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

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The discussion revolves around proving the equation curl curl f = grad (div (f)) - grad^2. Participants suggest using the definitions of inner and outer products, specifically the determinant form for the cross product, to establish the proof. There is also curiosity about the definition of the Laplace operator as the divergence of the gradient and its applicability to vector functions. Suggestions include evaluating del x F using determinant representation and expanding the left-hand side for clarity. The conversation emphasizes the complexity of the proof while encouraging collaborative problem-solving.
IgorM
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Can anyone help me proving this:

http://img88.imageshack.us/img88/3730/provei.jpg

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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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.
 
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! =)
 
IgorM said:
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.
 
This is the result:

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

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