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The Field-cooled Meissner experiment starts above the superconducting transition temperature with a weak, steady applied magnetic field already penetrating a solid sample. Cooling while keeping the applied magnetic field fixed tests whether the interior magnetic flux changes. This is the decisive contrast with cooling in zero magnetic field and switching the magnetic field on afterwards. The diagram shows the two histories for the same final applied magnetic field.

Meissner and Ochsenfeld reported magnetic expulsion in 1933, showing that the transition involves more than vanishing electrical resistance. A modern motivating version appears in MIT's lead-sample experiment: surround the sample with a solenoid supplying a steady magnetic field and a separate pickup coil connected to a recording instrument. Apply the magnetic field while the lead is still normal, then cool through its transition without changing the solenoid electric current. Expulsion produces a transient pickup-coil electric potential difference; warming through the transition produces the opposite sign as magnetic flux returns. MIT also checks the signal direction by switching off the applied magnetic field while the sample is normal. ^{[ref] [ref]}

Faraday's law of induction explains the readout. For a stationary $N$-turn pickup coil, with the same linked magnetic flux $\Phi$ per turn and a fixed orientation,Thus a transient measures a change in linked magnetic flux, not directly the static magnetic field at every point in the sample. For stipulated $N=100$ and $\Delta \Phi =-2\times 10^{-8}\mathrm{Wb}$ per turn, the signed pulse area is $+2\times 10^{-6}\mathrm{V}\mathrm{s}$. The apparatus geometry determines how a change in interior magnetic field contributes to that linked magnetic flux. Keeping the applied magnetic field fixed, repeating on warming, and checking a normal-state control make the expulsion interpretation testable; a stationary pickup coil has no sustained signal after the magnetic flux has stopped changing.

Why does the preparation history matter? In the stationary, ideal perfect conductor model, the interior electric field is zero. The differential form of Faraday's law of induction then givesIf the interior magnetic field was ${\mathbf{B}}_0$ when perfect conduction began, it remains ${\mathbf{B}}_0$; if it was zero, it stays zero when an external magnetic field is subsequently applied. Both a perfect conductor prepared in zero magnetic field and a sample exhibiting the Meissner effect can therefore exclude a newly applied weak magnetic field. Only the field-cooled comparison tests expulsion of an already present magnetic field. This argument concerns a stationary sample and the ideal zero-electric field model, not a moving conductor.

The equilibrium London equations supply the additional constraint. In a homogeneous region without penetrating flux tubes, they imply ${\nabla }^2\mathbf{B}=\mathbf{B}/{\lambda }_L^2$. For a half-space $z\ge 0$ with a magnetic field parallel to its flat boundary, fixed surface value $B_s$, and a bounded solution deep inside, this givesAt $z=5{\lambda }_L$ only $e^{-5}\simeq 0.00674$ of the surface magnetic field remains. The screening electric currents near the surface produce this reduction. Unlike the frozen interior magnetic field of a perfect conductor, this equilibrium profile does not retain an arbitrary initial ${\mathbf{B}}_0$. The extra constitutive relation, and not zero electrical resistance by itself, selects the screened state. Expulsion occurs during the transition, when transient electric fields can be present; the steady-state zero-electric field argument does not describe the process of establishing superconductivity. The diagram gives equilibrium end states, not transition kinetics. ^[ref]

These ideal comparisons concern a solid sample in equilibrium and a sufficiently weak magnetic field. Stronger magnetic fields can destroy superconductivity or permit magnetic flux to penetrate parts of a sample, so complete bulk expulsion is not a claim about every superconducting state. ^[ref] A hole in a superconducting loop can retain magnetic flux even while the material surrounding it screens its own interior; see flux conservation in a superconducting loop. The hole is not superconducting material, so its retained magnetic flux does not contradict the Meissner effect. ^[ref]

For a bulk type-II superconductor with negligible demagnetizing field, raising the applied magnetic field intensity past the lower critical value $H_{c1}$ favors a superconducting mixed state containing quantum vortex lines; superconductivity survives until the upper critical value $H_{c2}$. The magnetic field extends around each vortex over a penetration scale, not solely inside its core. Thus a nonzero average magnetic flux density can coexist with superconductivity. Below $H_{c1}$, a type-II superconductor also supports a Meissner state; a vortex-free state above the equilibrium entry field may instead be metastable. ^[ref]

