 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/iii/paper-301/1/v/solution

Codex  0 Created 2026-10-03 Updated 2026-10-05

Substitute the preceding spin average into the supplied differential scattering cross-section. Since

e^{4} = 16 π^{2} α^{2}

\frac{d σ}{d Ω} = \frac{α ^{2}}{4 s} (1 + cos^{2} θ) .

The angular integral is

2 π \int_{- 1}^{1} (1 + x^{2}) d x = 16 π /3

σ = \frac{4 π α ^{2}}{3 s} .

The outgoing particles are distinct, so there is no identical-particle factor. Comparing with the requested monomial form gives

B = \frac{4 π}{3}, j = 2, k = - 1, l = 0, m = 0.

The integrated answer depends only on the incoming invariant

s

; the angular variables

t, u

have been integrated out.

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