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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.

For $\delta \varphi =-iϵ\varphi$, the conserved current is $J^{\mu }=i({\varphi }^{\ast }{\partial }^{\mu }\varphi -\varphi {\partial }^{\mu }{\varphi }^{\ast })-2e{A}^{\mu }|\varphi {|}^2$. The conjugate field equations implywhich directly cancels the divergence of the explicit background term. Conservation requires no Lorenz gauge condition; constancy of total charge additionally requires vanishing boundary flux.

The Lorenz gauge condition isor ${\partial }_{\mu }{A}^{\mu }=0$ in covariant notation.

Start with the formally gauge-redundant path integral $Z=\mathrm{vol}(\mathcal{G}{)}^{-1}\int \mathcal{D}A{e}^{iS[A]}$. Use $A_{\mu }^{\theta }=A_{\mu }+{\partial }_{\mu }\theta$, and the generalized Lorenz gauge condition $F_{\omega }[A]={\partial }_{\mu }{A}^{\mu }-\omega$. In a perturbative neighborhood where residual zero modes are removed by boundary conditions, insert the Faddeev-Popov gauge-orbit identityThus ${\Delta }_{\mathrm{FP}}=\det \square$. Gauge invariance lets the $\theta$ integration factor out and cancel $\mathrm{vol}(\mathcal{G})$. Average over $\omega$ with the Gaussian integral weight $\exp [-i\int {\omega }^2/(2\xi )]$; the delta functional fixes $\omega =\partial \cdot A$ and givesRepresent the functional determinant by the independent Grassmann fields $\bar{c},c$. A field-independent determinant sign can be absorbed in normalization, so a convenient convention isThereforeThe Faddeev-Popov ghosts are free and decouple because $\square$ is independent of $A$. For a Non-Abelian gauge theory, an infinitesimal gauge transformation gives $\delta A_{\mu }^a=D_{\mu }^{ab}{\theta }^b$, so the Faddeev-Popov operator is ${\partial }^{\mu }{D}_{\mu }^{ab}[A]$ andIt contains a ghost-gluon vertex: these Faddeev-Popov ghosts do not decouple in covariant gauge fixing of Lorenz gauge type.

In a contractible region, the Maxwell equations $\nabla \cdot \mathbf{B}=0$ and the Poincaré lemma imply the existence of a magnetic vector potential $\mathbf{A}$ such that $\mathbf{B}=\nabla \times \mathbf{A}$. Faraday's law of induction then givesThe expression in parentheses is a gradient there, so a scalar electric potential $\varphi$ can be chosen withThese statements are local, or global when the domain has the required topology and the electromagnetic fields are regular. A nonzero flux through a topologically nontrivial closed surface can obstruct a single globally defined vector potential, even though local expressions remain valid. The electric potential and magnetic vector potential are not unique: $\mathbf{A}\mapsto \mathbf{A}+\nabla \chi$ and $\varphi \mapsto \varphi -{\partial }_t\chi$ leave both electromagnetic fields unchanged.

For the Lorenz gauge condition $\nabla \cdot \mathbf{A}=-c^{-2}{\partial }_t\varphi$, Gauss's law becomesThe Ampere-Maxwell law givesCancel the two gradient terms using the Lorenz gauge condition. HenceBoth potential equations have the same d'Alembert operator, with their respective charge density and current density sources. In a source-free region these are homogeneous wave equations with propagation speed $c$.

With charge density $\rho$, current density $\mathbf{J}$, and speed of light $c=({ϵ}_0{\mu }_0{)}^{-1/2}$, the four Maxwell equations in vacuum areThe electric potential and magnetic vector potential determine the electric field and magnetic field byThese definitions satisfy Gauss's law for magnetism and Faraday's law identically. The Lorenz gauge condition isSubstituting into Gauss's law and using the Lorenz gauge condition givesFor the Ampère-Maxwell equation, the double-curl identity givesThe two gradient terms cancel by the Lorenz gauge condition. Defining the d'Alembert operator with the sign convention $\square ={\nabla }^2-c^{-2}{\partial }_t^2$, the remaining equations areConversely, if the electric potential and magnetic vector potential satisfy these wave equations and the Lorenz gauge condition, the same substitutions in reverse give Gauss's law and the Ampère-Maxwell equation. The other two Maxwell equations follow identically from the definitions, proving equivalence within the Lorenz gauge.

