Chapter 1 — Summary, Toolkit and Problems

Part I ★ Summary & Problems Bettini pp. 63–69 · ~17 min read

  • two-body kinematics toolkit
  • Mandelstam invariants in practice
  • chapter summary

The chapter gave two ways to answer any kinematics question — boost everything, or compute an invariant. Thirty-eight problems are where the second stops being a preference and becomes the only practical route.

🎯 Why this matters

Three printed results in this chapter do not survive being redone. That is not a complaint about the book; it is why the solutions here are worked rather than quoted, and why any number you intend to rely on is worth recomputing.

Chapter 1 was the toolbox. This page is where you find out whether you can use it: the book’s own summary, the two-body kinematics formulary it gives you before the problems, and all 38 problems with worked solutions.

What you were supposed to take away

The book’s summary, expanded into what each item actually means and where it was built:

The chapter's summary (p. 69), with the section that owns each item
You should now haveWhat that means in practice§
the Lorentz transformations from general properties of space-timeDerived from five structural axioms — homogeneity, isotropy, relativity, the group property, causality — with light entering only at the end to fix c. The point is that this constrains ALL interactions, not just electromagnetism.1.1
mass, energy and momentum, their transformation properties, and the invariantsm² = E² − p² defines mass as the norm of a 4-vector; "relativistic mass" does not exist; the mass of a system is not the sum of its parts.1.2–1.5
the L and CM framesFixed target versus collider, and the quadratic penalty that decides which machine gets built.1.4–1.5, 1.12
SI and natural unitsħ = c = 1, and the ability to put the factors back. ħc = 197.3 MeV fm is the constant you will use most.1.6–1.7
cross-section, luminosity, decay rates, branching ratios, phase spaceσ is an effective area; Lσ is a rate; Γτ = ħ; the golden rule splits any rate into a matrix element and a phase-space volume.1.6–1.7
the basic aspects of a scattering experimentScattering is a Fourier transform: resolution costs momentum transfer, which costs beam energy. Form factors, Rutherford, Mott.1.8
the names of the particle types and of the four interactionsThe census: quarks, leptons, gauge bosons, hadrons; and why gravity never appears again.1.9–1.10
how charged particles and photons lose energy and are detectedBethe–Bloch and its 1/β² rise, bremsstrahlung and X₀, critical energy, λ₀, showers.1.11
the sources: cosmic rays, accelerators, collidersp = 0.3BR sizes every machine ever built; phase stability and stochastic cooling are what make them work.1.12
the basic detector types, tracking and calorimetryEleven instruments, each exploiting one of the energy-loss mechanisms above — and each dying, eventually, of rate.1.13a–d

If any row reads as unfamiliar, the problems below will find it — the topic filter maps onto these rows.

The two-body toolkit

Before the problems the book works out, once and for all, every kinematic quantity of a generic two-body scattering a+bc+da + b \to c + d in terms of the invariants. These are equations (P1.1)–(P1.15), and they are worth having at hand rather than re-deriving: most of the 38 problems are one substitution into one of them.

Everything rests on two facts from §1.4–1.5: ss, tt and uu are the same in every frame, and each of them can be written either from the initial state or from the final state.

Before the algebra, the picture the formulary is about — and the labelling the book uses without ever drawing it:

laboratory framea · E_a, p_ab at restcdθ_ac(P1.11)–(P1.15) live here:p_b = 0 makes t and u linear in E_d, E_ccentre-of-mass framea · E*_a, p*b · E*_b, p*cdθ*_ac(P1.2)–(P1.7), (P1.9)–(P1.10) live here:back to back, so one momentum p* serves all fours = (p_a + p_b)² · t = (p_a − p_c)² · u = (p_a − p_d)²the same three numbers in both panels — which is why you can compute them on the right and use them on the left

Draw this before starting any problem. The book’s solution to Problem 1.15 opens by saying so, and it is the single most useful habit in the chapter: nearly every sign error in relativistic kinematics is a mislabelled angle.