Real preparation histories may leave quantum vortex lines held in place by material inhomogeneities, retaining magnetic flux and producing hysteresis. This flux pinning restricts access to equilibrium: it is different from the arbitrary frozen interior magnetic field of the ideal perfect conductor model. Harvard's vortex-imaging work illustrates strong pinning to bulk disorder in a type-II superconductor. ^[ref] Separately, flux conservation in a superconducting loop and fluxoid and flux quantization in a superconductor explain why a hole can retain magnetic flux. Neither bore retention, vortex penetration nor weak screening in a thin sample should be silently identified with the equilibrium Planar London screening calculation.

Magnetic relaxation converts magnetic energy into material motion through the Lorentz force density, and dissipates the resulting kinetic energy through viscosity. In a perfect conductor, magnetic flux freezing preserves field-line connectivity throughout every smooth stage. A regular limiting state has zero velocity and is a magnetostatic equilibrium, but its magnetic field can contain current sheets. Conservation of one global magnetic helicity is weaker than preservation of the complete frozen field-line structure.

Compare two ideal models for a stationary, solid, simply connected sample. In the perfect conductor flux-history model the interior electric field is constrained to zero from the moment perfect conduction starts. Faraday's law of induction then givesConsequently the interior magnetic field retains its value at that moment. Cooling in an already applied magnetic field freezes a nonzero interior value; starting perfect conduction in zero magnetic field and then applying a magnetic field preserves zero. The zero initial interior magnetic field often assumed in perfect conductor wave-reflection problems is one particular preparation, not a general prediction of expulsion. ^[ref]

The Meissner state instead selects the screened equilibrium after either preparation, provided that equilibrium is accessible and the applied magnetic field lies in the Meissner state regime. The first London equation alone illustrates the missing information: at fixed ${\lambda }_L$, combining it with Faraday's law of induction givesThis permits an arbitrary time-independent integration field. The second London equation imposes the additional zero value of that bracket in the vortex-free local model.^[ref] Establishing superconductivity during cooling is not the fixed-parameter, zero-electric field model: transient electric fields accompany changing magnetic flux, and ${\lambda }_L$ itself depends on temperature. The equilibrium London equations predict final spatial screening, not the cooling rate or the time-resolved expulsion pulse.

Work in cylindrical coordinates and use the assumed independence of $\phi$ and $z$. In each material region, the zero curl of the static electric field givesThe tangential electric field is zero at a perfect conductor. The dielectric-interface boundary conditions make its tangential components continuous at each material interface, so $E_z=E_{\phi }=0$ throughout the gap. They also vanish in the empty interior and exterior for the regular solution with no applied external electric field. HenceThis specifies an axis of direction: the sign of $E_{\rho }$ depends on the sign of $V$.

For the static magnetic field, Gauss's law for magnetism impliesThus $\rho B_{\rho }$ is constant in each layer. Continuity of the normal magnetic field across every interface makes that constant global, and regularity at the axis makes it zero. Next use $\nabla \times \mathbf{H}=0$ away from the conducting surfaces. Its $\phi$ component gives $d{H}_z/d\rho =0$. There is no azimuthal surface current density, so $H_z$ is continuous across the conducting surfaces and the material interface. The decay of the exterior magnetic field at infinity therefore forces this global constant to zero. The only remaining component isThe magnetostatic Ampère law fixes its sign: positive inner electric current in the $+z$ direction produces positive $B_{\phi }$ for positive magnetic permeability.

Orient the unit normal $\mathbf{n}$ from $R_1$ to $R_2$ and write $\mathbf{H}$ for the magnetic field strength. For static electric fields, $\mathbf{E}=-\nabla \varphi$; the static Ampère's law gives $\nabla \times \mathbf{H}={\mathbf{J}}_f$. The curl product rule therefore gives the local Poynting flux and terminal power identityIntegrate over an oriented piece ${\Sigma }_R$ of the separating surface and apply Stokes theorem:Once all the wires are included, the first surface integral is ${\sum }_k{\varphi }_k{I}_k$: the electric potential is constant on each perfect conductor and the integral of its normal current density is its signed electric current. Equivalently, remove the wire cross-sections and apply Stokes theorem in the surrounding insulator. Each resulting hole has negative boundary orientation, and the magnetostatic Ampère law around it gives the positive contribution ${\varphi }_k{I}_k$.

