For a radial <null geodesic>, $k=0$ gives $\dot r^2=e^2$. Choose the affine parameter so that $r(0)=0$ and the outgoing branch has $\dot r=e>0$. Then
$$
r(\lambda)=e\lambda,
\qquad
\dot t=\frac e{1+r^2}.
$$
Using $dr/d\lambda=e$,
$$
t(\lambda)=\int_0^{r(\lambda)}\frac{dr}{1+r^2}
=\arctan r(\lambda).
$$
Consequently $r\to\infty$ while
$$
\boxed{t\longrightarrow\tau=\frac\pi2}.
$$
The conformal boundary is infinitely far away in affine parameter but is reached in finite static coordinate time.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 309 / 1 / e / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-309/1/e/solution