Chapter 6 — Summary and Problems · Elementary Particle Physics

The hard part of nearly every problem here is deciding which colour factor applies. After that it is Chapter 1 arithmetic, unchanged.

🎯 Why this matters

The strong interaction added a great deal of conceptual apparatus and almost no new machinery. That is worth noticing, because it means the difficulty of this chapter was never computational.

Twenty-eight problems, and they divide unevenly: about half are kinematics that would sit comfortably in Chapter 1, and the rest are the genuinely new material — colour factors, conservation laws with two new flavour quantum numbers, and the counting of states.

eight are DIS and elastic kinematics — 6.4, 6.5, 6.8, 6.9, 6.16, 6.19, 6.20, 6.22. Every one is $Q^2 = 4EE'\sin^2(\theta/2)$ and $x = Q^2/2m_p\nu$ rearranged, and 6.9 and 6.20 are the quantitative case for building a collider;
five are cross-sections and rates — 6.2, 6.3, 6.15, 6.17, and the R arithmetic threaded through the others;
five are colour bookkeeping — 6.1, 6.23, 6.24, 6.25, 6.26;
five are conservation laws and quantum numbers — 6.11, 6.12, 6.18, 6.27, 6.28;
four are wave functions and weak currents — 6.6, 6.7, 6.13, 6.14;
and 6.10 compares the two couplings, which is the chapter in one number.

The book’s Solutions section covers only 6.2, 6.5 and 6.8. The other twenty-five are worked here, and every number is computed in a snippet rather than asserted.

What the chapter established

The chapter summary (p. 272), with where each item was built

You should now have	What that means in practice	§
the evidence for colour, quark jets and gluon jets	R is three times Σz² at every energy and across three thresholds; jets follow 1 + cos²θ; a third jet appears 10 % of the time and its angular distribution makes the gluon a vector. All of it measured before any Lagrangian was written.	6.1
the nucleon's internal structure from deep inelastic scattering	F₂ almost independent of Q² means point-like constituents — Rutherford's argument again. And the quarks carry only half the momentum: the rest is gluons, found by subtraction.	6.2
the QCD Lagrangian and the colour charges	SU(3), exact, non-abelian. One commutator in the field tensor gives gluons that carry colour, couple to each other, and change everything downstream.	6.3
why only some colour combinations bind	Two channels out of four have a negative colour factor: the qq̄ singlet at −4/3 and the antisymmetric qq pair at −2/3. Hadrons are qq̄ and qqq because of a sign, not a rule — and the hyperfine ratio 4/3 confirms it to 2 %.	6.4
the evolution of α_s, and asymptotic freedom	Gluon loops carry the opposite sign to quark loops, the 33 beats the 2n_f, and 1/α_s slopes the other way from 1/α. The divergence moves from 10³⁵ GeV to 0.3 GeV — from harmless to being the reason hadrons exist.	6.5
the running of the quark masses	A confined particle has no rest frame in which to be weighed, so a quark mass carries a scheme and a scale. The b loses a third of its mass between its own scale and the Z.	6.6
the origin of the hadron mass	99 % of the proton is colour-field energy. Two lines of uncertainty-principle algebra give m_p ≈ 1 GeV and r_p ≈ 0.7 fm — and the Higgs supplies the other 1 %.	6.7
the quantum vacuum, and chiral symmetry	The vacuum fluctuates with pure glue at every scale, and its structure breaks a symmetry the Lagrangian has. The pion is the light debris — m² ∝ m_q, not m ∝ m_q, which is testable and works to 12 %.	6.8–6.9
lattice QCD, and the quark–gluon plasma	When α_s is too large to expand in, put spacetime on a grid — at a cost of (L/a)⁴. Three inputs give the whole light hadron spectrum, and the same method supplies the number that limits the muon anomaly.	6.10–6.11

Read the first two rows and then the third. Everything about colour was <strong>measured</strong> — R, the jets, the momentum deficit — before SU(3) appears on p. 240. That is the order the book chose and the order this chapter keeps, and it is why §6.1 and §6.2 come first even though §6.3 is logically prior.

