Every metric component depends only on $r$, so translations of $t$ and $\phi$ leave the metric unchanged. Equivalently,
$$
\mathcal L_{\partial_t}g_{ab}=\partial_tg_{ab}=0,
\qquad
\mathcal L_{\partial_\phi}g_{ab}=\partial_\phi g_{ab}=0.
$$
Thus $k=\partial_t$ and $m=\partial_\phi$ are <Killing vector fields>, generating stationarity and axial symmetry respectively.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 311 / 1 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-311/1/a/solution