After the shell, spherical symmetry and stationarity put the horizon at $r=2(M+E)$. Before the shell, an outgoing radial null geodesic in the mass-$M$ region has $r$ linear in an affine parameter. The normalizations $r(0)=2M$ and $r(1)=2(M+E)$ therefore give
$$
r(\lambda)=2M+2E\lambda
\qquad(\lambda<1).
$$
The degenerate intrinsic metric of the horizon has no $d\lambda^2$ term, so
$$
\boxed{
ds^2_{\mathcal H^+}=
\begin{cases}
4(M+E\lambda)^2d\Omega^2,&\lambda<1,\\
4(M+E)^2d\Omega^2,&\lambda>1.
\end{cases}}
$$
The two expressions agree at the shell, $\lambda=1$.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 311 / 2 / a / ii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-311/2/a/ii/solution