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Quick green function question

  1. Sep 11, 2006 #1
    I am demonstrating the mean value theorem, which says that for charge-free space the value of the electrostatic potential at any point is equal to the average of the potential over the surface of any sphere centered on that point. I have already found one way to do this, but would also like to show it using the following formula (which makes use of the Neumann boundry condition):

    [tex]\Phi(X) = < \Phi > + \frac{1}{4\pi\epsilon_0} \int_V \rho(x')G_N(x,x') \,d^3x. +\frac{1}{4\pi}\int_S \frac{\partial\Phi}{\partial n'}G_N \,da [/tex]

    The first term on the right hand side is the average of the potential over the surface - so of course I expect the other two to be zero. The second is obviously zero. The third can be written as:

    [tex]\frac{1}{4\pi}\int_S \nabla\Phi\cdot n' G_N \,da [/tex]

    [tex] \frac{1}{4\pi}\int_S (-E)\cdot n' G_N \,da [/tex]

    And I know the Green function is:

    [tex] G(x,x') = \frac{1}{\mid x-x' \mid} + F(x,x') [/tex]

    [tex]\frac{\partial G_N}{\partial n'}(x,x') = -\frac{4\pi}{S} [/tex]

    for x' on S.

    If I can argue [tex]G_N[/tex] is constant then I can slide it out of the integral and happily use gaus' law to make the third term zero. But... I don't understand Green functions well.

    That derivative is with respect to n. If I know that the Green function here is some constant on S, then the integral over that surface is also a constant on S. Is this true?

    We have to do the problems in front of the class. The professor does not allow us to not understand what we are writing. Any help would be appreciated!
  2. jcsd
  3. Sep 11, 2006 #2


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    Staff Emeritus
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    Normally one sets the Green function to zero on the boundary of a 3D problem. Nice presentation of the question.
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