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The electromagnetic four-potential has a vanishing temporal momentum and a Gauss law constraint in gauge theory. Gauge reduction leaves two transverse canonical pairs. Their canonical commutation relations give the photon creation and annihilation operators. Covariant Gupta-Bleuler quantization instead retains auxiliary polarization modes and selects the same physical state space by a subsidiary condition and null-state quotient.

For a massive test particle with electric charge $e$, choose the electromagnetic four-potential $A_t=-Q/r$ and proper time $\tau$. The cyclic coordinates $t,\varphi$ in the equatorial Lagrangian give the conserved quantities $E=mF\dot{t}+eQ/r$ and $l=r^2\dot{\varphi }$. Here $l$ is angular momentum per unit mass. Substituting these quantities into the four-velocity normalization $g(U,U)=-1$ givesThe energy includes the gauge-dependent electrostatic term; the displayed convention fixes that gauge by taking the potential to vanish at infinity.

The Dirac electromagnetic current is the electric charge times the Dirac current. Its coupling $-e{j}^{\mu }{A}_{\mu }$ to the electromagnetic four-potential is the matter interaction of quantum electrodynamics.

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.

For electric charge $q$ and electromagnetic four-potential $A_a$, the derivative ${\mathcal{D}}_a={\nabla }_a-iq{A}_a$ transforms covariantly under an electromagnetic gauge transformation.

On the equatorial path, the Lagrangian per unit mass isBoth $t$ and $\varphi$ are cyclic coordinates. The Euler-Lagrange equations therefore conserve their canonical momenta:ConsequentlyHere $E$ is the total conserved energy in the specified gauge of the electromagnetic four-potential. The paper's $l$ is the angular momentum per unit mass; the physical angular momentum is $ml$. The equatorial choice is consistent with the $\theta$ Euler-Lagrange equation, since its angular force term contains $\sin \theta \cos \theta =0$ at $\theta =\pi /2$ and the electromagnetic four-potential has no angular component.

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 free Maxwell Lagrangian $\mathcal{L}=-F_{\mu \nu }{F}^{\mu \nu }/4$ in mostly-minus signature. In terms of the spatial electromagnetic four-potential, define $\mathbf{E}=-\dot{\mathbf{A}}-\nabla A_0$ and $\mathbf{B}=\nabla \times \mathbf{A}$. Thus $\mathcal{L}=({\mathbf{E}}^2-{\mathbf{B}}^2)/2$. Taking the Cartesian components of $\mathbf{A}$ as coordinates gives the canonical momentaThis sign convention treats $A_i$ here as Cartesian spatial-vector labels; it is not a claim that covariant and contravariant four-vector components coincide.

The temporal momentum vanishes because no ${\dot{A}}_0$ occurs. This is the primary momentum constraint of the electromagnetic potential, so one cannot treat all four components as independent unconstrained oscillators. The Legendre transformation for the spatial components gives ${\Pi }_i{\dot{A}}_i-\mathcal{L}=({\mathbf{\Pi }}^2+{\mathbf{B}}^2)/2-\mathbf{\Pi }\cdot \nabla A_0$. After integration by parts, the Hamiltonian isSurface terms vanish for decaying fields or the corresponding regulated boundary conditions. The total constrained Hamiltonian also contains an arbitrary multiplier of ${\Pi }^0$. Preserving ${\Pi }^0=0$ under time evolution requireswhich is the Gauss law constraint in gauge theory. Both constraints are first-class constraints; they express the gauge invariance $\mathbf{A}\mapsto \mathbf{A}+\nabla \alpha$, $A_0\mapsto A_0-\dot{\alpha }$. The electric and magnetic fields are unchanged, so different gauge potentials need not describe different physical states.

