自由空間中のダイアディック・グリーン関数の対称性

4.7　自由空間中のダイアディック・グリーン関数の対称性

　自由空間中のダイアディック・グリーン関数

{\tilde{\tilde{G}}}_{e 0}

，

{\tilde{\tilde{G}}}_{m 0}

は，先に示したように，

\begin{array}{rcl} (1) & {\tilde{\tilde{G}}}_{e 0} (r, r^{'}) & = & (\tilde{\tilde{I}} + \frac{1}{k^{2}} \nabla \nabla) G_{0} (r, r^{'}) \\ (2) & {\tilde{\tilde{G}}}_{m 0} (r, r^{'}) & = & \nabla \times (\tilde{\tilde{I}} G_{0} (r, r^{'})) \end{array}

あるいは，

\begin{matrix} (3) & {\tilde{\tilde{G}}}_{m 0} (r, r^{'}) = \nabla G_{0} (r, r^{'}) \times \tilde{\tilde{I}} \end{matrix}

ここで，

G_{0} (r, r^{'})

は自由空間中のスカラ・グリーン関数を示し，

\begin{array}{rcl} G_{0} (r, r^{'}) & = & \frac{e^{- j k R}}{4 π R} \\ (4) & = & G_{0} (r^{'}, r) \end{array}

ただし，

\begin{array}{rcl} R & \equiv & | r - r^{'} | \\ (5) & = & \sqrt{(x_{1} - x_{1}^{'})^{2} + (x_{2} - x_{2}^{'})^{2} + (x_{3} - x_{3}^{'})^{2}} \end{array}

まず，

\begin{array}{rcl} \nabla G_{0} (r, r^{'}) & = & \nabla G_{0} (r^{'}, r) \\ = & \frac{\partial G_{0}}{\partial R} \nabla R \\ = & \frac{\partial G_{0}}{\partial R} (- \nabla^{'} R) \\ (6) & = & - \nabla^{'} G_{0} (r^{'}, r) \end{array}

ただし，

\begin{array}{rcl} \nabla R & = & \frac{r - r^{'}}{R} \\ (7) & = & - \nabla^{'} R \end{array}

さらに，

\begin{array}{rcl} \nabla \nabla G_{0} (r, r^{'}) & = & - \nabla \nabla^{'} G_{0} (r^{'}, r) \\ = & - \nabla (\frac{\partial G_{0}}{\partial R} (\nabla^{'} R)) \\ = & \nabla^{'} (\frac{\partial G_{0}}{\partial R} (\nabla^{'} R)) \\ (8) & = & \nabla^{'} \nabla^{'} G_{0} (r^{'}, r) \end{array}

ただし，

\begin{array}{rcl} \nabla^{'} \nabla R & = & \nabla^{'} (\frac{r - r^{'}}{R}) \\ = & - \nabla (\frac{r - r^{'}}{R}) \\ (9) & = & - \nabla \nabla R \end{array}

さて，

{\tilde{\tilde{G}}}_{e 0} (r, r^{'})

において，

r

，

r^{'}

を交換すると，

\begin{matrix} (10) & {\tilde{\tilde{G}}}_{e 0} (r^{'}, r) = (\tilde{\tilde{I}} + \frac{1}{k^{2}} \nabla^{'} \nabla^{'}) G_{0} (r^{'}, r) \end{matrix}

ここで，

\nabla \nabla G_{0} (r, r^{'}) = \nabla^{'} \nabla^{'} G_{0} (r^{'}, r)

より，

\begin{array}{rcl} {\tilde{\tilde{G}}}_{e 0} (r^{'}, r) & = & (\tilde{\tilde{I}} + \frac{1}{k^{2}} \nabla \nabla) G_{0} (r, r^{'}) \\ (11) & = & {\tilde{\tilde{G}}}_{e 0} (r, r^{'}) \end{array}

転置をとると，

\begin{array}{rcl} ({\tilde{\tilde{G}}}_{e 0} (r^{'}, r))^{T} & = & {(\tilde{\tilde{I}} + \frac{1}{k^{2}} \nabla^{'} \nabla^{'}) G_{0} (r^{'}, r)}^{T} \\ (12) & = & {\tilde{\tilde{G}}}_{e 0} (r^{'}, r) \end{array}

ただし，

(\tilde{\tilde{I}})^{T} = \tilde{\tilde{I}}

．また，

\begin{array}{rcl} (\nabla^{'} \nabla^{'} G_{0} (r^{'}, r))^{T} & = & {(\sum_{i = 1}^{3} \sum_{j = 1}^{3} \frac{\partial^{2} G_{0}}{\partial x_{i}^{'} \partial x_{j}^{'}} i_{i} i_{j})}^{T} \\ (13) & = & \nabla^{'} \nabla^{'} G_{0} (r^{'}, r) \end{array}

これより，次のように

{\tilde{\tilde{G}}}_{e 0}

についての対称性（symmetrical property）が得られる．

\begin{matrix} (14) & ({\tilde{\tilde{G}}}_{e 0} (r^{'}, r))^{T} = {\tilde{\tilde{G}}}_{e 0} (r, r^{'}) \end{matrix}

次に，

{\tilde{\tilde{G}}}_{m 0} (r, r^{'})

についても

r

，

r^{'}

を交換し，式

(2)

より，

\begin{array}{rcl} {\tilde{\tilde{G}}}_{m 0} (r^{'}, r) & = & \nabla^{'} \times (\tilde{\tilde{I}} G_{0} (r^{'}, r)) \\ = & \nabla^{'} G_{0} \times \tilde{\tilde{I}} \\ = & - \nabla G_{0} \times \tilde{\tilde{I}} \\ = & - \nabla \times (\tilde{\tilde{I}} G_{0} (r, r^{'})) \\ (15) & = & - {\tilde{\tilde{G}}}_{m 0} (r, r^{'}) \end{array}

ここで，ダイアディクス公式

(a \times \tilde{\tilde{F}})^{T} \cdot \tilde{\tilde{b}} = - {\tilde{\tilde{F}}}^{T} \cdot (a \times \tilde{\tilde{b}})

より，

\begin{aligned} (\nabla^{'} G_{0} \times \tilde{\tilde{I}})^{T} \cdot \tilde{\tilde{I}} = - {\tilde{\tilde{I}}}^{T} \cdot (\nabla^{'} G_{0} \times \tilde{\tilde{I}}) \\ (16) & (\nabla^{'} G_{0} \times \tilde{\tilde{I}})^{T} = - \nabla^{'} G_{0} \times \tilde{\tilde{I}} \end{aligned}

また，

{\tilde{\tilde{G}}}_{m 0}

の転置は，式

(3)

より，

\begin{array}{rcl} ({\tilde{\tilde{G}}}_{m 0} (r^{'}, r))^{T} & = & (\nabla^{'} G_{0} \times \tilde{\tilde{I}})^{T} \\ = & - \nabla^{'} G_{0} \times \tilde{\tilde{I}} \\ (17) & = & - {\tilde{\tilde{G}}}_{m 0} (r^{'}, r) \end{array}

したがって，次のように

{\tilde{\tilde{G}}}_{m 0}

についての対称性が得られる．

\begin{matrix} (18) & ({\tilde{\tilde{G}}}_{m 0} (r^{'}, r))^{T} = {\tilde{\tilde{G}}}_{m 0} (r, r^{'}) \end{matrix}