Chapter 9 — Summary and Problems

Part III ★ Summary & Problems Bettini pp. 430–434 · ~14 min read

  • chapter summary

Nothing here can be answered from memory. Every question descends from the same short structure, so a problem you cannot finish identifies exactly which part of it you have not understood.

🎯 Why this matters

A chapter that predicts can be examined in a way a chapter that catalogues cannot. Every one of these has an answer the theory fixes, which is why the set is both the longest and the most uniform in the book.

Thirty-six problems, the largest set in the book, and unusually well distributed:

  • eight are the electroweak bookkeeping — 9.5 to 9.8, 9.28, 9.29, 9.32, 9.33 — where you assign isospin and hypercharge and then use them as a checksum on processes that charge conservation cannot decide;
  • nine are widths and rates — 9.9 to 9.16, 9.23 — all of which reduce to the two master constants of §9.5 and the cZc_Z table;
  • six are experiment design — 9.1 to 9.3, 9.22, 9.24, 9.30 — asking what a beam, a detector or a machine has to deliver;
  • five are error propagation and counting — 9.18, 9.19, 9.21, 9.25, 9.31;
  • and the rest are the ones worth doing even if you skip the others: 9.20 on why the weak force is weak, 9.27 on a dark-matter line, 9.34 on flavour blindness, and 9.35–9.36, which arrive at the fact that a nucleus’s weak charge is essentially its neutron number.

The book supplies no solutions. All thirty-six are worked here.

What the chapter established

Every row traces to one angle, θ_W ≈ 29° — except the last two, which trace to a field value, υ = 246 GeV.
whathowthe number
The weak and electromagnetic interactions share a descriptionone rotation takes (W3,B)(W_3, B) to (γ,Z)(\gamma, Z)21 couplings → 2 constants
…and the angle is the same everywhereν–e scattering, atomic parity violation, Møller, LEP asymmetries, MW/MZM_W/M_Zsin2θW=0.232\sin^2\theta_W = 0.232 over 8 decades in QQ
The mediators exist, at the predicted massesa ppˉp\bar p collider, and a Jacobian edgeMW=80.377±0.012M_W = 80.377\pm0.012, MZ=91.1875±0.0021M_Z = 91.1875\pm0.0021
They carry weak charge themselvese+eW+We^+e^-\to W^+W^- diverges without the WWZWWZ vertexmeasured to 209 GeV
There are three families and no morethe height of the Z resonanceNν=2.984±0.008N_\nu = 2.984 \pm 0.008
The masses come from a broken symmetrya scalar doublet with a degenerate vacuumυ=246\upsilon = 246 GeV; 4 d.o.f. → 3 eaten + 1 H
…and the scalar exists, with the right couplingsγγ\gamma\gamma and ZZ4ZZ^*\to4\ell, then couplings vs massMH=125.25M_H = 125.25 GeV, μ=1.002±0.057\mu = 1.002\pm0.057

“21 couplings → 2 constants”, drawn: every fermion’s pair of Z couplings, from one angle

g_Lg_R−0.5−0.25+0.25+0.5+0.25−0.25ν(+0.500, 0)e μ τ(−0.268, +0.232)u c t(+0.345, −0.155)d s b(−0.423, +0.077)g_L = T₃ − Q sin²θ_W g_R = − Q sin²θ_WTwo integers you already know, and ONE measured number. Nothing else is free.Turn sin²θ_W and all four points slide along fixed lines — the neutrino cannot move at all, since Q = 0.

Supplied — the chapter’s first row claims “21 couplings → 2 constants” and the book leaves it as a sentence. Each fermion type couples to the Z with two independent strengths, left and right, and a pre-electroweak theory would have had to measure every one of them. Here all eight numbers on the plot come from gL=T3Qsin2θWg_L = T_3 - Q\sin^2\theta_W and gR=Qsin2θWg_R = -Q\sin^2\theta_W with sin2θW=0.232\sin^2\theta_W = 0.232 — the weak isospin and charge are known integers, so one measured angle fixes the lot.

That is what unification buys, and it is why measuring sin2θW\sin^2\theta_W in eight different places over eight decades in QQ (row 2) is a test rather than a repetition. The neutrino sits on the gR=0g_R = 0 axis for a reason no measurement could change: it has no electric charge, so the Qsin2θWQ\sin^2\theta_W term vanishes and a right-handed neutrino would couple to nothing at all.

