Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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.

    Attached Files:

    Last edited: Jul 24, 2008
  2. jcsd
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?
Draft saved Draft deleted

Similar Discussions: Vector analysis for a variational problem
  1. Vector Problem (Replies: 2)