Solve Reimann Curvature: Diff Geom Tips for Computing 256 Components

[Kevin McHugh]

Should I post in Diff Geometry? I searched that forum, and did not see what I was looking for. I want to compute all 256 components of the curvature tensor. Do I start with the equation of geodesic deviation in component form, or can I go straight to the definition of the components in terms of connection coefficients? What are the first steps to embarking on this adventure?

 

[stevendaryl]

Should I post in Diff Geometry? I searched that forum, and did not see what I was looking for. I want to compute all 256 components of the curvature tensor. Do I start with the equation of geodesic deviation in component form, or can I go straight to the definition of the components in terms of connection coefficients? What are the first steps to embarking on this adventure?

When you say "compute", do you mean, to come up with numerical values, as opposed to symbolic expressions? The actual numerical value for the components will depend on which geometry you are starting from.

As to that number, 256, that's an over-estimate. Many of the components are equal, because of symmetries. There are only 20 independent components.

 

[pervect]

Should I post in Diff Geometry? I searched that forum, and did not see what I was looking for. I want to compute all 256 components of the curvature tensor. Do I start with the equation of geodesic deviation in component form, or can I go straight to the definition of the components in terms of connection coefficients? What are the first steps to embarking on this adventure?

Many of the components are zero. I think there's only about 21 unique non-zero components buried in the mass of 256.

If you want to do it by hand, I'd recommend looking at Wald's "General Relativity" section 3.4, pg 47, "Methods for computing curvature". There's a straightforward section on how to do it in a component basis, and a discussion of a much easier to carry-out method based on curvature two-forms which is unfortunately less straightforward :(, even though it may be less work in the long run.

However, if you just want the answers, I'd suggest using a program like Maxima (which is free). There are some guides to the unfortunately not-quite-trivial matter of making Maxima's notation mach the notation in GR textbooks. This guide is written by Chris Hillman here on PF. https://www.physicsforums.com/threads/brs-using-maxima-for-gtr-computations.378991/

The Bel decomposition might also help in sorting out the vast number of components of the Riemann into something more useable and physically meaningful. Wiki's description of it is unfortunately too terse to be really useful, https://en.wikipedia.org/wiki/Bel_decomposition. There's a section on it in MTW but I don't recall exactly where it's buried, and it's not named as the Bel decomposisiton. Look for the keywords "electrogravitic tensor", though, if you want to hunt it down.

Wikki's description of the Bel decomposition does illustrate how one can reduce the 256 element Riemann into 3 16 element tensors based on a time-like congruence that can be taken as representing an observer. The three tensors in the GR decomposition are the electrogravic tensor, the magnetogravitic tensor, and the topogravitic tensor. (Some non-GR applications might have 4 tensors in the decomposition). The total number of components is greatly reduced, the size of each component is much better, and the physical meaning is a lot clearer, too. jIf you look at a stationary observer in a Schwarzschild metric, you can get a result as simple as an electrogravitic tensor which is numerically equal to the topogravitic tensor, and a zero magnetogravitic tensor. There are only 3-4 nonzero compoents in the electrogravitic (and numerically equal topogravitic) tensor, and the former can be readily physically interpreted as tidal forces.

 

[PeterDonis]

I think there's only about 21 unique non-zero components buried in the mass of 256.

It's actually 20, as stevendaryl said.

 

[pervect]

It's actually 20, as stevendaryl said.

Unless there is a typo in MTW, it should be 21 components. I believe only 20 of themn are independent as Steve said, however. But if you want to write all of the components down on paper, I believe there's 21 of them. See MTW pg 348 "He [Cartan] went on to package the 21 components of $R_{\mu\nu\alpha\beta}$ into six curvature two-forms." I'm a bit hazy on the constraint, I think it was one of the Bianchi identities. As I recall you can only independently specify 20 of the 21 components, when you do that, the identity sets the value of the last component.

 

[PeterDonis]

Unless there is a typo in MTW, it should be 21 components.

I'll have to check my copy when I get a chance; I don't remember where the 21 came from.