s=(Ea+mb)2pa2=ma2+mb2+2EambEa=sma2mb22mb\htmlClass{t-s}{s} = \left(\htmlClass{t-Ea}{E_a}+\htmlClass{t-mb}{m_b}\right)^2 - \htmlClass{t-pa}{p_a}^2 = \htmlClass{t-ma}{m_a}^2 + \htmlClass{t-mb}{m_b}^2 + 2\,\htmlClass{t-Ea}{E_a}\htmlClass{t-mb}{m_b} \quad\Longrightarrow\quad \htmlClass{t-Ea}{E_a} = \frac{\htmlClass{t-s}{s} - \htmlClass{t-ma}{m_a}^2 - \htmlClass{t-mb}{m_b}^2}{2\,\htmlClass{t-mb}{m_b}}
(P1.1)

Read left to right for the CM energy of a fixed-target machine; read the arrow for the beam energy you need to reach a given s. Both directions get used constantly.

Every symbol, one at a time

Hover or tap a symbol above — it lights up in the equation and its meaning, units and type appear here.

The formulary — (P1.2) to (P1.15), for a + b → c + d
Eq.ResultWhat it is for
P1.2–P1.3E_b = (s + m_b² − m_a²)/2√s ; E_a = (s + m_a² − m_b²)/2√sCM energies of the two incoming particles. Note they sum to √s, as they must.
P1.4p_a = p_b = √(E*² − m²)The common CM momentum of the initial state — equal and opposite by definition of the frame.
P1.5–P1.7E_c = (s + m_c² − m_d²)/2√s ; E_d = (s + m_d² − m_c²)/2√s ; p_c = p_d
P1.8t = m_c² + m_a² + 2p_a p_c cos θ_ac − 2E_a E_c = m_d² + m_b² + 2p_b p_d cos θ_bd − 2E_b E_dThe momentum transfer written two ways — through the a→c vertex or the b→d vertex. Both are the same invariant.
P1.9–P1.10cos θ_ac = (t − m_a² − m_c² + 2E_a E_c)/(2p_a p*_c)The scattering angle, extracted from t. See the warning below about how P1.9 is printed.
P1.11–P1.12in L (p_b = 0): t = m_b² + m_d² − 2m_b E_d ⇒ E_d = (m_b² + m_d² − t)/2m_bThe recoil energy of the target from the momentum transfer alone. This is the elastic-scattering workhorse of §1.8.
P1.13E_c = m_b + E_a − E_d = (s + t − m_a² − m_d²)/2m_bThe scattered particle's lab energy. Used in Problem 1.17 to get the opening angle without a single Lorentz transformation.
P1.14–P1.15u = m_d² + m_a² + 2p_a p_d cos θ_ad − 2E_a E_d ; in L, u = m_b² + m_c² − 2m_b E_cThe third invariant, and its simple lab form. With P1.13 it gives s + t + u = Σm² immediately.

Six of these are the same formula with the labels permuted — which is the point. <strong>Learn the pattern, not the fourteen equations.</strong> Every energy is (s ± m² ∓ m²)/2√s in the CM, or (something − t)/2m_b in the lab, and every momentum is √(E² − m²).

⚠️ Equation (P1.9) as printed

The book prints

cosθac=tma2mc2+2mamc2papc,\cos\theta^*_{ac} = \frac{t - m_a^2 - m_c^2 + 2\,m_a m_c}{2\,p^*_a p^*_c},

with masses in the last term. Its neighbour (P1.10) has energies in the same slot, and so does the equation it is derived from. Solving (P1.8) for the cosine gives

2papccosθac=tma2mc2+2EaEccosθac=tma2mc2+2EaEc2papc,2 p_a p_c \cos\theta_{ac} = t - m_a^2 - m_c^2 + 2E_aE_c \quad\Longrightarrow\quad \cos\theta^*_{ac} = \frac{t - m_a^2 - m_c^2 + 2E^*_aE^*_c}{2\,p^*_a p^*_c},

so the printed 2mamc2m_am_c should be 2EaEc2E^*_aE^*_c. The two coincide only when both particles are at rest, which is exactly the case where the angle is undefined.