The numbers, all at once

every numeric answer in the chapter's problems

import numpy as np
GF, s2, MZ, MW, me, u = 1.1663788e-5, 0.232, 91.0, 80.0, 0.5109989e-3, 1.66053907e-27
cz = lambda i3, q: i3 - s2*q
K = GF*MZ**3/(3*np.sqrt(2)*np.pi); Ge = GF*MW**3/(6*np.pi*np.sqrt(2))
Gn = K*.25; Gl = K*(cz(-.5,-1)**2+cz(0,-1)**2)
Gu = 3*K*(cz(.5,2/3)**2+cz(0,2/3)**2); Gd = 3*K*(cz(-.5,-1/3)**2+cz(0,-1/3)**2)
Gh = 2*Gu+3*Gd; G3 = 3*Gn+3*Gl+Gh
GW = 3*Ge + 3*Ge*(.974**2+.99**2+.224**2+.22**2); H = 389.379

Ne = 547e3/u/2; sg = 1.7e-45*24; phi = 1/(4*sg*Ne)
print(f"9.1   duty cycle {2*6e-3/14.4:.1e}; fluence {phi:.1e} /m^2 per pulse; I = {phi*3.2**2:.1e} nu/pulse")
print(f"9.2   theta_max = {np.sqrt(2*me/20)*1e3:.1f} mrad at 20 GeV; need ~0.7 mrad; iron excluded")
mpi, mmu = .13957039, .1056584; Es = (mpi**2-mmu**2)/(2*mpi); g = 200/mpi; b = np.sqrt(1-1/g**2)
print(f"9.3   E* = {Es*1e3:.1f} MeV; E_max = {g*Es*(1+b):.1f} GeV = {g*Es*(1+b)/200:.3f} E_pi; E_min = {g*Es*(1-b)*1e6:.0f} keV")
print(f"      theta*=50 mrad -> theta_lab = {np.arctan(np.sin(.05)/(g*(np.cos(.05)+b)))*1e6:.1f} urad")
print( "9.4   sigma(nu)/sigma(nubar) = 3 EXACTLY, from angular momentum alone")

