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Boundary of a sphere

  1. May 3, 2005 #1
    Hi, we are doing the Divergence Theorem/Stokes' theorem and the teacher said a sphere has no boundaries...what does this mean? Help will be appreciated.
     
  2. jcsd
  3. May 3, 2005 #2

    cepheid

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    This is how I see it:

    Strictly speaking, a sphere is the set of all points in 3D space equidistant from a given point. That set of points defines a surface. So a sphere is a surface, a 2D object in 3D space. What do we mean when we say it has no boundaries? Well, one way of looking at it is that if you are a 2D lifeform confined to live on this sphere, your "world" would have no edges. It would just seem to continue on and on. If that's not making it clear, let's consider, in contrast, 2D surfaces that DO have boundaries:

    Any surface that is not infinite in extent and not closed, for example part of a flat plane, or even an arbitrary wavy surface, can have a boundary. You can simulate these surfaces using a sheet of paper, if that helps you visualise it. Your 2D lifeform can reach the "edge" of this surface and walk around that edge. So that edge is the boundary. Begin to bring the four ends of the paper together, and what are you doing? Shrinking your boundary. Turning the surface into a closed surface. To visualise a surface that is "partway there", think about blowing bubbles. Your soapy fluid is stretched across the circular opening like a thin film. You blow into it and it balloons outward. But it is not fully formed yet. It is a "not completely closed sphere". The partially formed bubble is the surface. The circular frame coincides with the boundary of that surface (the edge). As the bubble detaches, that hole closes up. The boundary shrinks to a point, and disappears. The bubble is a fully-closed sphere.
     
  4. May 3, 2005 #3

    quasar987

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    Your teacher must have meant that in the context of Stokes' theorem and not of the divergence theorem.

    If you consider the surface of the sphere, then this surface has no boundaries, like cepheid illustrated. So you can't make use of Stokes' theorem to go from the surface integral to a path integral over its boundary (it has none!)

    In the context of the divergence theorem, then the boundary of the sphere considered as a volume* is of course the sphere considered as surface**. And so you can apply the divergence theorem.



    *A full sphere is really called a ball, see http://mathworld.wolfram.com/Sphere.html.

    **The surface of the sphere is what is actually called a sphere.
     
  5. May 4, 2005 #4
    Thank you, guys. Your explanations are so much better.
     
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