# Ricci scalars for space and spacetime, local and global curvature

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1. If Ricci scalar describes the full spacetime curvature, then what do we mean by $k=0,+1,-1$ being flat, positive and negative curved space?

2. Is $k$ special version of a constant "3d-Ricci" scalar?

3. What is the difference between the local and global spacetime curvature?

This post imported from StackExchange Physics at 2014-03-22 16:57 (UCT), posted by SE-user Winnie
asked Apr 17, 2013
$k$ is a constant appearing in the FLRW metric for a homogenous and isotropic universe, relating to the scalar curvature of spatial slices. $k$ is a special feature of certain cosmological models but the Ricci scalar exists for any spacetime.

This post imported from StackExchange Physics at 2014-03-22 16:57 (UCT), posted by SE-user Michael Brown

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The $k$ notation is generally used to describe Friedmann Robertson Walker cosmological models. These are built on the assumptions of homogeneity and isotropy. The spacetime can be described as being foliated by spatial slices of constant curvature. The k value is the sign of this spatial curvature if the {-1, 0, +1} convention is adopted. As the curvature is a constant, it makes sense to talk of its sign. Further details here.

This post imported from StackExchange Physics at 2014-03-22 16:57 (UCT), posted by SE-user twistor59
answered Apr 17, 2013 by (2,500 points)
is there any meaningful connection between K and the Ricci tensor? or is it overly complicated to use in the FRW metric?

This post imported from StackExchange Physics at 2014-03-22 16:57 (UCT), posted by SE-user Dmist
If I understood you correctly, the k value is simply the sign of R_ii? That is, for a homogenous and isotropic spacetime, when R_ii is constant.

This post imported from StackExchange Physics at 2014-03-22 16:57 (UCT), posted by SE-user Winnie
@Winnie yes, the sign of $^3R$

This post imported from StackExchange Physics at 2014-03-22 16:57 (UCT), posted by SE-user twistor59
@Dmist I guess you mean the full 4 dimensional Ricci tensor? I'm not sure that there is, I can't remember the expressions for the Ricci tensor for FRW off by heart. The full spacetime has an extrinsic curvature which describes how the three-slices are embedded, and this will complicate things.

This post imported from StackExchange Physics at 2014-03-22 16:57 (UCT), posted by SE-user twistor59
Being the scalar curvature, is k also preserved under the coordinate transformations? (Coordinate transformations of a FRW metric.)

This post imported from StackExchange Physics at 2014-03-22 16:57 (UCT), posted by SE-user Winnie
Under coordinate transformations which preserve the three-slices, but not under general coordinate transformations. The whole thing is slicing-dependent!

This post imported from StackExchange Physics at 2014-03-22 16:57 (UCT), posted by SE-user twistor59
@Twistor Thank you! So, it is possible to change intrinsic curvature of space with coordinate transformations?

This post imported from StackExchange Physics at 2014-03-22 16:57 (UCT), posted by SE-user Winnie
I'm getting an automatic message telling me to move this discussion to chat, but that you don't have enough rep yet to chat! Maybe you should raise separate questions, as this is veering off the original topic. But in the meantime... if by "intrinsic curvature" you mean the curvature scalar, then no, if you keep the slicing fixed (i.e. fix what you mean by "space"), then changing coordinates on that space will preserve the curvature scalar.

This post imported from StackExchange Physics at 2014-03-22 16:57 (UCT), posted by SE-user twistor59
Separate question here physics.stackexchange.com/questions/61414/…

This post imported from StackExchange Physics at 2014-03-22 16:57 (UCT), posted by SE-user Winnie

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