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How can a deSitter space have finite size?

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a deSitter space is a maximally symmetric solution of Einstein equations, I have some problem picturing one thing: this space is past and future (time) infinite but spatial slices have finite size, how can we have finite size in slicing at constant time, and infinite time size when slicing at constant radial coordinate if the solution is supposed to be symmetric in the plane (t,r).

This post imported from StackExchange Physics at 2014-05-01 12:18 (UCT), posted by SE-user toot
asked Jul 13, 2012 in Theoretical Physics by toot (445 points) [ no revision ]

1 Answer

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The reason is that the constant time slice ends on the cosmological horizon. If you look at deSitter space in t,r coordinates, the metric is (in appropriate units):

$$ ds^2 = - (1-r^2) dt^2 + {1\over 1-r^2} dr^2 + r^2 (d\theta^2 + \sin^2\theta d\phi^2) $$

The time coordinate freezes at r=1, so that the entire history of the region is contained in $0<r<1$. This is the reason for the finite volume. The distance between the spheres at $r$ and $r+dr$ is read off from the metric:

$$ ds = {dr\over \sqrt{1-r^2}}$$

and this has an inverse square-root singularity at r=1, which has a finite integral. So if you integrate for the volume, you multiply the radial distance by the area of the sphere at radius r, or $4\pi r^2$, to find

$$ V = \int_0^1 {4\pi r^2 \over \sqrt{1-r^2}} = \pi^2 $$

To do the integral, change variables to $u^2=1-r^2$ and it becomes $4\pi$ times the area of a quarter-circle of unit radius.

The result is finite because you are looking at a causal patch of deSitter space. The geodesics separate exponentially due to the positive curvature (in Euclidean positively curved space, they converge, but this is Minkowski signature). This means that the full space disconnects into regions which are not in causal contact with each other. Each such region is finite volume.

When you slice at constant r, you produce an infinite cylinder. This is not a contradiction.

The reason this is counterintuitive is because the t-coordinate is bad at r=1. The killing vector defining t becomes null at a finite distance away from the origin. The finite volume is the volume of the region where the t-coordinate is timelike and a killing vector, it is not the full volume of the space (in the GR maximally extended way of looking at things).

The Euclidean continuation of deSitter space is just the 4-sphere. This comes from changing the sign of the dt term in the metric. The physics of deSitter space is best thought of as a path integral on the sphere, not in some sort of maximal extension. This is a point of contention, as a lot of people don't like cutting off the universe at the visible universe, but this is the way suggested by the holographic principle. The periodicity of the deSitter space sphere in imaginary time tells you the Hawking temperature of the cosmological horizon, and the finite volume of the t=0 slice just becomes the volume of the 3-sphere section of the 4-sphere at imaginary time 0.

This post imported from StackExchange Physics at 2014-05-01 12:18 (UCT), posted by SE-user Ron Maimon
answered Jul 15, 2012 by Ron Maimon (7,535 points) [ no revision ]

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