The numbers, all at once

every numeric answer in the chapter's problems

import numpy as np
mp = 0.9382720882
q = {'u': 2/3, 'd': -1/3, 's': -1/3, 'c': 2/3, 'b': -1/3}
R = lambda fl: 3*sum(q[f]**2 for f in fl)

print("6.2  R = %.0f at 2.5 GeV (u d s), %.3f = 10/3 at 4 GeV (charm open)"
      % (R('uds'), R('udsc')))
print("6.3  jet angle at 20 GeV = %.1f-%.1f deg; N(90)/N(30) = %.3f"
      % (np.degrees(np.arctan(0.5/10)), np.degrees(np.arctan(1.0/10)),
         1/(1 + np.cos(np.radians(30))**2)))
nu = 25/(2*mp*0.2)
print("6.4  nu = Q^2/2 m_p x = %.1f GeV, E' = %.1f GeV" % (nu, 100-nu))
print("6.5  elastic means W = m_p, so Q^2 = 2 m_p nu and x = 1 exactly")
print("6.8  Q^2_max = 4E^2/(1 + 2E/m_p x) = %.1f GeV^2 at x = 0.2"
      % (4*100**2/(1 + 200/(mp*0.2))))
s = 4*30*820
print("6.9  HERA sqrt(s) = %.0f GeV; a fixed target would need %.1f TeV"
      % (np.sqrt(s), s/(2*mp)/1000))
for x in (0.4, 0.01, 0.0001):
    print("       x = %-7s Q^2_max = %9.1f GeV^2" % (x, x*s))
L, MZ, ainvZ = 0.200, 91.0, 129.0
for Q, nf in ((10.0, 4), (100.0, 5)):
    a_s = 12*np.pi/((33-2*nf)*np.log(Q**2/L**2))
    ainv = ainvZ - (6.667/(3*np.pi))*np.log(Q**2/MZ**2)
    print("6.10 Q = %3.0f GeV: alpha_s = %.4f, alpha = 1/%.1f, ratio = %.4f"
          % (Q, a_s, ainv, 1/ainv/a_s))
print("6.12 Q = I_z + Y/2 with Y = B + S = -1 forces I_z = +1/2 -> the Xi0")
print("6.15 sigma(mu mu) falls by 100 from 2 to 20 GeV; R goes 2 -> 11/3;")
print("       so sigma(hadrons) falls by only %.1f" % (100*R('uds')/R('udscb')))
Ep = lambda E, th, m: E/(1 + (E/m)*(1-np.cos(np.radians(th))))
print("6.16 E' = %.3f GeV; in the CM elastic scattering changes no energies"
      % Ep(5, 8, mp))
print("6.17 R_i = %.2e /s, dOmega = %.1e sr, dsigma/dOmega = %.1f b/sr, rate ~ 2e7 /s"
      % (1e-6/(2*1.602176634e-19), 1e-4/0.25,
         (2*82*(1/137.036)*197.3269804/40)**2/np.sin(np.radians(20))**4*1e-2))
for m, lab in ((mp, 'H'), (4*0.9315, 'He'), (56*0.9315, 'Fe')):
    print("6.19 %-2s target: E' = %.4f GeV" % (lab, Ep(1.0, 20, m)))
print("6.21 invert: E = E'/[1 - (E'/m)(1-cos th)] = %.1f GeV" % (12/(1-(12/mp)*(1-np.cos(np.radians(20))))))
Q2m = 4*15**2/(1+2*15/mp)
print("6.22 Q^2_max = %.1f GeV^2, proton recoil T = %.1f GeV (it takes nearly everything)"
      % (Q2m, Q2m/(2*mp)))
print("6.23 (a) +1/2  (c) -1/3 attractive   |   6.24 (b) +1/6 repulsive  (c) -1/3")
print("6.26 two gluons: L=0 -> 0++, 2++;  L=1 -> 0-+, 1-+, 2-+  (1-+ is exotic)")
M = 3.686
print("6.27 m(chi_c) = sqrt(M^2 - 2 M E_gamma) = %.3f GeV -> the chi_c0(3415), J^PC = 0++"
      % np.sqrt(M**2 - 2*M*0.26))
print("6.28 four P-wave states; only 0++ and 2++ go to two gluons (Landau-Yang)")

