How to compute $(\mathbf{A} \cdot \mathbf{\nabla})\mathbf{B}$?

Question

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.

Answer 1

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 $\mathbf{B}$.

Answer 2

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.