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Suppose $V$ is a Killing vector field, $F$ is an electromagnetic field tensor satisfying $dF=0$ and ${\mathcal{L}}_{V}F=0$, and a particle of nonzero mass $m$ obeys the Lorentz force equation ${\nabla }_{u}u=(q/m)F^a{}_b{u}^b$. Cartan's magic formula gives $d({\iota }_{V}F)=0$, so the local potential of a closed differential one-form proof gives $d\Phi ={\iota }_{V}F$ locally. The Killing equation eliminates $u^a{u}^b{\nabla }_a{V}_b$, and the force term gives $d(V\cdot u)/d\tau =(q/m)d\Phi /d\tau$. Therefore $mV\cdot u-q\Phi$ is locally conserved. The sign assumes contraction in the first argument of $F$. A global potential additionally requires the closed one-form to have zero periods; invariance alone is insufficient without the homogeneous Maxwell equations.

For identical particles of rest mass $m$ and electric charge $q$, the four-current is $j^{\mu }=(q/m)\rho u^{\mu }$. The matter stress-energy tensor obeys ${\nabla }_{\mu }{T}_{\mathrm{m}}^{\mu \nu }=F^{\nu \alpha }{j}_{\alpha }$, with the opposite force density in the electromagnetic stress-energy tensor. Antisymmetry of the electromagnetic field tensor makes its contraction with two four-velocities vanish, so mass continuity still holds. Removing that term leaves the displayed covariant Lorentz force law. The matter tensor alone is not divergence-free when an electromagnetic force acts.

For a constant electromagnetic field tensor, the relativistic Lorentz force is a constant-coefficient linear system on four-velocity, with solutionThe electric-type blocks of $F$ generate hyperbolic rotations, while magnetic-type blocks generate ordinary rotations.

A contravariant tensor transforms with one forward coordinate Jacobian for each upper index. A Lorentz four-vector is the rank-one example; the electromagnetic field tensor with two upper indices is a rank-two example. Lower-index covariant tensors transform with inverse Jacobians, and full contractions yield scalars. Index position specifies a transformation law rather than implying that the numerical components are invariant.

With convention ${\nabla }^{\mu }{F}_{\mu \nu }=-{\mu }_0{J}_{\nu }$, write the electromagnetic field tensor as $F_{\mu \nu }={\nabla }_{\mu }{A}_{\nu }-{\nabla }_{\nu }{A}_{\mu }$. Commuting the contracted covariant derivatives gives ${\nabla }^{\mu }{\nabla }_{\nu }{A}_{\mu }={\nabla }_{\nu }({\nabla }^{\mu }{A}_{\mu })+R^{\alpha }{}_{\nu }{A}_{\alpha }$. The Lorenz gauge condition removes the gradient-of-divergence term, leaving the displayed wave equation for a covector field. A locally inertial coordinate system does not eliminate the Ricci tensor term.

The dual electromagnetic field tensor is obtained by applying the Hodge star operator to the electromagnetic field tensor. For metric signature $(+---)$ and ${ϵ}^{0123}=+1$, define ${\tilde{F}}^{\mu \nu }=\frac{1}{2}{ϵ}^{\mu \nu \rho \sigma }{F}_{\rho \sigma }$. With $F^{0i}=-E_i/c$ and $F^{ij}=-{ϵ}_{ijk}{B}_k$, it has components ${\tilde{F}}^{0i}=-B_i$ and ${\tilde{F}}^{ij}={ϵ}_{ijk}{E}_k/c$. The homogeneous Maxwell equations become ${\partial }_{\mu }{\tilde{F}}^{\mu \nu }=0$.

Because the electromagnetic field tensor is the exterior derivative $F=dA$ of a potential, $dF=d^{2}A=0$. For a torsion-free connection this iswhich is the covariant form of the two homogeneous Maxwell equations.

