# What is the precise/defining mathematical behaviour for Meissner effect?

This video derives the Meissner effect $$\textbf{B}(x)=\textbf{B}(0)e^{-x/\lambda},$$ by reducing the equation, $$(\nabla^2-\lambda^{-2})\textbf{B}(\textbf{r})=0\tag{1}$$ into its one-dimensional counterpart $$(\partial_x^2-\lambda^{-2})B(x)=0\Rightarrow B(x)=B(0)e^{-x/\lambda}.\tag{2}$$

But shouldn't the actual solution to Eq.(1) be $$B\sim e^{-r/\lambda}/r\tag{3}$$ in three dimensions? See equation 3.68 here.

So the question is what is the precise/defining mathematical behaviour for Meissner effect? Is it given by (2) or is it given by (3)?

• Once again: Please do not rely on external links to make your question comprehensible, in particular, please explain all notation used even if one can try to guess it: E.g. what is $\lambda$? Also, why should the "actual solution" to eq. (1) be eq. (3)? What boundary, symmetry and/or regularity conditions are you assuming so that other possible solutions are discarded? Dec 28 '17 at 12:49
• The mathematical definition of the Meissner effect which also defines a superconductor is : perfect diamagnetism. Microscopically, Meissner effect is understood as discussed in an other post as the proportionality between the current and the vector potential. If you include this property (called London's constitutive relation) artificially into the Maxwell's equations, you get the London's equation you cite (in the case of stationary effects for the magnetic field only). Dec 29 '17 at 8:53

The point is the 3D equation is a PDE, which is therefore complicated, and doesn't really have a general solution. If you assume spherical symmetry, then the solution which doesn't diverge at $+\infty$ is given by $B\sim e^{-r/\lambda}/r$, which roughly gives the same kind of behaviour (the exponent is indeed $r/\lambda$, which is consistent with the OP's reference, but not with the equation written down by the OP as of when this answer was written). The blow up at $r=0$ should perhaps worry us.
But why should we assume spherical symmetry? There's actually no reason why we should. In fact it is often useful to write the equation in cylindrical coordinates, for instance in the study of (London) vortices: $$\left(\frac{d^2}{dr^2}+\frac{1}{r}\frac{d}{dr}-\frac{1}{\lambda^2} \right) B=0$$ under some appropriate conditions. This is a Bessel-type equation, and can be solved as $B\sim K_0(r/\lambda)$, where $K_0$ is one of the modified Bessel functions. Asymptotically, $$B \sim \begin{cases}\log{(\lambda/r)} & r \ll \lambda \\e^{-r/\lambda}/\sqrt{r} & r \gg \lambda\end{cases}$$ which gives a similar behaviour (this is a vortex: constant field inside the vortex, logarithmic decrease up to $\approx \lambda$ and then exponential decrease).