def W(x):
    c = lambda i3, q: i3 - x*q
    return (K*(c(-.5,-1)**2+c(0,-1)**2), 3*K*(c(.5,2/3)**2+c(0,2/3)**2), 3*K*(c(-.5,-1/3)**2+c(0,-1/3)**2))
for tag, x in (("9.9   sin^2 = 3/8: ", 3/8), ("9.10  sin^2 = 1/4: ", .25)):
    l, uu, dd = W(x); h = 2*uu+3*dd; t = 3*Gn+3*l+h
    print(f"{tag} Gl = {l*1e3:3.0f}, Gu = {uu*1e3:3.0f}, Gd = {dd*1e3:3.0f} MeV, Gh = {h:.2f} GeV")
    print(f"      Gamma_Z = {t:.2f} GeV,  Gnu/Gh = {Gn/h:.3f}" + ("   (vs 0.099 measured -> dead)" if x > .3 else ""))
print(f"9.11  BR(W -> e nu) = 1/9 = 11.1% by COUNTING; {Ge/GW*100:.1f}% from the widths")
print(f"9.12  g_Zee^2/g_Wenu^2 = {2*(cz(-.5,-1)**2+cz(0,-1)**2)/(1-s2):.3f};  rate ratio {Gl/Ge:.2f}")
print(f"9.13  g_Zuu^2/g_Wud^2 = {2*(cz(.5,2/3)**2+cz(0,2/3)**2)/(1-s2):.3f};  rate ratio {Gu/(3*.974**2*Ge):.2f}")
print(f"9.14  10 nb -> {10*(GW-3*Ge)/Ge:.0f} nb into hadrons ({10*GW/Ge:.0f} nb into everything)")
brs = [Gl/(N*Gn+3*Gl+Gh)*100 for N in (3,4,5)]
rat = [(G3/(N*Gn+3*Gl+Gh))**2 for N in (3,4,5)]
print(f"9.15  N=3,4,5: BR(mumu) = {brs[0]:.2f}, {brs[1]:.2f}, {brs[2]:.2f}%;  peak sigma ratios 1, .{rat[1]*1000:.0f}, .{rat[2]*1000:.0f}")
print(f"9.16  {12*np.pi*Gl*Gl/(MZ**2*G3**2)*H*1e3:.1f} nb at the Z peak;  {(4*np.pi/3)/MW**2*(3*.974**2*Ge)*Ge/GW**2*H*1e3:.1f} nb at the W peak")
E = np.sqrt(MZ**2+140**2)
print(f"9.17  BOTH electrons at {E/MZ*MZ/2:.1f} GeV, each {np.degrees(np.arctan(1/((E/MZ)*(140/E)))):.0f} deg from the beam")
E1, E2, th, dth = 60., 40., np.radians(140), np.radians(1)
m = np.sqrt(2*E1*E2*(1-np.cos(th)))
r = .5*np.sqrt((.15/np.sqrt(E1))**2 + (.15/np.sqrt(E2))**2 + (dth/np.tan(th/2))**2)
print(f"9.18  m = {m:.1f} GeV, dm = {r*m:.1f} GeV -- the ENERGY term dominates 4:1")
print(f"9.19  20% on sin^2 -> {80*.1:.0f} GeV on M_W;  1% on M_Z/M_W -> {2*(MW/MZ)**2*.01:.3f} on sin^2")
q2 = (.197)**2
print(f"9.20  the weak force is {1/(7.4**2*q2/(q2+MW**2)):.1e} times weaker at 1 fm -- ALL propagator")
EZ = np.sqrt(MZ**2+100**2)
print(f"9.21  x = {MZ/2000:.3f} each at rest;  {(EZ+100)/2000:.3f} and {(EZ-100)/2000:.3f} at p_L = 100 GeV")
print(f"9.22  {17e-36*1e32*86400:.0f} W pairs per day at LEP2")
G4 = 4*Gn+3*Gl+Gh; d = 1-(G3/G4)**2
print(f"9.23  one extra neutrino: +{(G4/G3-1)*100:.1f}% on Gamma_Z, -{d*100:.1f}% on the peak height")
print(f"9.24  {(5/d)**2:.0f} events for a 5 sigma exclusion -- a few hours of LEP")
Rs = [3.1*(Ge/GW)/(Gl/(N*Gn+3*Gl+Gh)) for N in (3,4,5)]
print(f"9.25  R = {Rs[0]:.1f}, {Rs[1]:.1f}, {Rs[2]:.1f} for N = 3, 4, 5;  R < 10.1 excludes N >= 4")
uL, uR, dL, dR = cz(.5,2/3), cz(0,2/3), cz(-.5,-1/3), cz(0,-1/3)
sn = (uL**2+dL**2) + (uR**2+dR**2)/3; sb = (uL**2+dL**2)/3 + (uR**2+dR**2)
print(f"9.26  CC ratio 3 exactly;  NC ratio {sn/sb:.1f}")
mz = 91.1876; mc = (136 + np.sqrt(136**2 + mz**2))/2
print(f"9.27  m_chi = 136 GeV (gamma gamma) or {mc:.1f} GeV (Z gamma)")
print(f"9.30  Glashow needed E_nu = M_W^2/2m_e = {MW**2/(2*me):.1e} GeV = 6 PeV")
print(f"9.31  lambda-bar = {197/16:.1f} fm >> R_A = 4 fm -> COHERENT; T_max = {2*16**2/70.7e3*1e3:.1f} keV")
print(f"9.32  c_Z^2 = {uL**2:.3f}, {uR**2:.3f}, {dL**2:.3f}, {dR**2:.3f};  isoscalar combination {sn:.3f}")
print(f"9.33  isoscalar combination {sb:.3f};  ratio {sn/sb:.1f}")
print( "9.34  exactly 1 -- the neutral current is flavour-blind")
print(f"9.35  c_A = +-1/2 ALWAYS;  c_V(e) = {cz(-.5,-1)+cz(0,-1):.3f}, nearly zero")
cVu, cVd = uL+uR, dL+dR
print(f"9.36  proton c_V = {2*cVu+cVd:+.2f}, neutron c_V = {cVu+2*cVd:+.2f} -> Q_W = -N, {abs((cVu+2*cVd)/(2*cVu+cVd)):.0f}:1")
print("\n(this snippet uses the book's sin^2 = 0.232 throughout; problems 9.32-9.36")
print(" state sin^2 = 0.23, which moves the last digit of c_V and the 9.36 ratio.")
print(" both are the book's own values, used where the book uses them.)")
prints
9.1   duty cycle 8.3e-04; fluence 3.7e+10 /m^2 per pulse; I = 3.8e+11 nu/pulse
9.2   theta_max = 7.1 mrad at 20 GeV; need ~0.7 mrad; iron excluded
9.3   E* = 29.8 MeV; E_max = 85.4 GeV = 0.427 E_pi; E_min = 10 keV
    theta*=50 mrad -> theta_lab = 17.4 urad
9.4   sigma(nu)/sigma(nubar) = 3 EXACTLY, from angular momentum alone
9.9   sin^2 = 3/8:  Gl = 103, Gu = 247, Gd = 309 MeV, Gh = 1.42 GeV
    Gamma_Z = 2.23 GeV,  Gnu/Gh = 0.116   (vs 0.099 measured -> dead)
9.10  sin^2 = 1/4:  Gl =  82, Gu = 275, Gd = 357 MeV, Gh = 1.62 GeV
    Gamma_Z = 2.36 GeV,  Gnu/Gh = 0.102
9.11  BR(W -> e nu) = 1/9 = 11.1% by COUNTING; 11.0% from the widths
9.12  g_Zee^2/g_Wenu^2 = 0.327;  rate ratio 0.37
9.13  g_Zuu^2/g_Wud^2 = 0.373;  rate ratio 0.44
9.14  10 nb -> 61 nb into hadrons (91 nb into everything)
9.15  N=3,4,5: BR(mumu) = 3.44, 3.22, 3.03%;  peak sigma ratios 1, .876, .773
9.16  2.1 nb at the Z peak;  8.8 nb at the W peak
9.17  BOTH electrons at 83.5 GeV, each 33 deg from the beam
9.18  m = 92.1 GeV, dm = 1.4 GeV -- the ENERGY term dominates 4:1
9.19  20% on sin^2 -> 8 GeV on M_W;  1% on M_Z/M_W -> 0.015 on sin^2
9.20  the weak force is 3.0e+03 times weaker at 1 fm -- ALL propagator
9.21  x = 0.045 each at rest;  0.118 and 0.018 at p_L = 100 GeV
9.22  147 W pairs per day at LEP2
9.23  one extra neutrino: +6.9% on Gamma_Z, -12.4% on the peak height
9.24  1622 events for a 5 sigma exclusion -- a few hours of LEP
9.25  R = 9.9, 10.6, 11.3 for N = 3, 4, 5;  R < 10.1 excludes N >= 4
9.26  CC ratio 3 exactly;  NC ratio 2.4
9.27  m_chi = 136 GeV (gamma gamma) or 149.9 GeV (Z gamma)
9.30  Glashow needed E_nu = M_W^2/2m_e = 6.3e+06 GeV = 6 PeV
9.31  lambda-bar = 12.3 fm >> R_A = 4 fm -> COHERENT; T_max = 7.2 keV
9.32  c_Z^2 = 0.119, 0.024, 0.179, 0.006;  isoscalar combination 0.308
9.33  isoscalar combination 0.129;  ratio 2.4
9.34  exactly 1 -- the neutral current is flavour-blind
9.35  c_A = +-1/2 ALWAYS;  c_V(e) = -0.036, nearly zero
9.36  proton c_V = +0.04, neutron c_V = -0.50 -> Q_W = -N, 14:1

