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A Dot Del

  1. Feb 20, 2007 #1
    Why is it that

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


    edit: sorry about that. fixed the typo.
    Last edited: Feb 21, 2007
  2. jcsd
  3. Feb 20, 2007 #2
    I assume there is some kind of typo in this question?
  4. Feb 20, 2007 #3
    Did you mean [tex]\vec{A}\cdot\nabla \neq \nabla\cdot\vec{A}[/tex]?

    If so then evaluate them both, what do you notice?
  5. Feb 21, 2007 #4


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    The first is a differential operator, while the second is a number (value of a function).
  6. Feb 21, 2007 #5

    George Jones

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    Apply both sides to an arbitary function f = f(x,y,xz). What do you get?
  7. Feb 21, 2007 #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.
    Last edited: Feb 21, 2007
  8. Feb 21, 2007 #7


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    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.
    Last edited: Feb 21, 2007
  9. Feb 21, 2007 #8


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    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]
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