One consistent canonical quantization of the electromagnetic field first reduces this redundancy. Choose Coulomb gauge $\nabla \cdot \mathbf{A}=0$. In the absence of charges, the Gauss law constraint in gauge theory gives ${\nabla }^2{A}_0=0$; decay at infinity sets $A_0=0$. Residual spatially constant gauge parameters do not change $\mathbf{A}$. The remaining variables are the transverse pairs ${\mathbf{A}}^T,{\mathbf{\Pi }}^T={\dot{\mathbf{A}}}^T$. Their gauge reduction of the Maxwell Hamiltonian isThe two transverse coordinates for each nonzero momentum are the physical degrees of freedom. Treating the constraints as operator identities while retaining unconstrained componentwise commutators would be inconsistent: differentiating $[A_i^T,{\Pi }_j^T]=i{\delta }_{ij}{\delta }^3$ would give a nonzero right side even though ${\partial }_i{A}_i^T=0$.

The reduced Dirac bracket, promoted to a canonical commutation relation, instead gives the transverse equal-time commutator:whereIndeed the spatial transverse projector of a vector field $P_{ij}^T={\delta }_{ij}-k_i{k}_j/|\mathbf{k}{|}^2$ is symmetric, has $P^T\mathbf{k}=0$ and $(P^T{)}^2=P^T$. Projecting both entries of the unconstrained canonical bracket onto the transverse subspace therefore produces $P^T{\delta }^3$, equivalently the constrained Dirac bracket. The spatial zero mode is excluded by the decay condition; in a finite-volume regulator it must be treated separately.

Choose two orthonormal transverse photon polarization vectors ${\mathbf{ϵ}}_r(\mathbf{k})$, so thatThe canonical transverse photon field has mode expansionDifferentiate in time to obtain ${\Pi }_i^T$: the annihilation coefficient acquires $-i{\omega }_{\mathbf{k}}$ and the creation coefficient $+i{\omega }_{\mathbf{k}}$. The oscillator canonical commutation relations areSubstitution into $[A_i^T,{\Pi }_j^T]$ gives $i/2$ times the sum of the two Fourier kernels with phases $\pm i\mathbf{k}\cdot (\mathbf{x}-\mathbf{y})$. The polarization completeness relation and the evenness of $P^T(\mathbf{k})$ make that sum exactly $i{\delta }_{ij}^T$. This derives, rather than presumes, the correspondence between the oscillator and field commutators.

Substitute the mode expansions into $H_T$. Spatial integration sets paired momenta equal or opposite. The $aa$ and $a^{†}{a}^{†}$ terms from the electric and magnetic energies cancel because ${\omega }_{\mathbf{k}}^2=|\mathbf{k}{|}^2$ and $\mathbf{k}\cdot {\mathbf{ϵ}}_r=0$. The remaining terms are harmonic-oscillator energies. Removing the state-independent zero-point energy by normal ordering givesThe corresponding momentum operator is $\mathbf{P}={\sum }_r\int d^{3}k\mathbf{k}{a}_r^{†}{a}_r/(2\pi {)}^3$. These expressions imply $[H_T,a_r^{†}(\mathbf{k})]={\omega }_{\mathbf{k}}{a}_r^{†}(\mathbf{k})$ and $[\mathbf{P},a_r^{†}(\mathbf{k})]=\mathbf{k}{a}_r^{†}(\mathbf{k})$.

Define the Fock vacuum by $a_r(\mathbf{k})|0⟩=0$. Applying $a_r^{†}(\mathbf{k})$ creates a photon of energy $|\mathbf{k}|$, momentum $\mathbf{k}$, and one of two transverse polarizations. Its mass is zero because $k^2=0$. The relativistically normalized state is $|\mathbf{k},r⟩=\sqrt{2{\omega }_{\mathbf{k}}}{a}_r^{†}(\mathbf{k})|0⟩$, with inner product $2{\omega }_{\mathbf{k}}(2\pi {)}^3{\delta }_{rs}{\delta }^3(\mathbf{k}-\mathbf{q})$. Smearing these momentum eigenstates gives normalizable wave packets. Repeated commuting bosonic creation operators construct the symmetric multiphoton bosonic Fock space. Circular combinations of the two transverse photon polarization vectors transform with opposite phases under rotations about $\mathbf{k}$, giving the two physical helicities $\boxed{h=\pm 1}$.

