Maxwell stress divergence for classical fields

In a chart the inverse matrix of $g$ is $C^1$. Its Christoffel formula has continuous coefficients symmetric in its lower indices. Substitution into ${\mathrm{\nabla }}_a{g}_{bc}$ cancels the first metric derivatives and gives zero, so raising indices commutes with the connection on $C^1$ tensors. The antisymmetrized connection terms also cancel, giving ${\mathrm{\nabla }}_a{F}_{bc}+{\mathrm{\nabla }}_b{F}_{ca}+{\mathrm{\nabla }}_c{F}_{ab}=0$ from $dF=0$. Multiplying by $F^{ac}$ and exchanging $a,c$ in the third term gives $2F^{ac}{\mathrm{\nabla }}_a{F}_{bc}=\frac{1}{2}{\mathrm{\partial }}_b(F_{cd}{F}^{cd})$. Every term is continuous and uses only the stated first derivatives.

Raise the first stress index and differentiate by the product rule: ${\mathrm{\nabla }}_a(T^{\mathrm{E}\mathrm{M}}{)}^a{}_b={\mu }_0^{-1}[({\mathrm{\nabla }}_a{F}^{ac})F_{bc}+F^{ac}{\mathrm{\nabla }}_a{F}_{bc}-\frac{1}{4}{\mathrm{\partial }}_b(F_{cd}{F}^{cd})]$. The last two terms cancel by step 1.1. Inserting the sourced equation then gives $-F_{bc}{J}^c$; setting $J=0$ gives conservation. Metric compatibility identifies this with ${\mathrm{\nabla }}^a{T}_{ab}^{\mathrm{E}\mathrm{M}}$. Thus the identity holds at $C^1$ without applying a smooth-only stress theorem.

In the orthonormal basis, $F_{0i}=-e_i$ and $F_{ij}={ϵ}_{ijk}{b}_k$, so $F_{ab}{F}^{ab}=2(\mathrm{\mid }b{\mathrm{\mid }}^2-\mathrm{\mid }e{\mathrm{\mid }}^2)$. The $00$ component of $F_{ac}{F}_b{}^c$ is $\mathrm{\mid }e{\mathrm{\mid }}^2$, and its $0i$ component is $-e_j{ϵ}_{ijk}{b}_k=-(e\times b{)}_i$. Substitution into the stress formula gives its two displayed observer projections. All calculations are at a single point, require no derivatives of the basis, and apply also when either field is zero.