 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-4/37e/b/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

For a slowly moving particle in a static weak field, the spatial part of the geodesic equation reduces to

\frac{d ^{2} x ^{i}}{d t ^{2}} = - Γ^{i}_{00} .

To first order in the perturbation,

Γ^{i}_{00} = - \frac{1}{2} δ^{ij} \partial_{j} h_{00},

and hence

\frac{d v ^{i}}{d t} = \frac{1}{2} δ^{ij} \partial_{j} h_{00} .

Newtonian acceleration is

- δ^{ij} \partial_{j} Φ

, so

h_{00} = - 2Φ

. Substitution into part (a), followed by the Poisson equation

\nabla^{2} Φ = 4 π Gρ

, gives

κ ρ = - \nabla^{2} h_{00} = 2 \nabla^{2} Φ = 8 π Gρ .

h_{00} = - 2Φ, κ = 8 π G .

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