Why is $\vec{A}\cdot\nabla \neq \nabla\cdot\vec{A}$?

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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?
 
Swapnil said:
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?
 
The first is a differential operator, while the second is a number (value of a function).
 
Swapnil said:
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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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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ObsessiveMathsFreak said:
"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})

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})
 
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