In Feynman's Lectures on Physics (Volume 2, chapter 42) he states that Einstein's field equation is equivalent to the statement that in any local inertial coordinate system the scalar curvature of space (which he gives in terms of the radius excess of a small sphere) is proportional to the energy density at that point. Does anyone know of an elegant proof of this fact?
I have a half-elegant proof based on the following:
- Show that the $tt$-component of the stress-energy tensor, $T$, is equal to energy density in inertial coordinates (straightforward)
- Show that the $tt$-component of the Einstein curvature tensor, $G$, is proportional to the scalar curvature of space in inertial coordinates (horrific)
- Argue that since $T$ and $G$ are both symmetric 2-tensors, and symmetric 2-tensors form an irreducible representation of the lorentz group, if the $tt$-components are equal in every local inertial frame then the tensors themselves must be equal (straightforward)
The problem with the above is that the only way I can find to prove step 2. is to explicitly calculate the radius excess in terms of the metric in local inertial coordinates. This requires finding the geodesic paths of length $r$ to order $r^3$, expanding the metric to third order at the surface of the generated sphere, and calculating the surfacearea (or volume, I found area easier). The integral for the surface area ends up with about 40 terms to take care of (which do not seem to be easily reducible through e.g. symmetry considerations), and took me several pages to do. Since the relationship between scalar curvature and the curvature tensors is quite fundamental I expect there must be a more direct, abstract argument, but I cannot find it.