Nothing later depends on it — Problem 1.17’s solution uses the correct relation in the equivalent form t=2mp2+2p2cosθ2E2t = 2m_p^2 + 2p^{*2}\cos\theta^* - 2E^{*2} — but the formula as printed will not reproduce that problem’s answer.

⚙️ Engineer’s bridge — why invariants beat transformations

Problem 1.17 is deliberately solved twice: once by boosting every 4-vector into the CM and back, and once by computing ss and tt and substituting. The second route is shorter, and the reason will be familiar.

A Lorentz transformation is a change of basis. Carrying a problem through one means tracking every component of every vector through a matrix multiply, and every component is an opportunity to drop a sign. The invariants are basis-independent scalars — the checksums of the problem. You compute them once, in whichever frame is easiest, and they are then valid everywhere.

It is the same instinct as working with a norm, a determinant or a trace instead of the matrix entries: if the answer you want is a scalar, find a scalar route to it. And when you do need both routes, they are a free cross-check — which is precisely how the book uses them here.

Where it breaks: an invariant route exists only when the answer is a scalar, and a great deal of what this book measures is not. Anything angular — the 1+cos2θ1+\cos^2\theta distribution of §5.7, a Dalitz plot, a forward–backward asymmetry — is frame-dependent by construction, so you must name the frame and do the transformation. The rule is therefore narrower than it sounds: prefer invariants for quantities, and expect to work in a specified frame the moment the question is about a direction.

Reproduce it

import numpy as np

mp = 0.93827209                                   # GeV
p1, th = 3.0, np.radians(10.0)                    # Problem 1.17
E1 = np.hypot(p1, mp)
s  = (E1 + mp)**2 - p1**2
print(f"P1.1 check: s = {s:.4f} GeV^2 from (E_a+m_b)^2-p_a^2, "
      f"and E_a back out = {(s - 2*mp**2)/(2*mp):.4f} GeV")

rs   = np.sqrt(s)
Ecm  = rs/2                                       # all four particles equal mass
pcm  = np.sqrt(Ecm**2 - mp**2)
t    = 2*pcm**2*(np.cos(th) - 1)                  # elastic, equal masses

bad  = (t - 2*mp**2 + 2*mp*mp)     /(2*pcm*pcm)   # (P1.9) as printed
good = (t - 2*mp**2 + 2*Ecm*Ecm)   /(2*pcm*pcm)   # with energies
print(f"P1.9 as printed (masses)  : cos theta* = {bad:+.4f} -> {np.degrees(np.arccos(bad)):.1f} deg")
print(f"P1.9 with energies (right) : cos theta* = {good:+.4f} -> "
      f"{np.degrees(np.arccos(good)):.1f} deg   <- the input angle")

u = 4*mp**2 - s - t                               # from the identity, then verify
print(f"s + t + u = {s+t+u:.4f} GeV^2 ;  sum of m^2 = {4*mp**2:.4f} GeV^2"
      f"   -> identity holds")
prints
P1.1 check: s = 7.6593 GeV^2 from (E_a+m_b)^2-p_a^2, and E_a back out = 3.1433 GeV
P1.9 as printed (masses)  : cos theta* = -0.0152 -> 90.9 deg
P1.9 with energies (right) : cos theta* = +0.9848 -> 10.0 deg   <- the input angle
s + t + u = 3.5214 GeV^2 ;  sum of m^2 = 3.5214 GeV^2   -> identity holds
t = 0u = 0u > 0 — unphysicalt (GeV²)u (GeV²)−4−200−2−4θ* = 180° (backscatter)θ* = 0 (forward)θ* = 90°: t = u = −2.07Problem 1.17, θ* = 10°t = −0.031, u = −4.106Elastic pp atp = 3 GeV/cs = 7.659 GeV²Σm² = 3.521 GeV²t + u = −4.138= Σm² − sThe line is fixedby the beam energy.The angle picks thepoint on it.Two numbers, andthe event is fullydetermined.

s+t+u=m2s+t+u=\sum m^2 is not a formula to memorise, it is a constraint: it collapses three invariants to two. Fix ss and every possible event of this reaction lies somewhere on one straight line; the requirement that both tt and uu be negative cuts that line down to the coloured segment, whose two endpoints are forward and backward scattering. The elastic peak everyone measures sits at the crowded right-hand end — Problem 1.17’s 10° is already almost on top of the forward endpoint.