(this snippet uses the book's sin^2 = 0.232 throughout; problems 9.32-9.36
state sin^2 = 0.23, which moves the last digit of c_V and the 9.36 ratio.
both are the book's own values, used where the book uses them.)

Three of those repay a second look.

9.20 is the chapter in one line. The weak coupling is larger than the electric one, and at 1 fm the weak force is three thousand times weaker. Every bit of that is the propagator: q2/(q2+MW2)q^2/(q^2+M_W^2) against 1/q21/q^2. “Weak” was never a statement about a charge.

9.35 finds a near-zero that is not an accident. The electron’s vector coupling to the Z is cV=12+2sin2θW=0.04c_V = -\tfrac12 + 2\sin^2\theta_W = -0.04, vanishing exactly at sin2θW=1/4\sin^2\theta_W = 1/4 — the same special point that makes §9.4’s ratio RR equal 1. Nature sits a few per cent away from it, and half the precision electroweak programme consists of measuring how far.

9.36 ends somewhere surprising. Because the proton’s vector weak charge is +0.04+0.04 and the neutron’s is 0.50-0.50, a nucleus’s weak charge is essentially N-N — it counts neutrons, by more than an order of magnitude. And because the effect adds coherently over them, a part-per-million parity violation becomes measurable in a heavy atom. That is why atomic parity-violation experiments use caesium and thallium, and it is a tabletop measurement of the same angle the LHC measures.

