Proving Identity of $\nabla \times (\nabla \times F)$

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In summary, the conversation involved a question about proving an identity involving vector fields using different methods. The conversation delved into the use of definitions and notation, and various attempts were made to solve the problem. Ultimately, the use of summation notation and moving terms around allowed for the identity to be proven.
  • #1
Benny
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Hi, I am having an immense amount of trouble trying to show that the following identity is true.

Q. Let F be a C^2 vector field. Prove that

[tex]\nabla \times \left( {\nabla \times \mathop F\limits^ \to } \right) = \nabla \left( {\nabla \bullet \mathop F\limits^ \to } \right) - \nabla ^2 \mathop F\limits^ \to [/tex]

Firstly, I know that this can be done by using the definitions to find the ith component of the LHS. However, the use of this method requires something to be noted about the curl of a vector field and since I don't understand how one can just somehow notice that, I will not use this method.

I attempted this question by using:

[tex]
\nabla \times \mathop F\limits^ \to = \sum\limits_{i = 1}^3 {\mathop {e_i }\limits^ \to \times \partial _i \mathop F\limits^ \to }
[/tex]

[tex]
\nabla \bullet \mathop F\limits^ \to = \sum\limits_{i = 1}^3 {\mathop {e_i }\limits^ \to \bullet \partial _i \mathop F\limits^ \to }
[/tex]

[tex]
\nabla ^2 \mathop F\limits^ \to = \sum\limits_{i = 1}^3 {\partial _i ^2 \mathop F\limits^ \to }
[/tex]

where e_i is the usual basis vector, the d_i is the ith first order partial derivative.

Using the above I proceeded as follows.

[tex]
\nabla \times \left( {\nabla \times \mathop F\limits^ \to } \right) = \nabla \times \left( {\sum\limits_{i = 1}^3 {\mathop {e_i }\limits^ \to \times \partial _i \mathop F\limits^ \to } } \right)
[/tex]

[tex]
= \sum\limits_{j = 1}^3 {\sum\limits_{i = 1}^3 {\left( {\mathop {e_j }\limits^ \to \times \left( {\mathop {e_j }\limits^ \to \times \partial _j \partial _i \mathop F\limits^ \to } \right)} \right)} }
[/tex]

[tex]
= \sum\limits_{j = 1}^3 {\sum\limits_{i = 1}^3 {\left( {\mathop {e_j }\limits^ \to \bullet \partial _j \partial _i \mathop F\limits^ \to } \right)} } \mathop {e_i }\limits^ \to - \sum\limits_{j = 1}^3 {\sum\limits_{i = 1}^3 {\left( {\mathop {e_j }\limits^ \to \bullet \mathop {e_i }\limits^ \to } \right)} } \partial _j \partial _i \mathop F\limits^ \to
[/tex]

Where I have used the vector triple product.

[tex]
= \sum\limits_{j = 1}^3 {\sum\limits_{i = 1}^3 {\left( {\mathop {e_j }\limits^ \to \bullet \partial _j \partial _i \mathop F\limits^ \to } \right)} } \mathop {e_i }\limits^ \to - \sum\limits_{j = 1}^3 {\sum\limits_{i = 1}^3 {\partial _j ^2 \mathop F\limits^ \to } }
[/tex] since the second sum kind of disappears when i and j are not equal and when i and j are equal the dot product is equal to one.

[tex]
= \sum\limits_{j = 1}^3 {\sum\limits_{i = 1}^3 {\left( {\mathop {e_j }\limits^ \to \bullet \partial _j \partial _i \mathop F\limits^ \to } \right)} } \mathop {e_i }\limits^ \to - \nabla ^2 \mathop F\limits^ \to
[/tex]

At this point I'm having trouble trying to get a div in there somewhere. Is there anything I can do with the stuff inside the remaining double sum? Any help would be good thanks.
 
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  • #2
notice that the LHS equals to:

[ F'yz - F'zy , F'xz - F'zx , F'xy - F'yx ]
in differential math, F'ab = F'ba (where a,b = x,y,z)
so you get [0 , 0 , 0] on the LHS.
hope this helps...
 
  • #3
Benny, your notation is not familiar to me. (What is the first term in your summation for the curl?) What would your notation for the gradient of scalar field be? Are you familiar with the alternating epsilon notation?


greytomato said:
notice that the LHS equals to:

[ F'yz - F'zy , F'xz - F'zx , F'xy - F'yx ]
in differential math, (where a,b = x,y,z)
so you get [0 , 0 , 0] on the LHS.
hope this helps...

If the LHS were identically zero, then (for example) the usual derivation to obtain the wave equation from the Maxwell Equations would fail.

Note that F'ab = F'ba (which appears to be the equality of mixed partials) is true when the object whose mixed-partial-derivatives are being taken is a scalar field.
 
