# A Dot Del

Why is it that

$$\vec{A}\cdot\nabla \neq \nabla\cdot\vec{A}$$

?

edit: sorry about that. fixed the typo.

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I assume there is some kind of typo in this question?

Why is it that

$$\vec{A}\cdot\nabla \neq \vec{A}\cdot\nabla$$

?

Did you mean $$\vec{A}\cdot\nabla \neq \nabla\cdot\vec{A}$$?

If so then evaluate them both, what do you notice?

dextercioby
Homework Helper
The first is a differential operator, while the second is a number (value of a function).

George Jones
Staff Emeritus
Gold Member
Why is it that

$$\vec{A}\cdot\nabla \neq \nabla\cdot\vec{A}$$

?

edit: sorry about that. fixed the typo.

Apply both sides to an arbitary function f = f(x,y,xz). What do you get?

"Dot Del" is what is technically referred to as an "abuse of notation". Of course, some people consider del to be an abuse of notation in itself.

You see, by itself $$\nabla$$ is just $$(\frac{\partial}{\partial x_1},\frac{\partial}{\partial x_2}, \ldots , \frac{\partial}{\partial x_n})$$. It's nice but all the derivatives have the same coefficient on them (i.e. one).

To allow for more general operators we use $$A \cdot \nabla$$ to stand for $$(a_1 \frac{\partial}{\partial x_1} , a_2 \frac{\partial}{\partial x_2} , \ldots , a_n \frac{\partial}{\partial x_n})$$ where $$A = (a_1,a_2, \ldots, a_n)$$. It's confusing because usually the dot product is associative, but now we're demanding it no be for $$\nabla$$.

I'm not a great fan for this notation myself, but since I've got nothing better to offer, I guess I'll just have to live with it.

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arildno
Homework Helper
Gold Member
Dearly Missed
Abuse??
When A has the interpretation of a velocity field, then $\vec{A}\cdot\nabla$ has the very nice interpretation of the convective derivative operator.

The only "abuse" I'm able to see is that the "dot product" is defined for vectors, while the $\nabla$ operator isn't a vector at all.
However, then we must agree that $\nabla\cdot\vec{A}$ is an equally abusive notation.

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HallsofIvy
You see, by itself $$\nabla$$ is just $$(\frac{\partial}{\partial x_1},\frac{\partial}{\partial x_2}, \ldots , \frac{\partial}{\partial x_n})$$. It's nice but all the derivatives have the same coefficient on them (i.e. one).
To allow for more general operators we use $$A \cdot \nabla$$ to stand for $$(a_1 \frac{\partial}{\partial x_1} , a_2 \frac{\partial}{\partial x_2} , \ldots , a_n \frac{\partial}{\partial x_n})$$
No, it's not. $A \cdot \nabla$ is
$$(a_1 \frac{\partial}{\partial x_1} + a_2 \frac{\partial}{\partial x_2} + \ldots + a_n \frac{\partial}{\partial x_n})$$