# How to use index notation to prove one identity for $\textbf{a}\cdot\nabla\textbf{b}\times\textbf{c}-\textbf{c}\cdot\nabla\textbf{b}\times\textbf{a}$?

Try to show this identity \begin{align} \textbf{a}\cdot\nabla\textbf{b}\times\textbf{c}-\textbf{c}\cdot\nabla\textbf{b}\times\textbf{a}=[\nabla\cdot\textbf{b}\textbf{I}-\nabla\textbf{b}]\cdot(\textbf{a}\times\textbf{c}). \end{align} where $$\textbf{I}$$ is the identity matrix or 2nd order tensor $$\delta_{ij}$$. But when it is expanded with the Levi-Civita symbols, the next is completely lost. $$\epsilon_{jkl}a_ic_l\partial_ib_k-\epsilon_{jmn}c_pa_n\partial_pb_m$$ Are there important tricks I didnot know to use index notation correctly?

• $b$ should probably not have an index. Otherwise, how is the vector $\nabla\boldsymbol{b}$ defined ? Hint to solve the rest: $\boldsymbol{v}\times\boldsymbol{w}=\sum_{i,j=1}^3\epsilon_{ijk}v_iw_j\,.$ Then drop the $\sum\,.$ That's all index notation is about. Einstein summation convention. Jan 27, 2022 at 20:11
• @Kurt G. Think $\textbf{a}\cdot\nabla$ as a scalar then $\textbf{b}$ as a vector cross products $\textbf{c}$. Jan 28, 2022 at 10:42

I cannot make sense of this equation.

The expression $$\nabla\cdot\boldsymbol{b}\boldsymbol{I}-\nabla\boldsymbol{b}$$ can only be a rather clumsy way of writing the antisymmetric matrix:

$$B=\left( \begin{matrix}0&\partial_2b_1-\partial_1b_2&\partial_3b_1-\partial_1b_3\\ \partial_1b_2-\partial_2b_1&0&\partial_3b_2-\partial_2b_3\\ \partial_1b_3-\partial_3b_1&\partial_2b_3-\partial_3b_2&0\end{matrix}\right)\,.$$ It is easy to see that for any vector $$\boldsymbol{v}\,,$$ $$B\boldsymbol{v}=-(\nabla\times\boldsymbol{b})\times\boldsymbol{v}= \boldsymbol{v}\times(\nabla\times\boldsymbol{b})\,.$$ If my interpretation so far is correct it looks like you are trying to show $$(\boldsymbol{a}\cdot\nabla)\boldsymbol{b}\times\boldsymbol{c}- (\boldsymbol{c}\cdot\nabla)\boldsymbol{b}\times\boldsymbol{a}= B(\boldsymbol{a}\times\boldsymbol{c})\,,$$ which is the same as $$\tag{1} (\boldsymbol{a}\cdot\nabla)\boldsymbol{b}\times\boldsymbol{c}- (\boldsymbol{c}\cdot\nabla)\boldsymbol{b}\times\boldsymbol{a}= (\boldsymbol{a}\times\boldsymbol{c})\times(\nabla\times\boldsymbol{b})\,.$$ To see that this does not hold consider $$\boldsymbol{a}=\left(\begin{matrix}1\\0\\0\end{matrix}\right)\,,\quad \boldsymbol{b}=\left(\begin{matrix}x\\y\\z\end{matrix}\right)\,,\quad \boldsymbol{c}=\left(\begin{matrix}0\\1\\0\end{matrix}\right)\,.$$ Then $$(\boldsymbol{a}\cdot\nabla)=\partial_1$$, $$(\boldsymbol{c}\cdot\nabla)=\partial_2$$ so that $$(\boldsymbol{a}\cdot\nabla)\boldsymbol{b}=\boldsymbol{a}\,,\quad (\boldsymbol{c}\cdot\nabla)\boldsymbol{b}=\boldsymbol{c}\,.$$ Therefore the LHS of (1) is $$\boldsymbol{a}\times \boldsymbol{c}-\boldsymbol{c}\times\boldsymbol{a}= 2\boldsymbol{a}\times \boldsymbol{c}= \left(\begin{matrix}0\\0\\2\end{matrix}\right)\,.$$ On the other hand, $$\nabla\times\boldsymbol{b}=\boldsymbol{0}$$ so that the RHS of (1) is zero.

