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AdS Space Boundary and Geodesics

+ 6 like - 0 dislike
1025 views

I'm new to working with AdS space and am primarily concerned with black holes. I'm just playing round with the metric for AdS$_4$

$$ds^2=-f(r)dt^2+f^{-1}(r)dr^2+r^2d\zeta^2$$

for $f(r)=r^2+m $, $\space\space\space\zeta=d\theta^2+\sin^2\theta d\phi^2$.

My problem is trying to understand the boundary; specifically when considering particle trajectories:

  1. For null geodeiscs, I've read that they reach the boundary of AdS space, which seems to commonly expressed as saying they are represented as straight lines. I don't understand how these two phrases are the same and how to show that this is the case starting from the metric I've stated. Using constants of motion etc, and assuming a radial path, I find the equation

    $\frac{dr}{d\lambda}=k$, for $k$ constant.

  2. For timelike geodesics, I know they do not reach the boundary and equivalently I read that they are represented by the boundary of slices of the hyperboloid i.e. ellipses. Again, how do I show that this really represents timeline geodesics? As above (but $ds^2=1$ in this case) I find the equation

    $\Delta\tau=\log(r+\sqrt{k^2+m+r^2})\space \vert^b_{r_0}$ where $b$ is the boundary and $r_0$ the initial $r$.

I've been reading (as much as I can using the fairly limited coherent literature on the topic) and I can only find discussions on this matter, with some diagrams. None seem to go about this question the way I have above and consequently I'm thinking there must be something wrong with what I've done.

This post imported from StackExchange Physics at 2014-06-14 16:03 (UCT), posted by SE-user user13223423
asked Jun 4, 2014 in Theoretical Physics by user13223423 (45 points) [ no revision ]
Perhaps it is a better idea to split up your question in multiple separate questions. Right now, it's quite broad and vague...

This post imported from StackExchange Physics at 2014-06-14 16:03 (UCT), posted by SE-user Danu
I suggest you eliminate the part about general properties of AdS which is quite broad, and stick to the specific issue you have had in your calculations.

This post imported from StackExchange Physics at 2014-06-14 16:03 (UCT), posted by SE-user JamalS
Ok I've made it more specific

This post imported from StackExchange Physics at 2014-06-14 16:03 (UCT), posted by SE-user user13223423

2 Answers

+ 5 like - 0 dislike

As far as I could understand, it seems that you want to know whether timelike geodesics can reach the conformal boundary of AdS. If that's the case (please do confirm), the answer is no - no timelike geodesic can reach conformal infinity, it rather gets constantly refocused back into the bulk in a periodic fashion. You need timelike curves which have some acceleration in order to avoid this. Maximally extended null geodesics (i.e. light rays), on the other hand, always reach conformal infinity, both in the past and in the future. An illustration of these facts using Penrose diagrams can be found, for instance, in Section 5.2, pp. 131-134 of the book by S. W. Hawking and G. F. R. Ellis, "The Large Scale Structure of Space-Time" (Cambridge, 1973).

The detailed reasoning behind the above paragraph can be seen in a global, geometric way. In what follows, I'll largely follow the argument presented in the book by B. O'Neill, "Semi-Riemannian Geometry - With Applications to Relativity" (Academic Press, 1983), specially Proposition 4.28 and subsequent remarks, pp. 112-113. For the benefit of those with no access to O'Neill's book, I'll present the self-contained argument in full detail. I'll make use of the fact that $AdS_4$ is the universal covering of the embedded hyperboloid $H_m$ ($m>0$) in $\mathbb{R}^{2,3}=(\mathbb{R}^5,\eta)$

$$ H_m=\{x\in\mathbb{R}^5\ |\ \eta(x,x)\doteq -x_0^2+x_1^2+x_2^2+x_3^2-x_4^2=-m\}\ . $$

The covering map $\Phi:AdS_4\ni(t,r,\theta,\phi)\mapsto (x_0,x_1,x_2,x_3,x_4)\in H_m\subset\mathbb{R}^{2,3}$ through the global coordinates $(t\in\mathbb{R},r\geq 0,0\leq\theta\leq\pi,0\leq\phi<2\pi)$ is given by

$$ x_0=\sqrt{m(1+r^2)}\sin t\ ;$$ $$ x_1=\sqrt{m}r\sin\theta\cos\phi\ ;$$ $$ x_2=\sqrt{m}r\sin\theta\sin\phi\ ;$$ $$ x_3=\sqrt{m}r\cos\theta\ ;$$ $$ x_4=\sqrt{m(1+r^2)}\cos t\ .$$

