# Infinite distance in finite time

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.

• @Danu Hmmm... not sure I agree. 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? I'm not sure what's going on..... Commented Oct 27, 2015 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
Commented Oct 27, 2015 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? Commented Oct 27, 2015 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? Commented Oct 27, 2015 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. Commented Oct 27, 2015 at 21:33

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 $[-1,1]$.