The same physical states can be obtained while retaining manifest Lorentz covariance through Gupta-Bleuler quantization. Add the Feynman gauge term $-({\partial }_{\mu }{A}^{\mu }{)}^2/2$ and quantize four auxiliary oscillator polarizations. Their algebra has an indefinite metric, including a negative-norm temporal excitation. Impose $({\partial }_{\mu }{A}^{\mu }{)}^{(+)}|\mathrm{phys}⟩=0$ and take the Gupta-Bleuler null-state quotient. Temporal and longitudinal excitations then do not survive as independent physical states. The result agrees with the positive, two-polarization space obtained by Coulomb gauge reduction; covariant gauge fixing is a different implementation of the same physical constraint, not four physical photon states.

For fields ${\Phi }_a$ with a first-derivative Lagrangian density, suppose an infinitesimal global transformation at fixed coordinates is $\delta {\Phi }_a=\alpha {\Delta }_a$ and $\delta \mathcal{L}=\alpha {\partial }_{\mu }{K}^{\mu }$, with constant infinitesimal $\alpha$. Noether theorem gives the on shell conserved currentThis follows by varying the Lagrangian, integrating the terms with ${\partial }_{\mu }{\Delta }_a$ by parts, and setting the Euler-Lagrange expressions to zero. If spatial boundary flux vanishes, $\int d^{3}x{J}^0$ is a conserved Noether charge. Spacetime symmetries are covered by including the induced field variation at fixed coordinates and the corresponding total derivative $K^{\mu }$.

An example of a theory with a Lorentz four-vector is the source-free Maxwell field with electromagnetic four-potential $A_{\mu }$. Under an Abelian gauge transformation,The change in $F_{\mu \nu }$ is ${\partial }_{\mu }{\partial }_{\nu }\chi -{\partial }_{\nu }{\partial }_{\mu }\chi =0$, so the Maxwell Lagrangian is gauge-invariant. For $A_{\mu }=(\Phi ,-\mathbf{A})$ define $E_i=F_{0i}$ and $F_{ij}=-{ϵ}_{ijk}{B}_k$. With signature $+---$,The minus sign in $-F^2/4$ gives the positive electric kinetic term and positive physical field energy; reversing it would reverse the Hamiltonian's sign for the physical modes. The normalization $1/4$ accounts for antisymmetry and yields the conventional equations. Varying $A_{\nu }$ gives ${\partial }_{\mu }{F}^{\mu \nu }=0$.

For time translation at fixed coordinates take ${\Delta }_{\nu }={\partial }_0{A}_{\nu }$. Since $\alpha$ is constant, $\delta F_{\mu \nu }=\alpha {\partial }_0{F}_{\mu \nu }$, and hence $\delta \mathcal{L}=\alpha {\partial }_0\mathcal{L}$. Thus $K^{\mu }={\delta }^{\mu }{}_0\mathcal{L}$. The momentum derivative isNoether theorem therefore gives the canonical stress-energy tensor's time-translation currentHere ${\dot{A}}_i=E_i+{\partial }_i{A}_0$ was used. The canonical momentum of $A_0$ vanishes, while that of the lower spatial component $A_i$ is $E_i$: $A_0$ acts as a multiplier for Gauss law, not a propagating degree of freedom. The density above is not manifestly gauge-invariant, but its charge agrees with the physical energy on the constraint surface and with appropriate boundary conditions.

The gauge-covariant time translation instead has ${\Delta }_{\nu }={\partial }_0{A}_{\nu }-{\partial }_{\nu }{A}_0=F_{0\nu }$. It is time translation combined with a gauge transformation of parameter $\chi =-\alpha A_0$. Directly,using the definition of $F$, equivalently its Bianchi identity. The total derivative in $\delta \mathcal{L}$ is unchanged. The improved current isIt is the gauge-invariant Maxwell stress-energy tensor. In particular,The spatial current is the Poynting vector $\mathbf{E}\times \mathbf{B}$, so conservation gives ${\partial }_{t}u+\nabla \cdot (\mathbf{E}\times \mathbf{B})=0$. This is the familiar positive local electromagnetic field energy density, expressed entirely in measurable fields.

