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$\DYA{G}_{e0}$，$\DYA{G}_{m0}$は，先に示したように，
\begin{eqnarray}
\DYA{G}_{e0} (\VEC{r},\VEC{r}')
&=& \left( \DYA{I} + \frac{1}{k^2} \nabla \nabla \right) G_0 (\VEC{r},\VEC{r}')
\\
\DYA{G}_{m0}(\VEC{r},\VEC{r}')
&=& \nabla \times \Big( \DYA{I} G_0 (\VEC{r},\VEC{r}') \Big) \label{eq:gm00}
\end{eqnarray}
あるいは，
\begin{gather}
\DYA{G}_{m0}(\VEC{r},\VEC{r}')
= \nabla G_0(\VEC{r},\VEC{r}') \times \DYA{I} \label{eq:gm000}
\end{gather}
ここで，$G_0(\VEC{r},\VEC{r}')$は自由空間中のスカラ・グリーン関数を示し，
\begin{eqnarray}
G_0 (\VEC{r},\VEC{r}')
&=& \frac{e^{-jkR}}{4\pi R}
\nonumber \\
&=& G_0 (\VEC{r}',\VEC{r})
\end{eqnarray}
ただし，
\begin{eqnarray}
R &\equiv& |\VEC{r}-\VEC{r}'|
\nonumber \\
&=& \sqrt{(x_1-x_1')^2 + (x_2-x_2')^2 + (x_3-x_3')^2}
\end{eqnarray}
まず，
\begin{eqnarray}
\nabla G_0 (\VEC{r},\VEC{r}')
&=& \nabla G_0 (\VEC{r}',\VEC{r})
\nonumber \\
&=& \frac{\partial G_0}{\partial R} \nabla R
\nonumber \\
&=& \frac{\partial G_0}{\partial R} (-\nabla ' R)
\nonumber \\
&=& - \nabla ' G_0 (\VEC{r}',\VEC{r})
\end{eqnarray}
ただし，
\begin{eqnarray}
\nabla R
&=& \frac{\VEC{r} - \VEC{r}'}{R}
\nonumber \\
&=& -\nabla ' R
\end{eqnarray}
さらに，
\begin{eqnarray}
\nabla \nabla G_0 (\VEC{r},\VEC{r}')
&=& -\nabla \nabla ' G_0 (\VEC{r}',\VEC{r})
\nonumber \\
&=& -\nabla \left( \frac{\partial G_0}{\partial R} (\nabla ' R) \right)
\nonumber \\
&=& \nabla ' \left( \frac{\partial G_0}{\partial R} (\nabla ' R) \right)
\nonumber \\
&=& \nabla ' \nabla ' G_0 (\VEC{r}',\VEC{r})
\end{eqnarray}
ただし，
\begin{eqnarray}
\nabla ' \nabla R
&=& \nabla ' \left( \frac{\VEC{r} - \VEC{r}'}{R} \right)
\nonumber \\
&=& -\nabla \left( \frac{\VEC{r} - \VEC{r}'}{R} \right)
\nonumber \\
&=& -\nabla \nabla R
\end{eqnarray}
さて，
$\DYA{G}_{e0} (\VEC{r},\VEC{r}')$
において，$\VEC{r}$，$\VEC{r}'$を交換すると，
\begin{gather}
\DYA{G}_{e0} (\VEC{r}',\VEC{r})
= \left( \DYA{I} + \frac{1}{k^2} \nabla ' \nabla ' \right) G_0 (\VEC{r}',\VEC{r})
\end{gather}
ここで，
$\nabla \nabla G_0 (\VEC{r},\VEC{r}')
= \nabla ' \nabla ' G_0 (\VEC{r}',\VEC{r})$より，
\begin{eqnarray}
\DYA{G}_{e0} (\VEC{r}',\VEC{r})
&=& \left( \DYA{I} + \frac{1}{k^2} \nabla \nabla \right) G_0 (\VEC{r},\VEC{r}')
\nonumber \\
&=& \DYA{G}_{e0} (\VEC{r},\VEC{r}')
\end{eqnarray}
転置をとると，
\begin{eqnarray}
\Big( \DYA{G}_{e0} (\VEC{r}',\VEC{r}) \Big) ^T
&=& \left\{ \left( \DYA{I} + \frac{1}{k^2} \nabla ' \nabla ' \right) G_0 (\VEC{r}',\VEC{r}) \right\}^T
\nonumber \\
&=& \DYA{G}_{e0} (\VEC{r}',\VEC{r})
\end{eqnarray}
ただし，
$\big( \DYA{I} \big) ^T = \DYA{I}$．また，
\begin{eqnarray}
\Big( \nabla ' \nabla ' G_0 (\VEC{r}',\VEC{r}) \Big) ^T
&=& \left( \sum _{i=1}^3 \sum _{j=1}^3 \frac{\partial ^2 G_0}{\partial x_i' \partial x_j'} \VEC{i}_i \VEC{i}_j \right) ^T
\nonumber \\
&=& \nabla ' \nabla ' G_0 (\VEC{r}',\VEC{r})
\end{eqnarray}
これより，次のように$\DYA{G}_{e0}$についての対称性（symmetrical property）が得られる．
\begin{gather}
\Big( \DYA{G}_{e0} (\VEC{r}',\VEC{r}) \Big) ^T = \DYA{G}_{e0} (\VEC{r},\VEC{r}')
\end{gather}
次に，
$\DYA{G}_{m0} (\VEC{r},\VEC{r}')$
についても$\VEC{r}$，$\VEC{r}'$を交換し，式\eqref{eq:gm00}より，
\begin{eqnarray}
\DYA{G}_{m0}(\VEC{r}',\VEC{r})
&=& \nabla ' \times \Big( \DYA{I} G_0 (\VEC{r}',\VEC{r}) \Big)
\nonumber \\
&=& \nabla ' G_0 \times \DYA{I}
\nonumber \\
&=& -\nabla G_0 \times \DYA{I}
\nonumber \\
&=& -\nabla \times \Big( \DYA{I} G_0 (\VEC{r},\VEC{r}') \Big)
\nonumber \\
&=& -\DYA{G}_{m0}(\VEC{r},\VEC{r}')
\end{eqnarray}
ここで，ダイアディクス公式 $(\VEC{a} \times \DYA{F})^T \cdot \DYA{b} = -\DYA{F}^T \cdot ( \VEC{a} \times \DYA{b} )$より，
\begin{align}
&(\nabla ' G_0 \times \DYA{I})^T \cdot \DYA{I}
= -\DYA{I}^T \cdot ( \nabla ' G_0 \times \DYA{I} ) \nonumber \\
&(\nabla ' G_0 \times \DYA{I})^T = -\nabla ' G_0 \times \DYA{I}
\end{align}
また，$\DYA{G}_{m0}$の転置は，式\eqref{eq:gm000}より，
\begin{eqnarray}
\big( \DYA{G}_{m0} (\VEC{r}',\VEC{r}) \big) ^T
&=& \big( \nabla ' G_0 \times \DYA{I} \big) ^T
\nonumber \\
&=& -\nabla ' G_0 \times \DYA{I}
\nonumber \\
&=& - \DYA{G}_{m0} (\VEC{r}',\VEC{r})
\end{eqnarray}
したがって，次のように$\DYA{G}_{m0}$についての対称性が得られる．
\begin{gather}
\Big( \DYA{G}_{m0} (\VEC{r}',\VEC{r}) \Big) ^T = \DYA{G}_{m0} (\VEC{r},\VEC{r}')
\end{gather}