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Far-field dyadic Green's function

  1. Jun 24, 2013 #1
    1. The problem statement, all variables and given/known data
    I'm using the book "Principles of nano-optics" by Novotny and Hecht. I'm stuck a bit at understanding the derivation of the point spread function. It's just given as
    [tex]\mathbf{G} = \frac{\exp(i k_1 r)}{4 \pi r} \exp[-i k_1(x_0 x / r + y_0 y / r + z_0 z / r)]\\
    \qquad\times \left[ \begin{array}{ccc}
    (1 - x^2/r^2) & - x y/r^2 & -x z/r^2 \\
    - x y/r^2 & (1 - y^2/r^2) & -y z/r^2 \\
    - x z/r^2 & -y z/r^2 & (1 - z^2/r^2)
    \end{array}
    \right][/tex]
    in equation (D.4) in the book (if you have it).

    2. Relevant equations
    The far-field comes from the angular spectrum representation where the evanescent waves have been dropped. This is given as
    [tex]\vec{E}_\infty(s_x, s_y, s_z) = -2 \pi i k s_z \hat{\vec{E}}(k s_x, k s_y; z = 0) \quad ,[/tex]
    where E with the hat is the Fourier transform with respect to x and y of the field at z = 0. The vector s is just a unit vector indicating the direction.

    I know the dyad at z=0 is given as
    [tex]\mathbf{G}(\vec{r}, \vec{r}_0) = (\mathbf{I} + \frac{1}{k^2} \nabla \nabla) G_0(\vec{r}, \vec{r}_0)\quad,[/tex]
    where G0 is the scalar Green's function, known as
    [tex]G_0(\vec{r}, \vec{r}_0) = \frac{\exp(\pm i k \vert \vec{r} - \vec{r_0} \vert}{4 \pi \vert \vec{r} - \vec{r}_0 \vert ) } \quad ,[/tex]
    which is the Green's function for the Helmholtz equation.
    3. The attempt at a solution
    I've tried to do a direct Fourier transform of the dyad G, so I can substitute that into the far-field equation, but I've been unable to get that, because I can't get the Fourier transform of G0 with respect to x and y.
    I am able to get the Fourier transform of G0 with respect to x, y and z by using the Helmholtz equation and doing a Fourier transform on both sides. However, I need the Fourier transform to be only with respect to x and y but not z.
    Does anybody have an idea?
     
  2. jcsd
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