The problems

All 38, with a hint before the solution. The topic chips filter them; the progress bar and the “solved” ticks are stored in your browser.

📝 Chapter 1 problems

0/38 solved
  1. 1.1unitstheory
    Estimate the kinetic energy of a Boeing 747 at cruising speed and compare it with the energy released by a mosquito annihilating with an antimosquito.
    • M=400M = 400 t
    • v=850v = 850 km/h
    • mosquito mass 2.5\approx 2.5 mg (assumed — the book leaves it to you)
  2. 1.2kinematicstheory
    Three protons have momenta equal in magnitude and directions at 120°120° from one another. What is the mass of the system?
    • p=3|\mathbf p| = 3 GeV each
  3. 1.3unitstheory
    Compute the widths of the weak decays of the π±\pi^\pm, K±K^\pm and Λ\Lambda.
    • τπ=26\tau_\pi = 26 ns
    • τK=12\tau_K = 12 ns
    • τΛ=0.26\tau_\Lambda = 0.26 ns
  4. 1.4unitstheory
    Compute the lifetimes of the strongly decaying ρ\rho, ω\omega, ϕ\phi, KK^*, J/ψJ/\psi and Δ\Delta from their total widths.
    • Γρ=149\Gamma_\rho = 149 MeV
    • Γω=8.5\Gamma_\omega = 8.5 MeV
    • Γϕ=4.3\Gamma_\phi = 4.3 MeV
    • ΓK=51\Gamma_{K^*} = 51 MeV
    • ΓJ/ψ=93\Gamma_{J/\psi} = 93 keV
    • ΓΔ=118\Gamma_\Delta = 118 MeV
  5. 1.5scatteringtheory
    An accelerator produces a 20 GeV electron beam. Electrons scattered at θ=6°\theta = 6° are detected. Neglecting the recoil, what is the smallest structure in the proton that can be resolved?
    • E=20E = 20 GeV
    • θ=6°\theta = 6°
  6. 1.6kinematicstheory
    In a pppp collision the final state contains a particle of mass mm besides the two protons. (a) Give the threshold energy EpE_p and momentum ppp_p for a proton target at rest. (b) Give EpE_p^* and ppp_p^* for two protons colliding with equal and opposite velocities. (c) Evaluate both for a produced pion, and give the kinetic energy in case (a).
  7. 1.7kinematicstheory
    Consider γ+pp+π0\gamma + p \to p + \pi^0 on a proton at rest. (a) Find the threshold photon energy. (b) The Universe is filled with 3 K background radiation of photon energy Eγ1E_\gamma \approx 1 meV; find the minimum energy a cosmic-ray proton needs to induce π0\pi^0 photoproduction on it. (c) With σ=0.6\sigma = 0.6 mb just above threshold and a background photon density ρ108 m3\rho \approx 10^8\ \text{m}^{-3}, find the attenuation length.
  8. 1.8kinematicstheory
    The Universe is opaque to photons energetic enough that γγe+e\gamma\gamma \to e^+e^- can occur on a background photon. Compute the threshold energy against (a) the 3 K microwave background at Eγ1E_\gamma \approx 1 meV and (b) the extragalactic background light, taking λ=1 μ\lambda = 1\ \mum.
  9. 1.9kinematicstheory
    The Bevatron was designed to have enough energy to produce antiprotons. What is the minimum proton beam energy? Baryon number conservation forces the reaction p+pp+p+pˉ+pp + p \to p + p + \bar p + p.
  10. 1.10kinematicstheory
    At the LHC two proton beams of Ep=7E_p = 7 TeV collide head on. What beam energy would give the same centre-of-mass energy on a fixed hydrogen target? How does it compare with cosmic-ray energies?
  11. 1.11kinematicstheory
    A particle of mass MM decays to two bodies of masses m1m_1 and m2m_2. Give the energies and momenta of the products in the CM frame.
  12. 1.12kinematicstheory
    Evaluate the CM energies and momenta of the products of Λpπ\Lambda \to p\pi^- and ΞΛπ\Xi^- \to \Lambda\pi^-.
  13. 1.13kinematicstheory
    Find the energies and momenta of the products of Mm1+m2M \to m_1 + m_2 in the CM when m2=0m_2 = 0.
  14. 1.14kinematicstheory
