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I have several problems like these, I am learning about Levi-Civita and Kroenecker delta so those may be relevant, however, what I am really looking for is further discussion or good books/online tutorials that are specific to usage of the del operator.

I'm actually going to give you two problems because they're so well related anyway.

1. (this is p46 of Arfken & Weber 6th edition Mathematical Methods - its an example!)

[tex]\nabla[/tex](A.B) = (B.[tex]\nabla[/tex])A + (A.[tex]\nabla[/tex])B + Bx([tex]\nabla[/tex] x A) + A x ([tex]\nabla[/tex] x B)

2.

[tex]\nabla[/tex](A.B) = (A x [tex]\nabla[/tex]) x B + (B x [tex]\nabla[/tex]) x A + A([tex]\nabla[/tex] . B) + B([tex]\nabla[/tex] . A)

eijkeilm = delta jl delta km = delta jm delta kl

Ax(BxC) = B(A.C) - (C.A)B

K.. I just lost my post. But basically I got (immediately):

[tex]\nabla[/tex](A.B) = A.[tex]\nabla[/tex]B + B.[tex]\nabla[/tex]A

and I do not know how they came up with these cross products from there. For instance, B.[tex]\nabla[/tex]A came from Bi dA/dxi (these should be partial derivatives, curly d's, to be clear). Am I supposed to show (B.[tex]\nabla[/tex])A + Bx([tex]\nabla[/tex] x A) gives same answer in components as that? Eyeballing it seemed like it might, but when I tried it, I got Bx([tex]\nabla[/tex] x A) = 0.

Specifically, if you look at the writeup in Afken (p46), he cheats. He starts by applying the bac-cab rule to each of the cross products on the RHS, using the commutative property to move around the del operator. He then says he adds two resulting equations together, which normally I'd say, ok Afken, you're pulling it out of the hat but that seems fine, except - in this case it looks like a factor of 2 has been removed from the sum with no explanation. I'm sorry for not typing that out more clearly but the last time I did , the forum chewed it up. Hopefully someone has the book and can verify that mistake or explain why he's correct?.

## Homework Statement

I'm actually going to give you two problems because they're so well related anyway.

1. (this is p46 of Arfken & Weber 6th edition Mathematical Methods - its an example!)

[tex]\nabla[/tex](A.B) = (B.[tex]\nabla[/tex])A + (A.[tex]\nabla[/tex])B + Bx([tex]\nabla[/tex] x A) + A x ([tex]\nabla[/tex] x B)

2.

[tex]\nabla[/tex](A.B) = (A x [tex]\nabla[/tex]) x B + (B x [tex]\nabla[/tex]) x A + A([tex]\nabla[/tex] . B) + B([tex]\nabla[/tex] . A)

## Homework Equations

eijkeilm = delta jl delta km = delta jm delta kl

Ax(BxC) = B(A.C) - (C.A)B

## The Attempt at a Solution

K.. I just lost my post. But basically I got (immediately):

[tex]\nabla[/tex](A.B) = A.[tex]\nabla[/tex]B + B.[tex]\nabla[/tex]A

and I do not know how they came up with these cross products from there. For instance, B.[tex]\nabla[/tex]A came from Bi dA/dxi (these should be partial derivatives, curly d's, to be clear). Am I supposed to show (B.[tex]\nabla[/tex])A + Bx([tex]\nabla[/tex] x A) gives same answer in components as that? Eyeballing it seemed like it might, but when I tried it, I got Bx([tex]\nabla[/tex] x A) = 0.

Specifically, if you look at the writeup in Afken (p46), he cheats. He starts by applying the bac-cab rule to each of the cross products on the RHS, using the commutative property to move around the del operator. He then says he adds two resulting equations together, which normally I'd say, ok Afken, you're pulling it out of the hat but that seems fine, except - in this case it looks like a factor of 2 has been removed from the sum with no explanation. I'm sorry for not typing that out more clearly but the last time I did , the forum chewed it up. Hopefully someone has the book and can verify that mistake or explain why he's correct?.

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