The fermion electric dipole moment operator couples a fermion pseudotensor bilinear to the electromagnetic field tensor. Its nonrelativistic interaction is proportional to $\mathbf{S}\cdot \mathbf{E}$, so it is odd under parity and time reversal and signals CP violation.

Vary the antisymmetric auxiliary electromagnetic field tensor and the electromagnetic four-potential independently. The first variation sets $F_{ab}={\nabla }_a{A}_b-{\nabla }_b{A}_a$ algebraically, and variation of the potential gives ${\nabla }_b{F}^{ab}=0$. Substitution of the algebraic equation reduces the first-order action to the usual negative quadratic Maxwell action. Metric variation gives the same stress tensor after eliminating the auxiliary field, because the chain-rule term multiplying its variational equation then vanishes.

The Lorentz force is the force exerted by an electric field $\mathbf{E}$ and a magnetic field $\mathbf{B}$ on a body with electric charge $q$ and velocity $\mathbf{v}$. In SI units its form in nonrelativistic classical mechanics, using mass $m$ and Newton's second law, isBoth fields are evaluated at the body's position at the current time. The electric field contributes $q\mathbf{E}$ even at rest, whereas the magnetic force depends on velocity through a cross product. ^[ref]; ^[ref].

For a concrete charged particle in a uniform magnetic field, use nondimensionalization to stipulate $m=q=B=1$, magnetic field $\mathbf{B}=(0,0,1)$, zero electric field, initial position $(0,0,0)$, and initial velocity $(1,0,0)$. These are rescaled quantities, not an assertion that the mass, electric charge, and magnetic field have the same physical units. At the start, $(1,0,0)\times (0,0,1)=(0,-1,0)$, so the magnetic acceleration points toward $-y$. The resulting circular motion isAt $t=\pi /2$ the position is $(1,-1,0)$ and the velocity is $(0,-1,0)$; at $t=\pi$ the position is $(0,-2,0)$ and the velocity is $(-1,0,0)$. Thus the circle has center $(0,-1,0)$ and radius one. Viewed from $+z$ toward the $xy$ plane, the motion is clockwise. Changing only the electric charge to $q=-1$ gives position $(\sin t,1-\cos t,0)$ and reverses the bending to counterclockwise.

Adding initial velocity $v_z=0.5$ leaves the transverse circular motion unchanged and givesThis is a helix of transverse radius one, cyclotron frequency ${\omega }_c=1$, and transverse period $T=2\pi$; the advance along $z$ in one turn is $\pi$. The parallel component of velocity stays constant because the magnetic force has no parallel component: there is no magnetic acceleration along the magnetic field. If the initial velocity were entirely parallel to the magnetic field, the magnetic force would vanish and the path would be straight. ^[ref].

More generally, with zero electric field, no other forces, and a uniform constant magnetic field of magnitude $B>0$, decompose velocity into perpendicular and parallel components. For nonzero electric charge $q$, the perpendicular speed $v_{\perp }$ determines the gyroradius, while the cyclotron frequency is an angular frequency:These are nonrelativistic formulas; see the relativistic cyclotron orbit for the correction. The beauty is that increasing perpendicular speed increases the radius in exactly the proportion needed to leave the period unchanged. Conversely, known magnetic field, perpendicular speed, and measured radius determine the magnitude of the electric charge divided by mass. This makes the same geometry useful for comparing bodies with different electric charge and mass.

Two short checks explain the result. First, the cross product gives $\mathbf{v}\times \mathbf{B}=(-\sin t,-\cos t,0)$ for both stipulated positive-electric charge paths, exactly their acceleration. Second, the transverse magnetic force supplies the centripetal acceleration, so $m{v}_{\perp }^2/r=|q|v_{\perp }B$; for $v_{\perp }>0$ this gives the gyroradius, and ${\omega }_c=v_{\perp }/r$ gives the cyclotron frequency.