    In a monochromatic π\pi beam of momentum pπp_\pi some pions decay in flight as πμνμ\pi \to \mu\nu_\mu. We observe that in some cases the muons move backwards. Find the maximum pπp_\pi for which this is possible.
  15. 1.15kinematicstheory
    A Λ\Lambda of momentum pΛ=2p_\Lambda = 2 GeV decays as Λpπ\Lambda \to p\pi^-. In the CM the proton makes an angle θp=30°\theta^*_p = 30° with the Λ\Lambda direction. Find (a) the CM energies and momenta, (b) the Lorentz parameters of the L–CM transformation, (c) the lab energy and momentum of the π\pi, and the lab angle and momentum of the pp.
  16. 1.16kinematicstheory
    A ball collides elastically with an equal ball at rest. Compute the angle between the two final directions at non-relativistic speeds.
  17. 1.17kinematicstheory
    A proton of momentum p1=3p_1 = 3 GeV scatters elastically on a proton at rest; one proton comes out at θac=10°\theta^*_{ac} = 10° in the CM. Find (a) the kinematic quantities in the L frame, (b) those in the CM, (c) the angle between the final protons in L — is it 90°90°?
  18. 1.18detectorstheory
    A charmed meson D0D^0 decays as D0Kπ+D^0 \to K^-\pi^+ at d=3d = 3 mm from its production point. The total energy of the decay products is measured as E=30E = 30 GeV. How long did the DD live in proper time, and what is the π+\pi^+ momentum in the DD rest frame?
    • d=3d = 3 mm
    • E=30E = 30 GeV
    • mD=1.865m_D = 1.865 GeV
  19. 1.19detectorstheory
    A secondary monochromatic π\pi^- beam is produced at a target. At l=20l = 20 m from the target, 10 % of the pions have decayed. Find the momentum and energy of the pions.
    • l=20l = 20 m
    • 10 % decayed
    • cτπ=7.80c\tau_\pi = 7.80 m
  20. 1.20kinematicstheory
    A π\pi^- beam is stopped in liquid hydrogen, where π0\pi^0 are produced by charge exchange, π+pπ0+n\pi^- + p \to \pi^0 + n. Find the energy of the π0\pi^0, the kinetic energy of the neutron, the velocity of the π0\pi^0 and the distance it travels in one lifetime.
  21. 1.21scatteringtheory
    A 2 GeV electron beam hits an iron target (take pure 56^{56}Fe). How large is the maximum four-momentum transfer?
    • E=2E = 2 GeV
    • M56M \approx 56 GeV
  22. 1.22scatteringtheory
    Geiger and Marsden found that alpha particles bounced backwards off a thin foil 'not too infrequently'. Calculate the ratio between the scattering probabilities for θ>90°\theta > 90° and for θ>10°\theta > 10°.
  23. 1.23scatteringtheory
    A 6 MeV alpha beam of intensity Ri=103 s1R_i = 10^3\ \text{s}^{-1} crosses a 1 μ1\ \mum gold foil. Calculate the number of particles per unit time scattered at angles larger than 0.1 rad.
    • Z=79Z = 79, A=197A = 197
    • ρ=1.93×104 kg m3\rho = 1.93\times10^4\ \text{kg m}^{-3}
    • t=1 μt = 1\ \mum
  24. 1.24scatteringtheory
    Electrons of 10 GeV are scattered by protons initially at rest, at 30°30°. Find the maximum energy of the scattered electron.
  25. 1.25scatteringtheory
    If E=20E = 20 GeV electrons scatter elastically and emerge with E=8E' = 8 GeV, find the scattering angle.
  26. 1.26scatteringtheory
    Find the ratio between the Mott and Rutherford cross-sections for the same particles at the same energy at 90°90°.
  27. 1.27sourcestheory
    A particle of mass mm, charge q=1.6×1019q = 1.6\times10^{-19} C and momentum pp moves in a circular orbit at constant speed in a magnetic field B\mathbf B normal to the orbit. Find the relation between mm, pp and BB.
  28. 1.28unitstheory
    To measure the total π+p\pi^+p cross-section at 20 GeV, a 1 m liquid-hydrogen target is placed between two scintillation counters. Normalised to the same incident flux, N0=7.5×105N_0 = 7.5\times10^5 particles are counted with the target empty and NH=6.9×105N_H = 6.9\times10^5 with it full. Find the cross-section and its statistical error.
