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General Conditions for Stokes' Theorem

  1. Oct 15, 2013 #1
    What is the least restrictive set of conditions needed to utilize the formula ##\int\limits_{\Omega}\mathrm{d}\alpha=\int\limits_{\partial\Omega} \alpha##?
     
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  3. Oct 16, 2013 #2

    Ben Niehoff

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    I think the only conditions are those needed to define the integrals (i.e. the same kinds of conditions used to define 1-dimensional integrals). I don't think there are any extra geometrical conditions, provided you're on a differentiable manifold.

    You can even relax smoothness of the manifold (and the forms!) if you are careful about using Dirac delta functions. Generally, since a form is something you integrate, they should be thought of as distributions.

    The boundary operator can be used in a distributional sense as well. For example, the boundary of a sphere is zero. But it might be helpful to think of a sphere with a point removed, whose boundary is therefore a point; then you can use Stokes' theorem to integrate forms over the sphere.
     
  4. Oct 16, 2013 #3
    Could you please expand on this? Thinking of forms as distributions feels foreign, and I don't see where that line of thought would go.
     
  5. Oct 16, 2013 #4

    Ben Niehoff

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    Expand on it how? Surely you can figure out how to integrate something like

    [tex]\delta(x,y,z) \, dx \wedge dy \wedge dz[/tex]

    Do you have a specific question?
     
  6. Oct 16, 2013 #5

    Ben Niehoff

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    This might be a better example of what I'm talking about. Say we want to find the area of a sphere. The form we want to integrate is

    [tex]\omega = \sin \theta \, d \theta \wedge d \phi[/tex]
    Now, the sphere ##\Omega## is a closed surface, so ##\partial \Omega = 0##. However, the coordinate patch ##\tilde \Omega## covered by the coordinates ##\phi \in (0, 2\pi), \; \theta \in (0, \pi)## is not a closed surface, and is in fact contractible. We have that ##\partial \tilde \Omega## is the union of the north and south poles of the sphere, and a segment of a great circle that runs between them.

    Now, it so happens that

    [tex]\omega = d \big( - \cos \theta \, d \phi \big) = d \alpha[/tex]
    so we can use Stokes' theorem. So

    [tex]\int_{\tilde \Omega} \omega = \int_{\partial \tilde \Omega} \alpha = - \int_{\partial \tilde \Omega} \cos \theta \, d \phi[/tex]
    To integrate around the "cut" between the north and south poles, we draw a loop around it. On either side there is a vertical part where ##d \phi = 0##, and so these parts do not contribute. Then around the north and south poles, there are tiny circles, at which ##\cos \theta = \pm 1## and ##\phi## runs from 0 to ##2 \pi##. The tiny circles go opposite directions, so each part contributes positively:

    [tex]- \int_{\partial \tilde \Omega} \cos \theta \, d \phi = 2 \pi + 2 \pi = 4 \pi[/tex]
    So you see, if you are careful about how you cut up a manifold, you can apply Stokes' theorem in all sorts of situations.

    In this case, we took a closed surface and removed a set of measure zero to turn it into a surface with boundary. The reason this worked is because the form ##\omega## is smooth on the set of measure zero that we removed. If that were not the case (say ##\omega## had a delta-function-like contribution on the "cut"), then you would have to include an extra piece to account for that.
     
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