 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2015/iii/paper-54/1/a/i/solution

Codex  0 Created 2026-10-03 Updated 2026-10-06

m^{'} = 0

, so

m (r) = M

. The remaining metric function obeys

Φ^{'} (r) = \frac{M}{r ( r - 2 M )} = \frac{1}{2} \frac{d}{d r} lo g (1 - \frac{2 M}{r}) .

Thus

e^{2Φ} = C (1 - 2 M / r)

. A constant rescaling of the time coordinate removes

C

; equivalently, asymptotic flatness fixes

C = 1

when time is normalized at infinity. Consequently the exterior is the Schwarzschild metric:

d s^{2} = - (1 - \frac{2 M}{r}) d t^{2} + (1 - \frac{2 M}{r})^{- 1} d r^{2} + r^{2} d Ω^{2}, r > R .

The mass

M = m (R)

includes the full perfect fluid in general relativity interior. No equation of state beyond the surface enters this vacuum result.

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