The moving point charge has distributional charge density and current densityLetThe given retarded fundamental solution identity immediately implies that the retarded potentialssolve $\square \varphi =-\rho /{ϵ}_0$ and $\square \mathbf{A}=-{\mu }_0\mathbf{J}$ in the sense of distributions. Integrating the spatial Dirac delta function first givesand the analogous integral for $\mathbf{A}$ has numerator ${\mu }_{0}q{\mathbf{v}}_0(t^{\prime })/(4\pi )$.

Assume the physical trajectory has subluminal speed and the retarded time exists. At its root $t_r$, with ${\mathbf{n}}_r=\mathbf{R}(t_r)/R(t_r)$,The positive factor $\kappa =1-{\mathbf{n}}_r\cdot {\mathbf{v}}_0/c$ also proves uniqueness of retarded time for a subluminal source. The Dirac delta function substitution formula gives $\delta (g(t^{\prime }))=\delta (t^{\prime }-t_r)/\kappa$, soThese are the Lienard-Wiechert potentials. The denominator includes the time-substitution factor, rather than just the retarded distance. Away from the trajectory the electric potential and magnetic vector potential are ordinary smooth functions; on the trajectory their wave equations have the indicated distributional sources.

For completeness, the charge continuity equation ${\partial }_{t^{\prime }}\rho +{\nabla }^{\prime }\cdot \mathbf{J}=0$ also verifies the Lorenz gauge. Using $\nabla K=-{\nabla }^{\prime }K$ and ${\partial }_{t}K=-{\partial }_{t^{\prime }}K$, integration by parts in the retarded integrals giveswith the retarded boundary prescription. Thus the construction also has the required Lorenz gauge condition.

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.

Vary the Maxwell Lagrangian with respect to $A_{\nu }$. Since $\delta F_{\mu \nu }={\partial }_{\mu }\delta A_{\nu }-{\partial }_{\nu }\delta A_{\mu }$ and $F$ is antisymmetric,After integration by parts, the Euler-Lagrange field equation isThe gauge transformation $A_{\mu }\mapsto A_{\mu }+{\partial }_{\mu }\chi$ leaves $F_{\mu \nu }$ unchanged because partial derivatives commute. This gauge invariance means that potentials related in this way represent the same electromagnetic field; all four potential components are not independent physical photon degrees of freedom. In particular, the ungauged kinetic operator is singular and its temporal canonical momentum vanishes.

Choose the Lorenz gauge ${\partial }_{\mu }{A}^{\mu }=0$. The field equation then becomes $\boxed{\square A_{\mu }=0}$. Residual gauge transformations with $\square \chi =0$ preserve this condition. The correctly named Lorenz gauge condition concerns the divergence of the potential; Lorentz covariance is the separate transformation property of the equations.

For the alternative quadratic Lagrangian density, direct differentiation givesso its Euler-Lagrange field equation is also $\square A^{\nu }=0$. This density is the Feynman-gauge Maxwell kinetic density after a boundary-term subtraction. Indeed, with $D={\partial }_{\mu }{A}^{\mu }$,The divergence identity follows by expanding ${\partial }_{\mu }{K}^{\mu }$ and commuting its second derivatives. The quadratic density alone describes four wave equations; the gauge constraint must still be incorporated in the physical states. It is not an unconstrained four-positive-polarization replacement for Maxwell equations.

Using this specified boundary-term convention, the canonical momentum conjugate to $A_{\nu }$ isThe canonical commutation relations are thereforewith field-field and momentum-momentum brackets zero. Equivalently $[A_{\mu },{\dot{A}}_{\nu }]=-i{g}_{\mu \nu }{\delta }^3$. The temporal component has the opposite sign to the spatial components; replacing the metric by a positive Euclidean one would destroy these brackets.

The wave equation gives massless frequencies ${\omega }_{\mathbf{k}}=|\mathbf{k}|$. Hermiticity pairs its positive- and negative-frequency solutions, so the Heisenberg picture field has the expansionThis is a real field expansion for each component, but its oscillator metric is indefinite. The assumed brackets, with the dagger's vector index restored, areThey verify the photon oscillator completeness and canonical brackets. Inserting ${\Pi }^{\nu }=-{\dot{A}}^{\nu }$ gives at equal timesThe other brackets cancel between the two frequency contributions. Thus both the $2\omega$ normalization and the sign $-g_{\mu \nu }$ are needed.

