# What is the curvature of an empty universe?

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?

-
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 Brown Apr 23 '13 at 13:17
I mean completely empty $\rho_M=\rho_R=\rho_{\Lambda}=0$. – misi Apr 24 '13 at 2:53

## 1 Answer

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 eqution. So the two are the same space descibed by different co-ordinates. The hyperbolicity of the Milne universe is just down to taking different spatial slices, and it's Riemann tensor is everywhere zero like the Minkowski space. A quick Google found this article that goes into more detail.

-
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 2:49
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 5:43