Chapter 2 — Summary and Problems · Elementary Particle Physics

Nine of these problems are the same calculation. A two-body decay hands its daughters one determined momentum, and the rest is a question of which frame you ask in.

🎯 Why this matters

The book has stopped teaching kinematics and started assuming it. From here on these relations are used without comment, so a problem you cannot finish now is a page you will not be able to follow later.

Chapter 1 was a toolbox and its problems were exercises in using it. Chapter 2’s problems are different: most of them are the experiments of the chapter, done with numbers. Anderson’s track, Chamberlain’s counters, the bubble-chamber V⁰, the antiproton threshold — each appears again here as a calculation whose answer is the thing that convinced somebody.

What the chapter established

The chapter summary (p. 101), with where each item was built

You should now have↕	What that means in practice↕	§
the pions and the strange particles, mesons and hyperons, discovered in cosmic rays	Yukawa's range-is-a-mass argument; nine years of mistaking the muon for the pion, ended by Conversi–Pancini–Piccioni; and strangeness, one additive quantum number that explains both pair production and slow decay.	2.1–2.2
the measurements of the quantum numbers of the charged pion	Mass from geometry plus range, lifetime from a delay line, and spin from a reaction compared with its own inverse — three different tricks for reaching something unobservable.	2.3
the charged leptons e, μ and τ, and the three neutrinos	Thomson's null measurement; the τ found by looking for a conservation law being violated; and Reines and Cowan beating a "hopeless" cross-section with flux.	2.4
the Dirac equation and the bilinear covariants	Demanding first order in time forces anticommuting coefficients, hence matrices, hence four components — and spin, g = 2 and antimatter fall out uninvited. The five covariants are a complete basis for 4 × 4 matrices.	2.5
the discovery of the positron, the first antiparticle of a fundamental fermion	A lead plate that converts "which end came first" into "which end is more curved" — a sign ambiguity solved by adding a known asymmetry.	2.6–2.7
the beginning of high-energy accelerator experiments	The Bevatron built to a threshold computed from baryon-number conservation, and a spectrometer with four independent handles on one particle.	2.6–2.7
the discovery of the antiproton, the first antiparticle of a composite fermion	Why the Dirac argument did not settle it: the proton's g is 5.6, not 2, and the neutral neutron has a moment at all.	2.6–2.7
the Majorana equation describing a completely neutral spinor	Helicity and chirality have exactly opposite invariance and conservation properties; and a self-conjugate fermion cannot be massless, because γ⁵ is purely imaginary in the basis a real field requires.	2.8–2.9

Two threads run through all of it. <strong>Experimentally:</strong> every discovery here is a signature designed so that background cannot imitate it — a delayed coincidence, a non-coplanar pair, a plate that breaks a symmetry. <strong>Theoretically:</strong> a structural demand (first order in time) produced more physics than anyone put in.

One idea, used nine times

Nine of the nineteen problems are the same calculation. It is worth extracting once, because after this it will be reflex.

💡 What this really says — the two-body decay, and what a boost does to it

A particle of mass $M$ decaying at rest into $m_1$ and $m_2$ gives its daughters one determined momentum:

p^* = \frac{\sqrt{[M^2-(m_1+m_2)^2][M^2-(m_1-m_2)^2]}}{2M}, \qquad E^*_i = \frac{M^2 + m_i^2 - m_j^2}{2M}.

Now boost it. The daughter’s lab energy depends on the CM emission angle, and the two extremes are forward and backward:

E^{\max/\min} = \gamma\left(E^* \pm \beta p^*\right).

That single line answers Problems 2.2(b) and 2.19, and it has two consequences worth internalising:

A daughter can carry more momentum than the parent. In Problem 2.2 a 5 GeV kaon gives a 5.01 GeV muon, because the neutrino goes backwards and carries negative longitudinal momentum. Nothing is violated; momentum is a vector.
The energy range narrows to a fixed ratio. For an ultra-relativistic parent, $E^{\min}/E^{\max} = (m_\mu/m_\pi)^2 = 57.3\%$ for $\pi \to \mu\nu$ , whatever the beam energy. Problem 2.19 makes a 200 GeV pion beam into a 115–200 GeV muon beam, and the same ratio holds at 20 GeV or 2 TeV.

For a massless daughter pair the algebra collapses further: $4E_1E_2 = M^2$ regardless of the boost, which is Problem 2.3 and the basis of every π⁰ reconstruction in a calorimeter.

