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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.

With $D={\partial }_{\mu }{A}^{\mu }$, expand the Maxwell Lagrangian and its Feynman gauge term to findCommuting partial derivatives verifies the divergence identity. Both densities give $\square A^{\mu }=0$ under the usual variational boundary conditions. Their canonical momentum components differ: the original density gives $-F^{0\nu }-{\eta }^{0\nu }D$, whereas the subtracted density gives $-{\dot{A}}^{\nu }$. Thus a free-photon momentum expansion using the latter requires a stated boundary-term convention.

The source-free Maxwell Lagrangian gives $S=-\frac{1}{4}\int F_{\mu \nu }{F}^{\mu \nu }{d}^{4}x$ in mostly-minus Lorentzian signature. Its Euclidean action is the displayed positive quadratic functional of the electromagnetic field tensor. Variation, with boundary terms vanishing, yields ${\partial }_{\mu }{F}_{\mu \nu }=0$.

Use natural units $\hslash =c=1$ and the Minkowski metric $\eta =\mathrm{diag}(1,-1,-1,-1)$ throughout. An Abelian gauge transformation is $A_{\mu }\mapsto A_{\mu }+{\partial }_{\mu }\chi$. For a sufficiently regular $\chi$, commuting the partial derivatives gives $F_{\mu \nu }\mapsto F_{\mu \nu }$, so the Maxwell Lagrangian is invariant pointwise.

Vary the action with respect to the electromagnetic four-potential, using variations of compact support or vanishing on the boundary. The antisymmetry of the electromagnetic field tensor givesAfter integration by parts, $\delta S=\int d^{4}x({\partial }_{\mu }{F}^{\mu \nu })\delta A_{\nu }$. The principle of stationary action therefore gives the source-free Maxwell equations,In the Lorenz gauge, ${\partial }_{\mu }{A}^{\mu }=0$, this becomes $\square A_{\nu }=0$ with $\square ={\partial }_{\mu }{\partial }^{\mu }$. The PDF's name “Lorentz gauge” denotes the usual Lorenz gauge, named after Lorenz. Residual gauge transformations preserve it when $\square \chi =0$; this condition is not itself a complete removal of the gauge freedom.

Under a gauge transformation, the electromagnetic field tensor changes byThe partial derivatives commute for a smooth gauge parameter, or in the distributional formulation. Thus the Maxwell Lagrangian depends only on a gauge-invariant tensor:Strictly, the pointwise calculation requires the second derivatives to exist; smooth gauge parameters are the usual convention. Mere first differentiability in the wording can instead be understood through commuting distributional derivatives.

For the active Lorentz transformation of a vector field, coordinates are held fixed and $A^{\prime \mu }(x)={\Lambda }^{\mu }{}_{\nu }{A}^{\nu }({\Lambda }^{-1}x)$. Its field strength transforms as a two-index Lorentz tensor. The index contractions in the Maxwell Lagrangian are invariant by ${\Lambda }^T\eta \Lambda =\eta$, so the density transforms as a Lorentz scalar:For $\Lambda =I+\alpha \omega$, its inverse is $I-\alpha \omega +O({\alpha }^2)$. Expanding the coordinate argument therefore givesSet $v^{\mu }(x)=\alpha {\omega }^{\mu }{}_{\nu }{x}^{\nu }$. Part (iii) implies ${\partial }_{\mu }{v}^{\mu }=\alpha {\omega }^{\mu }{}_{\mu }=0$. The product rule then converts the variation into a total derivative:Its integral changes the action only by a boundary term, which is the condition needed for Noether's theorem.

For a translation with the sign used in the paper, the field variation at fixed coordinates is $\delta A_{\rho }=\beta {ϵ}^{\nu }{\partial }_{\nu }{A}_{\rho }$, and $\delta \mathcal{L}={\partial }_{\mu }(\beta {ϵ}^{\mu }\mathcal{L})$. Differentiating the Maxwell Lagrangian with respect to the field gradient givesThus Noether's theorem yields $j^{\mu }={ϵ}_{\nu }{T}^{\mu \nu }$, where the canonical stress-energy tensor of the Maxwell field isIt is conserved on the source-free Maxwell equations, but it is not generally symmetric. Indeed,need not vanish. For example, take $A_0=a{x}^1$, $A_2=b{x}^0$, and the other components zero, with nonzero constants $a,b$. This produces constant field strength and satisfies the source-free equations, but $T^{12}=0$ and $T^{21}=ab$.

Nor is it gauge-invariant: while $F$ and $\mathcal{L}$ are unchanged, a gauge transformation giveswhich is generally nonzero. These shortcomings motivate the improvement in part (vi); they do not contradict conservation of the canonical stress-energy tensor.