What is the significance of this curl product rule?

  • Thread starter Destroxia
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  • #1
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Homework Statement



Verify the identity:

## \nabla \times ( A \times B) = (B\bullet \nabla)A - (A\bullet\nabla)B + A(\nabla \bullet B)-B(\nabla\bullet A)##

My issue here is I don't understand the significance of why a term has B or A on the left of the dot product, and another has B or A on the right? (Eg ## B(\nabla)

Also, what is the significance of the ## \nabla ## being on the left, or right, of what it is multiplying? (Eg ## \langle \frac {d} {dx}, \frac {d} {dy}, \frac {d} {dz} \rangle \bullet \langle a_1, a_2, a_3 \rangle = \langle a_1, a_2, a_3 \rangle \bullet \langle \frac {d} {dx}, \frac {d} {dy}, \frac {d} {dz} \rangle##, or if the ## \nabla ## is on the right, does that leave the operator open to work on whichever comes AFTER the dot product?

Homework Equations




The Attempt at a Solution



I have my solution attached as a thumbnail, because it's pretty hefty.

problem.png


Why does my solution look different than the given solution at the top?
 

Answers and Replies

  • #2
andrewkirk
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does that leave the operator open to work on whichever comes AFTER the dot product?
Yes.
A somewhat less confusing - but perhaps less memorable - way to write the identity is:

$$\nabla\times (A\times B)\equiv B\cdot[\nabla A]-A\cdot[\nabla B]+(\nabla\cdot B)A-(\nabla\cdot A)B$$
the items in square brackets are vector fields and those in round parentheses are scalar fields. So the latter can be applied to vector fields by scalar multiplication.
 
  • #3
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Yes.
A somewhat less confusing - but perhaps less memorable - way to write the identity is:

$$\nabla\times (A\times B)\equiv B\cdot[\nabla A]-A\cdot[\nabla B]+(\nabla\cdot B)A-(\nabla\cdot A)B$$
the items in square brackets are vector fields and those in round parentheses are scalar fields. So the latter can be applied to vector fields by scalar multiplication.
Okay, I think I get it, but how come in my solution it seems like the A always comes first? I'm not sure what I did wrong, and why am I interpreting the ## \nabla ## as always being on the left?
 
  • #4
andrewkirk
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@RyanTAsher Your derivation goes wrong in the fourth line - the long one that you split across two lines. The elements of your first coordinate are all derivatives with respect to ##x##, which is the one coordinate with respect to which no derivatives should appear in the first coordinate of the curl.
 
  • #5
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@RyanTAsher Your derivation goes wrong in the fourth line - the long one that you split across two lines. The elements of your first coordinate are all derivatives with respect to ##x##, which is the one coordinate with respect to which no derivatives should appear in the first coordinate of the curl.
I'm clearly misinterpreting something then... I think I've fixed, but still not sure how this affects the order of the A, B, and also not sure of how to resolve this into the final answer. I've only done the first component.

hmmm.png
 

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