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Curl of (A*B)

  1. Nov 24, 2009 #1
    1. The problem statement, all variables and given/known data

    Prove that [tex]\nabla\times(A\times B)= (B.\nabla)A-(A.\nabla)B-B(\nabla.A)+A(\nabla.B)[/tex]

    2. Relevant equations

    bac-cab [tex]\nabla\times(A\times B)= (\nabla.B)A-(\nabla.A)B[/tex]

    3. The attempt at a solution

    I know that [tex]B(\nabla.A)=(\nabla.A)B[/tex] and [tex]A(\nabla.B)= (\nabla.B)A[/tex]

    So what about [tex](B.\nabla)A-(A.\nabla)B=?[/tex] Does it equal to zero? Or maybe bac cab is not related to this problem!
     
  2. jcsd
  3. Nov 24, 2009 #2
    When and how is this ["bac - cab"] equation valid? The above equation is valid only as long as "a" is....?
     
  4. Nov 24, 2009 #3
    So how can I prove it? Please help me!
     
  5. Nov 24, 2009 #4

    LCKurtz

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    Prove it by letting [itex]\vec A = \langle f,g,h\rangle,\ \vec B = \langle u,v,w\rangle[/itex] and just work out both sides to check they are equal. It's easy, just a little algebra.
     
  6. Nov 24, 2009 #5
    Another interesting way would be to do to use a gem of a trick that I learnt off of the Feynman Lectures. Refer to Feynman Lectures, Volume II, Lecture 27, Field Energy and Field Momentum.
     
  7. Nov 24, 2009 #6
    Thank you very much for your help. I actually proved it in the way you suggested me, but only for the x-component, and it was a lot of algebra!
     
  8. Nov 24, 2009 #7
    Thank you very much. I will try to study it.
     
  9. Nov 25, 2009 #8

    Pengwuino

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    Do you know index notation? Vector identities are quite easy with it.
     
  10. Nov 25, 2009 #9
    Please tell me about "index notation". I went to http://en.wikipedia.org/wiki/Index_notation but I didn't completely understand your point of view!
     
  11. Nov 25, 2009 #10

    Pengwuino

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    Gold Member

    http://www.physics.ucsb.edu/~physCS33/spring2009/index-notation.pdf [Broken]

    Give this a try. Unfortunately, when I learned it, it was during lectures and not in our textbook so I can't tell you what book you can learn it out of. This should be enough though.
     
    Last edited by a moderator: May 4, 2017
  12. Nov 26, 2009 #11
    Thank you very much. I downloaded it and I will read it.
     
    Last edited by a moderator: May 4, 2017
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