Write the Kruskal–Szekeres plane with $\hat r$ horizontal and $\hat t$ vertical. The relation
$$
\hat t^2-\hat r^2
=-e^{r/(2M)}(r-2M)
$$
shows that constant-$r$ curves are hyperbolae.

For $r>2M$, their right and left exterior branches satisfy
$$
\hat r^2-\hat t^2
=e^{r/(2M)}(r-2M)>0.
$$
For $0<r<2M$, their future and past interior branches satisfy
$$
\hat t^2-\hat r^2
=e^{r/(2M)}(2M-r)>0.
$$
The event horizons $r=2M$ are the null diagonals
$$
\boxed{\hat t=\pm\hat r},
$$
and the curvature singularities $r=0$ are the spacelike hyperbolae
$$
\boxed{\hat t^2-\hat r^2=2M}
$$
in the normalization stated in the question.

Constant Schwarzschild-$t$ curves are straight rays through the origin. In the exteriors their slopes obey
$$
\frac{\hat t}{\hat r}=\tanh\frac{t}{4M},
$$
while in the interiors
$$
\frac{\hat r}{\hat t}=\tanh\frac{t}{4M}.
$$
Thus the qualitative diagram is the usual four-region <Kruskal–Szekeres coordinates>[Kruskal diagram]: two exterior wedges separated from black-hole and white-hole interiors by the two null horizons, with spacelike singularities bounding the interior wedges.

Solution

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

 Solution

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