Strict charged-fluid symmetric reduction

Put $q(ϵ)=\sqrt{p^{\mathrm{\prime }}(ϵ)}\mathrm{/}(ϵ+p(ϵ))$. It is smooth and positive on $J$, and its integral over each compact interval between ${ϵ}_{\ast }$ and $ϵ$ is finite. For each $ϵ\in J$, choose a compact interval contained in $J$ whose interior contains both ${ϵ}_{\ast }$ and $ϵ$; [F2] on that interval gives $r^{\mathrm{\prime }}=q>0$. Since $r^{\mathrm{\prime }}=q$ is smooth, $r$ is smooth. For ${ϵ}_1<{ϵ}_2$, $r({ϵ}_2)-r({ϵ}_1)={\int }_{{ϵ}_1}^{{ϵ}_2}q(z)dz>0$, so $r$ is injective; [F3] gives a smooth local inverse and an open image near every point. Continuity and the intermediate value theorem make $r(J)$ an interval, so it is an open interval. The local inverses agree on overlaps by injectivity, giving a smooth global inverse $r(J)\to J$; thus $r:J\to r(J)$ is a smooth diffeomorphism. In particular ${\mathrm{\nabla }}_{a}r=c_s{\mathrm{\nabla }}_aϵ\mathrm{/}(ϵ+p)$ and this change of variables is reversible on its image.

At a point choose local coordinates with $g_{00}<0$ and positive spatial block $(g_{ij})$, with the time direction future. For $y^i=u^i$, the future solution of $g_{ab}{u}^a{u}^b=-1$ is $u^0=\frac{-g_{0i}{y}^i-\sqrt{(g_{0i}{y}^i{)}^2-g_{00}(1+g_{ij}{y}^i{y}^j)}}{g_{00}}\mathrm{.}$ The discriminant is positive, so this is smooth on its local state domain. Its three coordinate derivatives $e_A=\mathrm{\partial }u\mathrm{/}\mathrm{\partial }{y}^A$ are linearly independent and satisfy $g(u,e_A)=0$ by differentiating the normalization with the metric fixed; by [F4] they form a basis of $u^{\perp }$. In the principal part, ${\mathrm{\nabla }}_a{u}^b=e_A^b{\mathrm{\partial }}_a{y}^A$ up to terms involving first derivatives of the metric state, and ${\mathrm{\nabla }}_{a}r={\mathrm{\partial }}_{a}r$. Thus the energy row has coefficients $u^a$ on ${\mathrm{\partial }}_{a}r$ and $c_s{e}_A^a$ on ${\mathrm{\partial }}_a{y}^A$. Projecting the Euler row onto $e_B$ gives coefficients $c_s{e}_B^a$ on ${\mathrm{\partial }}_{a}r$ and $u^{a}g(e_B,e_A)$ on ${\mathrm{\partial }}_a{y}^A$. Therefore the four principal matrices are symmetric and their covector contraction on $(a,v)$ and $(a^{\mathrm{\prime }},v^{\mathrm{\prime }})$, with $v,v^{\mathrm{\prime }}\in u^{\perp }$, is exactly $\zeta (u)(a{a}^{\mathrm{\prime }}+g(v,v^{\mathrm{\prime }}))+c_s(a\zeta (v^{\mathrm{\prime }})+a^{\mathrm{\prime }}\zeta (v))$. The independent equation $u^a{\mathrm{\partial }}_{a}Z=0$ adds the positive scalar block $\zeta (u)$ and zero cross entries. Connection terms, derivatives of the metric-dependent coordinates, and the Lorentz force are lower order once the metric first derivatives are included as state variables.

Write $h=ϵ+p(ϵ)$, $D=u^a{\mathrm{\nabla }}_a$, $\vartheta ={\mathrm{\nabla }}_a{u}^a$, $a_b=D{u}_b$, and $f_b=F_{bc}{j}^c$. The product rule, metric compatibility, and $g(u,u)=-1$ give ${\mathrm{\nabla }}^a{T}_{m,ab}=u_b(Dh+h\vartheta )+h{a}_b+{\mathrm{\nabla }}_{b}p$, with $u^b{a}_b=0$. Contracting ${\mathrm{\nabla }}^a{T}_{m,ab}=f_b$ with $u^b$ therefore yields $-Dh-h\vartheta +Dp=-Dϵ-h\vartheta =0$, since $u^b{f}_b=0$ by antisymmetry of $F$. Applying $P_b{}^a={\delta }_b{}^a+u_b{u}^a$ to the same balance gives $h{a}_b+P_b{}^a{\mathrm{\nabla }}_{a}p=f_b$. Since $u(r)=c_{s}u(ϵ)\mathrm{/}h$, ${\mathrm{\nabla }}_{a}p=c_{s}h{\mathrm{\nabla }}_{a}r$, and $h>0$, these equations are equivalent to the displayed charged enthalpy equations; $u(Z)=0$ is unchanged. The force $f_b$ is algebraic in the state and contributes no principal derivatives.

Apply Gram–Schmidt to the frame in the Statement using the positive inner product $g{\mathrm{\mid }}_{u^{\perp }}$ from [F4], and at the point adjoin $u$ as the future time vector of a $g$-orthonormal frame. Write $\alpha =\zeta (u)>0$ and $\xi =\zeta {\mathrm{\mid }}_{u^{\perp }}$. Timelike character of $\zeta$ means $\mathrm{\mid }\xi \mathrm{\mid }<\alpha$, since $g^{-1}(\zeta ,\zeta )=-{\alpha }^2+\mathrm{\mid }\xi {\mathrm{\mid }}^2<0$. For $v\in u^{\perp }$, the elementary identity $\mathrm{\mid }\xi {\mathrm{\mid }}^2\mathrm{\mid }v{\mathrm{\mid }}^2-\xi (v{)}^2={\sum }_{i<j}({\xi }_i{v}_j-{\xi }_j{v}_i{)}^2\ge 0$ gives $\mathrm{\mid }\xi (v)\mathrm{\mid }\le \mathrm{\mid }\xi \mathrm{\mid }\mathrm{\mid }v\mathrm{\mid }$. Hence $Q_{\zeta }((a,v,z),(a,v,z))\ge \alpha ((\mathrm{\mid }a\mathrm{\mid }-\mathrm{\mid }v\mathrm{\mid }{)}^2+2(1-c_s)\mathrm{\mid }a\mathrm{\mid }\mathrm{\mid }v\mathrm{\mid }+z^2)$. Because $0<c_s<1$ and $\alpha >0$, this lower bound is positive for every nonzero $(a,v,z)$. This proves positive definiteness directly, without relying on the separate GR fluid positivity proof.