⚙️ Engineer’s bridge — the boost is a bilinear map, and it beams

Problem 2.8 asks for something that looks like bookkeeping and is not: an isotropic decay in the CM frame, transformed to the laboratory. The answer,

P(\cos\theta) = \frac{1}{2\pi}\,\gamma^{-2}\,(1 - \beta\cos\theta)^{-2},

is relativistic beaming, and you have met its structure before. Two features are worth naming:

It is a change of variables with a Jacobian, exactly like resampling a probability density. “Isotropic” is a statement about a measure, and a measure does not survive a nonlinear coordinate change unimodified. Anyone who has transformed a PDF, warped an image, or converted a spectrum from wavelength to frequency has divided by a Jacobian for the same reason.
The map $\cos\theta^* \mapsto \cos\theta$ is a Möbius transformation. $\cos\theta = (\cos\theta^*+\beta)/(1+\beta\cos\theta^*)$ is a bilinear map of the interval onto itself — the same family as an impedance transformation on a Smith chart, or a bilinear ( $z \leftrightarrow s$ ) transform in filter design. It compresses one end of the range and stretches the other, which is precisely what “beaming” means.

The physical consequence is everywhere in the rest of the book: at $\gamma = 10$ , half of an isotropic decay’s products land within about $1/\gamma \approx 6°$ of the forward direction. It is why forward calorimeters see most of the energy, why a neutrino beam is a beam at all, and why synchrotron light comes out in a searchlight cone.

Where it breaks: the $1/\gamma$ cone is a statement about ultrarelativistic emission, and it says nothing about the case the chapter actually cares about most. When the daughters are heavy and the decay is near threshold — a Λ giving up 97 % of its mass to a proton and a pion — the products are not ultrarelativistic in the parent frame and there is no forward cone at all, only a modest forward–backward asymmetry. Boost focusing is a limit, not a law, and it switches off exactly where two-body phase space is tightest.

γ = 1 (isotropic)
γ = 2
γ = 5
γ = 20

Problem 2.8 plotted: the same isotropic decay, seen from four laboratory frames. At γ = 1 it is flat by construction; at γ = 20 the forward direction is favoured over the backward one by a factor of (1+β)⁴/(1−β)⁴ ≈ 2.6 million. Nothing about the decay changed — only the frame.

The calculation nine problems share, as one picture

00.250.50.75100.250.5(m₁ + m₂) / M — how much of the parent’s mass the daughters already use upp* / Mone daughter masslessequal massesπ⁰ → γγ · 67.5K → μν · 235.5K⁰ → ππ · 206.0π → μν · 29.8Λ → pπ⁻ · 100.6

p* in MeV/c after each label. Every two-body decay in the book lands inside this wedge — there is nowhere else to be.

Supplied. Nine of the nineteen problems are the same formula, and plotted in scaled variables the formula is a single wedge with two edges: one daughter massless above, equal masses below. Both edges reach 0.5 at the left and zero at the right.

The right-hand end is the physically important part: as the daughters’ masses approach the parent’s, the momentum available collapses, which is why the Λ (p, π⁻ using 97 % of its mass) gives its daughters only 100 MeV/c while the far lighter K gives 235. That collapse is the phase-space throttle behind the neutron’s fifteen-minute lifetime (§1.10) and behind the π⁰’s speed: with two massless daughters it sits at the extreme left, p* = M/2, and nothing is holding it back.

The problems

All nineteen, with a hint before each solution. The book solves six of them (2.3, 2.4, 2.8, 2.13, 2.14 and 2.16); the other thirteen are written here, and every number in all nineteen was verified before it was committed.

