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My calculations tell me an empty universe has hyperbolic curvature. Is this correct? If it is, can anyone help me understand why this is intuitively?

Qmechanic
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misi
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    What do you mean by empty? Whether you mean no matter and/or no dark matter and/or no dark energy density makes a difference. – Michael Apr 23 '13 at 13:17
  • I mean completely empty $\rho_M=\rho_R=\rho_{\Lambda}=0$. – misi Apr 24 '13 at 02:53
  • I think that the use of the density parameter to detect curvature is using (strongly) the fact that the universe isn't empty. If it were, it would affect the other constants. For instance it seems you could choose the gravitation constant to be whatever you like without issue, since there would be nothing to attract anything else. – j0equ1nn Oct 18 '20 at 12:37

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I remember being confused by this, and thanks to help from this site I think I understand the problem (though I probably don't! :-).

If you take the FLRW metric and extrapolate to zero density you get the Milne metric, which is hyperbolic and maximally curved. However the Milne metric is equivalent to the Minkowski metric with a co-ordinate transformation, and the Minkowski metric is obviously also a solution to the vacuum equation. So the two are the same space desrcibed by different co-ordinates. The hyperbolicity of the Milne universe is just down to taking different spatial slices, and its Riemann tensor is everywhere zero like the Minkowski space. A quick Google found this article that goes into more detail.

John Rennie
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  • Right. So $k=-1$ is what my calculations give me. So what you (and the article) are saying is that although the space is hyperbolic, the Riemann scalar is $0$. Is this right? So the space is hyperbolic but the space time as a whole is flat? – misi Apr 24 '13 at 02:49
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    The Riemann tensor is everywhere zero. The space is flat but you're using a curved co-ordinate system. – John Rennie Apr 24 '13 at 05:43
  • So for clarity, this must be a statement about 2 local coordinate charts and furthermore they must have different ranges. Otherwise, it's non-sensical to claim in a Riemannian manifold can change curvature via a coordinate change. – zzz Mar 14 '18 at 02:13
  • @bianchira: strictly speaking it's a Lorentzian manifold not a Riemannian manifold. But yes, the curvature isn't changed simply because you choose different coordinates. – John Rennie Mar 14 '18 at 05:39
  • @JohnRennie correct me if I'm wrong, this whole discussion amounts to say that the spatial 3-curvature is not a quantity invariant for change of coordinates, while the 4-curvature is invariant under a general coordiante transformation, right? – AnOrAn Nov 13 '19 at 09:25
  • @AnOrAn correct – John Rennie Nov 13 '19 at 09:25
  • @JohnRennie So when we measure the 3-curvature of the universe we are measuring something that is valid only for a certain set of coordinates, comoving coordinates, then do you know what's the point of doing that? I mean a closed universe for us may be open for a non comoving observer right? – AnOrAn Nov 13 '19 at 09:31
  • This isn't really the place for prolonged discussions, but we could continue in the chat room if you want. – John Rennie Nov 13 '19 at 09:32