The spatial metric is stationary and only $\beta^r=b(r)=Mr/(r+M)^2$ is nonzero. The evolution equation therefore says
$$
2\alpha K_{ij}=(\mathcal L_\beta\gamma)_{ij},
$$
the <Lie derivative> of the spatial metric along the <shift vector>. For the radial component,
$$
2\alpha K_{rr}
=b\,\partial_r\gamma_{rr}
+2\gamma_{rr}\partial_rb
=-\frac{2M}{r(r+M)},
$$
so
$$
\boxed{K_{rr}=-\frac{M}{r^2}.}
$$
For the angular components,
$$
2\alpha K_{\theta\theta}
=b\,\partial_r(r+M)^2
=\frac{2Mr}{r+M},
$$
and spherical symmetry supplies
$$
\boxed{
K_{\theta\theta}=M,\qquad
K_{\phi\phi}=M\sin^2\theta.
}
$$
All off-diagonal components vanish.

Contracting with the inverse spatial metric gives the <mean curvature>
$$
\begin{aligned}
K&=\gamma^{rr}K_{rr}
+\gamma^{\theta\theta}K_{\theta\theta}
+\gamma^{\phi\phi}K_{\phi\phi}\\
&=-\frac{M}{(r+M)^2}
+\frac{M}{(r+M)^2}
+\frac{M}{(r+M)^2}.
\end{aligned}
$$
Thus
$$
\boxed{K=\frac{M}{(r+M)^2}.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 357 / 2 / d / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-357/2/d/solution