# Interesting Effect of Conformal Compactification on Geodesic

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1. Dec 29, 2015

### adsquestion

I'm trying to understand why timelike geodesics in Anti de-Sitter space are plotted as sinusoidal waves on a Penrose diagram (a nice example of the Penrose diagram for AdS is given in Figure 2.3 of this thesis: http://www.nbi.dk/~obers/MSc_PhD_files/MortenHolm_Christensen_MSc.pdf).

Bearing in mind that the Penrose diagram shows conformally compactified AdS in order to draw points at infinity, we need to be clear whether we are drawing the timelike geodesics of a) the UNCOMPACTIFIED AdS metric or b) the COMPACTIFIED AdS metric.

Considering each in turn:

a) UNCOMPACTIFIED AdS3 metric can be written in global coordinates as $$ds^2=L^2(-\cosh^2{\mu} dt^2 + d \mu^2 + \sinh^2{\mu} d \theta^2)$$
with $$t \in [-\pi,\pi), \theta \in [0,2 \pi), \mu \geq 0$$.
We can then substitute $\sinh{\mu}=\tan{\rho}$ to bring this to the form
$$ds^2=L^2(-\sec^2{\rho} dt^2 + \sec^2{\rho} d \rho^2 + \tan^2{\rho} d \theta^2)$$ with $$\rho \in [0,\pi/2]$$

Now let's suppose that we take a timelike observer ($ds^2=-1$) who is moving radially ($d \theta=0$). Their equation of motion will take the form
$$-1=L^2(-\sec^2{\rho} \dot{t}^2 +\sec^2{\rho} \dot{\rho}^2)$$.
We know that time translation invariance (stationary metric) gives rise to a conserved energy given by $E=-g_{tt} \dot{t} = L^2 \sec^2{\rho} \dot{t} \Rightarrow \dot{t} = \frac{E}{L^2 \sec^2{\rho}} = \frac{E \cos^2{\rho}}{L^2}$
And if we substitute this into the above equation of motion, we can rearrange to get $$-1=-\frac{E^2 \cos^2{\rho}}{L^2} + \frac{\dot{\rho}^2 L^2}{\cos^2{\rho}} \Rightarrow \frac{\dot{\rho}^2 L^2}{\cos^2{\rho}} = \frac{E^2 \cos^2{\rho}}{L^2}-1$$.
At this point we note that since we want to plot $\rho$ against coordinate time not proper time on the Penrose diagram, we use $\dot{\rho} = \frac{d \rho}{d t} \dot{t} = \frac{d \rho}{dt} \frac{E \cos^2{\rho}}{L^2}$ to find $\frac{d \rho}{dt} = \pm \sqrt{1- \frac{L^2}{E^2 \cos^2{\rho}}}$ and this can be integrated up (by substitution) to obtain:
$$\sin{\rho(t)} = \sqrt{1 - \frac{L^2}{E^2}} \sin{(t + t_0)}$$

This gives the sinusoidal behaviour that we see in the standard AdS Penrose diagram and it's also interesting to note that the greater the energy of the particle, the greater the amplitude of the trajectory and the closer the particle gets to infinity - this looks great and has all the properties we want BUT the problem is that we calculated it using the UNCOMPACTIFIED metric.

b) So let's see what happens when we use the geodesics of the COMPACTIFIED metric (recall the Penrose diagram requires using the COMPACTIFIED metric in order to be able to plot points at infinity at a finite point).

If we start with the UNCOMPACTIFIED metric in global coordinates $$ds^2=L^2(-\sec^2{\rho} dt^2 + \sec^2{\rho} d \rho^2 + \tan^2{\rho} d \theta^2)$$, we can multiply by the conformal factor $\Omega^2=\cos^2{\rho}$ to define the COMPACTIFIED metric $$\tilde{ds}^2 = \Omega^2 ds^2 = L^2(-dt^2 + d \rho^2 + \sin^2{\rho} d \theta^2)$$

Note that points at infinity of the original metric ($\mu=\infty$) now appear at finite position $\rho=\frac{\pi}{2}$ in the COMPACTIFIED metric which is what we need in order to plot the Penrose diagram). But what do the radial ($d \theta=0$), timelike ($ds^2=-1$) geodesics look like. Well their equation of motion looks like

$$-1=L^2(-\dot{t}^2 + \dot{\rho}^2)$$

As before this is a stationary metric so energy is conserved and given by $E=-g_{tt} \dot{t} = L^2 \dot{t} \Rightarrow \dot{t} = \frac{E}{L^2}$

