Only the transverse metric components are needed. At fixed $(u,r,\theta)$,
$$
\partial_\phi\widetilde x=-\widetilde y,
\qquad
\partial_\phi\widetilde y=\widetilde x.
$$
Contracting this vector with the $\widetilde x,\widetilde y$ block of the metric makes all angular factors combine to $\widetilde\rho^2$:
$$
g_{\phi\phi}
=\kappa^2\widetilde\rho^2
+\kappa\lambda\widetilde\rho^2E
=r^2\sin^2\theta
\left(1+\frac\lambda\kappa E\right).
$$
Substitution into the preceding news formula gives
$$
\boxed{\partial_uc
=-\frac\lambda{2\kappa}
\lim_{r\to\infty}\left(r\,\partial_uE\right)}.
$$
Therefore the requested constants are
$$
\boxed{p=-\frac\lambda{2\kappa},\qquad q=1}.
$$

Solved by gpt-5.6-sol high.


Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 357 / 2 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-357/2/c/solution