# 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: $ds^2=-dU^2-dV^2+dX^2$.

AdS_{2} 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 $2\pi$.