The Einstein Field Equations (ignoring cosmological constant) are:
$$ R_{\mu\nu} – \frac{1}{2}Rg_{\mu\nu} = \kappa T_{\mu\nu}$$
The trace of this equation gives $R – \frac{1}{2}4R = -R = \kappa T$.
Putting this back into the EFE, we get the "trace-reversed" form:
$$ R_{\mu\nu} = \kappa (T_{\mu\nu} – \frac{1}{2}Tg_{\mu\nu}) $$
I've looked at a number of sources for solving for the constant $\kappa$, and most use an argument that works in the trace-reversed form, but not in the "standard" form, and I can't figure out why. (An example is chapter 17 of "Gravitation" by Misner, Thorne and Wheeler.)
The argument uses these assumptions to force the EFE to match the Poisson Equation (using the $(+—)$ metric signature):
- Use "low-velocity limit" so the $00$-components of tensors dominate. In particular, $T^{00} = \rho c^2$.
- From geodesic equation, we get $\Gamma^k_{00} = \frac{1}{c^2} \partial_k \Phi$, and therefore $R_{00} = \frac{1}{c^2} \nabla^2 \Phi$
- Use weak field limit $g_{\mu\nu} = \eta_{\mu\nu} + h_{\mu\nu} $ where $h_{\mu\nu} << 1$
(ends up being $g_{00} = 1 + \frac{2\Phi}{c^2}$)
Now, supposedly in the trace-reversed form of the EFE, in the weak field limit, we can pretend the metric $g_{\mu\nu}$ is equal to the flat space metric $\eta_{\mu\nu}$, so $g_{00} = \eta_{00} = +1$ instead of $g_{00} = 1 + \frac{2\Phi}{c^2}$. From there, solving for $\kappa$ is easy:
$$T = T^\mu_\mu \approx T^0_0 \approx T^{00}g_{00} = \rho c^2 (+1) = \rho c^2$$
$$ R_{00} = \kappa (T_{00} – \frac{1}{2}T g_{00}) = \kappa (\rho c^2 – \frac{1}{2} \rho c^2 (+1)) = \kappa \frac{1}{2}\rho c^2 $$
Assuming Poisson's Equation $\nabla^2 \Phi = 4 \pi G \rho$ holds in the weak field limit, we get
$$ \frac{1}{c^2} \nabla^2 \Phi = \kappa \frac{1}{2}\rho c^2 $$
$$ \frac{1}{c^2} 4 \pi G \rho = \kappa \frac{1}{2}\rho c^2 $$
$$ \frac{8 \pi G}{c^4} = \kappa $$
Which is the expected constant of proportionality in the EFE. This $g_{00} = +1$ trick doesn't seem to work with the standard EFE form of $R_{\mu\nu} – \frac{1}{2}Rg_{\mu\nu} = \kappa T_{\mu\nu}$ (answer is off by a factor of 4), and I don't know the exact reasoning for this. Is it because $\kappa(\frac{2\Phi}{c^2})$ is of negligible size, so we ignore it in the trace-reversed form? That seems a strange assumption if we don't know the value of $\kappa$ ahead of time…
Wikipedia says this, but does not elaborate:
The trace-reversed form may be more convenient in some cases (for example, when one is interested in weak-field limit and can replace gμν in the expression on the right with the Minkowski metric without significant loss of accuracy).
Best Answer
I will use signature (+---). In the weak-field limit in which $\mathbf g \simeq \boldsymbol \eta + \mathbf h$, the components of the Ricci tensor are given by $$R_{\mu\nu} = -\frac{1}{2}\left(\square h_{\mu\nu} + \partial_\mu\partial_\nu h - 2\partial^\alpha \partial_{(\mu}h_{\nu)\alpha}\right)$$ Computing the trace to lowest order in $\mathbf h$ yields $$R = g^{\mu\nu} R_{\mu\nu} \simeq \eta^{\mu\nu}R_{\mu\nu} = -\square h + \partial^\mu\partial^\nu h_{\mu\nu}$$
Applying the condition that $\mathbf h$ be stationary $(\partial_0 \mathbf h\rightarrow 0)$, the $(00)$-component of the Einstein equation becomes $$R_{00} - \frac{1}{2}Rg_{00} = \kappa T_{00} \implies -\frac{1}{2}\square h_{00} + \frac{1}{2}\square h- \frac{1}{2} \partial^\mu\partial^\nu h_{\mu\nu} = \kappa T_{00}$$
On the other hand, if we assume the non-relativistic limit in which $T^{00}$ dominates over all of the other components of $\mathbf T$, the trace of the energy-momentum tensor is to lowest order given by $$T = g_{\mu\nu} T^{\mu\nu} \simeq \eta_{00} T^{00} = \rho c^2= T_{00}$$ and the $(00)$-component of the trace-reversed Einstein equation becomes (in 4 spacetime dimensions) $$R_{00} = \kappa(T_{00} - \frac{1}{2}T g_{00}) \rightarrow -\frac{1}{2}\square h_{00} = \kappa(T_{00} - \frac{1}{2}T_{00}) = \kappa T_{00}/2$$ $$\iff-\frac{1}{2}\square h_{00} = \frac{\kappa T_{00}}{2}$$
To see that these expressions are consistent with one another, note that the taking the trace of the full Einstein equation yields $$R - \frac{4}{2}R = \kappa T = -R = \square h - \partial^\mu\partial^\nu h_{\mu\nu}$$
The error is in assuming that $R = g^{\mu\nu} R_{\mu\nu} \approx R_{00} = -\frac{1}{2}\square h_{00}$. While $\mathbf T$ (and its trace) is dominated by its $(00)$-component, the same is not true of $\mathbf R$; indeed, $$R - R_{00} = -\frac{1}{2} \square h + \partial^\mu\partial^\nu h_{\mu\nu}$$
To see this even more explicitly, consider the case off a pressure-less fluid. In component form,
$$R_{\mu\nu} -\frac{R}{2}\pmatrix{1 & 0&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&-1} = \kappa \pmatrix{\rho c^2 & 0&0&0\\0&0&0&0\\0&0&0&0\\0&0&0&0}$$
In order for this equation to hold, each $R_{\mu\mu}$ must be equal to $-R/2$ (and hence $R=-\kappa \rho c^2$).