Geometric Interpretation of Einstein Tensor

[Markus Hanke]

Is there a simple geometric interpretation of the Einstein tensor ? I know about its algebraic definitions ( i.e. via contraction of Riemann's double dual, as a combination of Ricci tensor and Ricci scalar etc etc ), but I am finding it hard to actually understand it geometrically.

Specifically, what I am looking for is an interpretation along the lines of what Misner/Thorne/Wheeler do in "Gravitation" for the Riemann tensor, in terms of the "slots" of tensor understood as being a linear "machine". You take the tangent vector to a reference geodesic, insert it into both slots 1 and 3 of Riemann, and insert the separation vector to some neighbouring geodesic into slot 2; the result is a vector which signifies the ( covariant ) rate of change of your separation vector ( relative acceleration ) between the geodesics, with respect to your chosen time coordinate. Geodesic deviation, in other words.

Can something similar be done with the Einstein tensor ?

Apologies in advance if this question turns out to be either meaningless, imprecise, or trivial. Hopefully you can see what I am trying to get at.

 

[robphy]

Here's an old thread which asked a similar question... My reply gave some references... but no satisfactory answer.
https://www.physicsforums.com/threads/geometry-of-the-riemann-ricci-and-weyl-tensors.648969/

 

[Markus Hanke]

Thank you :) You are right in saying that no actual answer was arrived at, either on the thread you referenced, nor on any other thread on here ( that I can find ) asking the same thing. However, this set off bells in the back of my head, since I remembered long ago having come across an article to do with geometric explanations of the curvature tensors in GR; a Google search then revealed this article by Lee Loveridge :

http://arxiv.org/pdf/gr-qc/0401099v1.pdf

The answer is precisely equation (9) - contract both slots of the Einstein tensor with a unit time-like vector, and you get the scalar curvature in the corresponding spatial directions. So basically, once a time direction is chosen, the Einstein tensor is a measure of how the area of an infinitesimally small D-1 sphere differs from its counterpart in flat space ( eqn (8) in the same paper ). That seems like a pretty good geometric interpretation to me !

Is this an acceptable way to "visualise" the Einstein tensor ?

 

[robphy]

I think it's a good start.. but it's incomplete.
You get the diagonal components. Maybe one can transform to a basis of its eigenvectors, then transform back.
In addition, it's not clear to me how to see (in this visualization/interpretation) that "the Einstein tensor is divergence-free".

 

[martinbn]

I think it's a good start.. but it's incomplete.
You get the diagonal components. Maybe one can transform to a basis of its eigenvectors, then transform back.
In addition, it's not clear to me how to see (in this visualization/interpretation) that "the Einstein tensor is divergence-free".

It is symmetric and bilinear, so the non-diagonal elements are determined by the diagonal ones. $E(u,v)=\frac14[E(u+v,u+v)-E(u-v,u-v)]$

 

[PeterDonis]

it's not clear to me how to see (in this visualization/interpretation) that "the Einstein tensor is divergence-free".

The best geometric interpretation of this that I have seen is MTW's explanation that "the boundary of a boundary is zero", which is a general theorem in differential geometry. The divergence of the Einstein tensor can be interpreted, as MTW shows, as "the boundary of a boundary", so it must be zero by the theorem. I'm not sure how this fits in with the interpretation of the Einstein tensor itself that is given in post #3, though.

 

[robphy]

It is symmetric and bilinear, so the non-diagonal elements are determined by the diagonal ones. $E(u,v)=\frac14[E(u+v,u+v)-E(u-v,u-v)]$

Ah, yes... the polarization identity.

The best geometric interpretation of this that I have seen is MTW's explanation that "the boundary of a boundary is zero", which is a general theorem in differential geometry. The divergence of the Einstein tensor can be interpreted, as MTW shows, as "the boundary of a boundary", so it must be zero by the theorem. I'm not sure how this fits in with the interpretation of the Einstein tensor itself that is given in post #3, though.

Yes... algebraically. But how can this visualization suggest that?

 

[George Jones]

You might want to look at chapter 4, "The Geometry of Einstein's Equation" (pages 39, 40 for the Einstein tensor), from the book "Gravitational Curvature" by Frankel.

 

[ChrisVer]

can't you reach some interpretations by choosing such geometries that for example could bring the Einstein Tensor to be equal to the Ricci Tensor?? (vanishing Ricci scalar)...
In such a case, geometrically the ET and the RT would represent the same thing.

 

[Markus Hanke]

You might want to look at chapter 4, "The Geometry of Einstein's Equation" (pages 39, 40 for the Einstein tensor), from the book "Gravitational Curvature" by Frankel.

Thank you ! I'm not familiar with this text, but the reviews seems to be quite good. Would you recommend this book ( in general, not just in the context of this thread ) ?