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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.

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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 $\nabla^2 {\bf h}$.

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  • $\begingroup$ Sorry, I don't see where is $div(\delta h \times curl h)$ in my calculations, can you make it more explicit? $\endgroup$ Apr 22 at 12:25
  • $\begingroup$ 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. $\endgroup$
    – mike stone
    Apr 22 at 12:28

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