Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,047 questions , 2,200 unanswered
5,345 answers , 22,709 comments
1,470 users with positive rep
816 active unimported users
More ...

  AdS Space Boundary and Geodesics

+ 6 like - 0 dislike
3862 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
The argument in O'Neill's book actually comprises geodesics of all causal characters - timelike geodesics in the fundamental domain of $AdS_4$ behave as in my above comment, whereas null geodesics therein are one of the two connected components of the intersection of the hyperboloid with a null 2-plane through the origin. In particular, all null geodesics are also null geodesics in the ambient space $\mathbb{R}^{2,3}$. When lifted to the universal covering, complete null geodesics remain in a single sheet. Since this geodesic is complete but not closed, it must escape to conformal infinity.

This post imported from StackExchange Physics at 2014-06-14 16:03 (UCT), posted by SE-user Pedro Lauridsen Ribeiro
Yes sorry for having to retract the acceptance of your answer. My main problem is really the following: I have gone about this as I have always done in general relativity; that is, finding the conserved quantities from the Lagrangian of the line element and then finding the geodesics equations etc etc to end up with an equation for $dr/d\tau$. But either I've done it wrong or I can't interpret my final expression. On the other hand, now that I've read around the topic, it actually seems as though this isn't the way people find the geodesics…

This post imported from StackExchange Physics at 2014-06-14 16:03 (UCT), posted by SE-user user13223423
Instead they follow arguments similar to yours where you say "null geodesics therein are one of the two connected components of the intersection of the hyperboloid with a null 2-plane through the origin". However, $\textbf{why is this?}$. I really have no idea how to prove this is the case and even understand what this really means. Once I understand this, it makes my reasoning much more concise and elegant than long and boring calculations. (I don't have access to O'Neill's book and google misses those pages so I can't refer to it.)

This post imported from StackExchange Physics at 2014-06-14 16:03 (UCT), posted by SE-user user13223423
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
Most recent comments show all comments
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
+ 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

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\varnothing$ysicsOverflow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...