Electron–Positron Annihilation into a Muon Pair

α never appears in either answer. The angular distribution is fixed by spin and the hadronic ratio by counting colours, so both would survive a change in the strength of electromagnetism.

🎯 Why this matters

Dividing by that boringness is what turns R into a counting experiment. Everything common to numerator and denominator cancels, and what survives is a count of quarks and of colours.

Everything the chapter has built gets used at once, on the cleanest process in particle physics: $e^+e^- \to \mu^+\mu^-$ . Four leptons, no strong interaction anywhere, one diagram — and two results that fall out of the structure rather than the algebra.

Fig. 5.24 — the lowest-order diagram, and the only one

Click a vertex or an internal line.

One diagram at lowest order. Unlike Bhabha scattering (§5.6), the t channel contributes nothing — the initial and final states share no particles.

\frac{\mathrm{d}\sigma}{\mathrm{d}\Omega_f} = \frac{1}{(8\pi)^2}\,\frac{1}{\htmlClass{t-s}{s}}\,\frac{1}{4}\sum_{\text{spin}}\left|M_{fi}\right|^2, \qquad \frac{1}{4}\sum_{\text{spin}}\left|M_{fi}\right|^2 = (4\pi\alpha)^2\left(1 + \htmlClass{t-c}{\cos^2\theta}\right)

(5.44–5.45)

Bettini p. 205. The kinematic prefactor, and the dynamics quoted without proof — but both pieces can be checked, and are below.

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.

Where the angular shape comes from

Take the quantisation axis along the beam. Helicity conservation (§5.6) says the electron and positron spins must both be $+1/2$ or both $-1/2$ — never one of each — so the initial state has $m = \pm 1$ along the beam. The same argument on the muon axis gives $m' = \pm 1$ . Four combinations, and each is a rotation from one axis to the other.

Redrawn from Bettini Fig. 5.26. Double arrows are spin projections; note that in each panel the two initial spins point the same way, which is what helicity conservation forces for a particle–antiparticle pair. θ is the angle between the μ⁻ and the e⁻.

🔢 Worked example — the whole result, checked in twelve lines

Three claims made without proof in the book, and all three are arithmetic.

Reproduce it

import numpy as np

print("1. the four helicity configurations, squared and summed")
c = np.cos(np.radians([0, 30, 60, 90, 120, 150, 180]))
same, opp = 0.5*(1 + c), 0.5*(1 - c)          # d^1_{1,1} = d^1_{-1,-1} ; d^1_{1,-1} = d^1_{-1,1}
tot = 2*same**2 + 2*opp**2
print("   theta   sum|d|^2   1+cos^2   equal?")
for a, b, d in zip([0, 30, 60, 90, 120, 150, 180], tot, 1 + c**2):
    print(f"   {a:5d}  {b:9.6f} {d:9.6f}   {abs(b-d) < 1e-12}")

print("\n2. integrate dsigma/dOmega = (alpha^2/4s)(1 + cos^2 theta)")
al, hbarc2 = 1/137.035999166, 0.3894e6        # (hbar c)^2 in GeV^2 nb
x = np.linspace(-1, 1, 200001)
I = 2*np.pi*np.trapezoid(1 + x**2, x)
print(f"   int (1+cos^2) dOmega = {I:.6f}   exact 16pi/3 = {16*np.pi/3:.6f}")
print(f"   sigma * s = alpha^2/4 * {I:.4f} = {al**2/4*I:.6e}")
print(f"   4 pi alpha^2 / 3     = {4*np.pi*al**2/3:.6e}")
print(f"   in nb GeV^2:  {4*np.pi*al**2/3*hbarc2:.1f}     (the book prints 86.8)")

print("\n3. where the muons go")
lo = x <= 0
print(f"   theta = 0 vs 90 deg, dsigma ratio = {2/1:.0f}")
print(f"   fraction of the rate at theta >= 90 deg = "
      f"{np.trapezoid(1+x[lo]**2, x[lo])/np.trapezoid(1+x**2, x):.4f}   (exactly 1/2, by symmetry)")

