4
$\begingroup$

I understand why in anti-de Sitter (AdS) spacetime, null geodesics can reach spatial infinity in finite coordinate time, while timelike geodesics cannot reach spatial infinity at all, not even in infinite coordinate time (in the sense that any timelike geodesic's radius is bounded). But what about accelerated (i.e. non-geodesic) timelike trajectories? Can they reach spatial infinity in finite coordinate time? Finite proper time? In not, can they come arbitrarily close, unlike timelike geodesics?

Intuitively, it seems to me that the case of accelerated timelike trajectories is intermediate between the case of timelike geodesics and the case of null geodesics, because an accelerated timelike trajectory can "escape from the origin better than" any timelike geodesic by accelerating against the AdS gravitational attraction, but it can't "escape from the origin as well as" a null geodesic because it can never hit the speed of light. My intuition is that an accelerated timelike trajectory could reach spatial infinity, but only in both infinite proper time and infinite coordinate time.

But I'm not confident that I'm correct, because I don't have good intuition for the boundary behavior of AdS space. The fact that all timelike geodesics are bounded away from spatial infinity suggests to me that it's "harder" to reach spatial infinity in AdS space than in Minkowski space. But on the other hand, the fact that null geodesics can reach spatial infinity in finite coordinate time suggests that it's "easier" to reach spatial infinity in AdS space than in Minkowski space. Clearly these intuitions conflict with each other.

$\endgroup$
  • $\begingroup$ I have not seen the details on AdS, but are you sure it's timelike geodesics, or instead timelike trajectories? $\endgroup$ – Bob Bee Jan 9 '17 at 5:56
  • $\begingroup$ Related, but does not have the answer to your question. Still, maybe you can use what's there. That's in physics.stackexchange.com/questions/116813/… $\endgroup$ – Bob Bee Jan 9 '17 at 6:25
  • $\begingroup$ @BobBee ^^ What exactly do you mean by "it's"? I'm sure that timelike geodesics are bounded, and my question is whether or not all other timelike trajectories are as well. $\endgroup$ – tparker Jan 9 '17 at 7:14
  • $\begingroup$ I meant you don't need to label them accelerated. You mean any time like trajectories, and I was trying to confirm that's what you wanted. The reference in my other post may help with an approach, but does not answer it directly. $\endgroup$ – Bob Bee Jan 9 '17 at 18:14
  • $\begingroup$ @BobBee Yes, as I said in the OP, by an "accelerated" timelike trajectory I just mean a timelike trajectory that isn't a geodesic. $\endgroup$ – tparker Jan 10 '17 at 0:49
2
$\begingroup$

The issue of coordinate time vs. proper time/affine parameter is a bit of a red herring, since the former is a purely gauge (i.e. coordinate-dependent) quantity, while the latter is physical but uninteresting in the context of the asymptotic boundary. For instance, you say that null geodesics reach infinity in finite coordinate time, but for a given null geodesic I could of course choose a time coordinate that diverges when the null geodesic reaches infinity, making that geodesic reach the boundary in infinite coordinate time. Likewise, by definition any curve (of any signature) that reaches the asymptotic boundary does so in infinite proper time/distance/affine parameter.

Instead, the covariant observation is that null geodesics always reach the asymptotic boundary of AdS, while timelike ones never do. Your question can then be stated covariantly as: do there exist timelike curves that reach the asymptotic boundary of AdS? The answer is clear from the conformal diagram of AdS:

enter image description here

(ignore the two dots labeled $i^+$ and $i^-$; I just grabbed this picture off the web). The asymptotic (or conformal) boundary of AdS is the vertical line marked $r = \infty$ (the line $r = 0$ is just one possible choice of origin), and the line moving at 45$^\circ$ is a light ray (shown here reaching the boundary and then bouncing back in). With this picture, it's clear that we can draw a timelike trajectory (that is, a line whose angle with the vertical is always less than 45$^\circ$) to and from the conformal boundary, so indeed, there do exist timelike curves that do reach asymptotic infinity.

Edit: As Peter pointed out in a comment, a statement I made regarding proper time in my previous answer was slightly wrong. I believe the correct statement is that any timelike curve that reaches the asymptotic boundary must either do so in infinite proper time (e.g. a uniformly accelerated observer can eventually reach the boundary, but takes an infinite proper time to get there) or have divergent geodesic acceleration there (e.g. a timelike curve can get "close enough" to being null that it reaches the boundary in finite proper time, but it takes an infinitely large acceleration to do so). The physical interpretation, of course, is that no finitely-accelerated observer can reach the asymptotic boundary of AdS in finite proper time.

$\endgroup$
  • $\begingroup$ Why must it reach infinity in infinite proper time? There's time dilation to account for. $\endgroup$ – Peter Shor Aug 7 at 0:59
  • $\begingroup$ You're right; if a timelike curve approaches a null one sufficiently fast asymptotically, you can engineer the proper time along it to be finite. But I think its geodesic acceleration would then necessarily need to diverge as it reaches the boundary, so the curve would be singular in that sense. (A timelike curve with constant but finite acceleration, such as those generated by Rindler time translation in the AdS-Rindler wedge, would asymptotically approach a null curve but would still take infinite proper time to reach the boundary, I believe.) $\endgroup$ – Sebastian Aug 8 at 2:33
  • $\begingroup$ It would be very good to give a reference or calculation saying that a time-like curve with constant acceleration can reach the boundary. I don't see why that should be the case. Doesn't just hovering a constant distance from the origin require arbitrarily large acceleration as you approach the boundary? AdS space has some extremely non-intuitive properties, and you can't just use your intuition. $\endgroup$ – Peter Shor Aug 8 at 10:13
  • $\begingroup$ The "AdS-Rindler curves" I mentioned are such an example. These are the integral curves of a Killing vector field which leaves an AdS-Rindler wedge fixed; because they are generated by an isometry, their geodesic acceleration is constant, but from the structure of the AdS-Rindler wedge it's clear they still reach the asymptotic boundary. $\endgroup$ – Sebastian Aug 8 at 13:02
  • $\begingroup$ (Explicitly, the relevant portion of the AdS-Rindler metric is $ds^2 = -\xi^2 dt^2 + d\xi^2/(1+\xi^2)$, and the curves I'm talking about are curves of fixed $\xi$. These curves have unit tangent $u^a = (\partial_t)^a/\xi$, and it's easy to check that their acceleration $a^a = u^b \nabla_b u^a$ has constant norm along each such curve. That they reach the asymptotic boundary of global AdS as $t \to \pm \infty$ can be verified from e.g. equation (2.6) of arxiv.org/pdf/1211.7370.pdf .) $\endgroup$ – Sebastian Aug 8 at 13:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.