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Vector analysis notation question

  1. Aug 27, 2005 #1
    what exactly does

    [tex]
    (\mathbf{A} \cdot \nabla) \mathbf{B}
    [/tex]

    mean?


    is this the same as the divergence of A multiplied by B?

    if this was the case, why wouldn't it be written in a clearer notation?



    edit: i'm having trouble with the latex.

    further edit: note to self--use two spaces. :tongue:
     
    Last edited: Aug 27, 2005
  2. jcsd
  3. Aug 27, 2005 #2

    Hurkyl

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    I think you meant this:

    [tex]
    (\mathbf{A} \cdot \nabla) \mathbf{B}
    [/tex]

    (you don't need the dollar signs)


    Well, I'll give you a hint to help you get started on working it out yourself: (always better than having someone give you the answer! :smile:)

    [tex]
    \frac{d}{dx} x f(x) \neq x \frac{d}{dx} f(x)
    [/tex]

    (usually, anyways)
     
  4. Aug 27, 2005 #3
    oh, thanks.

    also, for some reason, it didn't like me using only one space between the symbols. :confused:
     
  5. Aug 27, 2005 #4

    Hurkyl

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    That could have been a trick of the browser: in mine, I generally have to manually reload a LaTeX image after changing it. Your browser may not have reloaded it after your first change, but for some reason did after changing it again.
     
  6. Aug 27, 2005 #5
    alright, i think i got it now...


    (ok, here goes the latex...)


    [tex]
    A_x \frac{\partial}{\partial x} \mathbf{B} + A_y \frac{\partial}{\partial y} \mathbf{B} + A_z \frac{\partial}{\partial z} \mathbf{B}
    [/tex]
     
    Last edited: Aug 27, 2005
  7. Aug 27, 2005 #6

    yeah, that seems to be it. i have to refresh the browser afterwards, apparently.
     
  8. Aug 27, 2005 #7

    Hurkyl

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    Almost right:

    [tex]
    A_x \frac{\partial}{\partial x} \mathbf{B} + A_y \frac{\partial}{\partial y} \mathbf{B} + A_z \frac{\partial}{\partial z} \mathbf{B}
    [/tex]

    But I'll chalk that up to a silly mistake, probably catalyzed by my hint!

    (Incidentally, B doesn't have to be a vector in this expression)
     
  9. Aug 27, 2005 #8
    and (for the sake of latex practice!) i'll compare it to

    [tex]
    ( \nabla \cdot \mathbf{A}) \mathbf{B}
    [/tex]


    [tex]
    ( \nabla \cdot \mathbf{A}) \mathbf{B} = ( \frac{\partial} {\partial x} A_x + \frac{\partial} {\partial y} A_y +\frac{\partial} {\partial z} A_z ) \mathbf{B}
    [/tex]
     
    Last edited: Aug 27, 2005
  10. Aug 27, 2005 #9

    Hurkyl

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    Oh, here's you're next challenge. :smile:

    What do you think:

    [tex]
    (\vec{A} \cdot \nabla)^2 f
    [/tex]

    means? (I'm assuming A is a constant)
     
  11. Aug 27, 2005 #10

    that's the same thing that i put in my post!

    either you caught the entry in a (most likely long) period of latex innacuracy, or that answer isn't right? :grumpy:


    so...before i get to work on that next one (which seems a bit harder...), please tell me if my answer is correct. :smile:
     
  12. Aug 27, 2005 #11

    ...i HOPE that it's just the partial derivatives being turned into second partial derivatives... :biggrin:

    otherwise... it's quite a big mess.


    i just don't feel *right* treating these operators like algebraic quantities... : /
     
  13. Aug 27, 2005 #12

    Hurkyl

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    Yah, I must have caught you before you got it all cleaned up. Yes, it's right!


    I'm probably going to log off soon after this, so I'll leave you with the punchline I wanted to work up to: the multivariable Taylor series!

    [tex]
    f(\vec{x_0} + \vec{h})
    = f(\vec{x_0}) + \frac{1}{1!} (\vec{h} \cdot \nabla) f(\vec{x_0})
    + \frac{1}{2!} (\vec{h} \cdot \nabla)^2 f(\vec{x_0})
    + \frac{1}{3!} (\vec{h} \cdot \nabla)^3 f(\vec{x_0})
    + \cdots
    [/tex]

    (Where this notation means you're supposed to do the differentiations first, then plug in [itex]\vec{x_0}[/itex])

    If I haven't messed that up, it works when f is a vector function as well. (Of course, f has to be analytic, or else it can't be given by a Taylor series)


    By the way, it just struck me, and I feel really silly not noticing this before: it seems to me that the operator [itex]\mathbf{A} \cdot \nabla[/itex] is just the directional derivative operator [itex]\nabla_{\mathbf{A}}[/itex]... does that sound right?
     
  14. Aug 27, 2005 #13

    Hurkyl

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    Yah, it is a disturbing (but cool) thing until you figure out what's going on. Bleh, it still seems disturbing!

    The trick is that multiplication of operators really means composition, and usually raising an operator to a power means you just multiply (i.e. compose) it by itself a bunch of times. But, fortunately, these operators are linear, so their multiplication distributes over addition!

    (And yes, it's a mess... which is why I'm only asking you to do ( )², and not ( )³. :smile:)
     
    Last edited: Aug 27, 2005
  15. Aug 27, 2005 #14

    yeah, i've seen this in arfken's book. :yuck:
     
  16. Aug 27, 2005 #15

    Hurkyl

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    You should see one of the things I'm trying to learn now... not only can you do algebraic operations with operators, but you can take their norms, do topology with them, apply all sorts of horrible functions to them (e.g. take e^T where T is some operator)... and you even have functions whose values are operators, and you can even do calculus on such terrible things!!!

    (I don't really think this is a horrible thing, I just like being overly dramatic!)
     
    Last edited: Aug 27, 2005
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