogonal to the B field direction (dip angle). so that a χ 2 estimator can be obtained analytically to yield the estimators â. thus allowing to indirectly measure s. with R = . The maximal distance of the arc from its cord is the so-called sagitta. the position resolution δy. i. From its definition. whose projection onto the plane orthogonal to the magnetic field is a circle of radius R: |p| cos λ (x − x0 )2 + (y − y0 )2 = R 2 . conveniently measured in radiation lengths X 0 . As shown in Problem 3.8) 0.6) with c = (2R)−1 . 0136 1 = . and is inversely proportional to B L 2 . this effect should be evaluated by considering the full spectrometer design. However.3 z B L βpT X0 1 0.3. the relative momentum resolution from √ MS alone does not depend on the momentum. Again. T. (3.8). Since the uncertainty from the position measurement and the MS inside the detector can be assumed to be independent. and m are used. and is inversely proportional to B L. It also depends on the traversed material as −1/2 ∼ X 0 . (2.3 B β L X0 and again the last equality assumes that the units GeV/c. and less in gas and low-Z materials. Therefore. T. the main features and the order-of-magnitude of the MS-induced uncertainty can be determined by considering a simple configuration where the particle track receives a deflection angle δθ = θT2 as given by Eq. and m are used. namely the multiple scattering (MS) inside the spectrometer. Therefore. This angle will induce a shift in the measured radius R such that: L L δθ = δ = 2 δ R. the overall momentum resolution is generally parametrised as: . see Prob- lem 3. At low momenta.0136 L = = δθ = z pT MS R L 0. R R δpT δR R pT 1 0.180 3 Accelerators and Experimental Apparatuses where the last equality assumes that the units GeV/c.9) 0. another effect usually dominates the momentum resolution. so the effect is more relevant in condensed and high-Z materials. the relative momentum resolution from the position measurements alone depends linearly on the momentum. the N + 1 measuring stations are assumed to be characterised by the same spatial resolution σ . where the relative energy resolution improves with energy. (3. the measuring stations are uniformly spaced. (2. 0.93). δpT δpT δpT = ⊕ = a pT ⊕ b.11) pT2 meas. the relative momentum resolution of tracking detectors gets worse at larger momenta. It can be shown [1] that the theory of χ 2 estimators provides the following relative momentum resolution: √ δ pT σ AN = .10) pT tot pT meas. pT MS where a and b are constants. In all cases. see Eq. Differently from calorimeters. They are illustrated in Fig.2. (3.3 |B| L 2 . Three configurations are worth being considered because of the possibility to determine analytically the corresponding momentum resolution. 3. In the first configuration. near to the magnet. 3. It can be shown [2] that a configuration like the one illustrated in the third cartoon of Fig. √ δ pT σ BN 256 = . we consider the case where the lever arm is L. thus showing that the estimate of Eq. N + 1 0.2 up to N ≈ 10. yields a relative momentum resolution: δpT 8σ 1 =√ . one may wonder what is the configuration that yields the smallest momentum resolution.13) pT2 MS 0. 3.8) applies to this configuration to better than 20% for typical values of N .4.8. compared to Eq.2.3 ÷ 1. and the other half are equally distributed at the rear and at the front of the magnetised volume. Given a fully magnetised spectrometer with lever arm L and N + 1 available stations.3. with half stations uniformly distributed at the front and rear of the spectrometer and the other half at the centre. i.e.4 smaller relative resolution for N ≈ 10. (3. (3. moving inside a static and uniform electric E and magnetic field B. mass m. Problems Problem 3.0136 CN = . (3. The answer is provided by the second layout of Fig. The effect of MS has been calculated as well [1].11).3 |B| β L X0 with C N = 1.15) pT2 meas. with B N = . (3.1 Tracking of Charged Particles 181 where A N is a function of N given by [1]: 720 N 3 720 AN = ≈ .3 |B| L 2 N +1 thus giving a factor of about 1. (3. and initial velocity v0 .3 |B| L A simple proof is proposed in Problem 3. A N /8 ≈ 0. . Finally. 0.12) (N − 1)(N + 1)(N + 2)(N + 3) N +5 √ Numerically.1 Determine the equation of motion of a classical point-like particle of charge q.14) pT2 meas. (3. giving δ pT 1 0. but the magnetised volume has a length of < L.8 ÷ 1.2. It can be shown that the momentum resolution from measurement uncertainty only achievable by such configuration is given by [1]: opt. where half of the stations are clustered around xc . the spectrometer is B not fully magnetised and the bending power is only |B| (bottom). and rear of the L spectrometre (centre).. In the last case. xc .. the spectrometer is immersed in a uniform magnetic field providing a total bending power of |B|L.16) described the Lorentz force. (3. and.17) and the . 3.17) dt which is a system of three coupled ODE. We choose the reference frame so that the z-axis is aligned along B.2 Examples of L particle tracking using N + 1 measuring stations over a total lever arm L.e. so that E = (E ⊥ . Equation (3.. (3. 0. The equation of motion r(t) is determined by Newton’s law: dp = F = q (E + v × B). Two layouts of the measuring stations are x0 x1 . 0. In the first B and second case. E ) with E ⊥ > 0. the measuring stations are equally distributed at the front and x0 xc xN rear of the magnetised N +1 N +1 N +1 N +1 volume.. and at the front and 4 4 L 4 4 rear of the spectrometer as to achieve a lever arm L B x0 xN Solution A classical particle of charge q and mass m moving inside a uniform electric field E superimposed to a uniform magnetic field B is subject to the classical force F = q(E + v × B). xN −1 xN deployed: uniformly spaced N +1 N +1 N +1 4 2 4 (top) and clustered at the front.16) The second term in the right-hand side of Eq. B = (0. |B|).182 3 Accelerators and Experimental Apparatuses Fig. middle. and the y-axis is parallel to E × B. i. We further define the origin such that at time t = 0 the particle is at the origin of the reference frame. (3. 24) ωB |B| |B| ωB . (3.20). we have an equation in x(t) alone: E ⊥ /|B| + v0y ẍ = −ωB2 x + (E⊥ + ωB v0y ) = −ωB2 x+ .19) m m m The first of Eq. ẏ(0) = v0y . y(0) = 0 ⎪ ⎩ eE m ⎪ ⎩ ⎪ ⎩ z̈ = m ≡ E ż(0) = v0z z(0) = 0 (3.20) ⎪ ⎩ ż = E t + v0z Inserting the second of Eq.1 Tracking of Charged Particles 183 boundary conditions can be written by components as: ⎧ ⎧ ⎧ ⎪ ⎨ẍ = m + m ẏ(t) ≡ E⊥ + ωB ẏ eE ⊥ e|B| ⎪ ⎨ẋ(0) = vx 0 ⎪ ⎨x(0) = 0 ÿ = − e|B| ẋ(t) ≡ −ωB ẋ .18) where we have defined: e |B| e E⊥ e E ωB ≡ . (3.23) v0y + E ⊥ /|B| Inserting this result into the second of Eq. we get: ⎧ ⎪ ⎨ẋ = E⊥ t + ωB y + vx 0 ẏ = −ωB x + v y0 (3. (3.21) ωB which can be integrated to yield: ⎛ ⎞ 2 1 E⊥ E ⊥ /|B| + v0y x(t) = ⎝− + v0y + v0x ⎠ cos(ωB t + α) + . (3. E ≡ . E⊥ ≡ . (3.22) ωB |B| ωB where α is defined such that v0x tan α = .20) into the first.18). (3.3. (3. Integrating once the three equations in the system (3.19) is also known as cyclotron frequency. we get: ⎛ ⎞ 2 1 E⊥ E⊥ v0 y(t) = ⎝ + v0y + v0x ⎠ sin(ωB t + α) − t− x. (3. A typical example is provided by the Time Projection Chamber discussed in Problem 2.42. the equation of motion of the particle on the x-y plane is a uniform circular motion with angular frequence ωB centred around the point: E ⊥ /|B| + v0y E⊥ v0x rc (t) = . (3. Another canonical example is a microstrip detector (see Problem 2. otherwise the trajectory goes through cusps. In a TPC.e.28) ż |B|2 0 Discussion The motion of charged particles inside a simultaneous static electric and magnetic field is a common situation in particle physics experiments. hence the particle momentum.20) to give: E 2 z(t) = t + v0z t.e.26) ωB |B| ωB which drifts with constant velocity along the direction of E × B. i.43) inversely polarised and immersed in an orthogonal magnetic field. the fields are parallel. i. the last component can be obtained by integrating the third equation of the system (3. but interact with the medium (the filling gas for TPC. namely: ⎛ ⎞ ⎛ ⎞ ẋ 0 E×B v = ⎝ ẏ ⎠ = ⎝ − |B| ⎠ = |E| . (3. the motion along the y direction reverts direction if at least one among |v0x | and |v0y | is larger than zero. a configuration that helps reducing the lateral diffusion of the charge. while the magnetic field allows to measure the track curvature. Along the direction of the magnetic field. E = 0. where the electric field is needed to drift the ionisation electrons to the multiplication region.− t− . i. In both cases. E ⊥ = 0. the mean velocity v is given by the time-average of v(t). see Problem 2. however. For the special case E = 0. see Problem 3.e.4.25) 2 Therefore. The velocities along the y-axis at the time when the motion along x reverts its direction are given by: 2 E⊥ 0 = ẋ(tn ) = + v y + v0x sin(ωB tn + α) 0 ⇒ tn = kπ − α |B| 2 E⊥ E⊥ ẏ(tn ) = ± + v y + v0x − 0 . (3.184 3 Accelerators and Experimental Apparatuses Finally.2. the . the motion is instead uniformly accelerated.27) |B| |B| Hence.0 . the equation of motion are not given by the solution found here because the electrons do not move in vacuum. (3. v0 = 0. and assuming that the particle is initially at rest. we can always assume ∂t A = 0.1 Tracking of Charged Particles 185 silicon crystal for the microstrip). (3. Since B is constant.29) 1 + ωB2 τ 2 |B| |B|2 where μe = eτ/m e is the electron mobility. respectively.6 of Ref.e. but the time- evolution law of Eq. See Chap. ΘL ∼ 10−1 . Computing explicitely the derivatives in Eq.2 Show that the trajectory of a charged particle of momentum p and charge q moving inside a uniform magnetic field B is a helix.17) maintains its form. (2. electrons) will drift with velocity μe ωB τ ωB2 τ 2 v= E+ E×B+ (E · B) B . (3. and τ is the mean time between two collisions with the medium constituents. the Lorentz angle turn out to be sizable. see Eq.31). with L(r. 4. What is the relation between the particle momentum and the radius of curvature of the trajectory? Discussion When dealing with relativitic particles. Using a typical value for electron mobility in silicium μe ∼ 103 V−1 cm2 s−2 and magnetic field intensities of order ∼1 T.8 of Ref. like pixels or microstrips. the drift velocity makes an angle ΘL with respect to the electric field direction such that: tan ΘL = ωB τ = μe |B|. we get: . the laws of kinematics change. 34. and φ(r) and A(r) are the scalar and vector potential of the electric and magnetic field. A simple friction model predicts that the charged particles (e. See also Chap. for E orthogonal to B.31) where ṙ ≡ v. [4]. The problem should be therefore treated with a proper kinetic theory of the interactions with the traversed material.g. i.112). (3. such that E = −∇φ and B = ∇ × A. Suggested Readings The notation has been adapted from Chap. 34. [3] for more details.6 of Ref. Problem 3. ṙ) = −mc 1 − 2 − q φ(r) + q ṙ · A(r). 2 dt ∂ ṙ ∂r c (3.30) In semiconductive devices. This can be for example proved by deriving the equation from the relativistic generalisation of the Euler-Lagrange equation for a point-like charge q and mass m moving in a classical electromagnetic field: d ∂L ∂L |ṙ|2 − = 0. In particular. [3] for more details on the motion of charged particles in gas uncer the combined effect of electric and magnetic fields. (3. this drift angle is also called Lorentz angle and has important implications on the position resolution of the detector.3. we have: x 1 y = − +C ⇒ x = − (x + K C) .186 3 Accelerators and Experimental Apparatuses d dpi ∂ Ai ( pi + q Ai ) = + + v j ∂i A j = −q ∂i φ − v j ∂i A j . Φ0 can be interpreted a posteriori as the angle in the x–y plane formed by the radial vector that connects the centre of the circumference to the initial position. (3. The constant K is defined as K ≡ |p|/q|B|. while λ ∈ [0. Eq.3. Integrating once the second of the system (3.17). dt 2 dt ds 2 ds ds 2 |p| ds (3. The boundary condition on r0 is subject to the constraint |r0 | = 1. Solution From Eq. = × B. measured with respect to the x-axis. 3. y (0) = − cos λ cos Φ0 . π/2] is the dip angle with respect to the x–y plane. dp = q E + q v × B. one can easily see that the momentum magnitude |p| is constant in time. so that we can directly get a geometrical parametrisation of the curve. provided that the relativistic definition of momentum is used. since d|p|2 dp = 2p · ∝ p · (p × B) = 0. dt dt ∂t d pi = −q ∂i φ − v j ∂i A j − ∂ j Ai = −q ∂i φ − v j εi jk [∇ × A]k = dt = −q (∂i φ − [v × (∇ × A)]i ) = q E i + q (v × B)i . y(0) = y0 ⎪ ⎩ ⎪ ⎩ ⎪ ⎩ z =0 z (0) = sin λ z(0) = z 0 (3. (3.33) dt dt Hence. (3. mγ |v|2 = q|v| × B.34) can be written by components as: ⎧ y ⎧ ⎧ qB ⎨x = |p| y ≡ K ⎪ ⎪ ⎨x (0) = − cos λ sin Φ0 ⎪ ⎨x(0) = x0 y = − q|p|B x ≡ − xK . We choose the reference frame like in Problem 3. see Fig.17) becomes: d 2r dr d 2r dr d 2r q dr mγ =q × B.34) where we have changed variable from the laboratory time t to the trajectory length s = |v| t.32) dt Hence.17) with E = 0.36) K K2 .1.35). (3. the relativistic extension of the time-evolution law for a point-like charge is identical to the classical expression of Eq. so that the system (3. (3.35) Here. (3. 3. we get: . Taking into account the boundary conditions in (3.3 Representation of y the trajectory r(s) of a (x0.1 Tracking of Charged Particles 187 Fig. with x respect to the x-axis B z with C a constant of integration. 3. hence x(s) satisfies the equation for a displaced harmonic oscillator. whereas Φ0 is the angle of the radial vector joining the centre of the circumference to r0 . y0) charged particle moving inside a uniform magnetic R field B. The angle λ is called Φ0 λ dip angle.35). (3. s cos λ x(s) = x0 + R cos − Φ0 − cos Φ0 . cos λ R .37) R where R ≡ K cos λ. Substituting this expression into the second of (3. and taking into account the boundary conditions for y: R s cos λ y = x = − cos λ cos − Φ0 .35). (3. while the z-component increments linearly with the curve length: the curve is therefore a helix of pitch p = 2π R tan λ. (3. s cos λ y(s) = y0 − R sin − Φ0 + sin Φ0 .e. i.41) tan λ .38) R The last component can be trivially integrated to give: z(s) = z 0 + sin λ s. for large momenta.39) The projection of the curve onto the x–y plane is a circle of radius |p| cos λ pT |R| = = .40) |q| |B| |q| |B| where pT is the transverse momentum. for |s| |R|. In particular. the trajectories in the x–z and y–z planes can be approximated by straight lines: sin Φ0 x(z) ≈ x0 + (z − z 0 ) tan λ cos Φ0 y(z) ≈ y0 − (z − z 0 ) (3. (3. giving a bending power B ≈ 12 T m uniformly with respect to the pseudorapidity η. In general.3 Show that the transverse momentum pT [GeV/c] of a particle of charge q = z e.45) . [5] for a more exact evaluation of B in the ATLAS toroid and CMS solenoid. like those generated at the inside of a long solenoid of internal radius Rcoil and intensity |B|. Problem 3.3 z B R.43) is called bending power.43) ρ ds |p| ds (|p|/q) The integral at the right-hand side of Eq.42) ds |p| ds cos2 λ The bending angle of the magnetic field is defined as the angle by which a charged particle is bent by the field all along its path. (3. (3. which in accelerator physics is also known as beam rigidity. The bending angle is therefore proportional to the bending power and inversely proportional to the ratio |p|/q. For uniform magnetic fields. as defined in Eq. [5]. the bending power from the interaction point up to the position of entrance in coil rcoil is given by: rcoil dr B = ds |B| cos λ = |rcoil ||B| cos λ = Rcoil |B|. See Ref. along a circular path of radius R [m] on the transverse plane. (3. Geometri- cally. however. the solenoid of the CMS experiment has an inner radius Rcoil ≈ 3 m and generates a magnetic field of intensity |B| ≈ 4 T. and it has to be computed numerically. (3.e. a measure of the particle resilence to be bent by a magnetic field. i. it corresponds to the integral along the path of the magnetic field component orthogonal to the trajectory. the bending power depends on the initial particle direction.e. Suggested Readings The formalism used in this exercise and its application in real experiments can be found in Ref. the magnetic field is no longer uniform and the bending power features a dependence on the direction η. (1. i. At large values of η.: 2 ds d r q dr B Δθ = = ds 2 = ds × B(r) ≡ . is given by the formula: pT = 0. moving inside a uniform magnetic field of intensity B [T]. (3.188 3 Accelerators and Experimental Apparatuses The radius of curvature ρ of the track along the trajectory is given by: 2 −1 −1 d r q dr |R| ρ(s) = 2 = × B = .44) 0 ds For example.161). 3.1 Tracking of Charged Particles 189 Solution From Eq. the photon energy is therefore given by: E γ ≈ 2 E e ≈ 2 |pe | = 96 MeV. we get: pT 1 |B| R = (z · e) · ·T · ·m = GeV/c GeV/c T m |B| R 2.48.40). see Problem 1.3 z . and with radius of curvature R = 20 cm. we see that the radius of curvature R of a particle moving in the plane orthogonal to B is related to the transverse momentum pT by the relation: pT = q |B| R. (3.49) We can thus verify a posteriori that the most probable interaction process for the incoming photon is indeed pair-production. 18211/2016 . (3.47) is valid to better than the permill level. To good approximation.48) Since the tracks are intially parallel.998 × 108 m/s e T m =z · = T m 109 e V |B| R = 0. A photon converts into a e+ e− pair. (3. Determine the energy of the incoming photon. are observed. 18211/2016 Problem 3. Two tracks on the plane orthogonal to the magnetic field. which is usually fine for most of the applications. Bando n.8 · 0. (3.46) by the units in which we wish to express the dimensionful quantities. the momentum transfered to the nucleus is of order |q| ∼ m 2e /E γ m e . The last approximation in Eq. and the kinetic energy |q|2 /2m N of the recoiling nucleus is negligible.3 · 0. (3.8 T.3 |B| R = 0. Solution The electron and positron have momentum |pe | given by Eq.4 A drift chamber is immersed in a magnetic field with intensity |B| = 0. (3. Bando n. namely |pe | = 0.2 MeV = 48 MeV. The tracks start parallel to each other.46) Dividing the two sides of Eq.45). (3.47) T m where we have used the fact that T = V m−2 s. (3. y (x) = . namely E max = 104. What was the largest electron energy attainable? Solution The effective radius of curvature at LEP was about (0. (3. Suggested Readings The energy record at LEP was achieved in 2000. thus outside the experimental value of 125 GeV established years later at the LHC [8]. see e.5 The LEP dipoles allowed for a maximum magnetic field |B| = 0. (3. Let’s denote the coordinate parallel to the electric field by y.6 A beam of momentum |p| = 2 GeV contains both kaons and pions. (3. Solution The kinetic energy acquired by the beam due to the work done by the electric field is negligible.52) m 2 m 2 |Δy (L)| = |p| |p|2 |p|2 2|p|3 .9 km. The beam passes between the plates of an electrostatic separator of length L. ÿ = .45): E max = |pmax | = (0. Determine the opening angle between the two beams emerging from the separator.51) 2|p| β |p| β The opening angle between the two beams at the exit of the separator is therefore given by: ⎛ ⎞ e |E| L ⎝ 1 + K − 1 + π ⎠ ≈ e |E| L (m 2K − m 2π ). y(t) = t . The electric field between the plates has uniform intensity |E| = 10 kV/cm.66 · 27/2π ) km ≈ 2. since the maximum centre-of-mass energy of 2 E max ≈ 209 GeV allowed the experiments to probe Higgs boson masses below about 115 GeV.5 GeV.3 · 0. (3. so we can assume |p| to be approximately constant during the motion.135 · 2. This limit had important implications on the reach for the Higgs boson later discovered at the LHC [6. 7].g. Problem 3. From Eq. such that y(0) = 0. we get: d(β y γ ) e |E| 1 e |E| 2 e |E| = m ≈ m γ β̇ y .32).190 3 Accelerators and Experimental Apparatuses Problem 3.9 × 103 ) GeV ≈ 117 GeV. and covered up to 2/3 of the 27 km-long storage ring. The maximum electron energy is therefore given by Eq.135 T. and was dictated by the limitations of the RF accelerators. dt mγ 2 mγ e |E| x 2 e |E| x y (x) = .50) The actual limit achieved at LEP was a bit smaller than this. the original CERN an- nouncement [9]. (3. However. The position of arrival y at the opposite side of the spectrometer is measured with an uncertainty σ y .4 of Ref. The beam to be composed of 12 C+ and 13 C+ ions in unknown proportion. Problem 3. 11.3.4 × 10−5 rad. Indeed. For larger beam momenta. as the numerical calculation also shows.g. Solution The kinetic energy of the ion at the entrance of the spectrometer is T = q V0 = |p|2 /2m. due to the ∼ |p|−3 scaling of the opening angle.54) qB qB .1 Tracking of Charged Particles 191 where we have expanded the square root to first order. What is the minimum resolution σ y necessary to identify the correct mass number to better than 3σ per incident ion? Discussion The use of accelerated beams and magnetic analysers for mass spectroscopy is a well established technology known under the name of accelerated mass spectroscopy (AMS).53) 2 · (2 GeV)3 Discussion Equation (3.52) shows that a static electric field can be used to separate particles of different mass but same momentum. electric fields more intense than a few tens of kV/cm can give rise to the formation of discharges (in air at standard pressure. followed by a magnetic analyzer of length L = 1 m and uniform magnetic field with intensity |B| = 1 kG. We can assume classical kinematics since q V0 = 1 MeV is small compared to the mass of carbon ions. the break- down voltage is about 30 kV/cm). which may have been for example selected by a magnetic field. (3. Numerically. Ref. An application of AMS is radiocarbon dating for assessing the age of organic samples. Positive ions are produced at rest at the grounded electrode and accelerated by the electric field towards the cathode. (3. [10]. Suggested Readings This topic is briefly discussed in Chap.40): √ |p| 2 m q V0 R= = (3. The radius of curvature R is then given by Eq. separation techniques based on the use of RF cavities become more efficient. we get: 106 eV/m · 1 m |Δy (L)| = · (0.23 GeV2 ) = 1.7 A mass spectrometer consists of a pair of electrodes kept at a potential difference |V0 | = 1 MV. [11] for more informations on the subject. this method becomes inefficient for momenta in excess of a few GeV. See e. 6 × 10−19 C 1 = −26 · 0. is small compared to the 12 C mass. (3.0 m. (3. (3.192 3 Accelerators and Experimental Apparatuses Fig. see Fig. the position resolution needs to satisfy: Δy 1 q Δm σy < = |B| L 2 = Nσ 4 Nσ 2 m V0 m 1 1. we have use the fact that one mole of 12 C weighs exactly 12 g.58) 4·3 2 · 2 × 10 kg · 10 V6 12 Bando n.1 cm.6 × 10−19 C 6. (3.56). Since Δm = 1 a. For 12 C+ . which is smalle compared to L.56) 2R 2 m V0 2 which is accurate to the percent level.u. (3. p2 13C+ V =0 V = −V0 Replacing the symbols by their values.0 × 1023 0.57) y 2 m 2 m In order to separate the two mass numbers to better than Nσ standard deviations per ion. 3.4. we can approximate the ion trajectory with a parabola to yield: L2 q |B| L 2 y≈ = .55) 1. we get for the 12 C+ population: 2 · 106 V 12 × 10−3 kg 1 R(12 C+ ) = = 5. Since L/R ≈ 0. thus justifying the use of Eq. (3.1 T In Eq. 3.1 T · 1 m2 · = 0.55).8 A tracking system consists of a pair of multiwire chambers separated by a distance d = 1 m. 13705/2010 Problem 3. The apparatus measures the lateral displacement y with respect to the initial beam position. y ≈ 10 cm. we can approximate: Δy 1 Δm y Δm ≈ ⇒ Δy ≈ . followed by a dipo- .4 Sketch of a mass E B spectrometer 12C+ Δy p1. and with spatial resolution σ = 150 µm.2. (3.2).2. The uncertainty on the bending angle is there- √ fore 2σb̂ = 4σ/L.3 B L).15) giving: δpT 8σ 1 8 · 150 × 10−6 pT =√ pT = √ = 5 × 10−4 pT . [2]. (3.60) pT MS pT MS X 0air 300 m . Solution The relative momentum resolution from the measurement error and from multiple scattering can be parametrised as in Eq. Using these values. [12] and Chap. We can therefore apply the result from Eq. 3. respectively.3. Neglect the effect of multiple scatterring inside the chambers. Assume a typical value of 100 µm and 1 m for the position resolution and lever arm. This angle is related to the momentum and bending power via Eq.59) pT N + 1 0.10): it is roughly constant at low momenta. before and after the magnet.59). we get: air Fe δpT δpT X 0Fe 1. while it grows linearly with pT at large momenta.3 B L 4 0.e. (3. in particular Chap. Solution The spectrometer design corresponds to the last of Fig. 8 of Ref.8 cm / = = = 8 × 10−3 . and by an identical pair of chambers.3. The effect of MS is more relevant in iron than in air because of the larger particle density and Z number. i. One can convince himself of this result by noticing that the system measures two segments. L = 2 d = 2 m.3 of Ref. The uncertainty on√ the slope √b̂ of each segment is given by the first of Eq.1 Tracking of Charged Particles 193 lar magnet with bending power B = 2 T m. Determine the relative resolution on the momentum transverse to the magnetic field as a function of the transverse momentum itself. 1N/R3/SUB/2005 Problem 3. the reader is addressed to dedicated textbooks. (3. Assume that both the dip angle of the impinging particle and the angular deflection induced by the magnet are small.3 · 2 · 2 where L is the total length of the spectrometer. thus giving δpT / pT2 = 4σ/(0. Suggested Readings For more information on the topic. in agreement with Eq. namely σb̂ = 2σ/d = 2 2σ/L. Bando n. (3.9 Discuss how the transverse momentum resolution δpT / pT for a charged particle measured by a magnetic spectrometer depends on the transverse momentum pT if the particle moves in air or inside iron. The radiation lengths for air and iron can be found in Table 2. respectively.43). (3. (3. 3. and pT is measured in GeV. up to momenta of about 100 GeV.0136 L 3/2 pT = √ .61) 8 δy β X0 Using typical values δy = 100 µm and L = 1 m. (3.10 A tracking system for measuring high-momentum muons consists of three chambers.63) 2 Fig. 3. y2 ). A given event will produce three spatial coordinates y = (y0 . see Fig. and denote the orthogonal coordinate by y. and σ2 = 100 µm. we define the new coordinate (y2 + y0 ) y1ext = . separated by a distance L = 1 m and with spatial resolution σ0 = 100 µm. 3. For later use. 13705/2010 Problem 3. δpT / pT ∼ b. We further chose the direction of the y-axis such that the concavity of the true muon trajectory is positive. σ1 = 50 µm. (3. Bando n.5 Illustration of the L L measuring system considered in the Problem y2 extrapolation B y1ext ≈s fit y0 y1 x0 x1 x2 . The chambers are located inside a uniform magnetic field of intensity |B| = 1 T. Consider muons of pT = 1 TeV that impinge almost normally to the chambers: what is their the charge misidentification probability? Solution Let the initial muon direction be aligned along the x-axis. so that δpT / pT ∼ a pT . y1 . multiple scattering is negligible in air for momenta in excess of a few GeV.5. whereas momentum resolution in iron is approximately constant.62) 1 GeV air Therefore. we get: 100 GeV Fe pT ≈ (3.194 3 Accelerators and Experimental Apparatuses The momentum at which the measurement error and multiple-scattering become of the same size is given by: z 0. (3.64) 4 where we have used the second of Eq. Since s is given by the difference between two gaussian and independent variables. Therefore: . where s coincides with the sagitta of the curve modulo terms of order L 2 /2R.3 · 1 · (1)2 μs = y1ext − y1 = − = = = 150 µm. (3. 4R 2R 2 pT 2 · 103 σ02 + σ22 σs = σ22 + = 87 µm.3 |B| L 2 0.2) to relate the variance of y1ext to the variance of y0 and y2 .3.1 Tracking of Charged Particles 195 It is easy to convince oneself that a sign-flip corresponds to s = (y1ext − y2 ) < 0. it will be also normally distributed with mean and standard deviation: (2L)2 L2 0. . e.65) where erf(x) is the error function defined by: x 2 dt e−t .65) by their numerical values. (3. Replacing the symbols in Eq. 1 0 (s − μs )2 1 μs Prob [s < 0] = √ ds exp − = 1 + erf − √ . 2π σs −∞ 2 σs2 2 2 σs (3.66) π 0 which is available in almost all standard libraries. the math library in Python.g. 2 erf(x) = √ (3. we get a misidentification probability: . . Particles enter the spectrometer almost perpendicular to the detector panes.67). It consists of three parallel planes of position detectors with spatial resolution σx = 100 µm. Bando n.67) 2 2 · 87 µm The analytical result can be cross-checked by a simple Monte Carlo simulation of the experimental setup.02)%.2% (3. 1 150 µm Prob [s < 0] = 1 + erf − √ = 4.4. separated by a distance a = 20 cm.11 A spectrometer measures the momentum of charged particles of unitary charge and momentum of a few GeV. 1N/R3/SUB/2005 Problem 3. we get Prob [s < 0]MC = (4. by using 1M toy events. Estimate the transverse momentum resolution at |p| = 2 GeV. For example. (3. see Appendix 3. in agreement with the analytical result of Eq. The planes are immersed in a uniform magnetic field B with intensity of 1 T.18 ± 0. parallel to the planes and orthogonal to the measured position x. the bending angle (3. a numerical simulation predicts that B(y) is approximately constant around B0 = 2 T within a tolerance of 10%. Does the answer change if the alignment between the two stations cannot be constrained to better than 5 mrad? Solution Given the approximate field map. a simple argument like the one used to derive Eq.68) within 20%.12 A magnetised iron slab of thickness d = 60 cm is saturated by a constant magnetic field whose force lines can be assumed to be fully contained inside the slab. 3. 1N/R3/SUB/2005 Problem 3. We can therefore use the analytical solution of Eq. Actually. Bando n. Let the other two dimensions of the slab be large compared to the thickness and let’s assume that the boundary conditions impose the magnetic fiel B to be given by B = B(y) ez . Determine how many vertical muons are needed to achieve the desired accuracy on the bending power by assuming a cosmic muon spectrum d Nμ /d E ∝ E −α with α = 2. where B(y) is an even function of y − d/2.3 · 2 · 0. (3. such that y = 0 (y = d) correspond to the upper (lower) faces. i = 1. 3.1% (3.22.3 |B| L 2 0.4)2 As we have seen. Furthermore.70) 5 .11) giving: √ δpT σ AN 100−6 · 8 · 1. (3.196 3 Accelerators and Experimental Apparatuses Solution The spectrometer configuration corresponds to the first of Fig.2 with N = 2. if we denote the three measured positions by yi .7. The muon momentum downstream of the slab is measured by an independent spectrometer of perfect resolution. for N = 2 we could have also easily guessed the extra factor of 1.3 · 1 · (0. (3. Indeed.22 σ. 2.8) would have given a result in agreement with Eq. (3.69) 2 2 in agreement with the Eq. where φ measures the angle on the transverse plane.22 = pT = · 2 = 4. The muon direction at the entrance and at the exit of the slab is measured by two identical stations of drift tubes with angular resolution σφ = 2 mrad.68) pT 0. (3.11). and ez is orthogonal to the transverse plane.43) is about 0. (3.6 Δφ = 72 mrad. the sagitta s and its uncertainty δs can be estimated by: y1 + y3 3 s ≈ y2 − ⇒ δs = σ ≈ 1. We wish to measure the bending power of the magnet to better than 1% by using cosmic muons with pT in excess of 5 GeV. The same arguments allows to approximate the trajectory by a parabola.34) and (3.3. we can approximate the finite sum at the right-hand side of Eq. lengths in b. the variance of the maximum-likelihood estimator is given by −1 1 σ2pT Δφ = . The z factor can be ±1 for positive/negative muons. and then generalise to the case where the curvature decreases throughout the trajectory because of a continuous energy loss.3 · z (s) − (0) = dr × B(r). Since B = B(y) ez .83).72) Equation 3. for almost vertical muons dr × B(r) = dy B(y) eφ .3%.74) i σi2 see Eq. From Eqs. and neglecting terms of order (Δφ)2 we get: pT Δφ = 0.3 · z dy B(y). (4. (3.1 Tracking of Charged Particles 197 hence the total track length inside the slab for vertical muons has length d to better than 0. Projecting both sides along eφ . Given N independent normal measurements with same mean pT Δφ and standard deviations σi . where eφ is the unit vector lying on the transverse plane and orthogonal to the projection of dr onto the transverse plane.74) by −1 1 σ2pT Δφ ≈ N dpT f ( pT ) . We first consider the ideal case that the energy loss is negligible compared to the muon momentum. (3. which is an acceptable approximation. We can therefore combine the measurements from N independent muon tracks into an estimator of the bending power. Each angular measurement comes with an uncertainty 2 0. and the magnetic field in T.75) σ p2T Δφ ( pT ) . it follows that: dr dr 0.73) βpT X0 pT where the first term on the right-hand side accounts for the measurement error from each DT station. (3. (3. and the second accounts for multiple-scattering across the iron slab.71) ds ds |p| where the momentum is measured in GeV.0136 d c2 σ pT Δφ = pT 2σφ2 + ≡ pT c1 + 2 (3. (3. (3. Notice that |p| is assumed to be constant.47).72 implies that for all muons the product pT Δφ is proportional to the bending power dy B(y). For N 1. satisfying the requirement: pTH dpT f ( pT ) = 1.77) is valid if the only uncertainty comes from DT measurement and from MS. Therefore: pT Δφ+ − pT Δφ− 0.76) pTL where pTL and pTH define the muon momentum acceptance in the sample. In order to remove the misaligne- ment bias from the measurements. but for our case a quick computation shows that the muon energy loss after traversing the slab is a non-negligible fraction of the initial muon energy.79) is equal to Eq. which alone would give a relative uncertainty on B of about 7%. hence larger than the target accuracy of 1%. In the presence of a systematic misalignment εφ = 5 mrad. pTH = ∞. X 0 = 1.77) and is now free from the systematic uncertainty due to the alignment. (3. we get: σ pT Δφ ≤ 10−2 ⇒ N 700. (2.08 T m.78) pT Δφmap The statistical scaling of the ML estimator (3. Indeed.3) with the values from Table 6. the variance of the new estimator (3.3 dy B(y) (3. the uncertainty on pT Δφ is limited by pTL εφ /0. the muon spectrum is so steeply falling that high.3 dy B(y) + pT εφ − = 2 2 = 0. we notice that the combination pT Δφ has a sign that depends on the muon charge.77) pTL (1−α) − pTH (1−α) pTL c2 + c1 pT2 The integral at the right-hand side of Eq. (3.pT events will have a negligible weight in Eq. (3. (3.198 3 Accelerators and Experimental Apparatuses where f ( pT ) ∝ d Nμ /d E is the differential distribution of muon energies that arrive at the detector. [3] we find: . The assumption that |p| can be treated as constant during the motion is certainly a valid approximation for tracking in air.3 = 0. By dividing the sample into a sub-sample of N /2 positive muons and one of N /2 negative muons.77) can be evaluated numerically. (3. σφ = 2 mrad. see Appendix 3.5.1 of Ref. from Eq. By assuming the power law d N /dpT ∝ pT−α .71) is required.8 cm).79) where we have assumed that the μ+ and μ− spectra are identical. however. Using the numerical inputs from the problem ( pTL = 5 GeV.3 dy B(y) + pT εφ + − −0. (3. A word of caution for Eq. Eq.77) becomes: −1 N (α − 1) pTH pT−α σ2pT Δφ ≈ dpT . We assume here that the detector acceptance is unity for all values of pT : although this is hardly true in practice.77). (3. (3. (3... (3.81) 1 − y pCT where C = d E/d x can be safely assumed to be constant for a given value of pT . the factor y C/ pT can be treated as a small perturbation and the integral can be expanded around y = 0 to yield: d C C 2 pT Δφ = 0.85) 0 pT 0 2 pT The correction at the right-hand side of Eq.83) B0 d 0 dy B(y) hence we can neglect it since it’s below the target resolution. we can parametrise the field as: 2 B0 d B(y) = B0 + y− + . the fourth and higher-order terms will give increasingly negligible contributions.82) 0 pT pT Given that B(y) is constant within 10%.6%. Eq.84) 2 2 Since δ B/B0 10%. Because of Eq. Following the assumption of the problem.85) is about 7% at pT = 5 GeV.81). (3. (3.3 · z dy B(y) 1 + y + y + . Since the energy loss per unit length is to a good approximation constant throughout the trajectory.45 MeV g−1 cm2 · 7. Even with this approximation.. . (3. and we can therefore neglect them.69 GeV.3 · z dy . (3. we can estimate the size of the O(y 2 ) term: d 2 2 0 dy B(y) y pCT (B0 d) 13 dpCT d ≈ ≈ 0.9 g cm−3 · 60 cm = 0. the generalisation of Eq.72) becomes: B(y) pT Δφ = 0. hence it gives a non-negligible contribution to the bending angle and should not be neglected. we see that the product pT Δφ is no longer proportional to the bending power alone. (3. Therefore. (3.72) can be now generalised to: . (3.80) so that neglecting the energy loss may result in a bias on the bending power in excess of the target accuracy of 1%. (3.1 Tracking of Charged Particles 199 ΔE ≈ 1.. but involves a combination of the bending power and of the first moment of the magnetic field. Under this assumption. one can easily verify that d d C 1Cd dy B(y) 1 + y = dy B(y) 1+ .3. the majority of accelerators in use nowadays are devoted to industrial and medical applications. where a similar analysis technique has been adopted to calibrate the magnetic field map in the return yoke of the CMS spectrometer.f. Particles accelerators can be classified in two main families depending whether the electric field that provides the acceleration is static or time-dependent. since any static electric field is irrotational: ∂B ∇ ×E=− ⇒ ds · E = 0 ⇔ no “circular” acceleration if B is static.6 The same accelerating field cannot be used again for the same particle. 3. .f. the latter are often to weak to be relevant for most of the applications requiring particles. like the Cockroft-Walton. is replicated n times in cascade. receiving a kick q Vr. cosmic radiation).3 · z dy B(y).86) 1 + 21 CpTd Suggested Readings This exercise is inspited to the work of Ref. at each turn. as experienced by the particle.87). so that E f = n Vr. . Time-dependent fields evade the bounds imposed by Eq. Linear accelerators operated with radiofrequency fields (LINAC) work like a chain of static accelerators of equivalent voltage Vr. ∂t (3. whose accelerating gradient. (3.200 3 Accelerators and Experimental Apparatuses pT Δφ = 0. etc. Besides being used for fundamental research. are naturally associated with the concept of linear acceleration: particles of charge q accelerate between two terminals kept at a voltage difference V0 and acquire an extra kinetic energy E f = |q V0 |. Ladderton. 3.87) Static accelerators are thus inherently limited in energy by the maximum electric fields V0 achievable in safe conditions in the laboratory before phenomena like spark formations and disruptive discharges occur. see Fig. This limitation is even more sever for fundamental research.. each particle visits several times the same point of the apparatus. Although numerous sources of natural radiation exist (e. X-tube.g.f. Van de Graaf. This can be realised in both linear and circular fashion. If the particle motion and the accelerating gradient are maintained with the appropriate . [13]. In circular accelerators. which is often conducted at the energy and intensity frontier. radioactive decays. Static accelerators.2 Accelerators Accelerating particles into beams with the desired energy and space-time structure is a primary need in experimental particle physics. and open the way to high-energy acceleration. (3. Problems Bando n. betatron. kept at an energy of 100 GeV. in which the accelerating field is 20 MV/m.13 Compare the energy lost by an electron initially at rest accelerated by a LINAC of length L = 10 km long. the energy transfer sums up constructively resulting in an overall energy increment. Does the result change if the particle were a proton? . and thus most of the exercises have a rather qualitative solution. with the energy lost by the same electron when. and synchrotron are based upon this concept. we propose the calculation of radiation loss in circular and linear colliders. it makes a full round of a circular accelerator of radius 10 km.6 Examples of a Van de Graaf (left) and Cockroft-Walton (right) accelerator hosted at the LNF site synchronisation. The cyclotron. 13705/2010 Problem 3. Among the few quantitative tests that a non-expert of accelerator physics should certainly know.2 Accelerators 201 Fig. 3.3. The purpose of this section is to make the reader familiar with the different solutions of accelerators. e.) loses energy in the form of electromagnetic radiation. the power loss per unit solid angle is described by Larmor’s formula dP e2 dP 2 e2 |v̇|2 = |v̇|2 sin2 θ. e2 has units of [kg m3 t−2 ]. Eq.92) as about 1. An accelerating field able to supply an energy per unit length comparable to this value.92) where rc = e2 /mc2 is the classical particle radius. We can now consider two cases: a linear acceleration. where ω is the revolution frequency Eq. etc.4. For a linear accelerator. dp/dt ∝ p. while |dp/dt| = ω|p|.91) where we have used again the relation d E = |v| d|p| ⇒ d E/d x = d|p|/dt. P= dΩ = . 2/3 (e2 /m 2 c3 ) (d E/d x)2 2 e2 (d E/d x) 2 (d E/d x) = = = . and the denominator at the right-hand side of Eq. rc = 2. For a circular accelerator.8 × 1012 MeV/cm. 3 m 2 c3 dt 3 m 2 c3 dt 3 m 2 c3 dx (3. (3.8 × 10−13 cm. The last equality has been proved in Problem 1. and a circular acceleration. (3. = γ 1 − β2 = = . i. d|p|/dt is approximately zero. (3. (3. The ratio between the power loss and the power supplied by the external accelerating field is therefore: Plin. The relativistic generalisation of the second of Eq. Here.g. i.90) 3 m 2 c3 dτ dτ 3 c where we have used the fact that d E = |v| d|p|. (3.89) becomes: . (3. it is independent of the particle energy and depends only on the external field. dp/dt · p = 0. a magnetic field. (d E/dt) (d E/d x) |v| 3 m c β2 4 3 β(mc2 /rc ) (3. Therefore.e. a RF wave.88) proceeds through the replacement [14]: 2 e2 |v̇|2 2 e2 dp dp 2 e2 dpμ dp μ P= = · → − = (3. For an electron. which is far above the maximum limit attainable in laboratories.89) 3 c3 3 m 2 c3 dt dt 3 m 2 c3 dτ dτ ! 