Meaning of Curvature Scalar (R) in GR & Its Evolution

[Magister]

What is the meaning of the curvature scalar (R) in GR? More precisely, what is the meaning of it's evolution? Why when we are concerning the solar system we take R to be small and when we are concerning the cosmological scales the we assume R to be large?

Thanks in advance.

 

[snoopies622]

I have the same question for the Ricci and Einstein tensors. I know how to compute them, but I don't have an intuitive sense of what they mean. They are sometimes described as being a kind of "average curvature" at a point, but that doesn't explain the difference between, say, G_\theta _\phi and G_\phi _r.

 

[Mentz114]

Surprisingly, the curvature scalar for the Schwarzschild metric is zero. There are non-zero components of the Riemann tensor. They are made from second derivatives of the metric and some of them represent tidal potentials.

In an expanding cosmological model, the curvature is inversely proportional to the square of the 'radius'. So it's high at the beginning and is decreasing as the cosmos expands.

 

[Magister]

In an expanding cosmological model, the curvature is inversely proportional to the square of the 'radius'. So it's high at the beginning and is decreasing as the cosmos expands.

But is that because as the universe expand, the curvature in at a point is getting smaller? If we look at the cosmos like a sphere, as the radius expands the curvature in the surface is getting smaller. Is this the interpretation one should give to R?

The thing that confuse me more is the fact that R=0 doesn't imply a flat manifold.

 

[Mentz114]

The thing that confuse me more is the fact that R=0 doesn't imply a flat manifold.

Yes, it's not very intuitive. To get a flat spacetime all the Christoffel symbols ( connections) have to disappear.

 

[robphy]

Yes, it's not very intuitive. To get a flat spacetime all the Christoffel symbols ( connections) have to disappear.

I agree that the issue of interpretation is not very intuitive...
but I hope there is a good interpretation to be found.Note that: in the plane, the Christoffel symbols for the polar coordinates are all not zero.

A flat space requires the Riemann Tensor to be zero (which can be formed from the Christoffel symbols and their derivatives). The scalar curvature is of course formed from the traces of Riemann.Possibly interesting reading: http://arxiv.org/abs/gr-qc/0103044

Note also that the curvature scalar is featured in the action for GR.

 

[Mentz114]

robphy,
so, both the plane in polar coords(PP), and the Schwarzschild metric have zero R and non-zero Christoffel symbols. How do we tell if either is curved ?

I notice that all the Ricci components are zero for both, but there are non-zero Riemann components for Schwarzschild, but not for PP.

M

 

[robphy]

robphy,
so, both the plane in polar coords(PP), and the Schwarzschild metric have zero R and non-zero Christoffel symbols. How do we tell if either is curved ?

I notice that all the Ricci components are zero for both, but there are non-zero Riemann components for Schwarzschild, but not for PP.

M

As suggested in my earlier post [and as you have done for your examples], compute the Riemann curvature tensor.

 

[Mentz114]

OK, so space-time is curved if there are tidal effects. Thanks for clearing that up.

 

[Stingray]

Just to add a little, the Ricci scalar is one of many curvature scalars that can be written down using the Riemann tensor. Einstein's equation tells you exactly what R is at every point. It always vanishes in vacuum, for example. The other curvature scalars do not necessarily have this behavior. Any one of them being nonzero tells you that the spacetime is curved. The reverse is not true, however. There are curved spacetimes (describing plane gravitational waves) where all standard curvature scalars vanish. The Riemann tensor still manages to be nonzero in these cases. There is therefore curvature.

 

[pmb_phy]

Yes, it's not very intuitive. To get a flat spacetime all the Christoffel symbols ( connections) have to disappear.

A vanishing Ricci tensor doesn't even imply a flat spacetime. In fact the Ricci tensor vanishes in the vacuum of a Schwarzschild spacetime.

Note - It should be noted that all of the components of the affine connection vanishing only means that you have a locally flat coordinate system. It does not mean that the spacetime is flat. This if a very important distinction.

Pete

 

[pmb_phy]

OK, so space-time is curved if there are tidal effects. Thanks for clearing that up.

Exactly. In fact the terms spacetime curvature and tidal gravity are exactly the same two things but described in different languages.

Pete