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On page 2 of this paper (http://arxiv.org/abs/1106.6073), Maldacena 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 Maldacena as to how the null geodesic looks. 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 Maldacena'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 $(\rho,t)$ Penrose 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!

I agree that if we compute the timelike geodesic with respect to the UNCOMPACTIFIED $AdS_3$ metric then we get a sinusoidal solution (see p10-11 of http://www.ncp.edu.pk/docs/snwm/Pervez_hoodbhoy_002_AdS_Space_Holog_Thesis.pdf for details) but the geodesics of the uncompactified space are NOT those of the compactified space - we can see in uncompactified AdS the timelike geodesics never reach infinity but in the compactified case they do.

It seems strange to me that timelike geodesics follow this trajectory that closely resembles the null trajectory and since all of the literature always plots the oscillatory motion I assume I have made a mistake however I cannot see it anywhere. Does anyone have any suggestions what I have done wrong?

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  • $\begingroup$ He's showing the geodesics of the uncompactified spacetime, but drawing them on the Penrose diagram of AdS. This is a very conventional way of depicting geodesics and other objects in AdS (and other spacetimes), as it allows you to see structure at a glance. Typically no one really cares about how things behave in the actual compactification, since that's an unphysical spacetime that's just used for drawing nice pictures. $\endgroup$ – Sebastian Feb 17 '17 at 13:59

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