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Continuous 2nd Partials a Substantial Requirement for Conservative Field?

  1. Jun 2, 2008 #1
    It can be shown that if F has continuous 2nd partials, then div curl F = 0. According to Stoke's Theorem, the work done around a closed path C is equal to the flux integral of curl F on a surface sigma that has C as its boundary in positive orientation. However, this integral is equal to the volume triple integral of div curl F. But if F has continuous 2nd partials, then div curl F = 0, and hence the work around C must be 0. Doesn't this show that F is conservative if F has continuous 2nd partials?
     
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  3. Jun 2, 2008 #2

    HallsofIvy

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    It isn't "F" that is "conservative"!

    A vector field [itex]\vec{f}[/itex] is said to be "conservative" (that's physics terminology; I prefer "is an exact derivative") if there exist a scalar function F, having continuous partials, such that [itex]\vec{f}= \nabla F[/itex]. IF there exist such an F, then yes, [itex]\vec{f}[/itex] is "conservative"!
     
  4. Jun 2, 2008 #3

    nicksauce

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    So you're saying that because

    [tex] \int_{\partial S}\vec{F}\cdot d \vec{l} = \int_{S} \nabla \times \vec{F} \cdot d \vec{S}[/tex]

    and because

    [tex] \int_{\partial V}\vec{F} \cdot d \vec{S} = \int_{V} \nabla \cdot \vec{F} dV[/tex]

    it follows that

    [tex] \int_{\partial V}\vec{F}\cdot d \vec{l} = \int_{V} \nabla \cdot \nabla \times \vec{F} dV = 0 [/tex]

    assuming F is smooth enough? The problem is that you can't go from the first theorem to the second. In the first S is an open surface (which has a boundary curve) and in the second S = delV is a closed surface (which does not have a boundary curve)
     
  5. Jun 2, 2008 #4

    nicksauce

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    The first integral in the third equation should be around delS not delV.
     
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