    • ρ=60 kg m3\rho = 60\ \text{kg m}^{-3}
    • l=1l = 1 m
  29. 1.29detectorstheory
    In the experiment of Chamberlain et al. in which the antiproton was discovered, the antiproton momentum was about p=1.2p = 1.2 GeV. What is the minimum refractive index needed to have the antiprotons above threshold in a Cherenkov counter? How wide is the Cherenkov angle if n=1.5n = 1.5?
  30. 1.30detectorstheory
    Two particles of masses m1m_1 and m2m_2 have the same momentum pp. Evaluate the difference Δt\Delta t in the times they take to cross a distance LL. With two scintillators measuring Δt\Delta t to 300 ps, how long must LL be to separate π\pi from KK at two standard deviations, if their momentum is 4 GeV?
  31. 1.31detectorstheory
    A Cherenkov counter containing nitrogen at pressure Π\Pi sits on a beam of momentum p=20p = 20 GeV. The index depends on the pressure as n1=3×109Π(Pa)n - 1 = 3\times10^{-9}\,\Pi\,(\text{Pa}). The detector must see the π\pi and not the KK. In which range must the pressure be?
  32. 1.32kinematicstheory
    Superman travels down an avenue at high speed. At a crossroads he sees the lights are green and continues, but is stopped by the police, who claim he crossed on red. Assuming both are right, what was Superman's speed?
  33. 1.33detectorstheory
    For the Cherenkov effect in water (n=1.33n = 1.33), determine (1) the minimum velocity for radiating, (2) the minimum kinetic energy for a proton and for a pion, (3) the Cherenkov angle for a pion of energy Eπ=400E_\pi = 400 MeV.
  34. 1.34detectorstheory
    A threshold Cherenkov counter contains N2_2 at a variable pressure, with n=1+3×109πn = 1 + 3\times10^{-9}\pi (pascals). A beam of π+\pi^+, K+K^+ and protons, all of the same momentum, crosses it. Knowing that the π+\pi^+ are above threshold for π5.2×103\pi \ge 5.2\times10^3 Pa, (a) find the momentum, (b) the minimum pressure at which the K+K^+ radiate, (c) the same for the protons.
  35. 1.35sourcestheory
    (1) What is the maximum energy of a cosmic-ray proton that remains confined in the Solar System (R=1013R = 10^{13} m, B=1B = 1 nT)? (2) What is it for the Milky Way (R=1021R = 10^{21} m, B=0.05B = 0.05 nT)?
  36. 1.36kinematicstheory
    Portable neutron generators use d+tn+4d + t \to n + {}^4He with deuterons accelerated to Td=130T_d = 130 keV. (1) Calculate the neutron kinetic energy. (2) With an isotropic production rate I=3×1010 s1I = 3\times10^{10}\ \text{s}^{-1}, what is the neutron flux at R=1R = 1 m? (3) Tagging the neutron by detecting the α\alpha, what time resolution is needed to locate a scattering nucleus to Δz=5\Delta z = 5 cm?
    • md=1875.6m_d = 1875.6, mt=2808.9m_t = 2808.9, mα=3727.4m_\alpha = 3727.4, mn=939.6m_n = 939.6 MeV
  37. 1.37scatteringtheory
    Neutrons of a few MeV are to be detected in a TPC containing 40^{40}Ar. If the energy is low enough the nucleus scatters coherently, as a single object. Taking a nuclear radius RA=4R_A = 4 fm, what is the minimum neutron kinetic energy needed to resolve the nuclear structure, and what is the maximum recoil energy of the nucleus at that limit?
    • RA=4R_A = 4 fm
    • mAr=37.2m_{\text{Ar}} = 37.2 GeV
  38. 1.38kinematicstheory
    Two photons of energies E1>E2E_1 > E_2 collide head on. If γ2\gamma_2 comes from a laser of wavelength λ=694\lambda = 694 nm, what is the minimum E1E_1 to produce an e+ee^+e^- pair? Compute the CM velocity at threshold as 1β1-\beta. What is the mass of the two photons if they move in the same direction?