The pullback of the ambient, flat pseudo-Riemannian metric $\eta$ defined above (with signature $(-+++-)$) by $\Phi$ after restriction to $H_m$ yields the $AdS_4$ metric in the form appearing in the question and in Pedro Figueroa's nice answer up to a constant, positive factor:

$$ds^2= m\left[-(m+r^2)dt^2+(m+r^2)^{-1}dr^2+r^2(d\theta^2+\sin^2\theta d\phi^2)\right]\ .$$

The conformal completion of $AdS_4$, on its turn, is obtained by means of the change of radial variable $u=\sqrt{m+r^2}-r$, so that $r=\frac{m-u^2}{2u}$, $dr=-\frac{1}{u}(\frac{m+u^2}{2u})du$ and $m+r^2=(\frac{m+u^2}{2u})^2$, yielding

$$ds^2=\frac{m}{u^2}\left[-\left(\frac{m+u^2}{2}\right)^2dt^2+du^2+\left(\frac{m-u^2}{2}\right)^2(d\theta^2+\sin^2\theta d\phi^2)\right]\ .$$

Conformal infinity is reached by taking $r\rightarrow+\infty$, which is the same as $u\searrow 0$. The rescaled metric $\Omega^2 ds^2$, $\Omega=m^{-\frac{1}{2}}u$ yields the three-dimensional Einstein static universe as the conformal boundary (i.e. $u=0$).

It's clear that $H_m$ is a level set of the function $f:\mathbb{R}^5\rightarrow\mathbb{R}$ given by $f(x)=\eta(x,x)$. Therefore, the vector field $X_x=\frac{1}{2}\mathrm{grad}_\eta f(x)=x$ (where $\mathrm{grad}_\eta$ is the gradient operator defined with respect to $\eta$) is everywhere normal to $H_m$ - that is, any tangent vector $X_x\in T_x H_m$ satisfies $\eta(X_x,T_x)=0$. Given two vector fields $T,S$ tangent to $H_m$, the intrinsic covariant derivative $\nabla_T S$ on $H_m$ is simply given by the tangential component of the ambient (flat) covariant derivative $(\partial_T S)^a=T^b\partial_b S^a$:

$$ \nabla_T S=\partial_T S-\frac{\eta(X,\partial_T S)}{\eta(X,X)}X=\partial_T S+\frac{\eta(X,\partial_T S)}{m}X\ .$$

The normal component of $\partial_T S$, on its turn, has a special form due to the nature of $H_m$ (notice that $\partial_a X^b=\partial_a x^b=\delta^b_a$):

$$ \eta(X,\partial_T S)=\underbrace{\partial_T(\eta(X,S))}_{=0\ ;}-\eta(S,\partial_T X)=-\eta(S,T)\ \Rightarrow\ \frac{\eta(X,\partial_T S)}{\eta(X,X)}X=\frac{\eta(S,T)}{m}X\ .$$

As such, we conclude that a curve $\gamma:I\ni\lambda\mapsto\gamma(\lambda)\in H_m$ ($I\subset\mathbb{R}$ is an interval with nonvoid interior) is a geodesic of $H_m$ if and only if $\frac{d^2\gamma(\lambda)}{d\lambda^2}(\lambda)\doteq\ddot{\gamma}(\lambda)$ is everywhere normal to $H_m$, that is,

$$\ddot{\gamma}(\lambda)=-\frac{1}{m}\eta(\ddot{\gamma}(\lambda),X_{\gamma(\lambda)})X_{\gamma(\lambda)}=\frac{1}{m}\eta(\dot{\gamma}(\lambda),\dot{\gamma}(\lambda))X_{\gamma(\lambda)}=\frac{1}{m}\eta(\dot{\gamma}(\lambda),\dot{\gamma}(\lambda))\gamma(\lambda)\ .$$

In particular, if $\eta(\dot{\gamma}(\lambda),\dot{\gamma}(\lambda))=0$, then $\gamma$ is also a (null) geodesic in the ambient space $\mathbb{R}^{2,3}$.

