# Product Rule

1. May 15, 2008

### qspeechc

Hello everyone. I'm trying to get my head around this product rule:

$$\nabla \times (A\times B) = (B\cdot \nabla )A - (A\cdot \nabla )B + A(\nabla \cdot B) - B(\nabla \cdot A)$$

Ok, we have this

$$\nabla = (\partial /\partial x,\partial/\partial y,\partial /\partial z)$$

and for dot products

$$a\cdot b = b\cdot a$$

Therefore in the product rule given above, is it not the case

$$(B\cdot \nabla )A = A(\nabla \cdot B)$$

and similarly, the other two terms on the RHS are equal?
Thank-you for your help.

2. May 15, 2008

### G01

Be careful. $$\nabla$$ is a vector operator not a vector. It will not commute the way you expect it to.

3. May 15, 2008

### HallsofIvy

Staff Emeritus
In fact, I would much prefer the notation $B\cdot(\nabla A)$ to $(B\cdot\nabla)A$.

4. May 15, 2008

### qspeechc

Aha! I think I get it now! The brackets were confusing, because usually we have to evaluate the stuff in the brackets first right?

Does this mean $$\nabla$$ always acts on the vector directly to its right?

5. May 15, 2008

### qspeechc

Hold on, this can't be right can it? Then we would have
$$\nabla \times (A\times B) = 0$$

wouldn't we? Can someone please tell me what $$(B\cdot\nabla)A[/itex] is? 6. May 15, 2008 ### dx First you evaluate [tex] \nabla A$$. You get a vector field, i.e. a vector at every point of space. Then $$B \cdot(\nabla A)$$ would be vector field whose value at any point is the dot of B with vector $$\nabla A$$ at that point.

Last edited: May 15, 2008
7. May 15, 2008

### qspeechc

I'm sorry for being extremely thick, but then doesn't that mean that

$$\nabla \times (A\times B) = 0$$?

This makes no sense, because it means the cross product of any two vectors has zero curl? Surely $$(B\cdot \nabla)\cdot A$$ is not the same thing as $$B\cdot (\nabla \cdot A)$$?

8. May 15, 2008

### dx

why does it mean that?

9. May 15, 2008

### Vid

The second operation not a dot product. del.A is a scalar. So (del.A)B is the scalar (d/dxAx + d/dyAy + d/dzAz) times the vector B = d/dxAx*B + d/dyAy*B + d/dzAz*B

Also, remember the del is an operator so del.A is not the same as A.del. A.del is still a scalar though being applied to B to it gets pretty messy looking.

http://mathworld.wolfram.com/ConvectiveOperator.html

Last edited: May 15, 2008
10. May 15, 2008

### dx

No, what I said was $$B \cdot (\nabla A)$$ is the same thing as $$(B \cdot \nabla)A$$

11. May 15, 2008

### qspeechc

Er, ok, what's the difference? A and B are vectors.

12. May 15, 2008

### qspeechc

Oh, ok, this site explains it to me. Thanks for all your help everyone, taking time to answer my stupid questions

13. May 15, 2008

### lurflurf

Some people do not like to take the gradient of a vector since it is a dyad, and it makes them feel ickky inside.

14. May 15, 2008

### lurflurf

This is a confusing aspect of vector calculus.
$$\nabla$$ acts to the right only.
One may define a bidirectional del/nabla
consider this bidirectional derivative
it is a good and correct habit when working with vectors to switch to bidirectional form
-hold all function left of operators constant
-change operators to birectional
-perfom manipulations ending with a form easy to conver to unidirectional form
-convert

recall this identity when working with products

$$\mathbf{(a\times\nabla)\times b+a\nabla\cdot b=a\times(\nabla\times b)+(a\cdot\nabla)b}$$

15. May 15, 2008

### Dick

It's only ickky if you don't know that there are things which aren't vectors or scalars. There are also tensors (or dyads). Then it could make you queasy.