0/36 solved
  1. 9.1neutrino beamstheory
    CHARM2 studied νμeνμe\nu_\mu e \to \nu_\mu e with a narrow-band beam of mean energy 24 GeV. The SPS gave two pulses 6 ms apart on a 14.4 s cycle. The useful target mass was 547 t with a square section 3.2 m on a side, containing equal numbers of protons and neutrons. Find the duty cycle. For one interaction every four pulses on average, what neutrino fluence and what intensity are needed?
    • σ/E=1.7×1045\sigma/E = 1.7\times10^{-45} m²/GeV
    • M = 547 t, l = 3.2 m
    • Z/A = 1/2
  2. 9.2kinematicstheory
    An electron of 20 GeV is detected by CHARM2, produced by elastic νe\nu e scattering. How large can its scattering angle be? What accuracy is needed in the direction measurement to verify it? Could the calorimeter be built of iron?
    • Eeθe22meE_e\theta_e^2 \le 2m_e
    • me=0.511m_e = 0.511 MeV
  3. 9.3beam designtheory
    A narrow-band νμ\nu_\mu beam is made from a monochromatic π+\pi^+ beam of 200 GeV decaying as π+μ+νμ\pi^+\to\mu^+\nu_\mu. Find the neutrino energy in the pion rest frame, the maximum and minimum laboratory energies, and the laboratory angle of a neutrino emitted at θ=50\theta^* = 50 mrad.
    • Eπ=200E_\pi = 200 GeV
    • mπ=139.57m_\pi = 139.57, mμ=105.66m_\mu = 105.66 MeV
  4. 9.4CC cross-sectionstheory
    Consider the charged-current cross-sections for neutrinos and antineutrinos on nuclei with equal numbers of protons and neutrons, neglecting all masses. Show that the neutrino total cross-section is three times the antineutrino one.
    • V−A: neutrinos are left-handed, antineutrinos right-handed
    • an isoscalar target
  5. 9.5quantum numberstheory
    Give the weak isospin, its third component and the weak hypercharge for eLe^-_L, νL\nu_L, uLu_L, dRd_R, and for their antiparticles.
    • YW=2(QIWz)Y_W = 2(Q - I_{Wz})
    • left-chirality fermions in doublets, right-chirality charged fermions in singlets
  6. 9.6quantum numberstheory
    Give the weak isospin, its third component and the weak hypercharge for μL\mu^-_L, τR\tau^-_R, tLt_L, bRb_R, and for their antiparticles.
    • YW=2(QIWz)Y_W = 2(Q - I_{Wz})
    • left-chirality fermions in doublets, right-chirality charged fermions in singlets
  7. 9.7conservation lawstheory
    Which of these are allowed and which forbidden, and why? WdLuˉLW^-\to d_L\bar u_L, WuLuˉLW^-\to u_L\bar u_L, ZW+WZ\to W^+W^-, WeRνˉRW^-\to e^-_R\bar\nu_R.
    • all gauge bosons have YW=0Y_W = 0
    • Table 9.1
  8. 9.8conservation lawstheory
    Which of these are allowed and which forbidden? dWuLd\to W^-u_L, ZeLeˉLZ\to e^-_L\bar e_L, WZWW^-\to ZW^-, WeLνˉLW^-\to e^-_L\bar\nu_L.
    • all gauge bosons have YW=0Y_W = 0
    • Table 9.1
  9. 9.9Z widthstheory
    A grand-unification theory based on SU(5) predicted sin2θW=3/8\sin^2\theta_W = 3/8, which experiment falsified. Find the partial widths of the Z in that hypothesis, and the ratio Γν/Γh\Gamma_\nu/\Gamma_h.
    • GFMZ3/(32π)=660G_F M_Z^3/(3\sqrt2\pi) = 660 MeV
    • cZ=IWzQsin2θWc_Z = I_{Wz} - Q\sin^2\theta_W
    • MZ=91M_Z = 91 GeV
  10. 9.10Z widthstheory
    Repeat for sin2θW=1/4\sin^2\theta_W = 1/4: the partial and total widths of the Z, and Γν/Γh\Gamma_\nu/\Gamma_h.
    • MZ=91M_Z = 91 GeV
    • GFMZ3/(32π)=660G_F M_Z^3/(3\sqrt2\pi) = 660 MeV
  11. 9.11branching ratiostheory
    Evaluate the branching ratio for We+νeW\to e^+\nu_e.
    • colour factor 3 for quarks
    • MW<mtM_W < m_t, so the top channels are closed
  12. 9.12couplingstheory
    Evaluate the ratio gZee2/gWeν2g_{Zee}^2/g_{We\nu}^2 and the decay-rate ratio Γ(Ze+e)/Γ(Weν)\Gamma(Z\to e^+e^-)/\Gamma(W\to e\nu).
    • gZ=(g/cosθW)cZg_Z = (g/\cos\theta_W)c_Z
    • gWeν=g/2g_{We\nu} = g/\sqrt2
    • sin2θW=0.232\sin^2\theta_W = 0.232
  13. 9.13couplingstheory
    Evaluate gZuu2/gWud2g_{Zuu}^2/g_{Wud'}^2 and Γ(Zuuˉ)/Γ(Wdˉu)\Gamma(Z\to u\bar u)/\Gamma(W\to \bar d u).
    • cZ(uL)=1/2(2/3)s2c_Z(u_L) = 1/2 - (2/3)s^2, cZ(uR)=(2/3)s2c_Z(u_R) = -(2/3)s^2
  14. 9.14resonancestheory
    Assume σ(uˉdeνˉe)=10\sigma(\bar u d \to e^-\bar\nu_e) = 10 nb at the W resonance. Evaluate the total σ(uˉdqqˉ)\sigma(\bar u d \to q\bar q) at resonance.
    • Breit–Wigner: σmaxΓinΓout/Γ2\sigma_{\max}\propto \Gamma_{\text{in}}\Gamma_{\text{out}}/\Gamma^2
  15. 9.15neutrino countingtheory
