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電磁場解析で使いそうな概念(4) ~スカラー波動方程式関係~

Last updated at Posted at 2019-11-04

調和振動する場におけるMaxwell方程式

 $\boldsymbol{E}(\boldsymbol{r},t)$および$\boldsymbol{H}(\boldsymbol{r},t)$を調和振動する波として,次のように表現する.

\begin{align}
  \boldsymbol{E}(\boldsymbol{r},t)  &=  E(\boldsymbol{r}) e^{i(\boldsymbol{k} \cdot \boldsymbol{r} -\omega t)}
      =  E(\boldsymbol{r}) e^{i \boldsymbol{k} \cdot \boldsymbol{r}} e^{-i \omega t}  =  \boldsymbol{E}_0(\boldsymbol{r}) e^{-i \omega t}  \tag{1}
  \\
  \boldsymbol{H}(\boldsymbol{r},t)  &=  H(\boldsymbol{r}) e^{i(\boldsymbol{k} \cdot \boldsymbol{r} -\omega t)}
      =  H(\boldsymbol{r}) e^{i \boldsymbol{k} \cdot \boldsymbol{r}} e^{-i \omega t}  =  \boldsymbol{H}_0(\boldsymbol{r}) e^{-i \omega t}  \tag{2}
\end{align}

 ここで,$\omega$は光の角振動数,$i$は虚数単位,$e$はネイピア数,$E(\boldsymbol{r})$および$H(\boldsymbol{r})$は位置のみに依存する振幅の大きさ,$\boldsymbol{E}_0(\boldsymbol{r})$および$\boldsymbol{H}_0(\boldsymbol{r})$は位置のみに依存する振幅ベクトルである.

 また,$\boldsymbol{k}$は波数ベクトルであり,その向きは入射方向を表し,大きさは平面波の波長と関係がある.1次元の場合は$k=\frac{\omega}{c}$の関係があるが,2次元の場合も$k_x = \frac{\omega}{c} \cos \theta,~ k_y = \frac{\omega}{c} \sin \theta$とすれば,成分を決められる.

 微分形のFaradayの法則の式$\nabla \times \boldsymbol{E}(\boldsymbol{r},t) = -\mu_0 \mu(\boldsymbol{r}) \frac{\partial \boldsymbol{H}(\boldsymbol{r},t)}{\partial t}$に式(1),(2)を代入すると,

\begin{align}
  \nabla \times \boldsymbol{E}_0(\boldsymbol{r}) e^{-i \omega t}  &=  -\mu_0 \mu(\boldsymbol{r}) \frac{\partial \boldsymbol{H}_0(\boldsymbol{r}) e^{-i \omega t}}{\partial t}  \nonumber
  \\
  \nabla \times \boldsymbol{E}_0(\boldsymbol{r}) e^{-i \omega t}  &=  i \omega \mu_0 \mu(\boldsymbol{r}) \boldsymbol{H}_0(\boldsymbol{r}) e^{-i \omega t}
\end{align}

 両辺から$e^{-i \omega t}$を消去して,

\begin{align}
  \nabla \times \boldsymbol{E}_0(\boldsymbol{r})  &=  i \omega \mu_0 \mu(\boldsymbol{r}) \boldsymbol{H}_0(\boldsymbol{r})  \tag{3}
\end{align}

 また,電流の無い系($\sigma(\boldsymbol{r})=0$)における,微分形のAmpereの法則の式$\nabla \times \boldsymbol{H}(\boldsymbol{r},t) = \epsilon_0 \epsilon(\boldsymbol{r}) \frac{\partial \boldsymbol{E}(\boldsymbol{r},t)}{\partial t}$に式(1),(2)を代入すると,

\begin{align}
  \nabla \times \boldsymbol{H}_0(\boldsymbol{r}) e^{-i \omega t}  &=  \epsilon_0 \epsilon(\boldsymbol{r}) \frac{\partial \boldsymbol{E}_0(\boldsymbol{r}) e^{-i \omega t}}{\partial t}
  \\
  \nabla \times \boldsymbol{H}_0(\boldsymbol{r}) e^{-i \omega t}  &=  -i \omega \epsilon_0 \epsilon(\boldsymbol{r}) \boldsymbol{E}_0(\boldsymbol{r}) e^{-i \omega t}
\end{align}

 両辺から$e^{-i \omega t}$を消去して,

\begin{align}
  \nabla \times \boldsymbol{H}_0(\boldsymbol{r})  &=  -i \omega \epsilon_0 \epsilon(\boldsymbol{r}) \boldsymbol{E}_0(\boldsymbol{r})  \tag{4}
\end{align}

