For radial light rays, the transformed metric gives
$$
0=-f\,dv^2+2\,dv\,dr
=dv(-f\,dv+2\,dr).
$$
The two families of <radial null trajectories in ingoing Eddington-Finkelstein coordinates> are therefore
$$
\boxed{v=\text{constant}}
$$
for ingoing rays and
$$
\boxed{\frac{dv}{dr}=\frac2f=\frac{2r}{r-2M}}
$$
for outgoing rays.

Now set $t_*=v-r$. Along ingoing rays,
$$
\boxed{\frac{dt_*}{dr}=-1},
\qquad t_*=-r+\text{constant}.
$$
Along outgoing rays,
$$
\boxed{\frac{dt_*}{dr}=\frac2f-1
=\frac{r+2M}{r-2M}},
$$
and integration gives
$$
t_*=r+4M\log\left|\frac{r}{2M}-1\right|+\text{constant}.
$$
Thus the ingoing rays are straight lines of slope $-1$ in the $r$-$t_*$ plane. Outgoing rays have positive slope outside $r=2M$, become vertical as they approach the horizon, and have negative slope inside it; the horizon itself is the limiting outgoing null ray.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 3 / 37E / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-3/37e/c/solution