index notation for $(A \times \nabla) \times B$

Question

I can't figure out this one, been trying for hours but always get wrong answer.

Simplify with index notation

$(A\times \nabla) \times B$

Can anyone just show me how to do it once? I think i will get the hang of it if i just see it.

Answer 1

I will use Einstein notation, that is, a product of two quantities with the same index means summation so $a_i b_i$ means $\sum_{i=1}^3 a_i b_i$ (I assumed the dimension of your vector space is $3$).

I will also use the Levi-Civita symbol $\varepsilon_{ijk}$ which has properties $\varepsilon_{123} = 1$ and $\varepsilon_{ijk} = - \varepsilon_{jik}$, and $\varepsilon_{ijk} = - \varepsilon_{ikj}$ (exchanging any two of its indices switches the sign).

Levi-Civita symbol can be used to express the components of cross product. If $ \mathbf{a} \times \mathbf{b} = \mathbf{c}$ then $i$-th component $c_i$ of vector $\mathbf{c}$ has

$$ c_i = \varepsilon_{ijk} a_j b_k $$

Levi-Civita symbol also has the property

$$\varepsilon_{ijk} \varepsilon_{imn} = \delta_{jm} \delta_{kn} - \delta_{jn} \delta_{km} \tag{1}$$

where $\delta_{ij}$ is Kronecker delta which has property

$$ \delta_{ij} a_j = a_i \tag{2}$$

So we come back to our example. Let's compute $i$-th component of $\mathbf{A} \times \pmb{\nabla}$:

$$ [\mathbf{A} \times \pmb{\nabla}]_i = \varepsilon_{ijk} A_j \nabla_k $$

Now we can compute $m$-th component of the whole vector $(\mathbf{A} \times \pmb{\nabla}) \times \mathbf{B}$ because we can view it as cross product of $\mathbf{A} \times \pmb{\nabla}$ and $\mathbf{B}$.

$$ [(\mathbf{A} \times \pmb{\nabla}) \times \mathbf{B}]_m = \varepsilon_{min} (\varepsilon_{ijk} A_j \nabla_k) B_n = - \varepsilon_{imn} \varepsilon_{ijk} A_j \nabla_k B_n = - (\delta_{jm}\delta_{kn} - \delta_{jn}\delta_{km}) A_j \nabla_k B_n = - \delta_{jm}\delta_{kn} A_j \nabla_k B_n + \delta_{jn}\delta_{km} A_j \nabla_k B_n = - A_m \nabla_k B_k + A_n \nabla_m B_n $$

where we used properties $(1)$ and $(2)$.

Note: In my opinion, it could be seen more easily without using index notation:

$$ (\mathbf{A} \times \pmb{\nabla}) \times \mathbf{B} = (\mathbf{A} \times \pmb{\nabla}) \times \overset{\downarrow}{\mathbf{B}} = - \overset{\downarrow}{\mathbf{B}} \times (\mathbf{A} \times \pmb{\nabla}) = - \mathbf{A} (\overset{\downarrow}{\mathbf{B}} \cdot \pmb{\nabla}) + \pmb{\nabla} (\overset{\downarrow}{\mathbf{B}} \cdot \mathbf{A}) = \pmb{\nabla} (\overset{\downarrow}{\mathbf{B}} \cdot \mathbf{A}) - \mathbf{A} \operatorname{div} \mathbf{B} $$

Answer 2

If $A$ is a constant vector, then $$ \begin{align} (A\times\nabla)\times B &=\left(\begin{bmatrix} a_1\\a_2\\a_3 \end{bmatrix} \times \begin{bmatrix} \partial_1\\\partial_2\\\partial_3 \end{bmatrix}\right) \times \begin{bmatrix} b_1\\b_2\\b_3 \end{bmatrix}\\ &=\begin{bmatrix} a_2\partial_3-a_3\partial_2\\ a_3\partial_1-a_1\partial_3\\ a_1\partial_2-a_2\partial_1 \end{bmatrix} \times \begin{bmatrix} b_1\\b_2\\b_3 \end{bmatrix}\\[3pt] &=\begin{bmatrix} a_3\partial_1b_3-a_1\partial_3b_3-a_1\partial_2b_2+a_2\partial_1b_2\\ a_1\partial_2b_1-a_2\partial_1b_1-a_2\partial_3b_3+a_3\partial_2b_3\\ a_2\partial_3b_2-a_3\partial_2b_2-a_3\partial_1b_1+a_1\partial_2b_1 \end{bmatrix}\\[3pt] &=\begin{bmatrix} a_3\partial_1b_3+a_1\partial_1b_1+a_2\partial_1b_2\\ a_1\partial_2b_1+a_2\partial_2b_2+a_3\partial_2b_3\\ a_2\partial_3b_2+a_3\partial_3b_3+a_1\partial_2b_1 \end{bmatrix} -\begin{bmatrix} a_1\partial_1b_1+a_1\partial_2b_2+a_1\partial_3b_3\\ a_2\partial_1b_1+a_2\partial_2b_2+a_2\partial_3b_3\\ a_3\partial_1b_1+a_3\partial_2b_2+a_3\partial_3b_3 \end{bmatrix}\\[6pt] &=\nabla(A\cdot B)-A\nabla\cdot B \end{align} $$ The last step assumes that $A$ is constant.