📝 Chapter 2 problems

0/19 solved

2.1kinematicstheory
Compute the energies and momenta in the CM system of the decay products of $\pi \to \mu + \nu$ .
2.2kinematicstheory
Consider the decay $K \to \mu + \nu$ . Find (a) the energies and momenta of the μ and the ν in the K rest frame; (b) the maximum μ momentum in a frame where the K momentum is 5 GeV.
2.3kinematicstheory
A $\pi^0$ decays emitting one gamma of energy $E_1 = 150$ MeV in the forward direction. What is the direction of the second gamma? What is its energy $E_2$ ? What is the speed of the $\pi^0$ ?
2.4relativitytheory
Two μ are produced by a cosmic-ray collision at an altitude of 30 km, with energies $E_1 = 5$ GeV and $E_2 = 5$ TeV. At what distance does each muon see the Earth's surface in its own rest frame? What distance does each travel, in the Earth frame, in one lifetime?
2.5relativitytheory
A $\pi^+$ is produced at an altitude of 30 km with energy $E_\pi = 5$ GeV. At what distance does the pion see the Earth's surface in its rest frame? What distance does it travel in the Earth frame in one lifetime?
2.6detectorstheory
A photon converts into an $e^+e^-$ pair in a cloud chamber with $B = 0.2$ T. Two tracks are observed with the same radius $\rho = 20$ cm, and the initial angle between them is zero. Find the energy of the photon.
2.7decaystheory
Given the lifetimes $\rho^0\!: 5\times10^{-24}$ s, $K^+\!: 1.2\times10^{-8}$ s, $\eta^0\!: 5\times10^{-19}$ s, $\mu^-\!: 2\times10^{-6}$ s, $\pi^0\!: 8\times10^{-17}$ s, identify the interaction responsible for each decay: $\rho^0 \to \pi^+\pi^-$ ; $K^+ \to \pi^0\pi^+$ ; $\eta^0 \to \pi^+\pi^-\pi^0$ ; $\mu^- \to e^-\bar\nu_e\nu_\mu$ ; $\pi^0 \to \gamma\gamma$ .
2.8kinematicstheory
For $\pi^0 \to \gamma\gamma$ in the CM the decay is isotropic. Give $P(\cos\theta^*, \phi^*) = \mathrm{d}N/\mathrm{d}\Omega^*$ . Then, in the L frame where the π⁰ travels along $z$ with momentum $p$ , write $P(\cos\theta, \phi)$ .
2.9detectorstheory
Chamberlain and co-workers used scintillators to measure the pion lifetime. Why did they not use Geiger counters?
2.10leptonstheory
Calculate the ratio between the magnetic moments of the electron and the μ, and between the electron and the τ.
2.11kinematicstheory
Repeat the antiproton threshold calculation of Problem 1.9 for target protons bound in a nucleus, with Fermi momentum $p_f = 150$ MeV. Use $p_p \simeq E_p$ for the incident proton.
2.12detectorstheory
We want a monochromatic beam of $p = 20$ GeV with a spread $\Delta p/p = 1\%$ . The beam is 2 mm wide, the magnet has a bending power $BL = 4$ T m, and the slit is $d = 2$ mm wide. Calculate the distance $l$ between magnet and slit.
2.13kinematicstheory
A hydrogen bubble chamber is exposed to a 3 GeV $\pi^-$ beam. An interaction is seen with all-neutral secondaries and two V⁰s pointing back to the primary vertex. For one of them the tracks measure $p^- = 121$ MeV, $\theta^- = -18.2°$ , $\phi^- = 15°$ and $p^+ = 1900$ MeV, $\theta^+ = 20.2°$ , $\phi^+ = -15°$ . What is the particle, given a ±4 % mass resolution?
2.14leptonstheory
Calculate the neutrino energy thresholds for (1) $\nu_e + n \to e^- + p$ , (2) $\nu_\mu + n \to \mu^- + p$ and (3) $\nu_\tau + n \to \tau^- + p$ .
2.15kinematicstheory
A photon of $E_\gamma = 511$ keV is scattered backwards by an electron at rest. What is the scattered photon's energy? And what if the target electron were moving against the photon with kinetic energy $T_e = 511$ keV?
2.16kinematicstheory
At the SLAC linear accelerator electrons reached $E_{el} = 20$ GeV. To produce a high-energy photon beam, a laser beam of wavelength $\lambda = 694$ nm was backscattered at 180° by the electron beam. What was the scattered photon energy?
2.17detectorstheory
In Anderson's event, a positive track emerged from the lead plate with measured momentum $p = 23$ MeV/c. Calculate its kinetic energy assuming it to be (a) a proton and (b) a positron.
2.18detectorstheory
In 1933 Blackett and Occhialini saw several $e^+e^-$ pairs in electromagnetic showers from cosmic rays in a Wilson chamber with $B = 0.3$ T. In one event both tracks described arcs of radius $R = 14$ cm. Calculate their energies.
2.19kinematicstheory
Consider a $\pi^+$ beam of momentum $p = 200$ GeV at a proton accelerator. A muon beam can be made by letting the pions decay in a vacuum pipe. Calculate the energy range of the muons.