    Take the number of light neutrinos to be 3, 4 or 5 in turn, changing nothing else. Evaluate for each the Z branching ratio into μ+μ\mu^+\mu^- and the ratio Γν/ΓZ\Gamma_\nu/\Gamma_Z, and the ratio of the peak hadronic cross-sections.
    • Γν=165\Gamma_\nu = 165, Γ=83\Gamma_\ell = 83 MeV, Γh=1.66\Gamma_h = 1.66 GeV
    • σpeak1/ΓZ2\sigma_{\text{peak}}\propto 1/\Gamma_Z^2
  16. 9.16resonancestheory
    Calculate σ(e+eμ+μ)\sigma(e^+e^-\to\mu^+\mu^-) at the Z peak and σ(uˉdeνˉe)\sigma(\bar u d\to e^-\bar\nu_e) at the W peak.
    • σpeak(e+e)=12πΓeΓμ/(MZ2ΓZ2)\sigma_{\text{peak}}(e^+e^-) = 12\pi\Gamma_e\Gamma_\mu/(M_Z^2\Gamma_Z^2)
    • σpeak(uˉd)=4π31MW2ΓudΓeνΓW2\sigma_{\text{peak}}(\bar ud) = \tfrac{4\pi}{3}\tfrac{1}{M_W^2}\tfrac{\Gamma_{ud}\Gamma_{e\nu}}{\Gamma_W^2}
    • 2c2=389.4  μ\hbar^2c^2 = 389.4\;\mub GeV²
  17. 9.17kinematicstheory
    A Z is produced at a ppˉp\bar p collider at s=540\sqrt s = 540 GeV, moving along the beams with pZ=140p_Z = 140 GeV. It decays as Ze+eZ\to e^+e^- with the electrons at 90° to the beams in the Z rest frame. Calculate the two electron energies in the laboratory.
    • MZ=91M_Z = 91 GeV
  18. 9.18error propagationtheory
    A Ze+eZ\to e^+e^- decay has calorimeter energies E1=60E_1 = 60 GeV and E2=40E_2 = 40 GeV, with resolution σE/E=0.15/E\sigma_E/E = 0.15/\sqrt E. The measured angle between the tracks is 140°±1°140°\pm1°. Find the error on mZm_Z.
    • M=2E1E2(1cosθ)M = \sqrt{2E_1E_2(1-\cos\theta)}
    • δMM=12(δE1E1)2+(δE2E2)2+(δθtan(θ/2))2\frac{\delta M}{M} = \frac12\sqrt{\left(\frac{\delta E_1}{E_1}\right)^2+\left(\frac{\delta E_2}{E_2}\right)^2+\left(\frac{\delta\theta}{\tan(\theta/2)}\right)^2}
  19. 9.19error propagationtheory
    Before their discovery the predictions were MZ=90M_Z = 90 and MW=80M_W = 80 GeV. If sin2θW=0.23\sin^2\theta_W = 0.23 with a 20 % uncertainty, what is the uncertainty on MWM_W? If instead MZ/MWM_Z/M_W is measured to 1 %, what is the uncertainty on sin2θW\sin^2\theta_W?
    • MW=37.3GeV/sinθWM_W = 37.3\,\text{GeV}/\sin\theta_W
    • sin2θW=1(MW/MZ)2\sin^2\theta_W = 1 - (M_W/M_Z)^2
  20. 9.20propagatorstheory
    At 1 GeV the weak charge gg is larger than the electric charge by 4π/sinθW7.4\sqrt{4\pi}/\sin\theta_W \approx 7.4. Why then is the electrostatic force between two electrons at 1 fm so much larger than the weak force?
    • c=197\hbar c = 197 MeV fm
    • MW=80M_W = 80 GeV
  21. 9.21parton kinematicstheory
    At the Tevatron (ppˉp\bar p, s=2\sqrt s = 2 TeV), what are the approximate momentum fractions of the annihilating quark and antiquark for a Z produced at rest? In what fraction are they sea quarks? And if the Z has longitudinal momentum 100 GeV?
    • s^=xqxqˉs\hat s = x_q x_{\bar q} s
    • MZ=91M_Z = 91 GeV
  22. 9.22luminositytheory
    LEP2 was designed to study e+eW+We^+e^-\to W^+W^-. If the cross-section at s=200\sqrt s = 200 GeV is 17 pb and the luminosity is 103210^{32} cm⁻²s⁻¹, how many events are produced per day?
    • 1 pb = 103610^{-36} cm²
    • 1 day = 86 400 s
  23. 9.23neutrino countingtheory
    What is the percentage change in the Z total width for one additional neutrino type? What is the change in the peak hadronic cross-section?
    • Γν=165\Gamma_\nu = 165 MeV
    • ΓZ=2.40\Gamma_Z = 2.40 GeV
    • σpeak1/ΓZ2\sigma_{\text{peak}}\propto1/\Gamma_Z^2
  24. 9.24statisticstheory
    Working at the Z with an e+ee^+e^- collider and assuming statistical uncertainties only, how many events are needed to exclude a fourth neutrino at five standard deviations?
    • one extra neutrino changes the peak cross-section by 12.4 %
  25. 9.25neutrino countingtheory
    UA1 and UA2 measured R=σWBR(Weν)σZBR(Ze+e)R = \dfrac{\sigma_W\,\mathrm{BR}(W\to e\nu)}{\sigma_Z\,\mathrm{BR}(Z\to e^+e^-)} from event counts alone, so that systematics cancel. Writing R=σWσZΓ(Wν)ΓWΓZΓ(Z)R = \dfrac{\sigma_W}{\sigma_Z}\cdot\dfrac{\Gamma(W\to\ell\nu)}{\Gamma_W}\cdot\dfrac{\Gamma_Z}{\Gamma(Z\to\ell\ell)}, evaluate R for 3, 4 and 5 neutrinos and set an upper limit on NνN_\nu from R<10.1R < 10.1 at 90 % CL.
    • σW/σZ=3.1\sigma_W/\sigma_Z = 3.1
    • Γ(Wν)=224\Gamma(W\to\ell\nu) = 224, ΓW=2035\Gamma_W = 2035 MeV
    • Γ(Z)=83\Gamma(Z\to\ell\ell) = 83 MeV, Γh=1662\Gamma_h = 1662 MeV
  26. 9.26cross-section ratiostheory
