Strict charged barotropic tensor system

Contract the matter balance in [F1] with $u^b$. Expanding the divergence gives $u^b{\mathrm{\nabla }}^a{T}_{m,ab}=-(u^a{\mathrm{\nabla }}_aϵ+(ϵ+p){\mathrm{\nabla }}_a{u}^a)$: the term $u^a{\mathrm{\nabla }}_{a}p$ from differentiating $(ϵ+p)$ cancels the contraction of ${\mathrm{\nabla }}_{b}p$, while $u_b{u}^a{\mathrm{\nabla }}_a{u}^b=\frac{1}{2}{u}^a{\mathrm{\nabla }}_a(g(u,u))=0$. The force contraction is $u^b{F}_{bc}{j}^c=cNZ{u}^b{F}_{bc}{u}^c=0$, since $F_{bc}=-F_{cb}$. Hence the displayed energy equation follows.

By [F2], $N^{\mathrm{\prime }}(ϵ)=N(ϵ)\mathrm{/}(ϵ+p(ϵ))$. The product and chain rules therefore give ${\mathrm{\nabla }}_a(N{u}^a)=N^{\mathrm{\prime }}{u}^a{\mathrm{\nabla }}_aϵ+N{\mathrm{\nabla }}_a{u}^a=\frac{N}{ϵ+p}(u^a{\mathrm{\nabla }}_aϵ+(ϵ+p){\mathrm{\nabla }}_a{u}^a)=0$ by step 1.1. Applying the product rule to $j^a=cNZ{u}^a$ gives ${\mathrm{\nabla }}_a{j}^a=cZ{\mathrm{\nabla }}_a(N{u}^a)+cN{u}^a{\mathrm{\nabla }}_{a}Z=0$, using [F1].

The matter balance in [F1] and the CP01 identity in [L1] have opposite exchange terms, so addition gives ${\mathrm{\nabla }}^a(T_{m,ab}+T_{\mathrm{E}\mathrm{M},ab})=F_{bc}{j}^c-F_{bc}{j}^c=0$. This is conditional on the exact draft supplier statement in [L1]; neither its draft status nor this conditional use establishes its acceptance.

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.

For the torus branch take the fixed flat background $\bar{g}=-d{t}^2+{\delta }_{ij}d{x}^{i}d{x}^j$ on $(-\tau ,\tau )\times {\mathbb{T}}_{2\pi }^3$; for the smooth branch shrink each initial chart to a relatively compact ball and use the same flat comparison metric there. At the initial slice set $g_{00}=-1$, $g_{0i}=0$, $g_{ij}=h_{ij}$, $p_{ij}=2K_{ij}$, $p_{00}=-2{\mathrm{tr}}_{h}K$, and $p_{0i}=h_{i\mathrm{\ell }}{h}^{jk}\mathrm{\Gamma }(h{)}^{\mathrm{\ell }}{}_{jk}$. Since $\bar{\mathrm{\Gamma }}=0$, the Christoffel formula gives ${\mathrm{\Gamma }}^0{}_{00}=-p_{00}\mathrm{/}2$ and ${\mathrm{\Gamma }}^0{}_{ij}=K_{ij}$, so [F11] gives $H^0=g^{00}{\mathrm{\Gamma }}^0{}_{00}+h^{ij}{\mathrm{\Gamma }}^0{}_{ij}=\frac{1}{2}{p}_{00}+{\mathrm{tr}}_{h}K=0$ and $H^{\mathrm{\ell }}=-h^{\mathrm{\ell }i}{p}_{0i}+h^{jk}\mathrm{\Gamma }(h{)}^{\mathrm{\ell }}{}_{jk}=0$ by direct contraction of the Christoffel formula. Set $q_{abi}={\mathrm{\partial }}_i{g}_{ab}$ from this initial metric. For torus data $h\in H^{s+1}$ implies $\mathrm{\Gamma }(h),p_{0i}\in H^s$, so the full metric jet lies in $H^s$. On a relatively compact smooth-data patch the corresponding range is compact. On ${\mathbb{T}}^3$ use its fixed global coordinate frame and apply Gram–Schmidt at each current positive spatial metric state; the resulting frame depends smoothly on that metric state on the open cone of positive spatial blocks, rather than serving as a rough coefficient fixed in $x$. On arbitrary smooth $\mathrm{\Sigma }$ use the same construction in local charts. Reconstruct $F_0$ uniquely from $(e,b)$ by [F12]. For the torus use the global fiber coordinates $y^i=u^i$ on the future unit-velocity hyperboloid; the normalization $g(u,u)=-1$ selects the positive root $u^0=\sqrt{1+h_{ij}{u}^i{u}^j}$ on the initial orthogonal slice, so these give a smooth state chart. On smooth patches use any local hyperboloid chart. Express ${ϵ}_0$ in enthalpy coordinate $r$. The CP02 Maxwell block is the current draft interface [L1]: its source-free matrices are symmetric with positive time matrix, and the algebraic current in [F1] changes only its zero-order term. The metric and fluid blocks are [F2] and [F3]. Since the stress, current and Lorentz force are algebraic, the combined principal matrices are block diagonal and symmetric. The initial time matrices are positive; compactness of the initial state range and continuity give a compact neighborhood on which their smallest eigenvalues have a common positive lower bound. The metric derivative relations propagate by the reduced equations, the velocity coordinates preserve $g(u,u)=-1$ and future orientation, and the enthalpy change is invertible by [F3]. The regularity $h\in H^{s+1}$ and $K,e,b,{ϵ}_0,u_0,Z_0\in H^s$ gives the complete first-order initial state in $H^s$.

