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London's first equation $$\frac{d}{dt}\vec{j}=\frac{n_se^2}{m}\vec{E}$$ where $j=-en_s\vec{v}_s$, $n_s$ is the number density of electrons that contribute to the supercurrent and $\vec{v}_s$ is their mean velocity, coupled with Faraday's law of induction, $$\vec{\nabla}\times\vec{E}=-\frac{\partial\vec{B}}{\partial t}$$ we promptly obtain $$\frac{\partial}{\partial t}\big(\vec{\nabla}\times\vec{j}+\frac{n_se^2}{m}\vec{B}\Big)=0.$$ This equation, only tells that $$\vec{\nabla}\times\vec{j}+\frac{n_se^2}{m}\vec{B}={\rm any~ function~ of~ position~ only,~in~ general.}$$ But the solution is taken to be $$\vec{\nabla}\times\vec{j}+\frac{n_se^2}{m}\vec{B}=0.$$ Please explain, why. I have followed the derivation of Aschroft and Mermin's solid state physics book. They just mention but did not show what fixes the right hand side to be zero. The answer to this question notes that it does not necessarily follow from Maxwell's equations; but I would like to know how it can be justified.

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    $\begingroup$ Possible duplicate...answer 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 physics.stackexchange.com/questions/405286/… $\endgroup$ Apr 19, 2020 at 8:51
  • $\begingroup$ Does this answer your question? London theory, an electromagnetic description? $\endgroup$ Apr 19, 2020 at 17:11
  • $\begingroup$ @MichaelSeifert If the 0 does not follow from Maxwell's equation, how does it follow or how is it justified? $\endgroup$ Apr 19, 2020 at 17:16
  • $\begingroup$ My recollection is that it comes from quantum mechanics. Specifically, I vaguely recall that there's a requirement that the canonical momentum $\vec{p} = m \vec{v} - e \vec{A}$ of the charge carriers must vanish in the ground state, and if you take the curl of this equation this is equivalent to the equation you're concerned about. But I'm not certain enough about this statement, or familiar enough with the underlying principles, to write an answer explaining it fully. $\endgroup$ Apr 19, 2020 at 17:27
  • $\begingroup$ I edited your question to make it clearer that you're looking for what the underlying reason for the vanishing of this quantity. Feel free to edit further and/or roll back my edit if you see fit. $\endgroup$ Apr 19, 2020 at 17:29

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In a superconductor the velocity of the condensate of paired electrons is given by $$ {\bf v}= \frac 1 {m^*}(\nabla \phi-e^*{\bf A}) $$ where $\nabla \equiv {\rm grad}$. Here $m^* \approx 2m_e$ and $e^*=2e$ are the effective mass and charge of the Cooper pairs and $\phi(x)$ is the local phase of the order parameter $\langle \psi \psi \rangle = |\psi|^2 e^{i\phi}$. On taking the curl of the equation for ${\bf v}$ we get $$ m^* {\boldsymbol{\omega}} +e^*{\bf B}=0, $$ where ${\boldsymbol{\omega}}= \nabla\times {\bf v}$ is the vorticity. We have used $\nabla \times (\nabla \phi)=0$ and $\nabla\times {\bf A}={\bf B}$. A similar but not identical equation holds for any inviscid charged fluid because changing ${\bf B}$ spins up the fluid via $$ \nabla\times {\bf E}= -\frac{\partial {\bf B}}{\partial t}, $$ but the inviscid charged fluid equation is only $$ m^* {\boldsymbol{\omega}} +e^*{\bf B}=\hbox{time independent}, $$ It is the zero on the RHS of the superconductor equation that is crucial for excluding (rather than merely trapping) the magnetic field. The zero comes from the fact that $\nabla \times (\nabla \phi)=0$, i.e the existence of a well-defined local phase for the order parameter.

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