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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2009/ib/paper-4/17c/solution

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Past exam of the mathematics course of the University of Cambridge / 2009 / ib / Paper 4 / 17C / Solution by

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

Choose the photon direction in the

x^{'}

–

y^{'}

plane. Its four-momentum in the stellar rest frame is

p^{' μ} = \frac{h ν ^{'}}{c} (1, cos θ^{'}, sin θ^{'}, 0) .

Let

β = v / c

,

γ = (1 - β^{2})^{- 1/2}

. The Lorentz transformation to

S

gives

p^{0} = γ (p^{'0} + β p^{' x}), p^{x} = γ (p^{' x} + β p^{'0}), p^{y} = p^{' y}, p^{z} = p^{' z} .

Using

p^{0} = h ν / c

and

p^{x} = (h ν / c) cos θ

yields the relativistic Doppler effect and relativistic aberration:

ν = γ ν^{'} (1 + β cos θ^{'}), cos θ = \frac{cos θ ^{'} + β}{1 + β cos θ ^{'}} .

For the oppositely directed transverse photons, their rest-frame spatial momenta cancel and their total energy is

E = 2 h ν^{'}

. Boosting this combined four-vector gives laboratory energy

γ E

p = \frac{γ E v}{c ^{2}} x = \frac{E v}{c ^{2} 1 - v ^{2} / c ^{2}} x .

The star loses rest mass

E / c^{2}

. At unchanged speed its momentum therefore decreases by

γ (E / c^{2}) v

, exactly the photon momentum. Its laboratory energy loss is likewise

γ E

. There is no rest-frame recoil from the symmetric emission, and constant speed is consistent with four-momentum conservation because the stellar mass decreases.

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