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The outward area element is $d\mathbf{S}=a^2\sin \theta d\theta d\varphi {\mathbf{e}}_r$. ThereforeSplit the sphere into the northern cap $0\le \theta \le \alpha$ and the southern cap $\alpha \le \theta \le \pi$. Their induced boundary orientations are $+C$ and $-C$, respectively. The Stokes theorem applies to ${\mathbf{A}}^{(+)}$ on the north and ${\mathbf{A}}^{(-)}$ on the south. Adding their fluxes gives $2\pi (1-\cos \alpha )+2\pi (1+\cos \alpha )=4\pi$, agreeing with direct integration.

There is no contradiction with the flux of a globally smooth curl through a closed surface: neither potential is regular on the entire sphere. On their overlap they differ by $2{\mathbf{e}}_{\varphi }/(r\sin \theta )=2\nabla \varphi$, a local gauge transformation. The longitude is not a globally single-valued scalar, so that difference can have nonzero circulation around $C$.

The magnetic vector potential is defined by $\mathbf{B}=\nabla \times \mathbf{A}$. The gauge transformationleaves both $\mathbf{B}$ and $\mathbf{E}=-\nabla \varphi -{\partial }_t\mathbf{A}$ unchanged. In magnetostatics choose the Coulomb gauge $\nabla \cdot \mathbf{A}=0$. Then Ampère's law and the curl-of-curl identity giveThe free-space Green function identity ${\nabla }^2|\mathbf{r}-{\mathbf{r}}^{\prime }{|}^{-1}=-4\pi {\delta }^{(3)}(\mathbf{r}-{\mathbf{r}}^{\prime })$ therefore givesThe boundary condition selects the homogeneous solution; the usual decay condition assumes suitably localized sources. An infinitely long static wire needs a reference for its logarithmically divergent potential, although its magnetic field is finite away from the wire.

For time-dependent sources use the Lorenz gauge, $\nabla \cdot \mathbf{A}+c^{-2}{\partial }_t\varphi =0$, and the outgoing retarded potentialThe scalar potential has the analogous retarded charge-density integral. The two gauges should not be conflated for a general time-dependent source.^[ref]

For the prescribed uncharged wire, $\rho =0$ and $\varphi =0$ is consistent. Translation invariance makes $\mathbf{A}=A_z(r,t)\hat{\mathbf{z}}$ with $\nabla \cdot \mathbf{A}=0$, so this particular solution satisfies both gauge conditions. With $s=z^{\prime }-z$,It is zero for $t<r/c$. For $t>r/c$, set $L=\sqrt{c^2{t}^2-r^2}$. The causal interval is $-L<s<L$, giving the switched infinite-wire retarded potentialDifferentiation gives the fields and Poynting vector, for $r>0$, $t>r/c$:Here $\mathbf{B}={\mu }_0\mathbf{H}$ and $B_{\varphi }=-{\partial }_r{A}_z$. All three fields vanish before arrival. For positive $I$, $\mathbf{E}$ opposes the imposed current and the radial energy flow after switching a line current points radially outward: $(-\hat{\mathbf{z}})\times \hat{\mathbf{\varphi }}=\hat{\mathbf{r}}$.

For the graph, let $u=ct/r$. Normalized to $E_{\ast }={\mu }_{0}Ic/(2\pi r)$, $H_{\ast }=I/(2\pi r)$ and $N_{\ast }={\mu }_0{I}^{2}c/(4{\pi }^2{r}^2)$, the curves areAt the wavefront, $|E_z|$ and $H_{\varphi }$ diverge as $(u-1{)}^{-1/2}$ and $N_r$ as $(u-1{)}^{-1}$. This abrupt line-current wavefront singularity reflects the ideal instantaneous turn-on of a line current; a finite rise time smooths the front. At late times,The vector potential grows logarithmically with time in this infinite-wire idealization even as the local magnetic field approaches its static limit. The divergent front should not be interpreted as a finite-energy pulse from a real instantaneous switching experiment.