Further reading

The book’s own list for this chapter, which is unusually good — nine of the fifteen entries are Nobel lectures by the people who built the instruments:

  • Wilson (1925), On the Cloud Method of Making Visible Ions and the Tracks of Ionising Particles§1.13b
  • Hess (1936), Unsolved Problems in Physics: Tasks for the Immediate Future in Cosmic Ray Studies§1.12
  • Blackett (1948), Cloud Chamber Researches in Nuclear Physics and Cosmic Radiation§1.13b
  • Glaser (1960), Elementary Particles and Bubble Chamber§1.13b
  • Alvarez (1968), Recent Developments in Particle Physics§1.13b
  • van der Meer (1984), Stochastic Cooling and the Accumulation of Antiprotons§1.12
  • Okun (1989), The concept of mass, Phys. Today June, 31 — the argument behind §1.4’s refusal to use “relativistic mass”
  • Bonolis (2005), Bruno Touschek vs. machine builders: AdA, the first matter-antimatter collider§1.12
  • Grupen & Shwartz (2008), Particle Detectors, CUP, and Kleinknecht (1998), Detectors for Particle Radiation, CUP — the two standard references behind §1.13a–d
  • Sci. Am. articles on the Tevatron (Lederman 1991), LEP (Meyers & Picasso 1990) and the SLC (Rees 1989)

Erratum — Eq. (P1.9) in the problem-section formulary

The CM scattering angle is printed as

cosθac=tma2mc2+2mamc2papc\cos\theta^*_{ac} = \frac{t - m_a^2 - m_c^2 + 2\,\mathbf{m_a m_c}}{2p^*_a p^*_c}

It should carry energies, 2EaEc2E^*_aE^*_c. Three independent checks agree:

  1. Its own neighbour. Eq. (P1.10), the same expression for the other pair, is printed with 2EbEd2E^*_bE^*_d in exactly that slot.
  2. The equation it comes from. (P1.8) is t=mc2+ma2+2papccosθac2EaEct = m_c^2 + m_a^2 + 2p_ap_c\cos\theta_{ac} - 2E_aE_c; solving it for the cosine returns +2EaEc+2E_aE_c, never a product of masses.
  3. Dimensions and limits. For massless particles the printed form would make cosθ\cos\theta depend on tt alone with no energy scale, which is wrong.

Feed Problem 1.17’s numbers through the printed version and it returns 91° for a configuration that had 10° put into it.

Erratum — two numbers in the book’s own Solutions section

The book solves some of these problems at the back (pp. 506–512). Two of its printed solutions do not survive being redone, so if you check your work against them, check these first. Both corrections are also in the solution notes of the problems above, where they belong; they are collected here because a reader comparing answers needs them before opening the problem, not after.

