 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/ii/paper-1/38a/a/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

Spherical symmetry puts any geodesic in a plane through the origin, chosen as

θ = π /2

E = (1 - r_{s} / r) \dot{t}, L = r^{2} \dot{ϕ} .

The null condition gives

\overset{r}{˙}^{2} = F (r) = E^{2} - (1 - r_{s} / r) \frac{L ^{2}}{r ^{2}} .

For

u = 1/ r

(d u / d ϕ)^{2} = \overset{r}{˙}^{2} / L^{2}

, hence

(\frac{d u}{d ϕ})^{2} = \frac{1}{b ^{2}} - u^{2} + r_{s} u^{3}, b = L / E .

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