 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2008/ib/paper-1/4c/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

Choose the transverse axis so that the photon travels in the

x y

p^{μ} = (E / c, E cos θ / c, E sin θ / c, 0)

v

along

x

, with

β = v / c

and

γ = (1 - β^{2})^{- 1/2}

, gives

E^{'} / c = γ (E / c - β p_{x}), p_{x}^{'} = γ (p_{x} - βE / c) .

∣ p^{'} ∣ = E^{'} / c

. Therefore

E^{'} = \frac{E ( 1 - β cos θ )}{1 - β ^{2}}, cos θ^{'} = \frac{c p _{x}^{'}}{E ^{'}} = \frac{cos θ - β}{1 - β cos θ} .

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