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I was reading in a paper (see 1st paragraph of introduction section in http://arxiv.org/pdf/1510.00709.pdf) that in AdS space, waves can reach the boundary in finite time and, since said boundary is timelike, they can be reflected back into the bulk.

I want to try to understand this statement. First of all, it is shown in the final answer of the following thread (AdS Space Boundary and Geodesics) that null rays take finite COORDINATE TIME to reach the AdS boundary but that this corresponds to infinite affine parameter. Q1: Does this mean an observer staying inside the bulk and measuring coordinate time could see the light ray go to infinity and back in finite time? But, if the light ray itself needs infinite affine parameter to reach r=infinity, how could it ever get back into the bulk? How are these two ideas consistent? I think I'm getting confused about coordinate time - I know there's no proper time for null rays but I still don't understand what's happening here.

Q2: Secondly, assuming our wave is travelling on a null geodesic, what does the boundary being timelike have to do with the ability to reflect the wave back into the bulk? What would have happened if it was a null boundary or a spacelike boundary? I have thought about this from the perspective of Penrose diagrams and, if I drew the boundary as null (at 45 degrees), then any null wave that hit it (coming in at 45 degrees), would be reflected and start travelling along the r=infinity surface, right? Similarly, if the boundary was spacelike, it would be drawn at >45 degrees to the vertical and I'm not sure any incoming 45 degree null wave would be able to end up back in the bulk. On the other hand, if it were timelike boundary and <45 degrees to vertical then I can see how null waves could continuously reflect off it. However, this "picture argument" needs tightening up - can anyone justify the physics? Why does the boundary being timelike allow for reflection?

Q3: How does one show the boundary is timelike? For this, I took AdS in Poincare coordinates and constructed the normal to surfaces of r=const. These are $n=g^{rr} \partial_r$ and so the components of this normal vector are $n^r=g^{rr}$. Then I compute the norm of this normal vector as $n^2=g_{\mu \nu} n^\mu n^\nu = g_{rr} g^{rr} g^{rr} = g^{rr} = \frac{r^2}{L^2} \rightarrow \infty$. We see that the norm is always positive for surfaces r=const and in fact, it actually becomes infinite on the boundary. This means the normal to the boundary is spacelike and I assume this means the tangent to the boundary is timelike i.e. the boundary is a timelike hypersurface. Is this correct?

Q4: If the above argument is correct, then why do I get a weird result when I apply it to Minkowski space? Here I write Minkowski in spherical polar coordinates and again the normal to surfaces r=const is $n=g^{rr} \partial_r$ and so $n^2=g_{rr}g^{rr}g^{rr}=g^{rr}=1$ and this is always equal to 1 regardless of what the value of r is on the hypersurface. This means the boundary to Minkowski has a spacelike normal and, by the above reasoning, this boundary would again be timelike. But we know from the Penrose diagram that the boundary of Minkowski is null with a null normal vector so what has gone wrong? And indeed, assuming my reasoning is wrong, how does one correctly show AdS has a timelike boundary and Minkowski has a null boundary?

Q5: In Minkowski, null waves take an infinite coordinate time to arrive at the boundary. Is this the reason there is no reflection back into the bulk or is it because the boundary itself is null? My argument using Penrose diagrams above would suggest that it's not possible for null boundaries even if the wave was able to arrive there in finite coordinate time, right?

Thanks very much for your help :)

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  • $\begingroup$ I don't understand your remarks about the "boundary" of Minkowski space. Minkowski space is topologically $\mathbb{R}^4$, it doesn't have a boundary (or the boundary is empty, whichever phrasing one prefers). $\endgroup$ – ACuriousMind Oct 6 '15 at 21:41
  • $\begingroup$ Minor comment to the post (v1): In the future please link to abstract pages rather than pdf files, e.g., arxiv.org/abs/1510.00709 $\endgroup$ – Qmechanic Oct 6 '15 at 21:56
  • $\begingroup$ @ACuriousMind He means conformal boundary. See for example physics.stackexchange.com/q/81131 $\endgroup$ – Jakub Oct 8 '15 at 10:21
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    $\begingroup$ @ACuriousMind I don't understand your point. Boundedness is not a topological property. The open ball is homeomorphic to $\mathbb R^n$ yet has a boundary. $\endgroup$ – a_a Dec 13 '15 at 0:33
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    $\begingroup$ @ulmo: Ah, I see how that is badly phrased. I was confused by talking about the boundary of Minkowski space when the question never embedded it into anything - and my point about $\mathbb{R}^4$ is just that one doesn't see immediately that there would be some "standard ambient space" in which to take the boundary. $\endgroup$ – ACuriousMind Dec 13 '15 at 0:41
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In general, coordinates mean almost nothing. And there are many conformally related metrics that are topologically compact that can contain a subset that has a metric (and domain, the subset) conformally related to a given spacetime. In other words you can put lots of different boundaries on a spacetime.

waves can reach the boundary in finite time

Reaching or not reaching something in finite coordinate is physically meaningless. For instance you can take Minkowski space $\{(t,x,y,z,): x,y,z,t \in \mathbb R\}=\mathbb R^4.$ And give it the Minkowski metric $ds^2=c^2dt^2-dx^2-dy^2-dz^2$ and you can't reach a border in finite time. But change to the coordinate $T=\arctan t$ and now you can and the border there is spacelike. Or you can instead let $X=\arctan x$ and now there is a timelike boundary a finite coordinate away. Coordinates by themselves have almost zero physical meaning. All you need is to have coordinate patches where you can have an open set and to have a locally one to one map between the coordinates and a neighborhood of events and to have a metric defined on your local coordinate system.

