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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]

The Meissner effect is the equilibrium expulsion of a sufficiently weak, static magnetic field from the bulk of a material exhibiting superconductivity. Expulsion includes a magnetic field that was already present before cooling: zero electrical resistance alone does not predict this. The Field-cooled Meissner experiment tests that distinction. A finite London penetration depth means that the magnetic field and screening electric current occupy a surface layer, rather than disappearing at a mathematical boundary.^[ref]

This makes superconductivity useful for shielding weak magnetic fields, and makes its magnetic response an independent test of the phase. The striking result is that the same screened equilibrium can be reached by two different preparation histories. The assumptions matter: Finite-thickness Meissner screening can leave an appreciable magnetic field throughout a small sample, and the Limits of Meissner screening distinguish quantum vortex penetration and retained magnetic flux from this equilibrium state.

Use SI units and normalize the condensate wavefunction so $|\psi {|}^2=n$. With mechanical momentum operator $-i\hslash \nabla -qA$, the charge current density isFor uniform $n$ and $\psi =\sqrt{n}{e}^{i\vartheta }$, this becomes $j=(nq/m)(\hslash \nabla \vartheta -qA)$.

Under the gauge transformation $A\mapsto A+\nabla \chi$, the wavefunction transforms as $\psi \mapsto e^{iq\chi /\hslash }\psi$, so $\vartheta \mapsto \vartheta +q\chi /\hslash$. The two added gradients cancel in $\hslash \nabla \vartheta -qA$, while $|\psi {|}^2$ stays fixed. Thus the superconducting current is gauge invariant. The same cancellation in the full displayed current proves it without requiring a constant amplitude.

For uniform carrier density in a region with smooth single-valued phase, taking a curl gives the London equation $\nabla \times j=-n{q}^{2}B/m$. In the time-independent setting Ampère's circuital law gives $\nabla \times B={\mu }_{0}j$, and Gauss's law for magnetism gives $\nabla \cdot B=0$. Thereforeso the Helmholtz equation and London penetration depth areThe equation has the modified, exponentially decaying Helmholtz sign. For a planar surface at $x=0$ with material in $x>0$, the bounded solution is $B(x)=B(0)e^{-x/\ell }$; the exponentially growing solution is excluded in the bulk. Hence a macroscopic region far from the surface has negligible field: magnetic flux is expelled from the bulk, the Meissner effect. The literal assertion of zero flux everywhere is the bulk limit: finite penetration remains within distance $\ell$ of a surface. Vortices or flux trapped through a hole require different phase or topology assumptions and are not ruled out by this smooth-phase derivation.

Write the superconducting order parameter as $\psi =\sqrt{n_s}{e}^{i\vartheta }$, with carrier mass $m$. Minimal electromagnetic coupling gives mechanical momentum $\hslash \nabla \vartheta -q\mathbf{A}$, so the current density isIt also follows by substituting $\psi$ into $(q\hslash /m)\mathrm{Im}({\psi }^{\ast }\nabla \psi )-(q^2/m)|\psi {|}^2\mathbf{A}$. In a gauge where the phase is constant, this reduces to the London equation $\mathbf{j}=-n_s{q}^2\mathbf{A}/m$.

Under $\mathbf{A}\mapsto \mathbf{A}+\nabla \chi$, the phase changes by $\vartheta \mapsto \vartheta +q\chi /\hslash$. The two gradient changes cancel, proving gauge invariance of the actual current. A formula omitting the phase term is valid only in its chosen gauge.

For a uniform superconductor without vortices, taking a curl gives $\nabla \times \mathbf{j}=-n_s{q}^2\mathbf{B}/m$. Combine this with the stationary Ampère's law $\nabla \times \mathbf{B}={\mu }_0\mathbf{j}$ and $\nabla \cdot \mathbf{B}=0$ to obtainFor a magnetic field parallel to the planar interface, translation symmetry reduces the equation to ${\mathbf{B}}^{\prime \prime }=\mathbf{B}/{\lambda }_L^2$. Boundedness and decay as $z\to \infty$, together with the tangential boundary value, give the Meissner effectHere ${\lambda }_L$ is the London penetration depth.

