The covariant field stress-energy tensor and its balance law

(Clause (a): symmetry, trace and regularity) Since $F^b{}_a={\eta }_{ac}{F}^{bc}$, the first term satisfies $F^{\mu \alpha }{F}^{\nu }{}_{\alpha }=F^{\mu \alpha }{\eta }_{\alpha \beta }{F}^{\nu \beta }$, and swapping the two factors and relabelling the dummy indices while using symmetry of $\eta$ gives $F^{\mu \alpha }{\eta }_{\alpha \beta }{F}^{\nu \beta }=F^{\nu \beta }{\eta }_{\beta \alpha }{F}^{\mu \alpha }$; the second term ${\eta }^{\mu \nu }$ is symmetric, so $T^{\mu \nu }=T^{\nu \mu }$. For the trace, ${\eta }_{\mu \nu }{T}^{\mu \nu }={\mu }_0^{-1}(F^{\mu \alpha }{F}_{\mu \alpha }-\frac{1}{4}{\eta }_{\mu \nu }{\eta }^{\mu \nu }{F}^{\alpha \beta }{F}_{\alpha \beta })$ and $F^{\mu \alpha }{F}_{\mu \alpha }=F^{\alpha \beta }{F}_{\alpha \beta }=F^2$, ${\eta }_{\mu \nu }{\eta }^{\mu \nu }=4$; hence ${\eta }_{\mu \nu }{T}^{\mu \nu }={\mu }_0^{-1}(F^2-F^2)=0$. Each component $T^{\mu \nu }$ is a finite sum of products of components of $F$ (a $C^1$ tensor field), so $T$ is a $C^1$ symmetric type-$(2,0)$ tensor field by [F5]; by the product rule [F6], ${\mathrm{\partial }}_{\mu }{T}^{\mu \nu }={\mu }_0^{-1}[({\mathrm{\partial }}_{\mu }{F}^{\mu \alpha })F^{\nu }{}_{\alpha }+F^{\mu \alpha }{\mathrm{\partial }}_{\mu }{F}^{\nu }{}_{\alpha }-\frac{1}{2}{F}^{\alpha \beta }{\mathrm{\partial }}^{\nu }{F}_{\alpha \beta }]$, in which the first factor is continuous by $(\mathrm{T}\mathrm{M}1)$ and the remaining derivative factors exist and are continuous by the smooth-branch hypotheses of [F1]; hence ${\mathrm{\partial }}_{\mu }{T}^{\mu \nu }$ is continuous on $U$.

(Clause (a): the transformation law) Let $x^{\mathrm{\prime }}=\mathrm{\Lambda }x+b$ be a proper orthochronous change of inertial chart. By [F4], $F^{\mathrm{\prime }\mu \alpha }{F}^{\mathrm{\prime }\nu }{}_{\alpha }={\mathrm{\Lambda }}^{\mu }{}_e{\mathrm{\Lambda }}^{\alpha }{}_f{F}^{ef}{\eta }_{\alpha \beta }{\mathrm{\Lambda }}^{\nu }{}_c{\mathrm{\Lambda }}^{\beta }{}_d{F}^{cd}={\mathrm{\Lambda }}^{\mu }{}_e{\mathrm{\Lambda }}^{\nu }{}_c{F}^{ef}{F}^c{}_f$, because ${\eta }_{\alpha \beta }{\mathrm{\Lambda }}^{\alpha }{}_f{\mathrm{\Lambda }}^{\beta }{}_d={\eta }_{fd}$ and $F^{ef}{\eta }_{fd}=F^e{}_d$; similarly $F^{\mathrm{\prime }}{}^2=F_{ab}^{\mathrm{\prime }}{F}^{\mathrm{\prime }ab}=F_{ab}{F}^{ab}=F^2$ by the two component laws and the inverse-matrix identity, and ${\eta }^{\mathrm{\prime }}=\eta$. Substituting in the definition gives $T^{\mathrm{\prime }\mu \nu }={\mu }_0^{-1}({\mathrm{\Lambda }}^{\mu }{}_e{\mathrm{\Lambda }}^{\nu }{}_c{F}^{ef}{F}^c{}_f-\frac{1}{4}{\eta }^{\mu \nu }{F}^2)={\mathrm{\Lambda }}^{\mu }{}_e{\mathrm{\Lambda }}^{\nu }{}_c{T}^{ec}$, the stated type-$(2,0)$ transformation law.