Reproduce it

import numpy as np
mpi, mpi0, mmu, me, mp, mK, mL, mKz = 139.57039, 134.9768, 105.6583755, 0.51099895, 938.27209, 493.677, 1115.683, 497.611

print("two-body decay p* and the boosted extremes:")
for name, M, m in (("pi -> mu nu  ", mpi, mmu), ("K  -> mu nu  ", mK, mmu)):
    ps = (M**2 - m**2)/(2*M)
    print(f"   {name} : p* = {ps:6.2f} MeV/c, E*_mu = {np.hypot(ps, m):7.2f} MeV")
psK = (mK**2-mmu**2)/(2*mK); EK = np.hypot(5000., mK)
print(f"   K at 5 GeV/c  : p_mu(max) = {EK/mK*(psK + 5000/EK*np.hypot(psK,mmu)):.0f} MeV/c"
      f"  -> MORE than the kaon's own momentum")
ps = (mpi**2-mmu**2)/(2*mpi); Es = np.hypot(ps, mmu)
EP = np.hypot(200e3, mpi); g, b = EP/mpi, 200e3/EP
print(f"   pi at 200 GeV : muons span {g*(Es-b*ps)/1e3:.1f} to {g*(Es+b*ps)/1e3:.1f} GeV"
      f" = {g*(Es-b*ps)/EP*100:.1f} % to {g*(Es+b*ps)/EP*100:.0f} %")
print(f"   and (m_mu/m_pi)^2 = {(mmu/mpi)**2*100:.1f} % exactly")

print("cosmic-ray survival from 30 km:")
for name, E, m, ctau in (("mu at    5 GeV", 5e3, mmu, 658.64), ("mu at 5000 GeV", 5e6, mmu, 658.64),
                         ("pi at    5 GeV", 5e3, mpi, 7.8045)):
    gm = E/m; L = np.sqrt(gm**2-1)*ctau/1000
    print(f"   {name}: gamma = {gm:7.1f}, decay length {L:9.1f} km -> "
          + (f"{np.exp(-30/L)*100:.1f} % arrive" if np.exp(-30/L) > 1e-3 else f"{np.exp(-30/L):.1e} arrive"))
print("   the lifetimes differ by 84x; the survival probabilities by 10^47")

cart = lambda p, t, f: np.array([p*np.sin(np.radians(t))*np.cos(np.radians(f)),
                                 p*np.sin(np.radians(t))*np.sin(np.radians(f)),
                                 p*np.cos(np.radians(t))])
P = cart(121., -18.2, 15.) + cart(1900., 20.2, -15.); Pm = np.linalg.norm(P)
Em = np.hypot(121., mpi)
print(f"V0 reconstruction (Problem 2.13), |p_V| = {Pm:.0f} MeV:")
for label, mplus, known, kn in (("as pi+ pi- ", mpi, mKz, "K0"), ("as p  pi-  ", mp, mL, "Lambda")):
    mV = np.sqrt((Em + np.hypot(1900., mplus))**2 - Pm**2)
    ok = "COMPATIBLE" if mV*0.96 <= known <= mV*1.04 else "INCOMPATIBLE"
    print(f"   {label}: m = {mV:.0f} MeV -> +-4 % gives {mV*0.96:.0f}..{mV*1.04:.0f}, "
          f"and the {kn} is {known:.1f}  {ok}")

bs = lambda Eg, Ee, pe: Eg*(Ee+pe)/(2*Eg + Ee - pe)
print("inverse Compton, backscattering off an electron:")
print(f"   511 keV photon, electron at rest        -> {bs(0.511, me, 0)*1e3:.0f} keV   (the photon loses)")
Ee = me + 0.511
print(f"   511 keV photon, electron T = 511 keV    -> "
      f"{bs(0.511, Ee, np.sqrt(Ee**2-me**2))*1e3:.0f} keV   (the photon gains)")
Eg, Eel = 1239.841984/694, 20e9
r = bs(Eg, Eel, np.sqrt(Eel**2-(me*1e6)**2))
print(f"   1.79 eV laser photon, electron at 20 GeV -> {r/1e9:.2f} GeV  (gain {r/Eg:.1e})")

