自由空間中の電流素子のスペクトル領域グリーン関数

自由空間中の電流素子のスペクトル領域グリーン関数の導出

　誘電体基板がなく，自由空間中の

x y

面上に電流源がある場合を考える．

\begin{matrix} (1) & {\tilde{Z}}^{(0)} = - \frac{1}{Y_{0} + Y_{0}} = - \frac{1}{2 Y_{0}} \end{matrix}

これより，

\begin{matrix} (2) & Z_{_{T E}}^{(0)} = - \frac{1}{2 Y_{0_{T E}}} = - \frac{1}{2 \frac{k_{z 0}}{ω μ_{0}}} = - \frac{ω μ_{0}}{2 k_{z 0}} \\ (3) & Z_{_{T M}}^{(0)} = - \frac{1}{2 Y_{0_{T M}}} = - \frac{1}{2 \frac{ω ϵ_{0}}{k_{z 0}}} = - \frac{k_{z 0}}{2 ω ϵ_{0}} \end{matrix}

よって，

\begin{array}{rcl} {\tilde{G}}_{x x}^{^{E J}} & = & \frac{1}{k_{t}^{2}} (k_{y}^{2} {\tilde{Z}}_{_{T E}}^{(0)} + k_{x}^{2} {\tilde{Z}}_{_{T M}}^{(0)}) \\ = & \frac{1}{k_{t}^{2}} {k_{y}^{2} (- \frac{ω μ_{0}}{2 k_{z 0}}) + k_{x}^{2} (- \frac{k_{z 0}}{2 ω ϵ_{0}})} \\ (4) & = & - \frac{k_{0}^{2} k_{y}^{2} + k_{z 0}^{2} k_{x}^{2}}{2 ω ϵ_{0} k_{t}^{2} k_{z 0}} \end{array}

ここで，

\begin{array}{rcl} k_{0}^{2} k_{y}^{2} + k_{z 0} k_{x}^{2} & = & k_{0}^{2} (k_{t}^{2} - k_{x}^{2}) + (k_{0}^{2} - k_{t}^{2}) k_{x}^{2} \\ (5) & = & k_{t}^{2} (k_{0}^{2} - k_{x}^{2}) \end{array}

より，

\begin{array}{rcl} {\tilde{G}}_{x x}^{^{E J}} & = & - \frac{k_{0}^{2} - k_{x}^{2}}{2 ω ϵ_{0} k_{z 0}} = \frac{k_{0}^{2} - k_{x}^{2}}{j ω ϵ_{0}} \frac{1}{j 2 k_{z 0}} \\ (6) & = & - \frac{j ω μ_{0}}{k_{0}^{2}} \frac{1}{j 2 k_{z 0}} (k_{0}^{2} - k_{x}^{2}) \end{array}

また，

\begin{array}{rcl} {\tilde{G}}_{x y}^{^{E J}} & = & {\tilde{G}}_{y x}^{^{E J}} = \frac{k_{x} k_{y}}{k_{t}^{2}} ({\tilde{Z}}_{_{T M}}^{(z)} - {\tilde{Z}}_{_{T E}}^{(z)}) \\ = & \frac{k_{x} k_{y}}{k_{t}^{2}} (- \frac{k_{z 0}}{2 ω ϵ_{0}} + \frac{ω μ_{0}}{2 k_{z 0}}) \\ = & \frac{k_{x} k_{y} (- k_{z 0}^{2} + k_{0}^{2})}{2 ω ϵ_{0} k_{t}^{2} k_{z 0}} = \frac{k_{x} k_{y}}{2 ω ϵ_{0} k_{z 0}} \\ (7) & = & - \frac{k_{x} k_{y}}{j ω ϵ_{0}} \frac{1}{j 2 k_{z 0}} = \frac{j ω μ_{0}}{k_{0}^{2}} \frac{1}{j 2 k_{z 0}} k_{x} k_{y} \end{array}

\begin{array}{rcl} {\tilde{G}}_{y y}^{^{E J}} & = & \frac{1}{k_{t}^{2}} (k_{x}^{2} {\tilde{Z}}_{_{T E}}^{(z)} + k_{y}^{2} {\tilde{Z}}_{_{T M}}^{(z)}) \\ = & - \frac{k_{0}^{2} - k_{y}^{2}}{2 ω ϵ_{0} k_{z 0}} = \frac{k_{0}^{2} - k_{y}^{2}}{j ω ϵ_{0}} \frac{1}{j 2 k_{z 0}} \\ (8) & = & - \frac{j ω μ_{0}}{k_{0}^{2}} \frac{1}{j 2 k_{z 0}} (k_{0}^{2} - k_{y}^{2}) \end{array}