Magnetic forces do no work: the instantaneous power of the magnetic part of the Lorentz force vanishes because a cross product is perpendicular to its first argument. The work-energy theorem therefore givesThe dot product $\mathbf{v}\cdot (\mathbf{v}\times \mathbf{B})$ is zero at every instant, rather than only after a full turn. With zero electric field and no other forces, the kinetic energy and speed remain constant; the electric field can increase or decrease kinetic energy according to the sign of $q\mathbf{E}\cdot \mathbf{v}$. In the two stipulated examples the constant kinetic energy is respectively $K=1/2$ and $K=5/8$ in rescaled units. The extra parallel velocity raises kinetic energy without changing the gyroradius or the cyclotron frequency.

For an electric example in rescaled units, stipulate $m=q=1$, zero magnetic field, constant electric field $\mathbf{E}=(2,0,0)$, no other forces, and rest at $t=0$. The Lorentz force gives acceleration $(2,0,0)$ and velocity $(2t,0,0)$, hence kinetic energy $K=2t^2$ and power $dK/dt=4t=q\mathbf{E}\cdot \mathbf{v}$. At $t=1$, the kinetic energy has risen from zero to two: the electric field has done positive work.

The existing electromagnetic field tensor formulation expresses the same interaction covariantly. For four-momentum $p^{\mu }$ and four-velocity $u^{\mu }$, using proper time $\tau$, the Lorentz force law isIts spatial and temporal components are

The electromagnetic field tensor action differs from the component-gradient potential action by $(2{\mu }_0{)}^{-1}\int {\partial }^b{A}^a{\partial }_a{A}_b{d}^{4}x$. Integration by parts expresses this term as a boundary integral plus $(2{\mu }_0{)}^{-1}\int ({\partial }_a{A}^a{)}^2{d}^{4}x$. With the Lorenz gauge and fixed or vanishing boundary data, the two actions therefore differ only by a boundary term. Varying the component-gradient action without restriction gives the potential wave equations; the Lorenz condition must be supplied to recover all Maxwell equations.

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 ${\partial }_{\mu }{F}_{\mu \nu }=0$.

The source-free electromagnetic Lagrangian density is quadratic in the electromagnetic field tensor. Its Euler-Lagrange field equation is ${\partial }_{\mu }{F}^{\mu \nu }=0$. Antisymmetry of $F$ makes the density invariant under gauge transformations, and contraction of its indices makes it a Lorentz scalar.

Perpendicular fields with $\mathbf{E}\cdot \mathbf{B}=0$ and $E^2-c^2{B}^2=0$ form a null electromagnetic field. The constant-field Lorentz-force generator is nilpotent, so its exponential terminates as a polynomial.

Both invariants of the electromagnetic field tensor vanish for this field class.

Covered by the official notes: definitions of four-force and four-velocity and the vanishing antisymmetric contraction (PDF p. 15; handwritten p. 10).^[ref]

To make the necessity statement precise, mass preservation must hold for every allowed timelike four-velocity, with the same field tensor and nonzero charge. It is not necessary for one prescribed trajectory alone. Use metric signature $(-,+,+,+)$, so $U_a{F}^a=-c^{2}dm/d\tau$. Put $H^{ab}=({\mathcal{F}}^{ab}+{\mathcal{F}}^{ba})/2$. The mass-preservation condition requires the quadratic form $H^{ab}{U}_a{U}_b=0$ for every timelike $U$. Every timelike vector can be rescaled to a physical four-velocity; by homogeneity the quadratic form vanishes on the open timelike cone. A polynomial vanishing on an open set is identically zero, so $H^{ab}=0$. Thus the electromagnetic field tensor must be antisymmetric under this universal quantifier. For a single velocity, the scalar contraction can vanish even for a nonzero symmetric part, which is why the quantifier is essential.

