In Tinkham's "Introduction to Superconductivity" Second Edition, Chapter 2.2.1, the author discusses the problem of a superconducting slab in a uniform external field. In the solution presented by the author, the magnetic flux density inside the superconductor is given by \begin{align} h = H_a \frac{\cosh(y/\lambda)}{\cosh(d/2\lambda)}, \quad -d/2<y<d/2, \end{align} where $d$ is the slab's thickness, $\lambda$ is the penetration depth and $H_a$ is the magnetic flux density of the external field. The author achieves this by setting the boundary condition $h(-d/2) = h(d/2) = H_a$ and using the London equations. It is not mentioned what is the solution for the field outside the slab, so I presume it is considered to be constant ($h=H_a$). What I find disturbing is that this would mean that the presence of the superconductor has no influence on the field around it, whereas I would expect it to bend the field lines as in the usual Meissner effect representations.

Moreover, in this paper, by Fiolhais and Essén, a similar problem is solved, but for a cylindrical superconductor, instead of a slab. In this paper a slightly different approach is used, where the boundary condition is set at infinity (${\bf B}(\rho=\infty) = {\bf B_0}$). This leads to a result where the field at the interface between the superconductor and vacuum is not ${\bf B_0}$ and neither is the field around it. This is more in line with the usual depictions of the Meissner effect.

My question is: which of these treatments is correct? Am I correct to understand that the field outside the superconductor is constant in Thinkham's solution? Also, Thinkham's notation seems outdated, is its ${\bf h}$ equivalent to the modern paper's ${\bf B}$ or does it represent something else?

Bonus Question: In the conclusion to their paper, Fiolhais and Essén state that their result is only valid for type II superconductors because of the smaller penetration depth compared to coherence length in type I superconductors. They also state that it's only valid below the lower critical field. Are these assertions also valid for Thinkham's solution?


1 Answer 1


It turns out that the answer is quite simple, but took me some time to figure this out. Here is an outline of my thought process:

  • Tinkham considers a system that is translationally invariant in two directions
  • The magnetic field must also obey this symmetry
  • Outside the superconductor, the magnetic field must also obey Laplace's equation:

\begin{align} {\bf \nabla}^2 {\bf B} = 0 \end{align}

  • The only solution to this equation that is translation invariant in $y$ and $z$ is

\begin{align} {\bf B}(x,y,z) = {\bf B}_0 (a + bx), \quad a,b \in R \end{align}

  • The linear part of the solution explodes at infinity. In order to satisfy the boundary conditions at infinity this part of the solution must be zero.
  • This leaves only the constant part left and after applying boundary conditions we set $a=1$. So, for a slab with finite thickness but infinitely long and wide, the only solution possible for the magnetic field outside the superconductor is \begin{align} {\bf B} = {\bf B}_0. \end{align}

In the case for a superconductor which one more finite dimension, such as a cylindrical one, translation invariance is broken in that dimension. This allows for different solutions to Laplace's Equation (such as $\sim \rho^{-1} \cos(\phi)$ in the cylindrical case). In these cases not only we can get a non-uniform field around the superconductor, but the value of the field at the interface can also be different from the value of the external field ${\bf B}_0$.

As for the bonus question, I'm not yet sure, but I don't see why those conclusions would not apply to Tinkham's case.


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