Understanding the Product Rule to \nabla × (A×B)

[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.

 

[G01]

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

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

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

 

[HallsofIvy]

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

 

[qspeechc]

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

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?

 

[qspeechc]

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

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?

 

[dx]

Can someone please tell me what (B\cdot\nabla)A[/itex] is?

<br /> <br /> First you evaluate \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.

 

[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)?

 

[dx]

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

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

why does it mean that?

 

[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

 

[dx]

Surely (B\cdot \nabla)\cdot A is not the same thing as B\cdot (\nabla \cdot A)?

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

 

[qspeechc]

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

 

[qspeechc]

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

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

 

[lurflurf]

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

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

 

[lurflurf]

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?

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
Dab=aDb=abD
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}

 

[Dick]

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

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.