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Trying to undestand Stoke's Thm

  1. Mar 27, 2006 #1
    I'm trying to understand Stokes' Thm better (Not just how to apply it, but when to apply it, and why it is so important). If someone could help me clear this up, that would be cool:

    Lets say we have a surface integral:
    [tex] \int \int_S \vec F \cdot d\vec S [/tex]
    and for some reason we can easily formulate a path around the surface.


    (I guess this next portion is the part I'm confused, or curious about)
    We would then be able to use Stokes' Theorem if:

    [tex] \oint \vec U \cdot d\vec r = \int \int_S \nabla \times \vec U \cdot d\vec S [/tex]

    IF we can come up with a vector that satisfies:
    [tex] \nabla \times \vec U = \vec F [/tex]

    right?

    Any insight on this would be awesome... I don't know why it's so hard for me to get a feel for the theorem. Thanks in advance.
     
    Last edited: Mar 27, 2006
  2. jcsd
  3. Mar 27, 2006 #2
    This is what I understand from reading D.J. Griffiths "Intro. to Electrodynamics", 3rd ed.:

    I think you have it backwards: Stoke's Thm, aka The Fundamental Theorem of Curls, is the statement that the flux "curling" through the surface is the same as the result of only travelling along the boundary of that surface. (In general, a surface integral depends on which surface you chose. But in the case of a curl it does not - no matter what surface you chose you get the same result since the surface has only 1 boundary.)

    E.g. in the case of the electrostatic field, since the potential around a closed loop is zero (this is also known as Kirchoff's voltage law) [tex]\oint \vec{E} \cdot d\vec{l} = 0[/tex]. As such, [tex]\nabla\times\vec{E} = 0[/tex].

    Finally, since [tex]\nabla\times\vec{E} = 0[/tex] then there is a scalar field, call it [tex]V(\vec{r})[/tex] which is related to [tex]\vec{E}[/tex] via the gradient. This result actually arises from Helmholtz's Thm (Appx. B, ibid). IOW, [tex]\vec{E} = -\nabla V[/tex].

    Did this help? ... if not I apologize.

    -LD
     
  4. Mar 27, 2006 #3
    If:
    [tex] \nabla \times \vec E = 0 [/tex]
    then [itex] \vec E [/tex] is a conservative field. Thus, the fundemental theorem of line integrals holds (if we define [itex] C [/itex] to be a smooth curve) since a potential function exists BECAUSE the curl is equal to 0.

    The fundemental theorem of vector calculus states that:
    [tex] \int_C \nabla f \cdot d\vec r = f(\vec r(b))-f(\vec r (a)) [/tex]

    Since [itex] \nabla \times \vec E = 0 [/itex] then:
    [tex] \vec U = \nabla E [/tex]

    Since we are traversing a closed path:

    [tex] U(\vec r(c)) - U(\vec r(c)) = 0 [/tex]
    Where: [tex] a = b = c [/tex]

    So without knowing if:
    [itex] \oint \vec E \cdot d\vec r = 0 [/itex] using Stokes' Theorem greatly simplifies the calculation since we know that the curl is equal to 0, we don't have to setup the limits of integration. So yeah, that seems like a good way to use the theorem.



    I'm not sure what you mean, with saying that I have something backwards though. What do I have backwards?


    I really do appreciate the help, so thankyou :)
     
    Last edited: Mar 27, 2006
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