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Green Functions and BC's

  1. Jun 26, 2008 #1
    Hey folks,

    I'm trying to get a handle on my old Nemesis, Green functions. I have a massless scalar field confined between two parallel plates separated by a distance a (in the z dimension) and the field satisfies Dirichel BC's. Basically I'm trying to work from line 1 of a book to line 2 (K. Miltons the Casimir Effect p23).

    'The Green function satisfies'


    "We introduce a reduced Green function g(z,z) according to the Fourier Transform"

    [tex]G(x,x')=\int\frac{d^dk}{(2\pi)^d}e^{i\vec{k}.(x-x')}\int\frac{d\omega}{2\pi}e^{-i\omega(t-t')}g(z,z') [/tex]

    This is all the book says so sorry of thats not much info. I'm fairly sure that [tex]\partial=\nabla+\frac{d}{dt}[/tex].

    What I want to understand (and see the math for) is how to get from line 1 to line 2. I'm pretty sure that it involves fourier transforms, but I would like to see it. Also, I don't understand the concept of a reduced green function. Can anyone either point me to a good reference, or better still explain how and why it is used.

    I hope someone can walk me through this.

  2. jcsd
  3. Jun 27, 2008 #2

    Ben Niehoff

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    I don't know all the details of that particular problem, but what I remember from electrostatics is that the "reduced" Green's function is essentially a separation-of-variables technique. If we have, for some linear operator L,

    [tex]LG(\vec x, \vec x') = \delta^3(\vec x - \vec x')[/tex]

    then we can write

    [tex]\delta^3(\vec x - \vec x') = \delta(x - x') \, \delta(y - y') \, \delta(z - z')[/tex]

    (or we could use some other coordinate system, such as [itex]r, \phi, \theta[/itex], transforming to it with the proper Jacobian).

    Given the boundary conditions, x and y are free, but z is bounded by the planes z=0 and z=a. So we can write

    [tex]\delta(x - x') = \frac{1}{2\pi} \int dk_x \, e^{ik_x(x-x')}[/tex]

    which is just an identity using Fourier transforms (the integral is over the entire real line). We can do likewise in y, because y also has no boundaries. This leaves us:

    [tex]LG = \frac{1}{(2\pi)^2} \int dk_x \, e^{ik_x(x-x')} \, \int dk_y \, e^{ik_y(x-x')} \, \delta(z - z')[/tex]

    My memory gets shaky at this point, but you might be able to see where to go. For [itex]L = \partial^2[/itex], L is pretty simple to invert for the two Fourier transforms. What's left is a function [itex]g(z, z')[/itex] that needs to be solved for.
  4. Jun 27, 2008 #3
    Thanks Ben. I think I actually realize what I don't understand now. Lets say our Geometry is in x,z and t. We can use techniques from Separation of variables to write our GF as:


    Where the [tex]\mu[/tex]superscript runs over x,z,t.

    Ok up to here, but then the second line of my first post implies that


    Ok, so here lies my problem. Why can I just assume that the reduced green function in x and t is just a delta function?
  5. Jun 27, 2008 #4

    Ben Niehoff

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    I may have made a slight mistake; I'm not sure. But you should have

    [tex]\partial^2 G = \delta(x - x') \, \delta(y - y') \, \delta(z - z') \, \delta(t - t')[/tex]

    rather than just G on the left hand side. Once you have that, you can choose to represent some of those delta functions by their Fourier transforms (or Bessel function series, or what-have-you) in the frequency domain.

    The reduced Green's function in a particular dimension is not a delta function; it is [itex]\partial^2g[/itex] that is a delta function.
  6. Jun 27, 2008 #5
    ok, I think it should be:

    [tex]G(x,xs')=\int\frac{d^dk}{(2\pi)^d}e^{i\vec{k}.(x-x')}\int\frac{d\omega}{2\pi}e^{-i\omega(t-')}g(z,z') [/tex]

    because the physics is contained in the z direction so we just expect plane wave solutions in the x and t so we can replace them with delta functions. Then we put this expression for G into the first equation G''.

    Thanks for your insight! :)
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