The reduced state has smooth coefficient maps on its open admissible state set, a compact initial range in that set, and a uniformly positive time matrix by step 1.1. Under A1 these are the hypotheses of [F5], which gives a local solution in the stated Sobolev class and a $C^1$ classical representative on a neighborhood of each initial point; compactness of ${\mathbb{T}}^3$ lets us shrink to one common cylinder. Since $s\ge 5$, put $q=s-1\ge 4$; then $q-2\ge 2>3\mathrm{/}2$, so [F15] with $m=2$ gives the state a representative continuous in time with two continuous spatial derivatives. The first-order system can be solved for ${\mathrm{\partial }}_{t}W=\mathcal{G}(t,x,W,{\mathrm{\partial }}_{i}W)$ because its time matrix is invertible; the coefficient map $\mathcal{G}$ is smooth on the admissible state set. The right side is continuously differentiable in the spatial variables, so ${\mathrm{\partial }}_{t}W$ and its spatial derivatives are continuous. Differentiating this same equation in time then shows ${\mathrm{\partial }}_t^{2}W$ is continuous as well. Thus the first-order state $W$ has the second derivatives required below. The relations ${\mathrm{\partial }}_{t}g=p$ and ${\mathrm{\partial }}_t{q}_i={\mathrm{\partial }}_{i}p$ propagate from their zero initial differences, so $q_i={\mathrm{\partial }}_{i}g$ and the metric regularity is $g\in L^{\mathrm{\infty }}{H}^{s+1}\cap C^0{H}^s$, ${\mathrm{\partial }}_{t}g\in L^{\mathrm{\infty }}{H}^s\cap C^0{H}^{s-1}$. Since both $p$ and the $q_i$ are $C^2$ components of this first-order state and are actual metric derivatives, every third metric derivative is continuous: pure spatial derivatives come from $q_i$, and derivatives with a time slot from $p$. Thus $g\in C^3$, while $F,ϵ,u,Z$ are $C^2$, and the original metric, Maxwell and fluid equations hold classically.

Put $U=dF$ and $V=d({\star }_{g}F)-{\mu }_0{\iota }_j{\mathrm{vol}}_g$. The sourced Maxwell evolution equations imply ${\iota }_{n}U={\iota }_{n}V=0$. By step 2.1, $g,F,j$ have enough regularity for $U,V$ to be $C^1$ and for $dU=d^{2}F=0$ and $dV=-{\mu }_{0}d({\iota }_j{\mathrm{vol}}_g)$ to hold pointwise. The exact coordinate calculation underlying [F9] is $d({\iota }_j{\mathrm{vol}}_g)={\mathrm{\partial }}_a(w{j}^a)d{x}^0\wedge \cdots \wedge d{x}^3=({\mathrm{\nabla }}_a{j}^a){\mathrm{vol}}_g$, where $w=\sqrt{\mathrm{\mid }\det g\mathrm{\mid }}$; it uses only first derivatives and so applies to these $C^1$ fields. The product and chain calculations in [F4] likewise use only first derivatives, giving ${\mathrm{\nabla }}_a{j}^a=0$. Therefore $dU=dV=0$.