prints

two-body decay p* and the boosted extremes:
 pi -> mu nu   : p* =  29.79 MeV/c, E*_mu =  109.78 MeV
 K  -> mu nu   : p* = 235.53 MeV/c, E*_mu =  258.15 MeV
 K at 5 GeV/c  : p_mu(max) = 5012 MeV/c  -> MORE than the kaon's own momentum
 pi at 200 GeV : muons span 114.6 to 200.0 GeV = 57.3 % to 100 %
 and (m_mu/m_pi)^2 = 57.3 % exactly
cosmic-ray survival from 30 km:
 mu at    5 GeV: gamma =    47.3, decay length      31.2 km -> 38.2 % arrive
 mu at 5000 GeV: gamma = 47322.3, decay length   31168.4 km -> 99.9 % arrive
 pi at    5 GeV: gamma =    35.8, decay length       0.3 km -> 2.4e-47 arrive
 the lifetimes differ by 84x; the survival probabilities by 10^47
V0 reconstruction (Problem 2.13), |p_V| = 1998 MeV:
 as pi+ pi- : m = 613 MeV -> +-4 % gives 589..638, and the K0 is 497.6  INCOMPATIBLE
 as p  pi-  : m = 1147 MeV -> +-4 % gives 1101..1193, and the Lambda is 1115.7  COMPATIBLE
inverse Compton, backscattering off an electron:
 511 keV photon, electron at rest        -> 170 keV   (the photon loses)
 511 keV photon, electron T = 511 keV    -> 841 keV   (the photon gains)
 1.79 eV laser photon, electron at 20 GeV -> 7.07 GeV  (gain 4.0e+09)

Erratum — two mislabelled components in the book’s solution 2.13

The book works Problem 2.13 at the back (p. 512), and the six Cartesian momentum components it lists carry the wrong superscript twice:

printed	value	should be
$p_x^-$	$121\sin(-18.2°)\cos15° = -36.5$	✓ negative track
$p_y^-$	$121\sin(-18.2°)\sin15° = -9.8$	✓ negative track
$p_z^-$	$121\cos(-18.2°) = 115$	✓ negative track
$p_x^+$	$1900\sin(20.2°)\cos(-15°) = 633.7$	✓ positive track
$p_y^-$	$1900\sin(20.2°)\sin(-15°) = -169.8$	$p_y^+$
$p_z^-$	$1900\cos(20.2°) = 1783.1$	$p_z^+$

The last two are computed from 1900 MeV and 20.2°, which are the positive track’s momentum and angle — the negative track has 121 MeV and −18.2°, and already has all three of its components listed above. So the labels say the negative track has six components and the positive track one.

Every number is right and the conclusion is unaffected: summing all six gives $|\mathbf{p}_V| = 1998$ MeV, and the two mass hypotheses come out at 613 MeV (incompatible with the $K^0$ ) and 1147 MeV (compatible with the $\Lambda$ inside ±4 %), exactly as the block above recomputes them from the same six numbers. It is a typesetting slip and nothing more — recorded because a reader checking their own reconstruction component by component will otherwise assume they have misassigned a track.

🔑 If you remember only three things

Anderson’s positron came off one photograph and four numbers. Redoing the arithmetic shows how little a discovery can require, and how much rested on each digit.
Isotropy is a statement about a frame. A decay uniform in the centre-of-mass arrives in the laboratory as a beam — the distribution changes shape while nothing about the decay does.
One determined momentum is what makes two-body decays useful. Mass measurements, resolution studies and calibrations all lean on there being nothing left to choose.

Where this goes next

Chapter 2 assembled the cast and the equation. Chapter 3 stops discovering particles and starts asking what is conserved — parity, charge conjugation, time reversal, isospin — and finds that two of the three are violated while their product survives. The θ–τ puzzle left open in §2.1–2.2 is resolved there, by giving up something nobody expected to lose.

✅ Check yourself — the two-body decay, nine times over

0/6 answered · 0 correct

1.Problem 2.2 gives a 5 GeV/c kaon a muon of 5.01 GeV/c. How can a daughter carry more momentum than its parent?
2.Problems 2.4 and 2.5 send a muon and a pion of the same 5 GeV energy down through 30 km of atmosphere. The lifetimes differ by a factor of 84. What happens to the survival probabilities?
3.Problem 2.19 finds that a 200 GeV pion beam gives muons from 114.6 to 200 GeV. What sets the lower end?
4.Problem 2.8 transforms an isotropic π⁰ → γγ decay into the lab and gets P(cosθ) ∝ γ⁻²(1 − βcosθ)⁻². What is that, structurally?
5.In Problem 2.13 the V⁰ reconstructs to 613 MeV under the π⁺π⁻ hypothesis and 1147 MeV under pπ⁻. With ±4 % resolution, what is it?
6.Problems 2.15 and 2.16 both backscatter a photon off an electron. At rest the 511 keV photon comes back at 170 keV; off a 20 GeV electron a 1.79 eV laser photon comes back at 7.1 GeV. What changed?