The Gupta-Bleuler formalism imposes the Lorenz gauge condition weakly on states:Here $(+)$ is the annihilation, positive-frequency part. Setting the full divergence to zero as an operator would conflict with the four-component canonical commutation relations. The subsidiary condition instead selects a physical pre-space inside the covariant photon Fock space, followed by a quotient of its null states.

Apply it to $|k,ϵ⟩={ϵ}^{\nu }{a}_{\nu }^{†}(k)|0⟩$. Commuting $k^{\mu }{a}_{\mu }$ through the creator gives a multiple of $-k_{\nu }{ϵ}^{\nu }|0⟩$. Hence the allowed photon polarization vectors obeyTake $k=(\omega ,0,0,\omega )$. The complete solution iswith complex coefficients. The last two vectors are transverse linear photon polarizations, and $(0,1,\pm i,0)/\sqrt{2}$ gives the two circular photon polarizations. The first vector is $k/\omega$, the allowed temporal-longitudinal combination. A purely spatial longitudinal vector $(0,0,0,1)$ by itself does not satisfy the subsidiary condition.

The oscillator brackets give, apart from the positive continuum normalization factor,Momentum eigenstates are distributionally normalized; a box or a wave packet makes the norm statement literal. The allowed longitudinal combination has zero norm because $k^2=0$, and it is orthogonal to every constrained state since $k\cdot ϵ=0$. It also gives a pure-gauge field wave: the field strength of $A_{\mu }\propto k_{\mu }{e}^{-ik\cdot x}$ vanishes. Thus $ϵ$ and $ϵ+ck$ represent the same physical photon polarization. The Gupta-Bleuler null-state quotient removes this null direction and leaves two positive-norm transverse photon photon polarizations. The negative-norm temporal oscillator of the unrestricted space is not an extra physical state. This is the transverse one-photon physical quotient.

Finally verify the two-photon fermion Ward identity directly for electron-positron annihilation into two photons. Define $D(r)=/r-m$ and the reduced Dirac propagatorThe equality follows from the Clifford algebra, since $(/r-m)(/r+m)=(r^2-m^2)I$. The incoming Dirac spinors obey $D(p)u(p)=0$ and $\overline{v}(q)(/q+m)=0$. On replacing one inserted photon polarization vector by $k$, the spinor factor in the two diagrams isAt the electron end, $/k=D(p)-D(p-k)$, soFor the other diagram, four-momentum conservation gives $p-k^{\prime }=k-q$. HenceThe two contributions therefore cancel exactly:Individual diagrams need not vanish; their sum implements the Ward identity. Linearity then gives invariance under $ϵ\mapsto ϵ+ck$, as required by gauge invariance. In particular, the null longitudinal photon polarization proportional to $k$ has zero physical production amplitude. Physical scattering produces the transverse photon classes, not an independent longitudinal massless photon.

The lower derivative index and the upper electromagnetic four-potential index are fully contracted. For a constant Lorentz transformation,HenceThis is a statement about Lorentz transformations, not electromagnetic gauge transformations. The quantity is not generally gauge invariant; setting it to zero is the Lorenz gauge condition, which is Lorentz covariant because zero remains zero in every inertial frame.

For a forward wave, the Faraday's law of induction fixes the relative amplitudes and signs:Thus the two scalar amplitudes in the source cannot both be positive when the stated propagation direction and axes are used. The supplied product magnitude fixes only an overall amplitude in the chosen field units.

One convenient potential representative isIndeed $-{\partial }_t\mathbf{A}$ gives the electric field and $\nabla \times \mathbf{A}$ gives the signed magnetic field above. Moreover $\nabla \cdot \mathbf{A}+c^{-2}{\partial }_t\phi =0$, the Lorenz gauge condition. The potentials are not unique because residual gauge transformations may preserve this condition.

Since $n>1$, the electromagnetic electromagnetic field invariants satisfy $\mathbf{E}\cdot \mathbf{B}=0$ and $B^2-E^2/c^2=(n^2-1)E^2/c^2>0$ away from field nodes. A boost along positive $z$ with speed $u=c/n$ givesHere ${\omega }^{\prime }={\gamma }_u(\omega -uk)=0$ and $k^{\prime }=(\omega /c)\sqrt{n^2-1}$. The wave becomes a static spatially varying magnetic pattern in this frame. This is possible because its medium wave four-vector is spacelike, $K^2=(1-n^2){\omega }^2/c^2<0$; it is not a vacuum-photon rest frame. The medium moves in the new frame, so its original rest-frame constitutive relation must not be reused there unchanged.

