Expand $(\vec{A}\times \nabla)\times \vec{B}$ using tensorial notation

Question

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.

Answer 1

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.

Answer 2

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.