London equation problems

Question

London's acceleration equation.

$$E=\frac{m}{ne^2} \frac{DJ}{dt}$$

is derived from the definition of current density and $F = ma$

However, why is the magnetic contribution of force ignored for the derivation?

As it is simply taken $F =eE$

Answer 1

It's because of the Meissner relation ${\boldsymbol \omega}m+ e{\bf B}=0$ which follows from taking the curl of Fritz and Heinz London's formula for the momentum $$ m {\bf v}= \hbar \nabla\left (\phi-\frac{e}{\hbar} {\bf A}\right). $$ Here ${\boldsymbol \omega}= \nabla\times {\bf v}$ is the fluid vorticity and $\phi$ is the superfluid order parameter phase. This means that when you use a vector identity to write the fluid dynamics Euler equation $$ \left(\frac{\partial {\bf v}}{\partial t}+ ({\bf v}\cdot \nabla) {\bf v}\right) = \frac{e}{m}({\bf E}+ {\bf v}\times {\bf B})- \nabla P $$ in the Bernouli form $$ \left(\frac{\partial {\bf v}}{\partial t}+ {\boldsymbol \omega}\times {\bf v}\right)= \frac{e}{m}({\bf E}+ {\bf v}\times {\bf B})- \nabla\left(P +\frac 12 |{\bf v}|^2\right) $$ the ${\bf v}\times {\bf B}$ force cancels against the vorticity term on the LHS.

It is the Messner relation that causes a magnetic field to be expelled from a superconductor. In the fluid dymamics of an inviscid charged fluid you can show that the quantity ${\boldsymbol \omega}m+ e{\bf B}={\rm constant}$ because changing the magnetic field causes an ${\bf E}$ field that generates vortcicity. What is special about the charged superfluid is that the constant is zero. This means that a ${\bf B}$ field comes with non-zero ${\boldsymbol \omega}$ and hence costs kinetic energy. If the constant where not zero, the magnetic field would just be trapped in the fluid, as it is in highly condutive plasmas.

Answer 2

Now that I think about this, the problem in some of the books involves confusion between total and partial derivatives. As the answer above, clearly points out, $\frac{dv}{dt}$ and $\frac{\partial v}{\partial t}$ are clearly not the same.

So, one could include $-e(\vec{v}\times\vec{B})$ in the acceleration equation, such that,

$$\frac{dv}{dt}=-\frac{e}{m}(\vec{E}+\vec{v}\times\vec{B})=-\frac{1}{ne}\frac{dJ}{dt}$$

However, using some fluid mechanics, and taking into account the Meissner effect, you could show that

$$\frac{\partial v}{\partial t}=-\frac{e}{m}E$$

This is exactly what the previous answer does.

What is interesting is that, in both cases, the curl of the derivative of $v$ is identical.

$$\vec{\nabla}\times\frac{\partial v}{\partial t}=-\frac{e}{m}\vec{\nabla}\times\vec{E}=\frac{e}{m}\frac{\partial B}{\partial t}$$

On the other hand,

$$\vec{\nabla}\times\frac{d v}{d t}=-\frac{e}{m}(\vec{\nabla}\times\vec{E}+\vec{\nabla}\times(\vec{v}\times\vec{B}))$$

The second term vanishes. This can be shown :

$$\nabla\times( v\times B)=v(\nabla.B)-B(\nabla.v)$$ The first term on the RHS is zero, due to the Maxwell equations. As for the second term,

$$v=-\frac{J}{ne}=-\frac{\nabla \times B}{\mu_0 ne}$$

Hence, $$\nabla.v \sim \nabla.(\nabla\times B)=0$$

Hence, while the terms $\frac{dv}{dt}$ and $\frac{\partial v}{\partial t}$are definitely not the same, their curls are. So, either way, one can derive (motivate) the second London equation from the first one.