# Geodesics of anti-de Sitter space

"Every timelike geodesic will cross the same point after a time interval of $$\pi$$" will be true if the half-period is $$\pi$$. You found the general solution for $$x(\tau)$$, namely $$x(\tau)=A\sin\tau+B\cos\tau$$ or, alternately, $$x(\tau)=A\sin{(\tau-\tau_0)}.$$ When $$\tau$$ increases by $$\pi$$, $$x$$ does come back to what it was, after a half-period.

But we want to show that, when $$x$$ comes back, $$t$$, and not just $$\tau$$, has increased by $$\pi$$. So what is $$t$$ doing?

When you substitute $$x(\tau)=A\sin{(\tau-\tau_0)}$$ into $$\frac{\dot{x}^2}{1+x^2}-(1+x^2)\dot{t}^2=-1$$ and solve for $$t$$, you get $$t(\tau)=\tan^{-1}{[\sqrt{A^2+1}\tan{(\tau-\tau_0)}]}+t_0.$$

To see what is going on here, let's take $$\tau_0$$ and $$t_0$$ to be zero (since they just represent uninteresting time translations) and look at the function $$\tan^{-1}{(\sqrt{A^2+1}\tan{\tau})}$$. Here is a plot of it when $$A=\sqrt{3}$$ (just an arbitrary value as an example):

But $$t$$ isn't really discontinuous like this. The arctangent function is multivalued, and we have to take the appropriate branch of it so that t increases continuously with $$\tau$$. This means we move up the second blue curve by $$\pi$$, the third blue curve by $$2\pi$$, etc. to get a continuous function $$t(\tau)$$ that looks like this:

The result is that whenever $$\tau$$ increases by $$\pi$$, so does $$t$$!

So, to summarize, the timelike geodesics are

\begin{align} x&=A\sin\tau \\ t&=\tan^{-1}{[\sqrt{A^2+1}\tan{\tau}]} \end{align}

where we have dropped the uninteresting time-translation constants.

When $$\tau$$ increases by $$\pi$$, $$t$$ also increases by $$\pi$$, and $$x$$ comes back to what it was. This is what you were trying to show.

First, the statement

will cross the same point after a time interval of $$\pi$$

is wrong. In the cited paper the actual statement

… each timelike geodesic which intersects the $$t$$ axis at the point $$t=t_0$$ intersects that axis again at $$t=t_0+\pi$$.

So the $$\pi$$ interval refers to passing through the $$x=0$$, the actual period for a massive particle moving along a geodesic (as in, not only position but also velocity of the particle is the same) is $$2 \pi$$.

To make the “focusing property” of AdS space intuitive let us recall the canonical embedding of AdS space into the ambient pseudo-Riemannian $$\mathbb{R}^{2,1}$$ space with two timelike and one spacelike coordinates: $$ds^2=-dU^2-dV^2+dX^2$$.

AdS2 is defined as a hyperboloid $$-U^2-V^2+X^2=-1$$. Internal static coordinates $$(t,x)$$ are connected with coordinates of ambient space via: $$(U,V,X) = (\sqrt{1+x^2}\cos(t),\sqrt{1+x^2}\sin(t),x) .$$ It is easy to see that the points with static coordinates $$(x_0,t_0)$$ and $$(x_0,t_0+2\pi)$$ are actually the one and the same. If we “unroll” the $$t$$ variable by making them distinct we actually go from AdS space proper to universal covering space of AdS. Timelike geodesics on AdS are the sections of hyperboloid by a timelike plane of an embedding space passing through the origin. To show that, one could start by showing that circle $$X=0$$, $$U^2+V^2=1$$ (or alternatively $$U=\cos \tau$$, $$V=\sin\tau$$, $$\tau$$ is proper time) is a geodesic and then use AdS isometries (which is a Lorentz group $$SO(2,1)$$ of an embedding space) to make this geodesic into all other timelike geodesics. Since these sections are closed curves (ellipses) (for the AdS space proper), or winding curves periodic in $$t$$ coordinate with a period $$2\pi$$ (for the covering space) we have proven the statement in question (with a correct period), without explicit calculations. Incidentally, the solution $$x(\tau) = A \sin(\tau) + B \cos(\tau)$$ becomes kind of obvious by way of embedding space, with $$A$$ and $$B$$ coming from Lorentzian transformations of $$U$$ and $$V$$.

The actual calculations in the OP's question for the geodesic equation are correct up until the last equation. One should remember, that the condition $$g(u,u)=-1$$ gives us dependence between $$A$$ and $$B$$ constant of the $$x(\tau)$$ and the energy constant $$E$$. Namely, $$1+A^2+B^2=E^2$$. As a result if we shift $$\tau\to \tau+\delta$$ to eliminate $$A$$, we could integrate $$\dot{t}=f(\tau)$$ to obtain $$\tan(t-t_0)=\frac{\tan(\tau)}{\sqrt{1+B^2}}.$$ We see, that the phase difference between $$t$$ and $$\tau$$ is never large and becomes zero after every $$\pi$$. And so $$x(t)$$ would also be periodic with a period of $$2\pi$$.