Electromagnetic stress-energy tensor

For metric signature $(-,+,+,+)$,In particular, $T^{00}=\frac{1}{2}({ϵ}_0{E}^2+B^2/{\mu }_0)$.

For a static rest-frame electromagnetic stress-energy tensor with no Poynting flux, a Lorentz transformation to a laboratory where it moves along $+x$ gives $u={\gamma }^2(u_0+{\beta }^2{T}_{\ast }^{xx})$ and $S_x={\gamma }^{2}v(u_0+T_{\ast }^{xx})$. Thus energy flux includes stress as well as rest energy density. In an ideal parallel-plate capacitor, $T_{\ast }^{xx}=+u_0$ for motion perpendicular to the electric field, but $T_{\ast }^{xx}=-u_0$ for motion parallel to it. The former has $S_x=2{\gamma }^{2}v{u}_0$; the latter has zero interior flux, with work at the moving plates transferring energy at the boundaries. The field subsystem need not have an integrated four-momentum independent of simultaneity surface; include material support stresses for a closed system.

With $(+---)$ signature and $c=1$, $T^0{}_0=(E^2+B^2)/(2{\mu }_0)$ while $T^0{}_i=-(\mathbf{E}\times \mathbf{B}{)}_i/{\mu }_0$. Raising the second index gives $T^{0i}=S_i$, where $\mathbf{S}$ is the Poynting vector, and symmetry gives the physical energy-flux components $T^{i0}$. The negative mixed component is a metric-sign effect, not reversed energy transport. In SI coordinates $x^0=ct$, $T^{i0}=S_i/c$.

For signature $(+---)$ in Gaussian electromagnetic units, $T^{ab}=(4\pi {)}^{-1}[-F^a{}_c{F}^{bc}+g^{ab}{F}_{cd}{F}^{cd}/4]$. The Electromagnetic Bianchi identity gives $F^{bc}{\nabla }_b{F}^a{}_c={\nabla }^a(F_{cd}{F}^{cd})/4$, so differentiating the stress tensor proves the displayed identity. Stress conservation implies the source-free Maxwell equations if the matrix $F^a{}_c$ is invertible; otherwise it only forces the field divergence into that matrix's kernel. Conversely source-free Maxwell equations always imply stress conservation.

With signature $(+,-,-,-)$ and $\mathcal{L}=-F_{\mu \nu }{F}^{\mu \nu }/4$, combine a time translation $\alpha {\partial }_0{A}_{\nu }$ with the gauge transformation of parameter $-\alpha A_0$. Its Noether current is $J^{\mu }=-F^{\mu \nu }{F}_{0\nu }-{\delta }_0^{\mu }\mathcal{L}$, the gauge-invariant energy current. The density is $(|\mathbf{E}{|}^2+|\mathbf{B}{|}^2)/2$. On the source-free field equations it differs from the canonical translation current by ${\partial }_{\nu }(A_0{F}^{\mu \nu })$, an antisymmetric superpotential divergence, so total energies agree when the boundary term vanishes.

For every null vector $k$, put $q^{\rho }=k_{\mu }{F}^{\mu \rho }$. Antisymmetry gives $q\cdot k=0$, so $q$ has nonnegative norm and