 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2014/iii/paper-50/4/a/solution

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

Vary the geodesic Lagrangian before imposing the normalization of proper time. With

u^{μ} = d x^{μ} / d τ

\frac{\partial L}{\partial u ^{α}} = - 2 g_{αν} u^{ν}, \frac{\partial L}{\partial x ^{α}} = - \partial_{α} g_{μν} u^{μ} u^{ν} .

g_{αν} \frac{d u ^{ν}}{d τ} + \partial_{ρ} g_{αν} u^{ρ} u^{ν} - \frac{1}{2} \partial_{α} g_{μν} u^{μ} u^{ν} = 0,

equivalently the affinely parametrized geodesic equation. For the static weak metric, its leading spatial connection coefficient is

Γ^{i}_{00} = - \frac{1}{2} g^{ij} \partial_{j} g_{00} = \frac{\partial _{i} Φ}{1 - 2Φ} ≃ \partial_{i} Φ.

All terms with two spatial velocities are suppressed by

v^{2}

. Therefore

d^{2} x^{i} / d τ^{2} ≃ - \partial_{i} Φ (d t / d τ)^{2}

Because

t

is cyclic, the same Euler-Lagrange equation gives

(1 + 2Φ) d t / d τ = constant

. On converting to coordinate time, the additional term

(d^{2} t / d τ^{2}) (d x^{i} / d t)

is of order

v^{2} \nablaΦ

and can be dropped. Hence

\frac{d ^{2} x}{d t ^{2}} = - \nablaΦ

to leading order, the Newtonian limit of general relativity. Restoring SI units gives

Φ = Φ_{SI} / c^{2}

, so the acceleration is

- \nabla (c^{2} Φ)

. Substituting $L = 1$ from proper-time normalization before varying would incorrectly erase the dynamics.

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