# Vector analysis for a variational problem

1. Jul 24, 2008

### Angelos K

I'm reading a proof that is via variation. $$\vec {\delta A}$$ stands for a variation of the vectorpotential $$\vec{A}$$. If I understand the argument correctly [many steps are presented as one] it means:

$$(\nabla \times \vec{A})* (\nabla \times \vec{\delta A}) = (\nabla \times \nabla \times \vec {A})*\vec{\delta A}$$.

I do not believe it. If I calculate the brackets on the left hand side individually I don't get any undifferentiated $$\vec{\delta A}$$. What am I doing wrong?

You can find the "whole" calculation on page 10 of the attached file.

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• ###### Superconductivity-2007-04.pdf
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Last edited: Jul 24, 2008