---
title: London Theory and the Meissner Effect
module: Superconductivity
moduleNumber: 10
lessonNumber: 2
order: 1002
summary: >
  A perfect conductor freezes the field it was cooled in; a superconductor
  expels it. The distinction needs a constitutive law beyond zero resistance —
  the two London equations — whose solution is exponential flux decay over the
  penetration depth. The same rigidity follows from a macroscopic condensate
  wave function, and the thermodynamics of the critical field fixes the
  condensation energy, the latent heat, and the specific-heat jump.
topics: [Superconductivity]
draft: false
sources:
  - book: Kittel
    ref: "Ch. 10 — Superconductivity"
  - book: Ashcroft & Mermin
    ref: "Ch. 34 — Superconductivity"
  - book: Hook & Hall
    ref: "Ch. 12 — Superconductivity"
---

The [phenomenology](/condensed-matter/superconductivity/superconductivity-phenomenology)
established two independent facts: below $T_c$ the resistance is exactly zero,
and the magnetic flux is expelled. Zero resistance alone does not produce flux
expulsion. A theory of the superconducting state must add a magnetic law to the
electrical one, and the London brothers supplied the simplest such law in 1935.
Its single new length scale, the penetration depth, is measured directly and
fixes the density of the current-carrying electrons.

## Perfect conductivity versus perfect diamagnetism

Take the electrical property at face value. Inside a perfect conductor a steady
electric field would drive an ever-increasing current, so in the steady state
$\vec E = 0$. Faraday's law then freezes the magnetic field,

$$
\nabla\times\vec E = -\frac{\partial \vec B}{\partial t} = 0
\quad\Longrightarrow\quad
\frac{\partial \vec B}{\partial t} = 0 .
$$

The interior field cannot change, but it can hold whatever value it had when the
sample became perfectly conducting. Cool a perfect conductor in an applied field
and the field stays trapped; cool it in zero field and then apply a field and the
interior stays field-free. History decides the outcome.

A superconductor behaves differently. Cooled through $T_c$ in a field, it expels
that field: the interior reaches $\vec B = 0$ regardless of the path taken to get
there. The final state is unique. This is a stronger statement than
$\partial\vec B/\partial t = 0$, and no purely electrical property implies it.

$$
% caption: Field-cooling contrast. A perfect conductor cooled in a field traps
% the flux it was cooled in; a superconductor cooled in the same field expels it,
% reaching B = 0 inside no matter what field it was cooled in.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  % --- perfect conductor: flux trapped ---
  \begin{scope}
    \draw[black, thick, fill=black!5] (0,0) circle (0.95);
    \foreach \y in {-0.55,-0.18,0.18,0.55}
      \draw[acc, ->] (-1.9,\y) -- (1.9,\y);
    \node[anchor=south] at (0,1.15) {perfect conductor};
    \node[anchor=north, black, font=\scriptsize] at (0,-1.2) {trapped};
  \end{scope}
  % --- superconductor: flux expelled ---
  \begin{scope}[xshift=5.4cm]
    \draw[acc, thick, fill=acc!10] (0,0) circle (0.95);
    \draw[acc, ->] (-1.9,0.55) .. controls (-0.7,0.55) and (-0.1,1.15) .. (0,1.35);
    \draw[acc, ->] (-1.9,-0.55) .. controls (-0.7,-0.55) and (-0.1,-1.15) .. (0,-1.35);
    \draw[acc, ->] (-1.9,0.18) .. controls (-0.85,0.18) and (-0.9,1.0) .. (0.8,1.1);
    \draw[acc, ->] (-1.9,-0.18) .. controls (-0.85,-0.18) and (-0.9,-1.0) .. (0.8,-1.1);
    \node[acc, font=\scriptsize] at (0,0) {$B = 0$};
    \node[anchor=south] at (0,1.5) {superconductor};
    \node[anchor=north, black, font=\scriptsize] at (0,-1.55) {expelled};
  \end{scope}
\end{tikzpicture}
$$

## The London equations

Model the superconducting electrons as a fraction with number density $n_s$ that
moves without dissipation. Each carries charge $-e$ and mass $m$ and accelerates
freely under a field,

$$
m\frac{\d\vec v_s}{\d t} = -e\vec E .
$$

With supercurrent density $\vec J_s = -n_s e\,\vec v_s$, this is the **first
London equation**,

$$
\frac{\partial \vec J_s}{\partial t} = \frac{n_s e^2}{m}\,\vec E .
$$

It describes perfect conductivity: a static field is not needed to sustain a
current, and any residual field accelerates the carriers. Taking the curl and
inserting Faraday's law $\nabla\times\vec E = -\partial\vec B/\partial t$,

$$
\frac{\partial}{\partial t}\left(\nabla\times\vec J_s\right)
  = -\frac{n_s e^2}{m}\,\frac{\partial\vec B}{\partial t}
\quad\Longrightarrow\quad
\frac{\partial}{\partial t}\!\left(\nabla\times\vec J_s + \frac{n_s e^2}{m}\vec B\right) = 0 .
$$

