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For signature $(+,-,-,-)$, translation invariance of the Maxwell Lagrangian gives this tensor. It is generally neither symmetric nor gauge invariant. The antisymmetric superpotential improvement $-{\partial }_{\rho }(F^{\rho \mu }{A}^{\nu })$ produces, on the source-free equations of motion, the symmetric electromagnetic stress-energy tensor ${\Theta }^{\mu \nu }=-F^{\mu \rho }{F}^{\nu }{}_{\rho }+{\eta }^{\mu \nu }{F}^2/4$, which is traceless in four spacetime dimensions.

Use the metric Levi-Civita connection and hold the covariant potential $A_a$ fixed when varying the metric. Symmetry of the connection cancels its terms in the electromagnetic field tensor:The Electromagnetic Bianchi identity follows from commuting partial derivatives, or $F=dA$ and $d^2=0$. Written covariantly it is ${\nabla }_c{F}_{ab}+{\nabla }_a{F}_{bc}+{\nabla }_b{F}_{ca}=0$, since total antisymmetrization also cancels the symmetric connection terms. Hence $\boxed{F_{[ab;c]}=0}$ without using any field equation.

For the potential variation, $\delta F_{ab}={\nabla }_a\delta A_b-{\nabla }_b\delta A_a$. Antisymmetry in the Maxwell action giveswhere the second equality is integration by parts followed by relabeling; compactly supported variations, or vanishing boundary variations, remove the surface term. Arbitrary $\delta A_a$ therefore give the source-free Maxwell equationsNo gauge fixing of $A$ is needed for this variational calculation.

In signature $(+---)$ define the Hilbert stress-energy tensor by ${\delta }_{g}S=\frac{1}{2}\int \sqrt{-g}{T}_{ab}\delta g^{ab}{d}^{4}x$. Since the independent covariant potential makes $F_{ab}$ metric independent,Combining these two terms yields the electromagnetic stress-energy tensorEquivalently $\delta S=-\frac{1}{2}\int \sqrt{-g}{T}^{ab}\delta g_{ab}$. The overall stress-variation sign has been matched to the signature and action. As a check, in an orthonormal frame $F_{cd}{F}^{cd}=2(B^2-E^2)$ and $T^{00}=(E^2+B^2)/(8\pi )>0$ for a nonzero field; copying a mostly-plus formula without changing conventions would fail this check.

Let $J^a={\nabla }_b{F}^{ab}$. Contract the Bianchi identity with $F^{bc}$ to obtain $2F^{bc}{\nabla }_b{F}_{ac}=\frac{1}{2}{\nabla }_a(F_{cd}{F}^{cd})$, hence $F^{bc}{\nabla }_b{F}^a{}_c=\frac{1}{4}{\nabla }^a(F_{cd}{F}^{cd})$. Differentiating the stress tensor cancels these derivative terms and gives the electromagnetic stress-energy divergence identityThus Maxwell's equation implies stress conservation for any field. Conversely, where the matrix $F^a{}_c$ is nonsingular, conservation implies $J^c=0$ by matrix inversion. When it is singular, conservation only says that $J$ lies in its kernel, so the converse cannot be inferred without the printed nonsingularity assumption.

For the first-order formulation, temporarily call the independently constructed potential curvature $f_{ab}(A)={\nabla }_a{A}_b-{\nabla }_b{A}_a$. In the first-order Maxwell action, vary the independent antisymmetric $F^{ab}$, with $F_{ab}=g_{ac}{g}_{bd}{F}^{cd}$. Its variation isso $F_{ab}=f_{ab}(A)$. Varying $A$ independently gives exactly ${\nabla }_b{F}^{ab}=0$, by the same integration-by-parts calculation as before. Eliminating the auxiliary field algebraically makes the integrand $f^{ab}{f}_{ab}-2f^{ab}{f}_{ab}=-f^{ab}{f}_{ab}$, recovering the original Maxwell action. The Bianchi identity then follows from $f=dA$. Metric variation also agrees after elimination: the chain-rule term from the eliminated field multiplies its already vanishing field equation. Thus the two actions give the same source-free electromagnetic dynamics and stress tensor. The Palatini procedure here means independent first-order field variables, not an additional variation of the spacetime connection.