For a closed separating surface the boundary term is absent. For a noncompact separating surface, take $R\to \infty$ with the usual isolated-circuit condition ${\oint }_{\partial {\Sigma }_R}\varphi \mathbf{H}\cdot d\mathbf{\ell }\to 0$. In either case,For a steady closed circuit the signed electric currents sum to zero, so shifting the reference of electric potential does not change the terminal power.

For a fully general noncompact surface, the hypotheses as printed need an additional condition at infinity. The exact result before imposing it isTime independence alone does not remove this boundary term or ensure a finite Poynting flux; for example, spatially constant crossed electric fields and magnetic fields carry a constant Poynting vector through an infinite plane. The stated terminal-power formula holds with the boundary condition specified above.

Use the real magnetic field $\mathcal{B}$ in the coupled equations; its leading periodic approximation will be $\mathrm{Re}(\mathbf{B}{e}^{i\omega t})$. In the conducting incompressible flow, the magnetohydrodynamic momentum equation, resistive induction equation and solenoidal magnetic-field constraint areHere $p$ is pressure, $\mathbf{j}$ is electric current density, and $\eta$ and $\nu$ are magnetic diffusivity and kinematic viscosity. The moving-conductor Ohm law gives $\mathbf{E}=-\mathbf{u}\times \mathcal{B}+\eta \nabla \times \mathcal{B}$.

In the insulating exterior, neglecting displacement current, the magnetic field is curl-free away from the wire and solenoidal. The imposed electric current density suppliesThe exterior electric field also satisfies Faraday's law. At the rigid wall, impose the no-slip boundary condition $\mathbf{u}=0$. At finite conductivity, the interface conditions for electromagnetic fields give continuous normal magnetic field, continuous tangential magnetic field for equal permeability and no prescribed singular surface current, and continuous tangential electric field. The normal electric current density is zero at an insulating wall; it is automatic for the present two-dimensional fields with current only along $z$. The induced fields decay far from the wire and wall, and initial conditions, or selection of the periodic state after transients, complete the specification. In the limiting perfect conductor approximation a surface current can emerge, permitting a tangential-field jump.

The fully coupled solution need not remain monochromatic: a harmonic magnetic field produces both a mean and a twice-frequency Lorentz force density, and the resulting velocity can generate further harmonics. The following parts consistently use the leading magnetic response with fluid motion neglected.

The Biot-Savart law gives the complex amplitude of the field from the real wire asAn oppositely directed image current $-J$ at $(0,b)$ givesThe method of images makes the normal oscillating magnetic field vanish at $y=0$, as required when the alternating component cannot penetrate a perfect conductor. The two tangential components reinforce one another there. ConsequentlyOnly the oscillating field is being excluded. A pre-existing steady magnetic field is not erased by the perfect conductor limit.

Let $\delta =\sqrt{2\eta /\omega }$ and $q=(1+i)/\delta$. Within the magnetic skin layer, differentiation across the layer dominates differentiation along the wall by the factor $b/\delta$. The leading resistive induction equation is thereforeBecause $\eta q^2=i\omega$ and $\mathrm{Re}q>0$, its decaying solution isThe solenoidal magnetic-field constraint determines the first smaller component,Corrections to $B_x$ from longitudinal diffusion have relative order $(\delta /b{)}^2$. Thus the zero normal boundary field from the perfect conductor calculation is a leading-order condition, not an additional exact condition at finite $\eta$. A smaller exterior-field correction matches $B_y$.

To leading order the electric current density is $j_z=qC{e}^{-qy}/{\mu }_0$. For two real harmonic fields, the period average of their product is half the real part of the product of one amplitude with the conjugate of the other. Thus the mean Lorentz force in a magnetic skin layer isIts normal component isThe first possible tangential contribution from the smaller $B_y$ vanishes: it is $-C{C}^{\prime }{e}^{-2y/\delta }\mathrm{Re}(q/q^{\ast })/(2{\mu }_0)$, and $q/q^{\ast }=i$. This cancellation matters in computing the tangential streaming flow.

