LondonModel

Maxwell equations

Here we show the constraints that Maxwell equations impose on magnetic field expulsion, leading to a finite length {$\lambda$}, required by supercurrents to shield the applied field. The core of this derivation is due to Heinz and Fritz London. The views of Fritz London on superconductivity were essential for the subsequent explanation by Bardeen, Cooper and Schrieffer, and also instrumental for a new, correct approach to the old issue of measure in quantum mechanics.

With boundary conditions at the sample edges and no static charges within, stationary Maxwell equations read:

{$$\begin{align*} \nabla\cdot \mathbf{ E}&=0 \qquad\qquad\qquad\qquad& \nabla\times\mathbf{ E}&= 0\\ \nabla\cdot\mathbf{B}&=0 \qquad\qquad\qquad\qquad& \nabla\times\mathbf{B}&= \mu_0\mathbf{j} \end{align*}$$}

and the electric field vanishes inside the metal. Only the last equation has sources, and we introduce the vector potential {$\mathbf{B}=\nabla\times\mathbf{A}$}, so that the LHS obeys the identity {$\nabla\times\mathbf B=\nabla\times\nabla\times \mathbf{ A}=\nabla(\nabla\cdot\mathbf{ A})-\nabla^2\mathbf{ A}$}. This expression simplifies with the choice of the Coulomb gauge ({$\nabla\cdot\mathbf{ A}=0$}), reducing to

{$$ \begin{equation} \nabla\times\mathbf B=-\nabla^2\mathbf A\end{equation}$$}

We will come baclk to this point.

Index

Geometrical considerations

The simplest geometry is that of an infinite superconducting semispace to the right, {$y>0$}, with an external induction field {$B_0\hat z$} from sources on the left (Fig. 1). It is easy to see that the solution to this problem is a supercurrent density

{$$\mathbf{j}(\mathbf{r})=-j_0 e^{- y /\lambda}\hat x$$}

for {$y>0$} and zero otherwise, with {$B_0=\mu_0\lambda j_0$}. The Ampere-Maxwell equation is satisfied with

{$$\mathbf{B}(\mathbf{r}) = B_0 e^ {-y/ \lambda} \hat z$$} as can be verified straightforwardly by calculating

{$$\left| \begin{matrix}\hat{x}&\hat{y}&\hat{z}\\ \partial_x&\partial_y & \partial_z \\ 0 & 0&B_0e^{-y/\lambda}\end{matrix}\right|$$}

This represents a surface current density decaying over a length {$\lambda$}, producing a surface magnetic field with the same penetration length.

Fig. 1 Infinite semispace

Another simple geometry is that of a superconducting cylinder inside an infinite conventional solenoid with parallel axes (Fig. 2). In the stationary case, no net charges. Outside the superconductor volume, the solution is straightforward in cylindrical coordinates {$\hat r, \hat \varphi,\hat z$}. If the conventional solenoid of radius {$R$} has an ideal surface current density {$\mathbf k(r)=\frac {B_0} {\mu_0} \delta(r-R)\,\hat \varphi$} circulating in the plane perpendicular to the solenoid axis, {$\hat z$}, the stationary solution is

{$$ \mathbf B(r) = B_0 \Theta(R-r) \hat z = \begin{cases} 0\qquad\qquad\quad \mbox{outside the solenoid} \\ B_0 \hat z \qquad\qquad \mbox{inside the solenoid} \end{cases} $$}

Index

London equation

This gives the boundary condition to our problem, which is now is to find self consistently how the supercurrents {$\mathbf{ J}_s(r)$} distribute themselves. F. and H. London showed that the Meissner effect is reproduced simply by supposing that

{$$ \begin{equation} \mathbf{A}(r)= - \mu_0\lambda^2 \mathbf{J}_s(r) \end{equation}$$}

This is the London equation, and it has profound implications, as discussed by P.W.Anderson. The vector potential in standard electromagnetism is not an observable quantity, because of the gauge invariance: adding the gradient {$\boldsymbol\nabla g(r)$} of any central function {$g$} to the vector potential produces the same {$\mathbf B$} field, i.e. the same physics. Yet, the current density is observable, thus also {$\mathbf A$} would become observable in superconductors, if the London equation were the final story (which is not the case), completely breaking the invariance. In a superconductor the choice of the vector potential is somehow limited, and the full gauge symmetry is broken, an issue that can be setted only by a quantum mechanical description Note.

The assumption of Eq. (3) is valdated by showing that it justifies the Meissner effect. Taking the rotor of the Ampere's law, Eq. (1) and applying the same identity to the double rotor, {$\nabla\times(\nabla\times\mathbf{B})=\nabla(\nabla\cdot\mathbf{B}) - \nabla^2\mathbf{B}$}, recalled below Eq. (1), we get

{$$ -\nabla^2 \mathbf{B} = \mu_0 \nabla\times \mathbf{J_s}, $$}

in the superconductor, which, by London Eq. (2), finally becomes

{$$ \begin{equation}\nabla^2 \mathbf{B} = \frac 1 {\lambda^2}{\mathbf B}. \end{equation}$$}

It is easy to recognize that the solution of this equation matching the cylindrical boundary conditions (Fig. 2) is a radial exponential decay of the magnetic field from its boundary value. With radial coordinate {$r$} centred on the superconductor axis, and cylinder radius {$r_0$}

{$$\begin{equation} \mathbf{B}(r) = B_0\, e^{-(r_0-r)/\lambda}\,\hat z \end{equation}$$}

satisfies Eq. (4). The supercurrents themselves will then be along the tangent unit vector {$\hat \varphi $} according to

{$$ \begin{equation} \mathbf{J}_s(r) = \frac {B_0} {\mu_0\lambda} \, e^{-(r_0-r)/\lambda}\,\hat \phi \end{equation}$$}

where {$B_0$} is the magnetic field produced by an external coaxial solenoid, homogeneous in the absence of the superconductor. In a similar fashion substituting (1) into the LHS of (2) and exploiting again the London equation, (3), we obtain a new equation identical to (4), valid for the vector potential, whose solution is directly obtained substituting (6) into (3):

{$$ \mathbf A(r) = -\lambda B_0 \, e^{-(r_0-r)/\lambda}\,\hat \phi$$}

Fig. 2 Blue dashed line: field profile inside the solenoid (black tube) and the superconductor (blue cylinder)

The value of the penetration depth {$\lambda$} was derived by London assuming that, upon application of an external field {$\mu_0 H$} the superconducting state retains its zero total momentum, the average value of {$\mathbf{p}=m\mathbf{v}+q\mathbf A$}. London had already the notion that the superconducting ground state was a true quantum state, hence he understood the vanishing of {$\langle \mathbf p\rangle$} as the zero expectation value of a observable, that cannot be altered by the application of a magnetic field. This value implies that the average carrier velocity must be {$\langle\mathbf v \rangle =- q\mathbf A/m$}. The classical definition of current density, {$\mathbf J_s=nq\langle v\rangle= - {nq^2}\mathbf A/m$}, provides the London equation (2) and fixes:

{$$\begin{equation} \qquad\qquad \lambda= \sqrt{\frac m {\mu_0 n q^2}} \end{equation}$$}

London assumed that the non-relativistic momentum {$\mathbf p$} is equal to {$m\mathbf v$} in zero magnetic field and to {${\mathbf p} = m{\mathbf v} + q{\mathbf A}$} (by the canonical minimal substitution), when the field is switched on. London was thinking in terms of electrons of charge {$q=-e$} and mass {$m_e$}. We now know that electron pairs are involved, {$q=-2e$} and {$m=2m_e$}.