An equation that is true is obtained by the Grassmann identity applied to the RHS of (1): $$\tag{2} (\boldsymbol{c}\cdot(\nabla\times\boldsymbol{b}))\,\boldsymbol{a}- (\boldsymbol{a}\cdot(\nabla\times\boldsymbol{b}))\,\boldsymbol{c} =(\boldsymbol{a}\times\boldsymbol{c})\times(\nabla\times\boldsymbol{b})\,.$$

If the matrix in the brackets on the RHS of your equation is meant to be $$(\nabla\cdot\boldsymbol{b})\boldsymbol{I}-\nabla\boldsymbol{b}$$ then it can be written as

$$C=\left( \begin{matrix}\partial_2b_2+\partial_3b_3&-\partial_1b_2&-\partial_1b_3\\ -\partial_2b_1&\partial_1b_1+\partial_3b_3&-\partial_2b_3\\ -\partial_3b_1&-\partial_3b_2&\partial_1b_1+\partial_2b_2\end{matrix}\right)\,.$$ According to your own separate answer: $$(\boldsymbol{a}\cdot\nabla)\boldsymbol{b}\times\boldsymbol{c}- (\boldsymbol{c}\cdot\nabla)\boldsymbol{b}\times\boldsymbol{a}= C(\boldsymbol{a}\times\boldsymbol{c})\,.$$

Let's take the first component to see if there is any story lying behind: \begin{align} &=\boldsymbol{\mathbf{e}}_1\cdot\bigl(\mathbf{a}\cdot\nabla\mathbf{b}\times\mathbf{c}-\mathbf{c}\cdot\nabla\mathbf{b}\times\mathbf{a}\bigr)\\ &=\boldsymbol{\mathbf{e}}_1\bigl(a_1\partial_1b_2c_3+a_2\partial_2b_2c_3+a_3\partial_3b_2c_3-a_1\partial_1b_3c_2-a_2\partial_2b_3c_2-a_3\partial_3b_3c_2\nonumber\\ &\phantom{=}-c_1\partial_1b_2a_3-c_2\partial_2b_2a_3-c_3\partial_3b_2a_3+c_1\partial_1b_3a_2+c_2\partial_2b_3a_2+c_3\partial_3b_3a_2\bigr)\nonumber\\ &=\boldsymbol{\mathbf{e}}_1\bigl(-\partial_1b_2(a_3c_1-a_1c_3)-\partial_1b_3(a_1c_2-a_2c_1)+(\partial_2b_2+\partial_3b_3)(a_2c_3-a_3c_2)\bigr)\label{eq:e1} \end{align} From the last term, $$\partial_2b_2+\partial_3b_3$$ is just the divergence missing the $$\partial_1b_1$$, so what we need is to add $$\partial_1b_1(a_2c_3-a_3c_2)$$ and subtract it. \begin{align} &\boldsymbol{\mathbf{e}}_1\cdot\bigl(\mathbf{a}\cdot\nabla\mathbf{b}\times\mathbf{c}-\mathbf{c}\cdot\nabla\mathbf{b}\times\mathbf{a}\bigr)=\boldsymbol{\mathbf{e}}_1\partial_jb_j(a_2c_3-a_3c_2)-\boldsymbol{\mathbf{e}}_1\partial_1b_j(\mathbf{a}\times\mathbf{c})_j\label{eq:e1_1} \end{align} The same can be done to other two components, therefore \begin{align} \mathbf{a}\cdot\nabla\mathbf{b}\times\mathbf{c}-\mathbf{c}\cdot\nabla\mathbf{b}\times\mathbf{a}=\left[(\nabla\cdot\mathbf{b})\mathbf{I}-\nabla\mathbf{b}\right]\cdot(\mathbf{a}\times\mathbf{c})\label{eq:D93} \end{align}

• Thanks for this clarification. Feb 1, 2022 at 18:32