Given $x\in H_m$, the linear span of $X_x=x$ and any tangent vector $T_x\neq 0$ to $H_m$ at $x$ defines a 2-plane $P(T_x)$ through the origin of $\mathbb{R}^5$ and containing $x$. In other words,

$$ P(T_x)=\{\alpha X_x +\beta T_x\ |\ \alpha,\beta\in\mathbb{R}\}\ , $$

and therefore

$$ P(T_x)\cap H_m=\{y=\alpha X_x+\beta T_x\ |\ \eta(y,y)=-\alpha^2 m+\beta^2\eta(T_x,T_x)=-m\}\ .$$

This allows us already to classify $P(T_x)\cap H_m$ according to the causal character of $T_x$:

  • $T_x$ timelike (i.e. $-k=\eta(T_x,T_x)<0$): we have that $m\alpha^2+k\beta^2=m$ with $k,m>0$, hence $P(T_m)\cap H_m$ is an ellipse;
  • $T_x$ spacelike (i.e. $k=\eta(T_x,T_x)>0$): we have that $m\alpha^2-k\beta^2=m$ with $k,m>0$, hence $P(T_m)\cap H_m$ is a pair of hyperbolae, one with $\alpha>0$ and the other with $\alpha<0$. The point $x=X_x$ belongs to the first hyperbola;
  • $T_x$ lightlike (i.e. $\eta(T_x,T_x)=0$): we have that $\alpha^2=1$ with $\beta$ arbitrary, hence $P(T_m)\cap H_m$ is a pair of straight lines, one given by $\alpha=1$ and the other by $\alpha=-1$. The point $x=X_x$ belongs to the first line. Notice that each of these lines is a null geodesic both in $H_m$ and in $\mathbb{R}^{2,3}$!

Moreover, $x=\gamma(0)$ and $T_x=\dot{\gamma}(0)$ define a general initial condition for a geodesic $\gamma$ starting at $x$. It remains to show that any curve that stays in $P(T_x)\cap H_m$ is a geodesic in $H_m$. This is clearly true for $T_x$ lightlike, since in this case we have already concluded that $\gamma(\lambda)=x+\lambda T_x$ for all $\lambda\in\mathbb{R}$. For the remaining cases (i.e. $\eta(T_x,T_x)\neq 0$), consider a $\mathscr{C}^2$ curve $\gamma$ in $P(T_x)\cap H_m$ beginning at $\gamma(0)=x$ with $\dot{\gamma}(0)=\dot{\beta}(0)T_x$ (we assume that $\dot{\gamma}(\lambda)\neq 0$ for all $\lambda$). Writing $\gamma(\lambda)=\alpha(\lambda)X_x+\beta(\lambda)T_x$, we conclude from the above classification of $P(T_x)\cap H_m$ that we can choose the parameter $\lambda$ so that

  • $T_x$ timelike: $\alpha(\lambda)=\cos\lambda$, $\beta(\lambda)=\sqrt{-\frac{m}{\eta(T_x,T_x)}}\sin\lambda$, so that $\eta(\dot{\gamma}(\lambda),\dot{\gamma}(\lambda))=-m$ with $\dot{\beta}(0)=\sqrt{-\frac{m}{\eta(T_x,T_x)}}$;
  • $T_x$ spacelike: $\alpha(\lambda)=\cosh\lambda$, $\beta(\lambda)=\sqrt{\frac{m}{\eta(T_x,T_x)}}\sinh\lambda$, so that $\eta(\dot{\gamma}(\lambda),\dot{\gamma}(\lambda))=+m$ with $\dot{\beta}(0)=\sqrt{\frac{m}{\eta(T_x,T_x)}}$.

In both cases, we conclude that

$$ \ddot{\gamma}(\lambda)=\frac{\eta(\dot{\gamma}(\lambda),\dot{\gamma}(\lambda))}{m}\gamma(\lambda)\ ,$$

i.e. $\gamma$ must satisfy the geodesic equation in $H_m$ with the chosen parametrization, as wished. Since any pair of initial conditions for a geodesic determines a 2-plane through the origin in the above fashion, we conclude that the resulting geodesic in $H_m$ will remain forever in that 2-plane. For later use, I remark that all geodesics of $H_m$ cross at least once the 2-plane $P_0=\{x\in\mathbb{R}^5\ |\ x_1=x_2=x_3=0\}$ - this can be easily seen from the classification of the sets $P(T_x)\cap H_m$. This allows us to prescribe initial conditions in $P_0$ for all geodesics in $H_m$.

