University Quantum Field Theory I · QED at Tree Level: the Lagrangian, Feynman Rules and Electron-Muon Processes · 9.6

The cross-section: 1 + cos²θ and σ = 4πα²/(3s)

This is the first QED number you can hold up against a detector. We carry the last topic's trace into the centre-of-mass frame with the muon mass kept, watch the angular distribution turn from isotropic into 1 + cos²θ, and follow σ from zero through a 1 μb peak to the 86.85 nb ÷ s law that PETRA tested.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

One virtual photon, two pointlike Dirac fermions, lowest order in α: that is the whole model, and in the centre-of-mass frame topic 9.05's trace collapses to one function of the muon speed β and the angle θ, (1/4)Σ|ℳ|² = e⁴(2 − β² sin²θ). Topic 5.04's formula adds one more β from phase space and nothing else. The shape is a statement about angular momentum.

Just above threshold the pair is born in an s wave whose spins carry the photon's polarisation, so the direction of flight is free, emission is isotropic and σ rises linearly in β. Far above threshold the vector coupling ties each muon's spin to its momentum, the beam's J_z = ±1 passes to the pair, and the distribution becomes 1 + cos²θ, twice as strong along the beam as across it. Integrated, σ = (4πα²/(3s)) β(3 − β²)/2 rises from zero, peaks near 1 μb at √s ≈ 0.25 GeV and settles onto 86.85 nb divided by s in GeV², the muon mass surviving only at order m_μ⁴/s².

The model stops in three places: slow muons attract, a correction of order α/β that wins at threshold; real and virtual photons add order-α corrections that Unit 12 computes; and at tens of GeV the Z enters. PETRA confirmed the 1/s law to a few per cent, and also saw a forward-backward asymmetry near −0.1 that no function of cos²θ can make.

Simple definition: The tree-level cross-section for e⁺e⁻ → μ⁺μ⁻ is the one-photon rate of muon pairs per unit flux: in the centre-of-mass frame dσ/dΩ = (α²β/(4s))(2 − β² sin²θ), which becomes (α²/(4s))(1 + cos²θ) and σ = 4πα²/(3s) far above threshold.
Example: At √s = 10.0 GeV, s = 100 GeV², so σ = 86.85 nb ÷ 100 = 0.869 nb, and the muon mass changes that by less than 10⁻⁷; per steradian, twice as many muons leave along the beam axis as at 90°.

Spin-averaged |ℳ|² in the centre-of-mass frame(1/4)Σ|ℳ|² = e⁴(2 − β² sin²θ)

Derived from topic 9.05's trace with p₁⋅p₃ = (s/4)(1 − β cos θ) and p₁⋅p₂ = s/2; writing 4m_μ²/s = 1 − β² absorbs the mass term into one function.

e² = 4πα; θ from the e⁻ to the μ⁻; β = √(1 − 4m_μ²/s) the muon speed; mₑ = 0; ℳ dimensionless for 2 → 2

Angular distribution at any muon speeddσ/dΩ = (α²β/(4s))(2 − β² sin²θ)

Derived from topic 5.04's |ℳ|²|p_f|/(64π²s|pᵢ|) with |p_f|/|pᵢ| = β. At threshold it is isotropic and proportional to β.

s in GeV², dσ/dΩ in GeV⁻² sr⁻¹; α²/4 × (ħc)² = 5.18 nb GeV² gives nb sr⁻¹

High-energy limitdσ/dΩ = (α²/(4s))(1 + cos²θ), for s ≫ 4m_μ²

Derived as β → 1. Even in cos θ, so tree-level QED predicts no forward-backward asymmetry at any energy.

same units; θ = 0 against θ = 90° in the ratio 2 : 1; the shape's corrections are of relative order m_μ²/s

Total cross-section with the muon mass keptσ = (4πα²/(3s)) β(3 − β²)/2

Derived by integrating over angles. It rises as β, an s wave, and with x = 4m_μ²/s differs from 4πα²/(3s) only by 3x²/8 at high energy.

σ in GeV⁻²; β(3 − β²)/2 = β(1 + 2m_μ²/s); valid from threshold, √s = 2m_μ = 0.2113 GeV

Point cross-section in nanobarns4πα²/(3s) = 86.85 nb × (1 GeV²/s)

Computed: 0.869 nb at √s = 10 GeV and 70.9 pb at 35 GeV, where PETRA's measured muon-pair cross-sections agreed to within a few per cent.