A perfect conductor fixes only the time derivative of the bracketed quantity, so
it permits any static field. The Londons imposed the stronger condition that the
bracket vanish outright, not merely its rate of change. That choice is the
**second London equation**,

$$
\nabla\times\vec J_s = -\frac{n_s e^2}{m}\,\vec B .
$$

> **Definition (London equations).** The superconducting electrons obey
> $\partial\vec J_s/\partial t = (n_s e^2/m)\vec E$ (perfect conductivity) and
> $\nabla\times\vec J_s = -(n_s e^2/m)\vec B$ (perfect diamagnetism). The second
> is a constitutive postulate, selected among the solutions of the perfect
> conductor because it alone reproduces the Meissner effect.

## The penetration depth

Combine the second London equation with the static Ampère law
$\nabla\times\vec B = \mu_0\vec J_s$ and $\nabla\cdot\vec B = 0$. Taking the curl
of Ampère's law,

$$
\nabla\times(\nabla\times\vec B) = \mu_0\,\nabla\times\vec J_s
  = -\frac{\mu_0 n_s e^2}{m}\,\vec B ,
$$

and using $\nabla\times(\nabla\times\vec B) = \nabla(\nabla\cdot\vec B) - \nabla^2\vec B
= -\nabla^2\vec B$, the field obeys a screened equation,

$$
\nabla^2\vec B = \frac{1}{\lambda_L^2}\,\vec B ,
\qquad
\lambda_L = \sqrt{\frac{m}{\mu_0 n_s e^2}} .
$$

For a superconductor filling the half-space $x>0$ with a field $B(0)$ applied
parallel to the surface, the bounded solution decays exponentially,

$$
B(x) = B(0)\,e^{-x/\lambda_L} .
$$

The **London penetration depth** $\lambda_L$ is the distance over which an
external field falls to $1/e$ of its surface value. For typical metals
$n_s \sim 10^{28}\text{–}10^{29}\ \text{m}^{-3}$ gives $\lambda_L \approx
20\text{–}50\ \text{nm}$. The field is not truly zero at the surface; it is
excluded from the bulk by screening currents flowing in the same surface layer,
since $\vec J_s$ satisfies the identical equation $\nabla^2\vec J_s =
\lambda_L^{-2}\vec J_s$.[^kittelpen]

[^kittelpen]: Kittel, Ch. 10, "London equation" and "penetration depth."

$$
% caption: An applied field parallel to the surface penetrates only a skin of
% thickness lambda_L, decaying as B(0) e^{-x/lambda}; the screening supercurrents
% occupy the same layer.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  % vacuum / superconductor divide
  \fill[acc!7] (0,0) rectangle (6.4,3.0);
  \draw[black, thick] (0,0) -- (0,3.0);
  \node[anchor=south east, black, font=\scriptsize] at (-0.05,2.7) {vacuum};
  \node[anchor=south west, acc, font=\scriptsize] at (0.15,2.7) {superconductor};
  % axes
  \draw[->, black] (0,0) -- (6.4,0) node[below] {$x$};
  \draw[->, black] (0,0) -- (0,3.3) node[left] {$B$};
  % decay curve
  \draw[acc, very thick, domain=0:6.0, samples=80, variable=\x]
    plot ({\x},{2.6*exp(-\x/1.4)});
  \node[acc, anchor=west] at (2.1,1.05) {decay};
  % lambda marker at x = 1.4 where B falls to 1/e
  \draw[black, dashed] (1.4,0) -- (1.4,{2.6*exp(-1)});
  \node[anchor=north, font=\scriptsize] at (1.4,0) {penetration depth};
\end{tikzpicture}
$$

