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

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Discussion Overview

The discussion centers on the mathematical expressions $\vec{A}\cdot\nabla$ and $\nabla\cdot\vec{A}$, exploring why they are not equivalent. Participants examine the nature of these expressions within the context of vector calculus, focusing on their interpretations and implications in differential operations.

Discussion Character

  • Technical explanation
  • Debate/contested

Main Points Raised

  • Some participants question whether there was a typo in the original question regarding the expressions.
  • It is noted that $\vec{A}\cdot\nabla$ is a differential operator, while $\nabla\cdot\vec{A}$ results in a scalar value when applied to a function.
  • One participant suggests applying both expressions to an arbitrary function to observe the differences in results.
  • There is a discussion about the notation, with some participants referring to "Dot Del" as an "abuse of notation," while others argue that both expressions involve non-standard uses of notation.
  • One participant emphasizes that when $\vec{A}$ represents a velocity field, $\vec{A}\cdot\nabla$ can be interpreted as the convective derivative operator.
  • Another participant challenges the interpretation of $\nabla$ as a vector, suggesting that the notation for both expressions could be considered abusive.
  • Clarifications are made regarding the mathematical representation of $A \cdot \nabla$ as a sum of partial derivatives rather than a straightforward dot product.

Areas of Agreement / Disagreement

Participants express differing views on the notation and interpretation of the expressions. There is no consensus on whether the notation is an abuse or on the implications of these expressions in vector calculus.

Contextual Notes

Some participants highlight the potential confusion arising from the notation and the interpretations of $\nabla$ and $\vec{A}$, indicating that the discussion may depend on specific definitions and contexts.

Swapnil
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Why is it that

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

?

edit: sorry about that. fixed the typo.
 
Last edited:
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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.
 
Last edited:
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.
 
Last edited:
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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