# Curl(A x B)

1. May 9, 2010

### SingBlueSilva

1. The problem statement, all variables and given/known data
I need to prove curl(A x B) = AdivB - BdivA + (B·∇)A - (A·∇)B
But I keep on getting confused in the numbers. I tried taking the cross product A x B and crossing that into the gradient, but I just get lost. I also tried going from the right side of the equation and get lost in there as well.

Can someone show me a clear way of doing this proof or put me in the right direction?

Thank you

2. May 9, 2010

### LCKurtz

There isn't a shortcut to these type of formulas. Just work both sides out. It might help slightly to avoid subscripts in the vectors so try letting

A = < u, v, w >
B = < r, s, t >

all functions of x, y, z, and work both sides. Hopefully you will see how to group things to show they are equal, assuming they are.

3. May 9, 2010

### vela

Staff Emeritus
Are you familiar with the Levi-Civita symbol εijk? Using it can make these sorts of proofs a lot easier.

4. May 9, 2010

### SingBlueSilva

No, I’m not familiar with that.

Just to be clear, curl(A x B) = AdivB - BdivA + (B·∇)A - (A·∇)B = 2AdivB - 2BdivA, correct?

5. May 9, 2010

### vela

Staff Emeritus
Bummer.
Nope.

6. May 9, 2010

### SingBlueSilva

scratch that last post, it was dumb.

Well I started by taking the left side of the equation and get to this:

[d/dy(us-rv)+d/dz(ut-rw)]i-[d/dx(us-vr)-d/dz(vt-ws)]j+[d/dx(ut-wr)-d/dy(vt-ws)]k

if I add d/dx(ur-ru) to the i component, d/dy(uv-vu) to the j component and d/dz(uw-wu) to the k component, I can come up with the first two terms. But I can’t figure out the rest.

7. May 9, 2010

### vela

Staff Emeritus
In the i term, for example, you should have terms like

r du/dx + s du/dy + t du/dz = (r d/dx + s d/dy + t d/dz) u

That's the x-component of (B⋅∇) A.