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

In summary, the conversation discusses a method for proving a given equation involving the Laplace operator and the divergence and gradient of a function. The conversation also touches on the definition and application of the Laplace operator on vectorial functions. It is suggested to use the determinant form for the cross product and the index notation for a simpler approach in proving the equation.
  • #1
IgorM
6
0
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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  • #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
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.
 
  • #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.
 

1. What is the proof for "curl curl f = grad (div (f)) - grad^2"?

The proof for "curl curl f = grad (div (f)) - grad^2" is known as the Helmholtz decomposition theorem. It states that any vector field can be decomposed into a sum of an irrotational vector field (grad (div (f))) and a solenoidal vector field (curl curl f). This proof is fundamental in the study of vector calculus and electromagnetism.

2. How is "curl curl f = grad (div (f)) - grad^2" used in physics?

"curl curl f = grad (div (f)) - grad^2" is used in physics to describe the behavior of electric and magnetic fields. It is a key component of Maxwell's equations, which govern the behavior of electromagnetic waves and are essential in understanding phenomena such as light, electricity, and magnetism.

3. What is the significance of the Laplacian (grad^2) in this equation?

The Laplacian (grad^2) in this equation represents the rate of change of a scalar field. In the context of "curl curl f = grad (div (f)) - grad^2," it represents the rate of change of the solenoidal component of the vector field. This component is crucial in understanding the behavior of electric and magnetic fields in physics.

4. Can "curl curl f = grad (div (f)) - grad^2" be applied to three-dimensional vector fields?

Yes, "curl curl f = grad (div (f)) - grad^2" can be applied to three-dimensional vector fields. In fact, it is most commonly used in three-dimensional space to describe the behavior of electric and magnetic fields. The equation can also be extended to higher dimensions, making it a fundamental tool in the study of vector calculus.

5. Is there a visual representation of "curl curl f = grad (div (f)) - grad^2"?

Yes, "curl curl f = grad (div (f)) - grad^2" can be visualized using vector field plots. The solenoidal component (curl curl f) appears as closed loops, while the irrotational component (grad (div (f))) appears as point sources or sinks. The Laplacian (grad^2) can also be represented by the density of the point sources or sinks. This visualization helps in understanding the behavior of the vector field and its components.

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