# Problem understanding the London Equation in superconductors

I'm facing quite a lot of issues trying to understand the reasoning, that leads to the London equations in superconductivity. In many places, the logic seems somewhat circular.

I've followed the excellent derivation given in this answer. In fact, this derivation is quite similar to the one I found in the book by Ashcroft and Mermin.

However, my first question is regarding the acceleration equation.

$$\frac{dJ}{dt}=-ne\frac{dv}{dt}=\frac{ne^2}{m}E$$

Here we have plugged in $$\frac{dv}{dt}=-eE$$, and in Ashcroft-Mermin, the author says, 'suppose an Electric field exists in the Superconductor'. My question here is, why should an electric field exist in the first place?

This answer then comes to my rescue. However, according to this answer, the electric field exists when we haven't reached the steady-state. Does this mean, the above equation holds only when we are not in the steady-state? As soon as the electrons reach a steady-state, this $$E$$ should vanish. So, what is the source of this $$E$$ in the above equation?

In the case of the Drude model, we provide an external electric field, and that explains the $$E$$ in the Drude model. In this case, however, we have only provided a constant magnetic field externally. So, where is this $$E$$ coming from?

My next question is that, if I assume that an electric field is momentarily created, and this creates the current, shouldn't I also assume that a momentary magnetic field was created? Shouldn't I consider the general case where $$\frac{dv}{dt}=-e(E+v\times B)$$?

This answer then explains why we don't use any contribution from the magnetic field.

However, here the reasoning seems a little circular to me. The above answer uses the Meissner relation $$\omega m+eB=0$$ and plugs this into the Euler equation for a fluid, to show that $$\frac{dv}{dt}=-eE$$, as the vorticity term cancels out the magnetic term.

But now if we plug $$\frac{dv}{dt}=-eE$$ and follow the derivation, we can show that $$\omega m+eB=0$$, as is done at the end of the first answer. Note that this is also assumed, as in reality, the RHS could be any constant. It is taken to be $$0$$ to explain the Meissner effect.

This reasoning doesn't make sense to me.

We use the Meissner relation, to show that $$\frac{dv}{dt}=-eE$$, and then use this equation, to show that the Meissner relation is true. This feels a lot like saying, if A is true, then B must be true. Since B is true, A must also be true.

Can someone help me understand where am I making a mistake?

If I am correct, is there any other way of reasoning, that does not lead to these apparent logical fallacies?

Is there any way of deriving the Meissner relation independently i.e. without considering $$\frac{dv}{dt}=-eE$$, since we use this Meissner relation later to show that $$\frac{dv}{dt}$$ is indeed equal to $$-eE$$? Of course, we can use $$\frac{dv}{dt}=-eE$$ to show that the Meissner relation is true, but the only way this would not be a logical fallacy would be to show that I can get the relation independently.

For example, let $$A$$ be the Meissner relation, and $$B$$ be the acceleration equation.

It has been shown that I can derive $$B$$ using $$A$$, and I can derive $$A$$ using $$B$$. The only way this is not circular reasoning is if I can derive one of them independently of the other.

EDIT:

I think I have found one of my mistakes, but I'm not sure. I'd be glad if someone verifies the following for me.

In the notes of my professor, what he does is take $$\frac{dJ}{dt}$$ as I have written in the first equation. However, I'm led to believe that this is incorrect. In fact, I should have $$\frac{\partial J}{\partial t}$$ instead.

Then my equation would be :

$$\frac{\partial J}{\partial t}=-ne\frac{\partial v}{\partial t}$$

Then I can show, that while $$\frac{dv}{dt}$$ is equal to $$-e(E+v\times B)$$, due to vorticity and Meissner effect, $$\frac{\partial v}{\partial t}$$ is equal to only the electric field part i.e. $$-eE$$.

Is this reasoning correct?

