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