2 " 2 e2 dp 2 2 d|p| 2 e2 6 2 = − β = γ β̇ − (β × β̇) 2 . i.e.89) becomes: 2 2 2 2 e2 2 dp 2 e2 dp 2 e2 dE Plin.88) dΩ 4π c3 dΩ 3 c3 where θ is the polar angle with respect to the particle acceleration v̇. Since d E/d x is proportional to the gradient of the accelerating field. energy loss in linear accelerators is always negligible. we assume the Heaviside–Lorentz units such that the first of Maxwell equation reads ∇ · E = 4πρ. For a classical particle. the field intensity should of order 1012 MV/cm.202 3 Accelerators and Experimental Apparatuses Discussion A particle of charge q that is accelerated by external forces (e. 8×10−15 GeVE 4 (3. .8 × 10−15 m · 104 m MeV2 = · 400 = 1.99) ΔE orbit = 7. (3.98) ΔE orbit If the accelerated were a proton. (3. (3. · = = = c 3 m 2e c4 dx 3 m e c2 dx 2 2.8 × 10−2 ΔE orbit = MeV = (100)4 = 880 MeV (3.96) would give ΔE LINAC = 8. (3.5 × 10−8 MeV (3. so that we can assume β = 1 since the beginning. Eq.95) (R/m) GeV protons Solution Assuming a constant accelerating field of intensity E = 20 MV/m over a distance L = 10 km.91) then gives: 2 2 L 2 e2 L dE 2 re L dE ΔE LINAC = Plin. The energy lost per orbit is given by 2π 2π R 4π e2 3 4 ΔE orbit = Pcirc.94) ω βc 3 R For an electron and proton beam.95) to give: 4 8.8 × 10−2 E 8. (3.2 Accelerators 203 2 e2 2 2 2 2 e2 c 4 4 Pcirc.3.0 × 10−10 MeV ΔE LINAC ⇒ ≈ 10.97) (R/m) GeV 104 Hence. Eqs.95) and (3. (3. we find that the ratio between the two energy losses is equal to ΔE LINAC ≈ 2 × 10−11 . = γ ω |p| = β γ . · = β γ .8×10−2 (R/m) electrons = 7. · = Pcirc.93) 3 m 2 c3 3 R2 where R is the radius of curvature of the circular orbit.8 × 10−11 MeV ΔE orbit hence the LINAC acceleration would result in a larger energy loss. we see that the electron becomes relativistic already after a few tens of centimeters. (3. we can use directly Eq.96) 3 0.94) takes the numerical values: ΔE orbit 4π 1 197 MeV fm E 4 1 ≈ 4 = MeV 3 137 R GeV mc /GeV 2 E 4 8.511 MeV m2 For the case of circular collider. Equation (3. the emitted power scales . Therefore: 45 GeV 4 938 MeV 4 ≈ 2 × 104 LEP-I/LHC power ratio = 45 GeV 4 100 m 2 7 TeV 0. Bando n.101) 510 MeV 27 km ≈ 830 LEP-I/ DAΦNE Energy supply and heat dissipation at LEP were one of the main challenge and ultimately limited the energy reach of LEP-II. (3. see also Problem 3. while the latter collides protons with a maximum beam energy of 7 TeV. (3.100) PB EB mA RA The CERN accelerators LEP-I and the LHC are hosted by the same tunnel of length L ≈ 27 km.93).5. 1N/R3/SUB/2005 Problem 3. the ratio between the power emitted by beam/accelerator A and the power emitted by a different beam/accelerator B is given by: 4 4 2 PA EA mB RB = (3. In particular. Suggested Readings More information on the electron synchrotron DAΦNE.14 What is the ratio between the power radiated by a LHC proton and en electron at LEP-I? And between the latter and an electron at DAΦNE? Solution Refering to Eq. The former used to collide electrons/positrons of energy E ≈ 45 GeV.93). can be found in Ref.511 MeV (3. hosted by the LNF. The DAΦNE collider located at the Frascati Laboratories collides electrons and positrons at an energy of about E = m φ /2 = 510 MeV and consists of two circular rings of length L ≈ 100 m.15 How does the power emitted by a relativistic particle of energy E and mass m moving in a circular orbit depend on the ratio E/m? Discuss an application or a consequence of this energy emission. Solution The power emitted by a charged particle moving along a circular trajectory in the form of syncrotron radiation is given by Eq.204 3 Accelerators and Experimental Apparatuses Suggested Readings The master reference for this problem is the classical textbook on electrodynamics by Jackson [14]. 18211/2016 Problem 3. [15] Bando n. 103) |B|ρ ∂ x which is the field derivative normalised to the beam rigidity |B|ρ. (3. which make electron beams appealing for a number of research applications.102) B y (x. y) = − q|p| ∂x x ∂ By ⇒ mγ ∂ By (3. Equation (3.e. rotated by 90◦ . On the other hand. [16]. This is normally achieved by RF cavities. Another advantage of syncrotron radiation is the natural development of a beam polarisation orthogonal to the orbital plane [14] and a reduced beam energy spread. y) = + q|p| mγ ∂ x y where the derivatives are computed at the centre of the quadrupole. Bando n. It can be proved that a series of identical quadrupoles of length . The latter can become challenging for accelerators operated at cryogenic temperatures (e. the enhanced emission of syncrotron radiation by lighter particles finds technological applications. for the production of syncrotron light as a diagnostic tool (like X-ray spectroscopy). The quadrupole strength is defined as 1 ∂ By K ≡ . e. Because of this scaling law.2 Accelerators 205 like ∼ (E/m)4 .g. like electrons or positrons. whose accelerating gradients are however limited in practice by power consumption. rotated by 90◦ one from the other. see e. [14].16 What is the role of a magnetic quadrupole in an accelerator? Discussion A magnetic quadrupole can be obtained by four dipoles. On the positive side. or. i. the lost power needs to be supplied at each turn as to maintain the particle orbit at the ring radius R. it is more expensive to maintain closed orbits for light particles.102) implies that the force exerted by a quadrupole is focusing in one direction and defocusing on the orthogonal direction. Ref. If the dipoles are aligned along the bisectrix of the x–y plane. guarantees . the LHC). y) = ∂x y Fx (x. and with alternated polarities (e. like protons or ions.g. but of alternated polarity. y) = ∂x x Fy (x. the radiation emitted by the particle needs to be removed by a cooling system. of the quadrupole strength. Suggested Readings The synchrotron radiation is discussed in detail in Chap. N-S-N-S). rather than for heavier ones.g. for a fixed radius R and beam energy E. For a reference on beam polarisation at LEP and its applications. The challenge is twofold: on the one hand.g. equivalently. depending on the sign of the derivative. 14 of Ref. 5N/R3/TEC/2005 Problem 3. symmetry dictates the magnetic field and Lorentz force experienced by a charged particle moving along the z-direction at the centre of the quadrupole to given by: ∂ By ∂ By Bx (x.3. Another application of quadrupoles is as the last beam focusing elements in the proximity of the interaction point. where f = ( K )−1 is the focal length of the quadrupole (strong focusing). 3.7 Example of a F O D O cell hosted at the LNF site: a magnetic dipole is followed by a pair of quadrupoles with alternated polarity. since it consists of a focusing (F) and defocusing (D) quadrupole. (3. See also Sect. The basic element of this lattice is called a F O D O cell. Modern synchrothrons are built upon the principle of strong focusing. 3.206 3 Accelerators and Experimental Apparatuses Fig. the reader is addressed to Ref. Solution Owing to the focusing/defocusing property of quadrupole fields. see Eq. can . see Fig. the latter find application as the building bricks of the magnetic lattice needed to maintain the beams in stable circular orbits. 30 of Ref. 5N/R3/TEC/2005 Problem 3. [3]. where the minimum beam size is usually desired.102). Suggested Readings For an introduction to the physics of accelerators.17 Define the emittance of a beam. for example provided by a dipole. separated by a region of no drift (O). Bando n. [17]. What is the normalised emittance εn ? What is the relation between emittance and brightness? Discussion The motion of a particle of momentum |p| and charge q.7. moving in a circular collider whose lattice consists of a series of focusing/defocusing and dipole elements. and by a RF cavity stability to the beam provided that the distance between two quadrupoles is less than 2 f . (3.104) can be proved as follows.104) ⎪ ⎩ ∂x z + x/ρ = 0 where the quadrupole strength K of Eq. as to re- flect the opposite behaviour of quadrupoles along orthogonal directions. For the x displacement. or in terms of infinitesimal variations. Consider for example the y displacement. The transverse motion consists of two decoupled equations. or. The co-rotating reference frame moves along the ideal trajectory and the position of the centre is parametrised by the path length variable s.3. hence z = −x/ρ. and z with respect to the ideal trajectory (a circle of radius ρ). The x-coordinate is usually taken as the outgoing radial direction of the co-rotating frame.2 Accelerators 207 be described the terms of the displacements x.107) βx and similar for y. The first two equations in (3. The amplitude functions βx. the transverse motion along x features equilibrium oscillations around the ideal trajectory with wavenumber ρ. y. (3. and |B| is the field intensity of the dipoles. The equation for the longitudinal motion can be instead proved by noticing that a displacement x from the ideal orbit will make the particle position along z move forward. and the z-direction is tangent to the ideal trajectory. consider the case that the trajectory differs from the ideal one by a constant displacement δ: the x displacement as a function of s is a projection of the displacement along a rotating radial vector: x = |δ| cos(s/ρ + φ0 ).106) 1 with : ψx = . a similar equation hold with inverted sign of K . The solution can be written in the form # x(s) = εx βx (s) cos(ψx (s)). the y-coordinate as the vertical direction. K x (s) = + |B|ρ ∂ x + ⎪ 1 ρ2 1 ∂ Bz y + K y (s) y = 0. The linearised equations read as [3] ⎧ 1 ∂ Bz ⎨x + K x (s) x = 0. (3. εx is constant and is fixed by the initial conditions. (3. Since these displacements are very small compared to the radius ρ. ρ is the radius of curvature of the ideal trajectory.102).105) ds |B| ρ where the magnetic field component in the y direction is given by Eq. d(d x/ds) = −(x ds)/ρ 2 . which are instances of Hills equation. the bending angle (3. K y (s) = − |B|ρ (3.y are periodic functions of s of period 2πρ. Even in the absence of quadrupoles. When the particle moves inside the quadrupole along an infinitesimal length ds.103) has been used to account for the focusing/defocusing effect of quadrupoles. For example. backward. by −(x/ρ) ds.43) along y is given by: dy (∂x Bz y) · ds dθ y = d = ≡ −K y y ⇒ y + K y y = 0 (3. The play both the role of an amplitude envelope along the circumference (the maximum displacement . and 2βx βx − βx2 + 4βx2 K x (s) = 4. the problem can be linearised. resulting in a system of differential equations. Here. e. one can easily show that .108) 0 βx After completing an integer number or turns. thus implying that the particle won’t necessarily find itself in the same displaced position after a full turn. and of local wavelength. After some algebra. The phase ψx at the position s along the circumference is given by s ds ψx (s) = ψx0 + .208 3 Accelerators and Experimental Apparatuses √ is bound by |x| ≤ maxs εx βx (s)). the phase has advanced by a number which is in general not a multiple of 2π . s = k 2πρ. i. (3. The latter is called beam emittance. which is the case if the beam particles are weakly interacting and if the beam radiation is negligible.y 2 ) ε0 = .110) βx √ √ i. whose principal axes are continuously stretched and rotated as s changes. In particular.y and of the initial phase φx. the semi-axes have lengths Δx = εx βx and Δx = εx /βx . . IUnder the assumption that the beam dynamics is described by a conserved Hamiltonian.y where now σx. one has: x. the motion is takes place along an ellipse of area π ε. the equation takes the nice form: x2 βx = 0 ⇒ + βx x 2 = εx .e. and is measured in units of (mm mrad). the Hamiltonian flow conserves the phase-space volume px. (3. x ). 1 + (−βx /2)2 β x 2 + 2 − x x x + βx x 2 = εx (3. (3. is also a time-invariant. (3.109) βx 2 In terms of the phase-space variables (x.111) βx.y 0 sampled from a statistical distribution. the area x.y . each particle will obey Eq. when βx = 0. Since the latter is proportional to d x x and dy y .y (2 σx. respectively.104) with a value of the invariant εx. Solution When the beam is made of N 1 non-interacting particles.y as the lengths that contain a certain number of standard deviations of the phase-space distribution. q) are coniugate variables (Liouville theorem). for a 87% fraction of the beam particles.y dqx. If the particle distribution in these planes is gaussian. y). For example. This condition is for example realised at the interaction points of circular colliders as to ensure the smallest beam sizes in the transverse plane. where( p.y ε0 in the displacement-divergence plane that contain a certain fraction of the beam particles. one can define the transverse beam sizes σx.y give the beam sizes in the displacement spaces (x. From Eq.2 Accelerators 209 In the relativistic limit. its emittance is εn . Liouville theorem states that p dq ∼ βγ d x x is conserved. We then have: 2ββ − β 2 − 4 = 0.116) β β∗ β∗ . Problem 3.e. (3. The brightness B of a beam is defined as the particle multiplicity per unit of emittance. [17]. The normalised emittance is the true time-invariant along the accelerator chain.3.e. so that β (0) = 0.114) then becomes: 2 s2 2 2s 2 β∗ + ∗ − − 4 = 0. (3.113) ε0 Suggested Readings For an overview on the physics of accelerators. therefore the beam emittance decreases like γ −1 for γ 1. in order to maximise the luminosity. (3. (3. For simplicity of notation.17. we suppress the coordinate index. Equation (3. After the acceleration. We chose the coordinate system such that s = 0 corresponds to the IP position.114) To solve this ODE.106). (3. Solution As discussed in Problem 3. the emittance gets shrunk by a factor γ .112) When a beam is created and injected into the accelerating stage. (3. and can be defined as the beam emittance for γ = 1. (3. we make the ansatz: β(s) = β ∗ + α ∗ s 2 . the α ∗ parameter is not independent from β ∗ .114) with the boundary condition β (0) = 0 implies that β (0) = 2/β ∗ . the beam optics is adjusted such that the amplitude function is at a minimum at the interaction point (IP). since Eq.18 Determine the functional form of the amplitude function β(s) in a region free of magnetic fields around the interaction point of a circular collider.115) where β ∗ = β(0).: N B= . we can write the ODE for β(s) in a region free of magnetic field by setting K (s) = 0. the reader is addressed to Ref. i. Actually.: εn = γ ε0 . i. A summary of the relevant notation can be found in the dedicated PDG review [3]. hence α ∗ = 1/β ∗ . (3.110).75 µm and β ∗ = 0. see Eq. approximately one order of magnitude smaller than the typical transverse momentum of bounded quarks inside the proton.55 m corresponding to pT = x |p| ≈ 200 MeV.19 Consider a LHC proton beam at s = 14 TeV. Discussion We can notice that at large distance from the IP. while the beam RMS increases linearly.117) β∗ which is a parabola in s. √ Problem 3. the beam size at the centre will shrink. The net effect depends on how the longitudinal beam size compares to β ∗ . By reducing the amplitude function at the interaction point (β ∗ ). 5N/R3/TEC/2005 Problem 3. the solution is indeed given by: s2 β(s) = β ∗ + .111). Estimate the size of the proton transverse momentum at the interaction point of the ring by assuming εn = 3. whose constant terms and curvature at the origin are not independent.118) β∗ γ β∗ 7 × 103 · 0.210 3 Accelerators and Experimental Apparatuses Hence. respectively.20 What are the betratron oscillations? . Bando n. (3. i.5 m. the design β ∗ at the LHC interaction points is about 0. while the size at the boundary of the luminous region will increase proportionally. (3.e. it follows that the phase-space points of the beam √ particles √in the displacement-divergence plane lie along ellipses with semiaxes ε β ∗ and ε/β ∗ . How does this number compare with the typical quark transverse momentum of order /2r p ∼ 2. The divergence at the IP is maximal and is given by ε εn 3. When the two beams have a finite crossing angle φ (assuming an angle of φ/2 of each beam direction with respect to the z-axis). thus reducing its contribution to the overall luminosity. the amplitude function grows quadrat- ically with s. An an example. The increase of the transverse beam size away from the IP is called hourglass effect. From Eq.75 µm x = = ≈ = 3 × 10−5 . the proton transverse momentum is given by |p|φ/2 ≈ 1 GeV. (3.55 m.5 GeV? And with the transverse momentum due to a beam-beam crossing angle μ = 300 µrad? Solution The interaction points correspond to local minima of the amplitude function β(s). provided that the motion is not relativistic. the cyclotron frequency ω B coincides with the revolution frequency ω of a charged particle moving in an orthogonal magnetic field.e. γ ≈ 1. (3.2 Accelerators 211 Solution The betratron oscillation.y (s). (3. which provides an envelope to the particle displacements from the ideal trajectory and plays the role of a local wavelength of the transverse motion. are the oscillation of beam particles in the transverse plane around the reference trajectory. the trajectory of the ideal particle that undergoes a perfectly circular and closed path. (3. the revolution frequency decreases. i. At relativistic energies. This condition is to be avoided to prevent local defects of the lattice to sum up constructively. i. i.120) dt mγ mγ γ Therefore. p Q x. dt mγ mγ dp e |B| e |B| ωB · er ≡ ω |p| = |p| ⇒ ω = = . thus called because they were first observed in betratrons. (3. where the revolution frequency ω is given by: dp e e |B| = p×B= |p| er . These oscillations are described by the amplitude functions βx. Bando n.3. a situation that can cause beam instabilities.32). From the equation of motion of Eq. but the velocity saturates to c.22. so that the revolution period scales like ∼|p|. then the particles find themselves in the same position after a certain number of turns.40).y = . see Problem 3. it follows that such a particle completes a full revolution in a time T = 2π/ω. 13153/2009 Problem 3.y = q with p and q integers. Bando n.21 What is the cyclotron frequency? Solution Th cyclotron frequency for a charged particle of mass m and electric charge e moving on a plane orthogonal to a uniform magnetic field of intensity |B| is defined as ω B = e |B|/m. see Eq.e.e. see Problem 3.y (s) are called beam tunes. 1 ds Q x.119) 2π βx. This can be intuitively understood by noticing that the the length of the circumference scales linearly with the particle momentum.17. 5N/R3/TEC/2005 . If the beam tunes are rational numbers. The number of cycles in the two transverse directions per turn. where an orthogonal and constant magnetic field bends the particles by 180◦ . Solution In a cyclotron. In a betatron. (3. a synchrotron. time-dependent otherwise.122) π R2 A guide field Bg orthogonal to the beam plane. In its simplest implementation. The polarity of the electric field in inverted at each half- turn. where charged particles are progressively accelerated and maintained in orbit by means of guide magnetic fields. the frequency of the accelerating potential is constant. since the radius of curvature changes proportionally to the momentum. 3. • f : constant for β 1. the adopted solution for high-energy accelerators. as of today. will maintain the beam at a fixed radius R provided that . it grows linearly with the particle momentum. Discussion The cyclotron. At higher energies. The synchrotron represents the historical evolution of the cyclotron and is.121) ∂t 2π R 2π R 2 1 with |Ḃa | = dσ · Ḃa (3. it consists of a yoke encircled by a toroidal beam tube of fixed radius. the radius of the D-shaped magnets determines the maximum energy. particles are accelerated by time-varying electric fields between the faces of two D-shaped magnets.120). Historically. the cyclotron has been the first accelerator used for particle physics. betatron. time-varying magnetic fields provide both the guide and the acceleration. (3. 1919). a cyclotron. Particles inside the tube are accelerated by a rotational electric field generated by the time varying magnetic field in the yoke: ∂Ba Φ̇(t) π R 2 |Ḃa | R ∇ ×E=− ⇒ |E| = = = |Ḃa |. 2. although the idea behind the betatron is even older (Wideröe. and the orbital radius change during the acceleration of a proton in: 1. Therefore. Furthermore. • R: time-dependent. the radiofrequency. The revolution frequency is given by Eq. we can summarise: • B: constant. hence it is approximately independent of the particle energy for β 1: for non-relativistic particle. and synchrotron are three different kinds of non-static circular accelerators.22 Explain how the magnetic field intensity.212 3 Accelerators and Experimental Apparatuses Problem 3. a betatron. the frequency needs to decrease in order to be always in phase with the arrival of the particle. Ap- proximately constant for γ 1. During the acceleration stage. Therefore. the field is maintained at a constant value. we can summarise: • B: time-dependent. a toroidal beam tube is immersed in a guide magnetic field generated by dipoles located all around the tube.123) 2 2 π R2 This implies a proportion of 2 : 1 between the accelerating and guide fields. When the ultimate beam energy is reached. • f : time-dependent. • f : constant. The guide field is incremented as to catch up with the increasing beam energy. • R: constant. In a synchrotron. and the radiofrequency is approximately constant for non-relativistic particles. the radius of the accelerated beam changes proportionally with the beam momentum. • R: constant. For electron/positron acceleration. For a given radius R. the latter is mostly limited by the maximum intensity of the guide field. the reader is addressed to Ref. . [17]. 18211/2016 Problem 3. the beam energy is progressively incremented by radiofrequency (RF) cavities synchronised with the arrival of the particles.2 Accelerators 213 R 1 1 ṗ = q|Ḃg |R = q|E| = q |Ḃa | ⇒ |Bg | = dσ · Ba (3. the magnetic field is constant.22. Bando n. see Problem 3. it must be a multiple of the beam revolution frequency. There- fore. In a cyclotron. while the magnetic field and the accelerating RF change with time until the largest energy is achieved.23 What are the main differences between a cyclotron and a synchrotron? What factors do limit the maximum energy achievable by each of them? Solution The main features of cyclotrons and synchrotrons have been discussed in Problem 3. The maximum beam energy is limited by the outer radius of the magnets and by the necessity of maintaining the RF in phase with the revolution frequency. it grows linearly with the particle momentum until the maximum beam energy is reached.13. Suggested Readings For an introduction to the physics of accelerators. we can summarise: • B: time-dependent. the limiting factor may come instead from the synchrotron radiation.3. In a synchrotron. the radius if fixed. see Eq. The main advantage of LPA is the intense accelerating field. How can you explain these peaks? . Solution Laser plasma acceleration consists in the acceleration of charged particles by the wakefield induced by a resonant short-pulse (pulse duration of order of the plasma frequency. for an ambient electron density of 1017 cm−3 .25 The bremsstrahlung spectrum emitted by a X-ray tube for medical applications features a Gaussian-like shape.124) to be compared with the RF electric fields of about 30 MV/m. (2. The main challenge are represented by the maximum luminosity achievable and by the overall power efficiency (luminosity as a function of electrical input power). 5N/R3/TEC/2005 Problem 3. Suggested Readings The state of the art in LPA technology has been summarised in the IFCA White Paper [18]. The largest accelerating gradients achievable by LPA would considreably reduce the length and costs of the accelerating facilities. For example. the reader is addressed to Ref. which can exceed the maximum electric fields of traditional RF cavities by about three orders of magnitude.22)). with a variety of peaks superimposed.24 Discuss a way to excite wakes in plasma acceleration? What would be the advantages of a plasma accelerator compared to traditional acceleration techniques? Discussion Wake-field acceleration is a technique for charged particle acceleration consisting in using the electric field induced by the passage of a moving charge (wakefield) as driving force to accelerate a beam of particles.214 3 Accelerators and Experimental Apparatuses Suggested Readings For an introduction to the physics of accelerators. The phase velocity of the plasma wave is approximately given by the group velocity of the laser. 18211/2016 Problem 3. the accelerating plasma electric field is of order E = 96 (n e /cm−3 )1/2 V/m ≈ 30 GV/m. [17]. (3. high-intensity laser source in an underdense plasma (n e ∼ 1017 cm−3 ). Bando n. Bando n. The spectrum of the emitted radiation is a continuous distribution.126) hence three times large than by using a single-ended accelerator at the same terminal voltage. L. Here.125) where −VI is the potential of the ion source.2 Accelerators 215 Solution X-ray tubes are devices where electrons are accelerated through an electric potential and impinge on a metallic anode emitting bremsstrahlung radiation. For more detailed information on negative ion beams. . becoming positive ions of charge +z e. they accelerate again towards another grounded electrode where they are finally extracted (usually through the use of a magnet analyser). (3. (3. At this point. Suggested Readings An instructive overview on electrostatic accelerators and on the concept of Tan- dem acceleration is presented in the lectures [19]. M levels). they encounter a charge stripper (usually made of gas or of a thin carbon foil). Solution For a He source.125) with V = 5 MV and VI = 0.3. 18211/2016 Problem 3. we therefore obtain a maximum energy Tmax = 3 e V = 15 MeV. featuring a number of peaks superimposed. The kinetic energy acquired by the ion when reaching the final electrode is therefore T = e VI + e V + e z V = e VI + e (1 + z) V. the maximum ion charge after the stripper is z = +2. where they get ionised. modulated at low frequency from the absorption by dead material. In a Tandem accelerator. negative ions of charge −e are produced at one electrode. and are accelerated towards the high-voltage terminal. and technological aspects of electrostatic accelerators. The latter correspond to excitation of the atomic levels of the atoms at the anode (K .26 What is the highest energy of an α particle produced by a 5 MV tandem accelerator? Discussion The concept of Tandem accelerator was invented in order to achieve higher beam energies compared to single-ended accelerators at the same maximum terminal voltage V . From Eq. [11]. (3. the reader is addressed to Ref. charge strippers. usually grounded or connected to a negative potential −VI . Bando n. t)] (3. We chose the reference frame such that the beam direction is aligned along the z-axis.8.216 3 Accelerators and Experimental Apparatuses 3. As discussed in Sect.129) Fig. y) δ2 (x. (1. 1.3. y). (3. y) is the surface density of the target at the point (x. The target is at rest and is characterised by a particle distribution n 2 (r). while the x. For the special case of uniform target density and uniform beam flux over a limited area. More specifically. Eq. as it is usually the case at colliders. (3.127) to a few special cases. y) while dΦ1 is the beam flux (number of particles per unit time) across an infinitesimal surface element d x d y centred around (x.128) where δ2 (x. Equation (3. Eq.11.127) where vrel is the relative velocity. Continuous Beam Against a Fixed Target Let’s assume that a direct current beam (1) with particle velocity v1 is directed onto a fixed target (2).8 Illustration of a x continuous beam of density n 1 (r) and velocity v1 n2(r) colliding against a fixed target of density n 2 (r) v1 n1(r) z . y) δ2 (x. and n 1. (3. y) = = dΦ1 (x. 3. y axes span the transverse plane. (3.134) becomes: L = Φ 1 δ2 . the luminosity is defined as the coefficient of proportionality between the measured event rate and the cross section. see Problem 1. The integration is performed over the volume around the interaction point. The relative velocity is vrel = |v1 |. the luminosity of a collider determines the event rate.288) implies: L (t) = vrel dr [n 1 (r. see Fig.127) becomes L = |v1 | d x d y dz n 1 (x. Assuming that the interacting particles are prepared in beams of fixed momentum. We can specialise Eq. 3. t) n 2 (r. y). y. y) n 2 (x.3 Luminosity and Event Rates The concept of luminosity is of cardinal importance in accelerator and collider physics.2 (r. The beam distribution along z is assumed to be uniform. t) are the particle densities of the colliding beams. z) = |v1 | d x d y n 1 (x. i. z) = Nb 0 = f coll d x d y dz ds n 1 (x. y. y) δ2 (x.3. 3. t) n 2 (x.3 Luminosity and Event Rates 217 Fig. We further assume that the bunches are equally spaced in time. −1 Nb f coll f coll L ≡ dt L (t) = f coll |v1 | d x d y dz dt n 1 (x. velocity v1 . Since the motion takes place along the z-axis.127) over many bunches Nb . (3. y) (3. z) = f coll d x d y δ1 (x. see Fig.127) becomes L (t) = |v1 | d x d y dz n 1 (x. (1. y. Eq. z. z − s) dz n 2 (x. We can however redefine the luminosity as the time average of Eq. z. and non-zero during the collision. z).9. (3. y. In the last equality. it’s zero before any bunch collision. the density n 1 must be a function of z − s. it suffices to integrate the particle density over a time large enough to contain the full bunch crossing through the target. z. so that the frequency f coll of bunch collision with the target is a constant.290). Bunched Beam Against a Fixed Target Let’s assume that a bunched beam (1) with bunch velocity v1 is directed against a fixed target (2). the luminosity here is a function of time.132) For the special case of uniform target and bunch density over a limited area.131) −1 The second equality holds if the beam structure is periodic with time period f coll : in this case.140) becomes: . This result is in agreement with Eq. y. For example.9 Illustration of a x bunched beam of bunch density n 1 (r). t) n 2 (x. (3. y. Equation (3. y. n2(r) colliding with a fixed target of density n 2 (r) with a v1 n1(r) z collision frequency f coll −1 |v1|frev where Φ1 is the beam flux across the surface of minimum area between the beam and the target transverse size.e. the variable s = |v1 | t has been introduced to make the integration variables homogeneous. so that: L = f coll d x d y ds n 1 (x.130) Strictly speaking. 3. y. s) n 2 (x. z) (3. y. With this choice: 2 φ φ vrel = c 2 cos − sin2 φ = c (1 + cos φ) = 2c cos2 (3. Since the beam structure is periodic. so that f coll is a constant. (3. We further assume that the beam are relativistic. 3. and divide the result by the time interval between two subsequent bunch crossings.134) c2 We can further simplify the expression by introducing the crossing angle φ. we have: φ L = 2 cos2 f coll d x d y dz ds n 1 (r. Writing explicitely vrel in terms of the beam velocities: (v1 × v2 )2 L = f coll (v1 − v2 ) − 2 d x d y dz dt n 1 (r. Again. Integrated Luminosity √ Since the statistical significance of a measurement grows like N .10. Putting everything together. Bunched Beam Against a Bunched Beam Let’s assume that two bunched beams (1) and (2) with bunch velocity v1 and v2 collide one against the other. colliding head-on with a v1 n1(r) n2(r) v2 z collision frequency f coll −1 s=0 |v1|fcoll L = f coll N1 δ2 .133) where N1 is the number of particles per bunch. we shall consider the time integral of L (t) for two bunches.136) 2 where we have again introduced the variable s = c t. if the transverse size of the target is smaller than the bunch. defined such that the angle between the two velocities is π − φ. see Fig. We further assume that the bunches are equally spaced in time. s). and velocities v1 and v2 .218 3 Accelerators and Experimental Apparatuses Fig. 3. at the end of the day what matters for the physics is the integrated luminosity defined as: . one should consider only those particles inside the smallest of the two areas. φ = 0 and we obtain the expected result vrel = 2c. t) n 2 (r. t).10 Illustration of two x bunched beams of bunch density n 1 (r) and n 2 (r). (3.135) 2 2 For back-to-back collisions. so that |v1 | = |v2 | ≈ c. s) n 2 (r. (3. This section collects a number of exercises that aim at familiarising with the concept of luminosity. For this reason. z.138) 4π σx σ y Generalise this expression to the case where the standard deviations of the two beams are different. bunch populations N1 and N2 .and y-axes. Solution The luminosity can be computed from Eq. the integral at the right-hand side of Eq. we chose the reference frame such that the bunches move along the z-axis.136) can be computed explicitely yielding: ds dr [ρ1 (r. s)] = ! " ! " ! " ! " (2π )−3 x2 y2 z2 s2 = 2 2 2 d x exp − 2 dy exp − 2 dz exp − 2 ds exp − 2 σ x σ y σz σx σy σz σz (2π )−3 √ √ √ 1 = ( πσx )( πσ y )( πσz )2 = (3.137) 0 where the integration is performed over the full run time Trun .3 Luminosity and Event Rates 219 Trun Lint = dt L (t) ≡ Trun L . s) ρ2 (r. The latter is limited by other contingent factors.139) ⎪ ⎩ρ2 (x. s) = (2π)− 2 2 y2 (z−s)2 σ x σ y σz exp − 2σx 2 − 2σ y2 − 2σz2 x (3.3. like the availability of the accelerating facility for the desired research program. Gaussian beam profile in three dimensions with standard deviations σx .136). and colliding head-on along the z-direction. etc. (3. the maximisation of L to reduce the overall run time is an important achievement for an accelerator. With this choice: ⎧ ⎪ 3 ⎨ρ1 (x. is given by the formula: N1 N2 f rev Nb L = . σz . We further assume that the centre of each bunch passes through the origin at the time s = 0.140) 2 2 2 σ x σ y σz 8π σx σ y . y. where a number of exact calculations can be performed for several cases of interest. turn-around vs beam life times. z. and much effort is usually devoted to this task. revolution frequency f rev . Problems Problem 3. while the transverse plane is spanned by the x. (3. To parametrise the bunch densities. (3. y. (3.27 Prove that the luminosity of a circular collider where two equally spaced and bunched beams with Nb bunches per beam. Particular emphasis is devoted to the luminosity of circular collider. s) = (2π)− 2 3 x2 exp − 2σ 2 − y2 − (z+s)2 σ x σ y σz x 2σ y2 2σz2 Assuming φ = 0 and relativistic bunches. power supply limitations. σ y . 113).141) 8π σx σ y 4π σx σ y It is interesting to notice that under these assumptions.136) then becomes: 1 N1 N2 f rev Nb L = 2 Nb f rev N1 N2 = . Using Eqs. (3. s)] = × (σ1 x σ2 x )(σ1 y σ2 y )(σ1 z σ2 z ) . the luminosity is independent of the bunch resolution along the direction of collision.220 3 Accelerators and Experimental Apparatuses Equation (3. s) ρ2 (r. (3. we can equivalently write the luminosity in a form that makes explicit its dependence on the accelerator parameters: N1 N2 f rev Nb L =γ # # .111) and (3. (3.140) now becomes: (2π )−3 ds dr [ρ1 (r. The integral of Eq.142) 4π βx∗ εxn β y∗ εny Let us now consider the more generic case that the beam dimensions are different. (3. ! " 1 1 1 1 1 1 d x exp − + 2 x 2 dy exp − + 2 y2 × 2 σx21 σx 2 2 σ y21 σy 2 2 . 1 1 1 σ2 z − σ12z dz ds exp − + z 2 − 2zs + s 2 = 2 σz21 σz22 σ22z + σ12z (2π )− 2 3 = × σ12x + σ22x σ12 y + σ22 y σ12z + σ22z . (3.142) when σ1 x = σ2 x and σ1 y = σ2 y . (3. we find that the luminosity is independent of the bunch RMS along z. 1 σ12z + σ22z (σ12z − σ22z )2 2 ds exp − 1− 2 s = 2 σ12z σ22z (σ1 z + σ22z )2 1 π = σ12z + σ22z = 2 π 3/2 3/2 σ1 x + σ2 x σ1 y + σ2 y σ1 z + σ2 z 2 2 2 2 2 2 2 1 = .143) 4π σ12x + σ22x σ12 y + σ22 y Putting everything together. Again. since .144) 2π σ12x + σ22x σ12 y + σ22 y which reduces to Eq. we get: N1 N2 f rev Nb L = . This could have been predicted a priori. (3. yielding: 2N1 N2 f rev Nb N1 N2 f rev Nb L = √ = (3.3.+ b ] ds dz I (z)[− 2c +s.148) 2( 12) σx σ y 2 12 σx σ y A comparison with Eq.142) shows that the luminosity for identical rectangular beams is again independent of the bunch size along z. with Jacobian (σ1 z σ2 z )/2. z.3 Luminosity and Event Rates 221 2 1 z−s z+s 1 L ∼ dz ds f f = dξ f (ξ ) . . (3. s) = 1 abc I (x)[− a2 .146) ρ2 (x.28 Determine the luminosity of a circular collider where two equally spaced and bunched beams with Nb bunches per beam. revolution frequency f rev . The integral at the right-hand side of Eq. s)] = 1 = d x I (x)[− a2 . collide head-on along the z-direction. The integral at the right-hand side of Eq. y. y.+ a2 ] I (y)[− b2 . s) = 1 abc I (x)[− a2 . Solution The luminosity is given by Eq.+ 2c +s] (3. y. and c. s) ρ2 (r. z. z) dimensions of size a. (3.+ 2c −s] = (a b c) 2 2 2 0 −c 1 c c 2 c c = ds s + − −s − + ds −s + − s− = a b c2 − 2c 2 2 0 2 2 2 1 c 1 = = . (3.+ b2 ] I (z)[− 2c −s.136) can be computed explicitely yielding: ds dr [ρ1 (r.+ 2c −s] where I is the index function. The technical note [21] contains a number of useful formula.+ a2 ] I (y)[− b2 . bunch populations N1 and N2 . and rectangular shape in the (x.145) σ1 z σ2 z σ1 z σ2 z 2 where we have just performed the change of variables ξ = (z − s)/σ1 z and ξ = (z + s)/σ2 z .+ b2 ] I (z)[− 2c +s.5%.147) a b c2 2 2ab We can express the √ beam sizes in terms √ of the RMS along the x and y components through σx = a/ 12 and σ y = b/ 12.+ 2c +s] I (z)[− 2c −s. Suggested Readings The lecture notes [20] represent a valid starting point for introducing the concept of luminosity. (3. (3. and differs from the result of Gaussian bunches by about 4. b.+ a2 ] dy I (y)[− b .145) is clearly independent of σi z .136) with the choice: ρ1 (x. Problem 3. (3. To summarise.4. provided that σ is replaced by the appropriate RMS. and σz = 8.7 µm. that was obtained under the assumption of Gaussian profiles.142) represents a valid approximation for reasonable distributions in the transverse plane. we can assume that that the angle φ between the two beams is in the x z-plane. is not a mere accident. one can expect that the value of L x 2 will not differ much from the result obtained under 3 the assumption of a parabolic distribution. a Gaussian. Evaluate the correction factor using the LHC design parameters φ = 285 µrad. Eq (3. which is 3/5 2 = 0. Suggested Readings The goodness of the RMS as a measure of the beam size in the calculation of luminosities has been studied in detail in Ref.11 Illustration of two x bunched beams of the same bunch density n(r) colliding at an angle π − φ φ z v1 s=0 v2 n1(r) n2(r) . it can be proved analytically [22] that this result holds for several # functional forms.g. The luminosity is given by Eq. σx = 16. Problem 3. z. [22] for some numerical results. see e. Indeed. s) = (2π) 2 exp − x+φ/22 − y 22 − (z+φ/2 −s) 2 σ x σ y σz 2 2σ x 2σ y 2σz (3. y.149) ⎩ρ (x. the value of the functional L x 2 is at a minimum for a parabolic distribution centred around x = 0 and defined on a compact support. s) = (2π)− 2 exp − x−φ/2 3 2 y2 (z −φ/2 +s)2 2 σ x σ y σz 2σ 2 − 2σ 2 − 2σ 2 x y z Fig.29 Consider the case of two identical bunched beams with Gaussian profile that collide with a crossing angle angle φ 1. Solution Without loss of generality.142). 3. etc. agrees numerically with the rectangular distribution. see Fig. Determine how the luminosity gets reduced with respect to the zero crossing angle. y. z.8 cm. 3. a rectangle. In particular.g. a triangle.222 3 Accelerators and Experimental Apparatuses Discussion The fact that the luminosity formula (3.). Table 10 of Ref. [22]. Since most of the reasonable assumptions for ρ(x) are of this functional form#(e.11. see Problem 4.136) with the choice: ⎧ −3 2 ⎨ρ1 (x. (3.268. and indeed one finds variations not larger than 5%. 3 Luminosity and Event Rates 223 Here.153) 4π σ y σz2 sin2 φ2 + σx2 cos2 φ 4π σx σ y σ 2 1 + σz2 tan2 φ 2 x 2 If we now make the assumption that φ 1 and that σz σx .e. (3.152) σz σz2 σz2 Restoring all the constants. φ ≈ 3 × 10−4 rad).150) x+φ/2 = −z sin φ2 + x cos φ2 x−φ/2 = +z sin φ2 + x cos φ2 The integrand at the right-hand side of Eq. i. (3.3. (3. σx2 σz2 σx2 σz2 σz σz2 (3. the other angle that sets the reference scale is the ratio . y. at the LHC. (3.151) The integration over s gives: ! φ " ! φ " s2 2xs sin √ x 2 sin2 ds exp − 2 + 2 = π σz exp 2 . and z: φ (2π )−3 √ √ L = 2N1 N2 f rev Nb cos2 π σ y π σz × 2 σ x σ y σz 2 2 2 ! " cos2 φ2 sin2 φ2 cos2 φ2 × d x dz exp −x 2 −z 2 + = σx2 σx2 σz2 N1 N2 f rev Nb cos φ2 N1 N2 f rev Nb 1 = = . which is typically the case at colliders.154) 4π σx σ y σx 2 Even though the crossing angle can be rather small in absolute scale (for example. (3. and integrating over x. then the result can be further simplified to give: ! "− 21 N1 N2 f rev Nb σz φ 2 L = 1+ .136) can be simplified to: ! ! " " 1 φ φ 2 φ φ 2 y2 exp − 2 x cos − z sin + x cos + z sin − 2 × 2σx 2 2 2 2 σz ! ! 