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I was given a task to prove $$(\vec{A}\times \nabla)\times \vec{B} = (\vec A \cdot \nabla)\vec B + \vec A \times \operatorname{rot} \vec B - \vec A \operatorname{div} B$$ using tensorial notation i.e. Kroneker delta and Levi-Civita symbol. Here are my wrong calculations: $$\epsilon_{ijk}A_j(\nabla \times B)_k = \epsilon_{ijk} A_j (\epsilon_{klm} \frac{\partial}{\partial x_l} B_m) = \epsilon_{kij}\epsilon_{klm}A_j \frac{\partial B_m}{\partial x_l} = (\delta_{il} \delta_{jm} - \delta_{im} \delta_{jl} A_j \frac{\partial B_m}{\partial x_l}) = A_j \frac{\partial B_j}{\partial x_l} - A_l \frac{\partial B_m}{\partial x_l} = A_j \frac{\partial B_j}{\partial x_l} - (A \cdot \nabla)B$$ It is not even close to what I wanted. I do not understand how I can ever derive three terms from initial expression. Please, help me out.

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    $\begingroup$ I'm not yet sure how to do it with the tensorial approach, I need to ponder it a bit, but, I hope you can see from my current half-answer why there are three terms. You get one from raw derivatives acting on $\vec{B}$ then the terms like $\partial_2A_3$ act on $\vec{B}$ under a product rule hence producing two terms; $2+1=3$. $\endgroup$ – James S. Cook Mar 1 '15 at 21:55
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    $\begingroup$ I think we need to antisymmetrize on the $\vec{A} \times \nabla$ term as to pick up the product rule. Unfortunately the usual formula $\vec{A} \times \vec{D}_k = \epsilon_{ijk}A_iD_j$ assumes that $\vec{A}$ and $\vec{D}$ have commuting components. Of course in this context that is false as $\vec{D}= \nabla$ is an operator so... we have to respect that. $\endgroup$ – James S. Cook Mar 1 '15 at 22:08
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Let ${\bf e}_i$ be unit vectors then $${\bf A}\times \nabla = [ \epsilon_{ijk}A_j\partial_k]{\bf e}_i$$

By the same formula it follows that $$({\bf A}\times \nabla)\times {\bf B} = \epsilon_{i'j'k'} (A\times\nabla)_{j'} B_{k'} {\bf e}_{i'} = \epsilon_{i'j'k'}[\epsilon_{j'jk}A_j\partial_k]B_{k'}{\bf e}_{i'}$$

Now by using the property $\epsilon_{ijk}\epsilon_{imn} = \delta_{jm}\delta_{km}-\delta_{jn}\delta_{km}$ we get

$$({\bf A}\times \nabla)\times {\bf B} = [\delta_{i'k}\delta_{jk'}-\delta_{i'j}\delta_{k'k}] A_j\partial_kB_{k'}{\bf e}_{i'} = [A_{k'}\partial_{i'}B_{k'} - (\partial_{k'}B_{k'})A_{i'}]{\bf e}_{i'}$$

The last term is simply $-{\bf A}(\text{div }{\bf B})$. Rewriting the first term it becomes

$$A_{k'}\partial_{i'}B_{k'}{\bf e}_{i'} = A_{k'}[\partial_{i'}B_{k'}-\partial_{k'}B_{i'}]{\bf e}_{i'} + A_{k'}\partial_{k'}B_{i'}{\bf e}_{i'} = ({\bf A}\times\text{rot }{\bf B}) + ({\bf A}\cdot \nabla){\bf B}$$

Putting it all togeather gives the desired result. The derivation of

$$({\bf A}\times\text{rot }{\bf B})\equiv {\bf A}\times (\nabla \times {\bf B}) = \ldots = A_{k'}[\partial_{i'}B_{k'}-\partial_{k'}B_{i'}]{\bf e}_{i'}$$

follows the same steps as above using the $\epsilon\epsilon = \delta\delta-\delta\delta$ formula.

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  • $\begingroup$ Oh, why is it $A_{k′} \partial _{i′} B_{k′} e_{i′}=A_{k′}[\partial _{i′} B_{k′} − \partial _{k′} B_{i′}]e_{i′}$. Why did you split it into two parts and changed subscripts? Thank you in advance. $\endgroup$ – Dankevich Mar 2 '15 at 19:19
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    $\begingroup$ @Dankevich What I did there was basically just to write $X = (X-Y) + Y$, i.e. I added and subtracted the same term (motivated by the answer). The $'Y'$ term I subtract is just $(A\cdot\nabla)B$. $\endgroup$ – Winther Mar 2 '15 at 19:25
  • $\begingroup$ Now I finally got it! Yay! $\endgroup$ – Dankevich Mar 2 '15 at 19:28
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Ok, so the question is what is meant by $\vec{A} \times \nabla$. Notice: $$ \vec{A} \times \nabla = \hat{x_1}(A_2 \partial_3-\partial_2 A_3)+\hat{x_2}(A_3 \partial_1-\partial_3 A_1)+\hat{x_3}(A_1 \partial_2-\partial_2 A_1)$$ when you feed this operator $\vec{B}$ under the cross product you have to think about product rules for the terms with partial derivatives to the left and component functions to the right.

It is not at all the same to start with $\nabla \times \vec{B}$. Even without the operator, the cross product is not associative. I haven't answered your question yet, but this is too long for a comment and I thought it would help for now.

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  • $\begingroup$ Haha, I understand you, but I simply can not write it like this, it will not count. And I still do not get how to do the first step formally. How do I deal with brackets in such cases, like in example I gave and in $\nabla \times (\vec A \times \vec B)$, for instance? It is rather not obvious:) $\endgroup$ – Dankevich Mar 1 '15 at 22:07
  • $\begingroup$ I'm sorry, I have trouble with this one. I need to stew on it a while, perhaps someone will swoop in and solve it before I finish this.... I haven't found the right way to write the operator expression in the Einstein notation just yet. $\endgroup$ – James S. Cook Mar 1 '15 at 22:17

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