Gauge freedom. The magnetic vector potential satisfies $\mathbf{B}=\nabla \times \mathbf{A}$. The transformationleaves $\mathbf{B}$ unchanged because the curl of a gradient is zero. For time-dependent fields, transform the electric scalar potential as $\varphi \mapsto \varphi -{\partial }_t\chi$ as well, so that $\mathbf{E}=-\nabla \varphi -{\partial }_t\mathbf{A}$ is unchanged. The scalar $\chi$ must have the regularity and single-valuedness appropriate to the physical region and its boundary conditions. Local values of $\mathbf{A}$ therefore depend on the gauge transformation.

Gauge and nonzero component. In the standard axisymmetric Coulomb gauge representative for the long-solenoid approximation,The divergence vanishes since $A_{\varphi }$ is independent of the azimuth. Inside, this agrees with the usual symmetric gauge representative $\mathbf{A}=\frac{1}{2}\mathbf{B}\times \mathbf{r}$. The Coulomb gauge condition alone does not force the other two components to vanish; the symmetry, regularity and customary boundary choice select this representative from the remaining gauge freedom.

Radial dependence. Write the interior magnetic field as $B_0\hat{\mathbf{z}}$, where $B_0={\mu }_{0}nI$ for a vacuum solenoid of solenoid turn density $n$. Its exterior magnetic field vanishes in the infinitely long idealization. Applying Stokes theorem to a coaxial circle gives $2\pi r{A}_{\varphi }(r)=\Phi (r)$. Therefore the Coulomb-gauge vector potential of a long solenoid isIt grows linearly inside, is continuous at $R$, and decays as $r^{-1}$ outside. Although its exterior curl is zero, its circulation is the enclosed magnetic flux $\Phi =B_0\pi R^2$. No regular single-valued gauge function removes it throughout the exterior while preserving this circulation.

Electric field of the rotating charged surface. For clarity retain absolute permittivity $ϵ$ and permeability $\mu$ in the cylinder formulae; setting them to unity gives the rationalized-unit convention in the question. If the stated unity means relative material constants in SI, use ${ϵ}_0,{\mu }_0$ instead. Assume prescribed charge densities in the laboratory frame and the nonrelativistic, steady-rotation model.

The axisymmetric surface charge is stationary as a distribution even while the material rotates. A Gaussian cylinder of length $L$ has enclosed charge zero for $r<R$ and $2\pi RL\sigma$ for $r>R$. Gauss's law therefore givesThe radial electric field jumps from zero to $\sigma /ϵ$ at the surface and then decays as $r^{-1}$. Its direction reverses for negative $\sigma$. The angular velocity does not enter this electric-field result for a specified surface charge.

Magnetic field of the volume-charged cylinder. A shell of radius $a$ and thickness $da$ has surface charge $\rho da$, azimuthal surface current $d{K}_{\varphi }=\rho \omega ada$, and the field of an infinite solenoid: $d{B}_z=\mu \rho \omega ada$ inside that shell and zero outside it. At radius $r<R$, only shells with $a>r$ contribute. Thus the magnetic field of a rotating charged solid cylinder isWith no applied background magnetic field, $\mathbf{B}=0$ outside. The profile is parabolic, largest on the axis and zero at the boundary. The sign follows $\rho \omega$. As a check, the current density is $J_{\varphi }=\rho \omega r$, and Ampère's circuital law gives $-d{B}_z/dr=\mu J_{\varphi }$, as required.

Electron impulse when the current is switched off. Choose positive $I$ so the initial magnetic flux is ${\Phi }_i={\mu }_{0}nI\pi R^2$ along $+z$. Faraday's law of induction gives, outside the solenoid,For an electron of electric charge $q=-e$, treat its position as fixed while the short pulse acts. The inductive impulse outside a switched solenoid isConsequently the nonrelativistic final velocity isThe induced electric field circulates in the positive azimuthal direction while a positive flux decreases, so the electron accelerates in the negative direction. Once the pulse has passed, there is no exterior field and it travels along the tangent at the impulse point, rather than continuing in a circular orbit. If the resulting momentum is relativistic, use ${\mathbf{v}}_f=c{\mathbf{p}}_f/\sqrt{m_e^2{c}^2+|{\mathbf{p}}_f{|}^2}$ instead. The sudden-switch result assumes negligible displacement during the pulse and the usual quasistatic electromagnetic induction approximation; “instantaneous” switching does not mean instantaneous propagation of fields to arbitrary distance.

