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Proof of the existence of a scalar potential

  1. Apr 6, 2009 #1
    1. The problem statement, all variables and given/known data
    Hi there - I'm wondering about how you can actually show the existence of a scalar potential for an irrotational vector field E - if [itex] \nabla \times E = 0[/itex] everywhere, then how does one show there exists a scalar potential [itex] \phi(x) [/itex] such that [itex]E=- \nabla \phi [/itex]?

    3. The attempt at a solution
    By Stokes' theorem we can see that [itex] \int_C E dx = \int_S \nabla \times E dS = 0 [/itex] everywhere so our integral is path independent, but does path independence necessarily prove the existence of a scalar potential?

    Thanks a lot, Mathmos6
     
  2. jcsd
  3. Apr 6, 2009 #2
    There is a well-known vector identity:

    [tex] \bigtriangledown \times \bigtriangledown\phi=0[/tex]
     
  4. Apr 6, 2009 #3

    Dick

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    Sure it does. Now pick a point A and define the potential at any point X as the path integral of E from A to X. It's well defined because of path independence.
     
  5. Apr 6, 2009 #4
    Ah, fair enough - thanks Dick!
     
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