2 2 "" 1 φ φ φ φ exp − 2 z cos + x sin − s + z cos − x sin + s = 2σz 2 2 2 2 ! " cos2 φ2 sin2 φ2 sin2 φ2 cos2 φ2 s2 2xs sin φ2 = exp −x 2 + −z 2 + − 2+ . z +φ/2 = +z cos φ2 + x sin φ2 z −φ/2 = +z cos φ2 − x sin φ2 . z) by ±φ/2 around the y-axis. we have indicated by x±φ/2 and z ±φ/2 the rotated of (x. namely: z +φ/2 = +z cos φ2 + x sin φ2 z −φ/2 = +z cos φ2 − x sin φ2 . and assume that the beams have an offset Δ with respect to the z-axis. Suggested Readings See the lecture notes [20] for an overview on the subject.12. if d1 = d2 nothing should happen since this case corresponds to a rigid translation of the two beams in the x z-plane. x+φ/2 = d1 − z sin φ2 + x cos φ2 x−φ/2 = d2 + z sin φ2 + x cos φ2 (3. using the LHC design parameters. (3.30 Consider the beam setup of Problem 3. The adimensional quantity φ2 σσxz is often referred to as Piwinski angle.12 Illustration of two x bunched beams of the same bunch density n(r) colliding n2(r) at an angle π − φ and with an offset Δ φ z v1 s=0 v2 n1(r) Δ . which can be large.224 3 Accelerators and Experimental Apparatuses σz /σx .156) Clearly. the integrand is proportional to: Fig. the luminosity gets reduced by about 16% compared to the case of perfectly head-on collisions. (3. How does the luminosity change as a function of the beam offset? Solution We can refer to Problem 3.e. (3. After collecting all terms that depend on x and z. we have: ! 2 "− 21 7. Problem 3.7 µm 2 i.151).29 for much of the discussion and for a good fraction of the calculations. as illustrated in Fig. For example. 3. 3. We need to modify Eq.155) 16.835.7 cm 285 × 10−6 1+ = 0. We can therefore expect the result to depend only on the relative offset Δ = |d1 − d2 | Referring to Eq.150) to include two offsets d1 and d2 along the axes orthogonal to the two beam directions. we can easily see that the integration over s and y is unaffected.29. We can therefore use the result Eq.157) σx2 σz2 σx The integration over x is straightforward and gives the same contribution as in Eq.154) and multiply the right-hand side by the offset-dependent corrections to give: ⎡ ⎤ ! 2 "− 21 N1 N2 f rev Nb σz φ ⎢ Δ sin2 φ2 2 Δ ⎥ 2 L = 1+ tan exp ⎣ 2φ − ⎦. (3.153).3. (3. the luminosity depends on the displacement like: .158) In particular.3 Luminosity and Event Rates 225 ⎡ 2 ⎤ cos φ2 d + d (d − d ) 2 exp ⎣− x ⎦× 1 2 2 1 + − σx 2σx 4σx2 ! " sin2 φ2 cos2 φ2 sin φ2 exp −z 2 + − z 2 (d2 − d1 ) (3. 4π σx σ y σx 2 sin cos2 φ 4σx4 σ 2 2 + σ 2 2 4σx2 x z (3. The integration over z gives the offset-dependent term in addition to the Piwinski term. for φ 1. Discussion From Eq. (3. (3. Δ2 L = L0 exp − 2 .159).159) 4σx where L0 is the luminosity for perfectly head-on and aligned collisions. it follows that the counting rate for any reaction occurring when the two beams interact will be given by . As of today. (3. Let the cross section for such destructive absorbtion be σ0 . . This method.31 In a fixed target collider.160) dt dt 4σx Hence. was first suggested by S. known as Van der Meer scan. van der Meer at the CERN Intersecting Storage Ring [23]. Interactions between the beam and the target remove beam particles. dN d N0 Δ2 = exp − 2 . it represents the principal method for an absolute luminosity calibration at hadron colliders. Problem 3. a monochromatic and continuous beam with flux Φin passes through a target at rest of unknown particle density. Determine the luminosity of the collider from the beam flux measured before and after the target. one can directly measure the transverse RMS σx by recording the counting rates at different values of Δ during a beam scan. if the interaction responsible for the beam absorbtion is the same that produces the signal events. The luminosity is therefore given by: Φin L = 1 − e−n 2 σ0 d ≈ Φin n 2 d (3. (3.161) The number of particles removed from the beam in an infinitesimal volume centred around the point (x. and J1 · (d x d y) is the infinitesimal flux across an area d x d y.g. y.135). The assumption of stationary density is clearly valid if the linear density along the beam direction is constant. z) n 2 (x. y. 1N/R3/SUB/2005 Problem 3. we get the pretty intuitive result: dN = L σ0 = Φin − Φout .161). (3. z) n 2 (x.166) σ0 where the last approximation is valid for d much smaller than the interaction length. (3. the flux decreases as a function of z according to an exponential law: Φ(z) = Φin exp [−n 2 σ0 z] . Equation (3. z). How many protons per bunch are there? What is the beam cross section in the interaction region? . and agrees with Eq. y.162) into (3.127) can be written as L = |v1 | d x d y dz n 1 (x. In particular. z)σ0 dz.163) σ0 σ0 σ0 where we have used the fact that J1 (r) = |v1 |n 1 (r) is the beam flux density. Bando n.226 3 Accelerators and Experimental Apparatuses Solution Let’s denote the beam velocity by v1 and the beam density by n 1 (x.164) dt i. For the case of a uniform target of thickness d. y.e.14.165) see e. y.32 Prove that the LHC luminosity is L = 1034 cm−2 s−1 knowing that the beam current is I = 0. (3. (3. y.162) * +. the observed event rate is equal to the flux absorbed from the beam. Problem 2.5 A and assuming 3000 bunches per beam. - 1/λ0 Plugging Eq. we get: 1 1 Φin − Φout L =− d x d y d(|v1 |n 1 ) = − dΦ1 = . (3. z) by the interaction with the target is: dn 1 = −n 1 (x. (3. z). Let the total pp cross section be σ pp .167) and (3. the number of particles removed from the bunch is given by .3 Luminosity and Event Rates 227 Solution The beam current is related to the total number of protons per beam by the relation: I I = e Nb N f rev ⇒ (N f rev Nb ) = . (3.y = 16.584 A) would give the well-known result of 1.167) e From Eqs. assumed large compared to the interval between two subsequent collisions. (3.167): I /e 0.245 kHz · 3000 A more refined calculation (Nb = 2801. which is related to the luminosity by Eq. we thus have: (N f rev Nb )2 (I /e)2 L = = .6 × 10−19 C N= = = 0. so that the overall collision frequency is less than the expectation from 25 ns spaced bunches. Notice that the collision frequency differs from the well-known value of 40 MHz. (3.9 × 1034 cm−2 s−1 . (3. The LHC revolution frequency is f rev = 11. we first derive the time evolution of the bunch population N . During a time interval dt.138). How does L change as a function of run time? Solution In order to determine the time evolution for the luminosity. I = 0. we have: (0. (3.170) f 11.168) 4π σx σ y f rev Nb 4π σx σ y f rev Nb The luminosity can be estimated if the collision frequency and the transverse dimensions of the bunches at the IP are known.33 A circular pp collider is characterised by a luminosity L0 at the start of the collision run.7 × 10−4 cm)2 · 11. Let’s denote the number of interaction points by n IP . so that several collisions take place and d N 1. The number of protons per bunch can be obtained directly from Eq.7 µm [24] for the RMS.3.138).5 A/1.245 kHz · 3000 (3. Using the design value σx. This is due to the fact that the LHC beams contain empty sections. 4π · (16. Problem 3.15 × 1011 protons/bunch. The colliding beams consist of Nb equally populated bunches (initial bunch population N0 ) with revolution frequency f rev .169) in agreement with the design value 1034 cm−2 s−1 .6 × 10−19 C)2 = = 0.9 × 1011 .5 A/1. (3. and that intersect in two points located along the ring. 245 kHz. other effects. determine the optimal run time of a collider. −1 −1 exponential law L (t) = L0 e−t/τ . given a beam lifetime τ and a turn-around time Tcycl .171) This last ODE can be trivially integrated to give: N0 L0 N (t) = ⇒ L (t) = . τ = 15 h. i. Using the exponential approximation. Solution Given a total operation time Ttot and a turn-around time of the machine Tcycl . 4π σx σ y N0 N0 Nb N0 dn dt Nb N0 − = . we can define the beam lifetime as the run time at which the luminosity gets√ reduced to a factor of 1/e from its initial value.34 Given a total operation time Ttot . including collision loss. the optimal run time Trun is defined as the run duration that maximises the integrated luminosity over the entire operation time. where τ = i τi is the total beam lifetime.e.8 h (3. (3.172) (1 + t/τnucl ) (1 + t/τnucl )2 We can evaluate numerically τnucl using the LHC design parameters (for two experiments): 2808 · 1.228 3 Accelerators and Experimental Apparatuses 2 f rev n IP dN N L0 σpp n IP d N = −dt N 2 σ pp . With this convention. τ = τnucl ( e − 1) ≈ 29 h. with n(t) = N (t)/N0 . = −dt . Problem 3. The time evolution of the luminosity can be often approximated by an . the reader is addressed to the technical report [24].15 × 1011 τnucl = = 44. see Problem 3.33. Discussion Besides nuclear collisions. τnucl ≡ n2 τnucl L0 σpp n IP (3. characterised by time scales τi .173) 1034 cm−2 s−1 · 100 mbarn · 2 In analogy with the exponential decay. Suggested Readings For an overview of the LHC design parameters. the run duration Trun that provides the largest total integrated luminosity. n(0) = 1. Estimate the maximum total integrated luminosity at the LHC using Trun = 4800 h (equivalent to 200 full-days). Trun can be determined from the condition: . contribute to beam losses. 3 Luminosity and Event Rates 229 ⎡ ⎤ ⎢ ⎥ .3. per run opt Trun opt opt Trun Trun + Tcycl 0= − ln 1 + . which is dictated by the injection cycle PS → SPS → LHC and by the energy ramping in the LHC.174) τ τ The last equation can be solved numerically. by using the LHC parameters Tcycl = 1. -* +.173)). we can estimate the integrated luminosity at the end of the operation time: . ∂ Trun ⎣ Trun + Tcycl 0 ⎦ ∂ Trun Trun + Tcycl opt * +. a numerical scan of Eq. (3. For example.5 h. With this value. (3.174) shows that the zero of the equation is opt obtained for Trun ≈ 5. τ = 15 h (to be compared to the value τnucl of Eq. (3. lum. Trun number of runs int.2 h. . ∂ ⎢ Ttot Trun ⎥ ∂ Ttot τ L0 (0)(1 − e−Trun /τ ) 0= ⎢ dt L (t)⎥ = . 5 × 1031 cm−2 s−1 .2 h 15 h (3. What is the probability of having zero interactions per bunch crossing? Discussion Equation (3. The two easiest ways to increase the luminosity are therefore to minimise the amplitude functions at the IP.35 Compute the average number of interactions per bunch crossing if the luminosity is L = 2. The number of bunches is limited by the minimum spacing of buckets along the beam. β ∗ and/or to maximise the number of particles per bunch. on the other hand the extra rate is fully concentrated in a small time window around each bunch crossing. which depends for example on the RF cavities. The beam energy γ is either limited by the accelerator or imposed by the physics program. While the luminosity clearly benefits from these two operations.175) This value is a crude approximation and should be taken as an order-of-magnitude estimate. the reader is addressed to the technical report [24]. The normalised emittance depends on the full acceleration cycle and is mostly fixed by technical limitations.142) encodes the recipes for maximising the luminosity of a given collider. 1N/R3/SUB/2005 Problem 3.5 h Lint = · 1034 cm−2 s−1 · 15 h · 1 − exp − ≈ 110 fb−1 . Suggested Readings For an overview of the LHC design parameters. 5. 4800 h 5. the total cross section σ = 20 mbarn and the bunches cross every 4 µs. This .5 h + 1. The revolution frequency is determined by the beam radius. Bando n. 18211/2016 Problem 3. i. f = Nb f rev using the notation of Eq.178) n! In hadron colliders. the number of antiprotons.36 An antiproton beam with momentum |p| = 6 GeV and total current I = 0.e.142).5%. hence: exp [−μPU ] μnPU P(n | μPU ) = ⇒ P(0.988 c. Calculate the revolution frequency. μPU = 27. The collision events are actually clustered in short time windows with duration of order Tcoll ∼ σz /c ∼ ns. (3. Solution The average rate of events is given by L σ .16 mA moves inside a circular ring of length L = 300 m. (3. (3.176) bunch crossing # bunch crossing / second f where f is the collision frequency. the number of pile-up events can be very large. At each round. and the integrated luminosity after T = 6 min. where the inelastic cross sections are large. We can compute the number of events per bunch crossing as: # collisions # collisions / second Lσ = ⇔ μPU = .177) The number of pile-up events will be distributed as a Poissonian variable with mean μPU .179) |p|2 + m 2p . the beam crosses a hydrogen gas target with a surface density δ = 1014 cm−2 . The collisions events occurring simultaneously in the same bunch crossing are called pile-up events. Bando n. What beam energy would it be needed if the same centre-of-mass energy had to be achieved by using a symmetric proton-antiproton collider? Solution The beam velocity is given by |p| |v| = = 0. which is expected to become as large as μPU ≈ 140 in the future high-luminosity LHC upgrade [25]. From the numbers given by the exercise: μPU = 1031 cm−2 s−1 · 20 mbarn · 4 µs = 2.230 3 Accelerators and Experimental Apparatuses would not be the case if the number of bunches or the revolution frequency could be increased. μPU ) = e−μPU ≈ 13. (3. at the LHC with design conditions. separated by longer inter-bunch intervals with no activity. For example.0 (3. 6 × 10−11 A 0.16 × 10−3 A L =Φδ = δ= · 1014 cm−2 = 1029 cm−2 s−1 .0 MHz Since the hydrogen target is at rest. so that the number of proton-iron interactions per second is: dN I d 1. The beam energy of circular collider operating at the same value of s is: 1 2 E circ = 2m 2p + 2m p (|p|2 + m 2p ) 2 . (3.3. 18211/2016 Problem 3.183) where we have used the conversion cm−2 = 10−24 barn−1 .182) e 1.6 × 10−19 C 17 cm . / 0 21 01 1 |p|2 1 E circ = mp + 1+ 2 = 1.6 × 10−19 C The integrated luminosity after T = 6 min is given by: Lint = L T = 1029 cm−2 s−1 · (6 × 60) s = 36 μb−1 . Since d λ.0 × 109 . the collider behaves like a fixed-target experiment.185) dt e λ 1.6 × 10−19 C Nb = = = 1.1 cm = = = 0.291). (3. The interaction length in iron is λ = 17 cm. (3. (3. see Eq.59 MHz. The proton flux corresponds to a current I = 0.16 × 10−3 A/1.290): I 0. (3.016 nA. Solution The interaction length is defined as the probability of interaction per particle and per unit length. (1.180) L 300 m We can use Eq.184) 2 2 mp Bando n.81 GeV. which is fine given the accuracy on the other data.1 cm. (1. Estimate the rate of charged and neutral pions emerging from the slab.0 MHz. The revolution frequency is therefore: |v| 3 × 108 m s−1 f rev = ≈ = 1.3 Luminosity and Event Rates 231 which we can approximate by c within 1%. The luminosity is therefore given by Eq.167) to determine the number of circulating protons in the beam: I /e 0. we can approximate this probability by d/λ.87) with m 1√= m 2 = m p . (3. (3. (1.181) f rev 1. The centre-of-mass energy is given by Eq.37 A 10 GeV proton beam crosses an iron slab of thickness d = 0. both the trigger and the detector are not sensitive to additional events. typically. the measured rate m in a non-paralysable system can be obtained like follows. Consider a time interval T max τ. Bando n.g.185).188) admits two solutions for ν at a given measured rate m. (3. The number of triggers m in this time interval will be given by the number of true events occurring in the same time window minus the number of true events occurring during the total dead time generated by the accepted triggers. in the assumption that the number of true events in a given time window is Poisson distributed. a pion. (3.e. a dead time τ . i. . etc. and a trigger efficiency ε (probability of accepting an event given that it has been accepted by the DAQ system). i. so that the dead time cannot be extended beyond the single-event processing time. we first notice that. This can be easily proved by noticing that the probability of zero events after a time Δt from a given event is e−ν Δt . sharing a sizable fraction of the incoming proton energy. 1N/R3/SUB/2005 Problem 3.e. during this latency. Only true events spaced by more than τ can generate a trigger. This is in contrast with the so-called paralyzable systems. The rate of events that need to be selected online is f = 5 kHz. ν −1 . so we can assume that the total rate of pions (neutral and charged) is equal to the interaction rate of Eq. p + p → p + n + π + .: m = ε ν e−(ν τ ) . where ν is the true event rate. case Eq. Determine the average data acquisition rate. p + n → p + n + π 0 . the time intervals between two subsequent events has an exponential distribution with decay constant ν. hence the measured rate is equal to the fraction of events such that Δt > τ . high-energy protons in matter first produce one energetic hadron.186) 1 + ντ For a paralyzable system. Given a true event rate ν..: εν m T = ν T − m T τ ν ⇒ m = ε m = . where the arrival of a new event before the system has finished processing the previous one extends the time window in which the system is not sensitive to additional events.38 An experiment with a direct-current beam is instrumented with a trigger system with efficiency ε = 20%.187) Differently from the non-paralyzable.35. The data acquisition system generates a dead time of T = 1 ms for each recorded event. (3. (3.232 3 Accelerators and Experimental Apparatuses As discussed in Problem 2. e. Discussion The data acquisition system illustrated in this problem is defined non-paralyzable because the system is not sensitive to additional events arriving after one trigger. 17 kHz.186). Suggested Readings An introduction to the concept of paralysable and non-paralysable DAQ systems can be found in Refs.3 Luminosity and Event Rates 233 Solution From Eq. where σacc is the cross section for producing an event that passes the selection. [2]. For example. (3. The second arises from the digitizer and affects only the rate of events that have fired the trigger. The DAQ system gives rise to two types of dead time: the first comes from the trigger system and affects the full collision rate. i.39 Collision events at a rate of 5 kHz are analysed to decide whether an event has to be recorded or not.3. Assuming both systems to be non-paralyzable (see Problem 3.2 · 5 kHz m= = 0. A more complete overview on the topic can be found in Ref. while σtot is the total cross section. εtrig can be the fraction of events that pass the online selection devised to identify signal events. we can visualise the rate reduction as a two-stage process: Trigger εtrig f Digitizer νtrig f −−−→ νtrig = −−−−→ νdig = . εtrig = σacc /σtot . Bando n. 1N/R3/SUB/2005 Problem 3. the rate in output to the digitizer needs to satisfy: . The decision takes a time Ttrig = 20 µs while the digitization takes Tdig = 1 ms.e.38).189) 1 + f Ttrig 1 + νtrig Tdig If one wants to maintain a fraction of lost events less than δ.188) 1 + 1 × 10−3 s · 5 × 103 s−1 It is interesting to notice that a paralzsable system operating with the same conditions would yield a trigger rate smaller by a factor of ≈25 compared to a non-paralyzable one. What rate of accepted events can be sustained if the dead time has to be maintained below δ = 20%? Solution Let’s denote the collision rate by f and the fraction of signal events that we wish to select online by εtrig . [26. 27]. (3. (3. where now the dead time refers by extension to the fraction of signal events lost because of dead time in the DAQ system. we can estimate the average data acquisition rate as: 0. The following mean rates are measured for the three signals: f A = 7. . 5 × 103 s−1 · 10−3 s 1 − 0. like those induced by beam halo particles.83. we thus get 1 0. at each bunch crossing. thus generating a random coincidence.234 3 Accelerators and Experimental Apparatuses ⎛ ⎞ νdig ε f 1 1 ≥ 1 − δ. A logic AND of the two signals measures the coincidence of the two signals in the same bunch crossing. or collisions with the rediual gas inside the beam pipe. corresponding to a dead time 1 δ =1− ≈ 0. A and B. are superimposed to the true events. are generated by the counters in the occurrence of an event.43 × 104 Hz.67 × 104 Hz. Bunches cross every T = 10 µs and. Background events.53 × 104 Hz. Bando n.05 − 0. (3. f B = 6.02 = 3%.40 Two scintillating counters are used to monitor the luminosity at a collider.193) 1 + f (Ttrig + Tdig ) If a smaller dead time is desired for the same efficiency. and f A∩B = 5. (3.191) and the accepted rate is: νtrig = εtrig f = 3% · 5 kHz = 150 Hz. [2]. Suggested Readings For an introduction to the queueing theory and to the use of trigger in HEP. two digital signals.2. possible solutions are to pipeline the events while waiting for a trigger accept.192) Notice that the largest dead time of the system is achieved by chosing εtrig = 1 (all triggering events are accepted). εtrig ≤ − .2 10−3 s (3. 1N/R3/SUB/2005 Problem 3. The two counters are located before and after the interaction point. ⎝ trig ⎠ ≥1−δ εtrig f 1 + f Ttrig 1 + εtrig f εtrig f T 1+ f Ttrig dig 1 1 δ Ttrig ≥ 1 − δ.190) 1 + f (Ttrig + εtrig Tdig ) f Tdig 1 − δ Tdig With the choice δ = 0. the reader is addressed to Ref. (3.2 2 × 10−5 s εtrig < − = 0. Determine the mean number of collisions per bunch crossing. or parallelise the event digitization by using multiple CPU’s. = + 1− · . we get: N · Prob [A] N · Prob [coll. f f .3. an absolute normalisation of the luminosity becomes possible. T T f T ν fA = ν + 1 − bA . and let us assume for simplicity that any collision produces a signal in A.] ν N · Prob [A] N · Prob [B] 1 = + 1− · . T T f T T N /T ν bA bB f A∩B = ν + 1 − . (3. and bB : ⎧ ⎪ ⎪ f = ν + 1 − ν b ⎪ ⎨ A f A ν f B = ν + 1 − f bB (3.195) f where bA is the background rate in counter A.] = ν/ f . and compute its probability as: Prob [A ∩ B] = Prob [coll. (3.] + (1 − Prob [coll. Solution Let’s first consider the signal in one of two counters. Since the acceptance of the counters is fixed. rate. say counter A. N · Prob [A ∩ B] N · Prob [coll. where ν is the coll. . and remembering that Prob [coll. (3. Similarly.194) by N /T . By measuring independently the individual counter rates and the coincidence rate. the luminosity is proportional to the rate of coincident events.]) · Prob bkg. If the cross section of the scattering process in the counter acceptance is also known (e. (3. we can consider the coincidence signal.] + (1 − Prob [coll. Because of background events.3 Luminosity and Event Rates 235 Discussion A standard technique for monitoring the luminosity at circular colliders is through the use of two symmetric counters located at both sides of the interaction point. Multiplying both sides of Eq. we can write a system of three equations in three unknowns ν. the rate of signal events can be disentangled from the false coincidences. false coincidences can be generated at random. bA .g.]) · Prob [A] Prob [B] . where f is the bunch crossing frequency.196) f f To summarise.] ν N · Prob bkg. Bhabba scattering for e+ e− machines). The probability of observing an event in A for a given bunch crossing is the sum of probabilities of two independent hypotheses: Prob [A] = Prob [coll.197) ⎪ ⎪ ⎪ ⎩ f A∩B = ν + 1 − ν bA bB .194) Let us now consider N 1 bunch crossings over a time T 1/ f . [28]. Appendix 1 The computer program below implements a Monte Carlo simulation of the experimental apparatus considered in Problem 3. (3.67 × 104 ) ν= Hz = 3. f 1 − νf f f A∩B − f A f B ν= .236 3 Accelerators and Experimental Apparatuses which can be used to determine the true collision frequency ν.199) The mean number of collisions per bunch crossing as defined in Eq. y2 ) are sampled from a Gaussian pdf centred around the true values yi = xi2 /2R and with standard deviation equal to the detector resolution.67 × 104 + 5. we obtain an estimate: (105 )(5. The measured points (y0 . 105 − 7. (3. The signed sagitta is computed from y2 + y0 s= − y1 .43 × 104 (3. (3.53 × 104 )(6.200) f 105 Hz Suggested Readings The luminosity measurement at LEP using two luminosity monitors mounted at small angles with respect to the beam is discussed in details in Ref. y1 .176) is given by: ν 3. ( f − ν) ( f A∩B − ν) = ( f A − ν) ( f B − ν) .198) f − f A − f B + f A∩B Replacing the symbols by the measured values.53 × 104 − 6. (3. Replacing the first two equations into the last one: ( f A − ν) ( f B − ν) f A∩B = ν + .43 × 104 ) − (7.201) 2 and a sign flip corresponds to the outcome s < 0.31 × 104 Hz. .331.10.31 × 104 Hz μPU = = = 0. pt=1000...0).0). B=1. s2=1.2f +/. s0=1.5*(y2+y0) .sqrt((1-mis)*mis/ntoys) # binomial pdf print ("Charge mis-id = (%.5*x*x/R # track sagitta\index{Sagitta} from measured position def sagitta(y0.0..y1...0.0. s1=0.0] for toy in xrange(ntoys): y0 = smear( track(x=0.y1. B=1.3 Luminosity and Event Rates 237 import math import random # the true particle positions. pt=1000. s1 ) y2 = smear( track(x=2. pt=1000. B=1. pt=1000.gauss(mu. pt=1000. s2 ) s = sagitta(y0.2f)"% (mis*100.%. B=1.0).3. B=1. sigma): return random.5e-04.12.202) pTL (1−α) − pTH (1−α) pTL c2 + c1 pT2 from Problem 3.y2): return 0.y2) outcomes[ int(s<0) ] += 1 mis = float(outcomes[1])/float(ntoys) err = math.0e-04. err*100)) ################ toys(ntoys=1000000) Appendix 2 The computer program below computes numerically the integral: −1 N (α − 1) pTH pT−α σ2pT Δφ ≈ dpT (3. s0 ) y1 = smear( track(x=1.0): R = pt/(0.0e-04): outcomes = [0. use MKS units and pT in GeV/c def toys(ntoys=100.3*B) return 0. .y1 # the measured particle positions def smear(mu..0.sigma) # generate ntoys pseudo-measurement.0. use MKS units and pT in GeV/c def track(x=1. 020 7. step = 1. Berlin.org/10. x_h=100.5)*step val = math.pow(2.0 n_step = int((x_h-x_l)/step) for s in xrange( n_step ): x = x_l + (s+0."N=".D. 1-alpha) .M. res. Phys. Lett. 1-alpha)) return 1.05 ) print"<sigma>=". Blum.1016/0029- 554X(63)90347-1 2. Lett. 9. ATLAS Collaboration. 336 (2007). New York.1088/1367-2630/9/9/336 6. Meccanica Analitica. Data Analysis Techniques for High-Energy Physics. Frühwirth et al. ATLAS and CMS Collaborations.0136. Electrostatic Accelerators (Springer. R.S. Phys.2) c2 = math. Particle Detection with Drift Chambers (Springer. Perkins. (Wiley.org/10. JINST 5. 2nd edn. L.7. Classical Electrodynamics. https://doi. -alpha)/(c2 + c1*x*x) integ += val*step integ *= (alpha-1)/(math.. S. W. D.math. step = 0. L.infn.114.com/cws/article/cern/28229 10. (Bollati Boringhieri. Hellborg (ed. 2000) 11. B 716.238 3 Accelerators and Experimental Apparatuses import math # the constants c1 = 2*math. 08. 24. https://doi. C 40.021 8.lnf.6 acc = 1. 2005) 12. 2008) 13. Marmi. alpha=2.1088/1748-0221/5/03/ T03021 14. Phys. Riegler. F. B 716. Patrignani et al.0e-03. Collaboration.0 ): integ = 0. 191803 (2015).M. Collaboration. w3. Particle data group. res/BP * math. Ragusa. Rolandi. https://doi. 2002) 5.8) BP = 2 * 0. Cambridge.org/10. J. https://doi.0e-02 def integrate(x_l=5. Jackson. -2) References 1. x_h = 300. http://cerncourier. Phys. 4th edn. Glusckstern. 08. W.S. (Cambridge University press.7.). Meth. 30 (2012). C. https://doi.191803 9. Lett. T03021 (2010). Chin. Phys. 1 (2012). C. 2nd edn. Nucl. 100001 (2016) 4. C.pow(x_h.org/10. 1103/PhysRevLett.2012. Fasano.1016/j. 114.H. Cambridge.pow(x./math.org/10.. Berlin. New J.0. 2000) 3.org/10.pow(x_l. Introduction to Hig Energy Physics.pow(acc. alpha=2.2012.1016/j. (Cambridge Press.pow(0./1. 381 (1963). Rev.sqrt(integ) ####################### x_l = 5.physletb. 1999) 15. https://doi.2)*(60.physletb. Instr.. 3rd edn. A. R. res = integrate(x_l=x_l. x_h=x_h.L.it/accelerators/dafne/?lang=en . Rolandi. R. CERN/MPS/DL 69-12 (1969) 23. 24 May–2 Jun 2005. E. Herr. Hinterberger. G. CERN-2015-005 (2015) 26. A. Germany. G. New York. Wilson. Leo.5170/CERN-2006-002. W. F. 4th edn. 361–378 (CERN-2006-002). E. http://www-bd. 56 (2011). B. CERN SL-97-72 (1997) . 15–26 Sep 2003. KEK-76-3 (1976) 22. World 4(1). van der Meer. Muratori. pp. 2010) 28. S. High Power Laser Technology for Accelera- tors. CERN-2004-003 (2004) 25. https://doi.95 20.5170/CERN-2006-012. (Springer.G.R.CERN Accelerator School: Intermediate Course on Accelerator Physics. T. 95-112 (CERN-2006-012). 1993) 27. vol.org/ 10. Techniques for Nuclear and Particle Physics Experiments.1088/2058-7058/4/1/20 17. 22 (1991). Blondel. W. Zeuthen. https://doi. Oxford. Knoll. LHC Design Report. Apollinari et al. Suzuki.org/10. (Wiley. H.CERN Accelerator School and KVI: Specialised CAS Course on Small Accelerators. published in ICFA Beam Dynamics Newslettern. 2001) 18. https://doi. Phys. The Netherlands. 2nd edn.References 239 16.361 21.F. Hereward. pp. CAS . CAS .org/10. Zeegse. An Introduction to Particle Acceleretors (Oxford University Press. Berlin. Radiation Detection and Measurement.gov/ icfabd/ 19. ISR-PO/68-31 (1968) 24.fnal. White Paper of ICFA-ICUIL Joint Task Force. Bravin et al. There is a vast literature of text books on probability. 4. the sample mean © Springer International Publishing AG 2018 241 L. No attempt is made here to be exhaustive in this respect. . statistics. On the one hand. [1]. .org/10.1 (Law of Large Numbers) Let {X 1 . A few concepts are of key importance in data analysis. see e. the properties of the likelihood function are illustrated by considering simple cases that can be handled analytically. and the construction of frequentist confidence intervals. only the probability of occurrence is prescribed. UNITEXT for Physics. Selected Exercises in Particle and Nuclear Physics.e. On the other hand. . Ref. A few concepts and classical results. most of the reactions studies in particle physics are associated with quantum- mechanical transition amplitude.1 Elements of Statistics The outcome of an experiment can be treated as a random variable. the presence of noise in any measurement introduces a certain degree of randomness.Chapter 4 Statistics in Particle Physics Abstract This chapter is devoted to the statistical treatment of experimental data. stand out for their pervasive presence and cardinal importance in the analysis of experimental data. the reader is addressed to dedicated books. . Bianchini. i. In this case. The “large number” theorems There are two main theorems one should always remember when dealing with large size data samples. the interpretation of an experiment has to be carried out according to the paradigms of statistics.g. This section collects a few outstanding results that are relevant for solving the proposed exercises. By its very nature. and data analysis applied to particle physics.1007/978-3-319-70494-4_4 . however. error propagation. For their proof. https://doi. but not the actual outcome of a given reaction. without the use of computer programs. Theorem 4. X N } be a sequence of N independent and identically distributed random variables with mean μ and variance σ 2 < ∞. Then. the outcome of a particle physics experiment is a to be considered as random process: the theory of probability thus provides the correct theoretical framework. After recalling a few outstanding theorems. For these reasons. The focus is then put on the combination of random variables. f. indicated by θ̂. and let f (X |θ ) be the p. . will ultimately have a Gaussian distribution. The likelihood function for point estimation Let X = {X 1 . An important application of the Large Number Law is the Monte Carlo method for numerical integration.d. The likelihood function of the data is defined by N L(X|θ) = f (X i | θ ).1) N converges to μ in the limit N → ∞. is a maximum of the likelihood function (4. . b] can be estimated by the sample mean g evaluated at points u i uniformly distributed in [a. This result guarantees that the repeated measurement of the same unknown quantity will provide asymptotically an estimator of the true value. but finite variance. .2 (Central Limit theorem) Let {X 1 . . the ML . Theorem 4. For the case that L is a differentiable function of θ .3) N 2 i σi converges to a normal distribution N (0. (4.242 4 Statistics in Particle Physics N Xi SN = i=1 (4. the p. 1) in the limit N → ∞. (4. Indeed. X N } be a sequence of N independent random variables with mean μi and variances σi2 . X N } be a set of N independent observations of a random variable X (in general. Under suitable hypothesis (Lyapunov condition).4). which is affected by a sufficiently large number of independent random fluctuations of a priori unknown distribution. . the integral of a function g over the interval [a. . The importance of the Central Limit theorem lies in its prediction that an experimental measurement. The domain of X is further assumed to be independent of θ . the Law of Large Number guarantees that: (b − a) N b b IN = g(u i ) → (b − a) du f (u) g(u) = du g(u).4) i=1 The maximum likelihood estimator of θ .f. . . of X which depends on a set of parameters θ . of the variable N (X i − μi ) Z N = i=1 (4.d. b].2) N i a a Another benefit of the MC integration compared to other approximate analytical methods is that the error on the integral scales like N −1/2 independently of the dimensionality. Considering for simplicity the one-dimensional case. X can a vector of random variables). but the global maximum will always asymptotically converge to the true value.e. it gets arbitrarily “close” to the true parameter value. • The covariance matrix of the ML estimator is ⎡ .4.1 Elements of Statistics 243 estimator is the root of the equation: N ∂ ln f (X i | θ) = 0 (4. i.5) ∂θ i=1 θ̂ In general. • a sufficient estimator (Darmois theorem). several roots may exist for finite N (local maxima).3 (Asymptotic properties of the ML estimator) The maximum likelihood estimator is asymptotically: • a consistent and unbiased estimator (Law of Large Numbers). • the most efficient estimator (Cramer–Rao inequality).e. i. it encodes all of the information on the parameters that the data contain. The ML estimator thus defined has the following asymptotic properties: Theorem 4. −1 ⎤ ⎡ . . θ = θ 0 . which can be simple if it consists of just one point. ⎤ ∂ 2 ln L(X|θ ) ∂ 2 ln f (X |θ) −1 V (θ̂)i j = −E ⎣ ⎦ = N −1 E ⎣ ⎦ = N Iθ −1 . Iθ−1 ).d. ξ (θ) = 0 with ξ is a vector of constraints of dimension r < k. ∂θi ∂θ j ∂θi ∂θ j (4. it consists of multiple simple hypotheses. which is often the case in particle physics where the prior knowledge is usually provided by some theoretical model. Thus. N (0. the maximum likelihood method for point estimation is the most used technique for the cases where a likelihood model of the data is available. e.g. Given two such hypotheses θ 0 and θ 1 .g. the likelihood-ratio test statistic λ is defined as: L(X|θ 0 ) λ= . the quantity N (θ̂ − θ ) is distributed with p.5) is also normally distributed with mean 0 and variance N Iθ .f. (4. and on its known asymptotic behaviour. e. (4. Owing to its attractive properties.6) √ where Iθ is the information matrix. or composite. Each point corresponds to a certain hypothesis. • The quantity at the left-hand side of Eq. The likelihood function for hypothesis testing Let θ represent a point of the k-dimensional parameter space on which we wish to make inference from the data.7) L(X|θ 1 ) The importance of the likelihood-ratio test statistic relies on its optimal properties to discriminate between two hypotheses. The 68.4% CL intervals on μ can be then determined as the points at which q = 1 and q = 4. Here. . the asymptotic distribution of the likelihood ratio ˆ L(X|θ̂) λ= (4. .3 and 95. θ̂ ) = 0. • asymptotically the most powerful test statistic to discriminate between two composite hypotheses (Wald’s theorem [2]). . See Ref. and X an array of independent observations of X . f (X |θ ). . θ̂ is the ML estimator obtained by maximising the likelihood with respect to all parameters. • If the maximum likelihood estimators of θ are given.4 (Properties of the likelihood-ratio test statistic) The likelihood-ratio test statistic is: • the best test statistic for simple hypotheses testing (Neyman–Pearson lemma). (4. (4. Problems Problem 4. .f. Discussion The maximum likelihood estimator is an instance of the so-called implicitly defined estimators.9) 1 q(μ) = −2 ln λ(μ) = −2 ln = L(X|θ̂1 . θ̂ˆ2 . while θ̂ˆ is the ML estimator obtained by maximising the likelihood in a subset of the parameter space defined by the r constraints ξ (θ) = 0 (conditional likelihood). . defined as the roots of the experimental observable: 1 ξ N (θ̂ ) = g(X i .10) N i . The special case ξ(θ1 ) = θ1 − μ = 0 gives: L(X|μ.244 4 Statistics in Particle Physics Theorem 4. θ̂ˆk ) (μ − μ̂)2 + O(N − 2 ). . In particular. .1 Let X be a random variable with p. θ̂k ) σμ̂ 2 where μ̂ is the ML estimator of μ.8) L(X|θ̂) is known (Wilks’ theorem [3]). Prove that the ML estimator θ̂ provides a consistent estimator of θ . [4] for more details. which is asymptotically Gaussian distributed around μ with standard deviation σμ̂ . the variable q = −2 ln λ is distributed like a χ 2 with as many degrees of freedom as the number of fixed parameters (r ). if the hypothesis under test is true.d. respectively. provided that: lim E ∂ξ N (θ ) > 0 (4. we thus have: ∂ ∂ f (X |θ ) 1 ∂ f (X |θ ) 0= d X f (X |θ ) = d X = f dX = ∂θ ∂θ f ∂θ ∂ ln f (X |θ ) =E (4. θ )] = 0. (4. one or multiple roots of the equation ξ N = 0 exist.5) divided by N . it must satisfy the normalisation condition d X f (X |θ ) = 1 (4.12) with respect to θ . 4. By the Law of Large Numbers. the Law of Large Numbers states that . Differentiating both sides of Eq. if we indicate by ξ N (θ ) the sum of Eq.12) with respect to θ .1 Illustration of an implicitely defined estimator. (4.10) converges to the true value θ . see Fig.1.f. provided that the derivative of ξ N is asymptotically non-zero where g satisfies the constraint E [g(X. the estimator defined by Eq. (4. we obtain: 2 ∂ ∂ ln f (X |θ ) ∂ ln f (X |θ ) ∂ ln f (X |θ ) 2 0= f (X |θ ) d X = + f dX (4. Solution Since f is a p.12) for any value of θ . 4.14) ∂θ ∂θ ∂θ 2 ∂θ Hence.11) N →∞ ∂θ This last condition guarantees that the first equation can be inverted to yield a solution.. One of such solutions will however converge to the true value θ for N → ∞.4.d.1 Elements of Statistics 245 ξN ∂ξN /∂θ θ̂0 θ̂1 θ Fig.13) ∂θ By differentiation twice Eq. (4. For finite N . Under this assumption. Suggested readings For more details on the proof. b) ⇒ . the likelihood function of the data is given by: 1 (ñ − s − b)2 1 (b̃ − b)2 L(ñ. in a channel where a signal of unknown yield s contribute.b ln L(ñ.11. b̃ | s.18) The constrained ML estimator b̂ˆ is obtained by fixing s and maximising − ln L with respect to b.2 An experiment measures ñ events. b̃ | s.15) N →∞ ∂θ 2 ∂ξ N ∂ ln f (X |θ ) ∂ ln f 2 lim =E = −E <0 (4. b) = ∂s. b) = √ √ exp − √ exp − 2π ñ 2ñ 2π σb 2σb (4. with ñ 1. Verify explicitly Wilks’ theorem by studying the behaviour of profiled likelihood ratio q(s). (4. b̃ | s.17) The ML estimators n̂ and b̂ of the signal and background yields are the solutions of the equation (ñ − s − b)2 (b̃ − b)2 ŝ = ñ − b̃ 0 = ∂s. yielding: b̃ + ñ−s b̂ˆ = σb2 ñ 0 = ∂b ln L(ñ. 7 of Ref. Thus. the ML estimator defined implicitely as the root of ξ N will asymptotically converge to the true value θ . the reader can refer to Chap.b + ⇒ 2ñ 2σb2 b̂ = b̃ (4.246 4 Statistics in Particle Physics ∂ ln f (X |θ ) lim ξ N (θ ) = E =0 (4. Solution For ñ 1.19) 1 σb2 + ñ1 . the number of√measured events can be assumed to be normally distributed with standard deviation ñ. Problem 4. [5]. see Problem 4. The channel is affected by a source of background for which an auxiliary measurement b̃ ± σb is available (b̃ can be assumed to be Gaussian distributed with standard deviation σb ).16) N →∞ ∂θ ∂θ 2 ∂θ The second condition guarantees that the first equation can be inverted to yield a solution. (N Iτ )−1 ). . in agreement with Wilks’ theorem. (4. The profile likelihood ratio q(s) is then given by: ⎡ ⎤ ˆ L(s. see Problem 4.3 An experiment measures the decay times of an unstable particle with lifetime τ . Problem 4. (4. (4. with variance n + σb2 . b̂) 2ñ 2σb2 2ñ 2σb2 2 2 ˆ 2 (ñ − s − b̂) (b̂ˆ − b̃)2 n̂ − s − b̃ n̂ − s − b̃ = + = + = ñ σb2 σb4 n σ12 + ñ1 σb2 n 2 σ12 + ñ1 b b 2 2 s − n̂ + b̃ ŝ − s = = .22) ∂τ 2 0 τ 3 τ 2 τ τ If the N measurements are all independent.9). i=1 τ τ (4.83). Notice that the effect of the uncertainty σb on the background yield is to broaden the negative log-likelihood function compared to the expectation from the sole Poisson statistics. t N }. the ML estimator τ̂ is distributed as N (τ.1 Elements of Statistics 247 which coincides with the combined ML estimator from two independent measurements. N Iτ ). Solution The decay times t are distributed according to the exponential law 1 −t f (t | τ ) = e τ ⇒ E [t] = τ. (4. Show that in the limit N → ∞.23) . b̂) ˆ 2 (ñ − s − b̂) (b̂ˆ − b̃)2 (ñ − ŝ − b̂)2 (b̂ − b̃)2 ⎦ q(s) = −2 ln = −2 − ⎣ − + + = L(ŝ.20) σb2 + n σb2 + n Since ŝ is Gaussian distributed (it is the linear combination of Gaussian distributed variables. ln L(X | τ ) = −N ln τ − . . the likelihood of the data X is given by: N N N 1 i=1 ti i=1 ti L(X | τ ) = f (ti | τ ) = N e− τ .9).21) τ The information Iτ is given by: 2 ∞ ∂ ln f (t | τ ) 2t 1 1 −t 1 Iτ = E − = − dt − + e τ = 2. (4. Denote the set of N measurements by X = {t1 . see Eq. Var [t] = τ 2 . whereas ∂τ ln L(X | τ ) is distributed as N (0. it follows that q(s) is distributed like a χ 2 with one degree of freedom. see Eq.4. produced at rest in the laboratory frame. . . where Iτ is the information. 248 4 Statistics in Particle Physics The ML estimator τ̂ is given by the roots of Eq. (4.22).27) ∂τ τ ∂ N Var ln L(X | τ ) = Var [z N ] = N Iτ (4. 1). N ∂ N ∂ N i=1 ti 0= ln L(X | τ̂ ) = ln f (ti | τ̂ ) = − + . Finally. Since the ML estimator is a linear function of z N . from Eq.f.23) we can see that N N √ ∂ N i=1 ti i=1 (t1 − τ) N ln L(X | τ ) = − + = = zN . (4. the quantity E [ f (X )] · Var [X ]. it will be also (asymptotically) normally distributed.e.24) N N Nτ zN In the last equation. [5] for further details on the information and its relation with the ML estimator. Var τ̂ = = . (4.d. (4. Estimate this bound. of X . i. 5 of Ref.28) ∂τ τ2 Suggested readings The reader is addressed to Chap. which the Central Limit theorem predicts to be asymptotically distributed as N (0. (4. with mean and variance: τ2 1 E τ̂ = τ.26) ∂τ τ τ2 τ2 τ The left-hand side of Eq. admits a lower bound. (4. Problem 4.5).26) is asymptotically normally distributed with mean and variance given by: √ ∂ N E ln L(X | τ ) = E [z N ] = 0 (4. (4.25) N N Iτ where we have made use of Eq.4 Show that for a given random variable X . . ∂τ i=1 ∂τ τ̂ τ̂ 2 N N ti τ i=1 (ti − τ ) τ̂ = i=1 = √ √ +τ (4.29) where f is the p. we have isolated the variable z N . the definition is consistent provided that: 3 1 1 K 5 1 10 10 K 52 J 2 K 5J 2 I = d x ρ0 (x) = dx − dx x2 − = = .1 Elements of Statistics 249 Discussion This exercise complements Problem 3. (4. We define the functional F as: ⎛ ⎞−5 ⎛ ⎞2 ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ F[ρ] ≡ ⎜ ⎝ d x ρ(x)⎟ ⎠ ⎜ ⎝ d x ρ 2 (x)⎟ ⎜ ⎟ ⎠ ⎝ d x x ρ(x)⎠ 2 (4. Solution Without loss of generality.4. K .30) I K J where I . the luminosity in such a case is proportional to d x n 2 (x). In particular. such that f = ρ d x ρ . Eq.28 by proving that the luminosity of a bunched collider whose bunch densities are independent is inversely proportional to the bunch RMS in the transverse plane. if we can consider the infinitesimal variation δρ at position x. K . (4. 