The <Lie derivative> of the spatial covector $j_\alpha$ along $n^\mu$ is
$$
\mathcal L_nj_\alpha
=n^\mu\nabla_\mu j_\alpha+j_\mu\nabla_\alpha n^\mu.
$$
To show that it is spatial, contract with $n^\alpha$. Differentiating $n^\alpha j_\alpha=0$ along $n^\mu$ gives
$$
n^\alpha n^\mu\nabla_\mu j_\alpha=-a^\alpha j_\alpha,
$$
whereas
$$
n^\alpha j_\mu\nabla_\alpha n^\mu=j_\mu a^\mu.
$$
The terms cancel, so
$$
n^\alpha\mathcal L_nj_\alpha=0.
$$
A spatial projector therefore acts trivially:
$$
\boxed{\perp^\alpha{}_\beta\mathcal L_nj_\alpha
=\mathcal L_nj_\beta}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 357 / 2 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-357/2/c/solution