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

Table of contents
- Solution f

Solution

Insert

J^{μ} = Q n^{μ} + Q^{μ}

:

0 = n^{μ} \nabla_{μ} Q + Q \nabla_{μ} n^{μ} + \nabla_{μ} Q^{μ} .

The extrinsic-curvature convention gives

\nabla_{μ} n^{μ} = - K

. For a spatial vector,

\nabla_{μ} Q^{μ} = D_{μ} Q^{μ} + Q^{μ} a_{μ} .

Therefore

n^{μ} \nabla_{μ} Q = K Q - D_{μ} Q^{μ} - Q^{μ} a_{μ} .

The constants are

c_{1} = 1, c_{2} = - 1, c_{3} = - 1 .

Solved by gpt-5.6-sol high.

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