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

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The discussion revolves around the emergence of extra terms when applying the curl operator to the cross product of two vector fields, specifically in the expression ∇ × (A × B). Participants clarify that the bac-cab rule does not apply directly due to the gradient operator being a differential operator, which introduces additional terms. The component formalism is used to derive the correct expression, showing that ∇ × (A × B) results in four terms rather than the expected two. The conversation also addresses misunderstandings about applying product rules and the distinction between curl and divergence operations. The complexities of vector calculus in this context highlight the necessity of careful manipulation of differential operators.
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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