0 . (4.30) gives: 4ρ0 x2 5 δF = + − F δρ d x. and J are not independent. Indeed. we then have the condition: 5J 21 5J 21 z2 4ρ0 x2 5 K 5 − − ≤x≤ + − =0 ⇒ ρ0 (z) = 4 I J I I K J J 0 otherwise (4. I . and X 2 is the variance of X .34) 4 I J . Indeed. we can assume E [X ] = 0. We then introduce the non- −1 normalised function ρ. where n is the bunch particle density in the transverse direction. and J are assumed to be finite.31) 9 with a > 0. (4. 4 I J 4 I 3I I 3 I2 K 5 1 K = d x ρ02 = d x ρ0 − d x x 2 ρ0 = K .33) However. 4 I J I 3I 3 I 23 4 I 3 3 3 K 5 1 K 5J 2 10 2 K 52 J 2 J = d x x 2 ρ0 = dx x2 − dx x4 = − = 5 . We shall prove that F has a minimum for the family of parabolic distributions 16 2 ρ0 (x) = max a 1 − x .32) K J J By requiring δ F = 0 for any δρ and d x. (4. the functional F is not bounded from above: for example. Discussion Consider n random variables x distributed with probability density function f (x). .30) satisfies Eq. we need to have: 3 52 32 I 2 J − 2 K −1 ≡ F[ρ0 ]− 2 = 5 1 1 ⇒ F[ρ0 ] = = 0. . . xn ) g(y) = dx f (x) δ(y − h(x)) = d x2 . . [6].38) −∞ . . Suggested readings This exercise is based on Ref.268.35) 3 53 A straightforward calculation shows that the solution (4. i. . . (4.37) where x̃1 are the roots of the equation y − h(x) = 0. but the variance is not. xn ) (4. The variable y is also a random variable. Solution A sample of data points distributed according to a generic p. (4. g of y is given by f (x̃1 . so that the functional F is infinite. The p. . If the numbers are randomly distributed on a different interval. 13153/2009 Problem 4.f.d.1] . for ρ0 (x) to be consistent.d. consider the Cauchy distribution C . one can always redefine x such that it is always contained within the [0. . x2 . ∂h x̃1 ∂ x1 (x̃1 . g(x) = I[0. f can be obtained by applying an analytical transformation to a sample of randomly distributed points x ∈ [0. Let y = h(x) be a generic function of x. Let’s define t F(t) ≡ dt f (t ). 1] interval. y ≡ F −1 (x). Bando n.5 Discuss a method to simulate a data sample distributed according to a given function f (x).35) for any a > 0. in this case.250 4 Statistics in Particle Physics Hence. .f. E [C ] is finite. with p. d xn . (4. x2 .072.36) However.e.d. Therefore: E [ f (X )] · Var [X ] ≥ F[ρ0 ] = 0.f. 1]. f. t ≥ t0 ] e−t1 /τ Prob [t ≥ t1 | t ≥ t0 ] = = −t /τ = e−(t1 −t0 )/τ .41) Prob [t ≥ t0 ] e 0 with t1 > t0 . By considering the variable y = σ z + μ.d. (4. what is the probability that it decays within the following time interval Δt? Solution The exponential law has “no memory”.6 If a particle with mean lifetime τ has not decayed after a time t. a cheaper method to generate the desired distribution relies on the Central Limit Theorem 4. i. (4. the probability that the particle decays at a time greater than t1 . a χn2 distribution with n degrees of freedom. Bando n. For the special case that f is Gaussian function of arbitrary mean μ and standard deviation σ .40) N /12 for uniformly distributed points xi is asymptotically distributed like N (0. we can then obtain the desired target distribution. 1). f . Examples relevant for HEP include the simulation of collision events and of particle interaction with the detector. and then considering the n sum y = i=1 z i2 .1 Elements of Statistics 251 By using Eq. Indeed: g (F(y)) d F = = f (y). (4.e.2 which guarantees that the variable N xi − N /2 z N = i√ (4.37). 1N/R3/SUB/2005 Problem 4. depends only on Δt and not on the . the probability of surviving over a time Δt = t1 − t0 . provided that the particle has not yet decayed at the time t0 . see Ref. For an overview on MC methods. and represents one of the most used tool to simulate multi-dimensional problems. [5]. If f ∼ e−x/2 x n/2−1 . it easy to verify that the sample of y values thus generated is distributed according to the target p. Hence. given that the particle has survived up to a time t0 . Suggested readings The method of generating events distributed according to a prior model by means of random numbers is called Monte Carlo generation. the above method canbe used to generate n independent variables z i .39) d F −1 dx dy where we have used the fact that g(x) = 1 and the chain rule ( f −1 ) = ( f )−1 . Indeed.4. can be computed by Bayes theorem: Prob [t ≥ t1 . If one of the two lifetimes is much larger than the other. g(t): ∞ 1 t t − τ1 − τ2 1 −t/τ2 t −t1 τ1 − τ1 g(t) = dt1 dt2 e 1 2 δ(t − t1 − t2 ) = e dt1 e 1 2 = τ1 τ2 0 τ1 τ2 0 1 ' τ −τ ( e−t/τ2 − e−t/τ1 −t 2 1 = e−t/τ2 1 − e τ1 τ2 = . which can be then inserted in the second equation giving .43) can be readily integrated yielding an exponential decay N A (t) = N0 e−λ A t . we can use Eq.d.d. see Problem 4. where t1 and t2 are independent random variables with exponential distribution with mean τ1 and τ2 . (4. Discussion The p. g(t) reduces to an exponential law with lifetime given by the maximum between τ1 and τ2 . Compute the activity of B after 1.8 A nucleus A undergoes α-decay to a nucleus B with a lifetime τ A = 2 min.f. We can therefore conclude that the probability of decaying in a time window Δt starting from any time at which the particle is still undecayed. whose joint p. of the variable t = t1 + t2 . 18211/2016 Problem 4.d. The nucleus B then decays to a nucleus C with lifetime τ B = 5 × 103 s. (4.37) to determine the p. Problem 4. (4.8. g(t) describes the differential time distribution of a stable isotope C produced from the nuclear decay chain A → B → C. Solution Since t is a known function of the random variables t1 and t2 . with N B (0) = 0 (4. is always given by 1 − e−Δt/τ .d. the p. and NC can be obtained analytically by solving a joint system of ODE that describes the detailed input-output balance: ⎧ ⎧ ⎪ ⎨ Ṅ A = −λ A N A ⎪ ⎨ N A (0) = N0 Ṅ B = +λ A N A − λ B N B .43) ⎪ ⎩ ⎪ ⎩ ṄC = +λ B N B NC (0) = 0 with λi = 1/τi .2 s and after 5 × 103 s. respectively. the p.d. (4. is also known. N B . Bando n. The first of Eq. As made clear by Eq.f.f.7 × 107 nuclei of type A.f.42).42) τ2 − τ1 τ2 − τ1 with t ≥ 0. At the beginning there are 2.f.f. Discussion The time evolution of the three populations N A .7 Determine the p. g(t) is not of exponential form.252 4 Statistics in Particle Physics intermediate time t0 .d. : N0 −t/τ B A B (t) = λ B N B (t) = e − e−t/τ A .54 × 102 Bq (4. almost all the decayed nuclei A have been transmuted into B. (4. τB − τ A τB − τ A (4. so that: N A (t) + N B (t) + NC (t) = N0 . the initial sample of nuclei A has fast transmuted into B: the activity of B is thus the same as if the sample at time t = 0 had been entirely made of nuclei B. The equation and the boundary conditions then fix the unknown coeffcients.44) τB − τ A In order to find N B (t).2 s τ A and t2 = 5 × 103 s = τ B τ A as: N0 t1 t1 N 0 t1 A B (t1 ) ≈ 1− −1+ ≈ = 0. The solution can be obtained by using the general algorithm for solving this type of ODE.f g(t) derived in Eq. (4.45) The activity A B of the nucleus B is defined as the rate of B → C decays in the sample.42) thus describes how the arrival of the nuclei C is distributed in time. (4. . This result is intuitive: after a time interval small compared to the lifetime of B.4.1 Elements of Statistics 253 a first-order polynomial ODE with a uniform term.48) τB e where we have expressed the results in the standard units of Bq = 1 Hz.d. we take a “probabilistic” approach where the time of a nuclear decay is treated as a random variable: the p. i.46) τB − τ A Given that τ B τ A . After a time interval t2 much larger than τ A .20 × 104 Bq. Solution Instead of solving Eq.e. The population NC (t) is thus proportional to the cumulative function of g evaluated at time t: t N0 NC (t) = N0 dt g(t ) = τ B 1 − e−t/τ B − τ A 1 − e−t/τ A = 0 τ B − τ A τ A e−t/τ A − τ B e−t/τ B = N0 1 + (4. or by trying an ansatz solution given by an arbitrary linear combinations of e−λ A t and e−λ B t functions.47) τB τB τA τB τ A N0 1 A B (t2 ) ≈ = 0. (4.43). and the activity of B is λ B λ A t1 N0 . the sample thus contains a population of B with size λ A t1 N0 . N B (t) = N0 − N A (t) − NC (t) = −t/τ A τAe −t/τ A − τ B e−t/τ B N0 τ B −t/τ B = N0 1 − e − 1+ = e − e−t/τ A . we can approximate the activities at the time t1 = 1. we first notice that the total number of nuclei N A + N B + NC must be conserved in time. d. From its definition.d.f. Given a random variable X with p. Sect. f (X ).g. f (x) = dt e−i t x φ X (t). −∞ 2π −∞ (4.51) 2π −∞ Hence. Problem 4.. Solution N Let’s define Z = i X i . in probability language.f.d.254 4 Statistics in Particle Physics Suggested readings For more details on the decay chain problem see e. the classical proof of the Central Limit theorem.g. We can apply Eq.d.49) The second equation shows how to revert back to the x space.f. Discussion This can be more elegantly proved by going into the Fourier-transform space. (4. (4.50) i=1 The method of characteristics is often used in probability to compute the p. by using the characteristic function of a p. of composite variables. . it also follows that the characteristic of the sum of N independent variables is given by the product of the N characteristic: N φ X i (t) = φ X i (t).11 of Ref. Notice that the mean and variance of Z could have been guessed a priori as a consequence of the linear dependence with respect to the N variables. 1.50) to give: ⎡ ⎤ ⎢ ⎛ ⎞ ⎛ ⎞ ⎥ ⎢ ⎥ N N 1 2 2 ⎢ N 1 N ⎥ φ Z (t) = φ X i (t) = exp i μi t − σi t = exp ⎢ ⎢i⎝ μi ⎠ t − ⎝ σi2 ⎠ t 2 ⎥ ⎥ 2 ⎢ 2 ⎥ i=1 i=1 ⎣ i ⎦ i μ σ2 +∞ 1 1 2 2 g(Z ) = dt e−i(Z −μ)t− 2 σ t = N (Z | μ.9 Prove that the sum of an arbitrary number of Gaussian random variables is still a Gaussian variable. or. see e. σ ). including discrete ones. (4. its characteristic φ X (t) is defined as +∞ +∞ 1 φ X (t) = E ei t X = d x ei t x f (x).f. g(Z ) is still a Gaussian p. [7]. the probability of observing k is given by k k e−(μ+ν) μm ν k−m Prob [k] = P(m | μ) P(k − m | ν) = = m=0 m=0 m! (k − m)! m k−m e−(μ+ν) (μ + ν)k k k! μ μ = 1− = k! m=0 m! (k − m)! μ + ν μ+ν =1 = P(k | μ + ν).g. respectively.53) e By making use of Eq.54) .1 Elements of Statistics 255 Problem 4.d.f. (4. (4.53). Problem 4. a signal process plus number of independent sources of background): the number of events counted in the bin of the observable under study is still Poisson-distributed with mean given by the sum of means of all contributing processes.10 Show that the sum of an arbitrary number of independent Poisson variables is still a Poisson variable.f. Under the assumption that m and n are independent. and their means by μ and ν.d. Solution It suffices to consider the sum of two variables: if their sum is still Poisson-distributed.11 Prove that a Poisson p. Discussion This fact has important implications in counting experiments where the channel under study receives contribution from several uncorrelated processes (e. k is a Poisson variable with mean μ + ν. Let’s denote these two variables by m and n. approaches a Gaussian p. the Poisson p. μ) + O(n − 2 ). becomes: 1 e n 1 μ P(n | μ) ≈ √ e−μ μn =√ e−(n−μ)+n ln n = 2π n n 2π n 1 n−μ (n−μ)2 −(n−μ)+n − n − 2n 2 +O( n12 ) √ = N (n | μ.4.d. 1 =√ e 2π n (4. (4. with the same mean and variance in the limit N → ∞.52) Hence. by recursive sum of pairs of variables we can prove that also the sum of an arbitrary number of such variables is still Poisson-distributed.f. We want to prove that k = m + n is Poisson-distributed with mean κ = μ + ν. Solution We can use Stirling formula to approximate the asymptotic behaviour of n!: √ n n n! ≈ 2π n . the .55) NB b Trun s2 For a given significance. since any Poisson process with large mean can be seen as the sum of an arbitrarily large number of Poisson variables with finite variance. The statistical significance for a counting experiment with large event counts is defined as the signal yields NS in units of the standard √ deviations of the background.. Bando n. such that n i 1. which for large event counts can be estimated by NB .12 An experiment selects signal events at a rate s and background events at a rate b.f. μ) for different values of μ. the data taking time scales with an increased signal rate more favourably than for the background: for a small relative increase s → (1 + εS )s. like the Pearson χ 2 test. see Problem 4. This proves very useful for statistical tests.10. μ) (dashed line) for different values of μ. Indeed. 1N/R3/SUB/2005 Problem 4.f. can be √ treated as gaussian variables with standard deviation σi = n i . 4.256 4 Statistics in Particle Physics Fig. Notice that we could have predicted the large sample behaviour of the Poisson p. √ Fig. the n! term is replaced by its analytical continuation Γ (x + 1) where we have Taylor-expanded the logarithm around μ/n ≈ 1.2 Comparison between P (n | μ) (solid √ line) and N (n | μ.2 compares P(n | μ) and N (n | μ. since the distribution of the test statistic becomes that of a χ 2 variable. 4. the bin contents n i of a histograms. Hence: NS s Trun b n≤√ =√ ⇒ Trun ≥ n 2 (4.d. Discussion Thanks to this result. For the Poisson p.d. by invoking the Central Limit theorem. the limiting distribution must be a Gaussian. For illustration. Solution Let’s denote the data taking time by Trun . Determine the data taking time needed to observe the signal with a statistical significance of n standard deviations. Let the sampling time be T . Solution Let’s divide the time window T into N equal subranges of duration τ = T /N .58) . the number of single-events in any of the subranges) given N trials. λ. if the flux is uniform. Bando n.13 Prove that the number of countings in a fixed time window T of a random process characterised by a uniform probability of occurrence per unit time λ is distributed according to the Poisson p.56) (N − n)! n! N N Taking the limit N → ∞ such that λT is finite. the interaction of each particle with the detector can be seen as a random process characterised by a uniform probability of occurrence per unit time. λ. Solution The number of events integrated over the sampling time T is distributed with a Poisson p. thus giving a Poisson distribution for the number of interactions in a fixed time window.57) n! n! with μ = λT = E [n] = Var [n]. (N − n + 1) +λT n lim Prob [n | N . 13153/2009 Problem 4.f. each with success probability λτ : n N! λT λT N −n Prob [n | N . (4. Indeed.d. The counting rate on a single channel is ν.e.1 Elements of Statistics 257 background is allowed to undergo a twofold relative increase without changing the significance for a given data taking time.13. For sufficiently large N . . T ] = e lim e N = N →∞ n! N →∞ Nn (λT )n e−λT μn e−μ = ≡ = P(n | μ).14 The surface of a given detector is irradiated by a unform flux. Determine the probability of having empty events and the probability of having single events. Problem 4. see Problem 4. T ] = 1− (4.d. The probability of observing n outcomes can be computed by using the Binomial law for n positive events (i. so that the probability of single-event in a given subrange will be given by λτ 1. we get: (λT )n −λT N (N − 1) . . Hence: Prob [n = 0] = P(0 | νT ) = e−νT Prob [n = 1] = P(1 | νT ) = e−νT (νT ) (4.f. while the probability of double-event (λτ )2 can be neglected. τ will become much smaller than λ−1 .4.. Vw = 0. (4. the use of event weights allows to modify the conditional distributions compared to the model used for the generatorion. the event yield Y and its variance σY2 are given by: N N 2 Y = i=1 wi σY i=1 wi N ⇒ = N . as we have already seen for the analogous problem of measuring the dead time of a paralyzable system. . i = 1. when the MC events consists of one or more correlated variables. like the simulation of scattering events at NLO accuracy. . the flux ν can be measured by simply inverting the first of Eq. the relative uncertainty is always equal or larger than for unweighted events. or statistical tools like sPlot [8].58).258 4 Statistics in Particle Physics By measuring the zero-event probability.d. 13153/2009 Problem 4. i. the events weights can be negative. respectively. .59) σY2 = i=1 wi2 Y i=1 wi If N 1.15 A Monte Carlo program simulates a scattering process. For example.e. The solution is however twofold. Bando n. As one can see. that dw f (w) = 1). although observables should be always associated with positive cross sections.58). (4. the Law of Large Numbers allows us to approximate the values of Y and σY2 by using the p. Solution The total event yield and the equivalent integrated luminosity of the sample are given by . . of which N+ events have weight w > 0 while N− of events with weight −w. N . The use of weighted MC events is pretty common in HEP analyses. Determine the equivalent integrated luminosity of this sample and the relative uncertainty on the simulated event yield.38. see Problem 3.f. a sample of N 1 events is generated. of the weights w. We can therefore write: 21 Y ≈ N E [w] = N μw σY 1 Vw ⇒ ≈√ 1+ 2 . Discussion When a sample consists of N weighted events each with weight wi . (4. f (w) (which we assume to be well defined.60) σY2 ≈ N E w2 = N (Vw + μ2w ) Y N μw where μw and Vw are the mean and variance of the weights w. Let the cross section of the simulated process be σ . (4. Also the single-event probability can be used to derive ν through the second of Eq. After running for some time.e. the equality being realised when wi ≡ w for any i. In some cases. i. what is the probability of observing at least one lepton if the branching ratio b → X ν is equal to 20%? Solution The b decays are independent from each other. (4. 1N/R3/SUB/2005 Problem 4. hence: Prob [n ≥ 1] = 1 − B(0 | N = 4. Besides the prompt lepton production b → X ν mentioned by the exercise. = = . Another distinctive feature of b-initiated jets is the presence of a charged lepton. ε = 0. (4. we can determine the relative uncertainty on the event yield: σY 1 1 =√ .1 Elements of Statistics 259 Y = wi = w (N+ − N− ) = w N (1 − 2 f ). the presence of a fraction f of negative-weight events increases the relative uncertainty of the simulated sample by a factor of (1 − 2 f )−1 compared to the Poisson expectation from unweighted events.64) Discussion Thanks to a lifetime of O(10−12 s).4. The number n of leptonic decays is therefore distributed according to a Binomial law with single-event probability ε = 0. Bando n.63) Y N 1−2f Therefore. rejection power against light jets as large as (1 − β)−1 ∼ 100 can be obtained for efficiencies in excess of 1 − α ∼ 50% (we have used the symbols α and β for the errors of first and second kind. (4. From Eq. i Y w N (1 − 2 f ) Lint.16 Given an event containing four b quarks.2) = 1 − (1 − ε)4 = 59%. (4. The mean and RMS of the weight distributions can be readily computed: μw = −w f + w (1 − f ) = w (1 − 2 f ) (4. respectively). B hadrons generated from the hadronisation of b quarks of sufficiently large momentum give rise to distinctive signatures that can help discriminate them from jets initiated by lighter quarks or gluons. an additional source of charged leptons arises from the semileptonic decay of charmed mesons produced in the decay chain of the initial . See Problem 2.62) Vw = w2 f + w2 (1 − f ) − w2 (1 − 2 f )2 = w2 1 − (1 − 2 f )2 .61) σ σ where we have introduced the fraction of negative events f = N− /N . With the use of modern vertex detectors.2.46 for more details.60). while the impurity B absorbes photons by emitting electrons.2706 + 0. 13705/2010 Problem 4.66) Notice that the result obtained by using a Binomial p. 11] Bando n. The pion i mis-identification probability f 1.d.17 A crystal contains two kinds of impurities A and B in equal amount.e given N = 200 incident photons is distributed according to a Binomial law with single-event probability ε. . agrees with the Poissonian estimate to better than 10−5 . What is the probability that at least three electrons are emitted? Solution Given an input photon.f. (4.1353 + 0. Suggested readings For more details on recent b tagging techniques at hadron colliders. denoted by PP.18 An experiment measures events containing a pair of particles reconstructed as electrons. exactly one (PF) and exactly zero (FF) reconstructed electrons are also measured. the branching ratio for producing at least one charged lepton is large as 35% [9]. we can approximate this distribution with a Poissonian of mean ε N . Since ε N = 2 is finite. the second consists of two mis-identified pions. Overall. containing. The impurity A absorbs photons without emitting electrons.2 for both reconstructed objects and for each event i is known from simulation. Problem 4.e. The collected data sample. the reader is addressed to Refs. ≥ 3 = 1 − P(i | εN ) = i=0 = 1 − (0. Let the dimension of the crystal be such that all entering photons get absorbed.01 (4.2707) ≈ 32%. Assume a prefect identification probability for real electrons.260 4 Statistics in Particle Physics B hadrons. The probability of emitting at least three electrons is therefore given by: 2 Prob n p. The absorbtion cross section of A is 99 times larger than the cross section for B. Two independent data samples. the probability that it gets absorbed by the impurity B is σB ε= = 0. Estimate the total background yield in the PP sample. is affected by two sources of background: the first consists of one real electron and one pion mis- identified as an electron. respectively. and let us assume that 200 photons impinge the crystal. [10.65) σA + σB The number of emitted electrons n p. 67) ⎪ ⎪ ⎪ ⎪ Prob [PF | π e] = 1 − f ai ⎪ ⎪ ⎩Prob [PP | π e] = f i a where the index a = 1.1 Elements of Statistics 261 Solution Under the assumption that the electron identification probability is perfect.70) The total background yield in the PP region is therefore given by: j fa f 1i f 2i ππ π e+π π ππ NPP | FF + NPP | PF − NPP | PF | FF = − .69). In particular. The probabilities for all possible combinations of wrong and correct identification of a background events are: ⎧ ⎪ ⎪ Prob [FF | π π ] = (1 − f 1i )(1 − f 2i ) ⎪ ⎪ ⎪ ⎪ ⎨Prob [PF | π π ] = f 1 (1 − f 2 ) + (1 − f 1 ) f 2 i i i i Prob [PP | π π ] = f 1 f 2 i i (4.69) j ∈ PF 1 − fa where the sum runs over the PF sample.68) and (4. the quantity fa j π e+π π NPP | PF = j . (4. but it underestimates the contribution from two-pion events.and two-pion events. (4. (4.and two-pion events.68) i ∈ FF (1 − f 1i )(1 − f 2i ) Since the PF sample contains both one. 2 refers to the particle not-reconstructed as an electron. the PF sample is contaminated by both one.71) j ∈ PF 1 − fa j i ∈ FF 1 − f 1i 1 − f 2i ππ πe . The contribution of two-pion events to the sum in Eq. the contribution from two-pion events would be overestimated because some events get double-counted. (4. The number of two-pion events in the PP sample can be therefore estimated as f 1i f 2i ππ NPP | FF = . whereas the FF sample contains only two-pion events. We can express the number of background events in the PP region per event falling in the FF or PF region by taking the ratio between the respective probabilities.69) can be however estimated from the FF sample to be f 1i f 2i f 2i f 1i f 1i f 2i ππ NPP | PF | FF = + = 2 1 − f i 1 − f i 1 − f 2i 1 − f 1i i ∈ FF 1 2 1 − f 1i 1 − f 2i i ∈ FF PF | FF PP | PF PF | FF PP | PF (4. by summing together Eqs. (4.4. provides the correct background yield in the PP sample from one-pion events. Equation (4. the gradient of the function at the mean value x0 . The linearised error propagation method is based on a first-order expansion of the function h around the experimental point x̄: z = h(x) = h(x̄) + ∂i h(x̄) (xi − x̄i ) + . Derive an expression for σz2 by assuming that h can be expanded to first order around (x0 .72) i Given the linear approximation of Eq. . −x0 /y0 ).2 Error Propagation Error propagation refers to the estimation of the variance of a random variable z which is a known funtion h of an array of random variables x. y0 ). Problems Problem 4. Solution For the case h = a x + b y.72). Var [z] from Eq. b). y0 is given by ∇h 0 = y0−1 (1. Then.72) is exact. . Equation (4. the linear expansion of Eq.73) then gives: . whose covariance matrix V is assumed to be known. y0 ) and covariance matrix V .262 4 Statistics in Particle Physics Notice that we made the assumption that the mis-identification probabilities factorise. y) is a pair of random variables with mean value (x0 . (4. This exercise illustrates an example of a data-driven technique for estimating the background contamination in events with mis-identified particles. y).73) then gives: σx2 ρσx σ y a σz2 = (a b) = a 2 σx2 + b2 σ y2 + 2 a b ρ σx σ y (4. the variance of z can be determined as: Var [z] ≈ ∂i h(x̄) Vi j ∂ j h(x̄) (4. For the case h = x/y.19 Consider a random variable z = h(x. The gradient of h is constant and given by ∇h 0 = (a. (4.73) ij Since V is a symmetric positive-definite matrix. specialise the result to two special cases: h = a x + b y and h = x/y. we get the result σz = (a σx ) ⊕ (b σ y ). 4.(4. where (x.73) is always positive. (4.74) ρσx σ y σ y2 b For the special case ρ = 0. For the special case ρ = 0.75) by applying the propagation of error to ln z = ln x − ln y and noticing that d ln z = dz/z.75) x0 y0 x0 y0 Notice that this result could have been obtained directly from Eq. the selection efficiency is estimated to be ε = 37%.3 × 10−4 . 12] for a more detailed discussion. For example.2 Error Propagation 263 1 σx2 ρσx σ y 1 σz2 = (1 − x /y ) = y02 0 0 ρσx σ y σ y2 −x0 /y0 2 1 x0 2 x0 = 2 σx − 2 ρσx σ y + σ y 2 2 y0 y0 y0 2 2 σ x σ y σx σy = z 02 + − 2ρ (4. selects N = 100 candidate events. the measured cross section is N − NB σ = . (4. (4.4.76) ε Lint where N B is the estimated background yield in the data sample and ε is the total efficiency (including the acceptance) of the experiment to the signal. Suggested readings There is a large number of textbooks dedicated to statistical methods for particle physics.20 After having collected an integrated luminosity Lint = 10 fb−1 .69 · (0. Bando n.0596 × 2) .76). (4. the related statistical uncertainty. (4. the branching ratio for B 0 → J/Ψ K S can be measured as: σ+ − π + π − N BR J/Ψ K 0 = = = σ BRπ π BR B0 + − + − ε Lint σB BRπ + π − BR+ − 0 100 = = 3. and the systematic uncertainty due to having simulated too few events? Assume σB0 ≈ 1 nb.77) 0. By using NMC = 1000 simulated events. the reader is addressed to Refs. Discussion Given a data sample of N scattering events corresponding to an integrated luminosity Lint . 1N/R3/SUB/2005 Problem 4. we get the result σz /z = (σx /x) ⊕ (σ y /y). Solution By using Eq. an analysis of B 0 → J/Ψ K S0 . [5. The latter is usually estimated by using a Monte Carlo simulation of the signal process. with J/Ψ → + − and K S → π + π − .37 · 10 fb−1 · 1 nb · 0. What are the measured BR. 78) BR J/Ψ K 0 N ε where the first term is the pure statistical uncertainty. CTF at Fermilab and CDHS at CERN. What is the combined result? Are the two measurements consistent with each other? Discussion Point-estimation from a combination of measurements with covariance matrix V can be performed by using the least squares method. More precisely. while the second represent a systematic uncertainty.02 and CDHS: 0. Bando n. See Problem 5. a channel which can be produced by both neutral B states. (4. The least squares estimator of the mean of an ensemble of measurements is the solution of the equation . the propagation of a physical neutral B meson initially produced as a B0 or B̄0 features time oscillations due to B0 − B̄0 mixing. (4.75): / 2 2 δBR J/Ψ K 0 1 δε ≈ √ + . which coincides with the ML estimator for normally distributed variables.295 ± 0.80) BR J/Ψ K 0 100 1000 · 0.7 × 10−4 [9]) plays an important role in B physics as a tool to measure one of the angles of the unitarity triangle.01.21 The ratio between neutral and charged current interactions from neutrinos on nucleus are measured by two independent experiments. The amplitude of such oscillations as measured via the J/Ψ K S0 decay.27 ± 0.43 for more details on this subject.79) ε NMC ε 2 2 NMC ε Putting everything together.37 The extra √ uncertainty due to the limited size of MC simulated events is therefore 11% − 1/ 100 ≈ 1%. (4.264 4 Statistics in Particle Physics where we have used the PDG values for BRπ + π − and BR+ − [9]. (4. Suggested readings The decay channel B 0 → J/Ψ K S0 (branching ratio 8. to be: CTF: 0. The latter be evaluated by using the expectation from a Binomial distribution: 2 δε NMC ε (1 − ε) (1 − ε) = = .37 ≈ + ≈ 11%. Under the assumption that the uncertainty on the integrated luminosity and on the inclusive cross section are negligible. the relative uncertainty on the branching ratio can be estimated from Eq. we thus have: 0 δBR J/Ψ K 0 1 1 − 0. provides a direct measurement of the C P-violating parameter sin 2β. 13705/2010 Problem 4. 290)2 t= = + = 1.81) ∂μ 2 ⎧ ⎨μ̂ = AT V −1 A−1 AT V −1 x = ij (V −1 )i j xi ⇒ (V −1 )i j ⎩σ −2 = AT V −1 A = V −1 ij (4. 1)T μ. V −1 is diagonal and the formula simplifies to the well-known result: xi i σ2 1 μ̂ = 1i . 13153/2009 Problem 4.2 Error Propagation 265 ∂ 1 0= (x − Aμ)T V −1 (x − Aμ) (4. Determine the precision on the mass measurement M.27 − 0. . .83): μ̂ ± σμ̂ = 0. 7 of Ref. For the special case of uncorrelated measurements.4. 1. From tabulated tables.22 An invariant mass distribution shows a Gaussian peak centered at a mass M. .84) The compatibility between the two measurements can be quantified by using the χ 2 test-statistic t defined as: xi − μ̂ 2 (0.85) i σi2 (0. (4. [5].25. the test statistic t is distributed like a χ 2 with n = 1 degree of freedom. Bando n. and has a width at half maximum equal to D.290)2 (0.290 ± 0.82) μ̂ ij ij where we have defined E [x] = Aμ = (1.01)2 If the measurements are nornally distributed. σμ̂2 = 1 . (4.02) 2 (0. the reader is addressed to Chap. . (4. . The two measurements are therefore well compatible with each other.83) i σi2 i σi2 Solution The combined measurement can be obtained by using the least square estimator of Eq. Suggested readings For a more formal treatment of methods for point-estimation. one can find that the p-value for this experiment is 26%. (4.009.295 − 0. it contains N events. 87) N Since σx̄ /σx 1. From this sample.0 (4.103).06 (68% CL) is obtained.266 4 Statistics in Particle Physics Solution For a Gaussian distributed variable. The latter is related to the standard deviation of the single-measurement. we can assume the mean value μ to be x̄. the relation between its standard deviation and FWHM is given by Eq. e. the likelihood of the two observations can be parametrised in terms of ρ and one of the two mean counting rates. a mean value of x̄ = 34.0 standard deviations from its mean is p = α/2 = 16%.g. the countings are NA = 200 and NB = 1000. This can be easily verified. see Problem 4.9. as: . What is the statistical uncertainty on the ratio R = NA /NB ? In case the observation time were reduced by a factor of 250. In a given time interval.35 N Bando n. by σx √ σx̄ = √ ⇒ σx = 0.86) N 2. how would you correct for it? Solution The variable R thus defined coincides with the ML estimator for the parameter ρ = μA /μB . What is the probability that a subsequent measurement results in a value x ≥ 37? Solution The variable x̄ is Gaussian distributed. Indeed.24 A radioactive source emits two kinds of uncorrelated radiation A and B.23 A sample of 2500 measurements of a quantity x are normally distributed. μB . This radiation is observed through a counter which is able to distinguish A from B. 18211/2016 Problem 4. hence its two-sided confidence interval at CL = 1 − α = 68% has a width of 2 σ .00 ± 0. (4. The combination of N independent measurementes gives an uncertainty on the mean: σ D σM = √ = √ . The probability p that a Gaussian variable of mean μ = x̄ fluctuates more than (37 − x̄)/σx = 1. Bando n.06 · 2500 = 3. do you think that the new ratio r = n A /n B would represent an unbiased estimator of the true ratio between the two activities? In case of negative answer. 13705/2010 Problem 4. σx . (2. For just one experiment. that follow either the frequentist or Bayesan approach.73). (4. and that the relative variations of the countings of order σA /A and σB /B around these values are small. the parameter μ is said to be contained inside a confidence interval Iμ (X) at a confidence level CL = 1 − α if: Prob μ ∈ Iμ (X) = α. respectively. (4.3 Confidence Intervals In the literature.90) R NA NB NA NB see Eq.91) irrespectively of the true (and possibly unknown) value of μ. this is only possible inference on the true value of the parameter μ that an experimental can provide. it is possible to estimate the variance of R by standard error propagation. In particular. NB . one can find two approaches to the definition of confidence levels. the mean counting rates for A and B will be of order 1 and 4.2 Error Propagation 267 e−μB (1+ρ) ρ NA μBNA +NB L(NA . [5]. The quantity α. However. which gives the result: / 2 2 / σR σA σB 1 1 = + = + = 7.7%. This assumption breaks up if at least one of the two countings is small. which reduces the uncertainty √ on the estimator of the mean number of decays n A and n B by a factor of 1/ N : for sufficiently large values of N . 7 of Ref. this is in general not the case. (4. if the observation time gets reduced by a factor of 250. In almost all the cases. 4.89) The asymptotic properties of the ML estimator [5] guarantee that R̂ is an unbiased estimator. Only the former is considered here. NB ) can be expanded around the mean values μA and μB .88) NA ! NB ! Setting the derivatives of L with respect to both parameters to zero. thus invalidating the estimate of Eq. which is coventional and should be chosen a priori. (4. Chap. This result assumes that the function R(NA .g. see e. According to the frequentist paradigm. . since the variance of the individual rates is known. one gets the expected result for the ML estimators: R̂ = NA /NB .90). the ratio r = n A /n B becomes again well behaved.4. if a large number of observations are made. (4. μ̂B = NB . ρ. μB ) = . (4. A possible way to overcome this problem is to repeat the measurements N 1 times. The variance of the estimator r will be larger. 9994. For a more refined discussion on the subject. the best estimate for the efficiency is ε̂ = 1 and the one-sided confidence interval at a given confidence level CL = 1 − α is given by ' 1 ( [0. ε) = ε M (1 − ε) N −M .e.0] N = 5000 .25 with M = N . [14]. 4. Suggested readings The concept of confidence interval is comprehensively illustrated in Ref.e.25 What is the correct statistical approach to assess the efficiency of a detector and of its related uncertainty? Solution The efficiency of a detector is defined as the number of output signals M per input events N .92) M! (N − M)! A frequentist confidence interval Iε at the confidence level CL = 1 − α can be built by using the Neyman construction [13]. 1. i. where ε̂ is the estimator of the unknown detector efficiency ε which is assumed to be independent of the time of arrival of the signals or their rate. Problems Bando n. (4. i.9418. for symmetric and double-sided intervals.e. i. The number of output signals for a fixed value of N is distributed according to a binomial law: N! B(M | N .0] N = 50 Iε = α N .93) [0. 1 = (4.268 4 Statistics in Particle Physics quantifies the amount of error of the first kind associated with the measurement. Solution Referring to Problem 4. Problem 4. for the cases N = 50 and N = 5000. as the interval of ε values for which the p-value corresponding to the measured value M is smaller than α (for one sided- intervals). 1. or α/2. [5. 12]. the probability that the true value does not lie within the confidence interval. 13153/2009 Problem 4. ε̂ = M/N . the reader is addressed to dedicated textbooks like Refs.26 Determine the one-sided 95% CI on the efficiency ε of a detector that reconstructs each of the N pulses sent in.3. An example of Neyman construction for a generic confidence interval is shown in Fig. Problem 4. (4. denoted by μul .95) n! . i. the likelihood of the data is given by exp [−μ] μn L(n | μ) = . Compare the 95% CL upper limit on μ as a function of the experimental outcome N obtained by using the Neyman construction with the confidence interval at the same confidence level obtained by using Wilks’ approximation of the negative log-likelihood function. For each value of μ.27 A counting experiment measures the number of outcomes of a process of unknown yield μ. The confidence interval [μ1 . x2 ] such that Prob [x ∈ [x1 .3 Confidence Intervals 269 Fig. one draws a horizontal acceptance interval [x1 .e. (4. such intervals contain the true value with a probability CL = 95% regardless of the true. Upon performing an experiment to measure x and obtaining the value x0 . μ2 ] is the union of all values of μ for which the corresponding acceptance interval is intercepted by the vertical line (from Ref. unknown value of ε.4.: N exp [−μul ] μnul α = Prob[n ≤ N ] = . Solution The upper limit at 1 − α = 95% CL.94) n=0 n! Under the assumption that only one process contributes. x2 ]|μ] = α. can be obtained by using the Neyman construction of confidence levels. 4.3 A generic confidence interval construction. one draws the dashed vertical line through x0 . [14]) By construction. the upper limit at 95% CL would be given by the root of the solution q(μul ) = 2.3 62.31 16 5 10.30 5.4 The negative log-likelihood ratio q(μ) from a Poisson process with mean μ for different values of the outcome N .4 0. one has: L(N | μ) μ q(μ) = −2 ln = −2 (N − μ) − 2 N log .75 3. Already for N = 10. 4. For illustration.3 3 50 63.1 5 20 29.3.1 The upper limits at 95% CL on the mean μ of a Poisson process as a function of the outcome N . the functions q(μ) are compared with their parabolic approximation from which one obtains the intuitive result for the ML estimator: μ̂ = N .1 reports the upper limits at 95% CL obtained by using the two alternative approaches. The values have been obtained by using the program reported in Appendix 4. obtained by using the classical Neyman construction (second column) and by using the asymptotic approximation of the negative log-likelihood ratio (third column) N Neyman Wilks Rel.. see Problem 4.96) L(N | μ̂) N If q(μ) were distributed like a χ 2 with one d. 13707/2010 .65 23 2 6. For illustration. (%) 1 4.11. Bando n.5 9. Indeed.1 28.6 1 100 118.70.f. Table 4. Fig.4 shows the function q(μ) for three particular values of N . for large values of N the Poisson distribution approaches a Gaussian.270 4 Statistics in Particle Physics Table 4.6 Fig.o. diff. and compares them to the respective parabolic approximation. which satisfies Wilks’ theorem by construction. 4.1 117. Wilks’ theorem states that the negative log-likelihood ratio q is asymptotically distributed like a χ 2 . For the experiment under consideration.0 16. (4.6 8 10 17. the asymptotic approximation is better than 5%. the number of detected particles is distributed according to a binomial p. import math # grid points for numerical scan nmax=1000 # the\index{Poisson density} Poisson p. the binomial distribution approaches a Poissonian of mean εN .d. def pois(n=1. For large values of N such that ε N is finite. We can therefore use the interval with boundaries ε N ± of N ε (1 − ε) as an estimator of the 68% quantile for ε. If the efficiency of conversion of each photon into photoelectrons is εQ = 20%.factorial(n) def get_CL_neyman(n=1. for a Poisson process of unknown mean value μ. n)/math. A numerical scan is performed in order to find the appropriate isocontour levels.982. (4. see Eq. CL=0. (4.4. what is the mean number of particles (and relative uncertainty) that are undetected? Solution The number of p.e.e. is therefore given by: ε = Prob n p.sqrt(n)+5 # a ’reasonable’ upper limit .f. see Problem 4. given N = 1000 incoming particles.e. with efficiency ε given by Eq. (4.97) Given N incoming particles.98) Appendix 1 This computer program computes the frequentist 95% CL upper limit (Neyman construction) and the approximate 95% CL upper limit from the asymptotic properties of the negative log-likelihood ratio (Wilks’ theorem). which is the condition for the particle to be detected.92).3 Confidence Intervals 271 Problem 4.95): alpha = 1-CL muL = n muH = n+3*math.exp(-mu)*math. (4.27.pow(mu. N̄ = N ε ± N ε (1 − ε) = 982 ± 4 at 68% CL. i..f. The probability of emitting at least one p. produced per incident particle is distributed like a Poissonian with mean εQ n γ .d.e. on average n γ = 20 photons are produced per incident particle.. see Problem 4.mu=1): return math. which approaches a Gaussian for large values √ ε N .28 In a Cherenkov counter.97) and variance N ε (1 − ε). > 0 = 1 − P(0 | εQ n γ ) = 1 − e−εQ n γ = 0.11. Ann. Pivk. Lond. Meth. Math. https://doi. Am. A 236. G.1103/ PhysRevD. W. Chin. H. P04013 (2013).705): muL = n muH = n+3.org/10. C 40. https://doi.R. https://doi. Stat.. 60–62. Cousin. Berlin.G. 54(3). ney[0]. R. (World Scientific. M. S.272 4 Statistics in Particle Physics step = (muH-muL)/nmax for mu_step in range(nmax): mu = muL+step*mu_step prob = 0.3873 .sqrt(n)+5. Protter. 2004) 2. 333–380 (1937) 14. Patrignani et al.org/10. chi2=2.57.S. C. J. Berlin.2*n*math..106 9. Wilks.org/10. Neyman. C. 50. 