The relationship between the metric and the coordinates is where the true physics is. And there are lots of different spacetimes with boundary that contain Minkowski space as a subset. And it doesn't mean anything physically if one of them has a timelike boundary and one of them have a lightlike boundary. They both contain Minkowski space and Minkowski space is geodesically complete.

and, since said boundary is timelike, they can be reflected back into the bulk.

Again, I can embedded Minkowski space into a spacetime with a boundary that has a timelike boundary, but it means nothing physically. So just saying you have a timelike boundary means nothing by itself.

null rays take finite COORDINATE TIME to reach the AdS boundary but that this corresponds to infinite affine parameter.

And coordinate differences mean nothing physically.

Q1: Does this mean an observer staying inside the bulk and measuring coordinate time could see the light ray go to infinity and back in finite time?

You don't measure coordinate time. A well designed clock (that is also a test object) measures proper time along a timelike curve. It doesn't know what coordinate system different people may or may not choose to use.

But, if the light ray itself needs infinite affine parameter to reach r=infinity, how could it ever get back into the bulk? How are these two ideas consistent?

There aren't two ideas. Coordinate times aren't physical.

Q2: Secondly, assuming our wave is travelling on a null geodesic, what does the boundary being timelike have to do with the ability to reflect the wave back into the bulk?

How does a boundary that can't affect your experiments have any ability whatsoever. It's a drawing. When you use the coordinate system with $T=\arctan t$ then you can make a surface of $T=+\pi/2$ but it is basically a surface of $t=+\infty$ and can't be reached or affect any experiments or ever come up in a prediction. It could be helpful for discussing or drawing things and for classifying or grouping things that are real curves in the actual space, but it is just wrong to talk as if things reach it.

What would have happened if it was a null boundary or a spacelike boundary?

By itself, it doesn't mean anything.

I have thought about this from the perspective of Penrose diagrams

You can draw multiple inequivalent Penrose diagrams that have the spacetime embedded inside. So don't read anything into it at all. Think of it as a tool that can help you understand not as an actual larger spacetime.

if I drew the boundary as null (at 45 degrees), then any null wave that hit it (coming in at 45 degrees), would be reflected and start travelling along the r=infinity surface, right?

Do you have a physical basis to say it would ever reach a boundary? For instance if you are using AdS for duality purposes, have you taken the correct notions of what is physical on the dual theory and mapped them over? If you are concerned about AdS as its own spacetime and a solution to Einstein's Equation have you showed that the paths taken by geodesics are the limiting case of a family of physical metrics with an actual metric there. Because when you take that limit the things in the limit are not asymptotically vacuum AdS in when the curves are at the boundary. So any analysis based on that will fail. Plus in the paper you cite the waves eventually have significant back reaction on the geometry, so the whole idea of curves on a background spacetime becomes physicslly meaningless.

Q3: How does one show the boundary is timelike?

Again, the larger spacetime can have boundaries, but it is possible to have the smaller (physical) spacetime not have boundaries and be complete. You are adding something to a spacetime that is already complete and you can add a single event or a timelike boundary or a spacelike boundary or add whole other spacetimes right next to it. Its just your choice of what larger spacetime to pretend tour already complete spacetime is sitting inside. Since you can put different boundaries on the space that AdS is a subset of, then AdS itself can have different boundaries.

Q4: If the above argument is correct, then why do I get a weird result when I apply it to Minkowski space?

You aren't making an argument. You are treating arbitrarily choices as if they were physical. And haven't even stated your criteria for physicality (regular, dual, test particle, etc.)

But we know from the Penrose diagram that the boundary of Minkowski is null with a null normal vector so what has gone wrong?

You know of a Penrose diagram with a lightlike boundary. It's totally wrong to say that diagram is the diagram because that wrongly implies there is just one.

And indeed, assuming my reasoning is wrong, how does one correctly show AdS has a timelike boundary and Minkowski has a null boundary?

Are you sure that's a question? Since we can embed AdS as a subset of a larger dimensional $\mathbb R^{2,3}$ space and that can have different boundaries I think the burden is on you to argue that it has to have only certain boundaries. The paper you cited wants boundaries it calls energy momentum conserving.

Q5: In Minkowski, null waves take an infinite coordinate time to arrive at the boundary.

In some coordinate systems. Not in others. So there is no physical content in that statement.