To compare the currents,The last term vanishes on the source-free Maxwell equations, leaving a stress-energy tensor improvement generated by a stress-energy superpotential. At the charge level,when the surface term vanishes and Gauss law holds. Thus the symmetries yield the same conserved energy under these conditions while the second supplies a gauge-invariant local density. Boundary terms would have to be retained if those boundary conditions were not imposed.

Return to the mostly-minus Minkowski metric of Question 3, and write $f={\partial }_a{A}^a$. The field $\xi (x)$ is assumed real and nonzero wherever its inverse occurs. Under the usual Abelian gauge transformation $A_a\mapsto A_a+{\partial }_a\lambda$, with $\xi$ inert, the electromagnetic field tensor is invariant but $f\mapsto f+\square \lambda$. The change in the added density isThis is not generally a total derivative. The action is not invariant under arbitrary gauge transformations. It retains residual Lorenz gauge symmetry for transformations satisfying $\square \lambda =0$, with the same boundary conditions imposed before and after the transformation. The action itself is undefined at $\xi =0$; that value can only be considered as a limiting gauge.

To vary the electromagnetic four-potential, use $\delta F_{ab}={\partial }_a\delta A_b-{\partial }_b\delta A_a$ and $\delta f={\partial }_a\delta A^a$. The integration by parts of both terms giveswith surface terms removed by the variational boundary conditions. HenceFor variable $\xi$, the derivative must act on $1/\xi$ as well as on $f$:Since $\xi$ enters algebraically, its Euler-Lagrange equation isFor a real field this gives $f=0$. Thus a dynamical gauge-fixing parameter imposes a constraint equation in field theory; it is not an ordinary propagating scalar. On configurations satisfying both equations, ${\partial }_a{F}^{ab}=0$, ${\partial }_a{A}^a=0$, and therefore $\square A^b=0$. There is no independent kinetic equation that determines $\xi$.

For the momentum-space equation at a prescribed constant $\xi \ne 0$, use $A_a(x)=\int d^{4}p{e}^{-ip\cdot x}{A}_a(p)/(2\pi {)}^4$. Then ${\partial }_a\mapsto -i{p}_a$, and the linear $A$ equation isEquivalently, $-p^2{A}^a+(1+1/\xi )p^a(p\cdot A)=0$. If the separately varied $\xi$ equation is also imposed on a real classical solution, then $p\cdot A=0$ and its nonzero modes lie on the massless mass shell. For constructing a full-field Green function, however, invert the fixed-background quadratic operator before imposing that on-shell constraint. Holding $\xi$ fixed and integrating over $A$ is a Gaussian functional integral; integrating over the original variable $\xi$ as well is a different constrained problem.

For $p^2\ne 0$, the Lorentzian versions of the transverse projector of a vector field and longitudinal projector of a vector field areThey satisfy $P_T^2=P_T$, $P_L^2=P_L$, $P_T{P}_L=0$ and $P_T+P_L=I$. The mixed-index operator isIts inverse follows by inverting these two scalar eigenvalues. Define $G_{ab}$ by $K^{ac}{G}_{cb}={\delta }_b^a$. The positive-sign covariant gauge-fixing inverse isThis is the algebraic inverse of the kinetic operator. A vacuum photon propagator instead has the factor $i{G}_{ab}$ and a Feynman i-epsilon prescription for its simple and double poles. If the name $G$ is used for the vacuum two-point function, include that factor $i$ consistently in its defining source equation. A retarded Green function would use different boundary conditions at the same poles. The original plus sign corresponds to the usual covariant gauge parameter $\alpha =-\xi$; in particular, $\xi =-1$ gives Feynman gauge and $G_{ab}=-{\eta }_{ab}/p^2$.

