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Q about Poisson eqn w/ Neumann boundary conditions as in Jackson

  1. Sep 26, 2015 #1
    I am reading Jackson Electrodynamics (section 1.10 in 3rd edition) and he is discussing the Poisson eqn $$\nabla^2 \Phi = -\rho / \epsilon_0$$ defined on some finite volume V, the solution using Greens theorem is

    $$\Phi (x) = \frac{1}{4 \pi \epsilon_0} \int_V G(x,x') \rho(x')d^3x' +\frac{1}{4 \pi } \int_{\partial V} G(x,x') \frac{\partial \Phi (x') }{\partial n'} - \Phi(x') \frac{\partial G (x, x') }{\partial n'} da'$$

    where G(x,x') is any Greens function

    He says that for Neumann boundary conditions i.e. $$\partial \Phi / \partial n$$ is given as an explicit function on the boundary, it might seem that the obvious solution would be to choose G so that on the boundary

    $$\frac{\partial G (x') }{\partial n'} = 0 $$

    However he says that this isnt so simple since by Gauss theorem

    $$\int_{\partial V} \frac{\partial G (x,x') }{\partial n'} da' = -4 \pi $$

    for any Green's function. This leads me to my first question. So what? The integral we are interested in is not

    $$\int_{\partial V} \frac{\partial G (x,x') }{\partial n'} da'$$

    but rather

    $$\int_{\partial V} \Phi(x') \frac{\partial G (x,x') }{\partial n'} da'$$

    which doesnt do anything funny through Gauss theorem, so

    $$\frac{\partial G (x') }{\partial n'} = 0 => \Phi(x') \frac{\partial G (x') }{\partial n'} = 0$$

    and so that term in the solution should just equal 0.


    He then goes on to say that what we should do is set

    $$\frac{\partial G (x') }{\partial n'} = - 4\pi/S $$

    where

    $$S =\int_{\partial V} da' $$

    so that

    $$ -\frac{1}{4 \pi } \int_{\partial V}\Phi(x') \frac{\partial G (x') }{\partial n'} da' = \int_{\partial V} \frac{\Phi(x')}{S} da' := <\Phi>_{\partial V}$$

    in other words it is the average value of phi on the boundary. He then says that in the limit that the surface goes to infinity this term goes to zero. I dont see why that should be the case, I can actually show explicitly functions that its average value on some surface does not depend on the "size" of the surface at all, and others which will blow up to infinity as the surface grows. What am I missing here?

    Finally, his opening line was about finite volumes how is the surface at infinity (unless he is talking about the weird mathematical cases where an infinite surface bounds a finite volume, but somehow I doubt it)


    TIA
     
  2. jcsd
  3. Sep 26, 2015 #2
    The Greens function represents the potential due to a point charge. Gauss's law says the the flux through the boundary is equal to the enclosed charge. In calculating the Greens function we need to make sure that it's consistent with Gauss. Therefore the integral of the flux through the bounary cannot be zero.
     
  4. Sep 27, 2015 #3
    I dont know if you are answering any of my questions, reinforcing any of my questions, or just giving an overall comment. In any case, in dealing with the equation strictly as a mathematical object, our intuition regarding phi being an actual potential and behaving like an Electrostatic force shouldnt impact the general soluttion
     
  5. Sep 27, 2015 #4
    My answer above addresses why [itex] \oint ds \frac{\partial G}{\partial n} \neq 0[/itex]. Note that I was talking about the Green Function [itex]G\left(x,x'\right) [/itex] and not the potential [itex]\phi \left(x\right) [/itex]

    In my opinion Jackson does a poor job of explaining Green functions. The key ingredient that he doesn't adequately explain in the first chapter is that [itex]G\left(x,x'\right) [/itex] is a specific function that solves the equation

    [itex] \nabla^2 G\left(x,x'\right) = C \delta(x-x'),[/itex]

    where C is a nonzero constant that depends on your normalization of G. (I like to set C to 1, but I think Jackson uses [itex] C = -4\pi[/itex]).