The parallel-field assumption is necessary for this one-dimensional Meissner solution. If the printed arbitrary constant ${\mathbf{B}}_0$ has a nonzero normal component, $\nabla \cdot \mathbf{B}=0$ forces that component to be constant across a translation-invariant half-space, whereas the London equation forces it to vanish. Thus no decaying planar Meissner state with that boundary value exists; one must change the geometry or allow a different magnetic state.

In the Landau-Ginzburg theory functional, $B^2/(2{\mu }_0)$ is the magnetic field energy density; the covariant-gradient term is the carrier kinetic energy including phase gradients and electromagnetic coupling. The terms $\alpha |\psi {|}^2+\beta |\psi {|}^4$ give the local condensation energy. With the normalization in the question, $n=|\psi {|}^2$ is the carrier density.

For a uniform field-free equilibrium, minimize $\alpha n+\beta n^2$ over $n\ge 0$. Stability for large $n$ requires $\beta >0$, and a nonzero minimum requires $\alpha <0$. ThereforeThe factor two follows from the question's quartic coefficient $\beta$, rather than a convention using $\beta /2$.

Let $D_A=-i\hslash \nabla -qA$. Under the gauge transformation,Its squared magnitude is unchanged, $|\psi |$ is unchanged, and $B=\nabla \times A$ is unchanged. Thus the full energy has gauge invariance.

For uniform $n$, write $\psi =\sqrt{n}{e}^{i\varphi }$. The kinetic density is $n(\hslash \nabla \varphi -qA{)}^2/(2m)$. Varying $A$ with compactly supported variations and integrating the magnetic term by parts givesAt a minimum its coefficient vanishes, giving the London equationTaking another curl, using $\nabla \cdot B=0$ and a smooth phase so $\nabla \times \nabla \varphi =0$, yieldsFor a planar surface this has the decaying solution $B(x)=B(0)e^{-x/{\lambda }_L}$ inside the material; the growing solution is excluded by boundedness in the bulk. Hence magnetic fields penetrate only a surface layer of thickness ${\lambda }_L$ and vanish deep in the macroscopic bulk, the Meissner effect. This conclusion uses the uniform-density, smooth-phase setting; a vortex has a singular phase and a nonuniform core and is outside those assumptions.

Varying $A_{\nu }$ gives the Maxwell equationsOn the vacuum manifold, choose unitary gauge so that $\varphi =v$. Then $\mathbf{J}=-2e^2{v}^2\mathbf{A}$ and the London equation isFor a static configuration, $\nabla \times \mathbf{B}=\mathbf{J}$ and $\nabla \cdot \mathbf{B}=0$, soThe magnetic field therefore decays exponentially inside the condensate. This is the Meissner effect, with penetration depth equal to the inverse gauge-boson mass.

Superconductivity is a phase of matter that supports steady electric current with zero bulk electrical resistivity and exhibits the Meissner effect: equilibrium screening of a sufficiently weak, static magnetic field from the interior. It occurs below a material-dependent superconducting transition temperature, provided the electric current and applied magnetic field are also sufficiently small. ^[ref] Bulk expulsion means that a sample much thicker than its London penetration depth has an almost field-free interior; the magnetic field need not vanish at its surface. Conventional BCS theory explains superconductivity through correlated Electrons forming Cooper pairs. ^[ref]

The two observations answer different questions. A resistance-versus-temperature experiment for superconductivity asks whether a steady electric current requires a measurable electric potential difference. A Field-cooled Meissner experiment asks whether a magnetic field already present in the material is expelled when it enters superconductivity. An ideal perfect conductor with zero electrical resistivity can preserve that pre-existing magnetic field; zero electrical resistance alone therefore does not establish the Meissner effect.

The usefulness follows directly from Joule heating, $P=I^{2}R$: maintaining a steady electric current through a segment with zero electrical resistance produces no resistive heating in that segment. This makes persistent electric currents attractive for producing steady magnetic fields. The beauty is that one phase has both a transport property and an equilibrium magnetic response, and the second cannot be inferred just by making an ordinary electrical resistance very small. The London equations encode this extra magnetic response. ^[ref]