# London theory, an electromagnetic description?

Currently I'm reading "Introducton to Solid-State Physics" by Charles Kittel, 8th edition and about superconductivity. I'm having a bit of trouble getting the whole plot, because as far as I'm concerned, the Meissner effect is an effect closely related to superconductors (let's assume a type II superconductor for the moment).

The magnetic field inside the superconductor, if we let the superconductor be a thin and long one, with axes parallel to the applied magnetic field $B_{ac}$ we can deduce that $B=0$ does not come from electromagnetic theory.

Later on I'm coming to the part where the London equations are actually trying to explain the effect of superconductivity once again, with electromagnetic theory, namely Maxwell's equations.

What am I missing? Where did I lose the plot on this one?

• Can you explain more explicitly what the argument you are having trouble with is? If you are dealing with magnetic fields it would be surprising is Maxwell's equations did not enter into the picture somewhere. May 11 '18 at 12:35
• I'm having problem with the book telling me that B=0 is not due to treating the superconductor as a medium of 0 resistivity. But when I read further into it, it seems like London thought you could describe the superconductivity and namely the Meissner effect with electromagnetic theory. Something seem weird about this, at least according to me May 11 '18 at 12:40

The London equation almost follows from the Drude model of conductivity and Maxwell's equations. Here's how.

In the Drude model, we assume that electrons moving through a metal are subject to two interactions. First, they experience a certain force $\vec{F}$, and accelerate in response to it. Second, there is a probability of an electron hitting a nucleus. When this happens, we assume that it stops dead. In an amount of time $\Delta t$, we assume that this probability is approximately $\Delta t/\tau$.

Based on these assumptions, it is not too hard to show that the total momentum $\vec{p}$ of a bunch of charge carriers in some volume of a metal is governed by the equation $$\frac{d \vec{p}}{dt} = - \frac{1}{\tau} \vec{p} + \vec{F}.$$ But the momentum of these charge carriers is proportional to the current density in the metal ($\vec{J} \propto \vec{p}$), and the force on them is proportional to the electric field ($\vec{E} \propto \vec{F}$). If we put in all the appropriate constants, this equation can be rearranged to become $$\frac{\partial \vec{J}}{\partial t} = - \frac{1}{\tau} \vec{J} + \frac{n q^2}{m} \vec{E},$$ where $n$, $q$, and $m$ are the number density, charge, and mass of the charge carriers (respectively). For a steady-state current ($\dot{\vec{J}} = 0$), this implies that $$\vec{J} = \frac{n q^2 \tau}{m} \vec{E},$$ which can be recognized as the microscopic version of Ohm's Law, with $\sigma = n q^2 \tau/m$.

A perfect conductor, however, is one where $\tau \to \infty$. This implies that $$\frac{\partial \vec{J}}{\partial t} = \frac{n q^2}{m} \vec{E},$$ If we take the curl of this equation, and apply Faraday's Law, we have $$\frac{\partial}{\partial t} (\vec{\nabla} \times \vec{J}) = \frac{n q^2}{m} \vec{\nabla} \times \vec{E} = - \frac{n q^2}{m} \frac{\partial \vec{B}}{\partial t}.$$ Thus, we must have $$\vec{\nabla} \times \vec{J} + \frac{n q^2}{m} \vec{B} = \text{constant with respect to t.}$$

So far we've used nothing but the Drude model and Maxwell's equations. The fundamental assumption of the London equation (and the part that does not follow from Maxwell's equations) is that the constant in this last equation is exactly zero: $$\vec{\nabla} \times \vec{J} + \frac{n q^2}{m} \vec{B} = 0.$$ Once we have this, we can then show that $\vec{B}$ nearly vanishes inside the bulk of a superconductor. If we take the curl of Ampere's Law, and assume static fields, it's not too hard to show that we have $$\nabla^2 \vec{B} = \frac{\mu_0 n q^2}{m} \vec{B}$$ which implies that $\vec{B}$ must either die off or grow exponentially inside a superconductor. Growing exponentially is right out, so it must be the case that magnetic fields decay as you move deeper into a superconductor. The characteristic length scale for this die-off is $\lambda = \sqrt{m/\mu_0 n q^2}$, and this works out to be on the order of a few dozen nanometers for most superconductors.

• Thanks a lot for this through answer, I saw in your description that you are a professor, it it possible to ask you for a recommendation of a book concerning Solid-State Physics that will be a bit more detailed than Charles Kittel's. Once again, thanks a lot for taking time to write me this answer- May 11 '18 at 13:15
• @JohnSkeet I liked Ashcroft and Mermin's 'Solid State Physics'. See physics.stackexchange.com/questions/22046/… for a list of recommendations. May 11 '18 at 13:47
• I'll look into that, thanks for the recommendation. May 11 '18 at 13:49
• @JohnSkeet: Yeah, solid-state isn't my field of expertise, so I don't have a better suggestion than Ashcroft & Mermin. May 11 '18 at 18:38

The first part of the argument is simply saying that the Meissner effect cannot simply be explain as a consequence of perfect conductivity. It is an independent physical phenomenon that has to be explained separately. It is not saying that the entire theory of electromagnetism cannot be applied to superconductors.

The second argument uses the London equations to explain how the Meissner effect does occur. As part of that argument it clearly must, describe the magnetic field. The mathematical description of that field is given, as always, by Maxwell's equations.

• If I would say that when you use "old" electromagnetic theory, using Ohm's law and Maxwell relations to deduce that $\frac {dB}{dt}=0$ but not that B=0, I've done the correct interpretation that the Maxwell relations are always true for electromagnetics, but Ohm's law is only working for "normal" conductors and not for superconductors? May 11 '18 at 13:00