Now we have complete knowledge of the geodesics in the fundamental domain $H_m$ of $AdS_4$. What happens when we go back to the universal covering? What happens is that the lifts of spacelike and lightlike geodesics stay confined to a single copy of the fundamental domain, whereas the lifts of timelike geodesics do not. To see this, we exploit the fact that translations in the time coordinate $t$ are isometries and the remark at the end of the previous paragraph to set $$\gamma(0)=X_x=x=(0,0,0,0,\sqrt{m})$$ in $H_m$ (i.e. $\gamma$ is made to start at $P_0$ with $t=0$), so that $$\dot{\gamma}(0)=T_x=(y_0,y_1,y_2,y_3,0)\ .$$ We also normalize $\eta(T_x,T_x)$ to $-m$, $+m$ or zero depending on whether $T_x$ is respectively timelike, spacelike or lightlike. Writing once more $\gamma(\lambda)=\alpha(\lambda)X_x+\beta(\lambda)T_x$, we use the classification of geodesics in $H_m$ by their causal character to write explicit formulae for $\gamma$:

  • $T_x$ timelike $\Rightarrow$ $\gamma(\lambda)=(\cos\lambda)X_x+(\sin\lambda)T_x$;
  • $T_x$ spacelike $\Rightarrow$ $\gamma(\lambda)=(\cosh\lambda)X_x+(\sinh\lambda)T_x$;
  • $T_x$ lightlike $\Rightarrow$ $\gamma(\lambda)=X_x+\lambda T_x$.

The above expressions show that, in the spacelike and lightlike cases, the last component $\gamma(\lambda)_4$ of $\gamma(\lambda)$ never goes to zero, which implies by continuity that the time coordinate $t$ stays within the interval $(-\frac{\pi}{2},\frac{\pi}{2})$, hence the lift of $\gamma$ to $AdS_4$ stays within a single copy of its fundamental domain. One also sees that the spatial components (1,2,3) of $\gamma(\lambda)$ go to infinity as $\lambda\rightarrow\pm\infty$, hence $u\rightarrow 0$ along these geodesics as $\lambda\rightarrow\pm\infty$. In the timelike case, the whole time interval $[0,2\pi]$ is spanned by $\gamma(\lambda)$ as $\lambda$ spans the interval $[0,2\pi]$. Since the curve is closed, its lift to $AdS_4$ spans the whole time line $\mathbb{R}$ as $\lambda$ does so. On the other hand, it's clear that in this case the spatial components of $\gamma(\lambda)$ just keep oscillating within a bounded interval of the coordinate $r$ - hence, the coordinate $u$ stays bounded away from zero. Therefore, a timelike geodesic $\gamma$ never escapes to conformal infinity.

This post imported from StackExchange Physics at 2014-06-14 16:03 (UCT), posted by SE-user Pedro Lauridsen Ribeiro
answered Jun 4, 2014 by Pedro Lauridsen Ribeiro (580 points) [ no revision ]
Most voted comments show all comments
This is exactly what I was looking for. I'll check out the reference as the null geodesics are primary concern. However, I will want to extend my discussion to the geodesics of massive particles. How would I show that the answer is no in this case? Either another reference or a quick explanation would be great.

This post imported from StackExchange Physics at 2014-06-14 16:03 (UCT), posted by SE-user user13223423
This is also briefly discussed in Hawking-Ellis, but a more direct argument goes as follows. If you consider the fundamental domain of $AdS_4$ as an embedded hyperboloid in $\mathbb{R}^{2,3}$, one can show that the timelike geodesics of the fundamental domain are the intersection of this hyperboloid with some 2-plane in $\mathbb{R}^{2,3}$ containing the origin, so that the resulting conic section is an ellipse. In particular, this timelike geodesic is closed. When lifted to the universal covering, it assumes the aforementioned periodic behavior.

