London equation and Maxwell equation

Question

I am reading Tinkham's Introduction to Superconductivity (2nd ed) (Amazon link).

On pages 4-5, they state that:"

The second London equation 1.4, when combined with the Maxwell equation ${\rm curl}\ h = \frac{4\pi J}{c}$ leads to: $$\nabla^2 h = \frac{h}{\lambda^2}$$

where equation (1.4) is:

$$\ h=-c\ {\rm curl} \ (\Lambda J_s )\tag{1.4}$$ $$\Lambda = \frac{4\pi \lambda^2}{c^2}$$

If I take a curl on (1.4) and equate it with Maxwell equation I get: $${\rm curl} \ h = - c \ {\rm curl}\ {\rm curl}\ (\Lambda J_s) = -c(\nabla(\nabla \cdot) -\nabla^2)\Lambda J_s=(\dagger);$$

I don't see how do they get $\nabla^2h =h/\lambda^2$. I do get: $\nabla^2 J =J/\lambda^2$, if the first term in $(dagger)$ is zero, and $J_s=J$.

So is this a misprint in the book, or am I mistaken?

Answer 1

Hints:

1. Given the Maxwell equation: $$ \nabla\times\mathbf h=\frac{4\pi}{c}\mathbf J $$ Take the curl of both sides, what do you get? How can (1.4) be used here?
2. Vector calculus tells you $$ \nabla\times\nabla\times\mathbf a=\nabla\left(\nabla\cdot\mathbf a\right)-\nabla^2\mathbf a $$ What assumptions must you make to get the London equation?