# Deriving London Equation from Helmholtz energy

I'm trying to derive the London equation of superconductivity following the book Ginzburg-Landau Phase Transition Theory and Superconductivity. So I arrived to:

$$\begin{gather} \mathcal{F} = \mathcal{F}_S + \frac{1}{2}\int_V\left(h^2+\lambda_L^2\left|\overrightarrow{\nabla}\times\overrightarrow{h}\right|^2\right)d\overrightarrow{r} \end{gather}$$

Now I have to minimize the free energy. $$\mathcal{F}_S$$ is a constant so it doesn't matter when taking variations. The problem I have is taking variations of the term $$\overrightarrow{\nabla}\times\overrightarrow{h}$$, I did this: $$\begin{gather} \delta\left(\left|\overrightarrow{\nabla}\times\overrightarrow{h}\right|^2\right) = \delta\left(\overrightarrow{\nabla}\times\overrightarrow{h}\cdot\overrightarrow{\nabla}\times\overrightarrow{h}\right)=2\delta\left(\overrightarrow{\nabla}\times\overrightarrow{h}\right)\cdot\overrightarrow{\nabla}\times\overrightarrow{h} \end{gather}$$

but I need to show that that is equal to $$-\nabla^2h$$, but I don't know how to prove it.

Use the equation $${\rm div}(\delta{\bf h} \times {\rm curl\,}{\bf h})={\rm curl\,}{\bf h}\cdot {\rm curl \,}\delta{\bf h}- \delta{\bf h}\cdot({\rm curl}({\rm curl\,} {\bf h}))$$ to integrate by parts, and then the usual expression relating $${\rm curl}({\rm curl\,} {\bf h})$$ to $$\nabla^2 {\bf h}$$.
• Sorry, I don't see where is $div(\delta h \times curl h)$ in my calculations, can you make it more explicit? Apr 22 at 12:25
• You are trying to integrate by parts to convert your ${\rm curl\,}{\bf h}\cdot {\rm curl \,}\delta{\bf h}$ into $\delta{\bf h}\cdot({\rm curl}({\rm curl\,} {\bf h})\propto \delta {\bf h}\cdot \nabla^2 {\bf h}$. The divergence is the integrated out part. Apr 22 at 12:28