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.

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


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

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):

enter image description here

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:

enter image description here

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.

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  • $\begingroup$ All good although $\sin (t + \pi) = - \sin (t)$, but A.V.S.'s answer covers that part, thanks! $\endgroup$ – 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 $2\pi$.

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