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Vector analysis for a variational problem

  1. Jul 24, 2008 #1
    I'm reading a proof that is via variation. [tex] \vec {\delta A} [/tex] stands for a variation of the vectorpotential [tex] \vec{A} [/tex]. If I understand the argument correctly [many steps are presented as one] it means:

    [tex] (\nabla \times \vec{A})* (\nabla \times \vec{\delta A}) = (\nabla \times \nabla \times \vec {A})*\vec{\delta A} [/tex].

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

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

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    Last edited: Jul 24, 2008
  2. jcsd
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