# Geodesics of anti-de Sitter space

It is said that (p. 9), given the anti-de Sitter space $$\text{AdS}_2$$, let's say in the static coordinates

$$ds^2 = -(1 + x^2) dt^2 + \frac{1}{(1+x^2)} dx^2$$

Every timelike geodesic will cross the same point after a time interval of $$\pi$$. That is, if $$(x_0, t_0) \in \gamma$$, then $$(x_0, t_0 + \pi) \in \gamma$$.

So I've been trying to find out how to show it. The non-zero Christoffel symbols are

$${\Gamma^x}_{xx} = - \frac{x}{1+x^2},\ {\Gamma^x}_{tt} = x + x^3, {\Gamma^t}_{xt} = {\Gamma^t}_{tx} = \frac{x}{1+x^2}$$

So that the geodesic equation is

$$\begin{eqnarray} \ddot{x}(\tau) &=& \frac{x}{1+x^2} \dot{x}^2 - \dot{t}^2 (x + x^3)\\ \ddot{t}(\tau) &=& -2 \frac{x}{1+x^2} \dot{x} \dot{t}\\ \end{eqnarray}$$

We also have the two following equality : the timelike geodesic is such that $$g(u,u) = -1$$

$$\frac{1}{(1+x^2)} \dot{x}^2 -(1 + x^2) \dot{t}^2 = -1$$

and since the metric is static, there is a timelike Killing vector $$\xi$$ such that $$g(\xi, u)$$ is a constant.

$$(1 + x^2) \dot{t} = E$$

or

$$\dot{t} = \frac{E}{(1 + x^2)}$$

This gives us

$$\dot{x}^2 = -(1 + x^2) + E^2$$

And so

$$\begin{eqnarray} \ddot{x}(\tau) + x &=& 0\\ \ddot{t}(\tau) &=& -2 x \dot{x} \frac{E}{(1 + x^2)^2}\\ \end{eqnarray}$$

Which gives us for a start that $$x(\tau) = A \sin(\tau) + B \cos(\tau)$$. Not quite periodic in $$\pi$$ (it should be $$2\pi$$ here), but more importantly this periodicity is in $$\tau$$ only and not in $$t$$, and it doesn't seem that $$t = \tau$$ in this scenario. Is there something wrong here or did I commit a mistake, either in interpreting the statement or the derivation here?

Given $$x(\tau) = \sin(\tau)$$, Wolfram Alpha gives out the following solution for $$t(\tau)$$, for instance :

$$t(\tau) = c_1 \tau + c_2 - \frac{1}{2\sqrt{2}} \arctan(2 \sqrt{2} \tan(\tau))$$

which doesn't seem to be particularly helpful here.

"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.

• All good although $\sin (t + \pi) = - \sin (t)$, but A.V.S.'s answer covers that part, thanks! – Slereah Nov 13 '18 at 8:27

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 $$2\pi$$.