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

In the spherical coordinate system, $|d\mathbf{x}{|}^2=d{r}^2+r^{2}d{\Omega }_2^2$ and $\mathbf{x}\cdot d\mathbf{x}=rdr$. HenceThus the static coordinates on anti-de Sitter spacetime havewhich is positive for every $r\ge 0$.

Affinely parametrized geodesics are the critical curves of the geodesic LagrangianFor $w=1+r^2$ and $w^{\prime }=2r$, the nonzero Christoffel symbols, up to symmetry in the lower indices, areThey follow either from the Euler-Lagrange equations or directly from the Levi-Civita connection formula.

The angular Euler-Lagrange equations are homogeneous in $\dot{\theta }$ and $\dot{\varphi }$. Therefore initial data with both angular velocities zero give the unique solution with constant $\theta$ and $\varphi$, so a radial geodesic remains radial.

Time-translation symmetry supplies the geodesic conserved quantity from a Killing vectorMetric compatibility makes the squared tangent norm another constant,For a proper-time parametrized timelike geodesic $k=-1$, for a null geodesic $k=0$, and for a unit-speed spacelike geodesic $k=1$. The constant $e$ is the conserved energy per unit mass associated with the static Killing vector ${\partial }_t$.

Use proper time $s$, so $k=-1$. The first integral becomesFor an outward geodesic starting at the origin,until it next reaches $r=0$. This occurs atThus 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 $T=\pi a$.

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 $\dot{r}=e>0$. ThenUsing $dr/d\lambda =e$,Consequently $r\to \infty$ whileThe conformal boundary is infinitely far away in affine parameter but is reached in finite static coordinate time.