For $\mathcal{L}(A_{\nu },{\partial }_{\mu }{A}_{\nu })$ define ${\Pi }^{\mu \nu }=\partial \mathcal{L}/\partial ({\partial }_{\mu }{A}_{\nu })$. A fixed-coordinate infinitesimal field variation $\delta A_{\nu }=\alpha {\Delta }_{\nu }$ satisfying $\delta \mathcal{L}=\alpha {\partial }_{\mu }{K}^{\mu }$ gives the Noether currentIndeed, the first variation issubtract the divergence of $K$ and use the Euler-Lagrange field equations. The components of the Lorentz four-vector are simply the field labels in this formula. For a transformation moving the coordinates, use its induced fixed-coordinate variation; this includes translations and Lorentz transformations. The spatial integral of $J^0$ is conserved when the spatial current has vanishing boundary flux.

With signature $(+,-,-,-)$, a gauge transformation is $A_{\mu }\mapsto A_{\mu }+{\partial }_{\mu }\chi$. The electromagnetic field tensoris unchanged, since mixed derivatives of $\chi$ commute. Thus $\mathcal{L}=-F_{\mu \nu }{F}^{\mu \nu }/4$ is gauge invariant. Write $A_{\mu }=(\Phi ,-\mathbf{A})$, so $F_{0i}=E_i$ and $F_{ij}=-{ϵ}_{ijk}{B}_k$. ThenThe overall minus sign produces a positive electric kinetic term and a positive physical Hamiltonian. Reversing it reverses the sign of the physical energy and gives the wrong positive-energy theory; the issue is the indefinite spacetime metric, not negative magnetic energy.

For ${\Delta }_{\nu }={\partial }_0{A}_{\nu }$, direct differentiation gives $\delta \mathcal{L}=\alpha {\partial }_0\mathcal{L}=\alpha {\partial }_{\mu }({\delta }_0^{\mu }\mathcal{L})$. Sincethe canonical energy-momentum tensor gives the conserved energy currentIts time component isbecause ${\partial }_0{A}_i=E_i+{\partial }_i\Phi$. Thus Noether's conserved canonical energy is $\int J_{\mathrm{can}}^0{d}^{3}x$. In pure, source-free electromagnetism the field equation gives $\nabla \cdot \mathbf{E}=0$, andFor sufficiently decaying fields, or boundary conditions removing this surface term,The canonical density is not locally gauge invariant, even though the total energy with these conditions is.

Now use the gauge-compensated time translation in electromagnetism,The extra term is a gauge transformation with the field-dependent parameter $\chi =-\alpha A_0$. It changes no field strength; therefore the combined variation still has $\delta F_{\mu \nu }=\alpha {\partial }_0{F}_{\mu \nu }$ and the same $K^{\mu }={\delta }_0^{\mu }\mathcal{L}$. The new current isIt is expressed entirely in gauge-invariant field strengths. In particularThis is the energy column of the electromagnetic stress-energy tensorin the chosen mostly-minus convention. Its conservation is the energy continuity equation.

The currents differ byon the source-free field equations. The final expression is an antisymmetric superpotential divergence, so it changes the integrated charge only by a boundary term. The compensating gauge transformation improves the local energy density without changing the physical conserved energy under the stated boundary conditions.

In an orthonormal frame, $F_{0i}=-E_i$ and $F_{ij}={ϵ}_{ijk}{B}_k$. The electromagnetic stress-energy tensor givesThe sign of $T_{0i}$ follows directly from $F_{0j}{F}_i{}^j=-E_j{ϵ}_{ijk}{B}_k$. Thusand $T_{00}\ge 0$. These formulas hold in every local Lorentz frame, so the criterion just proved establishes the dominant energy condition. Equality occurs for a nonzero null Maxwell field with $E^2=B^2$ and $\mathbf{E}\cdot \mathbf{B}=0$, whose energy current is null vector.