2nd edn. Oxford.2. 1993) 8.705) print n. C 71. Trans.08.S. https://doi. CERN/MPS/DL 69–12 (1969) 7. 356–369 (2005). Hereward.1088/1748-0221/8/04/ P04013 12. 2nd edn. J.org/10.1214/aoms/1177732360 4. G. chi2=2. Hackensack.D. 426–482 (1943). D 57.2005. 9. James.org/10. wil[0]. (Particle Data Group). mu=mu) if prob<alpha: break return [mu. 10.R. Probability Essentials. ATLAS Collaboration. Cowan. P. CL=0. prob] def get_CL_wilks(n=1. 2006) 6. https://doi. 3873–3889 (1998). G. 5. 100001 (2016) 10. 1554 (2011).M. Phys. Techniques for Nuclear and Particle Physics Experiments. Statistical Methods in Experimental Physics. Trans. Wald.95) wil = get_CL_wilks(n=n. (Springer. Royal Soc. Rev. Eur. P04008 (2016).nll] # run! for n in [1. 20. https://doi. A 555. Soc.. (wil[0]-ney[0])/ney[0] References 1. F. 2nd edn. Feldman. Phys. 1016/j. 100]: ney = get_CL_neyman(n=n. Le Diberder. Nucl. Phys.org/10.1140/epjc/s10052-011- 1554-0 5.2307/1990256 3. Jacod. Instrum. Statistical Data Analysis (Oxford University Press. Collaboration. step = (muH-muL)/nmax for mu_step in range(nmax): mu = muL+step*mu_step nll = -2*(n-mu) . 1998) 13.nima. J. https://doi. Cowan et al. A. Phil. Leo.*math. JINST 11. Ser.0 for x in range(n+1): prob += pois(n=x.log(mu/n) if nll > chi2: break return [mu. F.J.1088/1748-0221/11/04/ P04008 11.org/10. Math. JINST 8. (Springer. After introducing the concept of symmetry. in quantum-mechanical language. and ε 2 . The other independent vector with a defined 1 ↔ 2 symmetry is the relative momentum k = k1 − k2 . Solution The quantum-mechanical amplitude for the decay X → γ γ . We denote these vectors in the centre-of-mass by η. They are mostly concerned with the phenomenological predictions of Standard Model of particle interactions. the invariance of the Lagrangian L under a group of transformations gives rise to a conserved current. By Noether’s theorem.1 Conservation Laws The principles of quantum mechanics and the invariance of physics under transformations of the Poincaré group impose stringent constraints on the mathematical structure of a consistent theory. with J X = 1 must be a Lorentz-scalar and linear in the polarisation vectors of the particle X and of the two photons. UNITEXT for Physics. 5. Furthermore. Nature preferentially realises a number of additional symmetries which. for a massless spin-1 particle. Classifications of states in terms of their conserved quantum numbers proves of the greatest help to give a rationale to the experimental observations. to a conserved quantum number. which is of cardinal importance in the formulation of a good theory. ε1 . Bianchini. The two last sections are dedicated to flavour physics and to the Higgs boson.org/10. https://doi. provide further constraints. the Klein–Gordan equation © Springer International Publishing AG 2018 273 L. the proposed exercises will focus on the physics of electroweak and strong interactions. or.1007/978-3-319-70494-4_5 . Problems Problem 5. Selected Exercises in Particle and Nuclear Physics.Chapter 5 Subnuclear Physics Abstract The problems collected in the fifth chapter deal with particle physics from a more theoretical perspective.1 Show that rotation-invariance and Bose–Einstein symmetry implies that a spin-1 particle cannot decay into two photons. once introduced in the theory. ρ 0 . and fourth are odd under 1 ↔ 2 exchange.2) Since no amplitude consistent with the selection rules can be constructed. (ε1 × ε 2 ) · (η × k). (5. like φ 0 .3) 2 2 2 2 are allowed in electric dipole (E1) or magnetic dipole (B1) approximation? Discussion The electric and magetic dipole approximations consist in assuming an interaction Hamiltonian of the form .1) A term like (ε1 × k) · (ε2 × k)(η · k) is equivalent to the first of Eq. (5. as well as the radiative decay Z 0 → γ γ . J/Ψ .2 Which of the following electromagnetic transitions: 1+ 1− 3+ 1− → 0+ → 0− → 2+ → 1+ 1+ → 0+ . known as Yang’s theorem. to a pair of photons is forbidden. whereas Bose–Einstein symmetry requires the amplitude to be symmetric under photon label exchange. second. Discussion This result. see his original paper [1]. Bando n. 18211/2016 Problem 5. ω0 .274 5 Subnuclear Physics ∂ν2 Aμ = 0 implies ε · k = 0. has been advocated at the time when the first evidence of the 125 GeV Higgs boson was being gathered in the diphoton channel H → γ γ as a proof that the particle had to be a boson. (5. and furthermore J = 1. The third combination is identically zero since: j j (ε 1 × ε 2 ) · (η × k) = εi jk ε1 ε2k εilm ηl km = (δlj δkm − δkl δ mj )ε1 ε2k ηl km = = (ε 1 · η)(ε 2 · k) − (ε 2 · η)(ε 1 · k) = 0.1) modulo a function of k2 . the only combinations linear in the polarisation vectors and with a definite symmetry with respect to the two photon indices are: (ε1 · ε 2 )(η · k) (ε 1 × ε 2 ) · η. (ε 1 × ε 2 )(η · k) (5. the decay must be forbidden. since it decays to a pair of spin-1 particles. This theorem also implies that the decay of the neutral pseudo-vector mesons. Out of these four vectors. Suggested Readings This result was first formalised by Yang in 1950. Υ . The first. defined by the relation: | j1 m 1 | j2 m 2 = Cmj11 mj22 mJ 1 +m 2 | j1 j2 | J m 1 + m 2 . (5. (5. (5. [2] for a compendium of formulas. (5.6) J For the case j1 = 0. Δm J = 0. 1 and involve a change in parity: they are therefore compatible only with a E1 transition. (5. 18211/2016 . According to the Wigner–Eckart theorem.8) While r is a proper vector.5) reduced element j j J where Cm11 m22 M are the Clebsh–Gordan coefficients. and indeed: 1 Cmj11 j2 0 m2 0 = (−1) j1 −m 1 √ δ j j δm m . See Appendix B of Ref. j2 = 1.4) 2m Under rotations.7) 2 j1 + 1 1 2 1 2 Hence. the matrix elements between states with J = 0 vanish: the transition 0 → 0 cannot occur through dipole operators. Suggested Readings The Wigner–Eckart theorem is discussed in several textbooks of classical Quantum Mechanics. Solution The first and third transitions have ΔJ = 0. the matrix elements of a tensor operator Tqk between states of definite total momentum take the form: J m|Tqk |J m = CmJ qk mJ J ||T k ||J . The second one is forbidden for both transitions: it has to proceed through scalar or tensor operators. Bando n. ±1.1 Conservation Laws 275 e HE1 = e r · E. whereas B1 transitions do not. The fourth and fifth have ΔJ = 1 and do not change parity: they must be B1 transitions. L and S are pseudovectors. both r and L transform as vectors. The other selection rule for E1 and B1 transitions between states of definite total angular momentum is: ΔJ = 0. HB1 = − (L + gS) · B. the Clebsh–Gordan coefficients vanish because the tensor product of two states with total angular momentum zero does not overlap with states of angular momentum one. for vector operators (k = 1).5. This implies an important selection rule: E1 transitions change the parity of the state. ±1. like GUT theories where the quark and lepton doublets are embedded into multiplets of higher dimensions.22 MeV. the reader is addressed to Refs. Deuterium is a mass eigenstate of the strong Hamiltonian. ranging between 1033 and 1034 years at 90% CL. see Problem 2. Indeed. K − ) have been provided by the water-Cherenkov IMB-3 experiment at the Fairport Salt Mine [3]. depending on the channel under study [4]. The first state is antisymmetric under exchange p ↔ n. the second is symmetric.4 What is the total wavefunction of deuterium? The binding energy of deuterium is B D = −2. μ+ ) and a charged meson (π − . the lepton number L and baryon number B are not individually conserved. Discussion Lepton-number is an accidental symmetry of the SM Lagrangian. giving lower bounds on the lifetimes of order 1032 years at 90% CL. In several BSM extensions.3 Why is the decay n → p e− forbidded even without postulating lepton-number conservation? Solution A putative n → p e− decay would violate angular momentum conservation.29. ρ − . Suggested Readings For more details on the limits on the nucleon lifetimes. hence it must have a definite value of . one notices that a proton-neutron state can either have I = 0 (Ψ ∝ | pn − |np) or I = 1 (Ψ ∝ | pn + |np). i. The most stringent limits on lepton-number violating decays of neutrons into an antilepton (e+ . Solution The determination of the total wavefunction of deuterium (D) relies on symmetry arguments and experimental observations.276 5 Subnuclear Physics Problem 5.e. while the initial state has J = 1/2. In this case. see also Problem 5. First. the lepton number symmetry would be broken by the Majorana mass term. Another reaction which would bring to an evidence of lepton-number violation is the neutrinoless-double β decay (0νββ). a state made of two spin-1/2 particles can only have integer total angular momentum. The most stringent limits on lepton-violating proton decays come from the Super-Kamiokande experiment. bringing to stringent selection rules on the decays of hadrons. Problem 5. [3. invariance of the Hamiltonian under rotations.47. 4]. Argue why this implies that the mass of the neutron has to be fine-tuned compared to the proton and electron mass at the 10−3 level to allow for star nucleosynthesis. This reaction can take place provided that: 2 m p > m D + me = m p + mn − BD + me ⇔ (m n − m p + m e ) < 2. and −1 for S = 1.8 MeV compared to its nominal value. 2 Coming to the fine-tune question. S = 1: the symmetry is +1. Write down the . the fusion reaction would be endothermic. In this case.8 MeV.22 MeV. Since the deuterium wavefunction needs to be overall antisymmetric under p ↔ n (Fermi–Dirac statistics). One can also notice that a neutron mass smaller than 0. which is not possible since the strong dynamics conserves parity. or a pure 1 P1 state: mixing between the three would result in a state of undefined parity. hence: | pn − |np 3 ΨD = √ × α| S1 + 1 − α 2 |3 D1 . This inequality is indeed satisfied. L = 1. The reaction actually involves a neutral atom.4 MeV heavier than what it is. A pure L = 1 state is however discarded by the observation that the nuclear magnetic moment of deuterium agrees within ≈2% with the sum of the proton and nuclear magnetic moments. with dramatic implications for nucleosynthesis. Finally. We can therefore conclude that the neutron mass is fine-tuned for nuclear synthesis to better than O(1 MeV)/m n ≈ 0. since m n − m p + m e ≈ (1. which are not observed neither as stable particles. S = 0. Hence.1 Conservation Laws 277 I . 3. since the first exothermic reaction towards helium synthesis is: p p → D e+ νe . hence only the former is acceptable. Therefore.1%. For a pair of spin-1/2 particles.3 + 0. I = 0. deuterium can be either a superimposition of 3 S1 and 3 D1 . since P = (−1) L . the rest of the wavefunction needs to be symmetric. we notice that deuterium plays a key role in the nuclear synthesis occurring in stars. nor as resonances in nucleon-nucleon scattering. L = 0. Let us assume I = 1. which is the expectation from a S = 1 state (aligned spins). L = 2. The observation that J (D) = 1 implies three possible values for the angular momentum L: 1. one should expect two more degenerate states.5) MeV = 1. if only the neutron mass had been only 0. Problem 5. the symmetry is (−1) L (−1) S+1 . | pp and |nn.5 Consider a heavy nucleus ZA X decaying to a lighter nucleus ZA Y plus additional particles. 2. 1: the symmetry is +1 for S = 0. the observation of a finite quadrupole moment requires the presence of a small fraction of 3 D1 . would make the decay n → p e− ν̄e not possible. However.5. S = 1: the symmetry is +1. this binding energy is negligible compared to the other mass scales. since the electron orbitals must also adapt to the new nuclear configuration.12) M ( ZA X ) > M ∗ ( ZA−1 Y ) = M ( ZA−1 Y ) + BeK Discussion The EC is energetically favoured compared to the β + decay. This can be easily seen from the equations above. then also EC must be possible with larger Q-value.9) M ( ZA X ) > M ∗ ( ZA+1 Y ) + m e = M ( ZA+1 Y ) + Be + + β decay : ZA X → Z −1 Y e νe A (5. together with the equation for the Q-value: • β − decay • β + decay • α decay • electron capture (EC) Solution Let’s denote the atomic mass of the neutral atom by M . M ∗ .10) M ( ZA X ) > M ∗ ( ZA−1 Y ) + m e = M ( ZA−1 Y ) − Be + 2 m e α decay : ZA X → Z −2 Y α A−4 (5. since BeK ≈ 10 ÷ 100 KeV . We then have: β − decay : ZX → A − Z +1 Y e ν̄e A (5.11) M ( ZA X ) > M ( ZA−4 −2 Y ) + M (2 He) 4 EC : ZA X → Z −1 Y νe A (5. in the sense that if β + is allowed.24). see e.278 5 Subnuclear Physics reaction in terms of the initial nucleus and of its decay products for the following decays. Eq. (2. After a nuclear decay. the atom is in an excited or incomplete electronic configuration. and the binding energy of the atomic electrons by Be .g. Unless the electron belongs to a low-lying level (K -shell). Suggested Readings For an introduction to nuclear decays. . which are analysed in momentum p and polarisation ε. [5]. 1 of Ref.6 A π − beam on a polarised target produces Λ baryons. Bando n. Explain how this measurement allows one to test if the dynamics conserves parity. 1N/R3/SUB/2005 Problem 5. the reader is addressed to Chap. 2 m e ≈ 1 MeV. taken as quantisation axis. Indeed.e.1 Conservation Laws 279 Discussion A standard method to test the parity-conservation of strong interactions is through the detection of an asymmetry between the inclusive cross sections when the polarisations of the beam or of the target is flipped.5.13) Γ dΩ ∗ where α is determined by the V − A interaction. the centre-of-mass angle between the decay products and the baryon momentum.32. and indication of longitudinal polarisation would indicate that parity is violated by the production dynamics. On the contrary. Since the Λ and Σ baryons decay by the parity-violating electroweak interaction. if parity needs to be conserved. the produced Λ (Λ̄) developed a negative (positive) longitudinal polarisation. a leftover of the preferred helicity state of the left-handed . (5. Hence.13). the measurement of a left-right asymmetry with respect to a transverse polarisation. as it is the case for protons and neutrons. Solution The polarisation of the Λ baryons can be extracted from the distribution of the N π decay angle in the centre-of-mass according to Eq. Since the neutral-weak interactions are parity violating. like Λ or Σ. a longitudinal polarisation PL . When the polarity of the outgoing hadrons cannot be analysed through their decay products. Only transverse polarisation is allowed. is distributed as 1 dΓ = 1 + α PL cos θ ∗ . can be used to measure the level of longitudinal polarisation: the latter can be turned into a transverse polarisation by means of resonant magnetic fields. |PL | ≈ 0. with n̂ = . while PL is the longitudinal polarisation of the baryon. (5. If parity is conserved at the production level. (5.14) |p × p | is unaffected by the parity operation. their polarisation can be measured from the distribution of the decay products in the centre-of-mass. the Λ baryons should not be longitudinally polarised. a polarisation orthogonal to the scattering plane. which is instead compatible with invariance under parity. i. Suggested Readings The use of Λ and Λ̄ particles as polarimeters has been explored by the ALEPH Collaboration in e+ e− events at the Z 0 peak [6]. When strange baryons are produced. a polarisation onto the plane spanned by p and p changes sign under parity. p × p PT = S · n̂ . since n̂ → n̂. we have instead: ⎧ ⎪ ⎪ P = (−1) (−1) L = (−1) L+1 ⎪ ⎨ η P =−η̄ P p↔p ⇒ C P = (−1) S+1 . (5. with a width: m e α5 Γ (p-Ps → γ γ ) = = 8. for S = 0 one has C = +1 for the ground state L = 0. and C P eigenvalues for a particle-antiparticle bound state for both spin-0 and spin-1/2 particles. See Ref. particle and antiparticles are assigned opposite parity.16) In Eq. can only decay to two photons. hence it can only decay to 3γ with a smaller width .15) ⎪ ⎪ +1 = (−1) L C . motivates the existence of two positronium states of different lifetimes. (5. Since the photon carries C = −1. (5. we have: ⎧ ⎪ ⎪ P = (−1) L ⎪ ⎨ p↔p ⇒ C P = +1. which relates the properties of transformation of particle- antiparticle bound states in a given configuration of spin S and orbital angular momentum L to the spin-statistics properties of the constitutents.7 Determine the C. Solution We can use Landau’s rule. C = (−1) L ⎩ ⎪ (1)↔(2) † p↔p † b ↔d For a spin-1/2 particles-antiparticle state. Problem 5. Equation (5.16). for Dirac fermions.280 5 Subnuclear Physics s (s̄) quark. the ortopositronium (“orto” = parallel spin) has C = −1. Indeed.16) applied to the positronium. P. ⎪ ⎪ −1 = (−1) L C (−1) S+1 . For a spin-0 particle- antiparticle state. the parapositronium (“para” = antiparallel spins). which are q q̄ bound states. [7] for a broader discussion on the experimental tests of parity violation.0 × 109 s−1 . C = (−1) L+S ⎩ ⎪ (1)↔(2) p↔p χ↔χ̄ s↔s (5.17) 2 For S = 1. we have used the fact that. Discussion Among the others. these properties imply selection rules on the allowed decay of mesons. as the charge conjugation operation followed by a rotation of 180◦ by one of the two generators of isospin that are not diagonal (here taken to be I2 ). we can determine the P. By using these rules. and isospin I . G-parity is also violated. but violates isospin I . The latter is defined as G = C e i π I2 . G = (−1) L+S+I (5. J C P G (ρ 0 ) = 1−−+ . it behaves as a combination of ΔI = 1/2 and I = 3/2 operators. For a fermion-antifermion bound state with orbital momentum L. and. ρ 0 .18) 9π 2 The fact that the o-Ps → γ γ is forbidden is also a consequence of Yang’s theorem. As for isospin. Remember that SU (3) has rank N − 1 = 2. In particular. charge conjugation. Discussion The strong interaction conserves parity. The charged-current interaction does not conserve either of these symmetries. and G symmetry. I .1 Conservation Laws 281 2(π 2 − 9) m e α 6 Γ (o-Ps → γ γ γ ) = = 7. L 12 . at least at leading-order. and I numbers of the initial and final states. and the angular momentum of the pair with respect to the third pion. The electromagnetic interaction conserves parity.7. which gives to the assignments J C P G (η0 ) = 0+−+ .19) i. A particle with J = 1 decaying to three pions can thus have both values of parity. which are chosen to be I3 and Y . φ0 ) = 1−−− . (5. see Problem 5. C = (−1) L+S . For n = 3 pions. isospin. hence. although it preserves the third component. C. I3 . charge conjugation. spin S. we have: P = (−1) L+1 . it maximally violates P and C. As a consequence. G-parity. .5. Suggested Readings The reader is addressed to Ref. [8] for more application of these selection rules and for the derivation of the parapositronium decay width from the solution of the Schroedinger equation of the hydrogen atom. (5.e. but conserves C P.8 Analyze the decay of the η0 . φ 0 mesons into two and three pions in terms of the P. A final state with orbital momentum L has orbital parity (−1) L . A final state containing n pions has G = (−1)n . L 3 .1. and remembering that J C P (π ) = 0+− . G. Problem 5. ω0 .2 × 106 s−1 . see Problem 5. hence it admits two diagonal generators. C. one has to consider the angular momentum of each pair of pions in their centre-of-mass frame.9.20) see Problem 5. and J C P G (ω0 . For ω0 and φ 0 . since the final state would be antisymmetric for exchange of the two particles. 3 a Forbidden by Bose–Einstein symmetry Solution The results for the decay of the neutral pseudo-scalar and vector mesons into pions are reported in Table 5. conserves P but not G. G. since the symmetry of the orbital wavefuction is (−1) L = −1. φ 0 → π 0 π 0 π 0 (−1) = ±1 (−1) = (−1)0 (−1) = (−1)3 0 = 1. 2 ρ0 → π 0 π 0 π 0 (−1) = ±1 (−1) = (−1)0 (+1) = (−1)3 1 = 1. 2. • the decay of a vector meson with C = −1 into π 0 π 0 π 0 violates C. 2 ω0 . 2 ω0 . 1. instead. hence it has to be an electromagnetic interaction. 3 ρ0 → π 0 π 0 a (−1) = (−1)2 (−1)1 (−1) = (−1)0 (+1) = (−1)2 1 = 0. 2 η0 → π 0 π 0 π 0 (−1) = (−1)3 (−1)0 (+1) = (−1)0 (+1) = (−1)3 0 = 0. 1. 3 η0 → π + π − π 0 (−1) = (−1)3 (−1)0 (+1) = (−1)0 (+1) = (−1)3 0 = 0. 2.1 P. φ 0 → π + π − (−1) = (−1)2 (−1)1 (−1) = (−1)1 (−1) = (−1)2 0 = 0. Strong decays into π + π − π 0 are instead possible for ω0 and φ 0 . C. hence it cannot occur through neither the strong nor the electromagnetic interaction. φ 0 → π 0 π 0 a (−1) = (−1)2 (−1)1 (−1) = (−1)0 (−1) = (−1)2 0 = 0. 2 ρ0 → π + π − (−1) = (−1)2 (−1)1 (−1) = (−1)1 (+1) = (−1)2 1 = 0. 3 ω0 . . The main features to be noted are: • the η0 decay to two pions violates P. so it cannot be neither a strong nor an electromagnetic process. • a spin-1 meson cannot decay to π 0 π 0 . the PDG tables report the results of searches for decays that would violate some of the conservation laws expected from the SM. which explains the narrow width of the η0 meson compared to the ρ 0 . 3 ρ0 → π + π − π 0 (−1) = ±1 (−1) = (−1)1 (+1) = (−1)3 1 = 0. φ 0 → π + π − π 0 (−1) = ±1 (−1) = (−1)1 (−1) = (−1)3 0 = 0. 1. hence it is suppressed compared to the more likely ρ 0 → π + π − decay. Besides the measured decays. 1. In the latter case.1. Suggested Readings For this kind of problems. 1. 2 η0 → π + π − (−1) = (−1)2 (−1)0 (+1) = (−1)0 (+1) = (−1)2 0 = 0. this decay would also violate isospin. a valid reference is the dedicated summary tables of the PDG [9]. and I eigenvalues of the initial and final states for the decay of the neutral light mesons into two and three pions Decay P C G I η0 → π 0 π 0 (−1) = (−1)2 (−1)0 (+1) = (−1)0 (+1) = (−1)2 0 = 0. 3 ω0 . 2.282 5 Subnuclear Physics Table 5. 1. thus violating Bose–Einstein symmetry. the decay is electromagnetic. The decay into three charged pions. but not for ρ 0 because of G-parity. 2. 1. V -spin is conserved. the hadronic width turns out to be a factor of about 10 larger than the electromagnetic width responsible for J/Ψ → . charge conjugation C. Ξ ∗− . Solution Since π 0 → 3γ would be an electromagnetic interaction. while a two-gluon state has C = +1. while the final state has C = (−1)3 = −1. but violates isospin I . by single-gluon exchange diagrams because gluons are charged under SU (3)c . Bando n. is an instance of the so-called Okubo-Zweig-Iizuka rule (OZI). which can be explained within the QCD quark model as discussed above. while Σ − belongs to the V = 1/2 doublet (Σ − . which states that the decay width of a hadron whose Feynman diagram consists of unconnected lines is suppressed. 18211/2016 Problem 5. since Σ ∗− belongs to the V = 3/2 quadruplet (Δ− . at leading order. Σ ∗− . cannot be described. . Why is it so narrow? Solution The J/Ψ is a c c̄ bound state.5. The electromagnetic interaction conserves parity P. On the other hand. it should conserve C. Discussion The internal symmetries of the Hamiltonian determine selection rules on the decay of particles. However.1 Conservation Laws 283 Problem 5. but the decay width is suppressed by αs3 . I3 (remember the relation Q = I3 + Y/2). which have the same electric charge. ¯ Discussion The smallness of the J/Ψ decay width. since it is an internal symmetry of the (d. Strong decays into charmed mesons like J/Ψ → D D̄ are kinematically forbidden. Ω − ). Ξ − ). Here.9 Explain which selection rules forbid the electromagnetic decays π 0 → 3γ and Σ ∗− → Σ − γ . Two-gluon exchange is not possible either because the J/Ψ has C = (−1) L+S = −1. Overall. The Σ ∗− → Σ − γ decay would also be an electromagnetic process. Three-gluon exchange is possible. although it preserves the third component. like pions and kaons. Strong decays into lighter mesons. a q q̄ pair produced from the partonic sea is considered as unconnected. thus forbidding such a decay. s) quarks.10 The J/Ψ meson has a mass of 3096 MeV and a width of 100 keV. but it should violate V -spin. the neutral pion has C = 1. The ratio: .11 Motivate why the φ meson decays into three pions only 15% of the times. (5. The φ meson decay to a pair of neutral pions is forbidden by Bose–Einstein symmetry. the only combination is K S0 K L0 . see Problem 5. the symmetry of the orbital wavefunction is (−1) L = −1. is a pure K S0 K L0 state. Since the two kaons are in p-wave. yielding the physical ω0 and φ 0 states. (5. see Eq. [8]. see Table 5. Bando n. since C P|K S0 = +|K S0 and C P|K L0 = −|K L0 .8. hence the φ meson can only decay strongly to three charged pions.4. Solution Thanks to its large mass. Since the orbital parity is −1. This leaves the only combination: |K 0 K̄ 0 − | K̄ 0 K 0 √ . 2. φ → K S0 K L0 is the only allowed decay into neutral kaons. the final state needs to have also C P = +1. prove that: 1. the decay to kaons dominates with a branching ratio of about 84%. while the decay to charged pions is forbidden by G-parity conservation. Discussion Since the SU (3) flavour symmetry is broken by the much larger mass of the strange quark. the I = 0 octet and singlet pseudo-vector states mix together. For latter case. larger than for the ω0 and ρ 0 . 4 of Ref.284 5 Subnuclear Physics Suggested Readings A good introduction to this topic can be found in Chap. This can be also proved by noticing that C P is an almost perfect symmetry of the strong and weak Hamil- tonian. The OZI rule. and since the φ meson has C P = (−1) S+1 = +1. infact. however. The flavour part of the wavefunction needs to be antisymmetric as required by the Bose–Einstein statistics. in terms of the weak eigenstates. the decays φ → K + K − and φ → K 0 K̄ 0 are both kinematicall allowed. The antisymmetric K + K − and K 0 K̄ 0 states have I = 0. 1N/R3/SUB/2005 Problem 5.16).21) 2 which. Overall. as required by the isospin invariance of the strong interaction. although the Q-value is very small (about 30 MeV). suppresses this decay channel since it involves unconnected quark lines. The latter is almost entirely a s s̄ bound state. The width of φ decays into neutral kaons over the total width into kaons of any charge is about 0.3. while the remaining decays consist mostly of φ → K K . π − (d ū). the strong interaction satisfies ΔS = ΔI = ΔI3 = 0. In particular. Solution The seven decay reactions can be classified by using the selection rules imposed by the properties of invariance of the strong.39 [10]. (d) π − → μ− ν̄μ . Decays where ΔS = 0 or ΔI3 = 0 are mediated by the charged-current interaction. Suggested Readings This topic is well discussed by almost all textbooks on particle physics. and (g) Λ0 → p e− ν̄e . Classify the seven decays by interaction type by knowing that the quark contents of the hadrons taking place to the reaction chain are: Ξ ∗− (ssd). The results are reported in Table 5. 5 of Ref. Problem 5. followed by (b) K − → π − π 0 . electroweak (CC).5. 12]. and electromagnetic (EM) interaction. thus giving α = 0. which gives the only combination A ∝ ε · p∗ . The Coulomb barrier effect. (c) Σ 0 → Λ0 γ .23) |p∗± |2+ |p∗0 |2 where |p∗0 | = 107 MeV and |p∗± | = 125 MeV. (5. accounts for another 4%. Σ 0 (sdu). like Refs. (e) μ− → e− νμ ν̄e . [8. (f) π 0 → γ γ .43. K − (s ū). [11]. π 0 (u ū − d d̄). π − p → π 0 n and π − p → π − p at the Δ peak.12 The Ξ ∗− baryon can decay through the reaction (a) Ξ ∗− → K − Σ 0 . 13153/2009 Problem 5. ΔI = 0. where p∗ is the momentum in the centre-of-mass frame.1 Conservation Laws 285 Γ (φ → K S0 K L0 ) α= . . present in the K + K − decay only.2. the electromagnetic interaction satisfies ΔS = ΔI3 = 0. Suggested Readings See the original paper [10] on the measurement of the φ properties and the more extended discussion present in Chap. However.5 from isospin invariance. Bando n. the ratio α should be modified to: |p∗0 |2 α= = 0. the decay amplitude in the centre-of- mass frame must be a Lorentz-scalar and linear in the polarisation ε of the decaying meson. (5. p(uud).22) Γ (φ → K + K − ) + Γ (φ → K S0 K L0 ) would be 0.13 Determine the relation between the cross sections π + p → π + p. Hence. Λ0 (sdu). 3/2. namely: 3 3 |π + p = | . − − | .25) 3 3 9 9 . At the Δ peak. together with U -spin and V -spin. Sect.− 3 2 2 3 2 2 (5. we can write the tensor-product of a I N = 1/2 and Iπ = 1 state as a linear sum of states of total isospin Iπ N = 1/2. (5. the SU (3) flavour symmetry. + 2 2 − 1 3 1 2 1 1 |π p = | .g. Using the Clebsch–Gordan coefficients. − − | .− 3 2 2 3 2 2 2 3 1 1 1 1 |π 0 n = | . 44 of Ref. Indeed. which is at the basis of the observed pattern of hadrons. Solution Isospin invariance implies that the transition amplitude for a pion-nucleon scattering can be decomposed into a sum of amplitudes for different total isospin I . contains isospin I as a sub-symmetry. only the Iπ N = 3/2 amplitude contributes.2 Classification of Decay ΔS ΔI ΔI3 Interaction the Ξ ∗− decay chain proposed by the exercise Ξ ∗− → K − Σ 0 0 0 0 Strong K − → π− π0 −1 +3/2 −1/2 CC π − → μ− ν̄μ 0 −1 −1 CC μ− → e− νμ ν̄e 0 0 0 CC π0 → γ γ 0 −1 0 EM Σ 0 → Λ0 γ 0 −1 0 EM Λ0 → p e− ν̄e +1 +1/2 +1/2 CC Discussion The strong interactions are invariant under isospin transformations.24) from which we can deduce the proportion: σ (π + p → π + p) : σ (π − p → π 0 n) : σ (π − p → π − p) = √ 2 1 2 1 = |A 23 | : | 2 A 32 |2 : | A 23 |2 = 1 : : = 9 : 2 : 1.286 5 Subnuclear Physics Table 5. [9]. see e. d) and (ū.132) of Problem 5.5. Fermi and collaborators at the Chicago cyclotron in 1952.14 Express the isospin wavefunction of the charged and neutral pions in terms of the fundamental (u.15 Whay does the neutrinoless double-beta decay imply that the neutrinos are massive Majorana particles? Discussion A massive Majorana neutrino χ . d̄) representation. The pattern of measured cross sections enforced the interpretation of the resonance as a new I = 3/2 particle. assumed to be of left-handed chirality. Solution Refering to Eq. is described by a free Lagrangian: .7. |π 0 . √ . |d̄ u (5. The angular distribution of the nucleon-pion allowed to determine its spin.26) 2 ⎪ ⎪ 2 ⎪ ⎩ |ū u−| √ d̄ d 2 The three states are however not eigenstates of I3 .36. which differs from the more familiar composition of two spin-1/2 particles into a S = 1 state. 18211/2016 Problem 5. In this case. π 0 . using the fact that (π − . |π + = |ū d. Problem 5. The reader is encouraged to read their original paper [13]. Bando n.1 Conservation Laws 287 Suggested Readings The first evidence for the Δ resonances in pion-hydrogen scattering was established by E. the two states transform under the same representation of the SU (2) group. π + ) form an isospin triplet. see Problem 1. we see that the isospin triplet states made of a quark-antiquark pair are given by: ⎧ ⎪ |d̄ u+| √ ū d ⎪ ⎪ σiaj ⎨ 2 |π a = √ |q̄i q j = −i|d̄ u+i| √ ū d (5.27) 2 Notice the sign “−” in the π 0 state vector. The latter can be obtained by a simple rotation of the first two. giving the three pion states: − |ū u − |d̄ d |π . (5. the left-handed neutrinos are massless and lepton number is conserved. −σ ). • Any other combination between these two extreme cases gives rise to up to six distinct eigenvalues of the full neutrino matrix.e. If no right-handed neutrinos are introduced. mostly made of the weakly interacting νi . (5. In terms of a Dirac spinor χ D ≡ χ1 + (iσ2 )χ2∗ . • The elements of the matrix M R are much larger than the Yukawa mass term: there are three Majorana-type. See Ref. as demanded by the observation of neutrino oscillations. where now. one can construct a generic free-Lagrangian as 1 T Lfree M = iχi† σ̄ μ ∂μ χi + χi Mi j (iσ2 )χ j + h. [14] for more details. since under an arbitrary U (1) transformation.30) which is the Lagrangian for a massive Dirac neutrino and admits a conserved lepton- number associated with the transformation χ D → eiφ χ D . the mass term is not invariant under χ → eiφ χ . The special case M11 = M22 = 0. (iσ2 )χ2∗ transforms like a right-handed spinor. light neutrinos. and three Majorana-type. When two left-handed Weil spinors χ1. The Lagrangian does not admit a conserved neutrino number. (5. . Since the second option has the virtue of explaining the lightness of the neutrinos by assuming a large value for M R rather by postulating an unnaturally small value for the Yukawa coupling. lepton number is conserved. Lepton number is maximally violated.288 5 Subnuclear Physics 1 Lfree M = iχ † σ̄ μ ∂μ χ + m χ T (iσ2 )χ − m χ ∗ T (iσ2 )χ ∗ .28) 2 where σ̄μ = (1. it is also the more frequently considered. the Lagrangian takes the familiar form: Lfree D = i χ̄ D ∂/χ D − m χ̄ D χ D . M12 ≡ m > 0 gives rise to a conserved lepton-number associated with the transformation χ1 → eiφ χ1 and χ2 → e−iφ χ2 . (5. .29). The outstanding difference between Eqs. is a symmetric off-diagonal matrix.c. M R = 0: the neutrinos are Dirac particles in full analogy with the charged leptons and quarks. analogous to the one in Eq. If the three right-handed neutrinos are introduced. (5.29) i 2 where now M is 2 × 2 symmetric matrix.30) is that a new degree of freedom has been introduced to have Dirac-type neutrinos. heavy and approximately decoupled right-handed neutrinos.28) and (5. i. with lepton number violation. (5. one should distinguish between three cases depending on the value of the right-handed neutrinos mass term 1 M R Ni N j : 2 ij • The six-by-six mass matrix M.2 are available. the mass term in Eq. as is it the case for the neutrinos.32) i where Φ is a known phase space factor and M is the nuclear matrix element. we have: mν = Uei2 m i and Γ0νββ = |m ν |2 |M |2 Φ. the boost factor for neutrinos produced by standard reactions is so large that all (anti)neutrinos can be assumed to have a constant helicity h = −1 (+1). [15] for a more formal treatment of Majorana particles.17 The ρ + and K + mesons can both decay into π + π 0 . Indeed. The 0νββ decay of unstable nuclei would violate lepton number by two units.5. 18211/2016 Problem 5. hence this decay should occur. See Ref. the 0νββ amplitude must contain the propagator of a Majorana neutrino. spinors of a given chirality have a definite helicity in the limit β → 1. (5.31) p − m 2ν Since the weak eigenstates do not coincide with the mass eigenstates.31) should be replaced by a combination of masses of the three neutrino eigenstates. The latter is given by: m ν σ2 d 4 x ei px χ (x) χ (0) = − 2 .16 Is the helicity of a free neutrino produced in a weak decay conserved? Solution As shown in Problem 1. as shown by neutrino oscillations. (5. [14] provides a concise overview on this subject. Problem 5. Assuming m ν = O(eV).1 Conservation Laws 289 Solution The SM Lagrangian with Dirac-type neutrinos admits a conserved lepton number. If instead the neutrinos are of Majorana type. there is no conserved lepton number and this decay is allowed.6. Bando n. (5. a mixture of opposite helicity states. If the particle is massive and is produced with a given chirality. Since a Majorana neutrino with m ν = 0 would still result in a conserved lepton number. the spinor will contain. In terms of the PMNS matrix. Suggested Readings Ref. we should expect the amplitude to be proportional to m ν . in general. What is the total isospin of the π + π 0 state for the two decays? . conservation of isospin by the strong interaction implies I = 1. 2. the state I = 0 cannot contribute since I3 = +1 for a π + π 0 state. which does not conserve isospin. Suggested Readings Check the PDG reviews [9] for the decay modes and properties of the Δ and Λ baryons. hence. However. made of two opposite-sign tracks in the form of a “V”. Since the symmetry of the spatial component of the wavefunction is (−1) L = +1. 1.290 5 Subnuclear Physics Solution The composition of two isospin-1 particles can give I = 0. Solution The Δ resonance is a non-strange baryon which decays strongly to π − p. [11]. However. The Λ0 baryon has strangeness S = −1. While the Δ is usually referred to as a resonance rather than a particle. the final state is made of two bosons. Suggested Readings For more details on the ΔI = 1/2 rule for charged-current interactions. hence its wave function must be symmetric under exchange of the two particle indices.18 The Δ0 and Λ0 particles decay into π − p with a mean lifetime of about 10−23 s and 10−10 s. 13153/2009 Problem 5. respectively. which explains the smaller transition probability. explaining the smallness of the K + → π + π 0 decay probability compared to e. Bando n.g. hence I = 2. For the ρ decay. 7 of Ref. Discussion The I = 3/2 transition responsible for the K + → π + π 0 decay is suppressed. and. Motivate this large difference.9 cm. to be experimentally separated already at moderate energies since cτ ≈ 7. the much larger lifetime. The K + → π + π 0 is instead an electroweak transition. the reader is addressed to Chap. Its decay is mediated by the electroweak interaction. . the Λ0 is a so-called V-particle because its lifetime is long enough to allow its production and decay vertex. the isospin component has to be symmetric under π + ↔ π 0 exchange. K S0 → (π π ) I =0 . since ΔS = 1. see e. ū. and pentaquarks (q q q q q̄). s. . Suggested Readings For the experimental evidence of tetraquark-like states. tetraquarks (q q̄ q q̄). and vice versa. [16].1 Conservation Laws 291 Problem 5. The description of hadronic properties in terms of the quark model is well summarised in Sect. finding a structure in the J/Ψ p kinematics consistent with an intermediate u u d c c̄ resonance: a charmonium-pentaquark. . within the quark parton model. x + d x] of the parent hadron mass. The observation of pentaquarks has been established in 2015 by the LHCb Collaboration [16]: an amplitude analysis of the decay Λ0b → J/Ψ K − p has been performed.19 Prove that. published in Ref. baryons (q q q). if (n q − n q̄ ) is a multiple of three. Problem 5. BESIII. of which one is a heavy meson. (5. 15 of Ref. n ū . . has been reported by several experiments in recent years (Belle.g.) manifesting in the two-body invariant mass of meson pairs. the hadron charge Q is given by: 2 1 2 1 Q= n u − n d − n ū + n d̄ . CDF. anti-baryons (q̄ q̄ q̄). [9].33) Since (n d̄ − n d ) ∈ Z. Discussion A number of putative tetraquark states (X. then Q ∈ Z implies that (n q − n q̄ ) is also a multiple of three. Assume the proton to be made up of six different quarks: u. the requirement that the difference between the number of quarks and antiquarks bound in a hadron should be a multiple of three implies that hadrons must have an integer electric charge. d̄. s̄. is documented in Ref. Y. d. The observations of an exotic structure in the J/ψ p channel. [17]. Make use the proton quantum . then Q ∈ Z. D0.5. n d . 3 3 3 3 3Q = (2n u − n d ) − (2n ū − n d̄ ) = 2(n q − n q̄ ) + 3(n d̄ − n d ). respectively. consistent with two pentaquark resonances. Solution Let’s denote the multiplicity of u. the D0 evidence for a particle X (5586) decaying to Bs0 π ± . each quark is described by a parton density function (PDF) q such that q(x) d x gives the amount of quarks of type q sharing a fraction x ∈ [x. and n d̄ . Let n q = n u + n d and n q̄ = n ū + n d̄ be the quark and antiquark multiplicities. Indeed.20 Within the quark parton model. Vice versa. LHCb). all hadrons observed so far are composed of quarks and antiquarks in numbers such that (n q − n q̄ ) is a multiple of three: mesons (q q̄). ū. Then. Z . d. and d̄ quarks by n u . References to other searches can be found in the same article. 34) 0 0 0 where qV ≡ q − q̄. A paradigmatic example is provided by the average gluon momentum in hadrons. . Suggested Readings For a broader overview on sum rules in QCD. (5. the reader is addressed to dedicated monographs like Refs. we get the following system of equations: ⎧ 1 ⎪ ⎪ Q = d x 23 u(x ) − 13 d(x ) − 13 s(x ) − 23 ū(x ) + 13 d̄(x ) + 23 s̄(x ) = 1 ⎨ p 0 B p = 0 d x 13 u(x ) + 13 d(x ) + 13 s(x ) − 13 ū(x ) − 13 d̄(x ) − 13 s̄(x ) = 1 1 ⎪ ⎪ 1 ⎩ S p = 0 d x −s(x ) + s̄(x ) = 0 from which we can derive the relations: ⎧ 1 1 ⎪ ⎪ 0 u(x ) − ū(x ) d x = 0 u V (x ) d x = 2 ⎨ 1 1 0 d(x ) − d̄(x ) d x = 0 dV (x ) d x = 1 (5. (b) baryon number B p = +1. sV (x ) d x . Discussion Relations between the moments of the PDF for different types of partons.35) ⎪ ⎪ ⎩ 1 1 0 s(x ) − s̄(x ) d x = 0 sV (x ) d x = 0 Notice that this result agrees with the expectation from the flavour SU (3) model of hadrons. which can be inferred from lepton DIS data even in the absence of tree-level gluon interactions with leptons. are known as sum rules.5. d x x g(x) ≈ 0. (c) strangeness S p = 0. which assigns the u u d flavour content to the proton. like Eq. and prove useful in constraining parton-related observables that cannot be directed measured.292 5 Subnuclear Physics numbers (a) electric charge Q p = +1. baryon number. (5. [18. and strangeness. 19]. to predict the values of: 1 1 1 u V (x ) d x .35). dV (x ) d x . Solution From the constraints on the electric charge. 5. . which can be seen as the force mediator between the matter particles of spin-1/2. and therefore it enters only through the interaction with matter fields. we denote the gauge fields associated to the three sub- groups of Eq. while the subscript corresponds to the hypercharge. 1)−2/3 . a = 1. 1)−2 }. 2.37) as G aμ . . 8. In addition to the gauge bosons. Ref. see e.2 Electroweak and Strong Interactions The standard model of electroweak and strong interactions (SM) is a relativistic quantum field theory invariant under the gauge group: G SM = SU (3)C × SU (2) L × U (1)Y . [20]. respectively. The Lagrangian density of the gauge fields can be expressed in a compact form in terms of the field strengths: G aμν = ∂μ G aν − ∂ν G aμ + gs f abc G bμ G cν a Wμν = ∂μ Wνa − ∂ν Wμa + g2 εabc Wμb Wνc (5. g2 ) have been introduced.38) where the charges under G SM have been explicitated: the first (second) number in parenthesis indicates under which representation of the color (isospin) group the fields transform. or families.2 Electroweak and Strong Interactions 293 5. The right–handed neutrino is however sterile because it does not couple with the . In the SM. In a gauge theory. . 