(Clause (b): the components) Substituting [F2] into the definition: $T^{00}={\mu }_0^{-1}(F^{0i}{F}^0{}_i-\frac{1}{4}{\eta }^{00}{F}^2)={\mu }_0^{-1}(\mathrm{\mid }E{\mathrm{\mid }}^2\mathrm{/}{c}^2+\frac{1}{4}\cdot 2(\mathrm{\mid }B{\mathrm{\mid }}^2-\mathrm{\mid }E{\mathrm{\mid }}^2\mathrm{/}{c}^2))={\mu }_0^{-1}(\frac{1}{2}\mathrm{\mid }E{\mathrm{\mid }}^2\mathrm{/}{c}^2+\frac{1}{2}\mathrm{\mid }B{\mathrm{\mid }}^2)={\epsilon }_0\mathrm{\mid }E{\mathrm{\mid }}^2\mathrm{/}2+\mathrm{\mid }B{\mathrm{\mid }}^2\mathrm{/}(2{\mu }_0)=u$, using ${\eta }^{00}=-1$, ${\epsilon }_0=({\mu }_0{c}^2{)}^{-1}$ and [F3]. Next $T^{0i}={\mu }_0^{-1}{F}^{0j}{F}^i{}_j={\mu }_0^{-1}(E_j\mathrm{/}c){ϵ}_{ijk}{B}_k=(E\times B{)}_i\mathrm{/}({\mu }_{0}c)=S_i\mathrm{/}c=c{g}_i$, because ${\eta }^{0i}=0$ and [F3]. Finally $T^{ij}={\mu }_0^{-1}(F^{i0}{F}^j{}_0+F^{ik}{F}^j{}_k-\frac{1}{4}{\delta }_{ij}{F}^2)$; here $F^{i0}{F}^j{}_0=(-E_i\mathrm{/}c)(E_j\mathrm{/}c)=-E_i{E}_j\mathrm{/}{c}^2$ and $F^{ik}{F}^j{}_k={ϵ}_{ikm}{ϵ}_{jkn}{B}_m{B}_n={\delta }_{ij}\mathrm{\mid }B{\mathrm{\mid }}^2-B_i{B}_j$ by [F2], so with $F^2=2(\mathrm{\mid }B{\mathrm{\mid }}^2-\mathrm{\mid }E{\mathrm{\mid }}^2\mathrm{/}{c}^2)$, $T^{ij}={\mu }_0^{-1}(-E_i{E}_j\mathrm{/}{c}^2+{\delta }_{ij}\mathrm{\mid }B{\mathrm{\mid }}^2-B_i{B}_j-\frac{1}{2}{\delta }_{ij}(\mathrm{\mid }B{\mathrm{\mid }}^2-\mathrm{\mid }E{\mathrm{\mid }}^2\mathrm{/}{c}^2))=-{\epsilon }_0(E_i{E}_j-\frac{1}{2}{\delta }_{ij}\mathrm{\mid }E{\mathrm{\mid }}^2)-{\mu }_0^{-1}(B_i{B}_j-\frac{1}{2}{\delta }_{ij}\mathrm{\mid }B{\mathrm{\mid }}^2)=-{\sigma }_{ij}$, again by ${\epsilon }_0=({\mu }_0{c}^2{)}^{-1}$. This is the displayed component identification, and the spatial block is minus the Maxwell stress of [F3].