This post imported from StackExchange Physics at 2014-06-14 16:03 (UCT), posted by SE-user Pedro Lauridsen Ribeiro
A good reference that proves the above assertions in detail is the book by B. O'Neill, "Semi-Riemannian geometry - With Applications to Relativity" (Academic Press, 1983). See Proposition 4.28 and subsequent remarks, pp. 112-113.

This post imported from StackExchange Physics at 2014-06-14 16:03 (UCT), posted by SE-user Pedro Lauridsen Ribeiro
I've read around the topic a bit now and understand the idea much more. However, I'm still struggling to mathematically show that a massless particle reaches the boundary in AdS$_4$ while a massive particle doesn't. Hawking and Ellis (and some other papers) provide a nice discussion and diagrams which I can follow, but none really mathematically show me the reasoning; especially for the metric I'm using.

This post imported from StackExchange Physics at 2014-06-14 16:03 (UCT), posted by SE-user user13223423
@user13223423 It has just came to my attention, and I apologize in advance if my impression is mistaken - are you sure your form of the metric is correct? Shouldn't $f(r)$ be $r^2+m$ $(m>0)$ and $d\zeta^2=d\theta^2+\sin^2\theta d\phi^2?$ As far as I remember, $f(r)=r^2-m$ in your formula (with $\sin$ instead of $\sinh$) yields the four-dimensional de Sitter ($dS_4$) metric.

This post imported from StackExchange Physics at 2014-06-14 16:03 (UCT), posted by SE-user Pedro Lauridsen Ribeiro
Most recent comments show all comments
I've edited the post to show what I achieve using the Lagrangian, constant of motion, setting $\dot{\phi}=\dot{\theta}=0$ and $ds^2=0,1$ respectively for the null case and timeline case.

This post imported from StackExchange Physics at 2014-06-14 16:03 (UCT), posted by SE-user user13223423
@user13223423 I've added enough details to my answer to make it self-contained.

This post imported from StackExchange Physics at 2014-06-14 16:03 (UCT), posted by SE-user Pedro Lauridsen Ribeiro
+ 4 like - 0 dislike

I'll start from scratch. To obtain the $AdS_4$ metric one takes $\mathbb{R}^{2,3}$ and embed the quadric $$-(x^0)^2+(x^1)^2+(x^2)^2+(x^3)^2-(x^4)^2=-1$$ where the 1 in the right hand side may be any positive constant. The solution (or parametrization) $$(x^0)^2+(x^4)^2=\cosh^2\sigma,\hspace{0.25in}(x^1)^2+(x^2)^2+(x^3)^2=\sinh^2\sigma$$ with $\sigma\in\mathbb{R}^+$, is known as global coordinates (as it covers the whole of the quadric) and the induced metric takes the form $$ds^2=-\cosh^2\sigma\,dt^2+d\sigma^2+\sinh^2\sigma\,d\zeta^2\tag{1}$$ with $t\in\mathbb{R}$ for the universal cover (so that no closed timelike curves arise) and $d\zeta^2$ the metric of $S^2$. With the change of coordinates $r=\sinh\sigma$ the metric takes the form $$ds^2=-f(r)\,dt^2+f^{-1}(r)\,dr^2+r^2\,d\zeta^2,\hspace{0.25in}f(r)=1+r^2\tag{2}$$ with $r\in\mathbb{R}^+$, which is just similar to what you wrote and how it's usually written (you may specify which coordinates you're working on), so I'll just stick to it.

So, what you want to do is to check what happens with geodesics. To these means, due to rotational or spherical symmetry, you can just fix to any angles the sphere $S^2$, so that $d\zeta^2=0$, it doesn't matter which one you take, it'll be the same; when people visualize this in a Penrose diagram, they say every point in the diagram represents $S^2$.

For null geodesics, as $ds^2=0$, taking an affine parameter $\lambda$, from (2), $$(1+r^2)\dot{t}^2=(1+r^2)^{-1}\dot{r}^2\tag{3}$$ Also, as $\partial_t$ (meaning the vector with components $V^t=1$, $V^i=0$) is a global Killing vector, you've got the constant $$V_t\dot{t}=g_{t\alpha}V^\alpha\dot{t}=-(1+r^2)\,\dot{t}\equiv-E\tag{4}$$ (which being a $t$-translational invariance, may be regarded as conserved energy). This way, then, using (3) and (4), $$E^2=\dot{r}^2$$ and then for outgoing light rays, $$r=E\lambda$$ and $\lambda\to\infty$ as $r\to\infty$, which is ok, since it means the space is geodesically complete, anyhow, using this solution and (4), you get $$t=\arctan(E\lambda)$$ so that for $\lambda\to\infty$, $t=\pi/2$, so that it takes a finite coordinate time for a lightray to reach infinity.