α = 1/137.036, (ħc)² = 0.38938 GeV² mb, 1 nb = 10⁻³³ cm²

Coulomb factor for slow muons (quoted)S = (πα/β) / (1 − e^(−πα/β)) ≈ 1 + πα/(2β)

Quoted, not derived, and neglected in this lesson: 1.12 at β = 0.1, 1.02 at β = 0.5. As β → 0, Sβ → πα, so σ starts at 2π²α³/s ≈ 67 nb, not zero.

dimensionless; multiplies σ near threshold; β the muon speed in the centre-of-mass frame

The trace in the centre-of-mass frame

Work in natural units, ħ = c = 1, with the metric (+, −, −, −). Topic 9.05 left the spin average, with mₑ set to zero, as (1/4)Σ|ℳ|² = (8e⁴/s²)[(p₁⋅p₃)(p₂⋅p₄) + (p₁⋅p₄)(p₂⋅p₃) + m_μ²(p₁⋅p₂)]. In the centre-of-mass frame each beam carries energy √s/2 along z, and the μ⁻ leaves at angle θ to the e⁻ with the same energy and momentum β√s/2, where β = √(1 − 4m_μ²/s). Then p₁⋅p₃ = p₂⋅p₄ = (s/4)(1 − β cos θ), p₁⋅p₄ = p₂⋅p₃ = (s/4)(1 + β cos θ) and p₁⋅p₂ = s/2. The bracket becomes (s²/8)(1 + β² cos²θ + 4m_μ²/s), and writing 4m_μ²/s = 1 − β² collapses it to (1/4)Σ|ℳ|² = e⁴(2 − β² sin²θ), derived here. It is dimensionless, as a 2 → 2 amplitude must be, and the energy enters only through β.

Phase space adds one β

Topic 5.04's centre-of-mass formula is dσ/dΩ = |ℳ|²|p_f|/(64π²s|pᵢ|). With mₑ = 0 the incoming momentum is √s/2 and the outgoing one β√s/2, so their ratio is β: the slower the final pair, the less phase space it has. With e⁴ = (4πα)² = 16π²α² this gives dσ/dΩ = (α²β/(4s))(2 − β² sin²θ), derived. Dimensions: α is a pure number and 1/s is GeV⁻², an area per steradian once multiplied by (ħc)² = 0.3894 GeV² mb. Far above threshold β → 1 and dσ/dΩ = (α²/(4s))(1 + cos²θ). The prefactor α²/4 is 5.18 nb GeV², so at √s = 10 GeV the muons arrive at 0.104 nb sr⁻¹ along the beam and 0.052 nb sr⁻¹ at 90°. Keeping the electron mass would change all of this only by terms of order mₑ²/s, two parts in 10⁵ even at the muon threshold.

Reading the shape: isotropic at threshold, 1 + cos²θ above

Split the distribution as 2 − β² sin²θ = (1 + cos²θ) + (1 − β²) sin²θ. The first piece is all that survives at high energy. The second is (4m_μ²/s) sin²θ: not the trace's m_μ²(p₁⋅p₂) term alone, which is the isotropic 4m_μ²/s, but that term less the (4m_μ²/s) cos²θ by which β² cos²θ falls short of cos²θ. It is the rate for muons of equal helicity, J_z = 0 along their line of flight, which needs the mass to flip a helicity and vanishes for the massless muons whose amplitudes topic 9.07 takes one by one. At threshold the two pieces add to 2 at every angle: the pair is born in an s wave, ³S₁, whose spins carry the photon's polarisation while the direction of flight is free, so emission is isotropic. At high energy the vector coupling locks each spin to its momentum, the beam's J_z = ±1 is handed to the pair, and the muons favour the beam axis two to one over 90°. The ratio of dσ/dΩ at 90° to that at 0° is (2 − β²)/2, a speedometer for the muons: 0.857 at √s = 0.25 GeV, 0.522 at 1 GeV. It approaches 0.5 only as m_μ²/s, so the shape still feels the muon mass after σ has stopped feeling it.

Integrating: why the mass correction to σ is second order

Over angles, ∫dΩ (2 − β² sin²θ) = 2π ∫(2 − β² + β² cos²θ) d(cos θ) = (8π/3)(3 − β²), so σ = (4πα²/(3s)) β(3 − β²)/2, derived. Since 3 − β² = 2 + 4m_μ²/s, the factor is also β(1 + 2m_μ²/s). Near threshold σ ≈ (2πα²/s)β, linear in β: an s wave, whose cross-section grows with phase space alone, whereas a P wave would start as β³. Far above threshold put x = 4m_μ²/s. Then β ≈ 1 − x/2 − x²/8 and the factor 1 + x/2 cancel at first order, leaving σ = (4πα²/(3s))(1 − 3x²/8 + …): the phase space the mass takes away, the extra (1 − β²) sin²θ in |ℳ|² gives back. At √s = 0.5 GeV the true σ is 1.3% below 4πα²/(3s), and at 1 GeV only 0.08% below. Between threshold and that regime σ must turn over, rising from zero, peaking a few tens of MeV above threshold and then falling as 1/s.

