 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ii/paper-2/36d/a/solution

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

L = \frac{1}{2} [- f \dot{t}^{2} + f^{- 1} \overset{r}{˙}^{2} + r^{2} \dot{θ}^{2} + r^{2} sin^{2} θ \dot{ϕ}^{2}] .

E = f \dot{t}, L_{z} = r^{2} sin^{2} θ \dot{ϕ}, L^{2} = r^{4} (\dot{θ}^{2} + sin^{2} θ \dot{ϕ}^{2}) .

The constant norm of the tangent is

g_{ab} \overset{x}{˙}^{a} \overset{x}{˙}^{b} = - κ

, where

κ = 1

for a timelike geodesic parametrized by proper time and

κ = 0

for a null geodesic. Substitution gives

- κ = - \frac{E ^{2}}{f} + \frac{r ˙ ^{2}}{f} + \frac{L ^{2}}{r ^{2}},

and hence

\overset{r}{˙}^{2} + V (r) = E^{2}, V (r) = f (r) (κ + \frac{L ^{2}}{r ^{2}}) .

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