# London equations and flux pinning

I am having some conceptual issues with the London equations, namely the equation for the magnetic field and penetration depth.

$$\frac{d\vec{J}}{dt} = \frac{ne^2}{m}\vec{E}$$

Behavior of magnetic fields in perfect conductors

$$\frac{d \nabla × \vec{J}}{dt} = \frac{ne^2}{m}\nabla × \vec{E}$$

$$\frac{d \nabla × \vec{J}}{dt} = - \frac{ne^2}{m} \frac{\partial \vec{B}}{\partial t}$$

$$\nabla × \vec{J} = -\frac{ne^2}{m}\vec{B}$$

Substituting maxwells equations we obtain.

$$\nabla^2 \vec{B} = \frac{\mu_0 n e^2}{m} \vec{B}$$

The solution to such an equation is a exponential decay of the magnetic field, or B=0. Which makes sense as when there is no external magnetic field B=0, but when there is a changing external magnetic field, an induced current opposes this magnetic field, causing the decay.

Thought experiment:

what if I were to manually "spawn" a constant magnetic field, and then "spawn" a superconductor.

There would be no changing magnetic field meaning no current to oppose the existing magnetic field, so the B field must be the external field.

Any increase or decrease to this B field now, would "lock " the magnetic field inside the superconductor to be the original external B field!

Or similar scenario, I have a constant magnetic field, and THEN i decrease the temperature of the conductor so that its a superconductor, the magnetic field should still be constant, as there are no induced currents.

Problems:

The London equations in their present form allow for no such constant solution, almost as if they inherently assume the magnetic field always starts from zero, and then from that 0 field, any increase or decrease in field strength generates a opposing field within the superconductor.

I have a feeling that this incompatibility with constant fields have something to do with

$$\frac{d \nabla × \vec{J}}{dt} = - \frac{ne^2}{m} \frac{\partial \vec{B}}{\partial t}$$

And the assumption that when integrating, c= 0

Not assuming this:

$$\nabla^2 \vec{B} = \frac{\mu_0 n e^2}{m} \vec{B} + \vec{c}$$

I feel that this equation makes more physical sense as per my thought experiments, as this equation allows for such constant solutions to exist.

However I'm not sure what boundary condition I would need to find what the constant is, if C mirrored the right term but negative, then constant solutions could exist ( although would ruin the equation for decaying solutions)

I assume C depends on the initial B fields value at t=0

So my question is:

Taking my thought experiments into account. Why do no constant solutions exist in the London equations? As the induced currents only occus when a changing magnetic field is in play. A constant one should not effect anything

I'm going to focus on this aspect of the question (which to me is more physical than "spawning" a conductor), and give a physical, not mathematical, answer:

Or similar scenario, I have a constant magnetic field, and THEN i decrease the temperature of the conductor so that its a superconductor, the magnetic field should still be constant, as there are no induced currents.

The short answer is that your assumption that there are no induced currents in this case is incorrect; there are induced currents, and they result in canceling the magnetic field inside the superconductor.

One way to understand what happens during a superconducting phase transition is the onset of superdiamagnetism.

Recall that ordinary diamagnetism is a quantum mechanical effect where the magnetic moments of individual atoms line up to oppose an applied magnetic field. This occurs because individual electrons in an atom are essentially following current loops with a small resistance. In the presence of an applied field, these loops precess around the applied field, in turn generating an opposing magnetic field. Normally, when electrons are bound to their nuclei, this effect is present, but extremely small (while diagmanetism is present in any material, if any paramagmentism or ferromagnetism is present in an ordinary material, it will swamp the diagmagnetic contribution).

However, in a superconductor, where the electrons can flow without resistance through the material, one finds there is perfect diamagnetism (or superdiamagnetism). An applied field inside a superconductor immediately generates large surface currents which oppose the applied field. This leads to the superconductor expelling the magnetic field. If a magnetic field is present in the material as it is being cooled, the expulsion of the magnetic field will occur immediately after the material is cooled below the critical temperature.

The the London equation correctly captures the behavior of the superconducting state described above, although it is an effective long-distance description, and does not include all the microscopic details.

• Do we use $E$ in the London equation because of these induced currents? Is that the origin of the $E$ field in the superconductor ? I know that in a steady state, the electric field inside a superconductor must be $0$. So, a superconductor in a magnetic field is not in it's steady state ? Jun 9, 2022 at 7:52