Use proper time $s$, so $k=-1$. The first integral becomes
$$
\dot r^2=e^2-1-r^2=A^2-r^2,
\qquad A=\sqrt{e^2-1}.
$$
For an outward geodesic starting at the origin,
$$
r(s)=A\sin s
$$
until it next reaches $r=0$. This occurs at
$$
\boxed{T=\pi}.
$$
Thus every <radial timelike geodesic in anti-de Sitter spacetime> through the origin returns after the same proper time; restoring anti-de Sitter radius $a$ gives $T=\pi a$.

Solution

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

 Solution

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