= Solution
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.
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