The official notes state only ${\mathcal{F}}^{\prime }=\Lambda \mathcal{F}{\Lambda }^T$ (PDF p. 15, handwritten p. 10); they do not evaluate the requested coefficients. For a boost along positive $x$, use ${\Lambda }^0{}_0={\Lambda }^1{}_1=\gamma$, ${\Lambda }^0{}_1={\Lambda }^1{}_0=-\gamma v/c$, with the transverse diagonal entries equal to one. Multiplying the specified electromagnetic field tensor givesThe corresponding Lorentz transformation of electromagnetic fields isConsequentlyThese coefficients depend on the frame speed, but are constant for the fixed boost.

For a torsion-free connection, the connection terms in the antisymmetrized covariant derivative cancel: $F_{\mu \nu }={\partial }_{\mu }{A}_{\nu }-{\partial }_{\nu }{A}_{\mu }$. Thus the electromagnetic field tensor is the exterior derivative of the four-potential. It is a tensor, is invariant under $A_{\mu }\mapsto A_{\mu }+{\partial }_{\mu }\chi$, and obeys the homogeneous Maxwell equations ${\nabla }_{[\alpha }{F}_{\beta \gamma ]}=0$. The covariant divergence equation is the generally covariant form of the local source equation; use its minus sign as specified in the paper.

Expand that equation before imposing the Lorenz gauge condition:The curvature commutator on a covariant tensor givesHere the contraction $g^{\mu \rho }{R}^{\alpha }{}_{\mu \rho \nu }=-R^{\alpha }{}_{\nu }$ accounts for the second sign reversal. Hence in Lorenz gaugeThis curved-spacetime electromagnetic potential wave equation contains a Ricci tensor term even in a locally inertial coordinate system: a connection can vanish at one point while its derivatives and curvature remain nonzero.

Use $g_{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$ throughout. Vary the Maxwell action with respect to the electromagnetic four-potential. Antisymmetry of the electromagnetic field tensor gives $\delta \mathcal{L}=-F^{\mu \nu }{\partial }_{\mu }\delta A_{\nu }$. After integration by parts, the Euler-Lagrange field equation isA gauge transformation $A_{\mu }\mapsto A_{\mu }+{\partial }_{\mu }\chi$ leaves $F_{\mu \nu }$ unchanged because partial derivatives commute. It therefore leaves the Maxwell action and observable electromagnetic fields unchanged: different potentials represent the same physical field. This redundancy matters for canonical quantization: before gauge fixing, the momentum conjugate to $A_0$ vanishes and the four potential components cannot be treated as four independent physical oscillators.

Choose the Lorenz gauge $D={\partial }_{\mu }{A}^{\mu }=0$, attainable by solving $\square \chi =-D$ with suitable boundary data. The Euler-Lagrange field equation then reduces toResidual Lorenz gauge symmetry consists of transformations with $\square \chi =0$. The gauge condition and this residual redundancy ultimately leave two transverse polarizations of a free photon.

For the specified kinetic density ${\mathcal{L}}_C=-\frac{1}{2}{\partial }_{\mu }{A}_{\nu }{\partial }^{\mu }{A}^{\nu }$, one has $\partial {\mathcal{L}}_C/\partial ({\partial }_{\mu }{A}_{\nu })=-{\partial }^{\mu }{A}^{\nu }$. Its Euler-Lagrange field equation is therefore $-\square A^{\nu }=0$. This density is the Feynman-gauge Maxwell kinetic density after a boundary-term subtraction, sinceThe divergence follows by expanding both sides and commuting the second derivatives. Thus the wave equation follows from a Feynman gauge action; the additional physical-state condition will implement the Lorenz gauge. One must not impose $D=0$ inside the original action before varying it.

Take the covariant components $A_{\nu }$ as canonical coordinates. The canonical momentum computed from ${\mathcal{L}}_C$ isThe equal-time canonical Poisson brackets are $\{A_{\mu }(t,\mathbf{x}),{\Pi }^{\nu }(t,\mathbf{y})\}={\delta }_{\mu }{}^{\nu }{\delta }^3(\mathbf{x}-\mathbf{y})$, with coordinate-coordinate and momentum-momentum brackets zero. Applying canonical quantization givesIn particular $[A_{\mu },{\dot{A}}_{\nu }]=-i{g}_{\mu \nu }{\delta }^3$. The negative temporal sign is the source of the indefinite state-space norm in covariant photon quantization.

