# How does $(\vec{a}\cdot\vec{\nabla})\vec{p}=\vec{a}$?

Say we have a vector field $\vec{a}$ such that

$$\vec{a}=\begin{bmatrix}a_1(x_1,x_2,\ldots,x_n) \\ a_2(x_1,x_2,\ldots,x_n)\\ \vdots \\a_n(x_1,x_2,\ldots,x_n)\end{bmatrix}.$$

Then in an orthonormal basis the divergence $\vec{a}\cdot\vec{\nabla}$ is $$\vec{a}\cdot\vec{\nabla}=\frac{\partial[a_j(x_k)]}{\partial x_i}\vec{e_i}\cdot\vec{e_j}=\frac{\partial[a_j(x_k)]}{\partial x_i}\delta_{ij}=\frac{\partial[a_j(x_k)]}{\partial x_j}\left(=\frac{\partial a_j}{\partial x_k}\frac{\partial x_k}{\partial x_j}=\frac{\partial a_j}{\partial x_k}\delta_{kj}=\frac{\partial a_j}{\partial x_j}\right).$$

Einstein's summation convention has been applied here, therefore

$$\vec{a}\cdot\vec{\nabla}=\sum_{j=1}^n\frac{\partial a_j}{\partial x_j}.$$Let us also have a position vector denoted by $\vec{p}$ whose coordinates in the same orthonormal basis are given as $$\vec{p}=\begin{bmatrix}x_1 \\ x_2\\ \vdots \\ x_n\end{bmatrix}.$$

Now scaling the position vector by the divergence yields a new vector:

$$(\vec{a}\cdot\vec{\nabla})\vec{p}=\begin{bmatrix}x_1\frac{\partial a_j}{\partial x_j} \\ x_2\frac{\partial a_j}{\partial x_j}\\ \vdots \\ x_n \frac{\partial a_j}{\partial x_j}\end{bmatrix}.$$

Answer key shows $$(\vec{a}\cdot\vec{\nabla})\vec{p}=\vec{a},$$ or, in other words,

$$\begin{bmatrix}x_1\frac{\partial a_j}{\partial x_j} \\ x_2\frac{\partial a_j}{\partial x_j}\\ \vdots \\x_n\frac{\partial a_j}{\partial x_j} \end{bmatrix} = \begin{bmatrix} a_1(x_1,x_2,\ldots,x_n) \\ a_2(x_1,x_2,\ldots,x_n)\\ \vdots \\a_n(x_1,x_2,\ldots,x_n)\end{bmatrix}.$$

How?

• Note the standard use of \ldots and \vdots, as in my edit. (I use \ldots in things like $(a,\ldots,z)$ and \cdots in things like $a+\cdots+z$. I think if you just use \dots it will appear as \ldots in the former context and \cdots in the latter. If you just type "..." then, at least in using actual LaTeX rather than MathJax (which is what is used here) you will see $(a,\text{...},z)$ rather than $(a,\ldots,z)$, and that's one of the reasons why that form is deprecated. $\qquad$ Oct 5, 2016 at 16:46
• $(\vec a\cdot \nabla) \vec p= a_i\partial_i x_l\vec e_l=a_ie_i$ Oct 5, 2016 at 16:49
• $\vec a\cdot \nabla$ isn't a divergence, you get something like a weighted gradient if you apply it to another vector Oct 5, 2016 at 16:54
• math.stackexchange.com/questions/1382174/… Oct 5, 2016 at 17:08
• @tired: Well, damn those physicists and their notation. :) I simply assumed without much thought that since they are taking great lengths showing analogy with a dot product $$\vec{a}\cdot\vec{\nabla}=\vec{\nabla}\cdot\vec{a}.$$ Oct 5, 2016 at 17:11

Are you certain that the divergence is defined as a RHS operator and not $$\nabla \cdot \vec a$$