Antisymmetry gives $-F^{\rho \mu }{\partial }_{\rho }{A}^{\nu }=F^{\mu \rho }{\partial }_{\rho }{A}^{\nu }$. Combining this with the canonical stress-energy tensor removes the potential in favour of the electromagnetic field tensor:This is the electromagnetic stress-energy tensor in the $(+,-,-,-)$ convention. The product $F^{\mu \rho }{F}^{\nu }{}_{\rho }$ is symmetric in $\mu ,\nu$, since the contracted metric is symmetric. Its expression only in $F$ makes it gauge-invariant. These two properties hold without using the equations of motion.

For conservation, the source-free Maxwell equations are ${\partial }_{\rho }{F}^{\rho \mu }=0$. Hence the given correction equalson shell. This is an antisymmetric superpotential improvement: $B^{\rho \mu \nu }=-F^{\rho \mu }{A}^{\nu }$ is antisymmetric in $\rho ,\mu$, so ${\partial }_{\mu }{\partial }_{\rho }{B}^{\rho \mu \nu }=0$. Conservation of the canonical stress-energy tensor therefore implies ${\partial }_{\mu }{\Theta }^{\mu \nu }=0$. Equivalently, direct differentiation gives ${\partial }_{\mu }{\Theta }^{\mu \nu }=-F^{\nu }{}_{\rho }{\partial }_{\mu }{F}^{\mu \rho }$ after the homogeneous Maxwell equations cancel the field-strength derivative terms.

There are four conserved currents, one for each fixed $\nu$, with charges $P^{\nu }=\int d^{3}x{\Theta }^{0\nu }$. The Belinfante-Rosenfeld stress-energy tensor differs from the canonical tensor only by boundary contributions to these charges when the fields decay suitably.

Finally, in four spacetime dimensions,Thus all requested properties are

The trace of the electromagnetic stress-energy tensor in four spacetime dimensions isFor its covariant divergence, the source-free Maxwell equations eliminate the derivative of the first factor. Contracting the Bianchi identity ${\nabla }_{[a}{F}_{bc]}=0$ with $F^{ac}$ giveswhich cancels the derivative of the trace term. HenceFor $J^a=T^{ab}{V}_b$, stress-energy conservation and symmetry of $T^{ab}$ now imply

Use the electromagnetic field tensor convention stated above, and lower indices with the Minkowski metric $g=\mathrm{diag}(-1,1,1,1)$. Lowering a single time index changes its sign. Summing both orders of each antisymmetric index pair gives the Lorentz scalarFor $T^{00}$, $F_j{}^0=F^{j0}=-E_j/c$ and $g^{00}=-1$. Direct substitution in the printed electromagnetic stress-energy tensor yieldsFor a spatial index $i$, $g^{0i}=0$ and $F_j{}^i=F^{ji}=-{ϵ}_{ijk}{B}_k$. HenceThus the full top row isHere $u_{\mathrm{EM}}$ is the electromagnetic energy density and $\mathbf{S}$ the Poynting vector, using ${ϵ}_0{\mu }_0{c}^2=1$. By the symmetry $T^{0i}=T^{i0}$, these components describe both energy flux divided by $c$ and $c$ times the electromagnetic momentum density $\mathbf{g}=\mathbf{S}/c^2$. Keeping the factor $c$ is essential because the time coordinate is $ct$.

Let starred quantities denote the capacitor's rest inertial frame, $u_0={ϵ}_0{E}^2/2$, $\beta =v/c$, and $\gamma =(1-{\beta }^2{)}^{-1/2}$. Work in the uniform interplate region of an ideal parallel-plate capacitor, neglecting its fringing field of a capacitor. To transform from that rest inertial frame to a laboratory in which the capacitor moves along $+x$, use the positive-sign Lorentz transformationFor a purely electric field, the spatial block of the electromagnetic stress-energy tensor isThis is the negative of the Maxwell stress tensor: it has $-u_0$ along the electric field and $+u_0$ perpendicular to it. The stress contribution to energy transport in a moving capacitor is essential to the Lorentz transformation of the energy density and flux.

