# what does $(A\cdot\nabla)B$ mean?

I was studying a physics book and I saw this expression $$(A\cdot\nabla)B$$ where $A$ and $B$ are vectors. What's the definition of this?

I've also seen this in some identities

• Note that $(A\cdot \nabla)\phi = A\cdot (\nabla \phi)$ is the directional derivative of the scalar field $\phi$ in the direction of $A$. Your expression is just the vector analog of this (which is also usually called the directional derivative).
– user137731
Commented Jul 1, 2016 at 20:33

The idea is that, say, $A=(A_1,A_2,A_3)$ and $\nabla = (\partial/\partial x_1,\partial/\partial x_2,\partial/\partial x_3)$, so formally $$A\cdot\nabla=A_1\frac{\partial}{\partial x_1}+A_2\frac{\partial}{\partial x_2}+A_3\frac{\partial}{\partial x_3},$$so$$(A\cdot\nabla)B=A_1\frac{\partial B}{\partial x_1}+A_2\frac{\partial B}{\partial x_2}+A_3\frac{\partial B}{\partial x_3}.$$

• Your $B$ is a scalar field in the last equation. Since OP requires $B = (B_1, B_2, B_3)$, that last part should be more like $(A\cdot\nabla)B=(A_1\frac{\partial B_1}{\partial x_1}+A_2\frac{\partial B_1}{\partial x_2}+A_3\frac{\partial B_1}{\partial x_3}, A_1\frac{\partial B_2}{\partial x_1}+A_2\frac{\partial B_2}{\partial x_2}+A_3\frac{\partial B_2}{\partial x_3}, \ldots)$
– csha
Commented Aug 24, 2018 at 10:24
• Your version is the same as mine. For example, since $B=(B_1,B_2,B_3)$, it's clear that $\frac{\partial B}{\partial x_1}= \left( \frac{\partial B_1}{\partial x_1}, \frac{\partial B_2}{\partial x_1}, \frac{\partial B_3}{\partial x_1}\right)$. Commented Dec 6, 2021 at 10:51

We can expand the vector notation for $\left(\vec A\cdot \nabla\right)\vec B$ as

$$\left(\vec A\cdot \nabla\right)\vec B=\left(A_x\frac{\partial \vec B}{\partial x}+A_y\frac{\partial \vec B}{\partial y}+A_z\frac{\partial \vec B}{\partial z}\right)$$

In tensor notation, this can be more compactly written as

$$\left(\left(\vec A\cdot \nabla\right)\vec B\right)_i=A_j\partial_j(B_i)$$

where summation over the index $j$ is implied.

If you're in $\Bbb{R}^3$ (usually the case in physics) and $\mathbf{A}$ is a vector and ${B}$ is a scalar

\begin{align}\mathbf{A}=&(A_x,A_y,A_z)\\\nabla=&(\nabla_x,\nabla_y,\nabla_z)\\(\mathbf{A}\cdot\nabla)=&A_x{\partial\over \partial{x}}+A_y{\partial\over \partial{y}}+A_z{\partial\over \partial{z}}\end{align}

And so

$$(\mathbf{A}\cdot\nabla){B}=A_x{\partial{B}\over \partial{x}}+A_y{\partial{B}\over \partial{y}}+A_z{\partial{B}\over \partial{z}}$$

• Not really. $A\cdot\nabla$ is not quite what you say. There's only one cdot in $(A\cdot\nabla)B$. Commented Jul 1, 2016 at 20:05
• this isn't (A⋅∇)⋅B this is : (A⋅∇)B Commented Jul 1, 2016 at 20:05
• Sigh. Your expression for $A\cdot\nabla$ is simply wrong. You don't have to fix it if you don't want to, but it's wrong. You statement regarding what $(A\cdot\nabla)B$ means if $B$ is a scalar is also simply wrong. Commented Jul 1, 2016 at 20:26
• You're right ! Now that I look at it on my desktop and not my smartphone. I will fix Commented Jul 1, 2016 at 20:41
• Much better. Of course now it's identical to an earlier answer... Commented Jul 1, 2016 at 20:50