I Geometric Interpretation of Einstein Tensor

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1. Apr 7, 2016

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.

2. Apr 7, 2016

3. Apr 7, 2016

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 ?

4. Apr 7, 2016

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".

5. Apr 7, 2016

martinbn

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

6. Apr 7, 2016

Staff: Mentor

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.

7. Apr 7, 2016

robphy

Ah, yes... the polarization identity.

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

8. Apr 7, 2016

George Jones

Staff Emeritus
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.

9. Apr 7, 2016

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.

10. Apr 8, 2016

Markus Hanke

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 ) ?