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Classical electrodynamics - Green function

  1. Sep 13, 2008 #1
    Hello,

    I don't fully understand the meaning of Green function, and how one should use it. According to Jackson's "Classical Electrodynamics" - 'the method of images is a physical equivalent of the determination of the appropriate F(x, x') to satisfy the boundary conditions'.
    Where Green function is: G(x, x') =1/|x-x'| + F(x, x').
    moreover, 'the variable x' refers to the location P' of the unit source, while the variable x is the point P at which the potential is being evaluated'.

    For example, given that the potential is zero on an infinite plane, apart from a disk of radius a where the potential is constant, V, what is the appropriate Green function?

    The answer is [(x-x')2+(y-y')2 +(z-z')2]-0.5 - [(x-x')2+(y-y')2 +(z+z')2]-0.5

    according to the method of images, I think there should be two charges, Q and -Q, symmetrically situated above and below the plane. I belive that this is exatly what the above mentioned Green function describes, but here is what I don't understand:

    If the Green function should describe the two charges, then the coordinates x', y' and z' should be constans. But eventually, they are variables of integration...
    This is truly confusing, could someone please clarify this point for me?

    Thanks! :-)
     
  2. jcsd
  3. Sep 13, 2008 #2

    gabbagabbahey

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    when the charges are point charges, then [tex](x'_i,y'_i,z'_i)[/tex] does just represent the location of charge [tex]q_i[/tex]. But, when insead of point charges, you have an extended continuous charge distribution, [tex](x',y',z')[/tex] represents the location of the infinitesimal charge [tex]dq'[/tex], and integrating over [tex]dq'[/tex] is equivalent to adding up each infinitesimal piece. And for any [tex]f[/tex], [tex]\int{f(x',y',z')dq'}[/tex] is the continuous charge distribution analog to adding up [tex]\sum_{i}{f(x'_i,y'_i,z'_i)q_i}[/tex].
     
    Last edited: Sep 13, 2008
  4. Sep 14, 2008 #3
    Thanks for the reply, but this is still unclear to me. I know that in order to calculate the potential due to a continuous charge distribution the sum is replaced by an integral. But in the example I gave the charges in question are point charges!
    How come the answer includes x', y' and z' as variables of integration, as if we move these charges in space?
     
  5. Sep 14, 2008 #4

    Ben Niehoff

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    x', y', and z' are continuous variables, but the charge distribution is represented by a delta function. It's just for the sake of consistency: Green's method is always done using an integral. If you have a point charge q_1 at (1,0,0) and another point charge q_2 at (2,3,7), then your charge distribution is given by

    [tex]\rho = q_1 \delta(x - 1)\delta(y)\delta(z) + q_2 \delta(x-2)\delta(y-3)\delta(z-7)[/tex]

    When you integrate this over all space, of course the integral transforms directly into a sum.

    Basically what you are doing with Green's method is convolution, only in three dimensions. For every point in the source distribution, you add a 1/r potential from that point, proportional to the charge density at that point.
     
  6. Sep 14, 2008 #5

    gabbagabbahey

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    Is this an example in Jackson? If so, let me know what section it is in and I'll take a look at my copy...it should help me to answer your question better.

    EDIT- nm...see above ^^^
     
  7. Sep 14, 2008 #6
    Hey, thanks again, Ben and gabbagabbahey.

    Ben, I think you gave me the answer I was looking for, but I still don't fully grasp the matter. I was confused because of the quote I brought in the first message on this thread, where the Green function is described as the mathematical equivalent to the images method. I thought it should describe the image charges we imagine in order to satisfy the boundary conditions, but actually the charge distribution does that. So what is exatly the purpose of the Green function? why convolution is necessary here?
    What is the difference between the "Green method" and simply finding the appropriate charge distribution and using the formula
    [tex]\phi[/tex]=1/(4pi[tex]\epsilon0[/tex])[tex]\int\rho(x')/|x-x'|d3x'[/tex]
    for the potential?

    gabbagabbahey, I'm not sure wether the problem I've mentioned is taken from Jackson or not, but my question is more general, it isn't just about that specific problem. If you could check out the example for using Green function at the begining of section 2.6 and clarify it a little bit it would be very helpful as well. :)
     
  8. Sep 15, 2008 #7
    anyone?...

    please?
    :blushing:
     
  9. Mar 23, 2010 #8
    well the definition of the green function is the following laplacian(G)=d(x-x')
    there's nothing more

    we use green function because we can use the second identity of green to find a expression for the potential in terms of the distribution of charge, the values of the potential and the normal derivative of the potential at the surface. The green function is just a method to solve this value problems, like we only need that laplacian(G)=d(x-x') that mean that we can choose
    G=1/lx-x'l - F(x.x') where Laplacian(F)=0

    Then we can use F to eliminate one of the terms in the surface integral (jackson equation 1.42)
    that mean that if i know the potential at the surface i choose G zero at the surface and if i know the normal derivative of the potential at the surface I choose the normal derivative of G equal to zero. The point is that if I find G that satisfy one of this boundary condition and laplacian(G)=d(x-x'), then the problem is done. The problem is that G is not that easy to find.

    Then there is not mystery, the Green function is just part of a method to solve a boundary condition problem using the second green identity. Some authors relate F with the charge distribution outside the volume, but really that is not important for the application of the method. My recommendation is to look for some Math-book that use green function to solve general non-homogeneous differential equations (Arfken have a section after frobeniuos method).
     
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