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Helmholtz theorem

  1. Jul 26, 2006 #1
    Hi everyone!

    This question has me a little stumped.
     

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  2. jcsd
  3. Jul 27, 2006 #2

    quasar987

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    Let's work on this together. On my side, I "simplified" the 2 original integrals to

    [tex]=\frac{1}{4\pi}\int_V \frac{\nabla _s \cdot \vec{F}(\vec{x}_s)}{R^2}\hat{R} \ d^3x_s - \frac{1}{4\pi}\int_V \frac{\nabla _s \times \vec{F}(\vec{x}_s)}{R^2} \times \hat{R} \ d^3x_s[/tex]

    Anyone else care to contribute?
     
    Last edited: Jul 27, 2006
  4. Jul 27, 2006 #3

    quasar987

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    On the other hand, take just the first integral. Isn't it just 0?

    [tex]\nabla_t \int_V \nabla_s \cdot \left( \frac{\vec{F}(\vec{x}_s)}{R} \right) \ d^3x_s = \nabla_t \int_{\partial V} \left( \frac{\vec{F}(\vec{x}_s)}{R} \right)\ \cdot \ \hat{n} \ d^2x_s \rightarrow \nabla_t 0=0[/tex]

    just by taking V sufficiently large.

    ?!
     
    Last edited: Jul 27, 2006
  5. Jul 27, 2006 #4
    See the attached file.

    Pete
     

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  6. Jul 28, 2006 #5

    quasar987

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    Did you write this Pete?

    There's something I don't get at all:

    Equation (10c): " [itex]\nabla[/itex] and [itex]\nabla '[/itex] are related by [itex]\nabla = -\nabla '[/itex] "

    How can they be related since the primed coordinates and the non-primed are not related?
     
  7. Jul 28, 2006 #6

    quasar987

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    Also, in the last paragraph of page 3,

    "Regarding the first integral on the right; as the radius of the surface increases as r then the area of the surface increases as r². However the integrand decreases as r³. Therefore as we let the radius go to infinity we see that the first integral vanishes."

    The author seems to be making the assumption that F decreases as r², something that was not mentionned in the original statement of the theorem.

    It's a detail important to mention imo.
     
  8. Jul 29, 2006 #7

    siddharth

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    In this case, it's because the operator is being used on [tex] \frac{1}{|r - r'|} [/tex]. So, the derivates of [itex] r - r' [/itex] wrt to primed coordinates, are the negative of the derivates wrt to unprimed.

    I don't see how it how they are related like that in any other case, except when they operate on functions of [itex] r - r' [/itex].

    Yeah, I think that the argument should be the other way around. If that integral should converge, then F should decrease as 1/r^2 or faster.
     
    Last edited: Jul 29, 2006
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