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?
