The metric tensor is unitless. That can be seen from the fact that $g_{\mu\nu}v^\mu v^\nu$ gives the square of the four-vector length of $v$, and thus has the unit of $v^2$.

The scalar curvature is a contraction of the Ricci tensor. A contraction doesn't change the units. Also the Ricci tensor is a contraction of the Riemann tensor.

The Riemann tensor is made of coordinate derivatives of the connection coefficients, which are made of coordinate derivatives of the metric. Since each coordinate derivative adds a unit $m^{-1}$, the Ricci tensor and curvature scalar have both unit $\mathrm{m}^{-2}$.

The cosmological constant then of course also has to have the unit $\mathrm m^{-2}$, so that the units match.

$T$, the stress-energy tensor, has the unit of energy density, or pressure (both are actually the same unit, if you look closer), that is, $\mathrm J/\mathrm m^3$ or $\mathrm N/\mathrm m^2$.

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.

## Best Answer

In practice, given a stress-energy tensor $T_{\mu\nu}$, we may attempt to find solutions to the Einstein field equations using perturbation theory. The basic idea is to expand around a known solution $g_{\mu\nu}$ by a perturbation $h_{\mu\nu}$. In the case of a flat background,

$$\delta G_{\mu\nu} = 8\pi G \delta T_{\mu\nu} = \partial_\mu \partial_\nu h - \partial_\mu \partial_\alpha h^\alpha_\nu -\partial_\nu \partial_\alpha h^\alpha_\mu + \partial_\alpha \partial^\alpha h_{\mu\nu} - \eta_{\mu\nu} \partial_\alpha \partial^\alpha h + \eta_{\mu\nu} \partial_\alpha \partial_\beta h^{\alpha \beta}.$$

In some cases, one may solve the equations exactly, or more typically, employ numerical methods. Another alternative approach when faced with a stress-energy tensor is to try to determine the symmetries the metric may have, and then plug in an

ansatzfor the metric to yield a set of differential equations which may be more tractable, analytically and numerically.Yet another alternative is to generate new solutions from old ones through the introduction of pseudopotentials, a method due to Harrison, Eastbrook and Wahlquist, called the method of prolongation structures.

There are several other methods such as those that rely on Lie point symmetries of differential equations. There is also a Backlund transformation, which relies on identifying a simpler differential equation whose solution satisfies a condition involving the solution to the harder problem.

These methods are too involved to present here and require a significant background. They are explained in

Exact Solutions to the Einstein Field Equationsby H. Stephani et al.Addressing your other question, if given a metric $g_{\mu\nu}$, of course one can compute $T_{\mu\nu}$ through the Einstein field equations, you just plug it in and tediously compute all the curvature tensors. There are strictly speaking probably some requirements on the functions in $g_{\mu\nu}$, but you can get away with most things, even distributions, such as a delta function, which may lead to a stress-energy tensor describing a brane.