No inertial frame makes the magnetic field vanish throughout a nontrivial wave: that would turn the positive invariant $B^2-E^2/c^2$ into a nonpositive one. Both fields vanish at isolated nodes already, which is distinct from removing a field everywhere.

The Lorenz gauge condition is $\nabla \cdot \mathbf{A}=-c^{-2}{\partial }_t\varphi$. Substitute the electric potential and magnetic vector potential into Gauss's law:Next the Ampère-Maxwell equation, the curl of the curl identity, and the Lorenz gauge condition giveThe gradient terms cancel. Since ${\mu }_0{ϵ}_0{c}^2=1$, the time component obeys $\square (\varphi /c)=-{\mu }_{0}c\rho$, so togetherThe four-potential can be chosen to transform as a four-vector. To make the argument rigorous, choose the retarded potential with no freely added homogeneous solution. If $\square G_{\mathrm{ret}}(x)={\delta }^{(4)}(x)$, this choice isThe retarded Green function is a Lorentz scalar under proper orthochronous Lorentz transformations, and $d^{4}y$ is invariant. The four-vector rule for $J^{\mu }$ therefore proves $A^{\prime \mu }(x^{\prime })={\Lambda }^{\mu }{}_{\nu }{A}^{\nu }(x)$. More generally, transform any homogeneous initial data by this same rule. Both the wave equation and the Lorenz gauge condition are then preserved.

Qualification. The wave equation by itself cannot prove the transformation rule for every independently chosen gauge potential: the d'Alembert operator has a nontrivial kernel of a linear map. For a concrete counterexample, zero sources and zero fields admit $A^{\mu }=0$ in one inertial frame and $A^{\prime \mu }=(0,C,0,0)$ in another. Both satisfy the wave equation and the Lorenz gauge condition, but the latter is not the Lorentz transformation of zero. It is a residual Lorenz gauge symmetry, generated by $\chi =C{x}^{\prime }$. Thus the usual assertion is a covariant choice of four-potential, with a consistent gauge transformation and homogeneous data, rather than a uniqueness result from the displayed equation.

Let $s=k^{\mu }{x}_{\mu }$. The Lorenz gauge condition on the plane electromagnetic wave givesThe vacuum wave equation gives $k_{\mu }{k}^{\mu }=0$ for a nonzero amplitude; equivalently $k^0=\omega /c$ and $|\mathbf{k}|=\omega /c$. Use the gauge transformation $A^{\prime \mu }=A^{\mu }+{\partial }^{\mu }\chi$ withBecause $\square \chi =0$, this is a residual Lorenz gauge symmetry and $k_{\mu }{ϵ}^{\prime \mu }=0$. For a propagating plane electromagnetic wave, $k^0\ne 0$, so take $\lambda =-{ϵ}^0/k^0$. ThenThe second equality follows from $-k^0{ϵ}^{\prime 0}+\mathbf{k}\cdot {\mathbf{ϵ}}^{\prime }=0$. The electric field and magnetic field are unchanged by this gauge transformation. Thus Lorenz gauge permits a representative in radiation gauge; Lorenz gauge alone does not require the spatial amplitude of every representative to be transverse.

Treat $\varphi$ and ${\varphi }^{\ast }$ as independent fields, keep $A_{\mu }(x)$ as the prescribed background, and use the source's interaction with no added $e^2{A}^2|\varphi {|}^2$ term. The momentum derivatives areVarying ${\varphi }^{\ast }$ and $\varphi$ gives respectivelyNo Lorenz gauge condition has been assumed. The background divergence term must be retained unless such a condition is separately imposed.

The canonical momenta conjugate to the two scalar fields areso $\dot{\varphi }={\pi }^{\ast }-ie{A}_0\varphi$ and ${\dot{\varphi }}^{\ast }=\pi +ie{A}_0{\varphi }^{\ast }$. Define the physical three-vector potential by $A_{\mu }=(A_0,-\mathbf{A})$. The Legendre transform in the proper canonical variables isEquivalently the first product is $\pi {\pi }^{\ast }+ie{A}_0({\varphi }^{\ast }{\pi }^{\ast }-\varphi \pi )+e^2{A}_0^2|\varphi {|}^2$. The last quadratic term arises from replacing velocities by canonical momenta even though the original Lagrangian has no quadratic interaction. It is not permission to modify the specified Lagrangian. An explicitly time-dependent background generally makes the scalar-field energy time dependent.