86.85 nanobarns over s, and what a collider counts

To compare with a measurement, restore ħ and c: 4πα²/(3s) = (4π/3)(1/137.036)² × 0.38938 GeV² mb ÷ s = 86.85 nb × (1 GeV²/s), computed. The 1/s is forced by dimensions: α carries none, so once √s dwarfs every mass, s is the only scale from which an area can be built. The numbers fall quickly: 9.65 nb at √s = 3 GeV, 0.869 nb at 10 GeV and 70.9 pb at 35 GeV. At a luminosity of 10³¹ cm⁻² s⁻¹, the order PETRA reached, 70.9 pb = 7.09 × 10⁻³⁵ cm² gives 7.1 × 10⁻⁴ pairs per second, about 61 a day. Because the muon behaves as a point charge, this cross-section carries no form factor, and that makes it the yardstick of e⁺e⁻ physics: hadron production is quoted in units of it, the ratio R = σ(e⁺e⁻ → hadrons)/σ(e⁺e⁻ → μ⁺μ⁻), named here only.

Where tree level stops: slow muons, radiation and the Z

Three effects are left out, each named here and none derived. Slow muons attract: their Coulomb interaction multiplies σ by S = (πα/β)/(1 − e^(−πα/β)) ≈ 1 + πα/(2β), quoted from the Sommerfeld treatment of a non-relativistic pair. It exceeds 1.1 only for β below 0.12, within 1.5 MeV of threshold, but there it wins: Sβ → πα, so σ starts not at zero but at 2π²α³/(4m_μ²) ≈ 67 nb. Real and virtual photons add corrections of order (α/π) ln(s/mₑ²), about 5% at √s = 35 GeV, their size set by the cut on radiated photon energy; Unit 12 computes them. And at tens of GeV the Z enters. PETRA, running at √s up to 46.8 GeV, found σ(μ⁺μ⁻) equal to 4πα²/(3s) within a few per cent once radiative corrections were applied, quoted here as measured. The Z changes that σ by under half a per cent at 35 GeV, yet it produces a forward-backward asymmetry near −0.1 there, which no tree-level QED term can give.

Change one variable at a time

Make the relationship visible.

Interactive model

Collision energy √s0.300 GeV

Marker angle cos θ0.50

Start at √s = 0.212 GeV: the left curve is flat and low, an s wave. Raise √s and σ climbs to about 1 μb near 0.25 GeV before following the dashed 1/s. By 0.5 GeV σ is within 1.3% of it, yet on the left the forward value is still 9% low.

MUON SPEED β0.710

σ AT TREE LEVEL854.9 nb

σ ÷ 4πα²/(3s)0.886

dσ/dΩ AT THE MARKER66.3 nb sr⁻¹

Live interpretationMUON SPEED β: 0.710. σ AT TREE LEVEL: 854.9 nb. σ ÷ 4πα²/(3s): 0.886. dσ/dΩ AT THE MARKER: 66.3 nb sr⁻¹

Catch the common trap

Explain before calculating.

Misconception check

Students who meet PETRA's measured forward-backward asymmetry, quoted here as about −0.1 at √s ≈ 35 GeV, often look for it in a term this calculation dropped: the muon mass, the recoil, a higher power of β. None can produce it. The tree-level result 2 − β² sin²θ contains cos θ only as cos²θ, at every speed and for every mass, so it is unchanged by θ → π − θ and its asymmetry A_FB = (F − B)/(F + B) is exactly zero; topic 11.09 traces that to charge conjugation. An odd power of cos θ needs interference between amplitudes that behave oppositely under C. QED supplies that only at order α³, through two-photon exchange and the interference of initial- and final-state radiation, of order +1 per cent depending on cuts. The large negative part is γ–Z interference, the weak interaction showing itself more than 50 GeV below the Z peak.

At √s = 0.300 GeV, with m_μ = 0.10566 GeV kept and mₑ neglected, the tree-level σ(e⁺e⁻ → μ⁺μ⁻) equals 4πα²/(3s) multiplied by which factor?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyAn e⁺e⁻ collider runs at √s = 12.0 GeV and records an integrated luminosity of 50.0 pb⁻¹. At tree level, how many μ⁺μ⁻ pairs are produced, and how many of them have the μ⁻ in the central region 60° ≤ θ ≤ 120°?