• For derivations of the London equation, see: physics.stackexchange.com/a/405294/226902 , physics.stackexchange.com/q/544978/226902 , Jun 9, 2022 at 9:32
• @Quillo I've gone through these answers and that is where my doubts come from. I was hoping someone could clear these for me. Jun 9, 2022 at 9:34
• "Does this mean, the above equation holds only when we are not in the steady-state?" Why do you think this is true? E=0 would be Indicative of the situation you're describing, whilst also satisfying the above equation. Plugging in E = 0 to this equation yields the same form of solution, where dj/dt = 0 in this case. The equation mentioned is a generalisation of the situation, with the specific solution you're commenting on, is when E=0 Jun 11, 2022 at 19:26

First question: "Does this mean, the above equation holds only when we are not in the steady-state?". For the superconductor, when you are in the steady-state $$E=0$$, the current is constant. In a normal conductor (with finite resistance) you still have to apply $$E$$ to obtain $$J$$ that is constant in time and not zero (in this dissipative case, the steady-state for $$E=0$$ would be $$J=0$$).

Second question: "So, what is the source of this $$E$$ in the above equation?"/"So, where is this E coming from?". $$E$$ is an external field, that can be used to set electrons into motion. As long as it is switched on, you can not be in the steady-state (unless you accelerate electrons so much that you break the superconducting state and dissipation sets in). You switch it on, create a persistent current, and switch it off. the persistent current will stay there (this is what you call "steady-state").

Third question: Meissner relation independently arises as a corollary of the fact that the velocity of the electrons is related to the phase of the complex order parameter. To have $$B$$ you have to break this fact locally, introducing quantized magnetic flux tubes.

• This makes sense, but can you clear a few additional questions. You mention that $E$ is an external field. However, all I have done is given a constant magnetic field, and the superconductor expels this field. Where is the $E$ coming from exactly? The only way I can think about this is, that when we give the magnetic field to the superconductor, it disrupts the steady-state, and create an electric field for a moment. This $E$ causes currents that expel the magnetic field. The $E$ dies down quickly, but the currents persist. Jun 9, 2022 at 13:54
• Another possible explanation could be, the moment I turn on the magnetic field, there is a change in flux. So, in a sense, the magnetic field does change from being $0$ to be some finite $B$. Perhaps, this is what induces the $E$ in the superconductor. Jun 9, 2022 at 13:57
• In any case, this $E$ vanishes quickly, as the system comes into a steady state. However, the initial current persists and is nothing but $J=-nev$. Taking partial derivatives on both sides, I suppose we can show the London equations. Anyway, is the above argument, related to the origin of $E$, correct ? Jun 9, 2022 at 14:00
• Sorry but I do not understand your problem with $E$. Any $E$ can generate a current $J$. Typically it is an applied external field, because this si the way to generate currents. Ofc may be generated as a by-product if a time varying $B$ field, if you like. Jun 9, 2022 at 22:23
• Suppose I'm trying to show the Meissner effect, in a superconductor. I apply a magnetic field to this superconductor, and the field lines are expelled. Note that I have only provided a magnetic field. In the London equations, it is assumed that an electric field momentarily arises in a superconductor, according to Ashcroft Mermin. That is how they derive the first London equation, and using it they derive the second one. I want to know why does a $E$ momentarily arises with the superconductor, since we have not provided any external $E$ to it. Jun 10, 2022 at 5:35

If your question is about the dynamics of the Meissner effect (i.e. how superconductors expel magnetic fields), it is a pretty controversial issue and AFAIK still unsettled.

• Very interesting references, thank you :-) They help to understand the real question. It is worth mentioning, however, that Hirsch has some very particular and sometimes controversial views on superconductivity: more on his webpage jorge.physics.ucsd.edu/jh.html Jun 13, 2022 at 17:46
• @Quillo Concerning Hirsch's views I suggest to read carefully this very recent report: nature.com/articles/d41586-022-03066-z .There is something rotten in the HTS kingdom. Sep 28, 2022 at 4:51