Types" of Space-Time Curvature in GR

[Drakkith]

When we talk about space-time curvature or the curvature of space, how many different "types" of curvature are there according to GR?

For example, the rounded surface of a cylinder is curved in only 1 dimension, while the other is flat. For a sphere, both dimensions of the surface are curved. Since space-time is 4 dimensional, I assume the curvature extends to all 4 dimensions?

Also, I've read about intrinsic and extrinsic curvature. Which one does GR predominately deal with?

 

[Matterwave]

GR, in the covariant formulation, predominantly deals with intrinsic curvature. This is because the 4-dimensional space-time manifold is not, as far as we know, embedded into a higher dimensional space.

However, in the 3+1 formulation of GR where you split your space-time manifold into a foliation of Cauchy surfaces, you can naturally talk about the extrinsic curvature of each Cauchy surface embedded into space-time.

The (local) intrinsic curvature of space-time is calculated through the Riemann tensor, which has 20 independent components in general. So I guess you can say there are 20 "types" of curvature? I don't really know what you mean by "types" though.

 

[Drakkith]

So I guess you can say there are 20 "types" of curvature? I don't really know what you mean by "types" though.

Well, for example, we commonly talk about the "curvature of the universe", and how it can be flat, hyperbolic, or spherical. I assume this curvature is different in some way than the curvature of space-time near a large, massive object like a star. But, what's the difference?

 

[pervect]

When we talk about space-time curvature or the curvature of space, how many different "types" of curvature are there according to GR?

For example, the rounded surface of a cylinder is curved in only 1 dimension, while the other is flat. For a sphere, both dimensions of the surface are curved. Since space-time is 4 dimensional, I assume the curvature extends to all 4 dimensions?

Also, I've read about intrinsic and extrinsic curvature. Which one does GR predominately deal with?

Your example of a cylinder having two curvatures is the extrinsic curvature. GR is concerned mainly with the intrinsic curvature of the cylinder. For any two dimensional surface, there are two principal extrinsic curvatures. The intrinsic curvature is equal to the product of the two principal extrinsic curvatures, which is a single number. This number is 0 in the case of the cylinder because the Gaussian curvature is proportional to the product of the principal curvatures, and one of the principal curvatures is zero for the cylinder.

See for instance http://en.wikipedia.org/w/index.php?title=Gaussian_curvature&oldid=598900344

In GR, the curvature is generally taken to be described completely by the Riemann curvature tensor. This is a tensor with 4 indices. In two dimensions, each index can take one of two values, in four dimensions, each index can take one of 4 values.

Thus in 2 dimensions, the Riemann has 2 x 2 x 2 x 2 = 8 components. However, there is a great deal of symmetry in the Riemann (formally described by the Bianchi identities), and there is only one degree of freedom in the tensor.

The only non-zero components in the 2d case would be $R_{0101} = R_{1010} = -R_{0110} = -R_{1001},$ so there is only one degree of freedom in the GR case, thus specifying the Gaussian curvature, a single number, specifies the whole tensor.

In four dimensions, the Riemann has 4 x 4 x 4 x 4 = 256 components, but there are 20 degrees of freedom http://mathworld.wolfram.com/RiemannTensor.html. So one needs to specify 20 unique and independent numbers at a point in a general space-time to fully describe its curvature.

Using the Bel decomposition, http://en.wikipedia.org/w/index.php?title=Bel_decomposition&oldid=597299599, given a space-time split, we can refine the 20 components of the Riemann into smaller tensors which provide more physical insight. There are:

6 components for the electrogravitic tensor, a 3 x 3 tensor which is basically equivalent to the Newtonian idea of the tidal force tensor. It's typically given the symbol E to represent it's similarity to the electric field in electromagnetism. The six degrees of freedom represent three degrees of freedom for the rotation of the tensor, and 3 values which represent the magnitude of the tidal force along each principal axis of the tensor.

8 components for the magnetetogravitc tensor, which describes frame dragging effects. If you have a static spacetime, this tensor is zero. This tensor is typically given the symbol B to represent its similarity to the magnetic field in electromagnetism.

6 components for the topogravitic tensor, which describes the purely spatial part of the space-time curature. As I recall the topogravitic tensor is equal to the electrogravitic tensor if you have a vacuum space-time. Its usually given the symbol L, and it doesn't have any electromagnetic equivalent as electromagnetism doesn't warp space.

Thus for a typical static Schwarzschild metric, B=0 and L=E, so all you need to specify is E, and that can be specified by the orientation of the three principal axes of the tensor (which takes three numbers to specify) and the magnitude of the tidal forces along said axes (which also takes three numbers to specify).