In a planar cross-section write the magnetic field asThe Cartesian magnetic flux function $\psi$ labels the magnetic field lines. Ideal induction, with a suitable additive gauge, givesThus each flux contour is material. Incompressibility also preserves the area inside any material contour. Neither a magnetic island's flux distribution nor the connections of a separatrix can be freely altered during relaxation.

A configuration with two or more magnetic islands separated by an X-type saddle illustrates the mechanism. As the magnetic islands reshape to lower magnetic energy, the arms of the separatrix can be pressed together. Magnetic reconnection would change the topology, so it is excluded in a perfect conductor. The X-type structure can instead flatten into an extended interface, with oppositely directed tangential fields on its two sides. The transition shrinks and the electric current density concentrates into a current sheet. Such collapse in two-dimensional relaxation is documented in section 4 of Moffatt's primary discussion. It is a possible asymptotic outcome, not a finite-time singularity assertion or a property of every planar field.

The local equations show precisely how the interface can support equilibrium. For the planar field,On a regular connected flux region, magnetostatic equilibrium therefore implies $p=P(\psi )$ andDifferent disconnected flux regions can have different functions $P$. Their limiting fields need not join with continuous first derivatives of $\psi$ across a common separatrix; continuity of total pressure allows the tangential-field jump.

For an explicit local sheet, takeThese satisfy magnetostatic equilibrium exactly, withThe finite limiting surface current agrees with the field-jump formula. This is a local equilibrium demonstration of the sheet balance, not a claim that this particular family is the full ideal relaxation trajectory from the given initial data.

Strictly planar fields of the form above have zero magnetic helicity: choose $\mathbf{A}=\psi {\mathbf{e}}_z$, giving $\mathbf{A}\cdot \mathbf{B}=0$. Thus this strictly planar example interprets the last part as a separate illustration of topology-constrained relaxation. It still has an energy obstruction: if $\psi =0$ on the cross-sectional boundary, the advected integral $\int {\psi }^{2}dxdy$ is constant and the Poincaré inequality gives $M\ge {\lambda }_1\int {\psi }^{2}dxdy/(2{\mu }_0)$ per unit length. If the earlier nonzero-helicity condition is retained instead, a possible extension is a field depending on only two coordinates but permitting an axial component, with periodic boundary conditions in the invariant direction. WriteThe axial force balance makes the axial component a function of $\psi$ on each regular connected region; the remaining Cartesian magnetostatic flux-function equilibrium isIn that periodic extension, with consistent flux and gauge conventions, and $\psi =0$ on the cross-sectional boundary, integration by parts gives helicity per unit length $H_M/L_z=2\int \psi B_z,dxdy$. It can be nonzero. The same separatrix-collapse mechanism still concentrates current in a sheet. Three-dimensional linkage is therefore unnecessary for sheet formation; planar frozen flux-contour topology already supplies the constraint.

In vacuum without sources, Maxwell equations are $\nabla \cdot E=\nabla \cdot B=0$, $\nabla \times E=-\dot{B}$, and $\nabla \times B=\dot{E}/c^2$. For $E^{\prime }=cB$, $B^{\prime }=-E/c$, both divergences remain zero, andThis proves the quarter-turn electromagnetic duality transformation.

A perfect conductor has zero electric field in its bulk. Continuity of tangential electric field therefore gives $n\times E=0$ at its surface. Faraday's law makes a nonzero-frequency magnetic field vanish in that bulk, and continuity of its normal component gives $n\cdot B=0$ for the wave field. An independently trapped static magnetic flux is not excluded by perfect conductivity alone; the stated condition applies to the oscillating field under consideration.

For amplitudes with dependence $e^{i(kz-\omega t)}$, let ${\gamma }^2={\omega }^2/c^2-k^2$ and suppose $({\Delta }_{\perp }+{\gamma }^2)\psi =0$. With $k\ne 0$ and $\omega \ne 0$, define the transverse-magnetic family of scalar-potential conducting waveguide modes byThe electric divergence is ${\Delta }_{\perp }\psi +ik{e}_z=0$; the magnetic divergence is zero because the transverse field is a rotated gradient. Direct differentiation gives $\nabla \times e=i\omega b$ and $\nabla \times b=-i\omega e/c^2$, where longitudinal differentiation means multiplication by $ik$. Thus all vacuum equations hold, with $\boxed{{\gamma }^2={\omega }^2/c^2-k^2}$.

