Trajectories in AdS
On page 2 of this paper (http://arxiv.org/pdf/1106.6073v2.pdf), <PERSON> explains (and has a very nice picture) showing the trajectories that a timelike and null particle would take in AdS space.
Of course, what this diagram really shows is the conformal compactification of AdS - hence the cylinder. To see this, I can consider the metric of, say, $AdS_3$: $ds^2=L^2(-\cosh^2{\mu} dt^2 + d \mu^2 + \sinh^2{\mu} d \theta^2)$ where $\mu \in [0, \infty)$ is the radial variable of $AdS_3$.
To conformally compactify, we substitute $\sinh{\mu}=\tan{\rho}$ and find $ds^2=L^2(-\sec^2{\rho} dt^2 + \sec^2{\rho}d \rho^2 + \tan^2{\rho} d \theta^2) = \frac{1}{\cos^2{\rho}} ( - dt^2 + d \rho^2 + \sin^2{\rho} d \theta^2)$ where $\rho \in [0, \frac{\pi}{2}]$ is the new angular variable
We can then consider the compactified metric $\hat{ds}^2 = \cos^2{\rho} ds^2 = ( - dt^2 + d \rho^2 + \sin^2{\rho} d \theta^2)$ which is exactly the cylinder metric with radial boundary at $\rho=\frac{\pi}{2}$ representing $\mu=\infty$ of the original $AdS_3$.
Now my concern is that conformal compactification preserves angles (whether curves are spacelike, timelike etc) but it doesn't preserve distances and so we need to be careful whether we are plotting geodesics of $AdS_3$ or of its compactification.
I agree with <PERSON> as to how the null geodesic looks. 474
Indeed, if I set $ds^2=0$, and supress the motion in the angular direction with $d \theta=0$ then I can rearrange the compactified metric to find (I denote affine parameter with $\lambda$):
$0=L^2(-(\frac{dt}{d \lambda})^2+(\frac{d \rho}{d \lambda})^2) \Rightarrow \frac{d \rho}{dt}=\pm 1 \Rightarrow \rho = \pm (t+t_0)$. In other words, a future directed null ray would travel on a straight line of gradient 1 as shown in <PERSON>'s picture.
However, I disagree with his trajectory for the timelike particle. Here, I have $ds^2=-1$ and so, again supressing the motion in the angular direction with $d \theta=0$, I can rearrange the compactified metric to find (now denoting proper time by $\tau$):
$-1=L^2(-(\frac{dt}{d \tau})^2+(\frac{d \rho}{d \tau})^2) \Rightarrow \frac{d \rho}{d \tau}= \sqrt{-\frac{1}{L^2}+\frac{E^2}{L^4}}$ where I have used that the timelike Killing vector $\partial_t$ generates a conserved energy $E=-g_{tt} \frac{\partial t}{\partial \tau} \Rightarrow \frac{\partial t}{\partial \tau} = \frac{E}{L^2}$.
To plot this curve on the $(\rho,t)$ <PERSON> diagram, I use that $\frac{d \rho}{dt} = \frac{d \rho}{d \tau} \frac{d \tau}{dt} = \sqrt{-\frac{1}{L^2}+\frac{E^2}{L^4}} \frac{L^2}{E}= \sqrt{1-\frac{L^2}{E^2}} \Rightarrow \rho(t) = \sqrt{1-\frac{L^2}{E^2}} t$ which is again a linear trajectory as opposed to the oscillatory one that he draws! 394