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Is there a Schwarzschild-de Sitter solution with the correct topology?

+ 2 like - 0 dislike
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The topology of spatial slices of de Sitter is $S^3$. However, the topology of spatial slices of the space-time normally given as "Schwarzschild-de Sitter" (for instance on wikipedia) is $S^1\times S^2$. Similarly, in the cosmological region (where $r$ is large), the metric has a translation symmetry in the $t$ variable and an $SO(3)$ group of rotations in the spherical direction. This is completely unlike the $SO(4)$ symmetry group for spatial slices of de Sitter. Furthermore, for $r$ fixed, if one takes the mass to zero and $r$ sufficiently large, one obtains a metric $-(r^2-1)^{-1} dr^s + (r^2-1) dt^2 +r^2(d\theta^2+\sin^2d\phi^2)$. Thus, in this limit and recognizing $r$ as time coordinate and $t$ as a space coordinate, one finds the limiting spacelike hypersurfaces either are isometeric to $R\times S^2$ or (with an identification in $t$) to $S^1\times S^2$. Since the limiting behaviour can be misleading, let me state my question as follows. Are there Schwarzschild-de Sitter solutions such that any spacelike hypersurface for which $r$ is bounded and which cannot be extended in the cosmological region must have the topology of the a $3$-ball (i.e. $S^3$ minus a point) instead of $S^1 \times S^2$?

asked Dec 17, 2016 in Theoretical Physics by anonymous [ revision history ]
recategorized Dec 18, 2016 by Dilaton

I recommend reading chapter 4, specifically section 4.3 of the Griffiths & Podolský Exact space-times monography. The short answers is that the spherical coordinates do not cover the whole space-time and you need to use different charts to cover it.

Thanks Void. It's a good point that the chart with spherical coordinates is somewhat misleading and one should think of the conformal diagram. The standard diagram for de Sitter is a square with scri plus and scri minus as the top and bottom sides and the north and south pole as the sides. Diagonally across this square are the cosmological horizons given by the domain of dependence of scri plus excluding the two poles. In the conformal diagram for Schwarzschild-de Sitter, one sees a similar cosmological horizon. If I've understood correctly, this means that each of the (infinitely many) expanding regions of Schwarzschild-de Sitter is like a de Sitter with a black hole at the north and south poles. This explains why the region between has the topology of an interval cross $S^2$ instead of the topology of a $3$-ball. My question remains: Is there an explicit solution where the spatial slices outside the horizon have the topology of 33-ball.

1 Answer

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One explicit form where you can see the topology of the spatial slices in spherical coordinates as $S^3$ is $r=\sin \chi$.

But

  1. These are different spatial slices than in the dS FLRW form!
  2. The global topological consideration should not be done without understanding the whole analytical extension, the whole of Schw.-dS is larger than simple dS!

I am not aware about any "asymptotically FLRW" set of coordinates describing Schw.-dS but this chart would be very ugly because it wouldn't be stationary or homogeneous/isotropic on the spatial slice. Again I recommend Griffiths & Podolský on references on various coordinate charts.


It may sound pedantic but I would also like to stress that these coordinate charts are not different "solutions" but simply coordinate charts mapping the unique solution in different ways. I assume you know about this fact but seeing how many people nowadays do not understand the fleeting nature of coordinates in relativity, I think one should be careful in the language.

answered Dec 18, 2016 by Void (1,505 points) [ no revision ]

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