Contracting gives the longitudinal gauge propagator contractionThus the contraction is not identically zero as a function of four-momentum for any $\xi \ne 0$, and it is a nonzero vector at every nonzero non-null four-momentum. A particular component may vanish when $p_b=0$. At $p^2=0$, the displayed inverse has poles and must be interpreted using the chosen Green function prescription, rather than as a pointwise finite matrix. In the limiting covariant Landau gauge, $\xi \to 0$, the longitudinal part vanishes. A covariant photon Green function may have a longitudinal component even though physical photon polarizations are transverse.

Specify the Minkowski metric convention ${\eta }_{\mu \nu }={\eta }^{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$, with $x^0=ct$ and ${\partial }_0=c^{-1}{\partial }_t$. For the electromagnetic four-potential, $A_{\mu }={\eta }_{\mu \nu }{A}^{\nu }=(\varphi /c,-\mathbf{A})$. The electromagnetic field tensor isUsing $\mathbf{E}=-\nabla \varphi -{\partial }_t\mathbf{A}$ and $\mathbf{B}=\nabla \times \mathbf{A}$, its two component matrices areChanging metric/sign conventions changes the component convention, so stating it is necessary. Under a Lorentz transformation, the tensor law is $F^{\prime \mu \nu }(x^{\prime })={\Lambda }^{\mu }{}_{\rho }{\Lambda }^{\nu }{}_{\sigma }{F}^{\rho \sigma }(x)$. For the stated Lorentz boost, put $\beta =v/c$ and $\gamma =(1-{\beta }^2{)}^{-1/2}$:Multiplication $F^{\prime }=\Lambda F{\Lambda }^T$ gives, for example, $F^{\prime 02}=-\gamma E_y/c+\gamma \beta B_z$ and $F^{\prime 12}=-\gamma B_z+\gamma \beta E_y/c$. Reading off all entries proves Lorentz transformation of electromagnetic fields:For the wire, the lab line charge is zero and the current is $I=nqu+(-nq)(-u)=2nqu$. At points off the wire, write $s^2=y^2+z^2>0$. Gauss's law and Ampère's law giveThe boost leaves $y,z$ unchanged, so the field transformations giveIts physical source is the relativistic charge density of counterstreaming beams. Their boosted number densities are $n_{+}^{\prime }=\gamma n(1-vu/c^2)$ and $n_{-}^{\prime }=\gamma n(1+vu/c^2)$, by transforming each charge-current four-vector. Thus ${\lambda }^{\prime }=q(n_{+}^{\prime }-n_{-}^{\prime })=-2\gamma nquv/c^2=-\gamma vI/c^2$. A line charge has radial field ${\lambda }^{\prime }(0,y,z)/(2\pi {ϵ}_0{s}^2)$; using ${\mu }_0{ϵ}_0{c}^2=1$ reproduces the boxed field. Oppositely moving populations contract differently under the boost, so neutrality in one frame does not imply neutrality in the other. The ideal wire's fields are singular on its axis.

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.

Write the electromagnetic field tensor as $F^{\mu \nu }={\partial }^{\mu }{A}^{\nu }-{\partial }^{\nu }{A}^{\mu }$. Substitution into the Covariant Maxwell equation with the minus-plus-plus-plus metric givesChoose the Lorenz gauge ${\partial }_{\mu }{A}^{\mu }=0$. Since ${\partial }_{\mu }{\partial }^{\mu }={\nabla }^2-c^{-2}{\partial }_t^2$, the electromagnetic four-potential obeysConvolution with the outgoing retarded Green function gives the retarded electromagnetic potential

For the point particle, ${\tau }_{\ast }(x)$ is its unique retarded proper time, characterized byThus $y({\tau }_{\ast })$ is where the particle's worldline intersects the past light cone of the observation event $x$; the displayed expression is the covariant form of the Lienard-Wiechert potentials.

On the $z$-axis put $D=\sqrt{R^2+z^2}$. The retarded time is $t_{\ast }=t-D/c$. For the circular orbit, the spatial velocity at that time isBecause the displacement from the retarded source point to the observation point is orthogonal to this velocity, $|R^{\nu }{\dot{y}}_{\nu }|=\gamma cD$, while ${\dot{y}}^{\mu }=\gamma (c,\mathbf{v})$. The factors of $\gamma$ cancel, leaving