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Related to stokes theorem

  1. Apr 5, 2010 #1
    1. The problem statement, all variables and given/known data

    [tex] \nabla \times f \vec{v} = f (\nabla \times \vec{v}) + ( \nabla f) \times \vec{v} [/tex]

    Use with Stoke's theorem

    [tex] \oint _C \vec{A} . \vec{dr} = \int \int _S (\nabla \times \vec{A}) . \vec{dS} [/tex]

    to show that

    [tex] \oint _c f \vec{dr} = \int \int _S \vec{dS} \times \nabla f[/tex]

    2. Relevant equations


    I think the only that you really need to know is the the scalar triple product.

    3. The attempt at a solution

    Ok, I allowed

    [tex] \vec{A} = f \vec{v} [/tex]

    And subsituted into Stokes' theorem

    [tex] \oint _C f\vec{v} . \vec{dr} = \int \int _S (\nabla \times f \vec{v}) . \vec{dS} [/tex]

    which gives

    [tex] \oint _C f\vec{v} . \vec{dr} = \int \int _S f ((\nabla \times \vec{v}) + ( \nabla f) \times \vec{v} ). \vec{dS} [/tex]

    My problem is that I got to this stage, and thought that this would work really great if

    [tex] \nabla \times \vec{v} = 0 [/tex].

    So I got stuck on this for a while, looked at my textbook, and they said that we let

    [tex] \vec{A} = f \vec{v} [/tex] where [tex] \vec{v} [/tex] is a CONSTANT VECTOR.

    Hence, the curl of a constant vector is zero, and the RHS becomes

    [tex] \int \int _S f ( \nabla f) \times \vec{v} ). \vec{dS} [/tex]

    which we rearrange with the triple scalar product identity to give

    [tex] \int \int _S f ( \vec{dS} \times \nabla f) ) .\vec{v} [/tex]

    So now the thing looks like this:

    [tex] \oint _C f\vec{v} . \vec{dr} = \int \int _S f ( \vec{dS} \times \nabla f) ) .\vec{v}
    [/tex]

    and cancelling [tex] \vec{v} [/tex] from both sides as it's constant, gives us:

    [tex] \oint _c f \vec{dr} = \int \int _S \vec{dS} \times \nabla f[/tex]

    The problem

    My problem is that this restricts the kind of vector field [tex] \vec{A} [/tex] can be, as it has to be a product of some function times a constant vector.

    Does this make the derived identity non-universal, or is this a necessary step in order to arrive at the identity?
     
    Last edited: Apr 5, 2010
  2. jcsd
  3. Apr 5, 2010 #2

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    The vector identity you proved involves only [itex]f[/itex]. So, who cares what you restricted [itex]\textbf{A}[/itex] to be? The identity is valid for any scalar function [itex]f[/itex], and that's all you were asked to prove.
     
  4. Apr 5, 2010 #3
    Yeah you're right. I don't know why I got hung up on this, I guess I was reading too much into it.
     
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