prints

6.2  R = 2 at 2.5 GeV (u d s), 3.333 = 10/3 at 4 GeV (charm open)
6.3  jet angle at 20 GeV = 2.9-5.7 deg; N(90)/N(30) = 0.571
6.4  nu = Q^2/2 m_p x = 66.6 GeV, E' = 33.4 GeV
6.5  elastic means W = m_p, so Q^2 = 2 m_p nu and x = 1 exactly
6.8  Q^2_max = 4E^2/(1 + 2E/m_p x) = 37.5 GeV^2 at x = 0.2
6.9  HERA sqrt(s) = 314 GeV; a fixed target would need 52.4 TeV
     x = 0.4     Q^2_max =   39360.0 GeV^2
     x = 0.01    Q^2_max =     984.0 GeV^2
     x = 0.0001  Q^2_max =       9.8 GeV^2
6.10 Q =  10 GeV: alpha_s = 0.1927, alpha = 1/132.1, ratio = 0.0393
6.10 Q = 100 GeV: alpha_s = 0.1319, alpha = 1/128.9, ratio = 0.0588
6.12 Q = I_z + Y/2 with Y = B + S = -1 forces I_z = +1/2 -> the Xi0
6.15 sigma(mu mu) falls by 100 from 2 to 20 GeV; R goes 2 -> 11/3;
     so sigma(hadrons) falls by only 54.5
6.16 E' = 4.753 GeV; in the CM elastic scattering changes no energies
6.17 R_i = 3.12e+12 /s, dOmega = 4.0e-04 sr, dsigma/dOmega = 25.5 b/sr, rate ~ 2e7 /s
6.19 H  target: E' = 0.9396 GeV
6.19 He target: E' = 0.9841 GeV
6.19 Fe target: E' = 0.9988 GeV
6.21 invert: E = E'/[1 - (E'/m)(1-cos th)] = 52.5 GeV
6.22 Q^2_max = 27.3 GeV^2, proton recoil T = 14.5 GeV (it takes nearly everything)
6.23 (a) +1/2  (c) -1/3 attractive   |   6.24 (b) +1/6 repulsive  (c) -1/3
6.26 two gluons: L=0 -> 0++, 2++;  L=1 -> 0-+, 1-+, 2-+  (1-+ is exotic)
6.27 m(chi_c) = sqrt(M^2 - 2 M E_gamma) = 3.416 GeV -> the chi_c0(3415), J^PC = 0++
6.28 four P-wave states; only 0++ and 2++ go to two gluons (Landau-Yang)

⚙️ Engineer’s bridge — the collider argument, in two problems

Problems 6.8 and 6.9 are the same calculation on two machines, and put side by side they are the clearest statement in the book of why colliders exist.

6.8, fixed target: a 100 GeV electron beam on hydrogen reaches $Q^2_{\max} = 37$ GeV². The limiting factor is the denominator $1 + 2E/m_px$ , which for these numbers is about a thousand — almost the entire beam energy goes into moving the proton rather than resolving it.

6.9, collider: 30 GeV on 820 GeV reaches $Q^2 = 39\,000$ GeV² at the same $x$ . A thousand times more, from beams whose energies are of the same order.

The reason is a scaling law worth internalising: for a collider $s = 4E_1E_2$ , and for a fixed target $s \approx 2Em$ . The collider’s reach grows as the product of the beam energies; the fixed target’s grows only linearly, and is throttled by a target mass you cannot change. To match HERA on a fixed target you would need 52 TeV of electrons.

The engineering shape of this is familiar: you are comparing a design whose figure of merit is a product against one where a fixed constant sits in the denominator. Improving both sides of a product beats improving one side of a sum, every time, and it is why the difficult, expensive option — storing and colliding two beams — won.

Where it breaks: improving both sides of a product beats improving one side of a sum only when both sides can actually be improved. For $p\bar p$ they cannot: antiprotons must be manufactured and stochastically cooled, and their supply — not the ring — sets the luminosity, which is precisely why the LHC gave up the antiproton advantage and collides $pp$ with two separate beam pipes and twin-bore magnets. The arithmetic favouring colliders is correct and says nothing about which beams you can obtain; the engineering constraint that decides real machines sits entirely outside it.

fixed target on hydrogen — Q²max ≈ 2Em_p x
collider against an 820 GeV proton — Q²max ≈ 4E·820·x
problem 6.8 (100 GeV fixed target) and 6.9 (HERA)

Both problems on one axis, at x = 0.4. The two lines have the same slope — both are linear in the electron energy — but the collider's sits nearly three decades higher, because its constant is 4 × 820 GeV rather than 2 × 0.94 GeV. The gap is the proton beam energy, and it is the entire argument. To reach HERA's point on the red line you would run off the right-hand edge of the plot.

