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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.