How to compute $(\mathbf{A} \cdot \mathbf{\nabla})\mathbf{B}$? I'm currently reading Intro to Electrodynamics by Griffiths, and in the maths section, there is the following problem:
"If $\mathbf{A}$ and $\mathbf{B}$ are two vector functions, what does the expression $(\mathbf{A} \cdot \mathbf{\nabla})\mathbf{B}$ mean?
(That is, what are its $x$, $y$, and $z$ components, in terms of the Cartesian components of $\mathbf{A}$, $\mathbf{B}$, and $\mathbf{\nabla}$?"
First, I naively thought it was just the divergence of $\mathbf{A}$ multiplied by $\mathbf{B}$, and that for some reason Griffiths wrote the divergence as $\mathbf{A} \cdot \mathbf{\nabla}$ this time, rather than $\mathbf{\nabla} \cdot \mathbf{B}$, which is the way i'm used to seeing it.
But when i looked in the solution manual, it says the answer is
$
\begin{align}
(\mathbf{A} \cdot \mathbf{\nabla}) &= \left(
A_x \frac{\partial B_x}{\partial x} + A_y \frac{\partial B_x}{\partial y} + A_z \frac{\partial B_x}{\partial z}
\right) \mathbf{\hat{x}} \\ &+ 
\left(
A_x \frac{\partial B_y}{\partial x} + A_y \frac{\partial B_y}{\partial y} + A_z \frac{\partial B_y}{\partial z}
\right) \mathbf{\hat{y}} \\ &+ 
\left(
A_x \frac{\partial B_z}{\partial x} + A_y \frac{\partial B_z}{\partial y} + A_z \frac{\partial B_z}{\partial z}
\right) \mathbf{\hat{z}}
\end{align}
$
I thought this question was weird because Griffiths hadn't used this notation yet until now, so I'm not sure why he thought I would be able to do this problem. I know it's not the typo or anything, because the next problem is similar, as it wants me to find $(\mathbf{\hat{r}} \cdot \mathbf{\nabla})\mathbf{\hat{r}}$
So, I guess my question is, what does this expression mean, and how do I calculate it? Obviously I'm not going to memorise this mess, and the notation seems to suggest a dot product is somehow involved.
 A: It means that the differential operator
$$
\mathbf{A} \cdot \nabla
= (A_x,A_y,A_z) \cdot (\partial_x,\partial_y,\partial_z)
= A_x \partial_x + A_y \partial_y + A_z \partial_z
$$
acts componentwise on the vector $\mathbf{B}$.
A: Think of the $\nabla$ symbol as the following vector:
$$\nabla = \left(\frac{\partial}{\partial x}, \frac{\partial}{\partial y},\frac{\partial}{\partial z}\right)$$
As you can see, this is not a normal vector, but a vector of differential operators. This is a little bit of abuse of notation, but it makes certain formulas much easier to express. Now, if we take $A\cdot \nabla$, we can calculate the resulting operator using the dot product:
$$A\cdot \nabla = A_x\frac{\partial}{\partial x} +A_y\frac{\partial}{\partial y} +A_z\frac{\partial}{\partial z}$$
As you can see, this is the operation that appears in every component of $(A\cdot \nabla)B$. Now, to apply $A\cdot \nabla$ to $B$, we simply transform each component of $B$ using this operator. For example, here is the x-component of $(A\cdot \nabla)B$:
$$(A\cdot \nabla)B_x= A_x\frac{\partial B_x}{\partial x} +A_y\frac{\partial B_x}{\partial y} +A_z\frac{\partial B_x}{\partial z}$$
Hopefully, this helps you understand the formula which Griffiths' gave for $(A\cdot \nabla)B$. I will leave it to you to derive the y-component and z-component of this vector.