A completely general space-time would need 20 numbers to specify, however.

 

[Drakkith]

Oh, wow, thanks, Pervect. I'll have some reading to do.

Edit: Quick question. For a cylinder, since the gaussian curvature is zero, does that mean that parallel lines on the surface neither diverge nor converge?

 

[Matterwave]

Well, for example, we commonly talk about the "curvature of the universe", and how it can be flat, hyperbolic, or spherical. I assume this curvature is different in some way than the curvature of space-time near a large, massive object like a star. But, what's the difference?

It's a little bit complicated. Really because there are just so many different tensors and scalars that refer to curvature. There is the most general tensor, the Riemann, then there are the various contractions of it such as the Ricci tensor, an the Ricci scalar, and there's also the trace-reverse of the Ricci tensor, the Einstein Tensor, as well as the trace-free version of the Riemann, the Weyl curvature tensor (there are yet others!). These are the "intrinsic" curvature tensors that we see often in GR.

The "curvature of the universe" that we talk about in the FRW metric actually is referring to the curvature of the Cauchy hypersurfaces which foliate the space-time. It is a property of the 3-dimensional hypersurfaces and as such is a 3-dimensional curvature, not a 4-dimensional one. In the case of the FRW metric, where the spatial hypersurfaces are maximally symmetric, we can divide the curvature into 3 broad classes of positive, negative, and 0. This is not generally the case if we don't have such nice properties of our space time. In general, as noted, one has to care about all 20 possible components of the Riemann tensor.

 

[Matterwave]

Oh, wow, thanks, Pervect. I'll have some reading to do.

Edit: Quick question. For a cylinder, since the gaussian curvature is zero, does that mean that parallel lines on the surface neither diverge nor converge?

The intrinsic curvature of a cylinder is 0, this indeed means that the geodesics on a cylinder remain parallel and do not converge or diverge. Geodesic deviation is a measure of the intrinsic curvature.

The fact that the extrinsic curvature is non-zero can be seen in the fact that the normal vectors to the cylinder do not remain parallel, but change as I move around the cylinder. How the normal vector behaves as we move around is a measure of the extrinsic curvature.

 

[Drakkith]

It's a little bit complicated. Really because there are just so many different tensors and scalars that refer to curvature. There is the most general tensor, the Riemann, then there are the various contractions of it such as the Ricci tensor, an the Ricci scalar, and there's also the trace-reverse of the Ricci tensor, the Einstein Tensor, as well as the trace-free version of the Riemann, the Weyl curvature tensor (there are yet others!). These are the "intrinsic" curvature tensors that we see often in GR.

Wow, that is complicated. Are all these tensors and scalars simply variations/simplifications of one "master" tensor/scalar? Or do they all represent something unique?

 

[Matterwave]

The "master" tensor is the Riemann. It has 20 independent components. The Riemann tends to be quite complicated since it's a rank 4 tensor. All the other tensors I mentioned are derived from this Riemann tensor in one way or another.

The Ricci tensor is 1 contraction of the Riemann tensor ($R_{\mu\nu}=R^\alpha_{\mu\alpha\nu}$), and it's important because its trace-reversed version, the Einstein tensor ($G_{\mu\nu}=R_{\mu\nu}-\frac{1}{2}Rg_{\mu\nu}$), is the tensor which shows up in the Einstein field equations, and is the tensor which is proportional to the stress-energy tensor ($G=8\pi T$ in geometrized units).

The Weyl tensor is the trace-free version of the Riemann tensor (the expression is a little complicated, so I won't reproduce it here). It encodes the curvature that can be present in the absence of matter. For example, the Schwarzschild solution is a vacuum solution of the Einstein field equations. This means that the Ricci tensor is 0, but the Weyl tensor is not 0.

Curvature in higher dimensions than 1 tends to become pretty complicated (even in 1 dimension, it can be a little complicated), so that's why there are all these tensors. They express different aspects of the total curvature.

 

[A.T.]

For a cylinder, since the gaussian curvature is zero, does that mean that parallel lines on the surface neither diverge nor converge?

When you unroll the cylinder you get a flat sheet on which parallel lines obviously stay parallel. The unrolling only affects the extrinsic curvature, while the intrinsic curvature (of zero) is preserved.

In other words: If you can unfold a surface into a flat sheet (without stretching and compressing distances within the surface), then it has no intrinsic curvature.

 

[Drakkith]

Awesome. Thanks guys. I'll read up more based on the answers I got here.