Because $t_a=\nabla_at$ is a gradient, $\nabla_at_b=\nabla_bt_a$. Therefore
$$
t^b\nabla_bt_a=t^b\nabla_at_b
=\frac12\nabla_a(t^bt_b)=0.
$$
Thus $t^a$ is an affinely parametrized <timelike geodesic vector field>, and because $t$ is constant on $S$, it emanates orthogonally from $S$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 311 / 3 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-311/3/a/solution