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Stokes theorem

  1. Apr 12, 2007 #1
    I was wondering as to how to prove stokes theorem in its general and smexy form.Also what is the intuition behind it(more important) aside from the fact that its a more general form of the other theorems from vector calculus?
     
  2. jcsd
  3. Apr 12, 2007 #2
    I don't mean for the following to sound offensive in any way. However, given what we've gathered of your current mathematical ability (an implicitly fallible process since we know you only through your posts here), attempting to understand a rigorous proof of Stokes' theorem may be biting off more than you can chew at the moment. For instance, do you know anything about cycle or boundary groups?

    Well, intuitively Stokes' theorem tells you that if you have some [itex]m[/itex]-dimensional manifold [itex]\mathcal{M}[/itex] (which satisfies some technical requirements) and some [itex](m-1)[/itex]-dimensional form [itex]\alpha[/itex] (again which satisfies some technical requirements), then the integral of the exterior derivative of [itex]\alpha[/itex] over [itex]\mathcal{M}[/itex] is equal to the integral of [itex]\alpha[/itex] over the boundary of [itex]\mathcal{M}[/itex]:

    [tex]\int_\mathcal{M} \,d\alpha = \int_{\partial\mathcal{M}}\alpha[/itex]

    Pretty straightforward, no? A major application of this in physics is then that if the boundary of your manifold is empty, the right hand side of this vanishes. This fact is often a life saver in practical calculations.
     
  4. Apr 12, 2007 #3
    It is probably beyond me at this moment as i don't know about those. I don't really mind. at least you were polite about it. so intuitively speaking it's just like the fundmanetal theorem?

    I don't think I'll be ready for awhile, but this boredom is really getting to me. I literally have NOTHING to do in my math classes. Ok well whatever.

    Thank you for your honesty.

    However despite the intuitvie appeal, it upsets me because unlike the itntuitive appeal of other theorems, it doesn't seem sufficient( I didn't care about the proof of green's theorem because it was very obvious). Maybe my understanding just isn't deep enough.
     
    Last edited: Apr 12, 2007
  5. Apr 15, 2007 #4

    mathwonk

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    the proof of greens theorem is all there is to proving stokes theorem.

    i.e. the proof of stokes is just a parametrized version of the proof of greens in a rectangle.

    it is indeed simply the FTC plus the trick of repeated integration.

    i.e. ftc is stokes in one dimension, and repeated integration gives the higher diml case by induction.

    parametrizing any manifold locally by maps from rectangles gives it on any manifold with boundary.


    rigorous patching of these parameter maps is usually done nowadays by a trick called partitions of unity. this is the deep idea that if you take a function that equals 1 on the interval [1,2], and goes down linearly to 0, at 0 and 3, and if you add it to a similar function that equals 1 on the interval [3,4], and goes down to 0, at 2 and 5, you get a similar function that equals 1 all the way from 1 to 4.

    If you keep this up, thus you have "partitioned" the unity function (the function 1) into several pieces.

    this is all in spivak's calculus on manifolds, together with the language of "chains" for parametrized rectangles, but I think lang makes it look easier, in his Analysis I, appendix.
     
    Last edited: Apr 15, 2007
  6. Apr 15, 2007 #5

    mathwonk

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    the "boundary" of a rectangle is just the 4 sides, each with some + and minus signs.

    a chain is a parametrized rectangle, and so the boundary of a chain is just the parametrized boundary of the rectangle. attaching a + or minus sign to each one allows you to add them, even getting other integers attached to each one.

    3 times a parametrized circle e.g. is like going around the circle three times. big whoop.
     
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