How can $(A \times \nabla) \times B$ be rearranged?

Given 3 vector spaces $$A$$, $$B$$ & $$C$$, I am able to derive that $$A \times (B \times C)$$ is equal to $$(A \cdot C)B-(A \cdot B)C$$.

However, I do not think I can apply this identity to $$(A \times \nabla) \times B$$, because I would get: $$(A \times \nabla) \times B = (A \cdot B)\nabla - (A \cdot \nabla)B$$ Which is clearly wrong, since $$(A \cdot \nabla)\nabla$$ is not a vector space.

Working with the Levi-Civita symbol yields:

$$\epsilon_{ijk}\epsilon_{jlm}A_l\partial_mB_k=A_k \partial_i B_k - A_i\partial_kB_k$$

Which doesn't seem to be particularly useful in this case.

• When you throw a nabla in there, the regular identities don't work, because nabla isn't a vector so much as it is an operator. For example: $$\nabla\times(\mathbf{A}\times\mathbf{B})=\mathbf{A}(\nabla\cdot\mathbf{B})-\mathbf{B}(\nabla\cdot\mathbf{A})+(\mathbf{B}\cdot\nabla)\mathbf{A}-(\mathbf{A}\cdot\nabla)\mathbf{B}.$$ – Adrian Keister Aug 12 at 16:24
• Yep & thanks for correcting typo. There is still a way to rearrange the expression I believe. – zabop Aug 12 at 16:25
• Yes, there is, as your expression is well-defined. You might consider going to the Levi-Civita notation and working it out yourself. – Adrian Keister Aug 12 at 16:26
• (Thanks! edited the OP) – zabop Aug 12 at 16:31
• Even if it did work, you applied the identity wrong. It would be $\nabla(\mathbf{A}\cdot\mathbf{B}) - \mathbf{A}(\nabla\cdot\mathbf{B})$, which isn't too far off from the proper answer. – eyeballfrog Aug 12 at 16:34

Here's an approach that I like. First, note that we can write $$\nabla \times F = \sum_{i=1}^3 e_i \times \frac{\partial F}{\partial x_i}$$ where $$e_1,e_2,e_3$$ denote $$\hat i, \hat j, \hat k$$, and $$\frac{\partial F}{\partial x_i} = \left(\frac{\partial F^{(x_1)}}{x_i},\frac{\partial F^{(x_2)}}{x_i}, \frac{\partial F^{(x_3)}}{x_i}\right).$$ With that in mind, we can write \begin{align*} (A \times \nabla) \times B &= \sum_{i=1}^3 (A \times e_i) \times \frac{\partial B}{\partial x_i} \\ & = \sum_{i=1}^3 \left[\left(A \cdot \frac{\partial B}{\partial x_i}\right) e_i - \left (\frac{\partial B}{\partial x_i} \cdot e_i \right)A\right] \\ & = \sum_{i=1}^3 \left(A \cdot \frac{\partial B}{\partial x_i}\right) e_i - A\sum_{i=1}^3\frac{\partial B}{\partial x_i} \cdot e_i \\ & = \sum_{i=1}^3 \left(A \cdot \frac{\partial B}{\partial x_i}\right) e_i - A(\nabla \cdot B) \end{align*} According to the other answer, it seems that the first term can be rewritten as $$A \cdot (\nabla B)$$.
We have (using Einstein summation notation): \begin{align*} (\mathbf{A}\times\nabla)_i&=\varepsilon_{ijk}A_j\partial_k \\ [(\mathbf{A}\times\nabla)\times \mathbf{B}]_{m}&=\varepsilon_{mil}\underbrace{\varepsilon_{ijk}A_j\partial_k}_{(\mathbf{A}\times\nabla)_i}B_l \\ &=-\varepsilon_{iml}\varepsilon_{ijk}A_j\partial_kB_l \\ &=-(\delta_{mj}\delta_{lk}-\delta_{mk}\delta_{lj})A_j\partial_kB_l \\ &=(\delta_{mk}\delta_{lj}-\delta_{mj}\delta_{lk})A_j\partial_kB_l \\ &=\delta_{mk}\delta_{lj}A_j\partial_kB_l-\delta_{mj}\delta_{lk}A_j\partial_kB_l \\ &=\delta_{mk}A_j\partial_kB_j-\delta_{mj}A_j\partial_kB_k \\ &=A_j\partial_mB_j-A_m\partial_kB_k, \end{align*} as you obtained (though with different dummy variables - an inconsequential difference). The final step is to recognize this in terms of vectors and the various products available to us (dot, scalar, cross). We have $$(\mathbf{A}\times\nabla)\times \mathbf{B}=\mathbf{A}\cdot(\nabla\mathbf{B})-\mathbf{A}(\nabla\cdot\mathbf{B}).$$ Now, the notation $$\nabla\mathbf{B}$$ requires some explanation. It's the Jacobian matrix $$\nabla\mathbf{B}=\left(\frac{\partial B_i}{\partial x_j}\right)_{ij},$$ a second-rank tensor, so that the dot product $$\mathbf{A}\cdot(\nabla\mathbf{B})$$ is still a vector, as we need (the result should definitely be a vector!).