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Question on Gradient

  1. Aug 15, 2013 #1
    If ##\vec F(x'y'z')## is function of ##(x'y'z')##. ##\nabla## is operator on ##(x,y,z)##.

    So:
    [tex]\nabla\left[\vec F(x'y'z') g(x,y,z)\right]=(\vec F(x'y'z') \nabla g(x,y,z)[/tex]
    or
    [tex]\nabla(\vec F g)=\vec F \nabla g[/tex]

    Am I correct?
     
  2. jcsd
  3. Aug 15, 2013 #2

    ehild

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    Is g(x,y,z) a scalar function?

    ehild
     
  4. Aug 15, 2013 #3

    vanhees71

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    Such tasks are most easily solved using the (Euclidean) Ricci calculus. Your question would read
    [tex]\vec{\nabla} \vec{F} g(\vec{x})=\partial_i(F_i g)=F_i \partial_i g=\vec{F} \cdot \vec{\nabla} g,[/tex]
    where [itex]\vec{F}=\text{const}[/itex], i.e., independent of [itex]\vec{x}[/itex].
     
  5. Aug 15, 2013 #4

    HallsofIvy

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    Yes, if x', y', z' and x, y, z are independent variables, that is correct.
     
  6. Aug 15, 2013 #5
    Yes.
     
  7. Aug 15, 2013 #6
    Thanks everyone, I just need to confirm this. I am not a math major, I just encountered all sort of math issues when I am studying antenna theory. This and Electromagnetics really put vector calculus through the ringer!!! I am sure I'll be posting many more of these kind of stupid questions.

    Thanks
     
  8. Aug 15, 2013 #7

    Ray Vickson

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    You would only be correct if you define what you mean by the product ##\vec{A}\vec{B}## for vectors ##A = \vec{F}(x',y',z')## and ##\vec{B} = \vec{\nabla}g(x,y,z)##. Presumably, you mean the outer product, which gives a 3x3 matrix with ##\vec{A}\vec{B}_{i,j} = a_i b_j.## (Of course, ##\vec{F}## could be a vector or other than 3 dimensions, so the matrix could be non-square.)
     
    Last edited: Aug 15, 2013
  9. Aug 15, 2013 #8
    g is only a scalar function of (x,y,z).
     
  10. Aug 15, 2013 #9

    lurflurf

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    ^Yes but F and ∇g are vectors so F∇g is the dyad product of F and ∇g.
     
  11. Aug 15, 2013 #10
    But my original question is
    [tex]\nabla(\vec F g)=\vec F \nabla g[/tex]

    It's only after the gradient that it become a vector ##\nabla g##, not before.

    That actually raise a funny question. If the result is ##\vec F \nabla g##, what is this? A vector multiplying a vector?
     
  12. Aug 16, 2013 #11

    Dick

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    It's a tensor product. It's an object with two indices.
     
  13. Aug 16, 2013 #12

    Ray Vickson

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    Yes, we all know that g is a scalar function.

    However, the question is whether or not YOU realize the issues. These are: (1) what do you mean by asking for the gradient of a vector function---that is, what do you mean by ##\vec{\nabla}\vec{G}?##; and (2) what do you mean by the product of two vectors that you wrote, namely, ##\vec{A} \vec{B},## where ##\vec{A} = \vec{F}(x',y',z')## and ##\vec{B} = \vec{\nabla} g(x,y,z)?##
     
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