On a wall parallel to $z$, $\psi =0$ makes both $e_z$ and the boundary-tangential derivative of $\psi$ vanish. It also makes $n\cdot b$ proportional to that tangential derivative, hence zero. This proves the Dirichlet boundary condition supplies both conductor conditions.

For a transverse-electric family, use duality and rescale the potential, or check directly thatsatisfies the same equations. If $t=\hat{z}\times n$ is the cross-sectional wall tangent, then $t\cdot e=n\cdot {\nabla }_{\perp }\psi$ and $n\cdot b=-(k/\omega )n\cdot {\nabla }_{\perp }\psi$. Consequently $\boxed{{\partial }_n\psi =0}$, a Neumann boundary condition, supplies both wall conditions. Duality exchanges the field constructions, but does not preserve electric-conductor boundary conditions without this new boundary test.

In SI units, the source-free Maxwell equations in vacuum areUse complex amplitudes with phase $e^{i(\omega t-k\cdot x)}$ and take real parts at the end. The electric divergence equation gives $k\cdot E_0=0$. Faraday's equation gives $k\times E_0=\omega B_0$, soThe magnetic divergence equation follows because $k\cdot (k\times E_0)=0$. Ampere's equation becomes $k\times B_0=-{\mu }_0{ϵ}_0\omega E_0$. Since $k\times (k\times E_0)=-|k{|}^2{E}_0$, a nonzero wave satisfiesThus both fields are transverse, mutually perpendicular for linear polarization, and obey the vacuum plane-wave dispersion relation.

For the interface let $n=e_z$ point from the minus side to the plus side and write $[E]=E_{+}-E_{-}$, similarly for $B$. Applying the integral Maxwell equations to a thin pillbox and a thin loop gives the electromagnetic boundary conditionsHere $j$ is the tangential surface current. The jump notation fixes the signs, including the direction of the normal.

For the perfect conductor, take the time-varying interior fields to vanish. Choose positive $k$ with $\omega =ck$. The incident electric field has complex form $E_0{e}_x{e}^{i\omega t+ikz}$. To cancel its tangential value at the surface, use the reflected field $-E_0{e}_x{e}^{i\omega t-ikz}$, whose wavevector points upward. The total vacuum fields are thereforeThe sign of each magnetic contribution follows from $B_0=k\times E_0/\omega$. At $z=0$, the electric field is zero and the magnetic field is tangential. The jump conditions yieldFor a real incident amplitude this is $j=(2E_0/{\mu }_{0}c)\cos (\omega t)e_x$. It supplies the reflected wave while satisfying the normal-field and tangential-electric conditions. Any independently imposed static interior field is separate from this oscillatory response.

In electrostatic equilibrium, the electric field inside a perfect conductor is zero. Apply the Faraday's law $\oint \mathbf{E}\cdot d\mathbf{l}=0$ to a thin rectangular loop crossing the surface, with its long sides tangential and its short sides shrinking to zero. It yields continuity of the tangential electric field. Since its interior value is zero, $E_x(0+)=E_y(0+)=0$. Earthing fixes the electric potential at the plane to zero. For completeness, a thin Gauss law pillbox gives $E_z(0+)=\sigma /{\epsilon }_0$; the normal component can be nonzero because of induced surface charge. These are the electrostatic boundary conditions at a conductor.

The vacuum Maxwell equations are $\nabla \cdot E=\nabla \cdot B=0$, $\nabla \times E=-B_t$ and $\nabla \times B=E_t/c^2$. Taking the curl of Faraday's equation and using $\nabla \times (\nabla \times E)=\nabla (\nabla \cdot E)-{\nabla }^{2}E$ gives $-{\nabla }^{2}E=-E_{tt}/c^2$. The same calculation for Ampère's equation gives ${\nabla }^{2}B=B_{tt}/c^2$. Hence each Cartesian component satisfies the electromagnetic wave equation.