(Clause (c): the divergence identity) By the product rule [F6], ${\mu }_0{\mathrm{\partial }}_{\mu }{T}^{\mu \nu }=({\mathrm{\partial }}_{\mu }{F}^{\mu \alpha })F^{\nu }{}_{\alpha }+F^{\mu \alpha }{\mathrm{\partial }}_{\mu }{F}^{\nu }{}_{\alpha }-\frac{1}{2}{F}^{\alpha \beta }{\mathrm{\partial }}^{\nu }{F}_{\alpha \beta }$, the last term from ${\mathrm{\partial }}_{\mu }({\eta }^{\mu \nu }{F}^2)={\mathrm{\partial }}^{\nu }(F^2)=2F^{\alpha \beta }{\mathrm{\partial }}^{\nu }{F}_{\alpha \beta }$ and the factor $-\frac{1}{4}$. The first term is $-{\mu }_0{j}^{\alpha }{F}^{\nu }{}_{\alpha }=-{\mu }_0{F}^{\nu \lambda }{j}_{\lambda }$ by $(\mathrm{T}\mathrm{M}1)$ of [F1] and $j_{\lambda }={\eta }_{\lambda \alpha }{j}^{\alpha }$. For the cancellation, contract $(\mathrm{T}\mathrm{M}2)$ in the form ${\mathrm{\partial }}_{\mu }{F}_{\nu \alpha }+{\mathrm{\partial }}_{\nu }{F}_{\alpha \mu }+{\mathrm{\partial }}_{\alpha }{F}_{\mu \nu }=0$ with $F^{\mu \alpha }$ and fix the free index $\nu$: the first term is $C_{\nu }:=F^{\mu \alpha }{\mathrm{\partial }}_{\mu }{F}_{\nu \alpha }$, the third is $F^{\mu \alpha }{\mathrm{\partial }}_{\alpha }{F}_{\mu \nu }=F^{\alpha \mu }{\mathrm{\partial }}_{\mu }{F}_{\alpha \nu }$ after relabelling $\alpha \leftrightarrow \mu$, which equals $C_{\nu }$ because $F^{\alpha \mu }=-F^{\mu \alpha }$ and $F_{\alpha \nu }=-F_{\nu \alpha }$, and the middle term is $-\frac{1}{2}{\mathrm{\partial }}_{\nu }(F^2)$, since $F^{\mu \alpha }{\mathrm{\partial }}_{\nu }{F}_{\alpha \mu }=\frac{1}{2}{\mathrm{\partial }}_{\nu }(F^{\mu \alpha }{F}_{\alpha \mu })=-\frac{1}{2}{\mathrm{\partial }}_{\nu }(F^2)$ (the two terms of the derivative of the product are equal after relabelling). Hence $2C_{\nu }-\frac{1}{2}{\mathrm{\partial }}_{\nu }(F^2)=0$, that is $C_{\nu }=\frac{1}{4}{\mathrm{\partial }}_{\nu }(F^2)$. Raising the free index, $F^{\mu \alpha }{\mathrm{\partial }}_{\mu }{F}^{\nu }{}_{\alpha }={\eta }^{\nu \rho }{C}_{\rho }=\frac{1}{4}{\mathrm{\partial }}^{\nu }(F^2)=\frac{1}{2}{F}^{\alpha \beta }{\mathrm{\partial }}^{\nu }{F}_{\alpha \beta }$, so the second and third terms of the product-rule expression cancel. Therefore ${\mu }_0{\mathrm{\partial }}_{\mu }{T}^{\mu \nu }=-{\mu }_0{F}^{\nu \lambda }{j}_{\lambda }$, and dividing by ${\mu }_0$ gives ${\mathrm{\partial }}_{\mu }{T}^{\mu \nu }=-f^{\nu }$ with $f^{\nu }=F^{\nu \lambda }{j}_{\lambda }$. Algebraically $f^0=F^{0i}{j}_i=(E_i\mathrm{/}c)J_i=J\cdot E\mathrm{/}c$ and $f^i=F^{i0}{j}_0+F^{ij}{j}_j=(-E_i\mathrm{/}c)(-c\rho )+{ϵ}_{ijk}{B}_k{J}_j=\rho E_i+(J\times B{)}_i$, so $f=(J\cdot E\mathrm{/}c,\rho E+J\times B)$; its reading as the electromagnetic four-force density delivered to matter uses the separately adopted clause (ii) of The Lorentz coupling: external force on a test charge and the continuum force and power densities [F7].

(Clause (c): component identities and comparison) Reading the balance of step 1.4 with $\nu =0$ and using $T^{00}=u$, $T^{i0}=S_i\mathrm{/}c$ from step 1.3 and ${\mathrm{\partial }}_0=c^{-1}{\mathrm{\partial }}_t$ gives ${\mathrm{\partial }}_{\mu }{T}^{\mu 0}=c^{-1}{\mathrm{\partial }}_{t}u+c^{-1}\mathrm{div}S=c^{-1}({\mathrm{\partial }}_{t}u+\mathrm{div}S)$, while $-f^0=-J\cdot E\mathrm{/}c$; multiplying by $c$ gives ${\mathrm{\partial }}_{t}u+\mathrm{div}S=-J\cdot E$, which is clause (a) of Local and fixed-volume electromagnetic energy balance [F7]. Reading it with $\nu =i$ and using $T^{0i}=c{g}_i$, $T^{ji}=-{\sigma }_{ij}$ gives ${\mathrm{\partial }}_{\mu }{T}^{\mu i}={\mathrm{\partial }}_t{g}_i-(\mathrm{div}\sigma {)}_i$ and $-f^i=-(\rho E_i+(J\times B{)}_i)$, that is ${\mathrm{\partial }}_{t}g-\mathrm{div}\sigma =-(\rho E+J\times B)$, which is clause (a) of Local and fixed-volume momentum balance for smooth sources [F7]; by clause (ii) of The Lorentz coupling: external force on a test charge and the continuum force and power densities those source terms are the power and force densities delivered to the matter model, so the balance reads as the field-matter balance in that interpretation.