As for timelike geodesics, you can proceed analogously, set e.g. $ds^2=-1$ (as the signature is -+++), then, if you will, using proper time $\tau$, $$\dot{r}^2+(1+r^2)-E^2=0$$ which for $r(0)=0$, $\tau\in(-\pi/2,\pi/2)$ (being $r>0$), $$r=\sqrt{E^2-1}|\sin\tau|$$ and thus $r$ is bounded. Now, I hadn't done this before, and I'm trying to figure out why e.g. if $E=\pm1$, then $r=0$ (if in general one set $ds^2=-\alpha$ with $\alpha>0$ one gets this for $E=\pm\alpha$), but the main thing here is that $r$ is bounded. You can verify this also with coordinate time using (4).

If as you say, you're concerned with black holes, maybe you could take the Schwarzschild-AdS metric: $f(r)=1+r^2-\frac{2M}{r}$ in (2) and try this same thing.

This post imported from StackExchange Physics at 2014-06-14 16:03 (UCT), posted by SE-user Pedro Figueroa
answered Jun 8, 2014 by Pedro Figueroa (85 points) [ no revision ]
Great answer, especially the last half which is a very nice way of showing them results. However, I'm only concerned with $\textbf{proper}$ time. Now you've shown me the coordinate time is finite, and it must be pretty easy to deduce the same for proper time from here right?

This post imported from StackExchange Physics at 2014-06-14 16:03 (UCT), posted by SE-user user13223423
Proper time is just a particular affine parameter, however you can't use it for null geodesics (see why not, e.g. here: physics.stackexchange.com/a/17539/24999). When people say null geodesics reach the boundary of AdS, they should specify they do it in a finite coordinate time.

This post imported from StackExchange Physics at 2014-06-14 16:03 (UCT), posted by SE-user Pedro Figueroa
That comment just helped me understand so much I've been missing for a long time! Just a couple more questions I have. Firstly, when you say it's straightforward to check the timelike case… I put $ds^2=1$ but, as you suggest, I can't suppress the transverse sphere and so I have a problem. Surely I can't use $t=\tanh \frac{\sigma}{2}$ in this case? And now I do need to think about if I'm trying to find coordinate time or proper time right? Lastly, how come "for lightlike geodesics, by spherical symmetry, to these means, you can just suppress the transverse sphere"?

This post imported from StackExchange Physics at 2014-06-14 16:03 (UCT), posted by SE-user user13223423
Ok, I've done some major edits, I hope it gets clearer. Also what I wrote as $t=\tanh(\sigma/2)$ was wrong and was meant to be $t=2\text{arctan}[\tanh(\sigma/2)]$.

This post imported from StackExchange Physics at 2014-06-14 16:03 (UCT), posted by SE-user Pedro Figueroa
Thanks a lot for this great answer. Mathematica tells me $r=\frac{\sqrt{E^2-1} \tan \tau}{\sqrt{\sec^2 \tau}}$ however (solving the differential equation). And I too am also wondering why $r=0$ when $E=\pm 1$...

This post imported from StackExchange Physics at 2014-06-14 16:03 (UCT), posted by SE-user user13223423
Yes, I might've oversimplified it; as $r>0$, I'm taking first e.g. $-\pi/2<\tau<\pi/2$ so $r=\sqrt{E^2-1}|\sin\tau|$, that's all. I'm guessing that this $|E|>1$ thing means the least energy a particle can have in these geodesics; in general it'd be $|E|>\alpha$ if one looks for timelike geodesics such that $ds^2=-\alpha$, but I'd try to discuss this with other people and I'll let you know; I hope that in general the process I did is clear.

This post imported from StackExchange Physics at 2014-06-14 16:03 (UCT), posted by SE-user Pedro Figueroa

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