    Calculate the ratio of the CC cross-sections of neutrinos and antineutrinos on nuclei with equal numbers of protons and neutrons, considering only valence quarks. Repeat for the neutral-current interactions.
    • the L+R configuration carries a factor 1/3
    • cZ=IWzQsin2θWc_Z = I_{Wz}-Q\sin^2\theta_W, sin2θW=0.232\sin^2\theta_W = 0.232
  27. 9.27kinematicstheory
    Dark matter might consist of 'neutralinos' χ, identical to their antiparticles, annihilating as χχγγ\chi\chi\to\gamma\gamma and χχZ0γ\chi\chi\to Z^0\gamma. A gamma telescope observes a monochromatic line at Eγ=136E_\gamma = 136 GeV. Find mχm_\chi in both hypotheses, assuming negligible initial kinetic energy.
    • MZ=91.19M_Z = 91.19 GeV
    • the pair annihilates essentially at rest, so s=2mχ\sqrt s = 2m_\chi
  28. 9.28colour and chiralitytheory
    (1) For a pair of quarks with colours R and B and a third quark G, is the force attractive or repulsive for the combinations RB + BR and RB − BR? (2) Is WbLcˉRW^-\to b_L\bar c_R allowed? (3) Is ZτRτˉRZ\to\tau^-_R\bar\tau_R allowed?
    • 33=63ˉ3\otimes3 = 6\oplus\bar3
    • Table 9.1
  29. 9.29colour and chiralitytheory
    Establish whether each is allowed, where the left superscript is colour and the right subscript chirality: (a) WBsLBuˉRW^-\to{}^Bs_L\,{}^B\bar u_R, (b) WBdRBuˉLW^-\to{}^Bd_R\,{}^B\bar u_L, (c) WRdLBuˉRW^-\to{}^Rd_L\,{}^B\bar u_R, (d) ZGuLGuˉRZ\to{}^Gu_L\,{}^G\bar u_R, (e) ZGuRGuˉRZ\to{}^Gu_R\,{}^G\bar u_R, (f) ZGuRGcˉLZ\to{}^Gu_R\,{}^G\bar c_L, (g) ZGtLGtˉRZ\to{}^Gt_L\,{}^G\bar t_R. And is 16(RqBqGq)\tfrac{1}{\sqrt6}({}^Rq\,{}^Bq\,{}^Gq - \ldots) bound?
    • gauge bosons carry no colour and no hypercharge
  30. 9.30historytheory
    In 1960 Glashow suggested establishing charged weak bosons through the resonant reaction νeeW\nu_e e^-\to W^-\to\ldots. Why did the idea not work in practice?
    • MW=80M_W = 80 GeV
    • me=0.511m_e = 0.511 MeV
  31. 9.31neutrino detectiontheory
    The 'boron neutrinos' from 8B2αe+νe^8\mathrm{B}\to2\alpha\,e^+\nu_e in the Sun reach Eν,max=16E_{\nu,\max} = 16 MeV. Consider elastic coherent scattering on a 76^{76}Ge nucleus of radius RA=4R_A = 4 fm. Is the resolving power enough to resolve the nuclear structure? What is the maximum recoil energy? How does the interaction probability depend on flavour?
    • c=197\hbar c = 197 MeV fm
    • m(76Ge)70.7m(^{76}\text{Ge}) \approx 70.7 GeV
  32. 9.32neutral currentstheory
    For neutral-current scattering of νμ\nu_\mu on the valence quarks of a nucleus, write the squared Z-charge factors of the quarks and their values at sin2θW=0.23\sin^2\theta_W = 0.23. Write the combination the total cross-section is proportional to, and evaluate it for the valence quarks and for a nucleus with equal numbers of u and d.
    • cZ=IWzQsin2θWc_Z = I_{Wz}-Q\sin^2\theta_W
    • L+R carries a factor 1/3
  33. 9.33neutral currentstheory
    Repeat problem 9.32 for antineutrinos.
    • the same four cZc_Z values
    • for an antineutrino the 1/3 moves to the LEFT-handed quarks
  34. 9.34neutral currentstheory
    Compare the neutral-current scattering of νe\nu_e and νμ\nu_\mu of the same energy on the valence quarks of a nucleus. Consider the ratio of the left- and right-quark contributions in the two cases. How large is the ratio between these ratios?
    • the Z couples identically to all neutrino flavours
  35. 9.35neutral currentstheory
    Find the relative amplitudes of the vector and axial couplings to the Z for νe\nu_e, ee^-, uu and dd, at sin2θW=0.23\sin^2\theta_W = 0.23.
    • cV=cL+cRc_V = c_L + c_R, cA=cLcRc_A = c_L - c_R
  36. 9.36atomic parity violationtheory
    Atomic electrons interact with the nucleus by photon exchange and, far more weakly, by Z exchange; the interference of the two gives observable parity violation. The main contribution comes from the product of the axial ELECTRON current and the vector QUARK currents. (a) Evaluate the axial Z-charges of the proton and the neutron. (b) Evaluate them for a nucleus with Z protons and N neutrons — is the larger contribution from protons or neutrons? (c) Explain why the product of the axial quark current and the vector electron current is less important.
    • cA(q)=IWzc_A(q) = I_{Wz}
    • cV(u)=+0.193c_V(u) = +0.193, cV(d)=0.347c_V(d) = -0.347
    • consider valence quarks only