## The rigid condensate wave function

The second London equation reads as an ad-hoc postulate, but it follows from a
single macroscopic wave function for the superconducting condensate,
$\psi = \sqrt{n_s}\,e^{i\theta(\vec r)}$, whose carriers have charge $q$ and mass
$m^\ast$. The gauge-invariant current of such a wave function is

$$
\vec J_s = \frac{n_s q}{m^\ast}\left(\hbar\nabla\theta - q\vec A\right) ,
$$

where $\vec A$ is the vector potential, $\vec B = \nabla\times\vec A$. Taking the
curl removes the phase gradient, since $\nabla\times\nabla\theta = 0$, and leaves

$$
\nabla\times\vec J_s = -\frac{n_s q^2}{m^\ast}\,\vec B ,
$$

the second London equation with $q\to -e$, $m^\ast\to m$. The physical content is
**rigidity**: the condensate phase resists deformation, so an applied field
generates exactly the screening current that cancels it in the interior. When the
pairing charge $q = -2e$ and mass $m^\ast = 2m$ are used, as
[Ginzburg–Landau theory](/condensed-matter/superconductivity/ginzburg-landau-theory)
requires, the penetration depth becomes $\lambda_L = \sqrt{m^\ast/(\mu_0 n_s q^2)}
= \sqrt{m/(\mu_0 n_p (2e)^2/2)}$ with $n_p = n_s/2$ the pair density, numerically
unchanged.[^amrigid]

[^amrigid]: Ashcroft & Mermin, Ch. 34, on the London rigidity and the macroscopic wave function.

## Thermodynamics of the transition

The superconducting state is the thermodynamically stable phase below $T_c$
because its Gibbs free energy is lower. Expelling a field costs energy, so the
free energy of the superconductor rises with applied field $H_a$. Per unit
volume, holding out a uniform field costs the magnetic work density
$\tfrac{1}{2}\mu_0 H_a^2$,

$$
g_s(T, H_a) = g_s(T, 0) + \tfrac{1}{2}\mu_0 H_a^2 .
$$

The normal-state free energy is nearly field-independent (its susceptibility is
tiny). At the critical field the two phases are in equilibrium,
$g_s(T, H_c) = g_n(T)$, which fixes the **condensation energy** — the free-energy
gain of the superconducting state at zero field,

$$
g_n(T) - g_s(T, 0) = \tfrac{1}{2}\mu_0 H_c^2(T) .
$$

The critical field is measured to follow a near-parabolic law,

$$
H_c(T) = H_c(0)\left[1 - \left(\frac{T}{T_c}\right)^2\right] ,
$$

which separates the superconducting and normal regions of the $(T, H)$ plane.

$$
% caption: The critical-field phase boundary. Below the parabola H_c(T) the
% material is superconducting; above it, normal. H_c falls from H_c(0) at T = 0
% to zero at T_c.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, black] (0,0) -- (6.4,0) node[below] {$T$};
  \draw[->, black] (0,0) -- (0,3.4) node[left] {$H$};
  % parabola H = H0 (1 - (T/Tc)^2), Tc at x=5.4, H0 at y=3
  \draw[acc, very thick, domain=0:5.4, samples=80, variable=\x]
    plot ({\x},{3.0*(1 - (\x/5.4)*(\x/5.4))});
  \fill[acc!10]
    plot[domain=0:5.4, samples=80, variable=\x] ({\x},{3.0*(1 - (\x/5.4)*(\x/5.4))})
    -- (0,0) -- cycle;
  \node[acc, anchor=west] at (0.6,1.0) {superconducting};
  \node[black, anchor=west] at (2.6,2.4) {normal};
  \draw[black, dashed] (5.4,0) -- (5.4,0.1);
  \node[anchor=north, font=\scriptsize] at (5.4,0) {$T_c$};
  \node[anchor=east, font=\scriptsize] at (-0.05,3.0) {$H_c(0)$};
\end{tikzpicture}
$$

The entropy of each phase is $S = -\partial g/\partial T$. Differentiating the
condensation-energy relation,

$$
S_n - S_s = -\mu_0 H_c\,\frac{\d H_c}{\d T} .
$$

Because $\d H_c/\d T < 0$, the superconducting state has the lower entropy: it is
the more ordered phase, as an ordered condensate should be. The **latent heat**
of the field-driven transition is

$$
L = T(S_n - S_s) = -\mu_0 T\,H_c\,\frac{\d H_c}{\d T} .
$$

At $T_c$ in zero field $H_c = 0$, so $L = 0$: the zero-field transition releases
no latent heat and is second order. In a finite field, where the transition
occurs at $T < T_c$ with $H_c \neq 0$, latent heat appears and the transition is
first order.

