# Curl of (A*B)

1. Nov 24, 2009

1. The problem statement, all variables and given/known data

Prove that $$\nabla\times(A\times B)= (B.\nabla)A-(A.\nabla)B-B(\nabla.A)+A(\nabla.B)$$

2. Relevant equations

bac-cab $$\nabla\times(A\times B)= (\nabla.B)A-(\nabla.A)B$$

3. The attempt at a solution

I know that $$B(\nabla.A)=(\nabla.A)B$$ and $$A(\nabla.B)= (\nabla.B)A$$

So what about $$(B.\nabla)A-(A.\nabla)B=?$$ Does it equal to zero? Or maybe bac cab is not related to this problem!

2. Nov 24, 2009

### WiFO215

When and how is this ["bac - cab"] equation valid? The above equation is valid only as long as "a" is....?

3. Nov 24, 2009

4. Nov 24, 2009

### LCKurtz

Prove it by letting $\vec A = \langle f,g,h\rangle,\ \vec B = \langle u,v,w\rangle$ and just work out both sides to check they are equal. It's easy, just a little algebra.

5. Nov 24, 2009

### WiFO215

Another interesting way would be to do to use a gem of a trick that I learnt off of the Feynman Lectures. Refer to Feynman Lectures, Volume II, Lecture 27, Field Energy and Field Momentum.

6. Nov 24, 2009

Thank you very much for your help. I actually proved it in the way you suggested me, but only for the x-component, and it was a lot of algebra!

7. Nov 24, 2009

Thank you very much. I will try to study it.

8. Nov 25, 2009

### Pengwuino

Do you know index notation? Vector identities are quite easy with it.

9. Nov 25, 2009

Please tell me about "index notation". I went to http://en.wikipedia.org/wiki/Index_notation but I didn't completely understand your point of view!

10. Nov 25, 2009

### Pengwuino

http://www.physics.ucsb.edu/~physCS33/spring2009/index-notation.pdf [Broken]

Give this a try. Unfortunately, when I learned it, it was during lectures and not in our textbook so I can't tell you what book you can learn it out of. This should be enough though.

Last edited by a moderator: May 4, 2017
11. Nov 26, 2009