Everything in chapter 9 descends from two numbers — and the three that do not are chapter 12’s problem

sin²θ_W = 0.232 (θ_W ≈ 28.8°)one rotation of (W₃, B) into (γ, Z)υ = 246 GeVthe field value in the vacuumg_L, g_Revery fermionM_W/M_Z= cos θ_WasymmetriesA_FB, A_LR, APVM_W, M_Zthe SCALE, not the ratio3 eaten d.o.f.→ W±, Z get massN_ν = 2.984from the Z width…and three things that need a THIRD input the model does not supply:M_H = √(2λ)·υ needs λ · every fermion mass needs its own Yukawa · neither is predicted — → ch12’s parameter count

Supplied. The summary table above says it in words — “every row traces to one angle, except the last two, which trace to a field value” — and the shape of the claim is worth seeing. θ_W fixes ratios and υ fixes the scale: the angle alone can never tell you how heavy a W is, and the vacuum value alone can never tell you how a fermion’s left and right couplings differ.

It also makes visible where the chapter stops. The Higgs mass and every fermion mass sit below the dashed line because each needs a coupling nobody can predict — 125.25 GeV is measured, not derived, and the same is true of all nine charged-fermion masses. That is precisely the bill §12.7 presents.

🔑 If you remember only three things

  • The arithmetic is the same each time. Charge assignments in, a coupling out, a rate from the coupling — thirty-six variations on three steps.

  • A wrong answer localises the error. The step that failed is identifiable rather than lost somewhere in the algebra, which is rare enough in a physics problem set to be worth exploiting.

  • The set is uniform because the theory is. Earlier chapters needed many kinds of problem because their physics was a catalogue; this one needs thirty-six of one kind.

Where this goes next

Chapter 10 is the hole this chapter left. §9.12 showed that the BEH mechanism gives neutrinos no mass at all, because the recipe needs a right-chirality partner that has never been observed. And neutrinos oscillate, so they have mass. Chapter 8’s formalism runs again — the same P(αβ)P(\alpha\to\beta), the same interference between mass eigenstates — but with mixing angles that are large rather than small, and with an origin of mass that is outside everything Chapters 7–9 built.

Chapter 11 turns to gravitational waves, and Chapter 12 collects what is left: the strong-CP problem, baryogenesis — where §8.11’s CP violation was shown to be far too small — dark matter, and the fact that §9.20’s 19 free parameters are 19 things nobody can explain.

Check yourself — Chapter 9 in review

0/6 answered · 0 correct

  1. 1.Problem 9.20: the weak coupling g is 7.4 times the electric charge, yet at 1 fm the weak force is thousands of times weaker. Where does the suppression come from?

  2. 2.Problems 9.4 and 9.26 both give exactly 3 for the CHARGED-current ν/ν̄ ratio, but the neutral-current ratio is 2.4. Why the difference?

  3. 3.Problem 9.24 finds that 1600 events suffice to exclude a fourth neutrino at 5σ, yet each LEP experiment collected four million. What does that tell you?

  4. 4.Problem 9.30 asks why Glashow's 1960 proposal to make a W in ν_e e → W failed. What is the answer?

  5. 5.Problem 9.35 finds c_V(e) = −0.04, nearly zero, while c_A(e) = −1/2 exactly. What is behind each?

  6. 6.Problem 9.36 ends with the nuclear weak charge being essentially −N, the neutron number. Why do the protons contribute so little?

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.