1)0 to accommodate massive neutrinos. (5. Following the standard notation. and two of the three adimensional coupling constants of the theory (gs . 2)−1 . 2)1/3 . d R ) of three different colors. each generator of the group is associated with a spin-1 gauge boson.37) Bμν = ∂μ Bν − ∂ν Bμ .36) where SU (3)C is the symmetry group associated with the strong force (colour). They can be conveniently arranged into a multiplet of fields: {Q(3. One should possibly add a right-handed neutrino N R (1. the spectrum of the theory accommodates all the elementary matter particles observed in experiments: quarks and leptons. d R (3. and SU (2) L × U (1)Y is the symmetry related to the electroweak interactions (isospin and hypercharge). The third coupling (g1 ) is associated with the abelian sub-group U (1)Y .g. . where f abc and εabc are the structure constants of SU (3)C and SU (2) L . a = 1. One family is composed by fifteen independent Weil spinors: • up and down left–handed quarks (Q) of three different colors. • down right–handed lepton (e R ). Wμa . • up and down left–handed leptons (L) . 3 and Bμ . These come in three replicas. e R (1. 1)4/3 . L(1. u R (3. (5. • up and down right–handed quarks (u R . there are 8 + 3 + 1 such gauge bosons. (5. with the inclusion of the Higgs doublet: φ+ φ1 + iφ2 Φ= = . only . respectively. the addition of a new SU (2) L -doublet of complex scalar fields. (5. (5. three more terms pop up: LY = −λe L̄ Φ e R − λu Q̄ iσ2 Φ ∗ u R − λd Q̄ Φ d R + h. A Lagrangian density of the gauge and of the matter fields of one family.40) 2 2 The matrices T a and σ a are a particular representation of the generators of SU (3)C and SU (2) L . only a few combinations of the fields are allowed.42) With a suitable unitary transformation in flavour space. so that the spectrum of the theory would consist of only massless particles. The λa matrices are taken to be the eight Gell-Mann matrices.36)). In the basis where the left-handed neutrino and electron are the first and second element of the isospin doublet. In the attempt to write all possible gauge-invariant and renormalizable combination of matter and gauge fields.36). (5.37). by three additional rotations of the Q. the second and third rows contain the kinetic lagrangian of the fermions plus the interactions with the gauge fields. is provided by: LSM = − 14 G aμν G aμν − 41 Wμν a Waμν − 41 Bμν B μν + (5. called the Higgs doublet.39) / L + i ē R D i L̄ D / Q + i ū R D / e R + i Q̄ D / u R + i d̄ R D / dR . (5. mass terms are excluded from the right-hand side of (5. σ a are chosen to be the three Pauli matrices. Notice that all fields are charged under U (1)Y . However. the λe matrix can be made diagonal.36) is therefore reducible to blocks. The Y matrices are diagonal in the multiplet space. u R . with the covariant derivative Dμ defined as σ · Wμ Y Dμ ψ ≡ ∂μ − i gs T · G μ − i g2 − i g1 Bμ ψ. (5. with the kinetic. and d R fields. renormalizable and invariant under an arbitrary local transformation of (5.294 5 Subnuclear Physics gauge fields (it is a singlet under transformations of the group in Eq. (5.41) φ0 φ3 + iφ4 with YΦ = 1.39) describes the dynamic of the gauge fields. Conversely. but only left-handed particles are charged under SU (2) L and only the quarks are charged under SU (3)C . The gauge symmetry dictates the structure of the Lagrangian: to achieve invariance under an arbitrary local transformation of Eq. (5. The representation in the multiplet basis of the group in Eq.39) for both the vector bosons and the fermions. triple and quartic self-interaction terms (the latter two are present only for the non-abelian groups). with the correct quantum numbers allows for additional gauge-invariant terms. Indeed. The first row of Eq.c. 44) 60π 3 where Φ = 0. However. the neutron width is proportional to Q 5 . 5. hence Γ must be proportional to another mass scale Δ to the fifth power. which provides the well-known result: G 2F m 5μ Γμ = . and tau decay via the weak interaction with mean lifetimes of 886 s.3. one can use Sargent’s rule.214) with proper modifications to account for the finite electron mass. (5.29 ps.29 MeV. The rotation will mix the SU (2) L quark doublets of the three generations. muon.82). the width must have the dimensions of an energy. Solution For the neutron.2 µs.43) 2 2 where the unitary matrix Vi j is called Cabibbo-Kobayashi-Maskawa matrix (CKM) and will be discussed in more detail in Sect.4. respectively: how can you explain the huge difference between their lifetimes? Discussion The three decays are mediated by the charged-current interaction.2 Electroweak and Strong Interactions 295 one between λu and λd can be diagonalised.47 is a numerical factor (that would be unity if the electron mass were neglected relative to Q) and θC is the Cabibbo angle. see Eq. the width can be computed exactly using the Fermi Lagrangian.5. 13707/2010 Problem 5. such that α = 1 gives a pure V − A interaction. Problems Bando n. (2. Hence. see Sect. 2. (5. where the ratio between V and A hadron currents is parametrised by a parameter α of O(1).21 The neutron. Hence. the neutron width can be calculated at LO: G 2F Q 5 Γn = cos2 θC (1 + 3α 2 )Φ. 5.45) 192π 3 . The other will remain non-diagonal.c. For the muon decay. (1. which is comparable to the neutron-proton mass difference. the decay widths must be proportional to G 2F . so that an additional rotation is needed to go into the canonical mass basis after EWSB. and 0. thus changing the charged-current Lagrangian into: g (1 − γ5 ) LCC = √ Vi j ū i γμ d j W + μ + h. with Q = m n − m p = 1. (5. (2. Using the effective Fermi Lagrangian of Eq. which can be described by the effective Fermi lagrangian of Eq.82). The hierarchy between the decay times reflects the hierarchy in the mass scale Δ. 7 × 10−9 : 2. [8. (5. from a pure Q 5 scaling. An instructive dissertation about the history of the charged-current interaction can be found in Chap. (5.296 5 Subnuclear Physics This time. For example. Bando n. while it is found to be roughly three orders of magnitude larger. with m μ = 106 MeV. while the latter occurs with a branching ratio of about 10−8 .46) 192π 3 Modulo phase-space factors. Ref.47) in fair agreement with the experimental result: 1 : 2.5 × 10−9 : 3. (5.44) is about 6. giving: 5 G 2F m 5τ Γτ ≈ 5 Γ (τ → e νe ντ ) = . 6 of Ref. For the tau decay. (5. Consider for example the decay B + → D 0 e+ νe and its light-flavour counterpart π + → π 0 e+ νe . The B mesons are fairly more long-lived particles compared to what one would expect with respect to charmed mesons. Hence: Γn Γn Q 5 6 Q 5 τn : τμ : ττ = 1 : : ≈1:6 : = Γμ Γτ mμ 5 mμ 1 : 1. we see that the numerical coefficient in Eq. [14] for more details on the neutron decay. of B mesons (τ ∼ 10−12 s). one would expect the B + lifetime to be of order 10−15 s.48) Γ (π + → π 0 e+ νe ) mπ+ − mπ0 . The former has a branching ratio of 10−1 . we can use the results from Problem 5. see e. Which are the dominant factors responsible for the large differences in lifetime? Solution The difference between the neutron and muon decay widths is discussed in Prob- lem 5. The naive ratio between the two decay widths would then be: 5 Γ (B + → D 0 e+ νe ) m B + − m D0 ≈ ≈ 5 × 1014 .6 × 10−16 . Suggested Readings The calculation of the muon decay with can be found in several textbooks on particle physics.21.22 Weak interactions are responsible for the decay of muons (τ ∼ 10−6 s).g. the width is proportional to m 5μ .28. If we want to relate all widths to the purely leptonic width of Eq.45). and of neutrons (τ ∼ 103 s). 1N/R3/SUB/2005 Problem 5.3 × 10−16 . The reader is addressed to Ref. the expression are pretty similar. [11]. (5. 12]. the Fermi Lagrangian of Eq. which involves the production of an off-shell W boson. How would you measure the latter? Solution The decay of the charged pion occurs through the charged-current interaction. The π 0 decay is instead an electromagnetic process which is not suppressed by the electroweak mass scale.8 m + + = −8 = 107 · ≈ 2 × 1011 .5 × 10−8 eV. provides a decay width [21] 2 + + G 2F 2 2 m 2μ Γ (π → μ νμ ) = cos θC f m mπ 1 − 2 2 = 2. The relation between the conservation of the axial current and the decay amplitude π 0 → γ γ can be traced to the fact that the pion fields are related to the divergence of the axial current by the relation: aμ aμ σiaj ∂μ J A = f π m 2π φπa (x). thus giving a factor |Vcb |2 ∼ 10−3 smaller width for B decays compared to the CKM-favoured d → u transition. which should be zero because of conservation of the vector and axial current in QCD (the symmetry would be exact in the limit of massless quarks).04.e. (5. More specifically.82). the breaking of a classical symmetry by quantum effects.2 Electroweak and Strong Interactions 297 while the observed ratio is: Γ (B + → D 0 e+ νe ) c τπ + · 10−1 7. (1. [11]. The amplitude is therefore suppressed by the small Fermi constant G F = 1. Suggested Readings An instructive introduction to the physics of B hadrons can be found in Chap. From this value.17 × 10−5 GeV−2 . i. with J A (x) = ψ̄i (x) γ μ γ 5 ψ j (x). (2. (5. where we have used Eq.8) to convert the result in SI. one gets τπ + = 25 ns.5 × 10−17 s). 11 of Ref. in this case by quantum fluctuations of the quark fields in the presence of a gauge field.5. (5.50) 8π π μ mπ μ where f π ≈ 130 MeV is the pion decay constant defined by 0|J A |π( p) = i p μ f π .49) Γ (π → π e νe ) 0 c τ B + · 10 5 × 10−4 m The difference is due to the magnitude of the CKM matrix element |Vcb | ≈ 0. Problem 5.6 × 108 s) is much larger than the lifetime of neutral pions (τπ 0 = 8. The amplitude for this decay can be related to the three- point axial-vector-vector amplitude.23 Provide a quantitative explanation of why the lifetime of charged pions (τπ + = 2. The fact that this decay occurs is a manifestation of the so-called chiral anomaly.51) 2 . can be related to the π 0 width. (5. since σ (γ A → π 0 X ) ∝ Z 2 Γ (π 0 → γ γ ) · BR(π 0 → γ γ ). which are limited by the granularity of the medium. Indeed. where the π 0 can be produced by either nuclear reactions (stars) or by the decay of stopped kaons via K + → π + π 0 .51) yields a cross section: 23 Z 2 α3 m 2π σ ≈ 1− 2 .988 [9].53) 64π 3 f π2 where NC is the number of coulors and eu.55) 3π 2 f π2 Eγ . the photo-production of π 0 ’s in the electromagnetic field of a heavy nucleus. with π 0 → γ γ and A is a nucleus of atomic number Z . The cross section for the reaction γ A → π 0 X .and down-type quarks. The produced π 0 had an average momentum of 235 GeV. an explicit calculation from Eq.e. 5. The second experimental technique makes use of the Primakoff effect. In order to increase the mean flight distance. assuming the pion momentum to be known by other means.54) with BR(π 0 → γ γ ) = 0. There have been four different experimental methods which have been utilised to measure the π 0 lifetime. (5. The early measurements of this kind studied the production and decay of neutral pions in emulsions. corresponding to an average flight distance d ≈ 50 µm. A better sensitivity is provided by taking into account the Dalitz decay π 0 → e+ e− γ (BR ≈ 1.2%). the neutral pions need to be produced at large boosts. The first method consists in a direct measurement of the size of the π 0 decay region by relating the decay length d to the π 0 lifetime by the relation d = (|p|/m π 0 ) cτπ 0 .76 eV. neutral pions were produced from the interaction of 450 GeV protons against a target consisting of two tungsten folis of variable distance.298 5 Subnuclear Physics The chiral anomaly provides a leading-order contribution to the decay width that can be parametrised via an effective Lagrangian α 1 Lchiral = εμνρσ F μν Fρσ φπ 0 . i. The latest and most sensitive of such measurements has been carried out by the NA-30 experiment at CERN [23].8 × 10−16 s. (5. see Fig. By increasing the distance between the two foils in a range from 5 to 250 µm. (5. the decay width becomes [22]: 2 α 2 m 3π 0 Γ (π 0 → γ γ ) = NC2 eu2 − ed2 = 7.1. corresponding to τπ 0 = 0. In this experiment. the rate of e+ e− conversion was changed because of the larger fraction of π 0 decay upstream of the second foil: the lifetime could be then measured from the rate of photon conversions as a function of the foil distance. (5.d are the electric charges of the up.52) 8π f π Making the quark content of the pion explicit. A are the vector and axial currents.g. where A is a nucleus with π0 γ atomic number Z γ∗ γ A. the two leptons . Z valid in the limit E γ m π . [22]. The third method exploited in experiments is based on measuring the cross section for e+ e− → e+ e− π 0 . the reader is addressed to Ref.5. Estimate the ratio between the two decay widths. [2]. Since the pion has J = 0. at the Frascati Laboratories and later exploited by several other experiments. Solution The leptonic decay of a charged pseudo-scalar meson can be described using the Fermi model of the charged-current interaction. The leptonic matrix element selects leptons of left-handed chirality. to be compared with Eq. (5. Suggested Readings For a comprehensive review on this subject. 5. where the π 0 is produced via photon-fusion. The fourth method is based on a measurement of the branching ratio of the radiative decay π + → e+ νe γ .53). giving an amplitude: GF A f i = √ 0|JA α (0)|π + μ+ νμ |JVα † (0) − JAα † (0)|0 .2 Electroweak and Strong Interactions 299 Fig. 13153/2009 Problem 5. as carried out in the PIBETA experiment.1 Leading-order γ diagram contributing to the reaction γ A → π 0 X . and which can be again related to Γ (π 0 → γ γ ). Problem 84 of Ref. Bando n.24 Explain why the decay of the charged pion into an electron is disfavoured compared to the decay into a charged muon. This method was pioneered in 1965 by Bellettini et al.56) 2 α where JV. The unknown hadronic matrix element is parametrised by the meson decay constant f π . see e. which can be related to π 0 → γ γ by isospin invariance. (5. See Ref [22] for a comprehensive review on this topic. as one can readily prove by noticing that: A f i ∝ ū ν /p (1 − γ5 )vμ = −m μ ū ν (1 + γ5 )vμ . This is reflected by the fact that the decay amplitude is proportional to m μ . χ1† σ χ2 ) see e. Ref. which is again proportional to |p∗ |2 . the first of these terms is zero since χ1 and χ2 must be orthogonal. [8]. and have a unique helicity in the limit β ∗ → 1. × 10−4 .57) where u ν and vμ are the Dirac spinors for particles and antiparticles.50). (1. [21]. This decay is therefore helicity-suppressed. see Problem 1. For a state with J = 0.300 5 Subnuclear Physics must have the same helicity in the centre-of-mass frame. [11]. Since the amplitude is between states of the same chirality.g. In Eq. [21] for a formal treatment of the charged pion decay. Indeed. see e.59) Γ (π + → μ+ νμ ) m 2μ (m 2π − m 2μ )2 Once a radiative correction of about −3.57). Chap.25 What is the LO expectation for the ratio R = σhad /σμμ for an e+ e− machine operating at a centre-of-mass energy of 6 GeV? What is the the LO expectation for the BR of the W boson into a lepton-neutrino pair? . we have used the anti-commutation relation {γμ . (5. the theoretical result agrees with the observed ratio of 1. (5. we should expect |A |2 to be proportional to the centre-of-mass momentum squared.23 × 10−4 [9]. it must also vanish in the limit β ∗ → 0. which is the only available vector. which must obey the Dirac equations /p ν u ν = 0 and ( /p μ + m μ )vμ = 0.g. (5.6. Bando n. the decay width must be proportional to the two-body phase-space. γ5 } = 0. Ref. This topic is usually well discussed by classical textbooks like e.9% is included [24]. yielding Eq. However. Finally. 6 of Ref. 0) (5. (5. Suggested Readings See Ref. left-handed leptons with |p∗ | > 0 are polarised.189). respectively. Hence. |p∗ |2 ∝ m 2π − m 2μ .28. The second must be contracted with p∗ . 1N/R3/SUB/2005 Problem 5. the low- energy matrix element expansion of operators like ψ̄γμ ψ and ψ̄γμ γ5 ψ between two spin-1/2 particles are: ū 1 γμ u 2 → (χ1† χ2 .58) ū 1 γμ γ5 u 2 → (0. see Eq. which would be opposite for particle and anti-particle. The decay width can be easily computed by using the standard covariant formalism.g. The ratio between the decay width to muons and electrons is thus given by: Γ (π + → e+ νe ) m 2e (m 2π − m 2e )2 = = 1. we have used the value αs (s = (6 GeV)2 ) ≈ 0. see Table 5.d. 9.3 (5. Table 5. the assumptions above are valid at s = 6 GeV.74 × 105 +2/3 .2 Electroweak and Strong Interactions 301 Discussion In the history of QCD.189) is the same for muons and for any of the quarks. (5. The NLO corrections to the ratio R gives a K factor [19]: αs (s) RNLO = RLO 1 + ≈ 3. Solution √ At LO.2.60) q=u.5.e. Fig. [18] d 3÷9 −1/3 u 1÷5 +2/3 s 75 ÷ 170 −1/3 c 1150 ÷ 1350 +2/3 b 4000 ÷ 4400 −1/3 t 1. i. which is in fair agreement with the LO estimate. and neglecting the mass of the produced quarks compared to their energy. (1. From Ref. so that the phase-space volume of Eq. then the LO ratio R is given by: 2 2 1 2 10 RLO = NC eq2 = NC 2· +2· − = NC ≈ 3.5.61). (5.2 of Ref.3 Mass and charge Flavour Mass (MeV) Charge (e) of the six quarks.s. the production of a μ+ μ− pair as well as of quark-antiquark pair at s = 6 GeV proceeds through a time-like photon in the s-channel. If n f quark flavours are kinematically accessible.61) π which agrees even better with the experimental value. m q ≤ √ s/2. see e. In obtaining the numerical value in Eq.g. [9]. The experimental outcome is R ≈ 4 [9].c 3 3 9 Given the experimental √ value for the quark masses. the experimental observable R = σhad /σμμ had a key role in providing direct evidence in favour of the quark model with NC = 3 and fractional electric charge. The cross section for e+ e− → q q̄ must be proportional to the number of coulors NC and to the charge squared of the quark. eq .3. c j=d.c. 3 of Ref.b . the interaction is similar to Eq. Bando n. again seen as a multiplet in quark-flavour space.μ..b=1 i=u. Solution With the only exception of the CKM matrix. (5.302 5 Subnuclear Physics Suggested Readings A formal treatment of R in perturbative QCD can be found in dedicated textbooks.s.c. = (1 − γ5 )ν Wμ+ + h.b 2 2 where δ ab makes evident that the charged-current interactions are diagonal in colour space. [19].s. the leptonic and hadronic widths are given by: Γlep = Γ (W → ν) = 3 Γ (W → ν) =e.c.64) Γhad = NC |Vi j |2 Γ (W → ν) = 2 NC Γ (W → ν) i=u. If we denote the LO width for the two-body decay by Γ (W → ν). the charged-current coupling of fermions is universal.τ 2 2 (5. 1N/R3/SUB/2005 Problem 5.63) a.62) modulo the insertion of the CKM matrix V : g LCC had = √ Q̄ V γ μ T − PL Q Wμ+ + h.τ (5. = 2 NC g = √ δ ab ū ia Vi j γ μ (1 − γ5 ) d bj Wμ+ + h.μ.t j=d. (5.c. like Chap. here collected into three SU (2) L multiplets.62) where T ± are the ladder operators of SU (2) L and PL is the left-handed helicity projector.26 What is the the LO expectation for the BR of the W boson into a lepton-neutrino pair? Discussion The coupling of the W boson to the leptons. lep √ γ 2 =e. For the quark doublets Q. is given by g g ¯ μ LCC = √ L̄ γ μ T − PL L Wμ+ + h.c. 62) and (5. (5.5. Solution We can use the charged-current Lagrangian of Eqs.62). Using Eq. (1. In the second of Eq. see Table 5.186). the contraction between the two tensors can be done trivially giving a factor of m 2W . (5. the decay width Γ (W + → + ν) is given by + + 1 |M |2 Γ (W → ν) = dΦ2 . while the polarisation tensor is symmetric. Hence: 1 g 2 m 2W dΩ ∗ g 2 m 2W G F m3 Γ (W + → + ν) = 2 = = √ W. (5. giving: 2 1 g 1 ∗ |M |2 = √ εμ (r )εν (r ) Tr /γμ (1 − γ5 )/ ν γν (1 − γ5 ) = 3 2 2 3 r 2 g 1 qμ qν = √ −gμν + 2 2 Tr /γμ ν/ γν (1 − γ5 ) = 2 2 3 mW 2 g 8 qμ qν μ ν μ ν μν = √ −gμν + 2 ( ν + ν − (ν)g ) (5. By using the fact that ν = m 2W /2.66) 2 mW 2J + 1 where the sum runs over the polarisations and spin states of the fermions.27 Determine the width of the W boson at LO. where and ν denote the up and down states of a generic SU (2) L doublet. and carried out the sum over the polarisation by using the completion relation. The trace containing γ5 gives a vanishing contribution because it is totally antisymmetric.63) to derive the amplitude M for the process W + → + ν. the branching ratio to any lepton is roughly 10%. Problem 5. (5. and made use of the unitarity of the CKM matrix. we have used the fact that PL / = / PR and PL2 = 1. Hence.67) 2 2 3 mW In the last equality. LO prediction is: ΓW = Γlep + Γhad = (2NC + 3)Γ (W → ν) = 9 Γ (W → ν) Γlep 3 1 BR(W → ν ) = = = (5.2 Electroweak and Strong Interactions 303 Differences in phase-space are here neglected since they give rise to small corrections. The amplitude squared can be readily obtained by using the Feynman rules corresponding to the Lagrangian of Eq.64).68) 2 mW 3 32π 48π 6 2π . (5. V −1 = V † . we have exploited the fact that the decay into final states with top quarks is kinematically forbidden.65) Γlep + Γhad 2NC + 3 3 Hence.3. 69) 2 8 m 2W to conveniently express the result in terms of the Fermi constant.70) 2 2π to be compared with the measured value of ΓW = (2.71). From Eq. we get the relation: Γ (τ → e νe ντ ) : Γ (τ → μ νμ ντ ) : Γ (τ → hadrons ντ ) = 1 : 1 : 3. (5.304 5 Subnuclear Physics where we have used the relation GF g2 √ = (5.28 Show that the branching ratios of the decay of τ leptons into muons.73) Γ (τ → π ντ ) cos2 θC 0. (5. (5.26.05 GeV. it is also evident that the decay width to strange mesons is Cabibbo- suppressed. What fraction of the hadronic decays do you expect to contain at least one kaon? Solution The ratio between the leptonic and hadronic widths of the τ lepton can be predicted by using the results for the decays of the on-shell W boson. Hence. (5. so that: Γlep 2 2 ≈ = . Finally. we can use Eq.71) Γhad NC (|Vud | + |Vus | ) 2 2 3 where we have used the fact that |Vud |2 ≈ 1 − |Vus |2 = cos2 θC . Indeed. and hadrons are in the approximate ratio 1 : 1 : 3. the electroweak τ decay can be treated as the two-stage reaction τ → W ∗ ντ .71) by roughly 20%.085 ± 0. Problem 5. discussed in Problem 5. (5. (5. W ∗ → anything. with a typical ratio: 2 Γ (τ → K ντ ) sin2 θC 0. .97 Only one hadronic decay out of twenty is expected to contain a kaon. (5. QCD corrections decrease the ratio in Eq.72) This expectation is in fair agreement with the observed result BR(τ → ν ντ ) ≈ 17%.22 ≈ = ≈ 5%.65) and get the result: 3G F m 3W ΓW = 9 Γ (W + → + ν) = √ = 2. electrons. the much smaller centre-of-mass energy m τ allows only two lepton flavours to contribute to the leptonic width and only the combination u d̄ and u s̄ to the hadronic width.042) GeV [9]. However. [25]. Production of a single top quark via t-. t W -.5.5) did not allow to produce t t¯ pairs from s-channel γ ∗ /Z 0 production. The single-top channel e+ e− → W − t b̄ is kinematically allowed. The sensitivity to m t came from electroweak precision observables [26] that could be measured to per mill accuracy [27]. How was it possible to determine the top quark mass at LEP? Solution The most abundant production mechanism of top quarks at hadron colliders is through the strong interaction q q̄. and s-channel is suppressed since it envolves an electroweak coupling and depends on the b quark PDF (for t. in particular the corrections to the ρ-parameter and to the Z 0 → b b̄ hadronic width: m 2W 3 G F m 2t ρ= = 1 + Δρ = 1 + √ + ··· m 2Z cos2 θW 8π 2 2 (5.74) 20 Γb = Γd 1 − Δρ + · · · 13 which are almost 1% corrections which could be measured at LEP. 18211/2016 Problem 5. The maximum LEP-2 energy of 209 GeV (see Problem 3.2 Electroweak and Strong Interactions 305 Suggested Readings For a modern review of τ lepton physics. we can estimate the momentum of the exchanged virtual photon γ ∗ necessary to resolve a distance d ∼ 1 fm to be . Bando n. the reader is addressed to Ref. Bando n. 18211/2016 Problem 5. [28] for a comprehensive overview on the electroweak precision tests (EWP).and t W -channel). Solution By using the uncertainty principle. Suggested Readings See Ref. g g → t t¯.29 Indicate at least one production process of the top quark at Tevatron and at LHC. but its cross section is small and no such events were observed.30 Estimate the minimum electron beam energy needed to study a system of linear dimension d = 1 fm. Notice that Eq. (5. (5. 1] quantifies the degree of inelasticity of the scattering (x = 1 corresponds to a purely elastic scattering). (5.76) If we denote the initial and final four-momentum of the system by P and P .79) together with Eq. the interaction with the virtual photon is suppressed at high energy by the lack of high- frequency modes. − ) ≡ (ν.78) or Eq. (5. while κ is the anomalous magnetic moment (κ = −1 for electrons and κ = 1. Depending on the energy/ angular acceptance of the experiment. the momentum transfer is simply given by geometry: |q| = E 2 + E 2 − 2E E cos θ . Such a current gives rise to the differential cross section: .2 (q 2 ) are the Lorentz-invariant form factors and paramerise the electric and magnetic dipole interaction with the photon.79 for protons).76) and Eq. (5.75) should be taken as an order-of-magnitude estimate rather than a lower bound: if the probed system has a smooth charge distribution. For example. q) (5. (5. if the system is a spin-1/2 particle of linear dimension d with no other intrinsic length scales. (5.79) where the small electron mass can be usually neglected. (5.77) 2Mν 2Mν where the Bjorken variable x ∈ [0.80) 2M where F1.77) we then have: |q|2 = ν 2 − q 2 = ν 2 + 2ν x M ⇒ |q| = ν (ν + 2 x M). (5. (5.75) determine the necessary beam energy. conservation of energy and momentum implies −q 2 W 2 − M2 P + q = P ⇒ =1− ≡ x. (5.78) If both the energy and polar angle θ of the scattered electron can be measured.306 5 Subnuclear Physics 1 |q| ≈ 2 GeV/c. the generic quantum-mechanical current Jμ for an elastic scattering in momentum-space can be parametrised as q ν σμν Jμ ( p. the four-momentum transfer q = − from the electron probe to the probed system of mass M is given by q = (E − E .75) 2d where one can use the relation c ≈ 200 MeV fm to convert a length into a momentum. p ) = ū( p ) F1 (q 2 )γμ + iκ F2 (q 2 ) u( p). Using the DIS notation. Eq. From Eq. albeit with a reduced margin. the final state produced by the binary parton interaction is characterised by zero transverse momentum (at LO) but a finite longitudinal momentum Pz and energy E. 1N/R3/SUB/2005 Problem 5.31 A centre-of-mass energy of 270 + 270 GeV as obtained at the proton- antiproton SPS collider at CERN was sufficient for producing W and Z 0 bosons.81) dΩ 4E 2 sin4 θ2 E 4M 2 2M 2 2 The form factors are proportional to the Fourier transform of the charge distribution of the hadron evaluated at the momentum transfer (in the centre-of-mass frame. (5. the total energy Pz and E are related to the parton fractions xa and xb by the relation: ⎧ √ ⎨(xa + xb ) s = E E ± Pz 2 √ ⇒ xa.b = √ . the form factors are concentrated at values |q| ∼ /d 1 and such that they vanish in the limit of large momentum transfers.2 Electroweak and Strong Interactions 307 dσ α 2 cos2 θ2 E 2κ F22 2 q2 2 θ = F12 −q − (F1 + κ F2 ) tan 2 .81). which is concentrated in the region |q| /d. With the convention that h a moves along the +z direction. Bando n. . thus suppressing the cross section as shown by Eq. Therefore.82) ⎩(xa − xb ) s = Pz s 2 √ The centre-of-mass energy ŝ of the partonic interaction is given by: √ √ ŝ = 2 p̂a p̂b = xa xb s. Solution The dominant W ± production channel at a proton-antiproton collider is via the Drell- Yan s-channel production: 1 Consider for example the case of a linear “box” of half-size d: the Fourier transform is proportional sin(d|q|/ ) to |q| . (5. (5.83) The centre-of-mass energy available for the partonic collision is therefore a factor of √ xa xb smaller than the collider energy. Why? Discussion When two hadrons h a and h b undergo a large-momentum √ transfer scattering in a symmetric circular collider at a centre-of-mass energy s m h . For smooth charge distributions characterised by a typical length scale d. q 2 = −|q|2 ). (5. the momentum transfer should not exceed a few times /d. for systems of dimension d.5. 88) .32 Sketch the structure functions x F3 (x). F2 (x). they are defined in terms of the hadronic tensor H μν (q. The reader is encouraged to go through the discovery paper published by the UA1 Collaboration [29]. d ū → W − (5.86) 2 2 thus giving an average partonic centre-of-mass energy of: √ √ √ ŝ ≈ xu p xd p s = 0. 1N/R3/SUB/2005 Problem 5. (5. and g(x) at Q 2 = 10 GeV2 .5 xd p ≈ · 0.25 · 0. Suggested Readings The first direct production of W ± bosons at a collider has been achieved at the proton- antiproton SPS synchrotron at CERN.85) from the area below the corresponding function. The average momentum fraction of the incoming proton (antiproton) taken by u V (ū V ) and dV (d̄V ) is given by: 1 1 xu p = xū p̄ = d x x u V (x). Bando n. Approximating these functions by triangles. Fi ).2 (right).12 · 540 GeV ≈ 95 GeV.308 5 Subnuclear Physics u d̄ → W + . we can estimate the integrals of Eq. They are different for electromagnetic (Fi ) and ν ν̄ charged-current interactions (Fi . by the relation: H μν ( p. (5. (5. 5.85) 0 0 Referring to Fig. To a more formal level. How do the structure functions evolve as a function of the Q 2 of the interaction? Discussion The structure functions Fi are adimensional functions that enter the theoretical γ expressions for DIS cross sections.25. xd p = xd̄ p̄ = d x x dV (x). q) = −F1 g μν + F2 p μ p ν + i F3 εμν σ τ p σ q τ . we can estimate: 1 1 xu p ≈ · 0. p). the scattering can involve two valence quarks. (5.84) In both cases. where q = − is the four-momentum transfer and p is the initial hadron four-momentum.87) which is just enough to produce both W ± and Z 0 bosons. (5. thus suppressing the q density at large x-values compared to low x-value. F3ν̄ (x) =2 u(x) − d̄(x) d. In the LO parton-model. they receive Q 2 - dependent corrections of O(αs ).5. valence quarks q of large momentum fraction x are resolved more and more into collinear q g pairs.ν. The middle equation is known as Callan-Gross relation and is valid for spin-1/2 quarks. The relative change of the structure function Fi with respect to Q 2 can be estimated to be: d Fi d Q2 ∼ αs 2 . u d. The difference between x F3 and F2 is therefore proportional to the antiquark density q̄.89) γ F3 (x) = 0 ν F3 (x) = 2 d(x) − ū(x) .2 Electroweak and Strong Interactions 309 Given that these functions are adimensional. and of any other intrinsic mass scale M 2 . (5. F1ν̄ (x) = u(x) + d̄(x) γ .38. the gluon and sea quark densities are enhanced at higher Q 2 .90) The structure function F2ν (x) is instead proportional to the sum of the quark PDF’s. is proportional to the difference between the quark and antiquark PDF: x F3ν+ν̄ = 2 x (u − ū + d − d̄) ∝ x qV .ν̄ F2 (x) = 2 x F1 (x) (5. u Here. u F1ν (x) = d(x) + ū(x). (5. .2 shows the functions x f (x.ν. they are functions of x = −q 2 /2 pq and are proportional to particular combinations of parton density functions evaluated at the Bjorken fraction x. u and d are the PDF for up.and down-type quarks in the proton. which make them evolve with energy as predicted by the Dokshitzer-Gribov-Lipatov-Altarelli-Parisi (DGLAP) evolution equations. The structure functions evolve with the Q 2 of the interaction according to the DGLAP evolution equations. they must depend on the ratio between q 2 . At NLO. see Problem 5. Solution The function x F3ν+ν̄ (x). see Problem 5. In turn. averaged for neutrino and antineutrino data. The LO prediction for the structure functions is: γ F1 (x) = 1 2 1 9 (d(x) + d̄(x)) + 4 9 (u(x) + ū(x)) d. Q 2 ) for different parton flavours and for two values of Q 2 [9]. pq.91) Fi Q At larger momentum transfer.38.ν̄ γ . Figure 5. 1N/R3/SUB/2005 Problem 5. u. 5. [9]) Suggested Readings The DGLAP evolution equations are discussed in detail in dedicated textbooks like Ref.310 5 Subnuclear Physics Fig. At one loop. s ≈ s̄.2 The bands are x times the unpolarized parton distributions f (x) (where f = u v .2 g2 . d. . the three RGE’s are given by: 2 The √ normalisation factor of 5/3 is conventional and is inspired by GUT theories. dv .118 (from Ref. with αs (M Z2 ) = 0.33 How do the electromagnetic and strong coupling constants evolve as a function of Q 2 ? Discussion The running of the coupling constants as a function of the renormalisation scale μ2 is described by a Renormalisation Group √ Equations (RGE).0 global analysis at scales Q 2 = 10 GeV2 (left) and Q 2 = 104 GeV2 (right). Bando n. and SU (3)C groups. there are three independent gauge couplings g1 ≡ 5/3 gY . 19]. and g3 associated with the U (1)Y . [18. g) obtained in NNLO NNPDF3. b = b̄. c = c̄. In the SM. SU (2) L . g. [30] for the full expression up to four-loops.92) d ln μ2 ⎪ ⎪ ⎩ 7 + 4π g3 where αi = gi2 /4π .92) gives rise to a solution of the form: α(μ20 ) α(μ2 ) = .97) 3π i mi . (5. (5. see Problem 5.2 × 1019 GeV.96) d ln μ2 3π i i where Ni is the number of coulours for particle i. with β0 = + 6×4π 19 g2 (5.93) μ2 1 + β0 α(μ20 ) ln μ20 If β0 > 0.95) 1 + gY2 /g22 with (gY /g2 )2 = tan2 θW ≈ 0. the solution is instead monotonically increasing and develops a Landau pole at a scale: 1 Λ = μ0 exp − . Solution An ODE like Eq. (5. At low energy. indeed far above the Plank scale M P = ( c/G N )1/2 = 1. and thus develops a Landau pole at an very large energy scale. (5. the one-loop RGE for α is given by: dα α2 2 = q Ni (μ − m i ).5. See e.2 Electroweak and Strong Interactions 311 ⎧ ⎪ ⎪ − 41 g dαi ⎨ 10×4π 1 = −β0 αi2 . (5. This is the case for the non-abelian groups like SU (3)C .94) 2β0 α(μ20 ) The electromagnetic coupling is given by gY e = g2 sin θW = . it also evolves like g1 . For U (1)Y . the solution gives rise to a monotonically decreasing coupling constant and to asymptotic freedom (a “free” theory at high energy). Ref. Since e is mostly given by gY . (5. This equation can be solved to yield: −1 α(μ20 ) 2 μ2 α(μ ) ≈ 2 α(μ20 ) 1− qi Ni ln 2 .30 at μ2 = m 2Z . where only the quark and charged lepton loops contribute.34. 118. C F . Solve the evolution equation at two-loop level with the boundary condition αs (Λ(1) QCD ) = ∞.99) 12π 24π 2 with TR . 1.187 GeV? 2. Discussion The introduction of a running coupling constant αs (Q 2 ) in the calculation of physical quantities removes the dependence on the unphysical renormalisation scale μ introduced by dimensional regularisation: the explicit μ-dependence of the physical amplitude is re-absorbed by defining a running coupling αs (Q 2 ) evaluated at the physical scale Q 2 : by using the functional form αs (Q 2 ) and a measurement at some reference scale Q 20 through a physical process. (5. Solve the evolution equation at one-loop level with the boundary condition αs (Λ(0) (0) QCD ) = ∞. respectively. moving from 1/137. giving dαs dαs d Q2 1 Q2 = −β0 αs2 . the running strong coupling can be predicted at any other scale Q 2 . Solution At the one loop-level. What is the value of ΛQCD given that αs (M Z ) = 0. [30]. when running from the electron mass up to the Z 0 mass. Suggested Readings A recent review of the RGE’s in the SM at the light of the recently discovered Higgs boson can be found in Ref.9. Problem 5. d − = −β0 d ln Q 2 d Q2 αs2 Q2 αs 1 1 Q2 − = β0 ln 2 (5.312 5 Subnuclear Physics For example. n f = 5.98).34 The evolution equation of the running strong coupling αs is given by: dαs 2 = β(αs ) = −αs αs β0 + αs2 β1 + · · · (5. β1 = . = −β0 .0 to about 1/128. (5. α increases by about 6%. while n f indicates the number of active quark flavours.100) αs (Q ) αs (Q 0 ) 2 2 Q0 . 2 and M Z = 91. only the β0 term is retained in Eq.98) d ln Q where the coefficients β0 and β1 are defined by: 11 C A − 4 TR n f 11 C 2A − (6 C F + 10 C A )TR n f β0 = . and C A the Dinkin index and the Casimir of the fundamental and adjoint representation. 105) 1 + β0 αs (Q 20 ) ln Q 20 where an arbitrary scale Q 20 is introduced in place of ΛQCD .101) ln(Q /Λ(0) 2 2 QCD ) By inverting the formula above evaluated at some reference scale Q 0 : 1 1 Λ(0) = Q 0 exp − . (5. B=− .106) d Q2 β0 −αs2 (1 + α) β0 s Next. (5.103) 12π 12π one obtains: 1 Λ(0) = 91. (5.118 together with the group-theoretical value 11 C A − 4 TR n f 11 × 3 − 4 × 1 ×5 β0 = = 2 = 0. (5. At the two loop-level: dαs 1 dαs Q2 = −β0 αs2 − β1 αs3 .5. (5. then: β0 αs (Q 2 ) = .101) is: αs (Q 20 ) αs (Q 2 ) = Q2 .118 Another popular way of writing Eq. (5.2 Electroweak and Strong Interactions 313 If we choose Q 0 ≡ Λ(0) (0) QCD such that αs (ΛQCD ) = ∞.2 exp − GeV ≈ 88 MeV.107) β0 αs2 (1 + ββ01 αs ) β0 αs2 αs 1 + ββ01 αs giving the solution: 2 β1 β1 A = 1.104) QCD 2 × 0.108) β0 β0 .610. (5.610 × 0.102) QCD 2β0 αs (Q 0 ) By using the measured value αs (M Z2 ) = 0. (5. C =+ . β1 = d ln Q 2 (5. we break the left-hand side of this equation into the sum of functions with simple primitive: 1 1 1 A B C − = − + + . (5. and work in the assumption that t is small (indeed.⎥ QCD αs (Q 2 ) = (5. modulo the factor β0 ≈ 0. we get: 2 1 dαs β1 dαs β1 dαs − 2 + − = d ln Q 2 β0 αs β0 αs β0 1 + ββ01 αs 1 1 β1 αs d + d ln = d ln Q 2 .108).111) αs (Q ) 2 t β0 β0 /β1 + t β0 1 + β1 t ln β02 /β1 +t β02 t where we also use the fact that x ln x → 0 for x → 0. (5. so it is of order 10−1 at the EW scale).112) t t we have: ⎡ ⎤ Q2 log log ⎢ Λ(1) ⎥ ⎢1 − β1 2 1 1 . By expanding at first order: −1 1 1 1 + C1 t ln C2 + C3 = 1 − C1 t ln + O(t).110) αs (Q 2 ) β0 β0 + β1 αs (Q 2 ) Λ(1) QCD 2 ! "−1 (1) 2 It is convenient to introduce the variable t ≡ ln Q 2 /ΛQCD .6. t coincides with αs at one-loop.109) β0 αs 0 β0 1 + ββ1 αs Q 0 2 Q 20 0 We now choose Q 0 ≡ Λ(1) (1) QCD such that αs (ΛQCD ) = ∞. αs (Q 2 ) = t . so that the second term at the left-hand side of Eq. then: 1 β1 β1 αs Q2 + ln = β0 ln . .314 5 Subnuclear Physics Replacing the constants with the values of Eq. β0 αs β0 1 + ββ01 αs 1 1 Q 2 β1 αs Q2 Q2 | Q2 + ln | = ln (5. (5. (5.111) can be neglected. In this case. (5.3 shows a summary of measurements of αs as a function of the energy scale Q. we can make the approximation: 1 β0 β1 t 1 1 ≈ − ln 2 .113) β0 log Q(1) 2 ⎣ 2 β02 log Q2 ⎦ ΛQCD Λ(1) QCD 2 Figure 5. 5. (5. g → g g. Problem 5. from a quark qi .35 Find the relation between the colour factors TR .116) . 4. C F . [18] for a formal treatment of renormalisation and for the derivation of the β-function in QCD. with i = 1. 8. and C A and the branching probabilities for q → q g. [9]) Suggested Readings The reader is addressed to Chap. . The respective degree of QCD perturbation theory used in the extraction of αs is indicated in brackets (from Ref. with a = 1.115) NC a i j NC a NC For the ga → gb gc splitting.3 Summary of measurements of αs as a function of the energy scale Q. which then transforms into a quark q j .2 Electroweak and Strong Interactions 315 Fig. Solution Consider the emission of a gluon ga . The colour part of the amplitude is proportional to A (qi → q j ga ) ∝ T jia . 2. 3. . . and g → q q̄. the colour part is proportional to: A (ga → gb gc ) ∝ f abc (5. .5.3 of Ref.114) Summing over the final colours and averaging over the initial ones: 1 a a∗ 1 NC C F ∗ AA = T ji T ji = Tr T T = a a = C F (5. which therefore implies: TR Tr 1 N A = C F Tr 1 NC ⇒ T R N A = C F NC . = 2 C = . the following group identities hold: TR NC 3 CA 2 N2 9 = 2 = . (5.120) ij aj The left-hand side of the two equations are identical upon summation over all indices. Tiaj T jia = C F δ ii (5. (5.119) NA a i j NA a NA The last equality has been obtained by using the relations: Tiaj T jia = TR δ aa . (5.121) Discussion For SU (3). tricks can be used to speed up the computation of colour factors. we have used the antisymmetry of the structure constants f abc and the fact that Tbca = −i f abc are the generators of the adjoint representation of SU (3).117) NA a Here.122) CF NC − 1 8 CF NC − 1 4 With the standard normalisation TR = 1/2. we then have C F = 4/3 and C A = 3. (5.118) Summing over the final colours and averaging over the initial ones we get: 1 a a∗ 1 NC C F ∗ AA = Ti j Ti j = Tr T T = a a = TR . When doing more involved calculations. (5.316 5 Subnuclear Physics Summing over the final colours and averaging over the initial ones: 1 1 1 a a AA∗ = ∗ f abc f abc = (−i f abc )(−i f acb ) = T T N A abc N A abc N A abc bc cb 1 = Tr T T a a = C A.123) TR . the colour part is proportional to: A (ga → qi q̄ j ) ∝ T jia . N A = NC2 − 1 = 8. In particular. For the ga → qi q̄ j splitting. with Casimir C A 1 N A . the antisymmetric tensor f abc can be always exchanged for traces of T a matrices using the relation: i T a T b − T b T a = i f abc T c ⇒ f abc = − Tr T a T b T c − T b T a T c (5. 5. colour octet quark-antiquark bound state |h.4 of Ref. This result depends on the Lorentz structure (ψ̄γ μ ψ) of the fermionic current in QED. while the sign “−” applies to q q̄ → q q̄. First. we recall that under the same infinitesimal transformation U (θ ) = 1 + iθa T a . Hence. the potential is predicted to be repulsive (attractive) for a quark-quark (quark-antiquark) interaction. a ∝ i j λiaj |q i q̄ j . [15].124) r where f is a colour factor which depends on the colour of the quark and antiquark states. We also prove that the three bound states above transform as singlets or octets. we can think of the QCD potential that binds quarks inside hadrons as being mediated by the exchange of a gluon. 3. θa .126) where f C is the colour factor that needs to be calculated.125) |r| where the sign “+” applies to the q q → q q case. The potential is given by: αs QCD Vqq = f . and is identical for the case of QCD.36 At the lowest order in QED. the discussion in Ref. colour singlet quark-antiquark bound state |h ∝ i j δi j |qi q̄ j .g. 2. The result from QED is the well-known Coulomb potential α V QED (r) = ± . (5.3. 