In the orthonormal frame choose the null vector $k=e_0+e_2$, which has a nonzero angular component. The curvature forms give
$$
R_{ab}k^ak^b
=R_{00}+R_{22}
=\frac12f''-\frac{f-1}{r^2}.
$$
For a null vector, the trace term in the <Einstein field equations> drops out, so the <null energy condition> implies $R_{ab}k^ak^b\geq0$. Therefore
$$
f''\geq\frac{2(f-1)}{r^2}.
$$

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 309 / 3 / d / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-309/3/d/solution