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  • $\begingroup$ Thanks for your reply. Could you please elaborate on how the null rays reflect off the AdS boundary and back into the bulk? Clearly I'm not understanding the physics of what's going on here. Thanks. $\endgroup$ – user11128 Oct 7 '15 at 19:58
  • $\begingroup$ @user11128 The boundary isn't physical $\endgroup$ – Timaeus Oct 7 '15 at 22:44
  • $\begingroup$ Your answer concentrates on what is not physical, but does not explain differences of AdS to Minkowski space. But AdS is indeed quite peculiar. If you're an observer and send pulse of light, it will go to infinity and back within a finite amount of your proper time. In the same way, when specifying boundary conditions for some additional field, you have to specify what happens at the boundary and cannot say that it just drops off at infinity like in Minkowski. $\endgroup$ – Jakub Oct 8 '15 at 10:52
  • $\begingroup$ @Jakub My answer is basically a long comment in the sense that I'm saying the endless talk about coordinates means nothing. $\endgroup$ – Timaeus Oct 8 '15 at 16:14
  • $\begingroup$ @Jakub Can you explain why it doesn't just drop off at infinity like in Minkowski? Why the bounce back? Thanks. $\endgroup$ – user11128 Oct 9 '15 at 8:22
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Q1: Does this mean an observer staying inside the bulk and measuring coordinate time could see the light ray go to infinity and back in finite time?

Yes.

But, if the light ray itself needs infinite affine parameter to reach r=infinity, how could it ever get back into the bulk? How are these two ideas consistent?

They aren't. What we have here is a "non-real solution". We see similar issues in some black hole hypotheses to do with infinite coordinate time and finite proper time.

I think I'm getting confused about coordinate time - I know there's no proper time for null rays but I still don't understand what's happening here.

Coordinate time is merely the number of reflections in my parallel-mirror light clock whilst I stay at home with everybody else. Your proper time is merely the number of reflections in your parallel-mirror light clocks during your exotic return trip. And even if you're moving at c and there is no proper time for you, your light can't go to infinity and back whilst mine flicks back and forth 31557600 times. Or some other finite number of times.

Q2: Secondly, assuming our wave is travelling on a null geodesic, what does the boundary being timelike have to do with the ability to reflect the wave back into the bulk?

Nothing. I'm afraid there are issues with Kerr black holes too. The coordinate speed of light at the event horizon is zero. And the Kerr black hole is rotating at a significant fraction of the speed of light.

What would have happened if it was a null boundary or a spacelike boundary? I have thought about this from the perspective of Penrose diagrams and, if I drew the boundary as null (at 45 degrees), then any null wave that hit it (coming in at 45 degrees), would be reflected and start travelling along the r=infinity surface, right?

Oh not Penrose diagrams.

enter image description here

Similarly, if the boundary was spacelike, it would be drawn at >45 degrees to the vertical and I'm not sure any incoming 45 degree null wave would be able to end up back in the bulk. On the other hand, if it were timelike boundary and <45 degrees to vertical then I can see how null waves could continuously reflect off it. However, this "picture argument" needs tightening up - can anyone justify the physics?

No. You can ask around all you like, you will not get any justification at all.

Why does the boundary being timelike allow for reflection?

It doesn't. There is no bouncing off the end of time. If I were you I would look instead into spacelike boundaries, as per the "hall of mirrors" mention in this article.

Q3: How does one show the boundary is timelike?

I don't think you can, not really. All you're doing with the maths is defining a boundary that doesn't actually exist. I'd go so far as to say the maths might be correct, but the physics isn't.

Q4: If the above argument is correct, then why do I get a weird result when I apply it to Minkowski space?

Because it isn't correct. Remember that the Anti-de Sitter space remains hypothetical, and that as far as we can tell the universe is a de-Sitter space. To be honest I only think the Ads has hung around because AdS/CFT correspondence is touted as something that "represents a major advance in our understanding of string theory and quantum gravity". As for what these major advances are, well, maybe you should ask a question about that.

Q5: In Minkowski, null waves take an infinite coordinate time to arrive at the boundary.

So they haven't got there yet, and they never ever will.

Is this the reason there is no reflection back into the bulk or is it because the boundary itself is null?

This is where it gets interesting. On the large scale, the universe is flat. And because it's been expanding for a finite time, space hasn't got there get either. It hasn't reached this mythical infinite location. The observable universe has a radius of circa 46 billion light years, and we have no evidence whatsoever that the universe is infinite. For myself I just don't see how an infinite universe can expand anyway, I think it goes totally against the grain of big bang cosmology. And what that says to me is this: the universe has a spacelike boundary. And there may be reflection into the bulk. That's what the hall-of-mirrors was all about in the Neil Cornish article. They didn't find it. But would we even recognise it if we saw it? Take a look at the Hubble ultra deep field, courtesy of NASA:

enter image description here

See the orange galaxy near the centre but up a bit and right a bit? Look across to the right and down a little. Another orange galaxy. I'm not saying this actually is evidence of some edge of the universe, I'm just trying to get you interested in something that I think might prove more rewarding.

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