    If you look at this equation, you'll note that the Green function [itex]G\left(x,x'\right) [/itex] represents the potential at the point [itex] \vec x [/itex] due to a point charge at [itex] \vec x' [/itex]. To determine [itex] \oint ds \frac{\partial G}{\partial n}[/itex] at the boundary integrate

    [itex] \nabla^2 G\left(x,x'\right) = C \delta(x-x'),[/itex]

    over the volume, and apply Gauss's law (Divergence theorem),

    which gives you

    [itex] \oint ds \frac{\partial G}{\partial n} =C [/itex]. Therefore [itex] \frac{\partial G}{\partial n} \neq 0 [/itex].

    You are missing the fact that G and [itex]\phi \left(x\right) [/itex] are not arbitrary functions but they are solutions to Poisson's equation. For a finite charge this prevents [itex]\left< \phi \left(x\right)\right> [/itex] from blowing up.

    He's just considering the limit where the boundary of the domain goes to infinity. So both the volume and the surface area are infinite. And he is arguing that in this limit the model is still well-behaved.
     
  6. Sep 29, 2015 #5

    I understand why [itex] \oint ds \frac{\partial G}{\partial n} \neq 0[/itex] I dont understand why [itex] \oint ds \frac{\partial G}{\partial n} \Phi(x) \neq 0[/itex] for [itex] \frac{\partial G}{\partial n} = 0[/itex] on the boundary.


    Yes an no, given [itex] \rho [/itex] arbitrary, I can make it so that [itex] \Phi [/itex] does blow up. Lets work backwards for a moment, set [itex] \Phi(r) = \alpha r^5 [/itex] then [itex] \nabla^2 \Phi = 30 \alpha r^3 = \frac{\rho}{\epsilon_0}[/itex]. Turning that around tells me if [itex] \rho = -\epsilon_0 \alpha 30 r^3 [/itex] then I know that [itex] \Phi = \alpha r^5 [/itex]. so given this rho I can tell you that alpha blows up, and it blows up faster than than the are element so taking the average value over any surface blows up to infinity as the surface goes to infinity. (just take the surface to be a sphere and you essential get a function cubic in r which must blow up as r goes to infinity)
     
  7. Sep 29, 2015 #6
    You're correct that we're interested in evaluating [itex] \oint ds \frac{\partial G}{\partial n}\phi [/itex], to do so we have to determine [itex] \frac{\partial G}{\partial n} [/itex]. Now we have some freedom in our choice, but we have to make sure that our choice of [itex] \frac{\partial G}{\partial n}[/itex] is consistent.

    If we chose G to be the function that satisfies the equation
    [itex]\nabla^2 G =\delta [/itex]

    then we have to make sure that our choice of [itex] \frac{\partial G}{\partial n} [/itex] is consistent with this equation.
    If you apply the divergence theorem you get [itex] \oint ds \frac{\partial G}{\partial n}=1[/itex]. Therefore our choice for [itex] \frac{\partial G}{\partial n}[/itex] has to satisfy this intergal equation. If it does not then it's not consistent. What happens if we choose [itex] \frac{\partial G}{\partial n}=0[/itex] as you suggest?

    Plugging this into the surface integral we get
    [itex] \oint ds \frac{\partial G}{\partial n}=\oint ds 0=0\neq 1[/itex]. Therefore the choice [itex] \frac{\partial G}{\partial n}=0[/itex] is not consistent with our definition of G.

    Now, we could ask is there a non-zero choice of [itex] \frac{\partial G}{\partial n}[/itex] that will set [itex] \oint ds \frac{\partial G}{\partial n}\phi =0 [/itex] for all [itex]\phi [/itex]? It turns out that there is not.

    You'll note that in your example you need an infinite amount of charge to get an infinite potential. So you need an unphysical charge to get an unphysical potential.

    I should amend my statement. He's arguing that the model is well behaved at infinity for any physical charge distribution.
     
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