Model the walls as a perfect conductor, with no oscillating fields in its interior. The tangential electric field is zero at the surface, otherwise it would drive unbounded conductor currents. The normal magnetic field is zero, since the divergence-free field has a continuous normal component and the interior oscillating field is zero. Thus $n\times E=0$ and $n\cdot B=0$. Normal $E$ and tangential $B$ may be nonzero because surface charge and surface current support their jumps.

The PDF writes the full magnetic field as $B=B_0(x,y)\hat{z}{e}^{i(kz-\omega t)}$. Taken literally, this cannot be a nonzero travelling wave: $\nabla \cdot B=ik{B}_z$, so for $k\ne 0$ it forces $B_z=0$. The subsequent requests require transverse components. The consistent intended reading is $B_z=B_0(x,y)e^{i(kz-\omega t)}$, with transverse fields determined by the Maxwell equations. Use that reading below.

All fields have the same $z,t$ factor. Put $\Delta =(\omega /c{)}^2-k^2$. The harmonic Maxwell equations are $\nabla \times E=i\omega B$ and $\nabla \times B=-i\omega E/c^2$. Their $x$ and $y$ components givetogether with ${\partial }_y{E}_z-ik{E}_y=i\omega B_x$ and $ik{E}_x-{\partial }_x{E}_z=i\omega B_y$. Substitution and solving, for $\Delta \ne 0$, givesThe corresponding electric components areFor a transverse electric mode of a rectangular waveguide, $E_z=0$. The electromagnetic wave equation becomesAt $x=0,a$, the tangential component $E_y=-i\omega {\partial }_x{B}_z/\Delta$ vanishes, so ${\partial }_x{B}_0=0$; at $y=0,b$, the corresponding condition $E_x=0$ gives ${\partial }_y{B}_0=0$. These are Neumann boundary conditions, and the normal-$B$ conditions give the same conclusions through the displayed transverse magnetic components. Separation giveswhere $m,n\ge 0$ are integers and $(m,n)\ne (0,0)$. The constant candidate is not a physical oscillating mode: its transverse derivatives vanish, so it has $E=0$ and Faraday's equation forces $B=0$ for $\omega \ne 0$. More generally a hollow single-conductor guide has no transverse-electromagnetic zero-cutoff mode: its transverse electrostatic potential is harmonic with the same constant boundary value, hence constant, leaving zero transverse field.

The dispersion relation is ${\omega }^2=c^2(k^2+{\kappa }_{mn}^2)$. The waveguide cutoff frequency for this mode is ${\omega }_{mn}=c{\kappa }_{mn}$; below it $k$ is imaginary and the field is evanescent. Since $a\ge b$, the lowest cutoff occurs at $(m,n)=(1,0)$:For a mode propagating in the positive $z$ direction ($k>0$), above its own cutoff ${\omega }_{mn}$, the phase velocity and group velocity areThese waveguide phase and group velocities have product $c^2$. The superluminal phase speed describes motion of a constant phase, while a narrow wave packet travels at the subluminal group velocity; it does not transmit information faster than light. At cutoff $k=0$ there is no travelling mode with nonzero longitudinal flux.

In a source-free vacuum, Maxwell equations areTaking a curl of the last two equations and using $\nabla \times (\nabla \times \mathbf{A})=\nabla (\nabla \cdot \mathbf{A})-{\nabla }^2\mathbf{A}$ gives the electromagnetic wave equationsSubstitution of the complex plane wave amplitudes into Maxwell equations, then taking real parts, givesFor a nonzero propagating wave these imply the dispersion relation $\boxed{{\omega }^2=c^2|\mathbf{k}{|}^2}$. Conversely choose any nonzero transverse complex $\mathbf{e}$ and set $\mathbf{b}=\mathbf{k}\times \mathbf{e}/\omega$ to obtain such a wave. The static uniform-field sector at $\omega =|\mathbf{k}|=0$ is separate from propagation.

At a perfect conductor, the oscillatory boundary conditions are $\mathbf{n}\times \mathbf{E}=0$ and $\mathbf{n}\cdot \mathbf{B}=0$. Surface charge and surface current allow nonzero normal electric and tangential magnetic fields. The magnetic condition here concerns the time-dependent wave field; a perfect conductor can retain a pre-existing static magnetic field.