💡 What this really says — the three problems worth doing even if you skip the rest

6.5 — showing that x = 1 is exactly elastic scattering. It converts the Bjorken variable from a definition into a piece of geometry: x is where you are in the physical region, and x = 1 is its edge. Every PDF vanishing at x → 1 follows, and so does the whole shape of Fig. 6.12.

6.10 — α/α_s at two scales. Two numbers, and they are moving towards each other. Everything Chapter 5 and Chapter 6 said about running couplings is in that convergence, and extrapolating it is what Chapter 10 does.

6.28 — which χ_c states can decay to two gluons. It uses the Landau–Yang theorem, which sounds like a technicality and is the reason the Higgs was known not to be spin 1 within days of its discovery. A conservation argument that costs three lines and rules out an entire hypothesis is the most economical thing in physics.

📝 Chapter 6 — all 28 problems

0/28 solved

6.1the gluon censustheory
How many gluons exist? Give the electric charge of each, and their strangeness, charm and beauty. What is the gluon spin? How many different quarks exist for every flavour, and what are their charges? Does QCD define the number of families?
- $3 \otimes \bar3 = 8 \oplus 1$
- the strong interaction is blind to flavour and to electric charge
6.2R across a thresholdtheory
Evaluate $R = \sigma(e^+e^-\to\text{hadrons})/\sigma(e^+e^-\to\mu^+\mu^-)$ at $\sqrt s = 2.5$ GeV and at $\sqrt s = 4$ GeV.
- $R = 3\sum_f z_f^2$ , summed over flavours with $2m_f < \sqrt s$
- $m_c = 1.27$ GeV
6.3jets and their angular distributiontheory
Consider $e^+e^-\to q\bar q$ at $\sqrt s = 20$ GeV. Give a typical jet opening angle in a two-jet event. If θ is the angle of the jet axis to the beams, what is the ratio of the counting rates at θ = 90° and θ = 30°?
- $\theta_{\text{jet}} \approx p_T/(\sqrt s/2)$ with $p_T \approx 0.5$ –1 GeV, Eq. (6.3)
- $\mathrm{d}\sigma/\mathrm{d}\Omega \propto 1 + \cos^2\theta$ , Eq. (6.8)
6.4DIS kinematicstheory
A 100 GeV electron beam hits liquid hydrogen. The energy $E'$ and direction of the scattered electron are measured. Find $E'$ for $Q^2 = 25$ GeV² and $x = 0.2$ .
- $x = Q^2/2m_p\nu$ , Eq. (6.16)
- $\nu = E - E'$ , Eq. (6.14)
6.5the elastic limittheory
What is the value of x in elastic scattering? Derive Eq. (1.83), $E' = E/[1 + (E/m)(1-\cos\theta)]$ , by taking the elastic cross-section as the limit of the inelastic one.
- $W^2 = m_p^2 + 2m_p\nu - Q^2$ , Eq. (6.12)
- elastic means the proton survives: $W = m_p$
6.6neutrino scattering on quarkstheory
Consider the scattering of $\nu_\mu$ and $\bar\nu_\mu$ by nucleons in the quark model. Considering the d, u and s quarks and antiquarks, write the contributing weak processes with a muon in the final state.
- the charged weak current raises or lowers the charge by one unit
- lepton number is conserved separately
6.7charm in neutrino scatteringtheory
As in the previous problem, but considering the c and c̄ quarks.
- the same charge bookkeeping
- charm is produced from strange, not from down, in the dominant channel
6.8the reach of a fixed-target experimenttheory
A 100 GeV electron beam hits liquid hydrogen. Find $Q^2$ as a function of the scattering angle θ and of x. What is the maximum momentum transfer for x = 0.2?
- $Q^2 = 2EE'(1-\cos\theta)$
- $E' = E - Q^2/2m_p x$
6.9why HERA was a collidertheory
At HERA an electron beam of 30 GeV meets a proton beam of 820 GeV. Calculate $\sqrt s$ and the energy a fixed-target electron beam would need to match it. Calculate $Q^2_{\max}$ for x = 0.4, 0.01 and 0.0001, and compare with Fig. 6.13.