One can check the sign using conserved canonical angular momentum: axial symmetry conserves $p_{\varphi }^{\mathrm{can}}=L_z+qr{A}_{\varphi }$. Initially $L_z=0$, and finally $A_{\varphi }=0$, so $L_z=q{\Phi }_i/(2\pi )=-e{\Phi }_i/(2\pi )$.

Interferometric measurement at constant current. Use a coherent electron beam split into two paths passing on opposite sides of a shielded long solenoid, then recombine them to form interference fringes. Keep the current constant during the passage and suppress leakage fields on the paths. In the Aharonov-Bohm effect, the phase difference due to the magnetic vector potential is the closed-loop integralThe overall sign reverses if the path orientation is reversed. The displacement in fringe periods is $\Delta \phi /(2\pi )$, so the measured fringe shift is linear in $I$ and repeats whenA gauge transformation adds a gradient whose closed-loop integral vanishes for a single-valued $\chi$. Endpoint changes in electron wavefunction phase cancel in the interference measurement. The measured Aharonov-Bohm phase is therefore gauge invariant: it determines the flux, or equivalently the circulation of $\mathbf{A}$. In the chosen Coulomb gauge this implies $A_{\varphi }(d)={\mu }_{0}nI{R}^2/(2d)$, but the experiment does not select or measure a unique local vector potential independently of gauge.

Far field of the rotating sphere. Interpret uniform charge as uniform volume charge of total $Q$, and take $\mathbf{\omega }=\omega \hat{\mathbf{z}}$. The magnetic dipole moment of a rigidly rotating charge distribution isIts axial component is $m_z=(\rho \omega /2)\int r_{\perp }^2{d}^{3}r$. For a solid sphere, $\int r_{\perp }^2{d}^{3}r=8\pi R^5/15$ and $\rho =3Q/(4\pi R^3)$, soThe magnetic field of a rotating charged spherical body is a static magnetic dipole field at large distances:In particular, $|\mathbf{B}|={\mu }_0|\mathbf{m}|\sqrt{1+3{\cos }^2\vartheta }/(4\pi r^3)$. Both the angular mean of the magnitude and its root mean square therefore scale as $\boxed{r^{-3}}$; the latter is $B_{\mathrm{rms}}=\sqrt{2}{\mu }_0|\mathbf{m}|/(4\pi r^3)$. The angular average of the vector is zero, which is different from average strength. Steady rotation leaves the charge and current distributions stationary, so this is magnetostatics, not a radiative $r^{-1}$ field. If uniform charge instead means a thin spherical surface, replace the dipole moment by $Q{R}^2\mathbf{\omega }/3$; the angular pattern and $r^{-3}$ scaling are unchanged.

The covariant derivative $D_{\mu }$ differentiates a field while preserving its transformation law under the local electroweak gauge symmetry $SU(2{)}_L\times U(1{)}_Y$. Here ${\partial }_{\mu }$ is the ordinary spacetime derivative and $\mu$ a Lorentz index. The gauge fields $W_{\mu }^j$, $j=1,2,3$, are the three SU(2) fields; $B_{\mu }$ is the U(1) field. The corresponding dimensionless coupling constant is $g$ for SU(2) and $g^{\prime }$ for U(1).

The ${\tau }_j$ are weak isospin generators in the representation of the differentiated field, with ${\tau }_j={\sigma }_j/2$ on a doublet and ${\tau }_j=0$ on an SU(2) singlet. The repeated index $j$ is summed. The generator $Y/2$ supplies the field's weak hypercharge in the convention $Q={\tau }_3+Y/2$, where $Q$ is its fractional electric charge. The factor $-i$ follows the convention of Hermitian generators. The two connection terms compensate the derivatives of spacetime-dependent gauge transformations.

Empirical equivalence must be specified relative to a domain and an account of what counts as observable. Two theories may fit all evidence collected so far while predicting different future results; that is ordinary underdetermination by a finite record. A stronger equivalence requires the same observable consequences for every permitted experiment in the stated domain. For stochastic predictions, matching a set of possible outcomes is not enough: the theories must assign the same outcome probability distributions under corresponding preparations and testing assumptions.

A clear physical example at the level of descriptions occurs in electromagnetism. Write the electric field and magnetic field using a scalar potential $\varphi$ and magnetic vector potential $\mathbf{A}$:For a smooth, single-valued function $\chi$, the gauge transformationleaves both fields unchanged: the curl of a gradient vanishes, and the mixed derivatives cancel in the electric-field expression. With the same charges, boundary conditions and force law, these descriptions give the same classical measurements. Agreement is structural, not an accidental fit to a few readings. However, many physicists regard such descriptions as representing one physical situation, so this alone is not an example of two genuinely incompatible physical theories. The smooth global transformation assumption also matters; equality of local fields is not an unrestricted license to identify every globally different potential configuration.

A second example exhibits possible surplus structure. For particles interacting by Newtonian forces depending only on relative positions, a common Galilean transformationleaves relative positions and accelerations unchanged. Accordingly, adding a privileged but completely uncoupled state of uniform rest does not alter any internal mechanical prediction. Rival choices of that unobservable rest state cannot be distinguished by these experiments. This is a deliberately specified completion of the mechanical model: it does not show that rotation or acceleration is undetectable, or that Newtonian mechanics and special relativity are empirically equivalent. The latter give different high-speed predictions. Whether undetectable absolute rest makes a genuinely different theory or a redundant representation is precisely one philosophical issue raised by the example.

Thus there are concrete cases of empirical redundancy, while the existence of equally well-motivated, fully distinct rivals with identical consequences across all physics is a much stronger claim. Historical theories often agree only in limited regimes or after additional assumptions. One must not convert that restricted agreement into an equivalence theorem for every possible test.

The philosophical relevance does not depend on finding a complete historical pair. A constructed rival can expose the logical limits of evidence, although an artificial extension need not deserve the same credibility as an economical account. If two specified models have equal likelihoods for every admissible data set $D$, thenfor finite positive prior odds. The Bayes factor is one; the data do not discriminate between them. Choices about simplicity or interpretation would be additional considerations. Constructive empiricism treats empirical adequacy as the relevant belief commitment in acceptance, while a scientific realist must explain what further warrants belief in the unobservable structure. The equivalence argument identifies a limit of evidential discrimination; it does not establish that every theoretical claim is equally good or that every unobservable claim is false.^[ref]

Empirical equivalence must be stronger than agreement with the evidence already collected: relative to a stated domain, empirically equivalent theories yield the same possible observational consequences. If they nevertheless make incompatible claims about unobservable entities, observations in that domain cannot alone determine which ontology is correct. This is a strong form of scientific underdetermination, directed against a unique inference from empirical success to full theoretical truth.

The scientific realist has several responses. One may deny that every formally constructed alternative is a serious physical rival, appeal to independently justified explanatory virtues, or restrict commitment to the structure genuinely shared by the alternatives. If two descriptions are connected by a change of variables or a gauge transformation, apparent difference may be representational rather than a disagreement about the world. Before calling such a pair a counterexample, identify what incompatible physical claims remain.

Actual and merely possible rivals have different evidential force. An actual, comparably developed rival demonstrates that the current inference is nonunique under specified criteria. A merely possible rival refutes a claim of logical entailment if its possibility is established, but does not automatically defeat ordinary fallible justification: otherwise any unexcluded sceptical possibility would defeat almost every empirical belief. A distant conceivable alternative need not deserve the same prior probability or explanatory standing as a well-tested account.

Consequently, empirical equivalence is a serious challenge to unrestricted, uniquely determined theoretical belief. It is not a deductive disproof of all scientific realism. A convincing argument must say whether the target is certainty or defeasible warrant, whether equivalence covers all observations or only present data, and why the rivals are physically distinct. Merely announcing that another story could be invented leaves these crucial questions unanswered.

The absence of magnetic monopoles is expressed by $\nabla \cdot \mathbf{B}=0$. In a regular simply connected region, a magnetic vector potential may be introduced so thatIts units are $\mathrm{T}\mathrm{m}=\mathrm{Wb}{\mathrm{m}}^{-1}$. It is not unique: adding a gradient leaves the magnetic field unchanged because $\nabla \times \nabla \chi =0$. In time-dependent electromagnetism,is a gauge transformation preserving both physical fields. Stokes theorem gives ${\oint }_C\mathbf{A}\cdot d\mathbf{\ell }={\int }_S\mathbf{B}\cdot d\mathbf{S}={\Phi }_B$ for a contour bounding an appropriate surface.

For localized steady currents, choose the Coulomb gauge $\nabla \cdot \mathbf{A}=0$. Substitution into Ampère's law and the curl-of-curl identity givesBoundary conditions are needed in addition to the gauge choice. For a uniform magnetic field one convenient choice is $\mathbf{A}=\frac{1}{2}\mathbf{B}\times \mathbf{r}$. For a magnetic dipole, $\mathbf{A}={\mu }_0\mathbf{m}\times \mathbf{r}/(4\pi r^3)$ outside its source. The magnetic vector potential simplifies field calculations and appears directly in the Lagrangian of a charged particle, $L=\frac{1}{2}m{v}^2+q\mathbf{v}\cdot \mathbf{A}-q\varphi$. In multiply connected regions, a field-free region can retain nontrivial circulation of $\mathbf{A}$, relevant to the Aharonov-Bohm effect; a local gauge argument cannot remove every global circulation.

Because the magnetic field obeys $\nabla \cdot \mathbf{B}=0$, it can be represented locally as the curl of a magnetic vector potential:This automatically enforces the absence of magnetic monopoles. It is also a global representation throughout ordinary all-space problems without monopole singularities. It is not unique: the gauge transformationleaves $\mathbf{B}$ unchanged because the curl of a gradient vanishes. In a time-dependent problem,makes the same transformation leave both physical fields unchanged and automatically satisfies Faraday's law.

In vacuum magnetostatics, choose the Coulomb gauge $\nabla \cdot \mathbf{A}=0$. Substitution into Ampère's law and the identity $\nabla \times (\nabla \times \mathbf{A})=\nabla (\nabla \cdot \mathbf{A})-{\nabla }^2\mathbf{A}$ givesfor a localized steady electric current density and a potential vanishing at infinity. Taking the curl recovers the Biot-Savart law. The magnetic vector potential converts a constrained vector field problem into componentwise Poisson equations once a gauge is chosen.

For a uniform magnetic field, $\mathbf{A}=\frac{1}{2}\mathbf{B}\times \mathbf{r}$ is a convenient choice. For a localized magnetic dipole moment $\mathbf{m}$, the leading far-field potential isAnother useful relation is Stokes theorem, which relates a closed-path integral to the magnetic flux through a spanning surface:The units of $\mathbf{A}$ are $\mathrm{T}\mathrm{m}$, equivalently $\mathrm{Wb}{\mathrm{m}}^{-1}$.

