 Solution

A gauge transformation isBecause partial derivatives commute, the added contribution to $F_{\mu \nu }$ isso the electromagnetic field tensor is gauge invariant.

DefineSince ${\partial }_0=c^{-1}{\partial }_t$ and $A_0=-\Phi /c$,

The identityfollows directly from $F=dA$. Its spatial and mixed components areFor the other two equations define the four-currentThen ${\partial }_{\nu }{F}^{0\nu }={\mu }_0{j}^0$ givesand ${\partial }_{\nu }{F}^{i\nu }={\mu }_0{j}^i$ givesThis is the Covariant Maxwell equation with the minus-plus-plus-plus metric.

Usingone findsThis is the electromagnetic energy density.

For a null vector, the trace term in $T^{\mu \nu }{k}_{\mu }{k}_{\nu }$ vanishes. PutAntisymmetry gives $q\cdot k=0$. A vector orthogonal to a null vector has nonnegative Minkowski norm, soExplicitly, in a frame with $k^{\mu }=\kappa (1,1,0,0)$,Hence the null energy condition for the electromagnetic field is strict whenever the contraction is nonzero: