Hi, I am having an immense amount of trouble trying to show that the following identity is true.(adsbygoogle = window.adsbygoogle || []).push({});

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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# Homework Help: Vector identity

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