Consider a conventional spin-singlet, weak-coupling, three-dimensional superconductor. Set $\hslash =k_B=1$ in the microscopic action and restore electromagnetic units below. An attraction $\lambda >0$ in an energy shell about the Fermi surface gives a Grassmann actionThe attractive sign permits pairing even when the single-particle Fermi gas is otherwise stable. A complex Hubbard–Stratonovich transformation introduces a pair field $\Delta$ proportional at the saddle to $\lambda ⟨{\psi }_{\downarrow }{\psi }_{\uparrow }⟩$. The action is then quadratic in fermions. Integrating them out gives the BCS pairing-field effective actionHere the trace uses one independent Nambu spinor pairing block; a fully doubled particle-hole representation instead carries its corresponding one-half factor. At a uniform saddle the Bogoliubov--de Gennes Hamiltonian can be written asThe negative branch describes the paired vacuum; excitations are Bogoliubov quasiparticles mixing particles and holes. Extremizing the effective action givesThe physical cutoff or a renormalized interaction is part of the microscopic model; an unregulated contact integral must not be used indiscriminately. With constant density of states $N(0)$ per spin and shell half-width ${\Omega }_c$, the zero-temperature weak-coupling solution is ${\Delta }_0\simeq 2{\Omega }_c{e}^{-1/[N(0)\lambda ]}$. The exponentially small scale follows from the logarithmic pairing susceptibility of states near the Fermi surface. For this ideal weak-coupling model, $2{\Delta }_0/(k_B{T}_c)\simeq 3.53$. The microscopic theory relates the gap, pairing and thermodynamic transition.^[ref]

Close to a continuous transition, for fields varying slowly compared with microscopic scales, expand the effective action in powers and gradients of $\Delta$. Its static part is the Ginzburg-Landau theory of superconductivity:The normalization of $\Delta$ fixes the coefficients; here $a(T)$ changes sign at $T_c$. The uniform minimum has $|{\Delta }_0{|}^2=-a/b$ for $a<0$, condensation free-energy density $-a^2/(2b)$, and characteristic coherence-length scale $\xi =\sqrt{K/|a|}$. Above the transition the minimum is at zero. The quartic stabilization and gradient cost are precisely the ingredients of Landau theory for a complex order parameter. This controlled near-transition expansion supplies a bridge between microscopic BCS theory and macroscopic magnetic response; its coefficients and validity depend on the material and fluctuation regime.^[ref]

Write $\Delta =(|{\Delta }_0|+\eta )e^{i\varphi }$. The radial fluctuation has positive local restoring curvature, whereas the phase has no local potential in a neutral theory with global $U(1)$ symmetry. Freezing the amplitude givesThe neutral case supports a gapless collective phase mode. In the charged case the electromagnetic field couples to this phase rigidity. Under $\mathbf{A}\mapsto \mathbf{A}+\nabla \chi$ and $\varphi \mapsto \varphi +q\chi /\hslash$, the displayed combination is invariant. Locally choosing a gauge with constant phase leaves a quadratic term in $\mathbf{A}$. Varying the free energy gives the transverse London current and magnetic screening scale,Combining this relation with static Maxwell equations gives exponential field penetration, the Meissner effect. Single-valuedness of the pair field around a loop with negligible current gives flux in units ${\Phi }_0=2\pi \hslash /|q|=\pi \hslash /e$. These are gauge-invariant physical consequences of the condensate stiffness.

For the dynamics, the neutral phase and density mode is reorganized by the Coulomb interaction and the dynamical electromagnetic field. In a bulk three-dimensional system the long-wave longitudinal mode has a plasma gap, rather than remaining the neutral sound mode. This is the Anderson-Higgs mechanism in a superconductor: a phase coordinate becomes part of the electromagnetic collective response instead of surviving as an independent massless physical particle. Local gauge transformations are redundancies of the description, so a literal breaking of a local redundancy is not an observable. Phase coherence, current response and magnetic screening provide the physical statements.^[ref]

A static Landau functional alone cannot determine that full dynamical spectrum. A two-dimensional film, for example, need not have the same long-wave plasma dispersion as a bulk charged system. Nor does a positive radial curvature automatically guarantee a sharp Higgs amplitude resonance: coupling to quasiparticles and other modes can damp it. The chosen mean-field phase is also a thermodynamic symmetry-breaking description; an isolated finite system can be formulated with definite particle number. The microscopic pairing saddle, the near-transition Landau expansion and the electromagnetic Anderson-Higgs response describe successive scales of the same superconducting state, with distinct assumptions at each step.

