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

  • Thread starter Swapnil
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In summary: NOT the same as (a_1 \frac{\partial}{\partial x_1} , a_2 \frac{\partial}{\partial x_2} , \ldots , a_n \frac{\partial}{\partial x_n}).In summary, the conversation discusses the notation of \vec{A}\cdot\nabla and \nabla\cdot\vec{A} and their differences. While the former is a differential operator, the latter is a numerical value of a function. The conversation also delves into the use of dot product and its association with \nabla. It is noted that some may consider this notation an "abuse" but it has its uses in interpreting velocity fields.
  • #1
Swapnil
459
6
Why is it that

[tex]\vec{A}\cdot\nabla \neq \nabla\cdot\vec{A}[/tex]

?

edit: sorry about that. fixed the typo.
 
Last edited:
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  • #2
I assume there is some kind of typo in this question?
 
  • #3
Swapnil said:
Why is it that

[tex]\vec{A}\cdot\nabla \neq \vec{A}\cdot\nabla[/tex]

?

Did you mean [tex]\vec{A}\cdot\nabla \neq \nabla\cdot\vec{A}[/tex]?

If so then evaluate them both, what do you notice?
 
  • #4
The first is a differential operator, while the second is a number (value of a function).
 
  • #5
Swapnil said:
Why is it that

[tex]\vec{A}\cdot\nabla \neq \nabla\cdot\vec{A}[/tex]

?

edit: sorry about that. fixed the typo.

Apply both sides to an arbitary function f = f(x,y,xz). What do you get?
 
  • #6
"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 [tex]\nabla[/tex] is just [tex](\frac{\partial}{\partial x_1},\frac{\partial}{\partial x_2}, \ldots , \frac{\partial}{\partial x_n})[/tex]. It's nice but all the derivatives have the same coefficient on them (i.e. one).

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

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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  • #7
Abuse??
When A has the interpretation of a velocity field, then [itex]\vec{A}\cdot\nabla[/itex] 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 [itex]\nabla[/itex] operator isn't a vector at all.
However, then we must agree that [itex]\nabla\cdot\vec{A}[/itex] is an equally abusive notation.
 
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  • #8
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 [tex]\nabla[/tex] is just [tex](\frac{\partial}{\partial x_1},\frac{\partial}{\partial x_2}, \ldots , \frac{\partial}{\partial x_n})[/tex]. It's nice but all the derivatives have the same coefficient on them (i.e. one).

To allow for more general operators we use [tex]A \cdot \nabla[/tex] to stand for [tex](a_1 \frac{\partial}{\partial x_1} , a_2 \frac{\partial}{\partial x_2} , \ldots , a_n \frac{\partial}{\partial x_n})[/tex]

No, it's not. [itex]A \cdot \nabla[/itex] is
[tex](a_1 \frac{\partial}{\partial x_1} + a_2 \frac{\partial}{\partial x_2} + \ldots + a_n \frac{\partial}{\partial x_n})[/tex]
 

1. What is the meaning of $\vec{A}\cdot\nabla$ and $\nabla\cdot\vec{A}$?

The operation $\vec{A}\cdot\nabla$ is known as the dot product, which is a way to combine two vectors to get a scalar quantity. The operation $\nabla\cdot\vec{A}$ is known as the divergence, which is a way to measure the outflow of a vector field at a given point.

2. Why is $\vec{A}\cdot\nabla$ not equal to $\nabla\cdot\vec{A}$?

These two operations are not equal because they involve different mathematical operations. The dot product involves multiplying the components of two vectors, while the divergence involves taking the partial derivatives of a vector field. Therefore, the order of the operations and the resulting quantities are different.

3. Can you give an example where $\vec{A}\cdot\nabla \neq \nabla\cdot\vec{A}$?

Yes, for example, let $\vec{A} = (x, y, z)$ be a vector field. Then, $\vec{A}\cdot\nabla = x\frac{\partial}{\partial x} + y\frac{\partial}{\partial y} + z\frac{\partial}{\partial z}$. On the other hand, $\nabla\cdot\vec{A} = \frac{\partial x}{\partial x} + \frac{\partial y}{\partial y} + \frac{\partial z}{\partial z} = 3$. These two operations are not equal, as the first operation gives a vector field while the second operation gives a scalar value.

4. Is there a physical significance to the difference between $\vec{A}\cdot\nabla$ and $\nabla\cdot\vec{A}$?

Yes, the difference between these two operations has important physical implications. The dot product represents the directional derivative of a vector field, while the divergence represents the net flow of the vector field at a given point. Therefore, these operations have different physical interpretations and are used in different contexts in physics and engineering.

5. Can $\vec{A}\cdot\nabla$ and $\nabla\cdot\vec{A}$ ever be equal?

Yes, under certain conditions, these two operations can be equal. For example, if the vector field $\vec{A}$ is a constant vector, then both operations will give the same result. Additionally, in some coordinate systems, such as orthogonal curvilinear coordinate systems, these operations can also be equal. However, in general, these operations are not equal and should not be treated as such.

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