Why Do Extra Terms Emerge in the Triple Vector Product with Del?

[ognik]

I know the bac-cab rule, but add $\nabla$ and it's not so clear ..

applying it to $\nabla \times \left( A \times B \right) = A\left(\nabla \cdot B\right) - B\left(\nabla \cdot A\right) ...$, not quite

Please walk me through why the other 2 terms emerge ?

 

[topsquark]

I know the bac-cab rule, but add $\nabla$ and it's not so clear ..

applying it to $\nabla \times \left( A \times B \right) = A\left(\nabla \cdot B\right) - B\left(\nabla \cdot A\right) ...$, not quite

Please walk me through why the other 2 terms emerge ?

bac-cab won't work with the gradient operator. I don't know if you have covered this before but I don't know of an easier way:

Write it in the component formalism:
\(\displaystyle \nabla \times \left ( A \times B \right )\)

This implies
\(\displaystyle \epsilon _{ijk} ~ \epsilon _{kmn} ~ \partial _{j} ~ (A_m ~ B_n)\)

Take the derivative:
\(\displaystyle = \epsilon _{ijk} ~ \epsilon _{kmn} \left [ ( \partial _{j} ~ A_m) B_n + A_m ( \partial _{j} ~ B_n ) \right ] \)

Note the following alteration in the first \(\displaystyle \epsilon\) factor: \(\displaystyle \epsilon _{kij} = \epsilon _{ijk}\) (\(\displaystyle \epsilon\) is antisymmetric in a switch of coordinates and we are making two here.)
\(\displaystyle = \left ( \epsilon _{kij} ~ \epsilon _{kmn} \right ) \left [ ( \partial _{j} ~ A_m) B_n + A_m ( \partial _{j} ~ B_n ) \right ] \)

\(\displaystyle = \left ( \delta _{im} ~ \delta _{jn} - \delta _{in} ~ \delta _{jm} \right ) \left [ ( \partial _{j} ~ A_m) B_n + A_m ( \partial _{j} ~ B_n ) \right ] \)

\(\displaystyle = \left [ ( \partial _{j} ~ A_i ) B_j + A_i ( \partial _{j} ~ B_j ) \right ] - \left [ ( \partial _{j} ~ A_j ) B_i + A_j ( \partial _{j} ~ B_i ) \right ] \)

Moving the vectors around in each term where needed:
\(\displaystyle = B_j ~ \partial _{j} ~ A_i + A_i ~ \partial _{j} ~ B_j - B_i ~ \partial _{j} ~ A_j - A_j ~ \partial _{j} ~ B_i \)

Which implies:
\(\displaystyle \nabla \times \left ( A \times B \right ) = ( B \cdot \nabla ) A + A (\nabla \cdot B) - B( \nabla \cdot A) - (A \cdot \nabla ) B\)

-Dan

 

[ognik]

Interesting & ta, but why can't we apply bac-cab - del is a vector?

 

[topsquark]

Interesting & ta, but why can't we apply bac-cab - del is a vector?

It's a vector, but it's also a differential operator. That means it has to have something to operate on. That is the source of the extra two terms in \(\displaystyle \nabla \times (A \times B)\) vs. \(\displaystyle A \times (B \times C)\).

-Dan

 

[ognik]

bac-cab won't work with the gradient operator. I don't know if you have covered this before but I don't know of an easier way:

Write it in the component formalism:
\(\displaystyle \nabla \times \left ( A \times B \right )\)

This implies
\(\displaystyle \epsilon _{ijk} ~ \epsilon _{kmn} ~ \partial _{j} ~ (A_m ~ B_n)\)
-Dan

Indulge me please, I think I am undoing some 'wrong knowledge' here.

If I do this using the determinant ('cos it is easier for me to see), then expanding over the top row I would get: $ \partial_x\left( A_yB_z -A_zB_y \right) - \partial_y\left( A_xB_z -A_zB_x \right) + \partial_z\left( A_xB_y -A_yB_x \right) $

Using the product rule: $ = A_y\partial_xB_z + \partial_x A_yB_z - A_z\partial_xB_y + \partial_xA_zB_y - ... $
Obviously you aren't then applying the operators to what follows them (if you did they'd all = 0) - why not please?

 

[topsquark]

Indulge me please, I think I am undoing some 'wrong knowledge' here.

If I do this using the determinant ('cos it is easier for me to see), then expanding over the top row I would get: $ \partial_x\left( A_yB_z -A_zB_y \right) - \partial_y\left( A_xB_z -A_zB_x \right) + \partial_z\left( A_xB_y -A_yB_x \right) $

Using the product rule: $ = A_y\partial_xB_z + \partial_x A_yB_z - A_z\partial_xB_y + \partial_xA_zB_y - ... $
Obviously you aren't then applying the operators to what follows them (if you did they'd all = 0) - why not please?

You've got the wrong product. Here you are using the "box product" formula:
\(\displaystyle \nabla \cdot (A \times B ) = \left | \begin{matrix} \partial _x & \partial _y & \partial _z \\ A_x & A_y & A_z \\ B_x & B_y & B_z \end{matrix} \right | \)

\(\displaystyle = \partial _x \left ( A_y ~ B_z - A_z ~ B_y \right ) - \partial_y \left ( A_x~ B_z - A_z ~ B_x \right) + \partial_z \left( A_x ~ B_y -A_y ~ B_x \right )\)

This is not the same as \(\displaystyle \nabla \times ( A \times B )\).

-Dan