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Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 53 / 2 / Solution by

Codex  0 Created 2026-10-03 Updated 2026-10-20

Use metric signature

(+ - - -)

and average over the four initial spin states, summing over final spins and the

N_{c} = 3

quark color charge states. The electrons carry no color charge, so their initial color charge average is trivial. A final color charge average would not give the inclusive relativistic cross-section requested.

Photon exchange gives the tree amplitude, up to an irrelevant overall phase,

M = \frac{e ^{2} Q _{q}}{s} δ_{ab} [\overset{v}{ˉ} (p_{1}) γ^{μ} u (p_{2})] [\overset{u}{ˉ} (q_{1}) γ_{μ} v (q_{2})], e^{2} = 4 π α .

Massless spin sums give

\sum u \overset{u}{ˉ} = \neq p

and

\sum v \overset{v}{ˉ} = \neq p

. Therefore

\overline{∣ M ∣^{2}} = \frac{N _{c} e ^{4} Q _{q}^{2}}{4 s ^{2}} tr (\neq p_{1} γ^{μ} \neq p_{2} γ^{ν}) tr (\neq q_{1} γ_{μ} \neq q_{2} γ_{ν}) .

The color charge multiplicity is

\sum_{a, b} ∣ δ_{ab} ∣^{2} = N_{c}

, not

N_{c}^{2}

. Each trace equals

4 (a^{μ} b^{ν} + a^{ν} b^{μ} - g^{μν} a \cdot b)

. Contracting them, the terms proportional to

(p_{1} \cdot p_{2}) (q_{1} \cdot q_{2})

cancel, leaving

32 [(p_{1} \cdot q_{1}) (p_{2} \cdot q_{2}) + (p_{1} \cdot q_{2}) (p_{2} \cdot q_{1})]

. Thus

\overline{∣ M ∣^{2}} = \frac{384 π ^{2} α ^{2} Q _{q}^{2}}{s ^{2}} [(p_{1} \cdot q_{2}) (p_{2} \cdot q_{1}) + (p_{2} \cdot q_{2}) (p_{1} \cdot q_{1})] .

In the centre-of-momentum frame every energy is

s /2

. Let

θ

be the angle between

p_{1}

and

q_{2}

. Then

p_{1} \cdot q_{2} = p_{2} \cdot q_{1} = \frac{s}{4} (1 - cos θ), p_{1} \cdot q_{1} = p_{2} \cdot q_{2} = \frac{s}{4} (1 + cos θ) .

Consequently

\overline{∣ M ∣^{2}} = N_{c} e^{4} Q_{q}^{2} (1 + cos^{2} θ)

.

The supplied two-body Lorentz-invariant phase space formula has a factor-of-two error. Deriving the normalization explicitly,

d Φ_{2} = \frac{d Ω}{32 π ^{2}}, incident flux = 4 p_{1} \cdot p_{2} = 2 s,

so the massless two-body scattering flux normalization is

\frac{d σ}{d Ω} = \frac{∣ M ∣ ^{2}}{64 π ^{2} s} = \frac{∣ M ∣ ^{2}}{128 π ^{2} p _{1} \cdot p _{2}} .

For example, integrating the spatial Dirac delta in

d Φ_{2}

[4 (2 π)^{2}]^{- 1} d k d Ω δ (s - 2 k)

; its radial integral supplies the factor

1/2

. The printed denominator

64 π^{2} p_{1} \cdot p_{2}

would double the relativistic cross-section and cannot reproduce the requested result with the standard scattering amplitude.

The correctly normalized differential and total results for electron-positron annihilation into a quark pair are

\frac{d σ}{d Ω} = \frac{3 α ^{2} Q _{q}^{2}}{4 s} (1 + cos^{2} θ), \int d Ω (1 + cos^{2} θ) = \frac{16 π}{3},

and hence

σ_{q} = \frac{4 π α ^{2}}{s} Q_{q}^{2} .

At energies where photon exchange and the parton description apply, sum over kinematically accessible quark flavors. The produced quarks undergo hadronization, so the leading inclusive hadronic rate is

σ_{had} = \frac{4 π α ^{2}}{s} \sum_{q} Q_{q}^{2}, R (s) = \frac{σ _{had}}{σ ( e ^{+} e ^{-} \to μ ^{+} μ ^{-} )} = 3 \sum_{q} Q_{q}^{2} .

This is the hadronic R ratio. Perturbative QCD adds a leading relative correction

α_{s} (s) / π

; thresholds, narrow resonances and, near the Z boson scale, electroweak exchange require additional treatment. The calculation is a photon-exchange, massless result from a tree-level Feynman diagram.

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