 Solution

ID: past-exam-of-the-particle-physics-course-of-the-university-of-oxford/2022/1/c/solution

s = (p_{1} + p_{2})^{2}

t = (p_{1} - p_{3})^{2} = q^{2}

and

u = (p_{1} - p_{4})^{2}

. Dropping the electron and muon masses gives

p_{1} \cdot p_{2} = p_{3} \cdot p_{4} = \frac{s}{2}, p_{1} \cdot p_{4} = p_{3} \cdot p_{2} = - \frac{u}{2} .

These identities follow by expanding the corresponding squared four-momenta, using

p_{i}^{2} = 0

and four-momentum conservation. Substitution into the supplied spin-averaged matrix element yields

\overline{∣ M ∣^{2}} = \frac{8 e ^{4}}{t ^{2}} (\frac{s ^{2}}{4} + \frac{u ^{2}}{4}) = 2 e^{4} \frac{s ^{2} + u ^{2}}{t ^{2}} .

The

1/4

initial spin average is already included in the given expression; no further averaging factor belongs here.

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