Maxwell stress algebra and covariant conservation

By antisymmetry of $F$, $F_{ac}{F}_b{}^c=F_{bc}{F}_a{}^c$, so $Q_{ab}=Q_{ba}$. Contracting gives $g^{ab}{Q}_{ab}=F_{ac}{F}^{ac}-\frac{1}{4}(4)F_{cd}{F}^{cd}=0$. Thus $Q$ is symmetric and trace-free in four dimensions.

Since $\mathrm{\nabla }$ is torsion-free, antisymmetrizing its derivatives cancels the symmetric Christoffel terms; therefore $dF=0$ from [F2] is exactly ${\mathrm{\nabla }}_a{F}_{bc}+{\mathrm{\nabla }}_b{F}_{ca}+{\mathrm{\nabla }}_c{F}_{ab}=0$. Contract with $F^{ac}$. The first and third terms agree after swapping $a,c$ and using $F^{ca}=-F^{ac}$ and $F_{cb}=-F_{bc}$; the middle term is $-\frac{1}{2}{\mathrm{\nabla }}_b(F_{cd}{F}^{cd})$. Hence $F^{ac}{\mathrm{\nabla }}_a{F}_{bc}=\frac{1}{4}{\mathrm{\nabla }}_b(F_{cd}{F}^{cd})$.

In the stated tetrad, $F_{0i}=-e_i$ and $F_{ij}={ϵ}_{ijk}{b}_k$ by [F3]. Raising indices with $\mathrm{diag}(-1,1,1,1)$ gives $F_{0i}{F}^{0i}=-e_i^2$ and ${\sum }_{i<j}{F}_{ij}{F}^{ij}=\mathrm{\mid }b{\mathrm{\mid }}^2$, so $F_{ab}{F}^{ab}=2(\mathrm{\mid }b{\mathrm{\mid }}^2-\mathrm{\mid }e{\mathrm{\mid }}^2)$. Substitution in $Q_{00}=F_{0c}{F}_0{}^c+\frac{1}{4}{F}_{ab}{F}^{ab}$ yields $Q_{00}=\frac{1}{2}(\mathrm{\mid }e{\mathrm{\mid }}^2+\mathrm{\mid }b{\mathrm{\mid }}^2)$.

Metric compatibility and the product rule give ${\mathrm{\nabla }}^a{Q}_{ab}=({\mathrm{\nabla }}^a{F}_{ac})F_b{}^c+F_{ac}{\mathrm{\nabla }}^a{F}_b{}^c-\frac{1}{4}{\mathrm{\nabla }}_b(F_{cd}{F}^{cd})$. The first term is $-{\mu }_0{F}_{bc}{J}^c$ by the assumed sourced Maxwell equation; the last two terms cancel by 1.2 after raising indices with $g$. Dividing by ${\mu }_0>0$ proves ${\mathrm{\nabla }}^a{T}_{ab}^{\mathrm{E}\mathrm{M}}=-F_{bc}{J}^c$.

For a spatial index $i$, $g_{0i}=0$ and $F_{0c}{F}_i{}^c={\sum }_j(-e_j){ϵ}_{ijk}{b}_k=-(e\times b{)}_i$. Thus $Q(n,{\mathfrak{e}}_i)=-(e\times b{)}_i$; division by ${\mu }_0$ gives the second observer formula. Setting $J=0$ in step 2.1 gives the source-free conservation statement.