 Solution

In the same Minkowski metric and electromagnetic units, $T^{\mu \nu }=-F^{\mu }{}_{\lambda }{F}^{\nu \lambda }+\frac{1}{4}{g}^{\mu \nu }{F}_{\alpha \beta }{F}^{\alpha \beta }$. Differentiating givesTo simplify the last two terms, contract ${\partial }_{\mu }{F}_{\nu \lambda }+{\partial }_{\nu }{F}_{\lambda \mu }+{\partial }_{\lambda }{F}_{\mu \nu }=0$ with $F^{\mu \lambda }$. Its first and third terms are equal, and its middle term is $-\frac{1}{2}{\partial }_{\nu }(F_{\alpha \beta }{F}^{\alpha \beta })$. Raising $\nu$ therefore gives $F^{\mu }{}_{\lambda }{\partial }_{\mu }{F}^{\nu \lambda }=\frac{1}{4}{\partial }^{\nu }(F_{\alpha \beta }{F}^{\alpha \beta })$. The cancellation leavesThis is the electromagnetic stress-energy divergence identity in rationalized units. As a sign check, $T^{00}=(E^2+B^2)/2$ is positive and $T^{0i}=(\mathbf{E}\times \mathbf{B}{)}_i$ is the Poynting vector. The integrated four-momentum $P^{\nu }=\int T^{0\nu }{d}^{3}x$ is conserved if the boundary flux vanishes. With matter sources, the electromagnetic field alone instead obeys ${\partial }_{\mu }{T}^{\mu \nu }=-F^{\nu \lambda }{J}_{\lambda }$: energy and momentum transfer to matter restore conservation of the total stress-energy tensor.