Feynman's statement of the Einstein Field Equations

Question

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 1

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.

Answer 2

Here is the easiest way to generalize that statement that I know of, which leverages the ADM formulation of relativity. I'm glossing over a lot below, and I might be off by some factors of 2 and some signs, but the argument below is generally correct.

In the neighborhood of an arbitrary point, pick an arbitrary timelike direction $t^{a}$ to be our time coordinate and a local spacelike surface $\Sigma$, which will have unit timelike normal $n^{a}$. In general, we have:

$$t^{a} = \alpha n^{a} + \beta^{a}$$

where $\alpha$ is called the lapse function, and $\beta$ is called the shift function. (and $\beta$ is tangent to $\Sigma$). Now, if we do this, it should be easy to show that $\gamma_{a}{}^{b} = \delta_{a}{}^{b} + n_{a}n^{b}$ is a projection operator onto $\Sigma$ and that $\gamma_{ab} = \gamma_{a}{}^{c}\gamma_{b}{}^{d}g_{cd}$, so we call $\gamma_{ab}$ the metric of $\Sigma$

Finally, we note that the surface has an Extrinsic curvature, which tells us how $\Sigma$ bends inside of our 4-d spaceitme (a cylinder in $\mathbb{R}^{3}$ is no intrinsic curvature, but it does have extrinsic curvature). It turns out that the extrinsic curvature is defined by $K_{ab} = -\gamma_{a}{}^{c}\nabla_{c}n_{b}$

All that is a long way to get to the fact that if you contract einstein's equation on $n^{a}n^{b}$, you get:

$$16\pi\rho = R - K^{ab}K_{ab} + K^{2}$$

, which is known as the Hamiltonian constraint. It turns out that there is always a choice of $\alpha$ and $\beta$ that can make $K_{ab}=0$ at an instant, and that since another contraction of Einstein's equation gives you

$${\dot \gamma_{ab}} = -2\alpha K_{ab} + \nabla_{(a}\beta_{b)}$$

this corresponds to choosing a locally Minkowski coordinate frame. In this frame, we then have:

$$16\pi \rho = R$$

which corresponds to Feynman's statement.

Answer 3

We know that Einstein tensor is given by

$G=R_{uv}-\frac{1}{2}g_{uv}R$

Now, assume an infinitesimally small spherical ball of uniform mass density at rest with respect to an observer at the center of the ball. We can define Minkowski's metric in this infinitesimal region of space such that the metric is given by:

$g_{uv}=g^{uv}=\begin{pmatrix}-1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{pmatrix}$

Now, since $R=g^{uv}R_{uv}$

$\Rightarrow R=g^{aa}R_{aa}$ if we use Minkowski metric.

$\Rightarrow R=-R_{tt}+R_{xx}+R_{yy}+R_{zz}$

Hence, Einstein tensor becomes

$G=R_{uv}-\frac{1}{2}g_{uv}[-R_{tt}+R_{xx}+R_{yy}+R_{zz}]$

Now, getting back to EFE

$R_{tt}+\frac{1}{2}[-R_{tt}+R_{xx}+R_{yy}+R_{zz}]=kT_{tt}$

$\Rightarrow G_{tt}=\frac{1}{2}[R_{tt}+R_{xx}+R_{yy}+R_{zz}]$

Now, since, $R_{aba}^b=g_{aa}g^{bb}R_{bab}^{a}$

So, $R_{tat}^a=-R_{ata}^t$

Hence, $G_{tt}=\frac{1}{2}[R_{xnx}^n+R_{yny}^n+R_{znz}^n]$, where $n$ can vary only over spatial values of co-ordinate axis, namely $n=x,y,z$.

This proves that $G_{tt}$ is proportional to the spatial curvature, because $[R_{xnx}^n+R_{yny}^n+R_{znz}^n]$ is proportional to the deviation of the sphere drawn in space from its euclidean counterpart.