print("\n4. the R steps that make Fig. 5.27 a staircase")
for lab, up, dn in [("below charm  (u,d,s)", 1, 2), ("above charm  (u,d,s,c)", 2, 2),
                    ("above bottom (u,d,s,c,b)", 2, 3)]:
    print(f"   {lab:26s} R = 3 x ({up} x 4/9 + {dn} x 1/9) = {3*(up*4/9 + dn/9):.3f}")

prints

1. the four helicity configurations, squared and summed
 theta   sum|d|^2   1+cos^2   equal?
     0   2.000000  2.000000   True
    30   1.750000  1.750000   True
    60   1.250000  1.250000   True
    90   1.000000  1.000000   True
   120   1.250000  1.250000   True
   150   1.750000  1.750000   True
   180   2.000000  2.000000   True

2. integrate dsigma/dOmega = (alpha^2/4s)(1 + cos^2 theta)
 int (1+cos^2) dOmega = 16.755161   exact 16pi/3 = 16.755161
 sigma * s = alpha^2/4 * 16.7552 = 2.230588e-04
 4 pi alpha^2 / 3     = 2.230588e-04
 in nb GeV^2:  86.9     (the book prints 86.8)

3. where the muons go
 theta = 0 vs 90 deg, dsigma ratio = 2
 fraction of the rate at theta >= 90 deg = 0.5000   (exactly 1/2, by symmetry)

4. the R steps that make Fig. 5.27 a staircase
 below charm  (u,d,s)       R = 3 x (1 x 4/9 + 2 x 1/9) = 2.000
 above charm  (u,d,s,c)     R = 3 x (2 x 4/9 + 2 x 1/9) = 3.333
 above bottom (u,d,s,c,b)   R = 3 x (2 x 4/9 + 3 x 1/9) = 3.667

Part 1 is the point of the section: the angular distribution is four rotation matrix elements, no dynamics required. Part 2 closes the chain — integrate the differential cross-section and Eq. (5.47) comes out, including its 86.8.

1 + cos²θ — the measured shape
2 × [½(1+cos θ)]² — same-helicity pairs
2 × [½(1−cos θ)]² — opposite

The two dashed curves are the two pairs of helicity configurations; each vanishes at one end, and each on its own is strongly asymmetric. Their sum is symmetric about 90° and never falls below half its maximum. The forward–backward symmetry is a prediction: any asymmetry measured here is evidence of something beyond a single photon — and at LEP it was, because the Z contributes too (§9.7).

⚙️ Engineer’s bridge — the 1/s law is dimensional analysis, and it tells you when you are done

Eq. (5.44)‘s $1/s$ needs no calculation. A cross-section is an area; in natural units an area is an inverse energy squared; and above every mass in the problem the only energy in the problem is $\sqrt s$ . Therefore $\sigma \propto 1/s$ , and the constant of proportionality is the only thing left to compute.

That is the same move as non-dimensionalising a system before solving it. Once the only scale is fixed, the form of the answer is fixed and the work reduces to a pure number — here $4\pi\alpha^2/3$ .

And the failure mode is the useful part. The argument assumed the mediator’s mass is negligible too. Put a heavy mediator in and there are now two energies, $\sqrt s$ and $M$ , so dimensional analysis alone no longer fixes the form — which is exactly the low-|q| plateau of §5.4’s propagator plot, and why the weak cross-section rises as $s$ instead of falling as $1/s$ until $\sqrt s$ reaches 80 GeV (§7.1). When a dimensional argument stops working, it is telling you a scale you had ignored has become relevant.

Where it breaks: dimensional analysis fixes the scaling and never the coefficient, and it is silent about everything that carries no dimensions. The $1+\cos^2\theta$ angular distribution — the part of §5.7 that proves the quark has spin ½ — cannot be obtained this way at all, because an angle is dimensionless and dimensional analysis has nothing to say about it. It also assumes only one scale is relevant, which is exactly the assumption that fails at the Z: two scales, and the clean $1/s$ becomes a resonance.