- head-on collider: $s = 4E_eE_p$
- fixed target: $s \approx 2E m_p$
- $Q^2_{\max} = x\,s$ (at y = 1)
6.10the two couplings comparedtheory
Evaluate the ratio $\alpha/\alpha_s$ at $Q^2 = (10\ \text{GeV})^2$ and at $Q^2 = (100\ \text{GeV})^2$ . Take $\Lambda_{\text{QCD}} = 200$ MeV, $\alpha^{-1}(M_Z^2) = 129$ and $M_Z = 91$ GeV.
- $\alpha_s = 12\pi/[(33-2n_f)\ln(Q^2/\Lambda^2)]$ , Eq. (6.55)
- $\alpha^{-1}(Q^2) = \alpha^{-1}(\mu^2) - (z_f/3\pi)\ln(Q^2/\mu^2)$ , Eq. (5.52)
6.11conservation laws in charm productiontheory
Which of these are allowed or forbidden by strong interactions, and why? (a) $\pi^-p \to \Lambda_c^+\pi^-$ , (b) $\pi^-p \to \Lambda_c^+D^-$ , (c) $\pi^-p \to \Lambda_c^+D^0$ , (d) $\pi^-p \to \Lambda_c^+D_s^-$ .
- initial state: $Q = 0$ , $\mathcal{B} = 1$ , $S = 0$ , $C = 0$
- $\Lambda_c^+ = udc$ ; $D^- = \bar cd$ ; $D^0 = c\bar u$ ; $D_s^- = \bar cs$
6.12identifying a baryon from its quantum numberstheory
A non-charmed baryon has strangeness $S = -2$ and electric charge $Q = 0$ . What are the possible values of its isospin $I$ and $I_z$ ? What is it usually called if $I = 1/2$ ?
- Gell-Mann–Nishijima: $Q = I_z + Y/2$ with $Y = \mathcal{B} + S$
- $\mathcal{B} = 1$ , $S = -2$ , $C = 0$
6.13a baryon wave function, factor by factortheory
The proton has $uud$ as valence quarks. Write down the wave function in its spin, isospin and colour factors, taking all orbital momenta to be zero.
- quarks are fermions: the total wave function must be antisymmetric
- $L = 0$ , so the spatial part is symmetric
6.14the Λ, and why it is differenttheory
As in the previous problem, but for the Λ hyperon.
- $\Lambda = uds$ , $I = 0$ , $J = 1/2$
- the same antisymmetry requirement
6.15point-like versus hadronic cross-sectionstheory
Consider (1) $e^+e^-\to\mu^+\mu^-$ and (2) $e^+e^-\to$ hadrons at $\sqrt s = 2$ GeV and 20 GeV. Find the ratio of process (1) at the two energies, the ratio of (2) to (1) at each, and the ratio of process (2) at the two energies.
- $\sigma(\mu\mu) = 86.8\ \text{nb}/s[\text{GeV}^2]$
- $R = 2$ below charm, $11/3$ above beauty
6.16elastic electron–proton scatteringtheory
We observe elastic scattering of $E = 5$ GeV electrons from protons at θ = 8° and measure their energy. (1) What is the expected value? (2) What is the scattered electron energy in the CM frame?
- $E' = E/[1 + (E/m_p)(1-\cos\theta)]$ , Eq. (1.83)
- in the CM the energies are unchanged by elastic scattering
6.17a Rutherford counting rate, end to endtheory
A beam of α particles of kinetic energy 10 MeV and intensity $I = 1$ μA hits a lead target ( $A = 207$ , $Z = 82$ , $\rho = 1.14\times10^4$ kg m⁻³) of thickness 0.2 mm. A detector of area 1 cm² sits 0.5 m beyond the target at θ = 40°. Find (a) the incident rate, (b) the solid angle, (c) the differential cross-section and (d) the counts per second.
- $\mathrm{d}\sigma/\mathrm{d}\Omega = \left(\dfrac{Z_1Z_2\alpha\hbar c}{4E}\right)^2\dfrac{1}{\sin^4(\theta/2)}$
- an α particle carries charge $2e$
6.18conservation laws in beauty productiontheory
Which of these are allowed or forbidden by strong interactions, and why? (a) $\pi^-p \to \Lambda_b^0K^0$ , (b) $\pi^-p \to \Lambda_b^0D^0$ , (c) $\pi^-p \to \Lambda_b^0B^0$ , (d) $\pi^-p \to \Sigma_b^-B^+$ , (e) $\pi^-p \to \Sigma_b^+B^-$ .
- initial state: $Q = 0$ , $\mathcal{B}_{\text{baryon}} = 1$ , $S = C = \text{beauty} = 0$
- $\Lambda_b^0 = udb$ , $\Sigma_b^- = ddb$ , $\Sigma_b^+ = uub$ — all with beauty $-1$
- $B^+ = u\bar b$ (beauty $+1$ ), $B^- = \bar ub$ (beauty $-1$ ), $B^0 = \bar db$ (beauty $-1$ )
6.19the target mass decides the energy losstheory
