The point $\hat t=\hat r=0$ is the bifurcation sphere $r=2M$. The observer starts there with $\dot r=0$, so its radial-geodesic equation
$$
-E^2+\dot r^2=-1+\frac{2M}{r}
$$
gives $E=0$. During infall,
$$
\dot r=-\sqrt{\frac{2M}{r}-1}.
$$
Put $x=r/(2M)$. The proper time to the singularity is
$$
\begin{aligned}
\tau_{\rm sing}
&=\int_0^{2M}\frac{dr}{\sqrt{2M/r-1}}\\
&=2M\int_0^1\sqrt{\frac{x}{1-x}}\,dx
=2M\frac{\pi}{2}.
\end{aligned}
$$
Thus
$$
\boxed{\tau_{\rm sing}=\pi M}.
$$
The normals of the geodesic foliation reach the physical singularity after finite coordinate time because unit lapse identifies coordinate and normal proper time. Nearby normals can also focus and form coordinate caustics. Consequently geodesic gauge is unsuitable for long-term black-hole evolution: the numerical slice encounters singular behavior in finite time rather than avoiding it.


Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 357 / 3 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-357/3/c/solution