Since $U,V$ are spatial three-forms, there are unique $C^1$ scalars $q,r$ with $U=q{\iota }_n{\mathrm{vol}}_g$ and $V=r{\iota }_n{\mathrm{vol}}_g$. Applying the same coordinate current-form identity to $qn$ and $rn$ yields $n(q)+q{\mathrm{\nabla }}_a{n}^a=0$ and $n(r)+r{\mathrm{\nabla }}_a{n}^a=0$. In a positively oriented normal frame, [F12] and [F13] give $F{\mathrm{\mid }}_{\mathrm{\Sigma }}={\iota }_b{\mathrm{vol}}_h$ and $({\star }_{g}F){\mathrm{\mid }}_{\mathrm{\Sigma }}={\iota }_e{\mathrm{vol}}_h$. The Riemannian coordinate divergence formula [F14] then gives $i^{\ast }U=({\mathrm{div}}_{h}b){\mathrm{vol}}_h$. Also $i^{\ast }({\iota }_j{\mathrm{vol}}_g)=-g(j,n){\mathrm{vol}}_h$, so $i^{\ast }V=({\mathrm{div}}_{h}e+{\mu }_{0}g(j,n)){\mathrm{vol}}_h$. The initial Maxwell constraints make both pullbacks zero; since $i^{\ast }({\iota }_n{\mathrm{vol}}_g)={\mathrm{vol}}_h$, this gives $q=r=0$ on the initial slice.

Since the lapse is positive, $n^0={\alpha }^{-1}>0$. The regularity established in step 2.1 makes $n$ at least $C^1$, so $n^i\mathrm{/}{n}^0$ is locally Lipschitz in the spatial variables and ${\mathrm{\nabla }}_a{n}^a$ is continuous; these properties also hold for every smooth reduced solution. Parametrize each normal curve by coordinate time $t$, so $\gamma (t)=(t,x(t))$, $d{x}^i\mathrm{/}dt=n^i\mathrm{/}{n}^0$ and $\dot{\gamma }=n\mathrm{/}{n}^0$. Existence and uniqueness in [F10], applied forward or backward on a sufficiently small compact coordinate cylinder, show that every point of a smaller neighborhood lies on exactly one such curve from the initial slice. Put $Q(t)=q(\gamma (t))$, $R(t)=r(\gamma (t))$ and $a(t)=(({\mathrm{\nabla }}_a{n}^a)\mathrm{/}{n}^0)(\gamma (t))$. The chain rule and step 4.1 give $Q^{\mathrm{\prime }}+aQ=0$ and $R^{\mathrm{\prime }}+aR=0$. Since $a$ is continuous, $I(t)=\exp ({\int }_0^{t}a(s)ds)$ is positive and $C^1$, and $(IQ{)}^{\mathrm{\prime }}=(IR{)}^{\mathrm{\prime }}=0$. The initial constraints give $Q(0)=R(0)=0$, so $Q=R=0$ for both positive and negative $t$. Thus $q=r=0$, hence $U=V=0$ and both sourced Maxwell equations hold.

Step 5.1 establishes the sourced Maxwell equations, so their classical regularity meets exactly the hypotheses of [L2]. The matter balance and advected-charge equation give the current lemma's unconditional conclusion ${\mathrm{\nabla }}_a{j}^a=0$; adding its matter-stress divergence $F_{bc}{j}^c$ to the CP01 electromagnetic divergence $-F_{bc}{j}^c$ gives ${\mathrm{\nabla }}^a(T_m+T_{\mathrm{E}\mathrm{M}}{)}_{ab}=0$. This establishes pre-Einstein conservation at the finite regularity already proved, without using the old smooth-only stress interface.