Plates parallel to $xy$. Choose ${\mathbf{E}}_{\ast }=E\hat{\mathbf{z}}$ and ${\mathbf{B}}_{\ast }=0$. The electromagnetic field tensor and electromagnetic stress-energy tensor areThe transformed nonzero independent electromagnetic field tensor components are $F^{03}=\gamma E/c$ and $F^{13}=\gamma \beta E/c$. Thus $\mathbf{E}=\gamma E\hat{\mathbf{z}}$ and $\mathbf{B}=-\gamma vE\hat{\mathbf{y}}/c^2$, consistent with $\mathbf{B}=\mathbf{v}\times \mathbf{E}/c^2$. Transforming the electromagnetic stress-energy tensor givesIn particular, the laboratory Poynting vector is $\mathbf{S}=2u_0{\gamma }^{2}v\hat{\mathbf{x}}$, and the electromagnetic energy density is $u=u_0{\gamma }^2(1+{\beta }^2)$. The ratio $S_x/u=2v/(1+{\beta }^2)$ is subluminal for $v<c$, but differs from the capacitor's velocity. The difference from a simple convective flux isThe extra term is $v{T}_{\ast }^{xx}$, the rest-frame stress multiplied by the velocity. This shows why treating stored field energy as an unstressed parcel moving rigidly with the plates misses part of the energy transport.

Plates parallel to $yz$. Now ${\mathbf{E}}_{\ast }=E\hat{\mathbf{x}}$. The rest tensors areThe electric field is parallel to the Lorentz boost, so $F^{01}=E/c$ remains unchanged and $\mathbf{B}=0$. The cancellation between $T_{\ast }^{00}$ and $T_{\ast }^{11}$ givesThere is no Poynting flux in the ideal interior in this orientation, although the region containing the electric field moves. The longitudinal negative stress contribution cancels the convective term: $S_x={\gamma }^{2}v(u_0+T_{\ast }^{xx})=0$. Field energy is removed at the receding boundary and supplied at the advancing boundary by work on the charged plates; it need not flow through the interior to accompany the moving region. At edges, omitted fringing fields provide additional local structure.

For a finite idealized interplate rest volume $V_0$, length contraction gives $V=V_0/\gamma$. The field-only integrated laboratory energies would consequently be $U_{xy}=\gamma (1+{\beta }^2)U_0$ and $U_{yz}=U_0/\gamma$, with $U_0=u_0{V}_0$. Their different transformation laws are not paradoxical: the field alone is not a closed system. Include the plates and the supports holding them apart, whose mechanical stresses balance the Maxwell stress tensor. The total closed-system four-momentum then has the ordinary Lorentz transformation law. This is the same distinction between a subsystem and a conserved whole that appears in b.

The Proca stress-energy tensor is the metric variation of the matter part of the Proca action. It consists of the electromagnetic stress-energy tensor plus the quadratic mass contribution $m^2(A_a{A}_b-g_{ab}{A}^2/2)$.

For a static rest-frame electromagnetic stress-energy tensor with no Poynting flux, a Lorentz transformation to a laboratory where it moves along $+x$ gives $u={\gamma }^2(u_0+{\beta }^2{T}_{\ast }^{xx})$ and $S_x={\gamma }^{2}v(u_0+T_{\ast }^{xx})$. Thus energy flux includes stress as well as rest energy density. In an ideal parallel-plate capacitor, $T_{\ast }^{xx}=+u_0$ for motion perpendicular to the electric field, but $T_{\ast }^{xx}=-u_0$ for motion parallel to it. The former has $S_x=2{\gamma }^{2}v{u}_0$; the latter has zero interior flux, with work at the moving plates transferring energy at the boundaries. The field subsystem need not have an integrated four-momentum independent of simultaneity surface; include material support stresses for a closed system.