Differentiating once more gives the heat capacities, $C = T\,\partial S/\partial T$,

$$
C_s - C_n = \mu_0 T\left[\left(\frac{\d H_c}{\d T}\right)^2
  + H_c\,\frac{\d^2 H_c}{\d T^2}\right] .
$$

At $T_c$ the second term drops ($H_c = 0$) and the heat capacity jumps by a finite
amount even though the latent heat vanishes — the signature of a second-order
transition. This is **Rutgers's formula**,

$$
\left.\Delta C\right|_{T_c} = C_s - C_n = \mu_0 T_c\left(\frac{\d H_c}{\d T}\right)^2_{T_c} .
$$

> **Worked example.** From the parabolic law, $\d H_c/\d T = -2H_c(0)\,T/T_c^2$,
> so at $T_c$ the slope is $-2H_c(0)/T_c$. Rutgers's formula then predicts
>
> $$
> \Delta C = \mu_0 T_c\left(\frac{2H_c(0)}{T_c}\right)^2 = \frac{4\mu_0 H_c(0)^2}{T_c} .
> $$
>
> The superconducting heat capacity is the larger one at $T_c$. Below $T_c$ it
> falls faster than the normal $\gamma T$ line and, once the energy gap is fully
> open, drops exponentially as $e^{-E_g/2k_B T}$, direct evidence for a gap in
> the excitation spectrum.

$$
% caption: Heat capacity through the transition. The superconducting branch jumps
% above the normal gamma-T line at T_c, then falls exponentially below it as the
% gap freezes out excitations; the jump with no latent heat marks a second-order
% transition.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, black] (0,0) -- (6.4,0) node[below] {$T$};
  \draw[->, black] (0,0) -- (0,3.4) node[left] {$C$};
  % normal line gamma T (dashed), through Tc
  \draw[black, dashed] (0,0) -- (6.0,2.4);
  \node[black, anchor=west, font=\scriptsize] at (5.0,2.35) {normal};
  % superconducting: exponential rise below Tc up to a jump above the line at Tc
  \draw[acc, very thick, domain=0.2:4.5, samples=80, variable=\x]
    plot ({\x},{2.6*exp((\x-4.5)/1.1)});
  % jump at Tc (x=4.5): normal value there is 1.8, sc value ~2.6
  \draw[acc, very thick] (4.5,2.6) -- (4.5,1.8);
  \draw[black, dashed] (4.5,0) -- (4.5,1.8);
  \node[anchor=north, font=\scriptsize] at (4.5,0) {$T_c$};
  \draw[acc, ->] (5.0,2.9) -- (4.55,2.4);
  \node[acc, anchor=west, font=\scriptsize] at (5.0,2.95) {jump};
  \node[acc, anchor=east, font=\scriptsize] at (2.4,0.7) {superconducting};
\end{tikzpicture}
$$

## The intermediate state

The parabola describes a long, thin sample aligned with the field, where the
field at the surface equals the applied field. For any other shape the field
crowds around the sample and the local surface field exceeds $H_a$. A sphere has
demagnetizing factor $N = \tfrac{1}{3}$, so the field at its equator is
$H_a/(1-N) = \tfrac{3}{2}H_a$. The equator reaches $H_c$ while the applied field
is still only $\tfrac{2}{3}H_c$, and the sample cannot remain wholly
superconducting, yet the interior field is too low to go entirely normal.

The resolution is the **intermediate state**: the sample breaks into alternating
laminae of normal and superconducting material, each thin enough that the field
in the normal sheets is exactly $H_c$. This geometric subdivision of a type I
superconductor is distinct from the vortex mixed state of a type II material,
which arises from the sign of the surface energy between the phases and is
treated in the [next lesson](/condensed-matter/superconductivity/ginzburg-landau-theory).

$$
% caption: The intermediate state of a type I superconductor: alternating normal
% and superconducting laminae, with the field passing through the normal sheets
% at exactly H_c and expelled from the superconducting ones.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[black, thick] (0,0) rectangle (6.0,2.4);
  % superconducting laminae (tinted)
  \foreach \x in {0.4,1.6,2.8,4.0,5.2}{
    \fill[acc!12] (\x,0) rectangle (\x+0.8,2.4);
  }
  % normal laminae carry field arrows
  \foreach \x in {1.2,2.4,3.6,4.8}{
    \draw[acc, ->] (\x,-0.5) -- (\x,2.9);
  }
  \node[acc, anchor=north, font=\scriptsize] at (0.8,-0.15) {super};
  \node[black, anchor=north, font=\scriptsize] at (2.4,-0.15) {normal};
  \node[anchor=south, black, font=\scriptsize] at (3.0,3.0) {through normal sheets};
\end{tikzpicture}
$$

London theory captures the electrodynamics of the superconducting state with one
empirical constant, $\lambda_L$, and the thermodynamics with one measured
function, $H_c(T)$. It says nothing about why the state forms or how deep the
condensation energy is. The order parameter $\psi$ introduced here as a device
becomes the central object of the phenomenological theory taken up next, and the
microscopic origin of the rigidity waits for
[BCS theory](/condensed-matter/superconductivity/bcs-theory).