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Proof: curl curl f = grad (div (f)) - grad^2

  1. Nov 19, 2011 #1
    Can anyone help me proving this:

    http://img88.imageshack.us/img88/3730/provei.jpg [Broken]

    And just for curiosity, is there a proof for why is the Laplace operator is defined as the divergence (∇·) of the gradient (∇ƒ)?
    And why it doesn't work on vetorial function.

    Thanks in advance, guys!
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Nov 20, 2011 #2


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    Hey IgorM and welcome to the forums.

    The easiest way I see proving this is to use the definition of the inner and outer products (inner = dot product, outer = cross product).

    Use the determinant form for the cross product on the LHS and then expand the RHS and they should come out to the be the same.
  4. Nov 20, 2011 #3
    I got stuck doing this, I tried to use curl twice on the left side to see if It would show as the right side.. but couldn't..

    Oh, thanks by the way! =)
  5. Nov 20, 2011 #4


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    Try evaluating the del x F first using the determinant representation, and then use the result of that and then use del x result to get your final result.

    It will probably be a little messy, but it shouldn't take you too long I think to expand out the LHS.
  6. Nov 20, 2011 #5
    This is the result:

    î(δyδxF2 - δy²F1 - δz²F1 - δxδzF3) + j(δx²F2 - δxδyF1 - δzδyF3 - δz²F2) + k(δxδzF1 - δx² - δy²F3 - δyδzF2)
  7. Nov 20, 2011 #6
  8. Nov 20, 2011 #7


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    The easiest way is to use index notation I think.
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