= Solution
Use <metric signature> $(+---)$ and the <curvature> sign fixed by the stated <Newtonian limit>. In flat spacetime the source-free <Maxwell equations> are $\partial_aF^{ab}=0$ and $\partial_{[a}F_{bc]}=0$. The <stress-energy tensor> in mixed form is $T^a{}_b=F^{ac}F_{cb}+\delta^a_bF_{cd}F^{cd}/4$. Its divergence is
$$
\partial_aT^a{}_b=(\partial_aF^{ac})F_{cb}+F^{ac}\partial_aF_{cb}+\frac14\partial_b(F_{cd}F^{cd}).
$$
Contracting the cyclic Maxwell equation with $F^{ac}$ gives $F^{ac}\partial_aF_{cb}=-\partial_b(F_{cd}F^{cd})/4$. The first term also vanishes, proving $\boxed{\partial_aT^{ab}=0}$. The component $b=0$ expresses local electromagnetic <energy conservation>, and the spatial components express <momentum conservation>. Integrating converts them into the usual energy and momentum flux balances.
In curved spacetime the <tensor> equations are
$$
\boxed{\nabla_aF^{ab}=0,\qquad\nabla_{[a}F_{bc]}=0.}
$$
They agree with the flat equations in coordinates of a <local inertial frame> at a point, where the connection vanishes, and covariance then extends them to arbitrary coordinates. <Antisymmetry> makes the homogeneous equation equivalent to $dF=0$. <Metric compatibility> permits the same contraction proof, now giving $\boxed{\nabla_aT^{ab}=0}$. In general,
$$
\partial_aT^{ab}=-\Gamma^a{}_{ac}T^{cb}-\Gamma^b{}_{ac}T^{ac},
$$
which need not vanish. Covariant local balance remains valid; ordinary coordinate components do not provide a universal conserved global energy-momentum vector in a curved geometry. Additional spacetime symmetries can furnish conserved currents. A locally inertial frame removes the connection at a point, not the <curvature> or the global obstruction.
The alternating array with fixed component $\epsilon_{0123}=1$ is appropriate in oriented inertial special-relativistic coordinates, whose proper <Lorentz transformations> have determinant one. It is not itself a covariant <tensor> under general coordinate changes. For an oriented <Lorentzian metric>, replace it by the metric <metric volume tensor> $\varepsilon_{abcd}=\sqrt{-\det g}\,[abcd]$. The genuine dual is
$$
\boxed{(*F)_{ab}=\tfrac12\varepsilon_{abcd}F^{cd}.}
$$
This is the <Hodge star>; the determinant factor, <orientation> and raising conventions are essential. Even in flat spacetime arbitrary curvilinear coordinates require the <metric volume tensor>, so the distinction is about coordinate covariance rather than <curvature> alone.
The isolated equation for $R_{00}$ singles out a coordinate component and cannot be a generally covariant law. Replacing it by a proportionality $R_{ab}\propto T_{ab}$ still fails: the <contracted Bianchi identity> gives $\nabla^aR_{ab}=\nabla_bR/2$, whereas stress-<energy conservation> requires $\nabla^aT_{ab}=0$. Such a law would force $R$ constant and hence impose an unphysical restriction on a varying <relativistic dust> density. Missing units or a coupling constant in the proposed proportionality would not repair this structural problem.
The <Einstein tensor> $G_{ab}=R_{ab}-Rg_{ab}/2$ is divergence-free. Therefore
$$
\boxed{G_{ab}=-8\pi G T_{ab}}
$$
is consistent with covariant stress-<energy conservation> in the source's <curvature> convention. Its trace gives $R=8\pi GT$ and hence $R_{ab}=-8\pi G(T_{ab}-Tg_{ab}/2)$. For slowly moving <relativistic dust> with $V^aV_a=1$, $T_{00}\simeq\rho$, $T\simeq\rho$, $g_{00}\simeq1$, so $R_{00}\simeq-4\pi G\rho$, exactly matching $-\Delta\phi$ and Poisson's equation.
Finally, electromagnetic stress-energy is trace-free: $T^a{}_a=F^{ac}F_{ca}+F_{cd}F^{cd}=0$ by <antisymmetry>. The traced field equation then gives $\boxed{R=0}$ for a purely electromagnetic source. This does not say $R_{ab}=0$ or that spacetime is flat; its trace-free <Ricci curvature> can still be nonzero.
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