 Solution

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

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

V (r) = L^{2} (\frac{1}{r ^{2}} - \frac{2 m}{r ^{3}} - H^{2}) .

It tends to

- \infty

r \to 0^{+}

, rises to one maximum, and tends to

- L^{2} H^{2}

r \to \infty

V^{'} (r) = 0

, so

- \frac{2}{r ^{3}} + \frac{6 m}{r ^{4}} = 0 ⟹ r = 3 m .

Since

V^{''} (3 m) < 0

, this photon sphere is unstable. Its energy satisfies

E^{2} = L^{2} (1/ (27 m^{2}) - H^{2})

 New to topics? Read the docs here!