The Meissner effect
Now we can describe some of the phenomena of superconductivity. First,
there is no electrical resistance. There’s no resistance because all
the electrons are collectively in the same state. In the ordinary flow
of current you knock one electron or the other out of the regular
flow, gradually deteriorating the general momentum. But here to get
one electron away from what all the others are doing is very hard
because of the tendency of all Bose particles to go in the same
state. A current once started, just keeps on going forever. It’s also easy to understand that if you have a piece of metal in the
superconducting state and turn on a magnetic field which isn’t too
strong (we won’t go into the details of how strong), the magnetic
field can’t penetrate the metal. If, as you build up the magnetic
field, any of it were to build up inside the metal, there would be a
rate of change of flux which would produce an electric field, and an
electric field would immediately generate a current which, by Lenz’s
law, would oppose the flux. Since all the electrons will move
together, an infinitesimal electric field will generate enough current
to oppose completely any applied magnetic field. So if you turn the
field on after you’ve cooled a metal to the superconducting state, it
will be excluded. Even more interesting is a related phenomenon discovered experimentally
by Meissner.11 If you have a piece
of the metal at a high temperature (so that it is a normal conductor)
and establish a magnetic field through it, and then you lower the
temperature below the critical temperature (where the metal becomes a
superconductor), the field is expelled. In other words, it starts
up its own current—and in just the right amount to push the field out. We can see the reason for that in the equations, and I’d like to
explain how. Suppose that we take a piece of superconducting material
which is in one lump. Then in a steady situation of any kind the
divergence of the current must be zero because there’s no place for it
to go. It is convenient to choose to make the divergence of $\FLPA$
equal to zero. (I should explain why choosing this convention doesn’t
mean any loss of generality, but I don’t want to take the time.)
Taking the divergence of Eq. (21.18), then gives that
the Laplacian of $\theta$ is equal to zero. One moment. What about the
variation of $\rho$? I forgot to mention an important point. There is
a background of positive charge in this metal due to the atomic ions
of the lattice. If the charge density $\rho$ is uniform there is no
net charge and no electric field. If there would be any accumulation
of electrons in one region the charge wouldn’t be neutralized and
there would be a terrific repulsion pushing the electrons
apart.12 So in
ordinary circumstances the charge density of the electrons in the
superconductor is almost perfectly uniform—I can take $\rho$ as a
constant. Now the only way that $\nabla^2\theta$ can be zero
everywhere inside the lump of metal is for $\theta$ to be a
constant. And that means that there is no contribution to $\FLPJ$ from
$p$-momentum. Equation (21.18) then says that the current
is proportional to $\rho$ times $\FLPA$. So everywhere in a lump of
superconducting material the current is necessarily proportional to
the vector potential:
\begin{equation}
\label{Eq:III:21:20}
\FLPJ=-\rho\,\frac{q}{m}\,\FLPA.
\end{equation}

Since $\rho$ and $q$ have the same (negative) sign, and since $\rho$
is a constant, I can set $-\rho q/m=-(\text{some positive constant})$;
then
\begin{equation}
\label{Eq:III:21:21}
\FLPJ=-(\text{some positive constant})\FLPA.
\end{equation}

This equation was originally proposed by London and
London13 to explain the
experimental observations of superconductivity—long before the quantum
mechanical origin of the effect was understood. Now we can use Eq. (21.20) in the equations of
electromagnetism to solve for the fields. The vector potential is
related to the current density by
\begin{equation}
\label{Eq:III:21:22}
\nabla^2\FLPA=-\frac{1}{\epsO c^2}\,\FLPJ.
\end{equation}

If I use Eq. (21.21) for $\FLPJ$, I have
\begin{equation}
\label{Eq:III:21:23}
\nabla^2\FLPA=\lambda^2\FLPA,
\end{equation}

where $\lambda^2$ is just a new constant;
\begin{equation}
\label{Eq:III:21:24}
\lambda^2=\rho\,\frac{q}{\epsO mc^2}.
\end{equation}
 We can now try to solve this equation for $\FLPA$ and see what happens
in detail. For example, in one dimension Eq. (21.23)
has exponential solutions of the form $e^{-\lambda x}$
and $e^{+\lambda x}$. These solutions mean that the vector potential must
decrease exponentially as you go from the surface into the
material. (It can’t increase because there would be a blow up.) If the
piece of metal is very large compared to $1/\lambda$, the field only
penetrates to a thin layer at the surface—a layer about $1/\lambda$
in thickness. The entire remainder of the interior is free of field,
as sketched in Fig. 21–3. This is the explanation of the Meissner
effect. How big is the distance $1/\lambda$? Well, remember that $r_0$, the
“electromagnetic radius” of the electron ($2.8\times10^{-13}$ cm),
is given by
\begin{equation*}
mc^2=\frac{q_e^2}{4\pi\epsO r_0}.
\end{equation*}

Also, remember that $q$ in Eq. (21.24) is twice the
charge of an electron, so
\begin{equation*}
\frac{q}{\epsO mc^2}=\frac{8\pi r_0}{q_e}.
\end{equation*}

Writing $\rho$ as $q_eN$, where $N$ is the number of electrons per
cubic centimeter, we have
\begin{equation}
\label{Eq:III:21:25}
\lambda^2=8\pi Nr_0.
\end{equation}

For a metal such as lead there are about $3\times10^{22}$ atoms
per cm$^3$, so if each one contributed only one conduction electron,
$1/\lambda$ would be about $2\times10^{-6}$ cm. That gives you the
order of magnitude.