A 1 GeV electron beam hits a liquid hydrogen target; a calorimeter measures the scattered electron energy at θ = 20°. Calculate it for elastic scattering. Repeat for liquid He and for iron (A = 56).
- $E' = E/[1 + (E/m)(1-\cos\theta)]$
- $m \approx A \times 0.9315$ GeV
6.20reconstructing an event at HERAtheory
At HERA, $E_e = 28$ GeV and $E_p = 820$ GeV. An electron scatters at θ = 120° with measured energy $E' = 223$ GeV. Calculate $\sqrt s$ and the kinematic variables $Q^2$ , $x$ , $\nu$ and $W$ .
- $s = 4E_eE_p$ for a head-on collider
- $Q^2 = 4E_eE'\sin^2(\theta/2)$ in the collider convention, with θ measured from the proton direction
6.21inverting Compton kinematicstheory
A monochromatic photon beam of unknown energy and known direction scatters on liquid hydrogen. The photons scattered at 20° are measured at $E' = 12$ GeV. What is the beam energy?
- $E' = E/[1 + (E/m_p)(1-\cos\theta)]$
6.22maximum momentum transfer and recoiltheory
For elastic scattering of a 15 GeV electron on a proton: (a) find $Q^2_{\max}$ and the corresponding proton recoil kinetic energy; (b) the same for E = 20 MeV on ⁵⁶Fe. How can the expression be simplified?
- $Q^2_{\max} = 4E^2/(1 + 2E/m)$ at θ = 180°
- $T_{\text{recoil}} = Q^2/2m$
6.23reading colour flow off a diagramtheory
Draw the gluon-exchange diagrams for these vertex pairs: (a) $R_q\to B_q$ , $B_q\to R_q$ ; (b) $R_q\to B_q$ , $\bar R_q\to\bar B_q$ ; (c) $R_q\to R_q$ , $\bar R_q\to\bar R_q$ . Specify which gluons are exchanged and the colour charges at both vertices. (d) Explain the meaning of the signs.
- colour lines are continuous through the diagram
- $\lambda_7 = \text{diag}(1,-1,0)/\sqrt2$ and $\lambda_8 = \text{diag}(1,1,-2)/\sqrt6$ in the (R, G, B) basis
- the colour factors of antiquarks are the negatives of the quarks'
6.24more colour flowtheory
Draw the gluon-exchange diagrams for: (a) $B_q\to G_q$ , $G_q\to B_q$ ; (b) $G_q\to G_q$ , $\bar R_q\to\bar R_q$ ; (c) $G_q\to G_q$ , $\bar G_q\to\bar G_q$ . Specify the gluons and the colour charges, and explain the signs.
- the colour factors of Eq. (6.34)
- antiquark factors are the negatives of the quark ones
6.25drawing strong-decay diagramstheory
Draw the lowest-order Feynman diagrams for (a) $\phi \to K^+K^-$ , (b) $\phi \to K^0\bar K^0$ , (c) $\phi \to \pi^+\pi^-\pi^0$ .
- $\phi = s\bar s$
- $K^+ = u\bar s$ , $K^- = \bar us$ , $K^0 = d\bar s$
- the OZI rule of §4.7
6.26the quantum numbers of a two-gluon statetheory
Consider a two-gluon system in a colour-singlet state whose colour wave function is symmetric. Write down the possible values of S, L, J, P and C up to L = 1.
- gluons are identical massless spin-1 bosons
- the total wave function must be symmetric under exchange
- $P = (-1)^L$ for two vector particles with intrinsic parities $(-1)(-1)$ ; $C = +1$ for two gluons
6.27identifying a state from one photontheory
At SPEAR, $e^+e^- \to \psi(2S) \to \chi_c\,\gamma$ , followed by $\chi_c \to \psi\,\pi\pi$ . The photon energy is $E_\gamma = 0.26$ GeV. Determine the χ_c mass and the possible values of its spin, parity, charge conjugation and isospin. Assume the radiative decay is E1.
- $m(\psi(2S)) = 3.686$ GeV
- E1: the photon carries one unit of angular momentum and negative parity
- $J^{PC}(\psi) = 1^{--}$
6.28which charmonium states can decay to two gluonstheory
Consider the P wave $c\bar c$ states, the χ_c. Establish their number and possible $J^{PC}$ . Which of them can decay into two gluons, and hence into hadrons?
- P wave: $L = 1$ ; $c\bar c$ has $S = 0$ or 1
- $P = (-1)^{L+1}$ , $C = (-1)^{L+S}$ for a fermion–antifermion pair
- a two-gluon state can be $0^{++}$ or $2^{++}$ (problem 6.26)

