div, grad, curl

**jedoob** · April 26th 2006, 04:55 AM

I have the following query:

I need to verfiy:

curl(AxB) = (B. $\nabla$ )A - B(divA) - (A. $\nabla$ )B + A(divB)

for A=(1,0,0) and B=(x,y,z).

I can do the left side and the div terms of the right side but I am unsure of how to go about working out the other two terms with the $\nabla$ in them.

Please could someone explain to me how to calculate these.

Thanks.

**topsquark** · April 26th 2006, 10:57 AM

Originally Posted by jedoob

I have the following query:

I need to verfiy:

curl(AxB) = (B. $\nabla$ )A - B(divA) - (A. $\nabla$ )B + A(divB)

for A=(1,0,0) and B=(x,y,z).

I can do the left side and the div terms of the right side but I am unsure of how to go about working out the other two terms with the $\nabla$ in them.

Please could someone explain to me how to calculate these.

Thanks.

By definition $\nabla$ is a vector. So:
$\nabla \equiv \left ( \partial _x , \, \partial _y , \, \partial _z \right )$

That means:
$A \cdot \nabla = \left ( 1 \partial _x + 0 \partial _y + 0 \partial_z \right ) = \partial _x$
(Recall that the dot product always produces a scalar quantity.)

and
$B \cdot \nabla = \left ( x \partial _x + y \partial _y + z \partial_z \right )$

Note that the order of these are important: $A \cdot \nabla \neq \nabla \cdot A$ .

That means that
$\left ( A \cdot \nabla \right ) B = \partial _x (x,y,z) = (1,0,0)$

and
$\left ( B \cdot \nabla \right ) A = \left ( x \partial _x + y \partial _y + z \partial_z \right ) (1,0,0) = 0$
since A is a constant vector.

-Dan

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