Solution 1.7(c), p. 506 prints the attenuation length as λ=1/(σρ)=5.6×1022\lambda = 1/(\sigma\rho) = 5.6\times10^{22} m =18= 18 Mpc, “(1 Mpc = 3.1×10213.1\times10^{21} m)”. Two things are wrong and they partly hide each other:

  1. With the problem’s own σ=0.6\sigma = 0.6 mb =6×1032= 6\times10^{-32} m² and ρ=108\rho = 10^8 m⁻³, 1/(σρ)=1.7×10231/(\sigma\rho) = 1.7\times10^{23} m, not 5.6×10225.6\times10^{22}. To get the printed number you need σ1.8\sigma \approx 1.8 mb.
  2. A megaparsec is 3.086×10223.086\times10^{22} m, not 3.1×10213.1\times10^{21} — the printed conversion is off by a factor of ten. It is what turns 5.6×10225.6\times10^{22} m into “18 Mpc”; the correct conversion of that same wrong length would be 1.8 Mpc.

Done consistently, the answer is 1.7×10231.7\times10^{23} m =5.4= 5.4 Mpc. The physics is unchanged — the GZK horizon is a few Mpc either way, which is the point of the problem — but a reader who trusts “1 Mpc = 3.1×10213.1\times10^{21} m” will carry a broken constant into every cosmological estimate afterwards. (The same solution also prints s=1.16s = 1.16 GeV² where it is 1.152, and hence Eγ=149E_\gamma = 149 MeV where it is 145.)

Solution 1.36(1), p. 510 prints pd=2mdTd=2×1875.6×0.13=61.25p_d = \sqrt{2m_dT_d} = \sqrt{2\times1875.6\times0.13} = 61.25 MeV. The radicand is right and the root is not: 487.7=22.1\sqrt{487.7} = 22.1 MeV. 61.25 is what you get from Td=1T_d = 1 MeV, and the problem says 130 keV. The conclusion — that the deuteron momentum is negligible, so the lab frame is the CM frame — is true either way, which is presumably why it survived proofreading.

The same solution writes Tα=(s+mα2mn2)/(2s)mnT_\alpha = (s + m_\alpha^2 - m_n^2)/(2\sqrt s) - \mathbf{m_n} and then correctly subtracts 3727.4, which is mαm_\alpha. The symbol is wrong, the arithmetic is right, and the printed 3.6 MeV is correct.

🔑 If you remember only three things

  • The formulary is the chapter, compressed. Nearly every problem reduces to s and t plus the two-body relations collected on this page — learn the shape of that list, not its entries.

  • Knowing a method is not the same as reaching for it. Everyone can recite that s is invariant; thirty-eight attempts is roughly what it takes before that becomes the first thing you write down.

  • Read the chapter’s own summary last, not first. “What you were supposed to take away” is a checklist, and a checklist is only useful to someone who has already tried and can tell which line they failed.

Where this goes next

Chapter 1 is finished. Chapter 2 stops building tools and starts using them: the discovery of the positron, the muon, the pion and the strange particles, all of it done with the cloud chambers, emulsions and counters of §1.13.

Check yourself — the toolkit, before you use it

0/6 answered · 0 correct

  1. 1.Fourteen formulas P1.2–P1.15 look like a lot to memorise. What is the pattern that makes them one formula?

  2. 2.Problem 1.6 asks for the threshold to make a particle of mass m in pp collisions. Why is the collider answer E* = m_p + m/2 so much better than the fixed-target E_p = m_p + 2m + m²/2m_p?

  3. 3.Problem 1.16 gives exactly 90° between two equal balls after an elastic collision, but Problem 1.17 gives 86° for two protons at 3 GeV. What changed?

  4. 4.Problem 1.7 finds that a cosmic-ray proton above ~7×10¹⁹ eV can photoproduce a pion on a 1 meV microwave-background photon, with an attenuation length of a few Mpc. What does that imply?

  5. 5.Problem 1.20 finds a π⁰ from stopped-pion charge exchange travels 5.3 nm in a lifetime. What is the practical consequence?

  6. 6.Equation (P1.9) is printed with 2m_a m_c where the derivation requires 2E_a E_c. How would you have caught this without an external source?

Study aid derived from A. Bettini, Introduction to Elementary Particle Physics, 3rd ed., Cambridge University Press 2024 — published Open Access under CC-BY-NC 4.0, DOI 10.1017/9781009440745. Not the book: an independently written interactive companion, figures redrawn.