Let $R=I-2\mathbf{n}{\mathbf{n}}^T$ be the orthogonal reflection across the boundary plane, so $\det R=-1$. The proposed reflected amplitudes are ${\mathbf{k}}^{\prime }=R\mathbf{k}$, ${\mathbf{e}}^{\prime }=-R\mathbf{e}$, and ${\mathbf{b}}^{\prime }=R\mathbf{b}$. For an orthogonal reflection, $(R\mathbf{a})\times (R\mathbf{d})=-R(\mathbf{a}\times \mathbf{d})$. HenceAlso $|{\mathbf{k}}^{\prime }|=|\mathbf{k}|$, so the reflected wave obeys the same dispersion relation. At $\mathbf{n}\cdot \mathbf{x}=0$, the incident and reflected phases coincide because ${\mathbf{k}}^{\prime }-\mathbf{k}$ is parallel to $\mathbf{n}$. Their summed amplitudes satisfyThe first has zero tangential part and the second zero normal part, proving both boundary conditions. Finally ${\mathbf{k}}^{\prime }\cdot \mathbf{n}=-\mathbf{k}\cdot \mathbf{n}>0$, as required for reflection into the vacuum.

In a vacuum with no charge or current, the Maxwell equations areFor the proposed plane electromagnetic wave, $\nabla \cdot B=0$ givesThe Ampère-Maxwell equation givesand substituting this into Faraday's law gives the dispersion relationThus both fields have transverse polarization relative to $k$, and the corresponding real electric field is

For incidence in the positive $x$-direction, choose the incident fieldsThe perfect conductor requires the tangential electric field to vanish at $x=0$. The reflected wave therefore hasThis is normal reflection of an electromagnetic wave from a perfect conductor. The magnetic field in $x\le 0$ isso its tangential value at the surface is

To first order in the applied fields, the carrier acceleration gives $m\dot{\mathbf{v}}=q\mathbf{E}$. The current density $\mathbf{j}=nq\mathbf{v}$ consequently obeys the first London equationTaking its curl and using Faraday's law yieldsFor an initially field-free, current-free preparation, the integration constant is zero. Equivalently, impose the second London equation appropriate to the Meissner state, $\nabla \times \mathbf{j}=-{ϵ}_0{\omega }_p^2\mathbf{B}$. In the final static state, Ampère's law and $\nabla \cdot \mathbf{B}=0$ then giveFor a planar half-space $z>0$ with a tangential surface field, the bounded solution isThe plasma-frequency form of London penetration depth therefore shows that an applied field is exponentially small several penetration depths inside the material.

The magnetic-history constant in an undamped plasma explains the required qualification. Undamped carrier motion alone fixes a time-independent value of $\nabla \times \mathbf{j}+{ϵ}_0{\omega }_p^2\mathbf{B}$; it does not force that value to vanish. A perfect conductor can retain a pre-existing field. The additional London equation describes London screening independently of that arbitrary ideal-conductor history. The original London paper explicitly distinguishes its description from acceleration theory.^[ref]

For the usual hollow rectangular electromagnetic waveguide, let the propagation axis be $z$ and the interior be $0<x<a$, $0<y<b$. A transverse magnetic polarization has $B_z=0$ and $E_z\ne 0$. At a perfect conductor wall the tangential electric field vanishes. Since $E_z$ is tangential to all four side walls, it satisfies a homogeneous Dirichlet boundary condition around the rectangle.

Separating the transverse Helmholtz equation therefore gives the transverse magnetic mode of a rectangular waveguideA zero value of either index makes $E_z$ identically zero, rather than producing a nonzero field uniform along that direction. The transverse electric and magnetic fields of this TM family are obtained from transverse derivatives of $E_z$, so they also vanish. Thus both TM indices must be nonzero: neither ${\mathrm{TM}}_{n0}$ nor ${\mathrm{TM}}_{0m}$ exists in this rectangular hollow-guide model. In contrast, transverse electric modes obey a Neumann boundary condition on $B_z$ and may have one zero index. The claim is about this waveguide geometry, not all possible guides.