Use International System of Units, with $c^{-2}={\mu }_0{\epsilon }_0$. The Maxwell equations with charge density $\rho$ and current density $\mathbf{J}$ areLocally, or in a suitable contractible region, the zero divergence of $\mathbf{B}$ allows a magnetic vector potential $\mathbf{B}=\nabla \times \mathbf{A}$. Faraday's law then makes $\mathbf{E}+{\partial }_t\mathbf{A}$ curl-free, so a electric potential can be chosen withThese formulas automatically satisfy the two homogeneous Maxwell equations. The local topology assumption rules out obstructions to expressing a closed field as a gradient or curl.

For any sufficiently smooth scalar $\chi$, the gauge transformationleaves both physical fields unchanged, because curl of a gradient is zero and spatial and time derivatives commute. In such a region this accounts for the freedom between potentials for the same fields.

The Lorenz gauge is $\nabla \cdot \mathbf{A}+c^{-2}{\partial }_t\Phi =0$. Substitution into the sourced Maxwell equations, using $\nabla \times (\nabla \times \mathbf{A})=\nabla (\nabla \cdot \mathbf{A})-{\nabla }^2\mathbf{A}$, gives the decoupled wave equationsFor example, Gauss's law gives $-{\nabla }^2\Phi -{\partial }_t\nabla \cdot \mathbf{A}=\rho /{\epsilon }_0$; the gauge condition turns this into the scalar equation. The remaining gauge freedom preserving the condition has $({\nabla }^2-c^{-2}{\partial }_t^2)\chi =0$.

Put $\theta =\omega t-Kz$ for the given wave. Direct differentiation gives $\nabla \cdot \mathbf{E}=0$ and $\nabla \times \mathbf{E}=K\mathbf{E}$. Integrating Faraday's law in time yields the wave's magnetic fieldOne could superpose a time-independent divergence-free and curl-free magnetic field; the expression displayed chooses no such extra background. It has zero divergence and ${\partial }_t\mathbf{B}=-K\mathbf{E}$, so the magnetic Gauss and Faraday equations hold. MoreoverSince the time derivative is nonzero, the vacuum Ampere-Maxwell law holds precisely whenThis verifies all four source-free equations, including both divergence constraints. The sign of $\omega /K$ selects the direction of propagation.

A wave has circular polarization when, at a fixed spatial point, its transverse electric vector has constant nonzero magnitude and rotates uniformly through a circle. Here $E_x^2+E_y^2=E^2$ and the angle changes at rate $\omega$, so the wave is circularly polarized. Its magnetic vector is perpendicular to the electric vector, has magnitude $|E|/c$ under the dispersion relation, and rotates with it. A handedness label additionally requires a convention for the observer's viewing direction.

The Lorenz-gauge potentials for a circularly polarized plane wave give a convenient choice. Take $\Phi =0$ and $\mathbf{A}={\omega }^{-2}{\partial }_t\mathbf{E}$. Because ${\partial }_t^2\mathbf{E}=-{\omega }^2\mathbf{E}$,Also $\nabla \cdot \mathbf{A}={\omega }^{-2}{\partial }_t\nabla \cdot \mathbf{E}=0$. Together with ${\partial }_t\Phi =0$, this proves these potentials satisfy the Lorenz gauge. They satisfy the source-free potential wave equations because every component has the same dispersion relation just obtained.

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.

For $\Delta \varphi =-i\varphi$ and $\Delta {\varphi }^{\ast }=i{\varphi }^{\ast }$, the interaction is invariant under the constant phase because every product has net zero phase. Applying the Noether formula to the two independent fields givesThe sign corresponds to the printed transformation $\varphi \to e^{-iϵ}\varphi$. Reversing the phase parameter would reverse the entire current, with no change in conservation.

The derivative products in the divergence of the first term cancel pairwise:The two field equations from part (c) implyMultiplication by $i$ exactly cancels the remaining background terms, proving ${\partial }_{\mu }{J}^{\mu }=0$ without requiring $A_{\mu }$ to be constant or divergence free. The global phase symmetry survives even though this truncated interaction is not exactly invariant under arbitrary local gauge transformations.

