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Can we make use of Greens function if there are no charges?

  1. Sep 10, 2015 #1
    I can't think of a situation where we can utilize greens function without the presence of a point charge. lets consider the following equation:

    [itex] \Phi=\frac{1}{4\pi \epsilon} \int dv \rho(x')G_{N} (x,x')+ \frac{1}{4\pi} \int da F_{s}(\rho , \phi) G_{N} + <\phi>_S [/itex]

    Here we see that a volume with no charge distribution could still have some potential (due to external fields) and we have green function contribution.
    The concept of green function and utilizing it to find ANY potential is very vague to me because we originally found green function using the concept of point charge, and I don't get how it would help us in finding any potential (even if there are no point charges). Any explanation is appreciated. Thanks
  2. jcsd
  3. Sep 11, 2015 #2


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    The boundary terms arise from inhomogeneous boundary conditions. Physically, you will need to induce certain charges on the boundary to keep these boundary conditions satisfied. You may see the Green's function contribution from the surface terms as accounting for these induced charges.
  4. Sep 12, 2015 #3
    Lets say we have the following conditions given and we want to find the greens function in cylindrical coordinates:

    ## \nabla^2 \phi =0## this will eliminate the volume charge term in the equation for ##\phi##
    we also have that ##\phi|_{s} = F_{s}(\rho , \phi)## at the surface.

    ##\phi=\frac{1}{4\pi} \int da F_{s}(\rho , \phi) G_{N} + <\phi>_{s} ##

    now for greens function do I consider the following or should I consider any further terms?

    ##G_{N}= \frac{1}{|\vec x -\vec x'|}##
  5. Sep 12, 2015 #4
    never mind I have found out where I was missing. your hint was very helpful. but there is one further question that pops out here. say we have written the greens function for such a problem in cylindrical coordinates, on the surface and considering the ##F_s = E_0## a constant, we arrive at the following equation:

    ##\phi=\frac{E_0}{4\pi} \int \frac{\rho ' d \rho ' d\theta'}{\sqrt(z^2 +\rho' ^2 -2\rho' z cos\theta' )} ##
    but as previously indicated don't have any charges here (just the boundary condition), hence what does the primed quantities indicate here if not the charge?
    thank you for your time again.
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