Equilibrium radiation and cavity frequencies. A blackbody absorbs all incident radiation at every frequency and direction. In thermal equilibrium its emission spectrum depends only on its temperature. Radiation inside an isothermal absorbing cavity, sampled through a small aperture that hardly disturbs equilibrium, realizes black-body radiation to a good approximation. Microscopic absorption and emission establish the equilibrium distribution; a perfectly reflecting wall is the convenient idealization used to count the modes once that distribution is specified.

In a rectangular electromagnetic cavity, the wave equation and perfect conductor wall boundary conditions select component wavenumbers $k_x=\pi l/L_x$, $k_y=\pi m/L_y$, $k_z=\pi n/L_z$. Combining the vacuum dispersion relation $\omega =c|\mathbf{k}|$ and $\nu =\omega /(2\pi )$ givesFor example, electric-field components can have the spatial dependenceFor three positive indices, Gauss's law requires $k_x{A}_x+k_y{A}_y+k_z{A}_z=0$, leaving two independent polarizations of an electromagnetic wave.

Index qualification. The original paper describes the indices as positive integers. That describes the three-nonzero-index family but omits valid electromagnetic modes with exactly one zero index, which have one polarization of an electromagnetic wave. For instance, with $n=0$, $l,m\ge 1$, only $E_z$ above remains; it is normal to the top and bottom faces and vanishes at the side walls. Two zero indices give no nontrivial cavity field. The omitted one-zero-index family does not affect the leading three-dimensional mode count, but is essential to the thin-cavity limit below. Retaining the paper's strict positivity restriction for every mode would instead give a thickness gap and would not yield its requested two-dimensional power law.

Three-dimensional mode count. A lattice cell in positive $\mathbf{k}$ space has volume ${\pi }^3/V$, where $V=L_x{L}_y{L}_z$. In a continuum approximation, a shell of radius $k$ and thickness $dk$ contains its positive octant, multiplied by two polarizations:Using $k=2\pi \nu /c$ givesFor a finite cavity the exact density of states is discrete; this formula is its bulk continuum limit, appropriate when many modes contribute and wall corrections are small.

Mean photon number. A mode is a quantum harmonic oscillator with thermal excitation energies $sh\nu$, $s=0,1,2,\dots$. The zero-point energy does not enter its thermal occupation. Put $q=e^{-h\nu /(k_{B}T)}$. Its partition function and mean occupation areThis is the Bose-Einstein distribution with zero photon chemical potential: equilibrium absorption and emission do not conserve photon number.

Thermal energy. Multiply the density of states by the mean excitation energy $h\nu ⟨s⟩$ and integrate:The substitution $x=h\nu /(k_{B}T)$ isolates all the dependence on cavity dimensions and temperature:Expanding $1/(e^x-1)={\sum }_{r=1}^{\infty }{e}^{-rx}$ gives an integral ${\sum }_{r}6/r^4=6\zeta (4)={\pi }^4/15$. HenceThis is the thermal energy only, excluding the zero-point contribution. The three-dimensional continuum regime requires $k_{B}T$ large compared with the geometric level spacings in all three directions.

Spectral sketch. The spectral energy density per unit ordinary frequency is Planck's law,At small frequency it rises as $8\pi k_{B}T{\nu }^2/c^3$, while at large frequency it decays exponentially. Raising temperature increases the spectrum at every fixed positive frequency, moves the peak to larger frequency in proportion to $T$, and increases the integrated area in proportion to $T^4$. A plot per unit wavelength has a different peak position because the integration variable changes.

Thin-cavity radiation. Let $A=L_x{L}_y$. The relevant low-temperature window isNonzero thickness indices are then thermally suppressed, while many in-plane modes are populated. The $n=0$ family has one polarization of an electromagnetic wave. Its allowed in-plane wavenumbers form a lattice of cell area ${\pi }^2/A$; a quarter of a circular shell therefore containsThus the dimensional crossover of thermal cavity radiation givesThe integral is $2\zeta (3)$ by the same geometric-series argument. A model with a different number of independent polarizations of an electromagnetic wave changes the prefactor but not the $T^3$ scaling. For a cavity with fixed finite $L_x,L_y$, cooling still further eventually freezes its lowest in-plane mode as well: the strict $T\to 0$ thermal energy is exponentially suppressed. The requested area law is the intermediate quasi-two-dimensional regime, not an arbitrarily low-temperature continuum limit.