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Question about topology in study of electricity

  1. Sep 21, 2007 #1
    This question pertains to physics, but has to do with the math. In order to find the flux of an electric field you can put a sphere around it and use that to find flux, since the amount leaving is the same at every point. My teacher said that if you put a cube around the field/charge, you would get the same amount of flux, but it would be a much more difficult computation. My question is, is the flux the same because the sphere and the cube have the same topology? what happens if you put the field inside a torus? Would the amount of flux calculated be different?
     
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  3. Sep 21, 2007 #2

    Hurkyl

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    For simplicity, suppose that the sphere is inside the cube. We can define the region R which lies between the sphere and the cube. R has two boundaries: the cube (oriented positively) and the sphere (oriented negatively) We have:

    flux through sphere + flux through R
    = flux through sphere + (flux through cube - flux through sphere)
    = flux through cube


    Intuitively, if there is no charge in R, then there is no source of electromagnetic flux. Therefore, the net flux through R must be zero, and so the flux through the sphere is equal to the flux through the cube.

    Rigorously, it depends on your definitions, but it essentially boils down to applying some form of Gauss's law to R; the net flux through the boundary is proportional to the net charge inside R.
     
  4. Sep 21, 2007 #3
    How do you determine the orientations? Are they assigned arbitrarily? Your explanation reminds me a bit of homology lol
     
    Last edited: Sep 22, 2007
  5. Sep 22, 2007 #4

    Hurkyl

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    Well, to do the problem properly, I simply need the sphere and cube to have the same orientation. I chose the orientation on my region so that it would agree with the cube's orientation and disagree with the sphere's orientation.

    Note that if the sphere and the cube had opposite orientations, the flux through the sphere would be the negation of the flux through the cube!


    As well it should! For example, if you're using point particles, then I believe that it can be fruitful to think of your particles as punctures in 3-space and flux as a linear functional on 2-chains.
     
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