1. Feb 6, 2017

### ShayanJ

I need to find the geodesics of AdS3 spacetime. But my searches have given me nothing. Can anyone give a reference where they're calculated?
Thanks

2. Feb 6, 2017

3. Feb 6, 2017

### ShayanJ

Interesting, but not what I needed. Thanks anyway.

P.S.
I can try to find a parametrization for curves in those pictures but I need a rigorous derivation of the geodesics.

4. Feb 7, 2017

### Ben Niehoff

You can easily get geodesics in $AdS_n$ by considering the embedding in $\mathbb{R}^{(n-1,2)}$,

$$X_{-1}^2 + X_0^2 - X_1^2 - X_2^2 - \ldots - X_{n}^2 = r^2.$$
Since $AdS_n$ is a maximally-symmetric space (its isometry group $SO(n-1,2)$ has the maximal number of generators $\frac12 n(n+1)$ ), it follows that the geodesic equation is also $SO(n-1,2)$-invariant. Generically, a solution of the geodesic equation (i.e., a geodesic) is defined by giving two points: the start and end points (caveat: not every two points on $AdS_n$ can be joined by a geodesic!). Given two points on the quadric surface above, together with the origin $(0,0,0,0,\ldots,0)$, one has a 2-plane, and you should be able to show that the intersection of this 2-plane with the quadric surface is in fact a geodesic. (In the same sense, the geodesics on a 2-sphere are given by the intersections of 2-planes through the origin of 3-space with that 2-sphere).

I can't quite come up with an argument right now that makes this fact "obvious", but it should be straightforward enough for you to write out the geodesic equation and find its first integrals. There should be enough first integrals (i.e. integration constants) to define the orientation of a 2-plane in $(n+1)$-dimensional space, and you should be able to derive an algebraic condition which is precisely finding the intersection of that 2-plane with the quadric surface that defines the embedding.

This post on Stack Exchange does a bit more work to show how the answer is obvious ;)