# A Green function issue: can bound. cond. be applied mode-wise?

1. May 30, 2017

### Twigg

Hello,

Before I begin, a lot of the math I try to describe in this post is stuff I worked out myself and have scribbled down. If something looks fishy or doesn't make sense, it could very well be totally wrong! Apologies for any confusions in advance.

Consider a slab of linear isotropic dielectric material in the lower half plane ($z < 0$) with vacuum in the upper half plane. After a temporal Fourier transform and some algebra, Ampere's law says $$[\nabla ^{2} + \epsilon( \omega) \frac{ \omega^{2}}{c^{2}}] \vec{E}(\vec{x},\omega) = c^{2} \mu_{0} \epsilon( \omega) \nabla \rho( \vec{x}, \omega) + i \mu_{0} \omega \vec{j}( \vec{x}, \omega)$$ where $\epsilon$ is the relative permittivity and the rest of the notation is standard. I was solving for the Green function $$[\nabla_{x'} ^{2} + \epsilon( \omega) \frac{ \omega^{2}}{c^{2}}] G_{\alpha\beta}(\vec{x} - \vec{x}',\omega) = -4 \pi \delta_{\alpha\beta} \delta (\vec{x} - \vec{x}')$$ and I had no problems until I started thinking about the boundary conditions on the Green function as $|z-z'| \rightarrow \infty$. Then I ran into two issues I wasn't 100% sure about.

After separating variables, I expand the Green's function in cylindrical coordinates $$G_{\alpha \beta} (\vec{x} - \vec{x}') = \frac{1}{2\pi} \sum_{m=-\infty}^{\infty} \int_{0}^{\infty} k_{\parallel} dk_{\parallel} e^{im(\phi - \phi')} J_{m}(k_{\parallel}\rho')J_{m}(k_{\parallel}\rho)G_{\alpha\beta}(z-z', k_{\parallel})$$ This results in the following ODE $$[\frac{\partial^{2}}{\partial z'^2} + (\frac{\omega^2}{c^2}\epsilon(\omega) - k^{2}_{\parallel})]G_{\alpha\beta}(z-z', k_{\parallel}) = -4\pi \delta_{\alpha\beta} \delta(z - z')$$ which I solved piecewise for the case where $z' > z$ (I call this piece $G^{+}_{\alpha\beta}$) and for the case where $z' < z$ (I call this piece $G^{-}_{\alpha\beta}$) to get $$G^{\pm}_{\alpha\beta}(z-z',k_{\parallel}) = A^{\pm}_{\alpha\beta} e^{+ik_{\perp}(z-z')} + B^{\pm}_{\alpha\beta} e^{-ik_{\perp}(z-z')}$$ where $k_{\perp} = [\frac{\omega^2}{c^2}\epsilon(\omega) - k^{2}_{\parallel}]^{1/2}$ evaluated on the principal branch. Due to the implicit boundary conditions associated with the Dirac delta, the above solution is subject to $$A^{+}_{\alpha\beta} - A^{-}_{\alpha\beta} = -(B^{+}_{\alpha\beta} - B^{-}_{\alpha\beta}) = -\frac{2\pi}{ik_{\perp}}\delta_{\alpha\beta}$$

This is where I started having issues. From what little I understand about the general theory of Green functions, the Dirichlet Green function is supposed to vanish at infinity. At first, I thought this means I would need to choose the coefficients $A^{\pm}_{\alpha\beta}$ and $B^{\pm}_{\alpha\beta}$ such that $G^{+}_{\alpha\beta}$ vanishes as $z \rightarrow -\infty$ and vice versa $G^{-}_{\alpha\beta}$ vanishes as $z \rightarrow \infty$. This led me to issue #1.

