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.


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 $\mathbf{B}$.


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.

  • 1
    $\begingroup$ @Christoffer One important consequence of this abuse of notation is that one cannot assume all of the usual properties of dot products apply (although some do still hold). In particular, don’t assume that $A\cdot\nabla$ means the same as $\nabla\cdot A$, but go back to what the notation literally means to decide whether it’s true (or if it even makes sense to write). $\endgroup$ – Erick Wong Jul 6 at 17:32

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