EM, dust and perfect-fluid energy inequalities

In a positively oriented orthonormal frame with future unit time vector $n=e_0$, the convention in [F1] gives $F_{0i}=-E_i\mathrm{/}c$, $F_{ij}={ϵ}_{ijk}{B}_k$, and $F_{ab}{F}^{ab}=2(\mathrm{\mid }B{\mathrm{\mid }}^2-\mathrm{\mid }E{\mathrm{\mid }}^2\mathrm{/}{c}^2)$. Substitution in the tensor formula yields $T_{\mathrm{E}\mathrm{M}}^{00}=u$ and $T_{\mathrm{E}\mathrm{M}}^{0i}={ϵ}_{ijk}{E}_j{B}_k\mathrm{/}({\mu }_{0}c)$. Symmetry follows because $g_{cd}{F}^{ac}{F}^{bd}$ is symmetric in $a,b$; its trace is ${\mu }_0^{-1}(F_{cd}{F}^{cd}-\frac{1}{4}\cdot 4F_{cd}{F}^{cd})=0$.

In a fluid rest frame $T^{ab}=\mathrm{diag}(ϵ,p,p,p)$. For a future timelike vector $v=(v^0,\mathbf{v})$, $v^0>\mathrm{\mid }\mathbf{v}\mathrm{\mid }$, its quadratic form is $T(v,v)=ϵ((v^0{)}^2-\mathrm{\mid }\mathbf{v}{\mathrm{\mid }}^2)+(ϵ+p)\mathrm{\mid }\mathbf{v}{\mathrm{\mid }}^2$. Thus $ϵ\ge 0$ and $ϵ+p\ge 0$ suffice for WEC; past timelike vectors give the same value as their negatives. Necessity follows by taking $v=(1,0,0,0)$, which forces $ϵ\ge 0$, and then $v=(1,r,0,0)$ with $r\uparrow 1$: if $ϵ+p<0$, the displayed value tends to that negative number, so is negative for some $r<1$. For a nonzero future null $\mathrm{\ell }$, $T(\mathrm{\ell },\mathrm{\ell })=(ϵ+p)\mathrm{\mid }\mathbf{\ell }{\mathrm{\mid }}^2$, so NEC is equivalent to $ϵ+p\ge 0$.

The mixed tensor of the perfect fluid is $T^a{}_b=\mathrm{diag}(-ϵ,p,p,p)$ in the same frame. Hence $-T^a{}_b{v}^b=(ϵv^0,-p\mathbf{v})$. If $ϵ\ge \mathrm{\mid }p\mathrm{\mid }$, then its time component is nonnegative and its squared norm is $-{ϵ}^2(v^0{)}^2+p^2\mathrm{\mid }\mathbf{v}{\mathrm{\mid }}^2\le -{ϵ}^2((v^0{)}^2-\mathrm{\mid }\mathbf{v}{\mathrm{\mid }}^2)\le 0$; it is therefore future causal or zero for every $v^0>\mathrm{\mid }\mathbf{v}\mathrm{\mid }$. Conversely the rest vector forces $ϵ\ge 0$, while requiring the vector to be causal for $(1,r,0,0)$ gives ${ϵ}^2-p^2{r}^2\ge 0$ for every $0<r<1$. Letting $r\uparrow 1$ yields $ϵ\ge \mathrm{\mid }p\mathrm{\mid }$. This proves DEC iff $ϵ\ge \mathrm{\mid }p\mathrm{\mid }$.

Put $a=E\mathrm{/}c$ and $b=B$. For any future unit observer used as the time vector of such a frame, the observer energy current $-T^a{}_b{n}^b$ has components $(u,(a\times b)\mathrm{/}{\mu }_0)$. The component identity $\mathrm{\mid }a\times b{\mathrm{\mid }}^2=\mathrm{\mid }a{\mathrm{\mid }}^2\mathrm{\mid }b{\mathrm{\mid }}^2-(a\cdot b{)}^2$ and $(\mathrm{\mid }a\mathrm{\mid }-\mathrm{\mid }b\mathrm{\mid }{)}^2\ge 0$ give $\mathrm{\mid }a\times b\mathrm{\mid }\mathrm{/}{\mu }_0\le \mathrm{\mid }a\mathrm{\mid }\mathrm{\mid }b\mathrm{\mid }\mathrm{/}{\mu }_0\le (\mathrm{\mid }a{\mathrm{\mid }}^2+\mathrm{\mid }b{\mathrm{\mid }}^2)\mathrm{/}(2{\mu }_0)=u$. Its time component is nonnegative and its squared norm is $-u^2+\mathrm{\mid }a\times b{\mathrm{\mid }}^2\mathrm{/}{\mu }_0^2\le 0$, so it is future causal or zero for every future unit observer. Every future timelike $v$ is a positive multiple of a future unit vector; the current scales by that same positive factor, proving DEC for all such $v$.

The trace is $-ϵ+3p$. The trace-reversed tensor $T_{ab}-\frac{1}{2}(g^{cd}{T}_{cd})g_{ab}$ has rest energy density $(ϵ+3p)\mathrm{/}2$ and pressure $(ϵ-p)\mathrm{/}2$. Applying the WEC equivalence from step 1.2 to this tensor gives SEC iff $ϵ+3p\ge 0$ and $ϵ+p\ge 0$.

For a future timelike $v$, DEC gives a future-causal $J^a=-T^a{}_b{v}^b$. In the rest frame of $v$, $v=(v^0,0,0,0)$ with $v^0>0$ and $J^0\ge 0$, so $T_{ab}{v}^a{v}^b=-g(J,v)=v^0{J}^0\ge 0$; this proves WEC for future timelike vectors, and $T(-v,-v)=T(v,v)$ proves it for past timelike vectors too. For a nonzero future null $\mathrm{\ell }$, vectors $\mathrm{\ell }+\delta n$ are future timelike for every $\delta >0$ and tend to $\mathrm{\ell }$, so continuity of the quadratic form proves NEC. Since $T_{\mathrm{E}\mathrm{M}}$ has zero trace by step 1.1, its trace-reversed tensor equals itself; SEC is therefore WEC.

For dust, write $ϵ=\rho c^2$ and $p=0$; then $T_m^{ab}=ϵU^a{U}^b\mathrm{/}{c}^2$ is the perfect-fluid tensor. Steps 1.2–3.1 give all four conditions because $\rho \ge 0$.

For a finite family, the future causal cone in an orthonormal frame consists of $(t,\mathbf{x})$ with $t\ge \mathrm{\mid }\mathbf{x}\mathrm{\mid }$. If $J_i=(t_i,{\mathbf{x}}_i)$ lie in this cone, then ${\sum }_i{t}_i\ge {\sum }_i\mathrm{\mid }{\mathbf{x}}_i\mathrm{\mid }\ge \mathrm{\mid }{\sum }_i{\mathbf{x}}_i\mathrm{\mid }$, so their sum is future causal. Since the DEC current is linear in $T$, finite sums preserve DEC. The WEC and NEC inequalities are linear in $T$, and trace reversal is linear in $T$, so finite sums preserve WEC, NEC and SEC as well. Therefore the Maxwell and nonnegative-dust conclusions add.