If we substitute this in we get $$-1=-\frac{E^2}{L^2} + L^2 \dot{\rho}^2 \Rightarrow \dot{\rho}^2 = \frac{1}{L^4} ( E^2 - L^2) \Rightarrow \dot{\rho} = \pm \frac{1}{L^2} \sqrt{E^2-L^2}$$

Again, in order to plot $\rho$ against coordinate time rather than proper time, we need to make use of $\dot{\rho} = \frac{d \rho}{dt} \dot{t} = \frac{d \rho}{dt} \frac{E}{L^2}$ which allows us to write
$\frac{d \rho}{d t} = \pm \sqrt{1-\frac{L^2}{E^2}}$ which integrates up directly to give

$$\rho(t) = \sqrt{1-\frac{L^2}{E^2}} t + t_0$$

Disappointingly, whilst this does have the property that increasing energy increases the amplitude of the geodesics, it does not have the sinusoidal behaviour that we see drawn on AdS Penrose diagrams and, furthermore, it appears that as time increases, the radial coordinate increases indefinitely i.e. $\rho \rightarrow \infty$ but this is not allowed since we know $\rho$ is bounded by $\frac{\pi}{2}$ which represents spatial infinity for AdS space.

QUESTION: Clearly what we really want to do is plot the geodesics of UNCOMPACIFIED AdS space on the Penrose diagram of COMPACTIFIED AdS space but this makes absolutely no sense to me and I can see absolutely no justification for doing this and yet authors continually do this in books an papers etc without giving any reason. Is there any explanation for this? Or is there perhaps a more general reason that I am missing to do with plotting of timelike geodesics in general spacetimes (does something happen to them under conformal compactification that alters the shape of their trajectory and thus means we need to use the information from the uncompactified metric)?

The best "reason" I've managed to come up with is that everything COULD be ok because I calculated the geodesics of the uncompactified metric using the $\rho$ coordinate which had finite range (same range as Penrose diagram) rather than the $\mu$ coordinate which has infinite range? BUT I didn't explicitly compactify in calculation a) and if I actually do the compactification I end up with straight line geodesics as in calculation b) and I'm really unsatisfied by that!

Thanks very much for any help!

Last edited: Dec 29, 2015
2. Dec 29, 2015

### bcrowell

Staff Emeritus
Penrose diagrams are just a tool for visualization. When we do a conformal transformation on the metric in order to compactify the space, the metric we get is an unphysical metric. Its timelike geodesics are of no interest, because they are unphysical. Note that you have infinitely many different conformal transformations you could have done. Each of these produces a different unphysical metric, with different unphysical timelike geodesics, and there is no reason to prefer one of these over another.

3. Dec 29, 2015

### adsquestion

Thanks for your reply.

This kind of makes sense. But what is the reason for drawing the uncompactified geodesics on the compactified metric diagram? Is it just because it's "the best we can do" in terms of visualisation? Or is there a deeper reason relating to the fact that my calculation which produced the sinusoidal version (with the uncompactified metric) used the same $\rho$ coordinate as appears in the compactified metric which the Penrose diagram is based on?

Secondly, when we consider the null geodesics; for the uncompactified metric, they take infinite affine parameter to reach spatial infinity but for the compactified metric, they can reach spatial infinity in finite affine parameter suggesting there is something genuinely different going on. Now I know that you said the compact metric is unphysical and of course in AdS space we would expect a light ray to take infinitely long to reach spatial infinity (geodesic completeness) but don't people say "light rays can reach infinity in finite affine parameter" for the compact metric? Doesn't AdS/CFT have some reliance on this in order to establish a causal connection between bulk and boundary?

4. Dec 29, 2015

### bcrowell

Staff Emeritus
Yes, that seems like a reasonable way of putting it. A good analogy might be map projections. On a map of the world like this one https://en.wikipedia.org/wiki/Goode...media/File:Goode_homolosine_projection_SW.jpg , we don't draw fictitious north-south lines that look vertical on the map. We draw north-south lines that are real.

The geometry of the compactified metric is unphysical, and that includes the affine geometry, i.e., the geometry of parallelism. For example, on the Penrose diagram for Minkowski space, we can draw in a Minkowski coordinate grid, and the lines don't look parallel. The fact that light rays can reach infinity in finite affine parameter, according to the unphysical geometry, is a property of the unphysical geometry, and it's ... unphysical.

Sorry, I don't know anything about string theory.

5. Dec 30, 2015

### adsquestion

Thanks very much again.

