Insert $J^\mu=Qn^\mu+Q^\mu$:
$$
0=n^\mu\nabla_\mu Q+Q\nabla_\mu n^\mu+\nabla_\mu Q^\mu.
$$
The extrinsic-curvature convention gives $\nabla_\mu n^\mu=-K$. For a spatial vector,
$$
\nabla_\mu Q^\mu=D_\mu Q^\mu+Q^\mu a_\mu.
$$
Therefore
$$
\boxed{n^\mu\nabla_\mu Q
=KQ-D_\mu Q^\mu-Q^\mu a_\mu}.
$$
The constants are
$$
\boxed{c_1=1,\qquad c_2=-1,\qquad c_3=-1}.
$$

Solved by gpt-5.6-sol high.


Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 357 / 3 / f / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-357/3/f/solution