(Clause (d): matter coupling, total conservation, finite flux and angular current) Let $T_{\mathrm{m}}$ be the adopted $C^1$ symmetric type-$(2,0)$ field of [F7] with ${\mathrm{\partial }}_{\mu }{T}_{\mathrm{m}}{}^{\mu \nu }=f^{\nu }$. Adding it to step 1.4 gives ${\mathrm{\partial }}_{\mu }(T+T_{\mathrm{m}}{)}^{\mu \nu }=0$ pointwise on $U$, and the total tensor is $C^1$ and symmetric. On every closed axis-parallel box $B={\prod }_{a=0}^3[c_a,d_a]$ with an open neighbourhood contained in $U$, the box and slice identities of [F8] apply to the total tensor because its divergence vanishes on the box: the box identity has zero left-hand side, and the slice balance expresses $Q^b(s_2)-Q^b(s_1)$ as minus the three side-flux differences, all side terms retained; write their sum as ${\mathcal{F}}^b(s_1,s_2)$, with the same signed faces as in the Statement. Thus $Q^b(s_2)-Q^b(s_1)=-{\mathcal{F}}^b(s_1,s_2)$ by clause (b) of [F8]. Equality of the two slice totals is therefore equivalent to ${\mathcal{F}}^b(s_1,s_2)=0$, and constancy throughout $[c_0,d_0]$ is equivalent to this equality for every $c_0\le s_1<s_2\le d_0$. The facewise-zero hypothesis in clause (c) of [F8] forces every side integral to vanish and hence is sufficient. It is not necessary: take the zero Maxwell fields and sources, and the constant adopted matter tensor $T_{\mathrm{m}}=\mathrm{diag}(0,P,0,0)$ with $P\mathrm{\ne }0$. It is $C^1$, symmetric and divergence-free; all $Q^b$ are zero, while for $b=1$ the integrals on the two $x^1$ side faces are both the nonzero number $P(s_2-s_1)(d_2-c_2)(d_3-c_3)$ and cancel in their signed difference. Zero divergence alone supplies the balance identity, not a guarantee that the net side flux vanishes. For the angular current $M^{\lambda \mu \nu }=x^{\mu }{T}_{\mathrm{t}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}}{}^{\lambda \nu }-x^{\nu }{T}_{\mathrm{t}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}}{}^{\lambda \mu }$, the product rule and the coordinate identities ${\mathrm{\partial }}_{\lambda }{x}^{\mu }={\delta }_{\lambda }^{\mu }$ and ${\mathrm{\partial }}_{\lambda }{x}^{\nu }={\delta }_{\lambda }^{\nu }$ [F6] give ${\mathrm{\partial }}_{\lambda }{M}^{\lambda \mu \nu }={\delta }_{\lambda }^{\mu }{T}_{\mathrm{t}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}}{}^{\lambda \nu }+x^{\mu }{\mathrm{\partial }}_{\lambda }{T}_{\mathrm{t}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}}{}^{\lambda \nu }-{\delta }_{\lambda }^{\nu }{T}_{\mathrm{t}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}}{}^{\lambda \mu }-x^{\nu }{\mathrm{\partial }}_{\lambda }{T}_{\mathrm{t}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}}{}^{\lambda \mu }=T_{\mathrm{t}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}}{}^{\mu \nu }-T_{\mathrm{t}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}}{}^{\nu \mu }=0$, the last equality by the symmetry of the total tensor established in step 1.1 and the vanishing divergence.

(Clause (e): status and scope) Clauses (a), (b), (c) and (d) are steps 1.1-1.4 and 2.1-2.2: the symmetry, trace and transformation law; the component identification with $u$, $S$, $g$ and $-\sigma$; the divergence identity with the algebraically evaluated four-force density; and the conditional total conservation, finite-box and angular conclusions with the adopted matter tensor. The argument used the smooth-branch hypotheses and the two tensor systems [F1], the dictionary and contractions [F2], the definitions of the energy flux and stress objects [F3], the component transformation law [F4], the tensor-field regularity criterion [F5], the product rule [F6], the independently proved local identities and the adopted continuum coupling [F7], and the finite-box theorem [F8]. The divergence is a product identity of $C^1$ objects: it is not asserted for the weak branch, at singular source support or for any product of distributions; symmetry, zero trace and the component display are algebraic; no matter tensor is derived from vacuum Maxwell, no radiation reaction or self-force is included, and no empirical claim is made. No choice principle is used.