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

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Discussion Overview

The discussion centers on the emergence of additional terms when applying the triple vector product involving the del operator, specifically in the context of the expression $\nabla \times (A \times B)$. Participants explore the implications of the bac-cab rule and the role of the gradient operator in this context, examining both theoretical and mathematical aspects.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants express confusion about why the bac-cab rule does not apply when the gradient operator is involved, suggesting that del is a vector but also a differential operator that requires something to operate on.
  • One participant provides a detailed component formalism to derive the expression for $\nabla \times (A \times B)$, indicating that it leads to additional terms compared to the standard bac-cab rule.
  • Another participant attempts to use a determinant approach to derive the same expression, questioning why the operators are not applied to the terms that follow them, indicating a potential misunderstanding of the product rule in this context.
  • There is a clarification that the product rule used in the determinant approach is not applicable to the triple vector product, highlighting a distinction between different vector operations.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the application of the bac-cab rule with the gradient operator, and there are competing views on the interpretation of the mathematical expressions involved.

Contextual Notes

Some limitations in understanding arise from the dependence on definitions of vector operations and the specific properties of the gradient operator, which are not fully resolved in the discussion.

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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