The spatial Fourier transform of the wave equation gives an oscillator of frequency $\omega =|\mathbf{k}|$ for each component. The Hermitian adjoint condition on the electromagnetic four-potential then yields the mode expansion of a free fieldThese are massless on shell modes. Consistently with the preceding equal-time canonical commutation relation,The normalization is the same as for the scalar modes, with $-g_{\mu \nu }$ replacing the species identity matrix.

Gupta-Bleuler quantization starts with the Fock vacuum annihilated by every $a_{\mu }$ and imposes only the annihilation, or positive-frequency part of a quantum field, of the Lorenz gauge condition:Imposing the entire Hermitian field divergence as a strong operator identity would be incompatible with the four-component canonical commutation relation. The weaker condition suffices: between two physical states its positive-frequency part vanishes on the ket, and its negative-frequency part vanishes on the bra. Hence $⟨{\mathrm{phys}}^{\prime }|{\partial }_{\mu }{A}^{\mu }|\mathrm{phys}⟩=0$.

For a one-photon state $|\mathbf{k},ϵ⟩={ϵ}^{\nu }{a}_{\nu }^{†}(\mathbf{k})|0⟩$, commuting the constraint through the creator givesThus the allowed photon polarization vectors satisfy $\boxed{k\cdot ϵ=0}$. Their state inner product, with the common momentum-normalization factor suppressed, is $⟨{ϵ}^{\prime }|ϵ⟩=-{ϵ}^{\prime \ast }\cdot ϵ$. Momentum eigenstates are delta-normalized; the same statements hold with finite norms for normalizable wave packets.

Rotate the spatial axes so that $k=\omega (1,0,0,1)$. The constraint is ${ϵ}^0={ϵ}^3$, so the complete allowed family isThe first vector is the allowed longitudinal gauge polarization, proportional to the null four-momentum $k$. Its state has zero norm because $k^2=0$; it is also orthogonal to every constrained state because $k\cdot ϵ=0$. A purely spatial vector $(0,0,0,1)$ alone is not an allowed physical polarization: its time component must accompany it to satisfy the constraint. The other two vectors are transverse polarizations, perpendicular to $\mathbf{k}$, with positive unit norms. For the displayed family,The timelike negative-norm excitation has been excluded by the constraint, and the remaining form is positive semidefinite. The Gupta-Bleuler null-state quotient identifies $ϵ$ and $ϵ+\lambda k$, removing the null direction. It leaves exactly two transverse physical photon polarizations with a positive inner product; their bosonic Fock space is the physical photon state space.

Finally use Feynman slash notation, $/a={\gamma }^{\mu }{a}_{\mu }$, and impose four-momentum conservation $p+q=k+k^{\prime }$. The external Dirac spinors obeySet $r=p-k$ and $r^{\prime }=p-k^{\prime }$. Replacing $ϵ$ by $k$ in the first term of the electron-positron annihilation into two photons amplitude, the rightmost contracted vertex obeysSince the Clifford algebra implies $(/r+m)(/r-m)=(r^2-m^2)I_4$, that term becomesFor the second term, four-momentum conservation gives $k=q+r^{\prime }$. The contracted vertex is now adjacent to the outgoing positron spinor on the left, soConsequentlyThe two terms cancel, proving the two-photon fermion Ward identityExchanging the two photon labels proves the same result for ${ϵ}^{\prime }=k^{\prime }$. By linearity the amplitude is invariant under $ϵ\mapsto ϵ+\lambda k$ on either external photon leg. Therefore the null longitudinal gauge polarization has zero production amplitude: physical scattering produces only the two transverse photon polarizations. This cancellation also shows why both photon orderings in the tree-level Feynman diagrams are essential; either term alone fails the Ward identity.

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.