# Vector analysis notation question

what exactly does

$$(\mathbf{A} \cdot \nabla) \mathbf{B}$$

mean?

is this the same as the divergence of A multiplied by B?

if this was the case, why wouldn't it be written in a clearer notation?

edit: i'm having trouble with the latex.

further edit: note to self--use two spaces. :tongue:

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Hurkyl
Staff Emeritus
Gold Member
I think you meant this:

$$(\mathbf{A} \cdot \nabla) \mathbf{B}$$

(you don't need the dollar signs)

Well, I'll give you a hint to help you get started on working it out yourself: (always better than having someone give you the answer! )

$$\frac{d}{dx} x f(x) \neq x \frac{d}{dx} f(x)$$

(usually, anyways)

Hurkyl said:
(you don't need the dollar signs)
oh, thanks.

also, for some reason, it didn't like me using only one space between the symbols. Hurkyl
Staff Emeritus
Gold Member
That could have been a trick of the browser: in mine, I generally have to manually reload a LaTeX image after changing it. Your browser may not have reloaded it after your first change, but for some reason did after changing it again.

alright, i think i got it now...

(ok, here goes the latex...)

$$A_x \frac{\partial}{\partial x} \mathbf{B} + A_y \frac{\partial}{\partial y} \mathbf{B} + A_z \frac{\partial}{\partial z} \mathbf{B}$$

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Hurkyl said:
That could have been a trick of the browser: in mine, I generally have to manually reload a LaTeX image after changing it. Your browser may not have reloaded it after your first change, but for some reason did after changing it again.

yeah, that seems to be it. i have to refresh the browser afterwards, apparently.

Hurkyl
Staff Emeritus
Gold Member
Almost right:

$$A_x \frac{\partial}{\partial x} \mathbf{B} + A_y \frac{\partial}{\partial y} \mathbf{B} + A_z \frac{\partial}{\partial z} \mathbf{B}$$

But I'll chalk that up to a silly mistake, probably catalyzed by my hint!

(Incidentally, B doesn't have to be a vector in this expression)

and (for the sake of latex practice!) i'll compare it to

$$( \nabla \cdot \mathbf{A}) \mathbf{B}$$

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

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Hurkyl
Staff Emeritus
Gold Member
Oh, here's you're next challenge. What do you think:

$$(\vec{A} \cdot \nabla)^2 f$$

means? (I'm assuming A is a constant)

Hurkyl said:
Almost right:

$$A_x \frac{\partial}{\partial x} \mathbf{B} + A_y \frac{\partial}{\partial y} \mathbf{B} + A_z \frac{\partial}{\partial z} \mathbf{B}$$

But I'll chalk that up to a silly mistake, probably catalyzed by my hint!

(Incidentally, B doesn't have to be a vector in this expression)

that's the same thing that i put in my post!

either you caught the entry in a (most likely long) period of latex innacuracy, or that answer isn't right? :grumpy:

so...before i get to work on that next one (which seems a bit harder...), please tell me if my answer is correct. Hurkyl said:
Oh, here's you're next challenge. What do you think:

$$(\vec{A} \cdot \nabla)^2 f$$

means? (I'm assuming A is a constant)

...i HOPE that it's just the partial derivatives being turned into second partial derivatives... otherwise... it's quite a big mess.

i just don't feel *right* treating these operators like algebraic quantities... : /

Hurkyl
Staff Emeritus
Gold Member
Yah, I must have caught you before you got it all cleaned up. Yes, it's right!

I'm probably going to log off soon after this, so I'll leave you with the punchline I wanted to work up to: the multivariable Taylor series!

$$f(\vec{x_0} + \vec{h}) = f(\vec{x_0}) + \frac{1}{1!} (\vec{h} \cdot \nabla) f(\vec{x_0}) + \frac{1}{2!} (\vec{h} \cdot \nabla)^2 f(\vec{x_0}) + \frac{1}{3!} (\vec{h} \cdot \nabla)^3 f(\vec{x_0}) + \cdots$$

(Where this notation means you're supposed to do the differentiations first, then plug in $\vec{x_0}$)

If I haven't messed that up, it works when f is a vector function as well. (Of course, f has to be analytic, or else it can't be given by a Taylor series)

By the way, it just struck me, and I feel really silly not noticing this before: it seems to me that the operator $\mathbf{A} \cdot \nabla$ is just the directional derivative operator $\nabla_{\mathbf{A}}$... does that sound right?

Hurkyl
Staff Emeritus
Gold Member
i just don't feel *right* treating these operators like algebraic quantities... : /
Yah, it is a disturbing (but cool) thing until you figure out what's going on. Bleh, it still seems disturbing!

The trick is that multiplication of operators really means composition, and usually raising an operator to a power means you just multiply (i.e. compose) it by itself a bunch of times. But, fortunately, these operators are linear, so their multiplication distributes over addition!

(And yes, it's a mess... which is why I'm only asking you to do ( )², and not ( )³. )

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Hurkyl said:
Yah, it is a disturbing (but cool) thing until you figure out what's going on. Bleh, it still seems disturbing!

The trick is that multiplication of operators really means composition, and usually raising an operator to a power means you just multiply (i.e. compose) it by itself a bunch of times. But, fortunately, these operators are linear, so their multiplication distributes over addition!

(And yes, it's a mess... which is why I'm only asking you to do ( )², and not ( )³. )

yeah, i've seen this in arfken's book. :yuck:

Hurkyl
Staff Emeritus