  • #4
[tex]
\nabla \times \mathop F\limits^ \to = \sum\limits_{i = 1}^3 {\mathop {e_i }\limits^ \to \times } \partial _i \mathop F\limits^ \to
[/tex]

So if i = 1 then e_i = e_1 = (1,0,0) and [itex]\partial _i \mathop F\limits^ \to = \partial _1 \mathop F\limits^ \to = \frac{{\partial \mathop F\limits^ \to }}{{\partial x_i }}[/itex].

The gradient of a scalar field would be [itex]\nabla f = \sum\limits_{i = 1}^n {\mathop {e_i }\limits^ \to } \partial _i f[/itex].

In any case I think I've got it now but it involved moving the squiggly d's around a bit.:bugeye:
 
  • #5
FYA
Allowing summation convention,
[tex]\nabla \times \left( {\nabla \times \mathop F\limits^ \to } \right) [/tex]
[tex]= \frac{\partial}{\partial x^{i}} \mathop {e_i }\limits^ \to \times \left( {\frac{\partial}{\partial x^{j}} \mathop {e_j }\limits^ \to \times \mathop F\limits^ \to } \right) [/tex]
[tex]= \frac{\partial}{\partial x^{i}} \frac{\partial}{\partial x^{j}} \left( { \mathop {e_i }\limits^ \to \times \left( { \mathop {e_j }\limits^ \to \times \mathop F\limits^ \to } \right) } \right) [/tex]
[tex]= \frac{\partial}{\partial x^{i}} \frac{\partial}{\partial x^{j}} \left( { \mathop {e_j }\limits^ \to \left( { \mathop F\limits^ \to \bullet \mathop {e_i }\limits^ \to } \right) - \mathop F\limits^ \to \left( { \mathop {e_i }\limits^ \to \bullet \mathop {e_j }\limits^ \to } \right) } \right) [/tex]
[tex]= \mathop {e_j }\limits^ \to \frac{\partial^{2} F_i}{\partial x^{i} \partial x^{j}} - \frac{\partial^{2} \mathop F\limits^ \to}{\partial x^{i} \partial x^{j}} \delta_{ij} [/tex]
[tex]= \mathop {e_j }\limits^ \to \frac{\partial}{\partial x^{j}} \left( { \frac{\partial F_i}{\partial x^{i}} } \right) - \frac{\partial^{2} \mathop F\limits^ \to}{\partial x^{i} \partial x^{i}} [/tex]
[tex]= \nabla \left( { \nabla \bullet \mathop F\limits^ \to } \right) - \nabla \bullet \left( { \nabla \otimes \mathop F\limits^ \to } \right) [/tex]
where
[tex] \nabla \bullet \left( { \nabla \otimes \mathop F\limits^ \to } \right) = \nabla^{2} \mathop F\limits^ \to [/tex]
 
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  • #6
I can't believe this thread was revived after well over a year. It brings back memories though. This was one of the problems I was struggling on while working through my problem booklet.

Since then I've learned to use the summtation convention and I can get the answer out in a few lines without going to double sums! Although I think the notation I use is slightly different to yours.
 
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  • #7
I happened to find this thread while searching for 'vector identity'.
I would appreciate if you show the slight difference for my study in your spare time.
Thank you in advance.
 
  • #8
There isn't much difference, it's just that I've been using cartesian tensor methods and an identity involving the epsilon and delta tensors to do problems with with cross products. I won't have much free time in the next few weeks so if you're interested in different notations my suggestion would be to just look them up in books.
 

FAQ: Proving Identity of $\nabla \times (\nabla \times F)$

What is the meaning of $\nabla \times (\nabla \times F)$?

$\nabla \times (\nabla \times F)$ is a mathematical operation that represents the curl of the curl of a vector field $F$. In other words, it is the curl of the curl of the gradient of a scalar field.

How do you prove the identity of $\nabla \times (\nabla \times F)$?

The identity of $\nabla \times (\nabla \times F)$ can be proven using the vector calculus identity $\nabla \times (\nabla \times F) = \nabla(\nabla \cdot F) - \nabla^2 F$, where $\nabla \cdot F$ represents the divergence of the vector field $F$ and $\nabla^2 F$ represents the Laplacian of the scalar field $F$.

What are some real-world applications of $\nabla \times (\nabla \times F)$?

$\nabla \times (\nabla \times F)$ has many applications in physics and engineering, such as in the study of fluid dynamics, electromagnetism, and heat transfer. It can also be used in computer graphics to simulate realistic fluid motion and in image processing to enhance edge detection.

Can $\nabla \times (\nabla \times F)$ ever equal zero?

Yes, $\nabla \times (\nabla \times F)$ can equal zero in certain scenarios, such as when the vector field $F$ is irrotational (has zero curl) or when the scalar field $F$ is harmonic (satisfies Laplace's equation).

Are there any alternative ways to express $\nabla \times (\nabla \times F)$?

Yes, $\nabla \times (\nabla \times F)$ can also be written as $\nabla \cdot (\nabla \times F) - \nabla^2 F$ or as $\nabla^2 (\nabla \cdot F) - \nabla \times (\nabla \times F)$, depending on the specific application and desired form.

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