The fact that the timelike geodesics are sinusoidal curves agrees with the fact that they are expected to be ellipses (which are parametrically sine curves) since the intersection of the AdS hyperboloid by a plane gives an ellipse as seen in Figure 11 of these nice notes: http://www.bourbaphy.fr/moschella.pdf
It's well know that the intersection of a sphere by a plane gives rise to the great circle geodesics but I was wondering if you have any insight as to why the intersection of the AdS hyperboloid by a plane only gives the elliptical timelike geodesics and not the null geodesics? Is this to do with the signature of the ambient space in both cases?

Secondly, how would the null geodesics appear on the hyperboloid?

Thanks again.

6. Dec 30, 2015

### bcrowell

Staff Emeritus
I don't understand why you're focusing so much on this fact, which is not even really true, for the reasons given in #2. If this fact holds, it holds for only one choice of the conformal transformation, and in any case it's of no physical interest.

7. Dec 30, 2015

### adsquestion

Sorry, I thought we agreed that timelike geodesics were sinusoidal when calculated with the full, uncompactified AdS metric?

8. Dec 30, 2015

### bcrowell

Staff Emeritus
Oh, sorry, I guess I misunderstood you. I thought you meant that timelike geodesics had a sinusoidal representation when represented on the Penrose diagram. So you're saying that the FLRW r coordinate is a sinusoidal function of the FLRW t coordinate? That would only be a one-parameter family of geodesics, but shouldn't we have a 2-parameter, since they can have an initial position and an initial velocity?

9. Dec 30, 2015

### adsquestion

Yep

I think this calculation assumes they all have initial position $\rho=0$ and then the initial velocity will depend only on energy E.

So I'm wondering about how, given that these sine curves are the ellipses we get by intersecting the AdS hyperbolic by planes, why does intersecting by planes (c.f. getting great circles on a sphere) only give timelike geodesics and not the null ones?

Secondly, the timelike geodesics are obviously reflected back by the negative curvature/potential wall of AdS and this means you get back to same position at the same time after you wrap the hyperbolic once ie a closed timelike curve. This is normally dealt with by unwrapping the hyperbolic along the time direction to produce a cylinder and then timelike geodesics return to same initial position at a later time ie we get rid of timelike periodicity. But what about the null geodesics? In AdS, we need to impose Dirichlet boundary conditions at spatial infinity to maintain global hyperbolicity and this will allow for reflection of null geodesics back to the same initial position (also see penrose diagram for image of this). Does this mean we have closed null curves as well?

Thank you!

10. Dec 31, 2015

### adsquestion

Actually, let me rephrase/simplify the questions I was asking in post #9 to the following:

1, Why does intersecting a surface with a plane give the geodesics?

2, Looking at Figure 2.3 of the link I provided in post #1, we see the Penrose diagram for AdS including the aforementioned sinusoidal timelike trajectories. Now if we look at the null trajectories, we see that a light ray emitted from $\rho=0$ (LHS) will travel to spatial infinity at $\rho=\frac{\pi}{2}$ (RHS) and if you do the calculation this takes infinite affine parameter. However, it is then reflected off spatial infinity (such reflective boundary conditions are needed to have global hyperbolicity) and will return to $\rho=0$ and again this return leg of the journey takes infinite affine parameter. This means that from the perspective of the light ray, infinite affine parameter has elapsed for both legs of the journey. However, interestingly, the light ray is able to get from $\rho=0$ to spatial infinity and back again within a finite coordinate time (a coordinate time $t=\pi$ elapses during this journey) but one might argue that coordinate time is unphysical and so there is no problem. But now consider the following dilemma: a timelike observer stuck at $\rho=0$ measures proper time being equal to coordinate time and so he is able to send out a light beam at $t=\tau=0$ and watch it go to spatial infinity and return back to him at $t=\tau=\pi$. In other words, he sees the light ray travel infinite distance in finite time - what is the resolution of this?
(This argument does seem to be supported by bottom of e.g. p11 of this thesis: http://www.ncp.edu.pk/docs/snwm/Pervez_hoodbhoy_002_AdS_Space_Holog_Thesis.pdf but I'm just not comfortable with it. On the other hand negatively curved spaces should have some weird properties I suppose.....)

Last edited: Dec 31, 2015
11. Dec 31, 2015

### bcrowell

Staff Emeritus
No, I don't think so. Aren't you describing a null geodesic that arrives back at the same position, but at a later time? That isn't a closed curve.

That is certainly a strange property! I don't have any good insight into it.

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