 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ii/paper-4/37d/iv/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

Move the positive cosmological constant to the source side and define

T_{α β}^{(Λ)} = - \frac{Λ}{8 π} g_{α β} .

T^{(Λ)} = - Λ/ (2 π)

. Its contribution to the trace-reversed

00

source is

T_{00}^{(Λ)} - \frac{1}{2} η_{00} T^{(Λ)} = - \frac{Λ}{8 π} .

\nabla^{2} Φ = 4 π ρ - Λ .

For a point mass,

\nabla^{2} r^{2} = 6

, so

Φ = - \frac{M}{r} - B r^{2}, - 6 B = - Λ, B = \frac{Λ}{6} .

The radial acceleration is

- \frac{d Φ}{d r} = - \frac{M}{r ^{2}} + \frac{Λ}{3} r .

Thus positive

Λ

supplies an outward acceleration and balances the point-mass attraction at the static radius

r_{s} = (\frac{3 M}{Λ})^{1/3} .

M / r ≪ 1

and

Λ r^{2} ≪ 1

, giving parametrically

M ≪ r ≪ Λ^{- 1/2} .

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