3. totally antisymmetric state builded with three quarks |h ∝ i jk εi jk |qi q j qk . Similarly. The only difference between the two theories arises from the colour structure of the quark-quark-gluon vertex and from the colour wavefunction of the bound state. Calculate the colour factor f for: 1. giving rise to the replacement α → f αs . [18] Problem 5.2 Electroweak and Strong Interactions 317 Suggested Readings The calculation of the colour factors in the q q̄ → g g amplitude provides an instructive example of coulour algebra. A guided calculation can be found in Sect. as anticipated. the interaction between an electron and a positron in a bound state is represented by a diagram where a photon is exchanged between a quark and an antiquark. see e. (5. Discussion The potential V (x) can be calculated from the Fourier-transform of the non- relativistic q q → q q amplitude corresponding to a single-gluon exchange. (5. quarks and antiquarks undergo the following transformation: . 1. (5.131) Remembering that Tacb = −i f bac are the generators of the adjoint representation of dimension NC2 − 1 = 8. T b ]lk qk q̄l = T jia qi q̄ j + iθb (i f abc Tlkc ) qk q̄l = T jia qi q̄ j + iθb (−i f bac Tlkc ) qk q̄l (5.134) .132) i. (5.130) For the meson octet: T jia qi q̄ j → T jia (δik + iθb Tikb )(δ jl − iθb Tlbj ) qk q̄l ≈ T jia qi q̄ j + iθb (T jia Tikb δl j − T jia Tlbj δik ) qk q̄l = T jia qi q̄ j + iθb (Tlia Tikb − Tlbj T jk a ) qk q̄l = T jia qi q̄ j + iθb [T a . It is easy to show that this holds true also for a finite transformation U .e. since: δ ji qi q̄ j → δ ji Uil Um∗ j ql q̄m = (U U † )lm ql q̄m = δ ji qi q̄ j . and so the sum is zero. (5. (5. since: εi jk qi q j qk → εi jk Uil U jm Ukn ql qm qn = | det U | εlmn ql qm qn = εi jk qi q j qk . (5. va transform like an octet of SU (3).133) and notice that the expression in parentheses is symmetric under exchange of any of the i jk indexes.129) It is easy to show that this holds true also for a finite transformation U .318 5 Subnuclear Physics qi → qi = (δi j + iθa Tiaj ) q j (5. we have: T jia qi q̄ j ≡ va → va + iθb Tacb vc .128) This allows us to prove that for the meson singlet: δ ji qi q̄ j → δ ji (δik + iθa Tika )(δl j − iθa Tlaj ) qk q̄l ≈ δ ji qi q̄ j + iθa (δ ji δ jl Tika − δi j δik Tlaj ) qk q̄l = δ ji qi q̄ j + iθa (Tlka − Tlka ) = δ ji qi q̄ j . for the baryon singlet we have: εi jk qi q j qk → εi jk (δil + iθb Tilb )(δ jm + iθb T jm b )(δkn + iθb Tkn b ) ql qm qn ≈ εi jk qi q j qk + εi jk θb (Tilb + T jm b + Tkn b ) ql qm qn . Finally.127) q̄i → q̄i = (δi j − iθa Tiaj ∗ ) q̄ j = (δi j − iθa T jia ) q̄ j (5. while the εi jk tensor is antisymmetric. modulo the replacement α → 4/3αs . (5.2 Electroweak and Strong Interactions 319 Solution • The wave-function of the bound state |h is given by |h = N δi j |qi q̄ j . a = N Tiaj |qi q̄ j . (5. with λiaj the eight Gell-Mann matrices. (5. The normalisation factor N can be readily obtained by requiring the normalisation condition: 1 1 = h|h = |N |2 δik δ jl δi j δkl = |N |2 Tr 1 NC ⇒ N = √ (5. the colour factor is: δi j δkl 1 a a 1 1 f = √ √ a a T T = T ji Ti j = Tr a T T a = C F NC = C F NC NC ki jl NC NC a N C ai jkl ai jkl (5.138) 2N 3 The factor f is positive: the resulting potential is identical to the one generated by a single-photon exchange between a fermion-antifermion pair in QED. The potential is therefore attractive.135) ij where δi j is the Kronecker tensor. With the usual normalisation TR = 1/2 of the Dinkin index: N2 − 1 4 f = CF = C =+ . • The wave-function of the bound state is given by |h. and C F is the Casimir of the fundamental representation. The normalisation factor N can be obtained by requiring the normalisation condition: ∗ 1 = h.136) i jkl NC Therefore. a = |N |2 Tiaj Tkla δi j δkl = |N |2 T jia Tiaj = |N |2 Tr T a T a .5.137) where we have used the identity C F ≡ a T a T a = C F 1 NC .139) ij where Tiaj = λiaj /2.140) TR . a|h. (5. i jkl ij 1 ⇒ N =√ . • The potential is generated by the exchange of a single gluon between two quarks.141) NC NC 6 We have made use of the completeness relation: 1 Tibj Tklb = TR δil δ jk − δi j δkl . which is repulsive for the QED case. (5. Therefore: 1 a a∗ b b 1 a a 1 f = T T T T = T T TR (δkl δi j − δki δ jl ) = TR bi jkl i j kl ki jl TR i jkl i j lk NC 1 TR 1 = (Tr T a )2 − Tr T a T a = − =− .lmn i jkl 1 N =√ . The wave-function of the bound state is given by |h = N εi jk |qi q j qk .320 5 Subnuclear Physics where we have used the normalisation condition Tr T a T b = TR δab . However. . (5.143) i jk The normalisation factor N can be obtained by requiring the normalisation condition: 1 = h|h = |N |2 εi jk εlmn δil δ jm δkn = |N |2 εi jk εi jk = |N |2 NC !. (5. one should remember that this negative sign is relative to the fermion-fermion amplitude. (5. i jk.144) NC ! Therefore: 1 1 fC = εi jk εlmn Tlia Tma j δkn = δil δ jm − δim δ jl Tlia Tma j = NC ! ai jklmn NC ! ai jlm C F NC 2 =− Tr T a T a = − =− . The colour factor is negative. (5.145) a NC ! 3 where we have used the relation i εi jk εimn = δ jm δkn − δ jn δkm . and therefore the resulting potential is attractive. the third one behaving like a spectator. The two quarks are in the antisymmetric anti-triplet state: 3 ⊗ 3̄ = 6 ⊕ 3̄.142) b NC The colour factor f is negative: the potential is therefore repulsive. the parton B will have four-momentum p B = p B + q. [15]. Discussion The kernel funcions. Problem 5. one can follow the original approach by Altarelli and Parisi [31] which makes use of old-fashioned perturbation theory. as the convolution of a universal function PB A (z). The goal is to factorise the infinitesimal cross section dσ for scattering of a lepton off the parton A which undergoes an almost collinear splitting A → B + C. 4. pC → z p A . Before interacting with γ ∗ .g. this process occurs at the second perturbative order.37 Calculate the Pqq (z) and Pgq (z) kernel functions using the explicit form of the related matrix elements. see Chap. The QCD potentials for bound states are also discussed in Ref. Solution We consider the DIS process where a lepton with four-momentum scatters against a parton A with four-momentum p A via a space-like photon γ ∗ carrying four- momentum q. Given that the matrix element for collinear splitting is divergent. in the form: ! α " dσ ( + A → + C) ∝ PB A (z) × dσ ( + B(z) → ) . (5. More details on colour algebra can be found in Ref.148) Eν − Ei .e. (5. or Altarelli-Parisi splitting functions. where C is the parton for which we will consider the collinear limit: p B → (1 − z) p A .7 of Ref. In old-fashioned perturbation theory. Chap. with an amplitude: MνiNR M fNR ν Mfi = dν . that gives the probability that parton B takes a fraction z of the incoming parton A.6 of Ref.5. the parton splits via A → B + C.146) i.2 Electroweak and Strong Interactions 321 Suggested Readings For a more exahustive discussion on the potentials generated by a single-exchange of the force mediator. 2 2 the cross section thus obtained will be proportional to an infinite factor dp⊥ /p⊥ which will be reabsorbed as a renormalisation of the parton pdf. 3. see e. s dz 2π (5. with the cross section for the DIS scattering + B → B . [18]. can be derived by using the covariant formalism in dimensional regularisation. [12]. [18]. Alternatively.147) After interacting with γ ∗ . f are eigenstates of H0 . The expression for the energies E B and E C is correct up to O( p2⊥ ).153) (2π )3 2E C (2π )3 2(1 − z)E A 16π (1 − z) 2 . 0). For the case of interest. (5.C |M A→B+C | 2 B |M+B→ | 2 |M+A→ +C |2 = 2 2 = p2⊥ p2 p2⊥ 4 zEA + 2z E A z E A + 2z E⊥A + (1 − z)E A + 2(1−z)E A − EA ABC |M A→B+C | B |M+B→ | 2 2 = . p⊥ 2z E A (5.C |M A→B+C | 2 × (5.152) ( p2⊥ )2 /(1 − z)2 where the factor E B /E A accounts for the incoming flux (dσ ( + B(z · p A ) ∝ 1/E B ).B. 0. 0. (5. E A .149) 2E B (E B + E C − E A ) √ where the factor 2E B accounts for the relativistic wave-function normalisation 2E used for the amplitudes at the numerator (using relativistic normalisation allows us to use Feynman rules for calculating specific amplitudes). −p⊥ ) 2(1 − z)E A The first equation fixes the reference frame. Inserting the expressions above into Eq. z E A . 0. The next step is to write the integration measure in terms of p⊥ and z: d 3 pC dp L d 2 p⊥ 1 dz = = d p2⊥ (5. and Mi→ν NR is the non-relativistic quantum- mechanical transition amplitude. (5. linear momentum is conserved at any vertex. the initial and final energies are equal.149). ν.150) p 2⊥ pC = ((1 − z)E A + . we can write: M A→B+C B R M+B→ R M+A→ +C = .151) ( p⊥ ) /(1 − z) 2 2 2 The infinitesimal cross section dσ is obtained by integrating over the phase space of the emitted parton: d 3 pC EB dσ ( + A → C + ) = × dσ ( + B(z · p A ) → )× (2π )3 2E C EA A. summing over the colour and spin polarisation of the final-state partons and averaging over those of the incoming parton: A. (1 − z)E A . pB = z E A + .322 5 Subnuclear Physics where i. The second and third express momentum conservation along the longitudinal direction: z is then the momentum fraction taken by the parton C and p⊥ is the transverse momentum and is treated as the perturbative parameter on which all energies are expanded.B. We use a coordinate system such that: p 2⊥ p A = (E A . 155) 2 p2⊥ Let’s now consider the two specific cases q → g q and g → q q̄.152) becomes: dσ ( + A → C + ) = 1 z (1 − z)2 = d p2⊥ dz |M A→B+C |2 × dσ ( + B(z · p A ) → ) 16π 2 (1 − z) ( p2⊥ )2 A. 2 p2 0 ⊥ 2π 2 p2 ⊥ (5. For gluon emission out of a quark. Comparing with Eq.158) (1 − z)2 2z 2z (1 − z)2 .154) where the coupling factor gs has been factored out of the amplitude squared. Eq.156) where the transverse tensor δi⊥j = δi j − pCi pC /pC2 has been introduced to project-out j only the physical polarisation states. ⎛ ⎞ μ μ = 2C F ( p B p νA + p νB p A − g μν p A p B ) ⎝ εμ εν∗ ⎠ = pol. and V A→B+C governs the pure collinear splitting.157) 2z E A z Putting everything together: 2 p2⊥ 2 p2⊥ 4 p2⊥ (1 + z)2 |Vq→gq | = 2C F 2 + = C F .C 2 1 d p⊥ αs z(1 − z) = dz |V A→B+C | × dσ ( + B(z · p A ) → ). (5. We have: (p A · pC )(p B · pC ) piA p B δi⊥j = p A · p B − j = pC2 (1 − z)E 2A z(1 − z)E 2A − p2⊥ p2⊥ = z E 2A − = + O( p2⊥ ) (1 − z)2 E 2A + p2⊥ (1 − z)2 p2⊥ p2⊥ δ ij ( p A p B )δi⊥j = 2 EA zEA + − z E 2A = (5. the amplitude squared is given by: 1 |Vq→gq |2 = CF Tr{γ μ /p B γ ν /p A }εμ εν∗ = 2 pol. j = 2C F (2 piA p B +δ ij (5.B. (5. with δi⊥j δ ji = 2. ( p A p B ))δi⊥j . q → g q. the splitting function can then be extracted: z(1 − z) PB A = |V A→B+C |2 .154). (5.2 Electroweak and Strong Interactions 323 Hence. (5.5. (5.163) 2z(1 − z) By using Eq.159) 1−z For gluon splitting.164) To summarise. g → q q̄.155). (5. j |Vg→q q̄ |2 = TR 2 pol.160) with δi⊥j = δi j − piA p A / p2A . we instead have: 1 Tr{γ μ /p B γ ν /pC }εμ εν∗ = 2TR (2 piB pC + δ i j ( p B pC ))δi⊥j . the splitting function can then be computed as: 1 + z2 Pqq (z) = C F .165) . Carrying out the contraction: j z E 2A (1 − z)E 2A piB pC δi⊥j = z E 2A (1 − z) − p⊥2 − j = −p⊥2 (5. the splitting function can then be computed as: Pqg (z) = TR (1 − z)2 + z 2 . (5.155). (5.324 5 Subnuclear Physics By using Eq. (5. (5.161) E 2A p2 p⊥2 δ ij ( p B pC )δi⊥j =2 zEA + ⊥ (1 − z)E A + − z(1 − z)E 2A + p⊥2 = 2z E A 2(1 − z)E A 1−z z 2p⊥2 = 2p⊥2 + +1 =− . the regularised (LO) Altarelli-Parisi splitting functions are: 1 + z2 1 + z2 3 (0) Pqq (z) = C F = CF + δ(1 − z) = Pq̄(0) q̄ (z) 1−z + (1 − z)+ 2 (0) Pgq (0) (z) = Pqq (1 − z) = Pg(0) q̄ (z) (0) (0) Pqg (z) = TR (1 − z)2 + z 2 = Pq̄g (z) (0) z 1−z 11 C A − 4 TR n f Pgg (z) = 2C A + + z(1 − z) + δ(1 − z).162) 2z 2(1 − z) 2z(1 − z) Putting everything together: (1 − z)2 + z 2 |Vg→q q̄ |2 = 2TR (2p⊥2 ) . (5. (1 − z)+ z 6 (5. 38 Calculate the Mellin moments of the regularized Altarelli-Parisi splitting functions.167) Suggested Readings This exercise is vastly inspired by the original paper by Altarelli and Parisi [31]. Problem 5.169) 2π 0 0 Taking the n − 1 moment on both sides of Eq.μ2 ) .169). 1].2 Electroweak and Strong Interactions 325 We have used the “+” prescription defined as: 1 F(z)+ = F(z) − δ(1 − z) dy F(y). (5. μ2 ) = d x x n−1 dz dy P (0) (z) f (y) δ(x − z y) 0 ∂μ 2π 0 0 0 1 1 ∂ αs μ2 2 f˜(n.5. (5. (5. (5. μ ) = 2 P (z) f ∂μ2 2π x z z 1 1 αs = dz dy P (0) (z) f (y) δ(x − z y).166) 0 such that: 1 1 x f (z) f (z) f (z) dz g(z) = dz [g(z) − g(1)] − g(1) dz x 1−z + x 1−z 0 1−z (5. we get: 1 1 1 1 ∂f αs d x x n−1 μ2 2 (x. μ2 ) = dz z n−1 P (0) (z) dy y n−1 f (y). (5. its nth Mellin moment is defined as: 1 c+i∞ 1 f˜(n) = dx x n−1 f (x) ⇔ f (x) = dn x −n f˜(n).170) ∂μ 2π 0 0 γ (0) (n) f˜(n.168) 0 2πi c−i∞ The Mellin moments allows to transform the integro-differential DGLAP evolution equation into a pure differential one: 1 ∂f αs dz (0) x μ 2 (x. Discussion Given a function f (x) with x ∈ [0. we have: 1 (0) γgg (n) = d x x n−1 Pgg (x) (5.326 5 Subnuclear Physics Solution For the case g → g(z) g(1 − z). m=1 m n(n − 1) (n + 1)(n + 2) 6 (5.172) 0 n n−1 m=1 m 1 1 1−x 1 d x x n−1 = d x (x n−2 − x n−1 ) = (5.173) 0 x 0 n(n − 1) 1 1 1 d x x n−1 x(1 − x) = d x (x n − x n+1 ) = (5.176) For the case q → q(z) g(1 − z). (5.177) n(n + 1) 2 m m=1 .174) 0 0 (n + 1)(n + 2) 1 d x δ(1 − x) = 1 (5. we have: (0) 1 1 + x2 1 1 + x2 γqq (n) = C F d x x n−1 = CF d x (x n+1 − 1) = 0 1−x + 0 1−x x − 1 n−2 1 = CF (xdx + · · · + 1)(1 + x 2 ) 01−x 1 ! " = CF − d x (x n−2 + · · · + 1) + (x n + · · · + x 2 ) = 0 1 1 1 = −C F + ··· + 1 + + ··· + = n−1 n+1 3 n−1 n 1 1 1 1 1 1 1 3 = −C F 2 + + − − 1 = −C F 2 + − − m n+1 n 2 m n+1 n 2 m=1 m=1 1 3 n 1 = CF + −2 .171) 0 The various terms that enter Pgg can be integrated to give: 1 1 x xn − 1 1 (x − 1)(x n−1 + · · · + 1) dx x n−1 = dx = dx = 0 (1 − x)+ 0 1−x 0 1−x 1 n 1 1 1 − d x (x n−1 + · · · + 1) = − + + ··· + 1 = − (5.175) 0 Putting everything together. we thus have: n 1 1 1 11 C A − 4 TR n f (0) γgg (n) = 2C A − + + + . Let’s study the implications of the DGLAP evolution functions to the non-singlet quark density q − q̄ ≡ q NS3 : .181) to: . ∂qi ∂ ln Q 2 = Pqq NS ⊗ qi + Pqq S ⊗ j q j + Pq̄q NS ⊗ q̄i + Pq̄q S ⊗ j q̄ j + Pqg ⊗ g ∂ q̄i ∂ ln Q 2 = NS Pqq ⊗ q̄i + S Pqq ⊗ j q̄ j + NS Pq̄q ⊗ qi + S Pq̄q ⊗ j q j + Pqg ⊗ g (5.178) n+2 n+1 n n(n + 1)(n + 2) For the case q → g(z) q(1 − z).182) ∂ ln Q 2 1 Since 0 d x qiNS (x) must be conserved (it is the amount of valence quarks qi in the 1 NS hadron). it follows that 0 dz Pqq = 0. taking the difference we have: ∂qiNS = Pqq NS ⊗ qiNS .2 Electroweak and Strong Interactions 327 For the case g → q(z) q̄(1 − z). (5. since: 3 for simplicity. we have: 1 (0) 1 + (1 − x)2 1 γgq (n) = CF dx x = CF n−1 d x 2x n−2 − 2x n−1 + x n = 0 x 0 2 2 1 n2 + n + 2 = CF − + = CF (5.181) S Given that Pqq and Pq̄q start only at O(αs2 ). (5.5.180) ∂ ln Q 2 = j Pq̄i q̄ j ⊗ q̄ j + j Pq̄i q j ⊗ q j + Pq̄i g ⊗ g Using flavour symmetry. (5. ∂qi ∂ ln Q 2 = j Pqi q j ⊗ q j + j Pqi q̄ j ⊗ q̄ j + Pqi g ⊗ g ∂ q̄i (5. we have: (0) 1 1 γqg (n) = TR (1 − x) + x = TR dx x n−1 d x 2x n+1 − 2x n + x n−1 = 2 2 0 0 2 2 1 n2 + n + 2 = TR − + = TR . the set of kernel functions at LO read as: Pqi q j (z) = Pq̄i q̄ j (z) ≡ Pqq NS (z) δi j + Pqq S (z) Pqi g (z) = Pq̄i g (z) ≡ Pqg (z) Pqi q̄ j (z) = Pq̄i q j (z) ≡ Pq̄q NS (z) δi j + Pq̄q S (z) which allows to simplify Eq. .179) n−1 n n−1 n(n + 1)(n − 1) We conclude by proving the necessity for the “+” prescription in the definition of the kernel functions. we absorb the coefficient αs /2π inside the splitting function. the splitting function Pqq (z) at order O(αs ) derived by using perturbation theory is infact a distribution. thus justifying the coefficient of the δ function in the Pgg splitting function.q̄ f 1 1 d x x Pgg + 2n f Pgq ⊗ g + d x x Pgq + Pqq ⊗ f =0 (5. [18].183) Hence. Suggested Readings More details on the Mellin moments applied to the DGLAP equation can be found in Ref. 0 0 ∂ ln Q 2 ∂ ln Q 2 f =q.186) 0 0 f Then. (5. (5. . d x x Pgq + Pqq = 0. by the law of convolutions.184) 1−z 1−z + (1 − z)+ 2 The last equality can be proved in the sense of distributions: 1 1 + z2 1 + z2 dz g(z) = dz [g(z) − g(1)] 0 +1−z 1−z g(z)(1 + z 2 ) − 2g(1) + g(1) − z 2 g(1) 1 + z2 = dz = dz g(z) + g(1) dz (1 + z) 1−z (1 − z)+ 1 + z2 3 1 + z2 3 = dz g(z) + dz δ(1 − z)g(z) = dz + δ(1 − z) g(z). (1 − z)+ 2 (1 − z)+ 2 (5.185) which proves the equality in the sense of distributions. and such that its integral is null. The requirement that the total hadron momentum is conserved further constrains the Pqg and Pgg splitting functions: ⎡ ⎤ ⎤ ⎡ 1 1 ∂f ∂g d x x ⎣g(x) + f (x)⎦ = const.187) 0 0 A quick calculation reveals that this is indeed the case.328 5 Subnuclear Physics d x x n [ f ⊗ g] (x) = d x x n f (x) · d x x n g(x). we must have: 1 1 d x x Pgg + 2 n f Pgq = 0. (5. ⇒ dx x ⎣ + ⎦ = 0. This amounts to replace: 1 + z2 1 + z2 1 + z2 3 CF → CF = CF + δ(1 − z) . 5. FCNC amplitudes would be identically zero if the quark masses were degenerate.1 There is no flavour transition in the lepton sector. called Pontecorvo-Maki- Nakagawa-Sakata matrix (PMNS). Equation (5. in full analogy to the quark sector.39) complemented with the Yukawa sector (5. we can state them in the form of theorems. (5. three neat consequences on the conservation of flavour arise.47. one has a contribution proportional to Vci Vti∗ : m i2 A = f Vci Vti∗ . the minimal extension of the SM which allows for neutrino masses introduces three right-handed neutrinos N R .b m 2W Fig. (5. whereas the charged-weak current mixes the isospin doublets thanks to the off-diagonal T ± operators. For example. consider the diagram shown in Fig. As seen in Problem 5. together with the unitarity of the CKM matrix. Theorem 5. there are no other observable lepton-flavour violating effects caused by the presence of such right-handed particles: this is a consequence of the large mass of the right-handed neutrinos that suppresses any lepton-flavour violating amplitude to a negligible level. which is an excellent approximation except when considering neutrino oscillations. Theorem 5. flavour transitions can only occur in the charged current interaction amplitude d j → W − u i (and its complex conjugate).4 that contributes to the FCNC amplitude t → c γ . which explains the suppression of flavour-changing neutral currents (FCNC) in hadronic decays: besides being loop- suppressed.62). 5.188) i=u. see Eq. The lack of flavour violation in the neutral-current interactions is a result of the neutral weak current being flavour-diagonal.42) should be then extended to include an additional term whose effect would be to misalign the lepton mass basis from the flavour basis by a the unitary matrix U .3 Flavour Physics From the structure of the SM Lagrangian of Eq. (5. This fact. is at the basis of the Glashow-Iliopoulos-Maiani mechanism (GIM).s. whose flavour dependence is encoded in the CKM matrix.4 Diagram W contributing to the FCNC t → c γ amplitude t di c Vti∗ Vci γ . Apart from neutrino oscillations.2 In the quark sector.5. Neglecting the neutrino mass.42).3 Flavour Physics 329 5. For each down-type quark di running in the loop. t) −Aμ̃ (−x. t) Z field Wμ± (x. Although this term is equivalent to a total derivative.3 Modulo a possible effect in the strong interactions (strong CP- problem).189) a which. Indeed. t) −χ̄ γμ̃ ψ (−x. t) χ̄ ψ(−x. To this purpose. The strong C P-problem is related to the unnatural suppression of a tree-level C P- violating gluon-gluon interaction term of the form: LC P−strong = θQCD εμνρσ G aμν G aρσ (5.s. thus giving a vanishing amplitude in the limit of identical quark masses (see the discussion in Ref. t) −G aμ̃ (−x. t) Higgs field ψ̄γμ χ (x. the only operator which is not invariant is exactly the one appearing in Eq. t) W ± fields h(x. C P violation can only take place to the extent that the CKM matrix is complex. Leaving aside the neutrino sector. like electric dipole moments. (5. one finds θQCD 10−9 . Every Lorentz index with a tilde requires an overall − sign if it is a space index or a + sign if it is a time index O C P O C P† Description G aμ (x.b Vci Vti∗ = 0. from the transformation properties under C P of the operators appearing in the SM Lagrangian. hence it would give vanishing contributions to scattering amplitudes. being renormalisable and of dimension d = 4.43) to the extent that V cannot be made real by any re-phasing of the quark fields. t) −χ̄ γμ̃ γ5 ψ (−x. t) Axial current ψ̄χ(x. [32] for further details). t) −Wμ̃∓ (−x. t) Fermion mass Fμν F μν Fμν F μν Field strength εμνρσ Fμν Fρσ −εμνρσ Fμν Fρσ Dual field strength . it also gives to non-perturbative effects that contribute to C P-odd observables.4 Transformation properties under C P of the operators appearing in the SM Lagrangian. as it is the case for three quark generations. t) Photon field Z μ (x. t) Gluon fields Aμ (x.4. t) −Z μ̃ (−x. reported in Table 5. would naturally find its place in the minimal SM Lagrangian. Theorem 5. From current limits.330 5 Subnuclear Physics The unitarity of the CKM matrix implies i=u. it is useful to remember Wolfenstein’s parametrisation of the CKM matrix Table 5. the only source of C P-violation comes from the charged-weak interaction. t) h(−x. t) Vector current ψ̄γμ γ5 χ (x. (5. the lepton charge is prescribed by the ΔS = ΔQ rule). they are members of the pseudo-scalar meson octet with isospin I = 1/2 and strangeness S = +1 and S = −1. A phase between the Bq operators. In the SU (3) flavour representation.4 provides the order-of-magnitude of the various entries in terms of the expansion parameter λ = sin θC ≈ 0. Bq . In the absence of weak interactions. the effective lagrangian can be schematically decomposed as the sum of three pieces [14]: Leff.39 Explain how the physics of K mesons has contributed to the comprehension of subnuclear physics in general.22. C P-violation has been observed from effects due to 4-fermion interactions that change strangeness and beauty. Problems Bando n. since only rephasing-invariant combinations of the CKM elements are truly observable. and C . semilept. Considering the case of ΔS = 0 transitions. With the phase convention C P |K 0 = | K̄ 0 . respectively.5. or via the semileptonic decays K 0 → π − + ν and K̄ 0 → π + − ν̄ (in the latter. = A (s̄ L γμ d L )(s̄ L γ μ d L ) + Bq (s̄ L γμ d L )(q̄ γ μ q) + C (s̄ L γμ d L )(¯ γ μ ) q ΔS=2 ΔS=1 ΔS=1.191) Violation of C P arises from an irreducible phase present among the coefficients A. which are responsible for the flavour-changing decay of mesons.3 Flavour Physics 331 ⎛ ⎞ 1 − λ2 /2 λ λ3 A(ρ − iη) V =⎝ −λ 1 − λ2 /2 λ2 A ⎠ + O(λ4 ) (5. An irreducible phase between A and any of the Bq is at the origin of the indirect violation of C P observed in the mass mixing of neutral kaons.190) λ A(1 − ρ − iη) −λ A 3 2 1 which. they would be stable. Experimentallty. 1N/R3/SUB/2005 Problem 5. besides making evident the presence of one irreducible phase. Discussion The K 0 and K̄ 0 are eigenstates of the strong hamiltonian. They are also one the antiparticle of the other. The weak interaction introduces ΔS = 1/2 and ΔS = 3/2 transitions that make the neutral kaons decay predominantly to two or three pions. is instead at the origin of the direct violation of C P. observed in the K 0 and B 0 decay. . the linear combinations 4 The location of the phase in the CKM matrix is purely conventional. 193). at leading order.193) 2 −1 1 2 2 where P is the projector to the strong eigenstates basis. M12 − i Γ212 M − i Γ2 0 m 2 − i Γ22 1 1 1 m1 + m2 Γ1 + Γ 2 with U = √ .2 0 . respectively. under the assumption that C P is conserved. and m i and Γi are the mass and width of the two mass eigenstates |K 1. Γ = .193). 0 (5. |K 20 = √ (5. (5. and then propagate according to the full Hamiltonian. 5. Hence. the physical state. like π − p → Λ0 K 0 .5. that propagates by virtue of Eq. they start their life as eigenstates of the strong Hamiltonian. by a diagram like the one shown in Fig. then the mass eigenstates must also be eigenstates of C P. (5. This hierarchy in lifetimes allows to distinguish between the two mass eigenstates by separating the short.and long-lived components of K 0 / K̄ 0 -initiated beams. the effective Hamiltonian matrix H˜ in the K 0 / K̄ 0 basis reads as: M − i Γ2 M12 − i Γ212 m 1 − i Γ21 0 H˜ = P (H0 + R ) P = = U −1 U. Owing to the different mass. respectively. tude K̄ 0 |H |K 0 = M12 by the relation Δm = 2|M12 |. oscillates between the K 0 / K̄ 0 states. with a total decay rate that is a factor of about 1. since the K 20 has C P = −1.192) 2 2 are eigenstates of C P with eigenvalue ±1. giving rise to a time-dependent transition amplitude with frequency in the rest frame given by Δm/. When neutral kaons are produced by the strong interaction.5 Leading-order Vis ui Vis∗ diagram contributing to the s d ΔS = 2 transition | K̄ 0 ↔ |K 0 ūj d¯ s̄ ∗ Vjd Vjs . This amplitude is described. Furthermore. 5.194) 2 where τ is the proper time. The CKM matrix elements are such that this amplitude is mostly given by the diagram corresponding to the exchange of charm quarks: Fig. it can only decay to three-pions or semileptonically. If the strong and weak Hamiltonian is postulated to be C P-invariant. M= .7 × 10−3 smaller than for K 1 .332 5 Subnuclear Physics |K 0 + | K̄ 0 |K 0 − | K̄ 0 |K 10 = √ . (5. the mass difference Δm is related to the ΔS = 2 ampli- As shown by Eq. Γ Γ 1 −i m 1 −i 21 τ −i m 2 −i 22 τ |Ψ (τ ) = √ e |K 1 + e 0 |K 2 . (5. (5.199) Hence. (5. (5. (5.5 and by ΔS = 1 transitions via | K̄ 0 → (2π ) I → |K 0 .198) 1+ε H˜12 has been introduced. (5. the effective Hamiltonian matrix of Eq.196) | p|2 + |q|2 These new states. Indeed. which is for example the case in Eq.. gluonic corrections to this amplitude are not under perturbative control.3 Flavour Physics 333 G 2F m 2c LΔS=2 ≈ (Vcs Vcd∗ )2 Fcc (d̄ L γ μ s L )(d̄ L γμ s L ). They are C P-eigenstates to the extent that | p/q| = 1. namely: 1 |K S. 1−ε H˜21 ≡ (5. (5. Solution The mixing of the two neutral kaons can be studied in a more general framework.193) must have identical diagonal elements H˜11 = H˜22 = H˜d by C P T - invariance. 5. invariance under C P of the Hamiltonian implies H˜12 = K̄ 0 |H˜ |K 0 = K̄ 0 |(C P)† H C P|K 0 = e2iδ K 0 |H | K̄ 0 = e2iδ H˜21 . with Fcc ≈ . which can be ultimately traced to the irreducible phase in the CKM matrix.L 0 = ≡ . are the eigenvectors of the full Hamiltonian. (5.L 0 = p|K 0 ± q| K̄ 0 . where the weak interaction determines the new mass eigenstates as a linear combination of K 0 and K̄ 0 .197) come out as eigenvectors of C P. This fully determines the eigenvectors in terms of the off-diagonal elements as: H˜12 |K 0 ± H˜21 | K̄ 0 (1 + ε)|K 0 + (1 − ε)| K̄ 0 |K S. and the states in Eq.191): an irreducible phase between the two. so that this operator can only provide an estimate of the mass difference [14].5. the off-diagonal elements of the Hamiltonian matrix of Eq. can thus make H˜12 and H˜21 not relatively real and turn K S0 and K L0 into an admixture of . (5.193) must be relatively real. The off-diagonal elements receive contributions from both virtual transitions via ΔS = 2 operators like the one depicted in Fig.192). see Eq. now defined K -short and K -long.195) π2 However. Assuming the more generic transformation property |K 0 → eiδ | K̄ 0 and | K̄ 0 → e−iδ |K 0 .197) |H˜12 | + |H˜21 | 2(1 + |ε|2 ) where the complex parameter ε defined by . see Eq. The results from the NA48 experiment at CERN gave a value of / / / η00 / / / ≈ 1 − 6 Re ε = (0. For example.202) Experimentally. Experimentally. one has: ΓS +ΓL Γπ + π − (τ ) ∼ e−ΓS τ + 2|η+− |e− 2 τ cos(φ+− − Δm τ ) + |η+− |2 e−ΓL τ . .201) The parameter ε controls the direct violation of C P. see Ref. thus offer- ing the opportunity to measure C P-violation. direct violation of C P would manifest itself through inter- fereing amplitudes sensitive to different combinations of the CKM elements. (5.203) Γ (K L0 → π − μ+ νμ ) + Γ (K L0 → π + μ− ν̄μ ) with an experimental result A L = (3. one finds |η+− | = (2. 5.334 5 Subnuclear Physics K 1 and K 2 . It can be related to the EWK phase of the amplitude for K 0 transitions to two-pion states in I = 0 and I = 2 configurations.7 ± 1. from an initial K 0 beam.201). where the relative orientation between the decay planes eπ and ee of the charged pions and of the electron-positron pair receives contributions from both C P-even and C P-odd amplitudes.011) × 10−3 and a consistent value for |η00 | [9]. This is for example the case of K L0 → π + π − e+ e− decays. (5. As discussed in Sect. Instead. (5.204) /η / ε +− thus giving a neat evidence of direct C P-violation through a non-zero value of ε . with a measured asymmetry [9] Neπ ·ee >0 − Neπ ·ee <0 A= = (13. η+− can be measured by studying the time-dependent decay rate of neutral kaons into π + π − .3. the fact that |ε| = 0 is an indirect manifestation that C P is broken. the parameter ε governs the amount of K 2 (K 1 ) present in K S (K L ).9950 ± 0.0007).06) × 10−3 [9]. because the mass eigenstates are not pure C P-eigenstates.5)%.32 ± 0. [11] for more details. The real part of ε can be measured from the time-integrated right-to- wrong sign ratio Γ (K L0 → π − μ+ νμ ) − Γ (K L0 → π + μ− ν̄μ ) AL = = 2 Re ε.200) Neπ ·ee >0 + Neπ ·ee <0 The traditional C P-violation parameters for the neutral kaons are defined as + − 0 0 π π |Hewk |K L0 π π |Hewk |K L0 η+− ≡ + − = ε + ε . which can be proved to be identical if C P is conserved by using unitarity and C P T invariance. The ε parameter can be instead constrained by measuring a deviation from unity of the double ratio of neutral-to-charged pion decay for K L0 and K S0 . With the phase-convention δ = 0. (5. This is experimentally the case. η 00 ≡ = ε − 2 ε π π |Hewk |K S0 π 0 π 0 |Hewk |K S0 (5. (5.232 ± 0. Considering an eigenstate of the free Hamiltonian of energy E p and momentum p. this is equivalent to a shift m → m (1 − δn).40 What is the K S0 regeneration? How can it be explained? Solution When two beams of K 0 and K̄ 0 of the same momentum |p| propagate through a thickness L of material. Hence. the reaction K̄ 0 p → π + Λ is an allowed ΔS = 0 reaction. In terms of the rest mass. [8]. Chapter 7 of Ref. which relates the forward scattering amplitude to the total cross section. Bando n. [11] is certainly a good starter.206) . Ref.3 Flavour Physics 335 Suggested Readings The physics of the K 0 / K̄ 0 system has been the object of an intensive theoretical and experimental research work over several decades.193) gets modified to: N βγ M − i Γ2 − 2π |p| f0 M12 − i Γ212 a − ε1 b ˜ H = ≡ N βγ ¯ M12 − i Γ212 M − i Γ2 − 2π |p| f0 b a − ε2 (5. This can be understood by the Optical theorem.5. (5. 1N/R3/SUB/2005 Problem 5. There are lots of textbooks where this topic can be studied in great details. (5. The reader is also addressed to study in more details the experimental setup used by the NA48 experiment to simultaneously measure the two-pion decays of both K S0 and K L0 . whereas the same is suppressed for K 0 . their phase at the exit has undergone a different shift because of the different elastic-scattering amplitude. since ΔS = +1. the extra phase taken up at the proper time τ because of elastic scattering with the nucleons is 2π N βγ τ |p|(n − 1) = f (0). the interaction with the medium induces a shift in the momentum |p| → (1 + δn) |p|. In terms of the Hamiltonian. The latter is necessarily different for K 0 and K̄ 0 : for example. the mass matrix of Eq. see e. since it provides a coincise overview of the underlying theory and summarises the main experimental results. For a wave propagating in matter with density N .205) |p| where f (0) takes different values for K 0 and K̄ 0 .g. this shift is equivalent to adding an extra momentum. |K 1. See Chap. which is more strongly peaked at zero for coherent regeneration. See Fig.336 5 Subnuclear Physics The eigenvalues and eigenvectors of this matrix are given by: λ1.208) Hence. [11] for more details.2 −a+ε2 ε1 + ε2 0 ±1 + 2r λ1. (5. The purely regenerated and diffractive components could be separated from an analysis of the angle between the incoming K 20 beam and the reconstructed π + π − momentum. the new eigenvectors can be decomposed in terms of the C P- eigenstates as: |K 10 = |K 10 + r |K 20 .6 for a graphical representation of the decay amplitude. Interestingly. an amount of K 10 of order r can be regenerated. (5.2 ∝ b ≡ 2 1 1 ε2 − ε1 π N βγ f 0 − f¯0 with r = − = . 5.207) 4b |p| m 1 − m 2 − 2i (Γ1 − Γ2 ) To first order in r .41 Consider the decays D 0 → K̄ 0 π 0 and D 0 → K 0 π 0 . Solution The flavour-changing decay D 0 → K̄ 0 π 0 proceeds at parton-level through the branchings c → s W + followed by W + → u d̄: the amplitude is therefore proportional to Vcs∗ Vud . even starting from a pure K 20 beam. (5.2 = a ± b − + O(εi ). See Fig. 7 of Ref.6 × 10−3 . The decay D 0 → K 0 π 0 requires instead the transitions c → d W + followed by W + → u s̄: the amplitude is therefore proportional to Vcd∗ Vus .6 for a graphical representation of the decay amplitude. Bando n. there is another sources of collinear K 10 : diffractive production via K 20 p → K 10 p.209) Γ (D 0 → K̄ 0 π 0 ) |Vcd∗ Vus |2 . 1N/R3/SUB/2005 Problem 5. 5. Suggested Readings The regeneration of the short-lived neutral kaon from a beam of pure K 20 passing through an iron plate was first observed at the Bevatron in 1960 [33]. Draw the Feyn- man diagrams of the two decays and estimate the ratio between the two decay amplitudes. |K 20 = −r |K 10 + |K 20 . The ratio between the two decay widths can be then estimated to be: Γ (D 0 → K 0 π 0 ) |Vcs∗ Vud |2 = = tan4 θC ≈ 2. 42 Why are the D 0 / D̄ 0 oscillations more difficult to be observed than for B 0 / B̄ 0 ? Solution The time-dependent oscillation in the decays of K 0 . Suggested Readings The doubly Cabibbo-suppressed decay D 0 → K + π − . hence oscillations slower and more difficult to be observed. the mass difference between the two D 0 - eigenstates is expected to be small. 5. which has its favoured- counterpart D 0 → K − π + in analogy with the decay considered in this exercise. (Vtb Vts∗ )2 ∼ λ4 . For K 0 . and B 0 (and their charge- conjugate states) is a consequence of mixing between the strong-eigenstates by the electroweak interaction.6 Partonic diagrams describing the decays D 0 → K̄ 0 π 0 and D 0 → K 0 π 0 The decay D 0 → K 0 π 0 is thus doubly Cabibbo-suppressed. D 0 .g. hence suppressed by a factor of λ4 ≈ 2 × 10−3 0 compared to e. Bs0 . the Bd case.5. The measured value of (0.019)% is in agreement with the expectation from CKM. and Bd0 .3 Flavour Physics 337 s d K̄ 0 K0 ∗ Vcs ∗ Vcd c d¯ c s̄ u u D0 D0 Vud π0 Vus π0 ū ū ū ū Fig.190). has been measured at the BaBar experiment at PEP-II [34]. The mass difference between the eigenstates Δm is determined by a box diagram that changes flavour by two units. the amplitude is proportional to (Vcs Vcd∗ )2 ∼ λ2 . (5. . 5. (5. analogous to the one depicted in Fig. Suggested Readings The first evidence for D 0 / D̄ 0 oscillations in the decay channel D 0 → K ∓ π ± has been established by the BaBar experiment at PEP-II [34]. (Vtb Vtd∗ )2 ∼ λ6 . the amplitude is instead ∗ 2 proportional to (Vcb Vub ) ∼ λ10 . Bando n.5. For the case of D 0 mixing. as discussed in Problem 5. 1N/R3/SUB/2005 Problem 5.210) respectively. see Eq. Therefore.303 ± 0.39. s. (5.212). Discussion Unitarity of the CKM matrix implies a number of relations between its elements. see Fig. Starting from the upper vertex and moving clockwise. only one is of phenomenological interest because the three terms are of comparable magnitude (of order λ3 ) and thus it can be experimentally verified to good accuracy: ∗ ∗ Vud Vub Vtd Vtb∗ Vud Vub + Vcd Vcb∗ + Vtd Vtb∗ = 0 ⇔ 1+ ∗ + = 0.212) Vcd Vcb Vcd Vcb∗ According to Eq.338 5 Subnuclear Physics Bando n. one has: Vi j Vik∗ = δ jk .211) i=u. so that: Fig. 5. it follows that −Vud Vub /Vcd Vcb∗ = ρ + iη: the low vertices of the unitarity triangle are therefore located in (0. (5.190). From Wolfenstein parametrisation of the CKM ∗ matrix in Eq. Considering for example the matrix by columns. (5. also called the unitarity triangle.t The case j = k gives three independent equations. while for j = k one has three equations in the complex space. η) plane (from Ref. (5.7. the three angles of the triangle are conventionally denoted by α. 0) and (1. and γ . the three terms at the left-hand side can be represented by as many vectors in the complex plane that define the contour of a closed triangle. respectively.c.43 What is the unitarity triangle? Indicate at least one process that allows to measure one of its angles and sides. b}. j. 18211/2016 Problem 5. β. equivalent to six independent equations in real space. 5. 0). k ∈ {d. Solution Out of the three equations in Eq. (5.211).7 The unitarity triangle in the complex (ρ. and the upper one in ρ + iη. [9]) 2 3 1 . 3 ◦ 73. which ∗ is proportional to Vub Vdu : the total phase is thus β + γ = π − α.214) can be traced back to the interference between two amplitudes: the first proceeds through the tree-level branching of the B 0 component: b̄ → c̄ W + followed by W + → c s̄. followed by b → u W − . The current world average is γ = +6. the reader is addressed to Chap. The current world +3.3 [9]. β = arg − . the angle γ can be measured from the interference between the CKM- favoured B + → D̄ 0 K + and the CKM-disfavoured B + → D 0 K + amplitudes in the resonant decays B + → K S0 π + π − K + . (5. The latter is instead related to the B 0 / B̄ 0 mass mixing through a calculable FCNC process. The former can be constrained by the measurement of Vub from the suppressed semileptonic decays B → X u ν.017 [9].213) Vud Vub Vtd Vtb∗ Vcd Vcb∗ Since these angles are related to an irreducible phase between the CKM elements. they must be related to C P-violating observables. and through the transition B 0 → B̄ 0 . The relative phase between the two is thus related to the angle β and gives rise to time-dependent oscillations of frequency Δm B . proportional to (Vtd Vtb∗ )2 .214) The mechanism responsible for Eq. proportional to Vud Vub . ∗ The two upper sides of the unitarity triangle have length proportional to |Vud Vub | ∗ and |Vtd Vtb |.3 Flavour Physics 339 ∗ Vtd Vtb∗ Vcd Vcb∗ Vud Vub α = arg − ∗ . The angle β can be measured from the time-dependent oscillations in the decay B 0 (t) → J/Ψ K S0 .2−7. Indeed. (5.44 Consider the two leptonic decays of the B mesons: . Indeed. such decays can proceed ∗ both through a tree-level transition b̄ → ū W + . Finally. Bando n. the second amplitude involves a transition B 0 → B̄ 0 by means of a ΔB = 2 effective operator proportional to (Vtd Vtb∗ )2 . the time-dependent amplitude squared for this decay is proportional to: | J/Ψ K S0 |H |B 0 (t) |2 ∝ e−Γ B |t| [1 − sin 2(β + φwk ) sin(Δm B t)] . 18211/2016 Problem 5.6−3. ρ ρ. The angle α can be measured by combining measurements of the time-dependent oscillations in the decays B 0 (t) → π π. ρ π.5 ◦ average is α = 87. 15 of Ref.0 [9]. The current world average is sin 2β = 0. (5. [11]. γ = arg − . and can thus be measured from the oscillation frequency in Bd0 and Bs0 decays. Suggested Readings For an overview of CKM-related measurements.691 ± 0. which is proportional to Vcd Vcb∗ . where B 0 (t) is an admixture of B 0 and B̄ 0 in analogy with the K 0 system.5. (5.5 × 10−3 ). the branching ratio is tiny compared to the semileptonic decay B → D ν .1) × 10−10 . we get an expected branching ratio of about 5 × 10−7 . respectively. only the decay B + → τ + ντ has been observed at the Υ (4S) factories.50).7 ± 0. it involves a transition from the third to the second generation. out of the three leptonic decays of the B + mesons.45 Discuss whether the partonic process b → s γ can be represented by a tree-level diagram in the SM. By using Eq.340 5 Subnuclear Physics B + → μ+ νμ B 0 → μ+ μ− (5. These two decays have been observed at the LHC with a significance in excess of 6σ and 3σ . namely BR(Bs0 → μ+ μ− ) = (3. modulo the replacement |Vud | → |Vub | and f π + → f B + . The decay B 0 → μ+ μ− is a FCNC process with |ΔB| = 1.0 × 10−6 s at 90% CL [9]. Bando n.5 × 10−6 s−1 . (5.216) Given that the B + lifetime is τ B + = 1.s → μ+ μ− decays is documented in Ref.50) with the appropriate replacements. i. Suggested Readings See the PDG review [9] dedicated to leptonic decays of charged pseudo-scalar 0 mesons. and since the decay is chirality-suppressed.1 ± 0. The observation of the rare Bd.215) Do you expect the two processes to have a similar branching ratio? If not. 18211/2016 Problem 5. (5. consistent with the current best upper limit of 1.2) × 10−9 (5. Presently.e. which one do you expect to be the largest? Solution The charged-current decay width for B + → μ+ νμ is analogous to the π + → μ+ νμ case of Eq. The measurements are in agreement with the SM expectation. Since the CKM element for b → u W − is small (Vub ≈ 4. [35].217) BR(Bd0 → μ+ μ− ) = (1. although more CKM-favourable compared to the leptonic decay. The expected SM branching ratio are however smaller than for the leptonic decay by at least two orders of magnitude. hence it is loop- suppressed. Solution . one gets: Γ (B + → μ+ νμ ) ≈ 3.6 ps. with f B + ≈ 190 MeV [9]. since the amplitude is proportional to Vtb Vts∗ . Since it has J PC = 1−− . the average βγ factor of B mesons in the laboratory frame is enhanced. For example. . Bando n. this amplitude can be studied from e. asymmetric e+ e− colliders with centre-of-mass.3 Flavour Physics 341 The partonic transition b → s γ is a FCNC process. In order to overcome this limitation. by using two beams of energies E 1 = 9 GeV and E 2 = 3 GeV.219) mΥ corresponding to an average decay distance of about 250 µm. By colliding beams of different energies E 1 and E 2 .15 ⇒ βγ ≈ 0.g. The latter is necessary to study C P violation by studying the time-dependent decay probability of the physical B 0 mesons as a function of its time of flight. the average γ factor of the B mesons in the laboratory frame is: . [11] for some more examples on this subject. Motivate this particular choice. the mean distance of flight in the laboratory for Υ (4S) produced at rest is β ∗ γ ∗ cτ ≈ 30 µm. but such that 2E 1 E 2 = m Υ . Experimentally. 