Issue #1: How should I apply the boundary conditions at $|z-z'| \rightarrow \infty$ to $G_{\alpha\beta}(z-z', k_{\parallel})$? I went back and forth on this. I'll try to summarize my different trains of thought briefly. In free space where $\epsilon(\omega) = 1$, then $k_{\perp}$ is real when $k_{\parallel}^{2} < \frac{\omega^{2}}{c^2}$ and $G_{\alpha\beta}(z-z', k_{\parallel})$ does not vanish as $|z - z'| \rightarrow \infty$. My first impulse was to conclude that the coefficients $A^{\pm}_{\alpha\beta}$ and $B^{\pm}_{\alpha\beta}$ must be zero when $k_{\perp}$ is purely real. But then I hesitated because the boundary condition states that $G_{\alpha\beta}(\vec{x} - \vec{x}')$ must vanish at infinity, but it does not explicitly require that $G_{\alpha\beta}(z-z', k_{\parallel})$ vanish at infinity. After all, $\int e^{ik(z-z')} dk = \delta(z-z')$, which goes to show that you can't always just look at the asymptotic behavior of the integrand and figure out what'll happen with the integral. Fine. Problem is, I can't evaluate $G_{\alpha\beta}(\vec{x} - \vec{x}')$, as defined in equation 3, unless I know how the coefficients $A^{\pm}_{\alpha\beta}$ and $B^{\pm}_{\alpha\beta}$ depend on $k_{\parallel}$, and I doubt I will be able to find a closed form even if I did know how the coefficients depended on $k_{\parallel}$. My next impulse was to treat the oscillatory cases where $k_{\perp}$ is real as "small enough" and instead use the boundary condition to remove any terms with complex $k_{\perp}$ that grow exponentially as $|z - z'| \rightarrow \infty$. This leaves me with a plausible candidate of $$G_{\alpha\beta}(z-z', k_{\parallel}) = \frac{2\pi}{ik_{\perp}} e^{i \mathrm{sgn}(\mathrm{Im}(k_{\perp})) k_{\perp} (z_> - z_<)}$$ where $z_>$ is the bigger of $z$ and $z'$ and likewise $z_<$ is the smaller of $z$ and $z'$, and $\mathrm{sgn}(x)$ is the signum function, which just spits out the sign of its argument (+1 or -1). Can anyone clarify?

Issue #2:
In this derivation, I never assumed causality as a constraint, so I was expecting two independent solutions, one advanced and one retarded. There should then be one degree of freedom in my answer, which would dictate what combination of the advanced and retarded solutions go into the general Green function, as in D'Alembert's formula. Equation 5 has 4 degrees of freedom, and Equation 6 constrains that down to 2 degrees of freedom. Then there are two additional boundary conditions: $G^(+)_{\alpha\beta} \rightarrow 0$ as $z \rightarrow -\infty$, and $G^-_{\alpha\beta} \rightarrow 0$ as $z \rightarrow \infty$. That would fully define the resulting Green function. What gives? Have I done goofed up? Are there redundant constraints?

2. May 31, 2017

### Twigg

Incidentally, it occurred to me that I did a silly. To solve the overarching problem I outlined above (solving Green function for slab of dielectric in the lower half plane), I can just use the usual Green function for the Helmholtz equation with boundary at infinity $$G_{\alpha\beta}(\vec{x} - \vec{x}') = A\delta_{\alpha\beta}\frac{e^{+ik|\vec{x} - \vec{x}'|}}{|\vec{x} - \vec{x}'|} + B\delta_{\alpha\beta}\frac{e^{-ik|\vec{x} - \vec{x}'|}}{|\vec{x} - \vec{x}'|}$$ with $A+B=1$ and $k = \frac{\omega}{c}n(\omega)$ where $n(\omega) = [\epsilon(\omega)]^{1/2}$ is the index of refraction, and match it to the boundary conditions for Maxwell's equations by throwing in some image sources with advanced and retarded parts.

Still, I can't say I understand everything I did wrong in the work I posted originally. To the best of my knowledge, I should have still come up with an integral expression for the Green function with boundary at infinity (equation 8). As I pointed out in issue #2, when I used the mode-wise boundary conditions $G^{\pm}_{\alpha\beta}(z-z', k_{\parallel}) \rightarrow 0$ as $z \rightarrow \mp \infty$, I wind up with 0 leftover degrees of freedom, whereas equation 8 has 1. However, not applying either of these two boundary conditions leaves me with 2 degrees of freedom. I could just use the limiting procedure most textbooks use and compare the result with the Poisson equation Green function when $k|\vec{x} - \vec{x}'| \rightarrow 0$ and that would get rid of one degree of freedom. I'm just not clear why applying the aforementioned boundary conditions gave a different result. Regardless, given that $$\frac{1}{|\vec{x} - \vec{x}'|} = \sum_{m = -\infty}^{\infty} \int_{0}^{\infty} dk_{\parallel} e^{im(\phi-\phi')} J_{m}(k_{\parallel}\rho) J_{m}(k_{\parallel}\rho') e^{-k_{\parallel}(z_{>} - z_{<})}$$ By comparison with equation 3, I should have gotten $$G_{\alpha\beta}(z-z', k_{\parallel}) = \frac{2\pi}{k_{\parallel}}e^{-k_{\parallel}(z_{>} - z_{<})}\delta_{\alpha\beta}(Ae^{+ik |\vec{x} - \vec{x}'|} + Be^{-ik |\vec{x} - \vec{x}'|})$$ where $k = [k_{\parallel}^{2} + k_{\perp}^{2}]^{1/2} = \frac{\omega}{c}n(\omega)$. Any ideas where I went wrong?