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

  • #1
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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!
Igor.
 
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  • #2
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.
 
  • #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! =)
 
  • #4
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! =)

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.
 
  • #5
This is the result:

î(δyδxF2 - δy²F1 - δz²F1 - δxδzF3) + j(δx²F2 - δxδyF1 - δzδyF3 - δz²F2) + k(δxδzF1 - δx² - δy²F3 - δyδzF2)
 
  • #6
Anyone?
 
  • #7
The easiest way is to use index notation I think.
 

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