\sigma\left(e^+e^- \to \mu^+\mu^-\right) = \frac{4\pi\alpha^2}{3s} = \frac{\htmlClass{t-n}{86.8\ \text{nb}}}{s\ \left[\text{GeV}^2\right]}

(5.47)

Bettini p. 206. One number, and it is the yardstick against which every e⁺e⁻ cross-section in the book is measured.

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 hadronic cross-section , and the plot to keep

Replace the muon pair by “anything made of quarks” and you get the quantity that Chapter 6 is built on.

e⁺e⁻ → hadrons
point-like 86.8 nb / s

Fig. 5.27, redrawn. The dashed line is Eq. (5.47) — the point-like μ⁺μ⁻ cross-section, a pure 1/s. The solid curve is the hadronic one: the SAME 1/s law, multiplied by R (the number of quark species times three colours, weighted by charge²), with a resonance spike wherever √s hits a real state. Continuum and R staircase are computed; peak heights are indicative rather than digitised.

💡 What this really says — three separate stories in one plot

This figure carries more of the book than any other, and it is worth reading three times.

The slope. Away from the spikes the hadronic cross-section falls as $1/s$ , exactly parallel to the dashed point-like line. Hadrons are produced by making a quark pair, and a quark pair is produced exactly like a muon pair — so the shape is a statement that quarks are point-like. Chapter 1’s Rutherford argument, run at much higher energy.

The gap between the curves. The vertical offset is the ratio $R$ , and it counts: how many quark flavours are light enough to be made, weighted by the square of their charge, times three for colour. That single factor of three is the cleanest measurement of the number of colours there is, and §6.4 is where the site does it.

The steps. $R$ is not constant. It rises from 2 to 10/3 when √s passes the charm threshold and to 11/3 past bottom, because a new quark pair becomes producible. Each step is a new flavour appearing — the plot is the particle census of Chapter 4, read off a cross-section.

And on top of all three, the spikes: every resonance of §4.5 and §4.9 and §4.10, plus the Z at 91 GeV, which is where Chapter 9 begins.

What the two curves have in common and where they differ

feature	μ⁺μ⁻ (dashed)	hadrons (solid)
energy dependence	1/s, exactly	1/s between thresholds — the same law, which is why quarks are point-like
normalisation	86.8 nb GeV², from α alone	R × 86.8, and R counts flavours × charge² × 3 colours
structure	none at all	a step at every quark-pair threshold, and a spike at every 1⁻⁻ resonance
what it measures	α, and nothing else	the number of quarks, their charges, and the number of colours

The muon curve is the <strong>control</strong>. Everything interesting about the hadronic one is a departure from it — which is why the book draws them on the same axes, and why R, their ratio, is what Chapter 6 actually plots.

🔑 If you remember only three things

The angular shape comes from spin, not from dynamics. Nothing about the strength of the interaction enters it, which is why the same shape appears wherever two spin-½ particles annihilate.
Measuring against a known process cancels the luminosity. You never have to know how many collisions took place, only how the two counts compare with each other.
Each step in R is a threshold opening. The plot is a census of quark flavours taken by turning the energy up until the next one can be made.

Where this goes next

§5.8 measures α by comparing this cross-section with the data as a function of energy — the deviations from a constant α are the running.
§6.4 turns the gap between the two curves into the R ratio and counts colours with it.
§7.1 is where the 1/s dimensional argument fails: a massive mediator introduces a second scale, and the weak cross-section rises with s until √s reaches the W mass.
§9.7 is the spike at 91 GeV. Its width counts the number of light neutrino species, and the answer is three.

✅ Check yourself — e⁺e⁻ → μ⁺μ⁻ and the hadronic cross-section

0/5 answered · 0 correct

1.Why does $\sigma \propto 1/s$ without any calculation being done?
2.The $(1 + \cos^2\theta)$ shape is the sum of four terms. What are they?
3.Why do the four configurations' amplitudes get squared and then added, rather than added and then squared?
4.In Fig. 5.27, what does the vertical gap between the hadronic curve and the dashed point-like line measure?
5.Away from the resonances, the hadronic curve is exactly parallel to the point-like one on the log–log plot. What does that parallelism prove?