general relativity - Feynman's statement of the Einstein Field Equations

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?

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I have a half-elegant proof based on the following:

1. Show that the $tt$-component of the stress-energy tensor, $T$, is equal to energy density in inertial coordinates (straightforward)



2. Show that the $tt$-component of the Einstein curvature tensor, $G$, is proportional to the scalar curvature of space in inertial coordinates (horrific)


3. 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.

Answer

The key to proving item 2 is to express the metric in Riemann normal coordinates, which is usually what is meant when you say you are working in a locally inertial coordinate system. In these coordinates, the metric is equal to the Minkowski metric at a point, the first derivatives of the metric at the point vanish, and the second derivatives of the metric are given by the Riemann tensor at that point. The explicit form of the metric components are then (see e.g. this document)

$$g_{\mu\nu} = \eta_{\mu\nu}-\frac13R_{\mu\alpha\nu\beta}x^\alpha x^\beta + ...$$

where the dots represent higher order corrections in the coordinate distance from the origin, $x^\alpha = 0$.

We need to compute the volume of a sphere of coordinate radius $r$. For this we need the spatial metric, which is $h_{\alpha \beta} \equiv g_{\alpha\beta}+u_\alpha u_\beta$, and $u^\alpha$ is tangent to the inertial observer, so points in the time direction. The spatial volume element comes from the determinant of $h_{ij}$ as a spatial tensor ($i,j$ are only spatial indices). We have

$$h_{ij} = \delta_{ij} -\frac13R_{i\mu j\nu}x^\mu x^\nu+...$$

and the first order correction to the determinant just adds the trace of this tensor,

$$\sqrt{h} = 1 + \frac12\left(-\frac13\delta^{kl}R_{kilj}x^i x^j\right). $$

It will be useful to work with spacetime indices in a moment, where the background spatial metric is given by $\delta_{\mu\nu} = \eta_{\mu\nu} +u_\mu u_\nu$, and its inverse is $\delta^{\mu\nu} = \eta^{\mu\nu}+u^\mu u^\nu$. Now the volume of the sphere is simply

$$V = \int d^3x \sqrt{h} = \int d^3x\left(1-\frac16 \delta^{\mu\nu}x^\alpha x^\beta R_{\mu\alpha\nu\beta}\right).$$

(the limit of integration is over a coordinate sphere centered at the origin).

The first term will give the flat space volume of the sphere, so we need to compute the second term to get what Feynman is calling the spatial curvature of space. Remember that the Riemann tensor is taken to be constant since it is evaluated at the origin. Also, when integrated over a spherical region, only the trace of $x^\alpha x^\beta$ contributes, the other parts canceling out, so we can replace $x^\alpha x^\beta \rightarrow \frac13 r^2 \delta^{\alpha\beta}$. So the integral we are computing becomes

$$\Delta V = -\frac16\frac{4\pi}{3}\delta^{\mu\nu}\delta^{\alpha\beta}R_{\mu\alpha\nu\beta}\int_0^{r_s} r^4 dr = -\frac{2\pi}{45} r^5\delta^{\mu\nu}\delta^{\alpha\beta}R_{\mu\alpha\nu\beta}.$$

The numerical coefficient is not important, we only care about the dependence on the Riemann tensor. Re-writing the $\delta$'s in terms of the background metric $\eta^{\mu\nu}$ and $u^\alpha$, we get

$$\delta^{\mu\nu}\delta^{\alpha\beta}R_{\mu\alpha\nu\beta}=(\eta^{\mu\nu}+u^\mu u^\nu)(\eta^{\alpha\beta}+u^\alpha u^\beta)R_{\mu\alpha\nu\beta} = R + 2 R_{\mu\nu}u^\mu u^\nu,$$

where $R$ is the Ricci scalar at the origin. Now we can easily check that this is proportional to the $uu$-component of the Einstein tensor (using $u^\mu u^\nu g_{\mu\nu} = -1$),

$$G_{\mu\nu}u^\mu u^\nu = \frac12(2 R_{\mu\nu}u^\mu u^\nu + R) \checkmark$$

Then from the rest of your arguments, we arrive at Feynman's conclusion: the energy density is proportional to the spatial curvature in all locally inertial frames.