I have some questions about anti-deSitter space, (note: I am not a physicist)

When describing deSitter space it is almost always mentioned that it has a positive cosmological constant and is therefore expanding, and in that regard is often compared with the long term future of our expanding universe (ie our universe is approaching deSitter) or the theoretical inflationary phase of the early moments after the bang.

Does that then mean that anti-desitter space is in some way contracting? What would happen to two observers at rest with respect to eachother over time? Would they get closer together?

Also it is often compared to those Escher prints, circle limit, with comments like, "the boundary is infinitely far away" but then also it is said that a light beam would reach the boundary in a finite time from the point of view of an observer at rest with respect to the source of the beam. I'm not sure how to make sense of this. Does this only apply to a light beam? Could something travelling slower than light hit this boundary in a finite time? In what sense is the boundary physical? To say that it can be reached by a light beam in a finite time gives it a different meaning than say, future timelike infinity in a deSitter universe. I am interested if anyone could explain the interesting properties of anti-desitter space and try to give a bit of background as to what is going on and how these interesting properties occur and how they relate to the negative cosmological constant.


2 Answers 2


Minkowski, de Sitter, and anti-de Sitter spacetimes all behave as though there is an acceleration between test particles that is equal to $Λ/3$ times the distance between them. The geodesics resemble solutions to the differential equation $x''(t) = \frac{Λ}{3} x(t)$, which is to say

$$\begin{align} x(t) &= A e^{t\sqrt{Λ/3}} + B e^{-t\sqrt{Λ/3}} & & Λ>0 \text{ (de Sitter)} \\ x(t) &= A \cos t\sqrt{-Λ/3} + B \sin t\sqrt{-Λ/3} & & Λ<0 \text{ (anti de Sitter)} \\ x(t) &= A + B\,t & & Λ=0 \text{ (Minkowski)} \end{align}$$

So in AdS spacetime, test particles tend to oscillate relative to each other, crossing paths repeatedly.

That isn't the same as saying that dS spacetime expands and AdS spacetime contracts. Expansion/contraction is the first time derivative of position, but the difference between dS and AdS is in the second derivative. You can have an expanding or contracting cosmology with a positive or negative $Λ$.

Another reason that these spacetimes can't be said to expand or contract is that they are totally symmetric, i.e., every point in them is equivalent to every other, so no property (like an overall size) can change over time.

You can indeed go "to infinity and back" in a finite time in AdS spacetime. You can think of it this way: if you look at the AdS solutions above, you'll see that if you toss a tennis ball away from yourself, it should return to you after a time $π\sqrt{-3/Λ}$ independently of how hard you threw it. Those solutions aren't really correct for large $A,B$, but that constant-time property does actually hold for arbitrarily large speeds in AdS spacetime. The harder you throw the tennis ball, the farther it gets before returning to you. In the $v\to c$ limit, its maximum distance from you goes to infinity. If you take that limit seriously, a light pulse should go all the way to infinity and return to you at the same time as the tennis ball.

  • $\begingroup$ I'd dispute that test particles "cross paths repeatedly"; aren't the points where they do so the Big Bang and the Big Crunch? $\endgroup$ Aug 14, 2023 at 17:20
  • $\begingroup$ @MichaelSeifert I was talking about empty AdS spacetime, where there is no beginning or end of time, and there are no distinguished re-crossing points (if you toss two tennis balls at different times, they reintersect your worldline at different times). In a $ρ>0, Λ<0$ FLRW cosmology, I think the big crunch would happen before the tennis ball had a chance to return. $\endgroup$
    – benrg
    Aug 14, 2023 at 23:42

Steven Vernau asked: "Does that then mean that anti-desitter space is in some way contracting? What would happen to two observers at rest with respect to each other over time? Would they get closer together?"

Yes. With the equations of motion where $\rm c=1$ and $\rm \dot{x}=dx/d\tau$

$$ \rm \ddot{t}=-\frac{2 \ \Lambda \ r \ \dot{t} \ \dot{r}}{\Lambda \ r^2-3} $$

$$ \rm \ddot{r}=\frac{\Lambda \ r \ \dot{r}^2}{\Lambda \ r^2-3}-\frac{1}{9} \ \Lambda \ r \ \dot{t} ^2 \left(\Lambda \ r^2-3\right) $$

$$ \rm \dot{t} = \frac{ 1}{\sqrt{1-\Lambda \ r^2/3 } \ \sqrt{1-\mu ^2 \ v^2)}} $$

$$ \rm \dot{r} = \frac{v \ \sqrt{1-\Lambda \ r^2/3}}{\sqrt{1-\mu ^2 \ v^2}} $$

and a negative $\Lambda=-1/3$ the $\rm \{ r, t \}$ spacetime diagram for the geodesics with initial conditions $\rm v_0= \{ \frac{9}{10} , \ \frac{3}{4} , \ \frac{1}{2} , \ \frac{1}{4} , \ \frac{1}{8} \}$ looks like this:

Ads geodesics plot

so they all return to $\rm r=0$ after the same timespan $\rm t=\pi \surd{|3/\Lambda|}$. When they reach the turnaround point at $\rm t/2$ they all are at rest relative to the observer at $\rm r=0$, so everything at rest relative to the observer will crunch into him after the half period, no matter how far away it is initially, and with the speed of light if the distance was initially infinite. That corresponds to a big crunch oscillation in an infinite space.

The $\rm r$ coordinate is not the physical distance measured with stationary rulers though (this coordinates show the true circumference $\rm C$ with $\rm r=C/2/\pi$), that would be $\rm \Delta R=\int_{r_1}^{r_2} \sqrt{-g_{rr}} \ dr$ with $\rm g_{rr}=-1/g_{tt}=-1/(1-r^2 \ \Lambda/3$), here with the coordinate $\rm r$ on the $\rm x$ and the measured distance $\rm R$ on the $\rm y$ axis:

AdS radial coordinate

$\rm r \to \infty$ still corresponds to $\rm R \to \infty$ though. With the time dilation $\rm d\tau/dt=\sqrt{g_{tt}} \ $ it's the other way around: stationary clocks tick faster relative to the observer the farther away they are, and infinitely fast at infinite distance (that effect is also relative and holds for every observer in the center of his own coordinate system).

In Schwarzschild spacetime clocks tick slower and rulers measure longer the closer to the black hole they get, and in AdS they do the opposite the farther away from the observer they get.

It takes a photon forever to make the finite distance to the black hole horizon in the frame of the Schwarzschild bookkeeper, and it takes it a finite time to get to infinity and back in the frame of the AdS observer.

  • $\begingroup$ Hey with what software did you create the graph? $\endgroup$
    – RC_23
    Aug 13, 2023 at 4:35
  • $\begingroup$ @RC_23 - click on the image and you'll see $\endgroup$
    – Yukterez
    Aug 13, 2023 at 4:37
  • $\begingroup$ The Hubble constant isn't imaginary. If you solve the Friedmann equations with $k=-1$ you'll get a sinusoidal real scale factor with a time-varying real Hubble parameter. There are no solutions with $k\ge 0$. $\endgroup$
    – benrg
    Aug 14, 2023 at 3:22
  • $\begingroup$ @benrg wrote: "The Hubble constant isn't imaginary" - in a way it is, since the period in imaginary time is π/H, so H=c√(Λ/3) needs to be imaginary for that. The big bang and big crunch in an infinite AdS universe happens all the time and simultaneously, not one after the other, you can throw the particle at any time and the outcome will always be the same. Therefore the metric is also independent of t, despite the period in imaginary time. $\endgroup$
    – Yukterez
    Oct 31, 2023 at 5:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.