# Curvature Scalar

1. May 22, 2008

### 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?

2. May 22, 2008

### 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 $$G_\phi _r.$$

3. May 22, 2008

### 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.

4. May 22, 2008

### Magister

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.

5. May 23, 2008

### Mentz114

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

6. May 23, 2008

### robphy

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.

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

7. May 23, 2008

### 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

8. May 23, 2008

### robphy

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

9. May 23, 2008

### Mentz114

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

10. May 23, 2008

### 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.

11. May 25, 2008

### pmb_phy

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

12. May 25, 2008

### pmb_phy

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

Pete