電場におけるスカラー波動方程式

 式(3)を変形すると,

\begin{align}
  \boldsymbol{H}_0(\boldsymbol{r}) = \frac{\nabla \times \boldsymbol{E}_0(\boldsymbol{r})}{i \omega \mu_0 \mu(\boldsymbol{r})}
\end{align}

 これを式(4)の左辺に代入すると,

\begin{align}
  \nabla \times \frac{\nabla \times \boldsymbol{E}_0(\boldsymbol{r})}{i \omega \mu_0 \mu(\boldsymbol{r})}  &=  -i \omega \epsilon_0 \epsilon(\boldsymbol{r}) \boldsymbol{E}_0(\boldsymbol{r})  \nonumber
  \\
  \frac{1}{i \omega \mu_0} \nabla \times \left\{  \frac{1}{ \mu(\boldsymbol{r}) }  \left[ \nabla \times \boldsymbol{E}_0(\boldsymbol{r}) \right]  \right\}  &=  -i \omega \epsilon_0 \epsilon(\boldsymbol{r}) \boldsymbol{E}_0(\boldsymbol{r})
  \\
  \nabla \times \left\{  \frac{1}{ \mu(\boldsymbol{r}) }  \left[ \nabla \times \boldsymbol{E}_0(\boldsymbol{r}) \right]  \right\}  &=  -i^2 \omega^2 \epsilon_0 \mu_0 \epsilon(\boldsymbol{r}) \boldsymbol{E}_0(\boldsymbol{r})
\end{align}

 また,$i^2 = -1$および真空中の光速が$c = \frac{1}{\sqrt{\epsilon_0 \mu_0}}$であることを使うと,

\begin{align}
  \therefore  \nabla \times \left\{  \frac{1}{ \mu(\boldsymbol{r}) }  \left[ \nabla \times \boldsymbol{E}_0(\boldsymbol{r}) \right]  \right\}
    -\frac{\omega^2}{c^2} \epsilon(\boldsymbol{r}) \boldsymbol{E}_0(\boldsymbol{r})
  =  \boldsymbol{0}  \tag{5}
  %\label{HelmholtzE0}
\end{align}

 ここで,$x$-$y$平面において($\boldsymbol{r}=(x,y,0)$),電場が入射面に対して垂直($z$方向)成分のみ持つ場合を$\boldsymbol{E}_0 = \left[ \begin{array}{ccc} 0& 0& E(\boldsymbol{r}) \end{array} \right] ^\text{T}$とおく.これを式(5)に代入すると,

\begin{align}
  &\nabla \times \left\{  \frac{1}{ \mu(\boldsymbol{r}) }  \left[ \nabla \times  
  \begin{Bmatrix}
    0\\  0\\  E(\boldsymbol{r})
  \end{Bmatrix}
  \right]  \right\}
    -\frac{\omega^2}{c^2} \epsilon(\boldsymbol{r})
  \begin{Bmatrix}
    0\\  0\\  E(\boldsymbol{r})
  \end{Bmatrix}
  \\
  =& \nabla \times \left\{  \frac{1}{ \mu(\boldsymbol{r}) } 
  \begin{vmatrix}
    \boldsymbol{i}&  \boldsymbol{j}&  \boldsymbol{k}\\
    \frac{\partial}{\partial x}&  \frac{\partial}{\partial y}&  \frac{\partial}{\partial z}\\
    0&  0&  E(\boldsymbol{r})
  \end{vmatrix}
   \right\}
    -\frac{\omega^2}{c^2} \epsilon(\boldsymbol{r})
  \begin{Bmatrix}
    0\\ 0\\ E(\boldsymbol{r})
  \end{Bmatrix}
  \\
  =& \nabla \times \left\{  \frac{1}{ \mu(\boldsymbol{r}) } 
  \begin{Bmatrix}
    \frac{\partial E(\boldsymbol{r})}{\partial y}\\
    -\frac{\partial E(\boldsymbol{r})}{\partial x}\\
    0
  \end{Bmatrix}
   \right\}
    -\frac{\omega^2}{c^2} \epsilon(\boldsymbol{r})
  \begin{Bmatrix}
    0\\ 0\\ E(\boldsymbol{r})
  \end{Bmatrix}
  \\
  =& \begin{vmatrix}
       \boldsymbol{i}&  \boldsymbol{j}&  \boldsymbol{k}\\
       \frac{\partial}{\partial x}&  \frac{\partial}{\partial y}&  \frac{\partial}{\partial z}\\
       \frac{1}{\mu(\boldsymbol{r})} \frac{\partial E(\boldsymbol{r})}{\partial y}&  -\frac{1}{\mu(\boldsymbol{r})} \frac{\partial E(\boldsymbol{r})}{\partial x}&  0
     \end{vmatrix}
     -\frac{\omega^2}{c^2} \epsilon(\boldsymbol{r})
     \begin{Bmatrix}
       0\\ 0\\ E(\boldsymbol{r})
     \end{Bmatrix}
  \\
  =& \begin{Bmatrix}
       0\\
       0\\
       -\frac{\partial}{\partial x} \left[ \frac{1}{\mu(\boldsymbol{r})} \frac{\partial E(\boldsymbol{r})}{\partial x} \right]  -\frac{\partial}{\partial y} \left[ \frac{1}{\mu(\boldsymbol{r})} \frac{\partial E(\boldsymbol{r})}{\partial y} \right]
     \end{Bmatrix}
     -\frac{\omega^2}{c^2} \epsilon(\boldsymbol{r})
     \begin{Bmatrix}
       0\\ 0\\ E(\boldsymbol{r})
     \end{Bmatrix}
  \\
  =& \boldsymbol{0}
\end{align}

