Anisotropic magnetic stress and null energy-current vector

For the pure magnetic case, the only independent nonzero component is $F_{12}=B_0$, so $F_{ab}{F}^{ab}=2B_0^2$. Substitution in [F1] gives $u=B_0^2\mathrm{/}(2{\mu }_0)$ and $T_{\mathrm{E}\mathrm{M}}^{ab}=\mathrm{diag}(u,u,u,-u)$ in the order $(0,x,y,z)$, with all off-diagonal entries zero. The rest observer $n={\mathrm{\partial }}_0$ has current $-T^a{}_b{n}^b=(u,0,0,0)$, which is timelike.

The mixed tensor is $T^a{}_b=\mathrm{diag}(-u,u,u,-u)$, so its two distinct eigenvalues each have eigenspace dimension two because $u>0$. For a perfect fluid, [F4] gives $T^a{}_b=(ϵ+p)U^a{U}_b\mathrm{/}{c}^2+p{\delta }^a{}_b$. This map sends the timelike vector $U$ to $-ϵU$ and every vector orthogonal to $U$ to $p$ times that vector; $U^{\perp }$ is three-dimensional by [F3] applied to $U\mathrm{/}c$. If $-ϵ=p$, the map is the scalar map $pI$ and all four eigenvalues coincide; otherwise its eigenvalue multiplicities are one and three. The magnetic tensor has neither spectrum, so it cannot be a perfect-fluid tensor in any observer frame.

For the crossed-field case, $F_{01}=-E_0\mathrm{/}c$ and $F_{31}=E_0\mathrm{/}c$ are the only independent nonzero components. They give $F_{ab}{F}^{ab}=0$ and $u=E_0^2\mathrm{/}({\mu }_0{c}^2)$. Direct substitution yields $T^{00}=T^{03}=T^{30}=T^{33}=u$ and all other components zero. Therefore $-T^a{}_b{n}^b=(u,0,0,u)$ for $n={\mathrm{\partial }}_0$, whose norm is $-u^2+u^2=0$ and whose time component is positive. The DEC current is future null, so its causal bound is saturated.