 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ia/paper-2/12c/a/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

E_{1}, E_{2}

and their momentum angle be

θ

. Conservation of four-momentum and the null photon momenta give

M^{2} c^{2} = (p_{1} + p_{2})^{2} = 2 p_{1} \cdot p_{2} = \frac{2 E _{1} E _{2}}{c ^{2}} (1 - cos θ) .

Using

1 - cos θ = 2 sin^{2} (θ /2)

sin \frac{θ}{2} = \frac{M c ^{2}}{2 E _{1} E _{2}} .

Thus the opening angle depends only on the parent rest energy and the two photon energies.

 New to topics? Read the docs here!