Superluminal speed in anti de sitter

Question

This a bit of an elementary question, but I would like to understand how one correctly computes velocities in anti de sitter. It is well known that photons, traveling on null geodesics, will actually hit the adS boundary in finite proper time. On a spacelike geodesics, say for example in global coordinates along the "radius" tangent vector $\partial/\partial r$, the proper distance between any point in the interior and the boundary is, as one would expect, infinite. Note that in these global coordinate, the photon hits the boundary also in finite coordinate time. So, naively, this would make particle on null geodesics go at superluminal speeds (to say the least!). Obviously, there is something wrong with this conclusion, but I have some trouble finding a concrete explanation of what is exactly going on.

Answer 1

There is, of course, no superluminal speed for photons in AdS spacetime, they travel at precisely the speed of light. Taking a look at the metric of AdS: $$ ds^{2}=-\left(1+\frac{r^2}{\ell^2}\right)\,dt^{2}+\frac {dr^2}{1+\frac{r^2}{\ell^2}}+r^{2}\,d\Omega _{n-2}^{2}, $$ one can notice that the $g_{tt}$ component diverges with growth of radial coordinate $r$. A so called fiducial observer, held static at large $r=\mathrm{const}$ would experience longer proper time per unit coordinate time.

Radial null geodesics are given by the equation: $$ \sqrt{1+\frac{r^2}{\ell^2}}\,dt = \frac {dr}{\sqrt{1+\frac{r^2}{\ell^2}}}, $$ which literally state that photon travel at the speed of light. And indeed the coordinate time $t$ for a photon to reach “infinity” is finite: $$ T=\int dt = \int_0^\infty \frac {dr}{1+\frac{r^2}{\ell^2}}=\frac{\pi \ell}{2}. $$