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

ognik
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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 ?
 
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ognik said:
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:
[math]\nabla \times \left ( A \times B \right )[/math]

This implies
[math]\epsilon _{ijk} ~ \epsilon _{kmn} ~ \partial _{j} ~ (A_m ~ B_n)[/math]

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

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

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

[math]= \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 ] [/math]

Moving the vectors around in each term where needed:
[math]= B_j ~ \partial _{j} ~ A_i + A_i ~ \partial _{j} ~ B_j - B_i ~ \partial _{j} ~ A_j - A_j ~ \partial _{j} ~ B_i [/math]

Which implies:
[math]\nabla \times \left ( A \times B \right ) = ( B \cdot \nabla ) A + A (\nabla \cdot B) - B( \nabla \cdot A) - (A \cdot \nabla ) B[/math]

-Dan
 
Interesting & ta, but why can't we apply bac-cab - del is a vector?
 
ognik said:
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 [math]\nabla \times (A \times B)[/math] vs. [math]A \times (B \times C)[/math].

-Dan
 
topsquark said:
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:
[math]\nabla \times \left ( A \times B \right )[/math]

This implies
[math]\epsilon _{ijk} ~ \epsilon _{kmn} ~ \partial _{j} ~ (A_m ~ B_n)[/math]
-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?
 
ognik said:
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:
[math]\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 | [/math]

[math]= \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 )[/math]

This is not the same as [math]\nabla \times ( A \times B )[/math].

-Dan
 

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