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Question on divergence/gradient theorem

  1. Oct 22, 2013 #1

    joshmccraney

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    hey pf!

    i had a subtlety about the divergence theroem/gradient theorem. specifically, the following was presented: $$ \iint_s p \vec{dS} = \iiint_v \nabla (p) dv$$

    i am familiar with the divergence theorem, but that is for a vector "dotted" with a surface element (flux) related to the divergence (expansion) through the volume. i believe the above is the gradient theorem. can anyone verify the legitimacy of this (and perhaps offer some physical intuition if it holds). i should say [itex]p[/itex] is not a constant.

    thanks!
     
  2. jcsd
  3. Oct 22, 2013 #2

    SteamKing

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  4. Oct 22, 2013 #3

    arildno

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    We may write:
    [tex]\vec{dS}=dS(n_{x}\vec{i}+n_{y}\vec{j}+n_{z}\vec{k})
    [tex]\int\int_{S}p\vec{dS}=\int\int_{S}pn_{x}dS\vec{i}+\int\int_{S}pn_{y}dS\vec{j}+\int\int_{S}pn_{z}dS\vec{k} (*)[/tex]
    Since the Cartesian unit vector can be extracted from the integrand!
    --
    But, we have:
    [tex]pn_{x}dS=(p\vec{i}\cdot\vec{dS})[/tex]
    And the first term in RHS in (*) is then simply, by the divergence theorem:
    [tex]\int\int_{S}pn_{x}dS\vec{i}=+int\int\int_{V}\frac{\partial{p}}{\partial{x}}\vec{i}dV[/tex]

    Do the similar thing to the other two terms on RHS in (*), and you get your identity.
     
  5. Oct 22, 2013 #4

    joshmccraney

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    thank you arildno!!!!! this makes a lot of sense now! youre brilliant!
     
  6. Oct 23, 2013 #5

    joshmccraney

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    on another note, arildno, did you just figure this out by thinking about what's going on or did you see it in a book before? either way, thanks again
     
  7. Oct 23, 2013 #6

    arildno

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    I noticed it a long time ago. I don't remember if it was in a book or on my own inspiration.
     
  8. Oct 23, 2013 #7

    vanhees71

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    I'd apply Gauß's theorem to the vector field
    [tex]\vec{V}(\vec{x})=\vec{a} p(\vec{x})[/tex]
    for a constant vector [itex]\vec{a}[/itex]. Then you have
    [tex]\vec{\nabla} \cdot \vec{V}=\vec{a} \cdot \vec{\nabla} p.[/tex]
    Then Gauß's theorem gives
    [tex]\int_V \mathrm{d}^3 \vec{x} \vec{a} \cdot \vec{\nabla} p = \vec{a} \cdot \int_V \mathrm{d}^3 \vec{x} \vec{\nabla} p = \int_{\partial V} \mathrm{d}^2 \vec{F} \cdot \vec{a} p = \vec{a} \int_{\partial V} \mathrm{d}^2 \vec{F} p.[/tex]
    Since this is true for all [itex]\vec{a}[/itex] you have proven your vector identity.
     
  9. Oct 23, 2013 #8

    arildno

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    Sure. It's matter of taste, really, which one approach is used.
    After all, my method is just as general as the one you present.
     
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