In the torus branch, the identity in F6 follows pointwise from the reduced equation and trace algebra because $g$ is $C^2$ and $T_m+T_{\mathrm{E}\mathrm{M}}$ is $C^1$. The contracted Bianchi identity used in the proof of F6 extends distributionally at this regularity: on a relatively compact chart, smooth local mollifications of $g$ converge in $C^2$ and remain Lorentzian after shrinking; their Einstein tensors converge uniformly, so their smooth contracted Bianchi identities pass to distributions. The covariant-derivative commutator used in F6 passes to distributions by the same local smoothing, since $H\in C^1$ and the connection is $C^1$. Thus taking the distributional divergence of F6, using the zero stress divergence from step 6.1 and the commutator identity, gives ${\mathrm{\square }}_g{H}_b+{\mathrm{Ric}}_{ba}{H}^a=0$ distributionally. Step 1.1 gives $H=0$ on the initial slice, hence all tangential derivatives vanish. In an initial orthonormal frame $(n,{\mathsf{e}}_i)$ the trace identity then gives $E(n,n)=\frac{1}{2}{\mathrm{\nabla }}_n{H}_n$ and $E(n,{\mathsf{e}}_i)=\frac{1}{2}{\mathrm{\nabla }}_n{H}_i$; [F7] makes these zero, so ${\mathrm{\partial }}_{t}H=0$ initially because ${\mathrm{\partial }}_t=n$ there. The universal Bianchi and trace-adjusted covector-divergence identities underlying [F6] depend only on the third metric jet, second harmonic-vector jet and first stress jet. At a fixed point these are exactly matched by smooth Taylor polynomials, with the polynomial metric Lorentzian nearby; their unrestricted coordinate identities therefore hold for the actual $C^3$ metric and classical fields too. Substitution of the actual reduced equation and conserved stress gives the same subsidiary equation, without replacing the solution by a smooth one.

Step 2.1 gives $g\in C^3$, so the coordinate components $H_b$ are $C^2$. The classical subsidiary equation from step 7.1 therefore has the coordinate form $g^{ab}{\mathrm{\partial }}_a{\mathrm{\partial }}_{b}H+C^a{\mathrm{\partial }}_{a}H+DH=0$, with bounded continuous matrix coefficients $C^a,D$ on a smaller closed time cylinder. Put $V={\mathrm{\partial }}_{t}H$, $W_i={\mathrm{\partial }}_{i}H$ and $Y=(H,V,W_i)$. Multiply the wave equation by $-1$, append ${\mathrm{\partial }}_{t}H=V$ and ${\mathrm{\partial }}_t{W}_i-{\mathrm{\partial }}_{i}V=0$, and multiply the latter three equations by $g^{ji}$. This is a closed homogeneous first-order system $A^0{\mathrm{\partial }}_{t}Y+A^k{\mathrm{\partial }}_{k}Y+LY=0$ with $A^0=\mathrm{diag}(I,-g^{00}I,(g^{ij}I))$. Its nonzero spatial blocks are $A_{VV}^k=-2g^{0k}I$, $A_{V,W_j}^k=-g^{kj}I$ and $A_{W_i,V}^k=-g^{ik}I$; thus each $A^k$ is symmetric. All undifferentiated $H$ terms, including the Ricci term and the connection terms from expanding the covector wave operator, are included in $LY$. On a short cylinder $A^0\ge aI$ for some $a>0$, the principal coefficients are $C^1$ with bounded derivatives, and $L$ is bounded. Since $Y\in C^1$, differentiate $\mathcal{E}(t)={\int }_{{\mathbb{T}}^3}{Y}^{\mathsf{T}}{A}^{0}Ydx$ and integrate the spatial terms by parts using periodicity to obtain ${\mathcal{E}}^{\mathrm{\prime }}=\int Y^{\mathsf{T}}({\mathrm{\partial }}_t{A}^0+{\mathrm{\partial }}_k{A}^k-L-L^{\mathsf{T}})Ydx$, hence $\mathrm{\mid }{\mathcal{E}}^{\mathrm{\prime }}\mathrm{\mid }\le C\mathcal{E}$. Step 7.1 gives $H(0)=V(0)=0$ and therefore $W_i(0)=0$, so $\mathcal{E}(0)=0$. Applying an integrating factor forward in time and then to the reversed-time inequality gives $\mathcal{E}=0$ on both halves of the cylinder. Positivity yields $Y=0$, in particular $H=0$. The residual identity F6 now gives $E_{ab}=0$, establishing the full Einstein equation.