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How can divF=0, but the volume integral of divF=/=0

  1. Oct 25, 2015 #1
    1. The problem statement, all variables and given/known data
    Basically, this is part C of a question where in part A we had to use the RHS of the divergence theorem below to calculate the LHS, and then in part B we had to calculate the divergene of F, which came to be 0. and part C asks us how can this be? Since in part A we used the LHS and shown that it does not equal 0.

    For reference, although I doubt it matters, this was for a sphere centered at the origin, and [itex]\vec{F}=\nabla \frac{1}{r}[/itex]

    I had no issues calculating divF or the RHS but just cannot get my head around how divF equals 0, yet the volume integral of divF does not.

    2. Relevant equations
    [tex]
    \int \int \int_V \nabla \cdot \vec{F} dV =\int \int_S \vec{F} \cdot d \vec{S}
    [/tex]

    3. The attempt at a solution
    I had no issues calculating divF or the RHS but just cannot get my head around how divF equals 0, yet the volume integral of divF does not.

    Is it simply because [itex]\vec{F}=\nabla \frac{1}{r}[/itex] and so is not defined at (0,0,0) ? I have tried searching google and cannot find much. I am not sure if this post is in the right place, it does refer to a coursework question, but theres no math involved in this question.

    Any help is much appreciated.
     
  2. jcsd
  3. Oct 25, 2015 #2

    SteamKing

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    Perhaps it would be more informative if you posted all parts of this problem you are working on. Right now, you have access to more information than we do.
     
  4. Oct 25, 2015 #3

    vela

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    Yes, that's pretty much it. The theorem only holds when you have a continuously differentiable vector field ##\vec{F}## defined on a neighborhood of ##V##. The vector field here isn't defined at ##\vec{r} = 0##, so it doesn't meet the requirements of the theorem.
     
  5. Oct 25, 2015 #4
    Right, I thought that was it. But because the RHS (in this case) equals [itex]4 \pi R^4 [/itex] is it mathmatically correct to say that [itex] \int \int \int_v \nabla \cdot \vec{F} dV = 4 \pi R^4[/itex] also? As that is what the question is essentially asking, that how can it be that the volume integral equals [itex]4 \pi R^4[/itex] when [itex]\nabla \cdot \vec{F} = 0 [/itex]

    Thanks!
     
  6. Oct 25, 2015 #5

    vela

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    Have you learned about the Dirac delta function?
     
  7. Oct 25, 2015 #6
    No not yet, I think he is covering something what he called "The Delta Function" on Tuesday. Maybe I should wait until then and then maybe I will get it?
     
  8. Oct 31, 2015 #7

    vanhees71

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    Just a hint: Think about the Coulomb field of a point charge sitting in the origin of your coordinate system!
     
  9. Oct 31, 2015 #8
    We ended up covering the delta function yesterday and it was explained to us. I calculated the RHS wrong anyway as I had a silly exponent mistake with the R's, they cancelled out and the answer was [itex]-4 \pi [/itex] . I had actually met the delta function last year, but it was only very briefly and only covered it in one dimension.
     
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