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It is shown in a previous thread (AdS Space Boundary and Geodesics) that it's possible for null rays to travel to infinity and back in AdS space in finite coordinate time. That is to say, an observer at r=0 would see it go away and come back in a finite time.

How can a light ray travel infinite distance in finite time without travelling superluminally?

Is it to do with the fact that "infinity" is really conformal boundary and so it's some messed up notion of infinite distance?

Or is it to do with there not being a notion of proper time for null rays and "finite coordinate time" not being the correct measurement? I mean locally we still have dx/dt=c by setting ds^2=0 so we can see it isn't breaking the speed of light!

Or is it some other reason to do with the negative curvature of AdS allowing this to happen?

Thanks.

Community
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user11128
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  • @Danu Hmmm... not sure I agree. We can define global coordinates for AdS (see sectino 2.1 of http://www.ncp.edu.pk/docs/snwm/Pervez_hoodbhoy_002_AdS_Space_Holog_Thesis.pdf). And then if we examine the propagation of the null ray as in section 2.5 of those notes, the light ray appears to have gone an infinite distance in finite time, right?

    I'm not sure what's going on.....

     – user11128 Oct 27 '15 at 16:13
  • I may have spoken too soon---I'm actually not sure that what I said was true. I'll delete my comment for now, although I'm not entirely sure that it's incorrect. – Danu Oct 27 '15 at 18:07
  • What is the purpose of asking questions about coordinate times? Physics isn't about coordinate times. It's easy to have an infinite amount of proper time in an finite amount of coordinate time, so who cares? – Timaeus Oct 27 '15 at 21:20
  • @Timaeus Ok. But surely I could manufacture a situation in which a timelike observer at r=0 emits a null ray which travels to conformal infinity and back in coordinate time t=pi. Meanwhile our observer just travels up the t axis of the Penrose diagram and so sees a proper time of pi elapse. In other words, a light ray has got to conformal infinity and back within finite proper time, no? – user11128 Oct 27 '15 at 21:29
  • @user11128 If you want to do that, ask that. Otherwise you are literally repeating questions that have been asked before. As you can tell by getting answers that are the same as answers to identical questions. For instance your first paragraph mentions finite coordinate time (which zero physicists give an iota of consideration for) and finite time. Which makes people think everything you ate asking about is equally unphysical. A spacetime without boundary has multiple conformal boundaries so multiple conformal infinites and many many coordinates. A physical question should be stated as such. – Timaeus Oct 27 '15 at 21:33
  • @Timaeus Sorry - I didn't realise the unphysical nature until you pointed it out. Let me ask you the physical question about the observer at r=0 who sees a light ray travel to conformal infinity and back in finite proper time. How is this consistent with speed of light bound? I'd say it's because conformal infinity is at a finite distance but then travelling to conformal infinity isn't the same as travelling to actual infinity and we can't really tell where the light ray "turns round" since conformal map between 2 coord systems isn't well defined here. Can you please explain what's going on? – user11128 Oct 27 '15 at 21:52
  • @user11128 if you have a spacetime with a spacelike geodesic of finite proper length that extends to the boundary of your spacetime your spacetime is geodesically incomplete. That's pretty unphysical right there regardless of whether you use words to call that endpoint a singularity or use words to call that endpoint an infinity. – Timaeus Oct 27 '15 at 21:55
  • @Timaeus Sorry, I'm still really confused. Much of the literature I've read has made a big deal of these light rays reaching "infinity" in finite time as meaning the boundary is causally connected to bulk and this being important for AdS/CFT so I really want to understand argument. Now I agree that if a null geodesic (haven't looked at eqns for spacelike sorry) can hit boundary in finite time then whole spacetime is unphysical but we can attach reflective boundary conditions to restore global hyperbolicity - won't this make a light ray going to infinity and back in finite proper time possible? – user11128 Oct 27 '15 at 22:04
  • @user11128 They might truly be making a big deal for a real reason. But if you first say finite coordinate time (something rather meaningless unless you spell out specific coordinates that are meaningful) and then say finite time (something vague) it makes it sound like you mean something meaningless every time you say something vague. And people like geodesically incomplete spacetime's sometimes but you do have to be honest. And sometimes there are finite proper length/time curves that aren't geodesics. And all light rays have zero proper time so you'll have to talk about affine parameters. – Timaeus Oct 27 '15 at 22:37
  • @Timaeus OK. Can a light ray, assuming reflective boundary conditions, travel to conformal infinity of AdS and back again in finite affine parameter?

    I'm pretty sure the answer would be yes. Can you offer some insight into how this is possible? I'm finding it difficult to visualise the difference between it going to actual infinity and conformal infinity?

     – user11128 Oct 27 '15 at 22:44

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You can always change the coordinates of a manifold, even flat space, to give the illusion of an infinite distance travelled in a finite amount of time. All you need is a coordinate transform that can map $I \rightarrow \mathbb{R}$. This is used a lot in conformal diagrams to put an entire spacetime within a finite diagram.

A common coordinate transform for this is the coordinate transform $x' = arctan(x)$ or equivalent, which maps $\mathbb{R}$ to $[-1,1]$.

That is why it can be tricky to just look at the coordinates to determine the speed of something.

Slereah
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  • Hi, I'm not sure I agree/understand how this would help. We can define global coordinates for AdS (see sectino 2.1 of ncp.edu.pk/docs/snwm/…). And then if we examine the propagation of the null ray as in section 2.5 of those notes, the light ray appears to have gone an infinite distance in finite time, right? Whilst they make different coordinate transformations for the spatial coordinates, they don't change the coordinate time t. So if you can reach the point rho=pi/2 in finite time then surely that's the same as reaching mu=infinity in finite time? – user11128 Oct 27 '15 at 16:41
  • Question: if you can use coordinate transforms to create impossible situations like transversing an infinite distance in finite time, then how do you determine which systems of coordinates are unphysical? – Fomalhaut Feb 06 '19 at 17:03