Check that the muon mass can be dropped: x = 4m_μ²/s = 0.04465/144 = 3.1 × 10⁻⁴, so σ differs from 4πα²/(3s) by 3x²/8 ≈ 4 × 10⁻⁸ and the shape is 1 + cos²θ to within 0.03%.
σ = 86.85 nb ÷ 144 = 0.6031 nb = 603.1 pb.
N = σ × ∫L dt = 603.1 pb × 50.0 pb⁻¹ = 3.016 × 10⁴ pairs; the units cancel, as an event count must.
The central region is |cos θ| ≤ 0.5. Weighting by 1 + cos²θ, the integral of (1 + c²) dc from −0.5 to 0.5 is 1 + 1/12 = 1.0833, against 8/3 = 2.6667 over the full range: a fraction 0.4063.
N_central = 0.4063 × 3.016 × 10⁴ = 1.225 × 10⁴. Isotropic emission would have put half, 1.508 × 10⁴, there, so the 1 + cos²θ shape moves about 2.8 × 10³ pairs toward the beams.

AnswerAbout 3.02 × 10⁴ μ⁺μ⁻ pairs, of which 1.23 × 10⁴ (40.6%) lie in 60° ≤ θ ≤ 120°, against 50% for isotropic emission.

MediumJust above threshold the muon mass matters. At √s = 0.400 GeV, with m_μ = 0.10566 GeV and mₑ neglected, find the muon speed β, the ratio of dσ/dΩ at θ = 90° to its value at θ = 0°, and σ in nanobarns, and compare each with the massless-muon prediction.

s = 0.1600 GeV² and 4m_μ²/s = 0.04465/0.1600 = 0.2791, so β² = 0.7209 and β = 0.8491.
From dσ/dΩ ∝ 2 − β² sin²θ, θ = 0° gives 2 and θ = 90° gives 2 − β² = 1.2791, so the ratio is 0.6395. Massless muons would give exactly 0.5: the mass fattens the distribution at 90° by 28%.
σ = (86.85 nb ÷ 0.1600) × β(3 − β²)/2 = 542.8 nb × 0.8491 × 1.1396 = 525.2 nb.
The massless 4πα²/(3s) is 542.8 nb, so the true σ is 3.2% lower. The leading estimate 3x²/8 with x = 0.2791 gives 2.9%; the rest is the next order in x.
Dimensions: α² is a pure number and 1/s is GeV⁻², an area once multiplied by (ħc)² = 0.3894 GeV² mb, which the 86.85 nb already contains.

Answerβ = 0.849; dσ/dΩ(90°) ÷ dσ/dΩ(0°) = 0.640 against 0.500 for massless muons; σ = 525 nb against 543 nb. The shape moves by 28% while σ moves by only 3%.

HardStarting from σ = (4πα²/(3s)) β(3 − β²)/2, find the centre-of-mass energy at which the tree-level e⁺e⁻ → μ⁺μ⁻ cross-section is largest, and its value in microbarns. Then use the quoted Coulomb factor S = (πα/β)/(1 − e^(−πα/β)) to estimate how much the neglected attraction of the slow muons changes it there.

Trade s for β using s = 4m_μ²/(1 − β²): σ = [4πα²/(3 × 4m_μ²)] × (1 − β²)β(3 − β²)/2. The prefactor is the massless formula at threshold, 86.85 nb ÷ 0.04465 = 1945 nb.
Maximise f(β) = (1 − β²)β(3 − β²) = 3β − 4β³ + β⁵. Setting f′ = 3 − 12β² + 5β⁴ = 0 gives β² = (6 − √21)/5 = 0.2835, the other root exceeding 1, so β = 0.5324 and f = 1.0363.
Then s = 0.04465/(1 − 0.2835) = 0.06232 GeV² and √s = 0.2496 GeV = 2.36 m_μ, only 38 MeV above threshold.
σₘₐₓ = 1945 nb × 1.0363/2 = 1008 nb ≈ 1.01 μb. Check by the direct route: 86.85 nb ÷ 0.06232 × 0.5324 × (3 − 0.2835)/2 = 1008 nb.
Coulomb factor: πα/β = 0.02293/0.5324 = 0.04306, so S = 0.04306/(1 − e^(−0.04306)) = 1.0217, close to 1 + πα/(2β) = 1.0215. The attraction raises the peak by about 2.2%, to about 1.03 μb, and pulls it slightly toward threshold, where S grows as 1/β.

Answerσ peaks at √s = 0.250 GeV (β = 0.532) at 1.01 μb at tree level; the neglected Coulomb attraction would raise it there by about 2.2%.