Keep the index convention of b: $A^{\mu }=(A_0,\mathbf{A})$, so $A_{\mu }=(A_0,-\mathbf{A})$ and the spatial gauge covariant derivative is $\nabla +ie\mathbf{A}$. The physical three-current is the spatial part of $J^{\mu }$:Equivalently, writing $\phi =R{e}^{i\theta }$ gives $\mathbf{j}=-2e{R}^2(\nabla \theta +e\mathbf{A})$. Under the gauge transformation in a, $\theta \mapsto \theta +\alpha$ and $\mathbf{A}\mapsto \mathbf{A}-\nabla \alpha /e$, so this expression is manifestly invariant. Switching to a convention with the opposite spatial gauge sign reverses the phase-gradient term, not the screening term proportional to $e^2$.

For static fields the spatial Ampère-Maxwell equation is $\nabla \times \mathbf{B}=\mathbf{j}$, while $\nabla \cdot \mathbf{B}=0$. Taking its curl yields the general magnetic-field equationNo uniform-condensate approximation has yet been used. For a varying scalar amplitude the gradient terms must be kept; simply replacing $|\phi {|}^2$ by a constant before deriving the current would conceal their origin. The homogeneous-condensate limit supplies the London screening equation in d.

Introduce three independent local parameters ${\alpha }_j(x)$. A consistent convention for the three Abelian gauge transformations isIndeed, $g(B_{\mu }^k-q{B}_{\mu }^{k+1})$ shifts by $-{\partial }_{\mu }({\alpha }_k-q{\alpha }_{k+1})$, cancelling the derivative of the scalar phase. Thus $D_{\mu }{\varphi }_k$ transforms by the same phase as ${\varphi }_k$. The two kinetic terms, the absolute-value potentials and all Abelian field strengths are invariant.

In units where the gauge coupling is $g$, the charge vectors under the three factors areLocally this is the gauge algebra of $U(1{)}^3$. To interpret it globally as three compact circle groups, the charges must belong to a compatible charge lattice: in the stated unit-charge convention integer $q$ is sufficient, while arbitrary irrational $q$ is not a single-valued compact-group representation. The local transformations and mass calculation remain meaningful for general positive real $q$ as a Lie-algebra calculation, or with noncompact Abelian factors.

A nonzero Higgs doublet has a real radial magnitude and three real angular coordinates. For a normalized vector $z=(z_1,z_2{)}^T$, the SU(2) matrix $U=\left(\begin{array}{cc}{z}_2^{\ast }& z_1\\ -z_1^{\ast }& z_2\end{array}\right)$ satisfies $U(0,1{)}^T=z$. Near the vacuum, the three broken generators can equivalently be parameterized by $e^{-i\mathbf{n}\cdot \mathbf{\sigma }+i{n}_3}$. A local gauge transformation removes these angular fields in unitary gauge. This coordinate choice is local about a nonzero vacuum and need not extend through zeros or topological defects.

Near a nonzero vacuum expectation value, the radial variable $h$ describes the Higgs mode and the angular variable $\pi$ describes the would-be Goldstone boson. For $D_{\mu }={\partial }_{\mu }+ie{A}_{\mu }$, their local gauge transformation is $h\mapsto h$, $\pi \mapsto \pi +f\alpha$, and $A_{\mu }\mapsto A_{\mu }-{\partial }_{\mu }\alpha /e$. The scalar kinetic term is $(\partial h{)}^2/2+(1+h/f{)}^2(\partial \pi +efA{)}^2/2$. These coordinates simplify gauge orbits but obscure ordinary power counting through higher-dimensional derivative vertices and fail at $\varphi =0$.

The constant mode of the worldline einbein is not removed by gauge transformations whose parameters vanish at the ends of a fixed interval. This Schwinger proper-time parameter remains integrated over after gauge fixing. For $\phi =(p^2+M^2)/2$, the geometric on-shell proper length is $M\int edt$.

A quark field is a Dirac field or a chiral Weyl spinor carrying the colour fundamental representation of $SU(3)$. Its three colour components transform as ${\psi }^i\mapsto U^i{}_j{\psi }^j$ under local gauge transformations. The colour index is independent of the Lorentz spinor index, and the weak representations of its left and right chiral components differ in the Standard Model.