Aside — three problem statements that are easy to misread

Three of these problems are easy to misread in ways that change every number, and all three caught me while writing the solutions:

6.10 asks for $Q^2 = (10\ \text{GeV})^2$ and $(100\ \text{GeV})^2$ — that is $10^2$ and $10^4$ GeV², not 10 and 100 GeV².
6.23(b, c) and 6.24(b, c) pair a quark with an antiquark, not two quarks. The overbars are small and the sign of every colour factor depends on them: the same colours attract for $q\bar q$ and repel for $qq$ .
6.18(c) and (e) hinge on the fact that $B^+ = u\bar b$ contains the antiquark while $B^- = \bar ub$ contains the quark, so their beauty is opposite to what the charge labels suggest.

None is a defect in the book — they are just places where careful reading is the whole problem.

Where a proton’s 938 MeV actually comes from

colour-field energy — 99.04 %u + u + d = 8.99 MeV — 0.96 %

Drawn to scale: the mass the Higgs mechanism supplies is the 5-pixel sliver at the left-hand edge.

the same sliver, magnified 100×:u — 2.16u — 2.16d — 4.67 MeV

Your mass is not the mass of your constituents. It is the energy stored in the field that confines them —

which is why E = mc² is not a footnote to this chapter but its subject, and why ch. 9’s Higgs explains only the sliver.

Quark masses are MS-bar at 2 GeV — a confined particle has no rest frame, so its mass carries a scheme and a scale (§6.6).

Supplied — the chapter’s summary row says “99 % of the proton is colour-field energy and the Higgs supplies the other 1 %”, and one per cent is a quantity prose cannot convey. To scale, the constituent masses are five pixels wide. §6.7’s two lines of uncertainty-principle algebra reproduce the other 99 % from Λ_QCD alone, with no quark mass anywhere in the calculation — which is the strongest statement in the chapter: the proton would weigh essentially what it weighs even if its quarks were massless. Everything you are made of gets its mass this way, and the mechanism chapter 9 spends a section on accounts for the sliver.

🔑 If you remember only three things

Half of a strong-interaction problem set is relativistic kinematics. The ratio is itself the lesson: new physics, old tools.
Bjorken x turns from a definition into a place. Showing that x = 1 is exactly elastic scattering is what fixes the meaning of the variable.
Reading the statement wrongly changes every number after it. Three of these are phrased in ways that invite exactly that, and no amount of correct algebra recovers from it.

Where this goes next

Chapter 7 is the third interaction, and it breaks the pattern the first two established. Electromagnetism and QCD are both parity-conserving, flavour-blind and mediated by massless bosons. The weak interaction is none of those things.

§7.1 — the Fermi theory, and where the 1/s dimensional argument of §5.7 fails because a massive mediator brings a second scale.
§7.2 — parity violation, which supplies the polarization and decay asymmetry that made §5.9b’s muon experiment possible.
§7.9 — quark mixing, where the flavour quantum numbers this chapter’s problems conserved so carefully stop being conserved.

✅ Check yourself — Chapter 6 review

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1.Problem 6.8 reaches Q² = 37 GeV² with a 100 GeV fixed-target beam; problem 6.9 reaches 39 000 GeV² with 30 GeV against 820 GeV. What is the scaling law behind the gap?
2.Problem 6.5 shows that elastic scattering is exactly x = 1. What does that tell you about parton distributions?
3.Problems 6.23 and 6.24 both ask for colour factors, and both contain a quark–antiquark case. Why does the antiquark change the sign?
4.Problem 6.11 asks which charm-production reactions are allowed. Why is π⁻p → Λ_c⁺D_s⁻ the instructive case?
5.Problem 6.28 finds that the χ_c1 cannot decay into two gluons. What general result is that, and where else does it matter?