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Gradient version of divergence theorem?

  1. Aug 10, 2015 #1
    So we all know the divergence/Gauss's theorem as
    [itex]∫ (\vec∇ ⋅ \vec v) dV = ∫\vec v \cdot d\vec S[/itex]

    Now I've come across something labeled as Gauss's theorem:
    [itex]\int (\vec\nabla p)dV = \oint p d\vec S[/itex]
    where p is a scalar function.

    I was wondering if I could go about proving it in the following way (replacing dot products with implied sums):
    With [itex]e_i := \hat e_i[/itex] and [itex] d\vec S := ds_1 \hat x + ds_2 \hat y + ds_3 \hat z = ds_i e_i [/itex],

    [itex] \oint p d\vec S = \oint p (e_i ds_i) = \oint (p e_i)(ds_i) = \oint \vec p \cdot d\vec s[/itex] (this p vector has scalar functional dependence still, it's just [itex](p)(\vec v)[/itex], a scalar times a vector, but still overall a vector in my mind)

    then applying divergence theorem and getting
    [itex]= \int (\vec \nabla \cdot \vec p) dV = \int \partial_i (p e_i) dV [/itex]

    and finally applying the product rule and the fact that [itex]e_i[/itex] is a unit vector
    [itex]\int (e_i \partial_i p + p \partial_i e_i) dV[/itex].

    The second term is zero, since it's a partial of a unit vector, which has no spatial dependence, leaving
    [itex]\int (e_i \partial_i p)dV = \int (\vec \nabla p) dV[/itex]

    Does that make sense? I think it seems to work out, but I'm concerned that it's flawed due to my free conversions between sums and vectors. It seems unnatural that I've said [itex]d\vec S =d\vec s = ds_i[/itex], despite defining them differently. One, I suppose has actual vector components, whereas the other is just a list of components.
     
    Last edited: Aug 10, 2015
  2. jcsd
  3. Aug 10, 2015 #2

    TMO

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    I believe the gradient version of the divergence theorem would be your typical statement that the integral of the path going through a potential is just the difference in potentials.
     
  4. Aug 11, 2015 #3
    You must be thinking of the fundamental theorem of calculus for gradients -- [itex]\phi (\vec b) - \phi (\vec a) = \int_a^b \vec \nabla \phi \cdot d\vec r [/itex]

    This was what was presented to me during a proof, and I had never seen it before
    [itex]\int (\vec\nabla p)dV = \oint p d\vec S[/itex]


    (oops, and I just realized I forgot to close the surface integral of divergence theorem in my original post :P)
     
  5. Aug 12, 2015 #4
    The usual proof of the scalar version of the divergence theorem involves replacing the vector field v by (p k), where p is a scalar field and k is an arbitrary constant vector. Then div v is equal to k.grad p. Since k is arbitrary it can then be removed from both sides giving the result.
     
  6. Aug 12, 2015 #5
    Ah, ok. The steps are essentially the same then. Just slightly different in how you get your constant vector. Thanks!
     
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