Exercise 13.12 from Gravitation by Misner Thorne and Wheeler

In exercise 13.12 the student is asked to show the relationship between the Ricci Curvature Tensor and the Curvature Scalar on one hand, and the Einstein Curvature Tensor, defined by the "contracted double dual" of the Riemann Curvature Tensor.

I noticed while trying to solve this problem that there was no help available online. Since I took the time to think about it, I felt the need to communicate my results. If you wish to take pot shots then please use the comments section to explain how I am all wet.

Here goes.

The double dual of Riemann, the dual on the first two indices and also the last two indices, is defined by MTW in equation 13.46 as

(1.1)

Here



is the Levi-Civita Tensor, whose definition and properties are laid out in Exercise 3.13 of MTW.

The Einstein Curvature Tensor is defined in equation 13.47 as the contraction

(1.2)

The Ricci Curvature Tensor and the Curvature Scalar are defined by equation 13.48

(1.3)

So the exercise consists of proving that the following relation holds among the above objects.

(1.4)

Which is equation equation 13.49 in MTW.

First off, MTW equation 13.46 says a bit more than what I wrote above in 1.1. It also says that

(1.5)

where the Permutation Tensor is defined by

(1.6)

This again follows MTW Exercise 3.13, which I may return to in another post someday. But for now I have fatter mice to catch.

Taking the contraction of 1.5 gives the Einstein Tensor in the following form.

(1.7)

Consider first those terms in 1.7 which have



or



So



(I shall discuss some of this fancy footwork with the indices in a moment but for now I am hot on the trail of my quarry.)

Next consider only terms in 1.7 which have



or



So



(The terms mentioned so far add up exactly to the second part of the right hand side of equation 1.4.)

These represent only two of six possible cases. The full set of possibilities may be described graphically as follows.



Where cases A and B have already been examined.

Case C



Case D



Case E



Case F is exactly like Case E and gives the same result.

So these four terms add up to the first part of the right hand side of equation 1.4. Taking into account all six terms gives equation 1.4 exactly.

Otherwise said, that would be two from column A @



and

four from column B @



for a grand total of



Which would be Q.E.D. more or less.

And now to return, for a moment, to the fancy footwork, involving indices, above. First off



Here I have used the symmetry of the Riemann Tensor on exchange of the first two indices with the last two and also the symmetry of the Metric Tensor. Contracting this gives



So all that switching around of indices above was actually legit.

O.K. So that's one problem in the bag!

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