18211/2016 Problem 5. It has a mass of 10. energy equal to the Υ (4S) mass are generally used. asymmetric e+ e− colliders (B-factories) have been built at√SLAC and KEK. the decay B + → K ∗+ γ .57. giving: . Solution The Υ (4S) is an excited bottomium state. (5. (5. . 5. The Q-value for both decays is small. ∗ |E 1 − E 2 | 2 γ =γ 2 + 1 = 1.46 In order to study C P violation in the B meson system. it can be produced via s-channel e+ e− scattering. TB Q Q = m Υ − 2 m B ≈ 20 MeV ⇒ β ∗γ ∗ ≈ = ≈ 0. 11 of Ref.5. Suggested Readings The reader is addressed to Chap.4. hence in the SM it cannot occur at tree-level. which is at the limit for the experimental detection of the B decay vertex.06.218) 2 mB mB Given that the B mesons have cτ of about 450 µm. It is generated at the one-loop level via a Penguin-diagram similar to the one shown in Fig.579 GeV and decays mostly to B + B − and B 0 B̄ 0 with nearly equal branching ratio. Bando n.342 5 Subnuclear Physics Suggested Readings The reader is addressed to Chap. if the PMNS matrix is non-diagonal. with A < e . with i = e. • Describe which are the irreducible backgrounds.8 A diagram γ contributing to the μ → e γ transition.8. Indeed. m ν 1 eV. (5. 15 of Ref. 5. A diagram like the one shown in Fig. 1N/R3/SUB/2005 Problem 5. τ . individual lepton numbers can be violated. [11] for more details on this subject. contribute by an effective operator: α mμ mν Lμ→e γ ≈ A (μ̄σμν e)F μν . Both can be broken in minimal extensions of the SM. However. are “accidental” global symmetries of the SM Lagrangian. Indeed. The crosses denote appropriate mass insertions W− W− μ− νμ N νe e− . μ. • Draw at least one diagram that can explain such decays.221) 32π / 2 mW / i Fig. the lepton number L and the individual lepton numbers L i . which is due to the largeness of the right-handed neutrino mass M R . as implied by the evidence of their oscillation. Discussion In the SM with Dirac-type neutrinos. the rate of decays that violate the individual lepton numbers are negligibly small due to the heaviness of the right- handed neutrino mass. where right-handed neutrinos mediate the transition between lepton families.47 The decays μ → e γ and τ → μ γ violate the conservation of the lepton number. (5.220) π m 2W m W The smallness of the observed neutrino mass. the branching ratio corresponding to this amplitude is: / /2 / 2 / 3α / ∗ m νi / BR(μ → e γ ) = / Uμi U ei / 10−50 . makes this amplitude totally negligible. 5. 39]. In extensions of the SM. The key experimental challenges are therefore the time resolution for measuring the coincidence and the energy/momentum resolution on both electrons and photons to reduce the contamination from radiative decays. where τ leptons are produced in pairs. 5. giving BR(τ → μ γ ) < 4. the transition is mediated by a right-handed neutrino. Here.5. In order to tag a signal event. in a two-body decay the electron and photon energies are fixed. the electron and photon are required to be in coincidence as to reduce the main background arising from the accidental pileup of a photon produced in a different muon decay μ → e νμ νe + γ (radiative decays) or from the interaction of the electron with the material upstream of the calorimeter. 39] for the details on searches for the lepton flavour violating decay τ → μ γ . where the photon and electron are instead produced in coincidence. [37]. The most stringent limits are obtained in e+ e− colliders. The second source of background is represented by a pure radiative decay.2 × 10−13 at 90% CL [37]. the amplitude receives additional. corresponding to BR(μ+ → e+ γ ) < 4. The main background arises from standard τ decays τ → μ νμ ντ where a photon accidentally overlaps with the muon. An analogous diagram holds for τ → μ γ .3 Flavour Physics 343 where Ui j are the appropriate elements of the PMNS matrix responsible for flavour mixing of neutrinos. Besides being back-to-back.4 × 10−8 at 90% CL [38]. in particular by the BaBar and Belle experiments [38.8. The tightest upper bound has been measured by the BaBar experiment at PEP-II. while in the ordinary three-body decay (Michel decay). . [38. contributions from diagrams where supersymmetric particles are exchanged in the triangle: the amplitude is no longer suppressed by the large mass scale M R . Tau decays via τ ± → μ± γ are best searched by producing τ leptons in high- energy collisions. See Ref. Muon decays via μ± → e± γ can be searched by stopping muon beams in matter and studying the kinematics of the electron and photon produced in their decay. The most recent results from the MEG experiment at PSI have been published in Ref. rather by the naturally smaller scale of supersymmetry breaking [36]. like the Minimal Supersymmetric Standard Model (MSSM). and potentially much larger. Suggested Readings Reference [14] provides an elegant and coincise theoretical overview on the subject. overall giving branching ratios of the order of 10−15 ÷ 10−14 . Solution A diagram contributing to the μ → e γ transition is shown in Fig. the same electron energy is only taken at the endpoint of the spectrum. The MEG experiment at PSI has put the most stringent upper limit on this decay. Expanding the Higgs doublet around the minimum. v/ 2) by a rotation of SU (2) L . for negative values of μ2 . It is only in a gauge theory that these scalar bosons get reabsorbed by the gauge bosons as their longitudinal degrees of freedom. By defining the vacuum expectation value of the Higgs doublet as 1/2 −μ2 v≡ . a Dirac mass term m i f¯ f .39) . (5. In order for the theory to be renormalizable. By virtue of the Brout-Englert-Higgs (BEH) mechanism [40. with the usual ∂μ ↔ Dμ replacement. (5. Φ ∗ ).4 Higgs Boson By adding a kinetic Lagrangian for Φ to the right-hand side of Eq. 41]. It still remains to generate masses for the fermions. (5. the vacuum state on which the expectation value of V (Φ. The neutral component of the doublet is chosen to develop v = 0. which singles-out a limited number of possible terms in V (Φ. Φ ∗ ).222) depends on the sign of μ2 . (5.222) where λ needs to be positive for the potential to be bounded from below and μ2 is a mass term for the Φ field. The most general scalar Lagrangian is then given by: L S = (D μ Φ † )(Dμ Φ) − μ2 Φ † Φ − λ(Φ † Φ)2 . with λi v mi ≡ √ .223) λ √ all states on which the expectation value of Φ is related to (0. Z 0 ). no operators with dimension d > 4 should be present at tree-level. sit on a minimum of the potential The ground state is clearly no longer invariant under an arbitrary transformation of the SU (2) L × U (1)Y sub-group: this symmetry of the Lagrangian has been spontaneously broken. If μ2 > 0. the scalar sector can be supplemented with a gauge-invariant potential V (Φ.222) becomes .344 5 Subnuclear Physics 5. The minimum of the scalar potential in Eq. Eq. and three Goldstone bosons appear in the spectrum. (5. (5. all generators of SU (2) L × U (1)Y but the combination T3 + Y/2 are broken. Φ ∗ ) is at a minimum is away from zero. Thus. so that the electric charge can be conserved. then the ground state |0 satisfies: 0|Φ|0 = 0. while the fourth (A) remains massless. More generally. (5. The ground state of the theory (vacuum) is defined as the state where the energy density is at a minimum.43): by replacing Φ with (0. On the contrary.224) 2 is generated for a fermion of type i. 1/ 2(v + H )). the BEH mechanism can give mass to three of the gauge bosons (W ± . This happens √ by virtue of the Yukawa interactions in Eq. the Higgs field propagates and interacts with the gauge fields. giving: 1 v= √ ≈ 246 GeV. and to the mass squared for gauge bosons. (5.48 Order by decreasing cross section value the following reactions. the Fermi constant G F . g H H V V = −2 2 ≡ 2 2 G F m 2V . At a hadron collider: • inclusive Z 0 production.227) ( 2 G F )1/2 The only unknown parameter of the Higgs sector is therefore the mass of the physical Higgs boson. • elastic pp scattering. . v v (5. and a cubic and quartic self–interaction with a vertex m 2H m 2H g3H = 3 and g4H = 3 .4 Higgs Boson 345 1 λ LH = (∂μ H )(∂ μ H ) − λ v2 H 2 − λ v H 3 − H 4 . (5.226) v v2 The μ and λ parameters of the bare Lagrangian can be traded for the Higgs boson mass and the vacuum expectation value (v).42) determine the strength of the interaction between the Higgs boson and the fermions: mf √ gH f f = ≡ ( 2 G F )1/2 m f .225) 2 4 √ which implies that the physical Higgs boson has a mass m H = 2 λ v.229) show that the Higgs boson couples to the SM particles with strength proportional to the particle mass for fermions. Problems Bando n.5. 1N/R3/SUB/2005 Problem 5. • inclusive b quark production. (5. • top quark pair-production. (5. The Higgs boson in the SM is a definite C P = 1 eigenstate.228) v while the couplings of the Higgs boson to gauge vectors can be extracted from the covariant derivative: m 2V √ m 2V √ g H V V = −2 ≡ −2( 2 G F )1/2 m 2V . The Yukawa couplings in Eq.229) Equations (5. 1. The latter can be put in relation with the W ± boson mass and with an other experimental observable. (5.228) and (5. • inclusive Higgs boson production. and is assigned the quantum numbers J C P = 0++ . 9 for different values of s.10. broken up by different production mechanism. and inclusive Higgs boson 10 5 5 10 -2 -1 events / sec for L = 10 cm s production (from Ref. In . although with different slopes.9.9 The theoretical proton .1 1 10 √s (TeV) 2. 5. [42]) 4 σb 4 10 10 3 3 33 10 10 jet 2 σjet(ET > √s/20) 2 10 10 σ (nb) 10 1 σW 1 10 0 σZ 0 10 10 jet σjet(ET > 100 GeV) -1 -1 10 10 -2 -2 10 10 -3 -3 10 σt 10 jet -4 σjet(ET > √s/4) -4 10 10 -5 σHiggs(MH=120 GeV) -5 10 10 -6 200 GeV -6 10 10 WJS2009 500 GeV -7 -7 10 10 0.346 5 Subnuclear Physics Fig. the following Higgs production mechanisms: • q q̄. Discussion √ As shown by Fig. the total proton-proton cross section varies slowly with s. t 6 6 10 10 quark production. At the LHC. The Higgs boson production cross sections increase with s. 5. inclusive 10 10 σtot b quark production. 5. which asserts that the total cross section 2 must be bounded from above by some constant √multiplied by ln s [44]. g g → H t t¯ • q q̄ → H Z 0 • gg → H • qq → Hqq • q q̄. g g → H b b̄ Solution The theoretical proton-(anti)proton cross section√for the processes listed in the exercise is shown in Fig. W ± and 10 7 7 10 Tevatron LHC Z 0 boson production. 5. is shown in Fig. The theoretical cross section for Higgs boson production. This is in agreement with the Froissart limit.(anti)proton cross sections proton-(anti)proton cross√ 10 9 9 10 sections as a function of s 8 8 for total scattering. the fast increase with energy is mostly due to the larger phase-space available for producing three massive particles. inside the running quark mass m q . [43]) QCD + H (NNLO pp → qq O EW) QCD + NL H (NNLO 1 pp → W D + NLO EW) (N NLO QC D in 4FS) pp → ZH NLO QC in 5FS.5. the higher-order corrections are important and can be re-asborbed. 1N/R3/SUB/2005 Problem 5. For quarks. Bando n. t-c (NLO Q pp → tH −2 10 6 7 8 9 10 11 12 13 14 15 s [TeV] particular. QCD H (NNLO pp → bb EW) NLO Q CD + −1 (NLO 10 pp → ttH h+s -ch) C D. the steeper increase of the vector-boson fusion (VBF) cross section is a reminiscence of the longitudinal boson scattering amplitude growing with energy: the partonic cross section increases like σ̂ ∝ ln ŝ. see e.4 Higgs Boson 347 Fig. 46]: 3 1 α m2 ŝ m2 σ̂ = 1 + H ln 2 − 2 + 2 H (5. Ref. [20] for a review on this subject. For example. for the case q q → H q q in weak-boson fusion. 5.231) 4 2π m 2H where NC is the number of colours. to a large extent. see e.49 What is the ratio BRb /BRτ for a Higgs boson mass of 125 GeV? Discussion The Born-level decay width of the Higgs boson to a fermion-antifermion pair is given by the general formula: 23 G F NC 4 m 2f ΓBorn (H → f f¯) = √ m H m 2f 1− . The latter is defined by a RGE: . [20] for a compendium of decay formulas. an approximate expression for the total cross section is given by [45.g.10 Standard Model LHC HIGGS XS WG 2016 Higgs boson production 102 M(H)= 125 GeV σ(pp → H+X) [pb] cross sections in LO EW ) D+N proton-proton collisions as a (N3LO QC pp → H function of the 10 centre-of-mass energy (from ) NLO EW Ref.g.230) 16 m 2W sin2 θW ŝ mH ŝ For t t¯ H . Ref. (5. Since γ0 > 0. can be expanded as a series in αs : 3 CF 1 γm (αs ) = αs γ0 + αs2 γ1 + · · · . [18]. (5. in fair agreement with the above estimate. [20]. the running mass m b (m 2H ) at the Higgs boson mass scale is a factor of about 0. (5. which. 2. whereas the running of the τ lepton mass between the pole mass and m H is negligible. Chap.235) Γ (H → b b̄) N C m 2 b (m H ) 3 4.7 = ≈ ≈ 0. in a mass-independent renormalisation scheme. 3. Solution For bottom quarks. (5. yields ratio of 0. Given Eq.18 GeV for the b quark pole mass.18 · 0.85 smaller compared to the pole mass m b (m 2b ).g. with γ0 = = . (5. see e.232) where γm is called anomalous dimension. [19].50 What is the BR and what the main background for H → γ γ ? .g. With the introduction of the running mass.3 of Ref. the running mass is a decreasing function of μ2 .231). the reader is addressed to Ref. Suggested Readings For a comprehensive review of the Higgs boson physics.4 of Ref.85 A complete calculation [43] of the partial widths. and using a value of 4.233) 4π π see e. we should therefore expected a ratio of: 2 Γ (H → τ + τ − ) 1 m 2τ 1 1.348 5 Subnuclear Physics αs (μ2 ) dm(μ2 ) γm (αs ) = −γm (αs ) m ⇒ m(μ ) = 2 m(μ20 ) exp − dαs . 1N/R3/SUB/2005 Problem 5.108.08. d ln μ2 αs (μ20 ) β(αs ) (5. the decay width becomes: 3 GF Γ (H → q q̄) = √ m H m q2 (m H ) 1 + Δqq + Δ2H . Bando n.234) 4 2π where Δqq and Δ H are additional radiative corrections of O(αs ). which includes higher-order corrections to both decays. Fig. The coupling to gluons is instead largely dominated by top quark loops. differing only on the coupling strength (gs → e) and on the colour algebra.27 × 10−3 at m H = 125 GeV [43]. (5. the leading-order amplitude receives contributions of comparable size from two interfering diagrams. Overall. The coupling to photons is mostly induced by loops of t quarks and of W boson. Since the spontaneous breaking of the SU (2) L × U (1)Y symmetry leaves unbroken both U (1)em and SU (3)C . increasing the decay width by a factor of about 12. see Eq.238) − 2τ + 3τ + 3(2τ − 1)arcsin2 τ τ −2 W Numerically.g. (5. The Lorentz structure of the top-mediated loop is identical between the H gg and H γ γ vertex. In the limit m t → ∞. However. For the photon case. √ +2 τ + (τ − 1)arcsin2 τ τ −2 t with Ai (τ ) = 2 √ (5.227).236) v v μν where v is the vacuum-expectation-value of the Higgs field.6 larger. the branching ratio into two photons turns out to be BR(H → γ γ ) ≈ 2. the H → γ γ amplitude receives large contributions from W loops. L H gg = cg G G .4 Higgs Boson 349 Discussion The Higgs boson is a gauge singlet. the effective operator generated by top quarks gives: e2 gs2 cγ = . Ref. [47]. the Higgs boson is neutral under the electromagnetic and strong interactions. It should be therefore suppressed by rougly a factor of (α/αs )2 ≈ 4 × 10−3 compared to the analogous H → g g decay. as discussed above.5. 128 2 π / 3 3 2 4m t 4m 2W / .19 × 10−2 . . A W is a factor of about 4. The dominant experimental backgrounds in proton-proton collisions arise from prompt and non-prompt di-photon production. cg = .237) 18 π 2 48 π 2 see e. Solution The H → γ γ decay is loop-mediated and it involves weak couplings. As such. (5. The resulting decay width is given by: / 2 2 //2 G F α 2 m 3H // 2 2 mH mH / Γ (H → γ γ ) = √ / NC At + AW / . They interfere destructively. it only couples to the photon and gluons via loop-induced operators of dimension five: H H a a μν L H γ γ = cγ Fμν F μν . Prompt photons can be either produced from hard parton interaction or from the fragmentation of final state partons. which should be compared to a branching ratio of BR(H → g g) ≈ 8. the generic cross section for exclusive production and decay channels is given by: σi Γ j σiSM Γ jSM κi2 κ 2j σ (i → H → j) ≡ (σi · BR j ) = = (1 − BRBSM ) ΓH Γ HSM j κ 2j BRSM j (5. Show that the κi modifiers extracted from a comprehensive set of cross section measurements are bounded from above. [20]. (5. The experimental measurement of the H → γ γ cross section at the LHC is documented in Refs. Indeed. These measurements can be related to their SM expectations via coupling strength modifiers κi . the Higgs boson cross section can only be measured in exclusive production and decay channels in the combination σi · BR j . (5. (5. but not from above. BRBSM ) space. it is easy to verify that a simultaneous measurements of multiple σi · BR j cannot constrain the κi into a closed confidence level. such that σi = κi2 σiSM and Γ j = κ 2j Γ jSM . a likelihood function of the σi · BR j measurements will admit a flat direction in the (κi .239) From Eq. κi2 κ 2j (σi · BR j ) ≤ (σi · BR j )SM = σiSM κi2 ⇒ κi ≥ . Indeed: κi2 κ 2j (σi · BR j ) = (σi · BR j )SM (1 − BRBSM ) j κ 2j BRSM j .239). a simultaneous re-scaling BRBSM + (λ2 − 1) κi → λκi and BRBSM → . Problem 5. 49].350 5 Subnuclear Physics Non-prompt and high.51 At a hadron collider. Hence. BRBSM . Conversely.240) λ2 with λ > 0. and by an unknown branching ratio into exotic particles. [48. Suggested Readings The basic formulas for the H → γ γ decay amplitude can be found in Ref. any of σi · BR j measurements yields a lower bound on κi and κ j . leaves all of the σi · BR j unchanged.241) κ 2j BRSM j σiSM .pT isolated photons mostly arise from π 0 → γ γ decays. Solution In terms of the strength modifiers. 240) is lifted. the total width can be measured as: ΓZ Γ SM · σ Z H (σ Z H )2 Γ ZSM (σ Z H · BR Z ) = σ Z H = σ Z H Z SM ⇒ ΓH = . or from off-shell interference with the background). BRBSM = 0. namely σi · BRBSM = σiSM · (κi2 BRBSM ).52 Show that. both the strength modifiers and the exotic branching ratio are clearly also limited from above. then the degeneracy remains. Suggested Readings For more details on the Higgs coupling extraction from LHC data. differently from what happens at a hadron collider. However. which removes the degeneracy of Eq. If the cross section σ Z H · BR Z can be also measured. the Higgs boson width can be measured in a model-independent fashion from the combination of two cross section measurements. so that the flat direction is again removed (for example. then an additional measurement is possible.243) .5.e. but assume that at least one of the κi is limited from above. The simplest one is clearly to assume no exotic decays. from the invariant mass distribution of the Higgs decay products. [50]. additional model-dependent assumptions are needed. If one wishes to set a bounded confidence interval on the strength modifiers in the lack of a direct measurement of the total width.242) Notice that this is not the case for a hadron collider. In this case. is the exotic decays can give rise to undetectable events.240). a physics-motivated case is provided by the assumption κ Z .4 Higgs Boson 351 Discussion If the total width Γ H can be constrained from above (for example. the reader is addressed to Ref. at an e+ e− collider operating above the Z H threshold. one can still allow for BRBSM ≥ 0. Alternatively. (5. (5. which can be detected through transverse momentum imbalance. i. the degeneracy of Eq. Problem 5. the inclusive Higgsstrahlung cross section σ Z H can be measured by studying the recoil mass distribution of the Z 0 : pe + + pe − = p Z + p H ⇒ s + m 2Z − 2 p Z ( pe+ + pe− ) ≡ m 2recoil = m 2H . Solution √ At a lepton collider with s > m H + m Z . (5. Alternatively. if one admits that the exotic decays consist of invisible particles only.W ≤ 1). ΓH σZ H ΓH (σ Z H · BR Z ) σ ZSM H (5. where the kinematics of the initial state is not known a priori. so that we can use a particular model (for example. see e. the inclusive cross section σW W H proves more difficult to be measured regardless of the Higgs decay channel because of the final-state neutrinos.g. Indeed. followed by H → W W ∗ . we have used the fact that σ Z H ∼ Γ Z H by first principles. Ref. Hence. However. 18211/2016 Problem 5. Discussion Notice that the key point here is to measure a channel where the same couplings are probed at the production and at the decay level.53 Which prospects would a muon collider offer and what are the main technological challenges for its realisation? Solution A muon collider operating at a centre-of-mass energy in excess of 100 GeV would offer the possibility of direct s-channel Higgs production. but becomes significant for μ-fusion. A summary of expected performances for the various options (CLIC. for unpolarised muon beams the cross section at the Higgs peak is given by: 16π 1 σ (s = m 2H ) = BRμ ≈ m 2H (2s1 + 1)(2s2 + 1) 4π ≈ · 2 × 10−4 · 0. Suggested Readings There is a growing literature about the potentialities of future lepton colliders on the Higgs boson physics. the SM). In a BSM perspective. Bando n.244) (125)2 A scan over the muon beam energy would allow to measure directly the Higgs boson width Γ H . 53]. one can still profit from multiple measurements in the Higgsstrahlung channel and derive σW W H from a suitable combination of them. the total width can be measured with the same precision as (σ Z H )2 /(σ Z H · BR Z ). a fine-step energy scan could allow to detect the presence of additional neutral resonances with mass splitting of order of the SM . the VBF production channel W ∗ W ∗ → H . In this latter case. (5. TLEP) can be found in Refs. The same argument can be therefore applied to e.389 mbarn ≈ 70 pb. CEPC.g. ILC. to get the coefficient of proportionality. so that one can relate Γi to σi without additional unknown couplings. [52. a reaction with a totally negligible cross section at a conventional e+ e− colliders. [51] for a future circular collider (FCC).352 5 Subnuclear Physics Here. 2% is still statistically limited.5. The finite muon lifetime sets constraints on the accelerating stage and provides a beam lifetime of order 1 msec. The energy resolution at E γ ∼ 60 GeV receives large contributions also from the noise and uniform terms: the electromagnetic calorimeters thus require good energy resolution. the H → 4μ one provides the smallest mass resolution. The muon momentum scale can be calibrated to high accuracy by using standard candles. Among the H → 4 channels. the mass resolution depends on the energy and angular resolution of the two photons. and the beam spread needs also to be of the same level to maintain the largest possible luminosity. the beam energy needs to be calibrated to better than 10−4 . . thus requiring the highest possible acceptance. The luminosity target requires to overcome the problem of generating and collecting high intensity muon beams (> 1012 muons/bunch) with reduced emittance.45). see Problem 5. In this respect. (3. see Eq. The main limitation comes from the limited statistics given that BR(H → 4μ) = 3. together with an efficient vertex identification capability. each contributing by a similar amount to the final accuracy. The main challenges posed by muon colliders are the luminosity and the energy spread of the beam. In the H → γ γ channel. The relative uncertainty of 0. Bando n. The angular resolution arises from the finite size of the calorimetric cells and from uncertainty on the vertex position in the presence of multiple pile-up events.24 GeV [54]. Given that synchrotron radiation is not a problem for muon beams of order 100 GeV. (3. 18211/2016 Problem 5.55.95).54 What aspects of a detector are crucial for a precision measurement of the Higgs boson mass? Solution The ATLAS and CMS combined measurement of the Higgs boson mass is mostly determined by the measurements in the H → γ γ and H → 4 channels. The result from the Run 1 of the LHC is m H = 125.09 ± 0. an appealing aspect of muon collider is the contained accelerator size: for a beam energy E = 60 GeV. large gains. see Eq. a fine granularity of the calorimeter. a radius R = 100 m would be sufficient for a magnetic field B ∼ 2 T. Suggested Readings For an overview on the subject. the reader can start from the PDG review [9] and references therein. are of key importance.2 × 10−5 [43]. and precise calibration.4 Higgs Boson 353 width. In order for a muon collider to serve as Higgs factory. 248) From Eq.247) The condition of identical photon energies in the laboratory frame implies that the opening angle between the two photons is given by Eq.354 5 Subnuclear Physics Suggested Readings See Ref. (5.08 rad = 61. 2 .249) 2 . (5.55 One of the possible channels for detecting a light Higgs boson H at the LHC is the rare decay H → γ γ . [54] for a discussion on the measurement of the Higgs boson mass at the LHC.6. |p| 1 γ = 1+ = 1.59). Assume a mass m H = 120 GeV and a longitudinal momentum along the beam axis |p| = 200 GeV.12. Solution The photon energy in the centre-of-mass frame is E ∗ = m H /2 = 60 GeV.246) The minimum opening angle corresponds to an emission at a polar angle θ ∗ = π/2 in the centre-of-mass frame. In the case of equally energetic photons.2 ∈ E ∗ γ − γ 2 − 1. the photon energy corresponding to an emission at an angle θ ∗ = π/2 is E 1 = E 2 = E ∗ . (1.245) mH γ2 The range of photon energy in the laboratory is given by Eq. 217] GeV.248). Indeed: E1 = E2 ⇔ E ∗ (1 + β cos θ ∗ ) = E ∗ (1 − β cos θ ∗ ) ⇒ θ ∗ = π/2 (5.247). see Problem 1. (5. (1. Bando n. we obtain a minimum opening angle: Δθ = acos 2β 2 − 1 = 1. β= 1− = 0. (5. 13705/2010 Problem 5.94.857.9◦ (5. By using Eq. compute the di-photon mass√resolution by using an electromagnetic calorimeter with energy resolution of 3%/ E and angular resolution of 5 mrad. The γ and β factors of the boost to the laboratory frame are given respectively by: . The invariant mass of the photon pair is given by: Δθ mγ γ = 2 E 1 E 2 (1 − cos Δθ ) = 2 E 1 E 2 sin (5. γ + γ 2 − 1 ≈ [16. which gives the same energy and polar angle in the laboratory to both photons. Compute the range of photon energies in the laboratory frame and the minimum opening angle between the two photons.109): ! " E 1. we estimate the relative uncertainty δm γ γ /m γ γ on the photon-pair mass as: δm γ γ 1 δ E1 1 δ E2 δ sin( Δθ 2 ) 1 σE ∗ δ(Δθ) = ⊕ ⊕ = √ ⊕ mγ γ 2 E1 2 E2 sin( Δθ 2 ) 2 E∗ 2 tan( Δθ 2 ) 1 1 σE ∗ σθ1 ⊕ σθ2 1 0.33. Pollard. (Springer. B 374.1016/0370- 2693(96)00300-0 7. 319–330 (1996). Cambridge. G. C. Phys.77.03 2 2 · (5 × 10−3 )2 2 = √ ⊕ = √ + = 0. see Problem 2. Di Giacomo. 2nd edn. C. A. 77. 242 (1960). 1993) 6. McGrew et al. Rev.M.R. 1992) 3. while the energy measurement contributes by 0. Techniques for Nuclear and Particle Physics Experiments. Paffuti.73).4 Higgs Boson 355 By using the linear error propagation of Eq. The combined LHC Run 1 measurement of the Higgs boson mass is mostly determined by the measurements in the H → γ γ and H → 4 channels.59%). Symmetry principles.N. Phys. Phys.59.5.1103/PhysRevD.27%.org/10.R. 052004 4.07221 [hep-ex] 5. 1980) . P. W. 052004 (1999). Yang. Rev. so one should consider the 3%/ E uncertainty as a lower bound. D 59. Besides. https://doi.org/10. References 1. Discussion The angular resolution in the di-photon mass arises from the finite size of the calorimetric cells and from uncertainty on the vertex position in the presence of multiple pile-up events. arXiv:1705.242 2.54 (5. Super-Kamiokande Collaboration (2017). Lett. Particle Physics (Cambridge University Press. [54] for a discussion on the measurement of the Higgs boson mass at the LHC. Leo. at the LHC experiments the energy resolution at E ∼ 60 GeV receives non-negligible contributions√ from the noise and uniform terms. Rossi. Suggested Readings See Ref. Problemi di Fisica teorica (Edizioni ETS. B. Berlin.65% 2 E∗ 2 tan( Δθ ) 2 2 60 4 · tan2 0. each contributing by a similar amount to the final accuracy.org/10. (4. https://doi..250) Notice that the angular measurement gives the dominant contribution to the mass uncertainty (0.1103/PhysRev. The ALEPH Collaboration. W. Gibson. https://doi. Phys. Boulder. 298 (1977). Phys.org/10. Cahn.W. 189–201 (2004). Phys. Parisi. Belle Collaboration. 2nd edn. The Experimental Foundations of Particle Physics. 2000) 9.L. Buttazzo et al.1016/0550- 3213(95)00208-A 37.K. 211802 (2007). 85.org/10. Nucl..1103/ RevModPhys. https://doi. H. Mod. Quantum Chromodynamics: High Energy Exper- iments and Theory (Oxford University Press. Rev. Phys. Dissertori.org/10. eConf C990809:157-173. 103 (1983). Peskin. https://doi. Phys.013 29.org/10.physrep. G. https://doi. 81–84 (2013).1016/0550- 3213(77)90384-4 32. Muller et al. Lett. Phys.V. R. Barbieri. Rev. CMS Collaboration and LHCb Collaboration. 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Dawson. 345 Cabibbo-suppressed decay. 351– 235. 310. 79 95. 215 Altarelli-Parisi splitting function. 153. 306. 345 Central Limit theorem. 98. UNITEXT for Physics. 2 Bremsstrahlung. 134. 99. 31. 188. 287. 127. 144. 102. 342–344. 56. 230. 145 Charge-conjugation operation. Bhabba scattering. 85. 298. 36. Centre-of-mass frame. 80. Bianchini. 242. 40. 297 Cabibbo-Kobayashi-Maskawa matrix. 152. 294 Callan-Gross relation. 217. 254 Binomial density. 256 Bending power. 161. 2. 311 79. 125. 170. 211 Cathode. 41. 209 Amplitude function. 91. 111. Selected Exercises in Particle and Nuclear Physics. 107. 311 Beam rigidity. 134 40. 73. 57. 26. 343 Bolometers. 49–51. 92. 123. 314. 79. 299. 1. 275. 205 Casimir. 8. 321 Beam tune. 332–336. 278. 35. 188. 260. 87. Bjorken fraction. 141. 75 285. Betatron. 355 353 AC-coupling. 55. 191 BEH mechanism. 10. 82. 172 Breit–Wigner. 69. 230. 326 Brightness. 264. 154. 187. 191 Branching ratio. 128. 234–236 ATLAS. 324 © Springer International Publishing AG 2018 359 L. 123. 12. 302. 193. 212 302. 171 Avogadro number. 109. 301. 7. 124. 37. 30. 215 Bucket. 356 Bevatron.Index A Bose–Einstein symmetry. 27. Acceptance (of a detector). 263. 198. 235 87. 340 BaBar. https://doi. 259. 116. 13 Bragg peak. 62. 58. 286. 89. 199 75. 260 Charge-conjugation and parity operation. 120 Anode. Boltzman constant. 87. 98. Bethe–Hadler. 214. 210 Bubble chamber. Boost. 331–334. 308. 269. 280 Characteristic function. 259. 81 70. 250 Belle. B 304. 67. 127. Centre-of-mass energy. 309.1007/978-3-319-70494-4 . 27. 22. 318. 254. 323. 278. 182. 50. 197. 339 Baryon number. 284 Aberration of light. 168. 338 Centre-of-mass momentum. 305. 229 Antiproton. 302 Binding energy. 354 Bethe formula. 196. 276. 80 Bunch crossing. 74. 169–171. 148. 190. 119. 9–11. 339. 108 C Cabibbo angle. 179. 87 Active transformation. 11. 170. 172. 75. 209. 355 Avalanche. 346 Cauchy density. 47. 162. 253. 229.org/10. 248. 76. 284. 231. 118–121. 17–22. 257. 252. 171. 124 Accelerated mass spectroscopy. 12. 323 Charge division. 155. 329. 288. 197. 168 Confidence interval. 144. 342 310. 12. 107. 131–134. 43. 251. 65. 36 119. 108. 302. 10. 20. 32. 93. 159. 64. 264. 147. 154 Descartes’ rule. 119. 352–354 Energy transfer. 135. 110. 82–93. 102. 211 Event shape variables. 158. 151–153. 116. 277 Dinkin index. 303. 332. 305. 228 Doppler broadening. 154. 123 Efficiency of a detector. 266–269. 142 Detailed balance. 152. 36. 295. 112. 301 Diethorn formula. 347 . 110. 247. 79. 278. 113. 333. 152.360 Index Charged-current interaction. 283. 200. 299. 301. 166. 235. Emittance. 198 Compton scattering. 3. 52. 143 Density correction. 222. 218. 193 355 Dokshitzer-Gribov-Lipatov-Altarelli-Parisi Cockroft-Walton. 336. Deep inelastic scattering. 244. 69 Exponential density. 311. 308. 267–271. 300. 148. 355. 95–103. 128 324. 121. 149. 102. 100 Crystal. 12 Compton peak. 146 DC-coupling. 81–83 Electric dipole. C P-violation. 229. 116. 67. 131– 307. 211–213. 45. 121. 37. 183. 293 Dalitz decay. 232–234. 260 Error propagation. 186. 90. 107. 149. 140. 201 equation. 276. 300. 112. 107. 134. 29. 79. 161. 306. 185 Cyclotron frequency. 78 Cherenkov threshold. 147 Crossing angle. Drift velocity. 177. 215 δ-ray. 339 134. 283. 289. 74. 185. 292. 300 F Dalitz plot. 301. 185. 209. 196 Electromagnetic. 337. 151. 278. 39. 168. 144. 332 Cosmic muons. 153 E Continuous slowing-down approximation. 306. Dip angle. 307. Decay width. Depletion region. 94 96. 229. 161–163. 287. 313. 149. 114. 287. 275. 156 Doppler effect. 287. 81. 280 Critical energy. 37. 65. 161 Dead time. 139. 355 216. 210. 262. 279 Chirality. 62. 352 Constant term. 232. 98. 163 Delta function. 306. 297–299. 137. 357 Cyclotron. C P T -invariance. 96. 188. 146. 111. 243 Electron capture. 327. 135. 252 D D0. 111 Cherenkov angle. 12. 342. 151. 136. 166. 78. 260. 233 Fake-rate. 349–351 292. 323. 314 CMS. 8. 347–349. 305. 281. 151. 351. 310. 136. 61 Factorisation scale. 158. 55. 336 310. 103. 152. 206. 233. 153. 47. 92. 156. 299. 159. 353 Confidence level. 145. Edge. 310. 147. 281. 167 258. 135. 63. 113. 228. 117 Drift tube. 335 276. 225–227. 289 Euler-Lagrange equation. 139. 165 138. 146. 258 Fermi constant. 288. 300. 91 Data acquisition. 312. 134. 300 Deuterium. 154 Electron scattering. 152. 110 Cherenkov radiation. 223 Electroweak interaction. 291. 172 Fano factor. 116. 170 Clebsh–Gordan. 285. 263. 48. 111. 304. 230. 183. 232. 294. 134. 268 Cosmic microwave background. 169. 39. Energy loss by collision. 76. Coulomb potential. Cross section. 202. 86 Chiral anomaly. 287. 64 Charge stripper. 40. 135. 299. 309. 128–130. 356 Cramer-Rao inequality. 92 Compensation. Energy gap. 117. 45. 283–285. 312. 185 157 Dynode. 330 Collision loss. 77. 140. 312 K -short. 140. 102. 80 Interaction length. 33. 341– 343 F O D O cell. 210. 303. 355–357 Helicity. 293 Ideal gas. 84. Higgsstrahlung. 209. 275. 204. 48. 84. 205. 190. 108. 12. 124. 347– Least squares estimator. 200. 254. 218 Linear accelerator. 79. 204. 331. 131. 298–300. 353. 127. 26. 200. 204. 29. 117. 147. 146. 34. 281. 25 Frascati Laboratories. 231 Fine-tuning. Fermi theory. Gauge boson. 93 Lifetime. 278. 296. 295. 307. 347 81. 71. Fixed target. 94. 51. 149. 96. 286. 43. 339. 54 Leakage current. 114. 242. 352 Infrared-safe. 286 K -long. Klein–Gordan equation. 284. 245 247. 166. 210. 283. 288. 333 Flavour chaning neutral current. 301. 99. 241–243. 279 Interaction point. 182 I LHCb. Heaviside function. 245. 161. Isospin. 61. 115 342. 217. 266 K G Kamiokande. 111. 297. 156. 41. 128 Flavour. 160. 206 J Form factor. Implicitely defined estimator. 296. 156 Landau pole. 44. 355 Information matrix. 283–289. 72 Limited streamer tube. 82. 109. 134. 345 331. 152. 139. 255 127. 167 Higgs boson. 307 Lepton number. 157. 152. 201. 290. 299 157. 228. 335 Grand Unified Theory. 275 331 Klein–Nishina formula. Laser plasma acceleration. 275. 292. 87–89. 166. 2 Jacobian peak. 301 JADE algorithm. 203 . 235 79. 136. 202 353. 125 KEK. 291. 157. 188. 278.Index 361 Fermi energy. 304 Hourglass effect. 301. 216. 56. 64 H L Hadronic calorimeter. 72 164. 86. 352. 291. 258 Helicity-suppressed decay. 116 G-parity. 259. 119. 8–10. 282. 44 Jacobian factor. 276. 227. 181. 292. 290. 149. 92. 21 Four-momentum. 206. 158. 147. 150. 178. 74–77. 354 289. 5 222. 144. 302. 171 Integrated luminosity. 65. 246. 231 Ionisation potential. 2 Kinetic energy. 251–253. 295. 52. 236. 344 Lever arm. 190. Gaussian density. Index of refraction. 342. 331. 214 304 Law of Large Numbers. 32. 140. 201. 37. 264 357 Left-handed particle. 92. 53. 343 Gamma factor. 96. 7 Impact parameter. 243 Likelihood function. 49. 335 Greisen–Zatsepin–Kuzminpin limit. 12. 216. 8–10. 294. 333. Jet algorithm. 98. 302. 278. 295 LEP. 167 Kinoshita–Lee–Nauenberg theorem. 122. 296. 69. 226–230. 99. 101 Lieanard formula. 278 Gain (of a detector). 314. 225. 313 Hadronic cascade. Heaviside-Lorentz units. 224. 34. 329. 12. 80. 179. 43. 150. 72–74 Full width at half maximum. 155 Large Hadron Collider. 346. 157–160. 215 Germanium. 210 Left-right asymmetry. 23. 63. 234. 82 Kurie plot. 121. 97. 189– 191. 51. 291. 69 Glashow-Iliopoulos-Maiani mechanism. 281 Hypercharge. 76. 54–60. 276. 189 Maximum energy transfer. 356 Lyapunov condition. 236. 114. 308 Pair-production. 89–91. 152 Primakoff effect. 266. 103 Pre-amplifier. 152. 219–222. 115 NA-7. 109. 177. 345 Neutral-current interaction. 74. 311 126–128. 90 Poisson density. 131–133. Okubo-Zweig-Iizuka rule. 294. 146. 270 Natural units. 162. 77. 249. 264 64. 139. 54–56. 167. 2. 298. 236. 25. 282. 6. 164–166. 52. 65 Photodetector. 224 N Pixel. 39–41. 77 172 PIBETA. 169. 97. 115. 354. 1. 264. 217. 25. 79. 4. 98 Paralyzable. 161. 63–67. 246–248. 77. 208. 357 Profile likelihood ratio. 13. 345 Particle identification. 3 Pentaquark. Natural width. 247 Non-paralyzable. 138. 230. 210. 149. 59. 40 O Luminosity. Phase-space measure. Parity operation. 164. 159–163. 208. 349 Monte Carlo. 195. 303.362 Index Liouville theorem. 167. 164. 9. 7. 68. 23. 169. 242. 1. 7. 85 Lorentz angle. 209 Lorentz group. 27. 125. 234– Opening angle. 172 Noble-gas detector. 327. 167 Multiwire proportional chambers. 115–117. 138– Mellin moments. 6 . 4 Penguin-diagram. 355. 4. 161. 84. 172. 160. 38 resonance. 300 Plank scale. 147. 184 Parton density function. 284. 3 Pontecorvo-Maki-Nakagawa-Sakata matrix. 336. 167 171. 111 88. 263. 187 Piwinski angle. 91. 45. 84. 153. 355 79. 51. 232. 233 Proper time. 169. 134. 98. 194. 92 344. 108. 293 Mobility. 58. 283. 163. 357 Pitch. 156– Multiple scattering. 73. 255–257. 152. 16. 166. 1. 5. 305. Mandelstam variable. 331 Positron emission tomography-magnetic Neutron moderation. 104. 4. 258. 313 NA-48. 190. Primary vertex. 110 Particle flow. 114 Mott scattering. 185 Normalised emittance. 178. 168 Newton’s method. 251. 112. 282 M P Magnetic dipole. 2. 8–10 Nucleosynthesis. 283 242–244. 85. 114. 286 216. 163. 206. 177 Passive transformation. 107. Parapositronium. 242 Ortopositronium. 232. 210. 224–231. 14–20. Minimum ionising particle. 308. 115. 95. 45. 89. 270 334 Mean excitation potential. 198 Photomultiplier. 167. 181. 110. 91. 172. 131–134. 55 157. 169 NA-30. 308 Lorentz transformation. 300 Noise. 241. 44. 264. 32. Photon conversion. 72. 55. 258. 166. 1 Narrow-width approximation. 278. 163. 279 Lorentz invariant. 116. 185 Phase-space. 125. 24. 263. 154. 330 140 Microstrip. 355 Pile-up. 109. 111. 147. 337 Plasma frequency. 355. 63. 193. 214. 301 Muon collider. 159 180. 343 MKS units. 2 Minkowski norm. 140 MEG. 293. 55. 96 Photoelectron. 163. 209 Non-relativistic normalisation. 33. Molier radius. 307. 258 Maximum likelihood estimator. 85 Mössbauer effect. 267. 198. 19. 45 Poincaré group. 165 Thrust. 89. 128. 119–122. 139. 93 Synchrotron radiation. 173. 145 Radiofrequency. 307 Time-pickoff methods. 173. 216 Relativistic normalisation. 291. 93–95 192. 15. 196. 354 R Standard pressure and temperature. 122 331. 172. 191–193. 319. 48. 64. 179. 22–25. 45. 227. 92. 127. 173 Rotation matrix. 23. 343 37. 22–25. 164. 9. 314 Tandem accelerator. Spectrometer. Trigger. Rutherford scattering. 86. 190. 51. 163– Q 169. 210. 311 Rapidity. 279 Silicon photomultiplier. 299. 289. 27 Scintillator. 191. 202. 345 Reverse bias. 293 Resonance. 146. 185 Shell correction. 63 189. 160 234 . 140. 63. 13 Transition radiation. 206. 295. 191 134. 154 SLAC. 141. 69 Quantum electrodynamic. 195. 355 Regeneration (of K-short). 153 Radioactive decay. 306. 214. 168 Quality factor. Radiocarbon dating. 338 Relative velocity. 332 213. 72 Revolution frequency. 36. Strong CP-problem. 130. 185 Quadrupole. 312. 88. 291. 167. 276. 309. 124. 127. 215 Resistive plate chamber. 301 Stochastic term. 342–344. 193. 87. 344. 58. 319. 115. 196. 303 Standard model of particle interactions. 114. 345 Time to digital converter. 286. 322 Spinor. 280. 116 354 Three-body decay. 202. 294 Recoil mass. 173. 118 Silicon. 179. 61. 97. Radiation length. 10 Time walk. 139. 201. 48. 194. 125. 6. 69. 355 Reduced mass. 314 Tracking. 45. 124. 212. 128. 177–180. 165. 173. 353 Supersymmetry. 168. 141. 324. 332. 72. 154. 144. 206 Range of a particle in matter. 37 Triangular function. 51. 72. 236 Transverse momentum. 195. 293. 324 Renormalisation group equation. 352. 113. 119 Structure function. 111. 278. 109. 255 Radiative decay. 10. 150 Stopping power. 29 Synchrotron. Time dependent oscillations. 285. 67. 83–86. 67 91. 205–207. 167. 114. 152. 104. 12. 54. 166. 114 Sphericity. 110. 115. 174 Right-handed particle. 279. 170 153 Stirling formula. 347. 211–213. 353 Scattering. θ-τ puzzle. 285. 98. 151. 341 221. 188. 170. Sargent rule. 205. 174 Running coupling constant. 219. Q-value. 8–10. 345 Rectangular density. 158. 48 Sagitta. 229–231 Time-of-flight. 317. 229 Strong focusing. 117. Time projection chamber. 166 Pulse-shape analysis. 205. 131. 313 T Renormalisation scale. 140 Semiconductor. 87. 117. 304. 171 Tetraquark. 177–182. 162. 284. 275. 111–113. 295. 331. 54. 8. 139. 45–51 Sum rules. 198 Transformation of velocities. 343 Quantum chromodynamic. 111. 195.Index 363 Pseudorapidity. 161. 25. 89. 290. 312. 161. 65. 61. 205. 232– 156. 292. 67. 323 200 Quantum efficiency. 12. 290. 302 Quark model. 213. 161. 314. 138 S Transverse mass. 138. 188 Secondary vertex. 310. Thomas cross section. 182. 343 ρ-parameter. 52. 310. 135. 11. 157–159. 74. 346. 167. 30. 104. 347 . Velocity selector. 21. 354 Two-body decay. 340. 340 V Vacuum expectation value. 285. 96 345 V -spin. 90. 347 Y Van de Graaf. 264. 258 85–87. 201 Yukawa’s interaction. 270 Wolfenstein’s parametrisation.364 Index Turn-around time. 10. 288 U W Unitarity triangle. Weighted Monte Carlo events. 79. 290. 246. 288 Wilks’ theorem. 25. 228 Vector-boson fusion. 55. 304. 349. 214 Unpolarised particle. 341 Wakefield. 29. 30. 331. 11. 277 U -spin. 354 Wigner–Eckart theorem. 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