 これより$z$成分において,次式が成り立つ.

\begin{align}
  \frac{\partial}{\partial x} \left[ \frac{1}{\mu(\boldsymbol{r})} \frac{\partial E(\boldsymbol{r})}{\partial x} \right]
  +\frac{\partial}{\partial y} \left[ \frac{1}{\mu(\boldsymbol{r})} \frac{\partial E(\boldsymbol{r})}{\partial y} \right]
  +\frac{\omega^2}{c^2} \epsilon(\boldsymbol{r}) E(\boldsymbol{r})
  =  0
\end{align}
\begin{align}
  \therefore \nabla \cdot \left[  \frac{1}{\mu(\boldsymbol{r})} \nabla E(\boldsymbol{r})  \right]  +\frac{\omega^2}{c^2} E(\boldsymbol{r})  =  0
  %\label{HelmholtzEz}
\end{align}

 この式をTransverse Magnetic(TM)偏光,またはE偏光などという.なお,この式はHelmholtz方程式の形をしている.

磁場におけるスカラー波動方程式

 式(4)を変形すると,

  \boldsymbol{E}_0(\boldsymbol{r}) = -\frac{\nabla \times \boldsymbol{H}_0(\boldsymbol{r})}{i \omega \epsilon_0 \epsilon(\boldsymbol{r})}

 これを式(3)の左辺に代入すると,

\begin{align}
  \nabla \times \left[ -\frac{\nabla \times \boldsymbol{H}_0(\boldsymbol{r})}{i \omega \epsilon_0 \epsilon(\boldsymbol{r})} \right]  &=  i \omega \mu_0 \mu(\boldsymbol{r}) \boldsymbol{H}_0(\boldsymbol{r})  \nonumber
  \\
  -\frac{1}{i \omega \epsilon_0} \nabla \times \left\{  \frac{1}{ \epsilon(\boldsymbol{r}) }  \left[ \nabla \times \boldsymbol{H}_0(\boldsymbol{r}) \right]  \right\}  &=  i \omega \mu_0 \mu(\boldsymbol{r}) \boldsymbol{H}_0(\boldsymbol{r})
  \\
  \nabla \times \left\{  \frac{1}{ \epsilon(\boldsymbol{r}) }  \left[ \nabla \times \boldsymbol{H}_0(\boldsymbol{r}) \right]  \right\}  &=  -i^2 \omega^2 \epsilon_0 \mu_0 \mu(\boldsymbol{r}) \boldsymbol{H}_0(\boldsymbol{r})
\end{align}

 また,$i^2 = -1$および真空中の光速が$c = \frac{1}{\sqrt{\epsilon_0 \mu_0}}$であることを使うと,

\begin{align}
  \therefore  \nabla \times \left\{  \frac{1}{ \epsilon(\boldsymbol{r}) }  \left[ \nabla \times \boldsymbol{H}_0(\boldsymbol{r}) \right]  \right\}
    -\frac{\omega^2}{c^2} \mu(\boldsymbol{r}) \boldsymbol{H}_0(\boldsymbol{r})
  =  \boldsymbol{0}  \tag{6}
  %\label{HelmholtzH0}
\end{align}

 ここで,磁場が入射面に対して垂直($z$方向)成分のみ持つ場合を$\boldsymbol{H}_0 = \left[ \begin{array}{ccc} 0& 0& H(\boldsymbol{r}) \end{array} \right] ^\text{T}$とおく.これを式(6)に代入すると,

\begin{align}
  &\nabla \times \left\{  \frac{1}{ \epsilon(\boldsymbol{r}) }  \left[ \nabla \times  
  \begin{Bmatrix}
    0\\  0\\  H(\boldsymbol{r})
  \end{Bmatrix}
  \right]  \right\}
    -\frac{\omega^2}{c^2} \mu(\boldsymbol{r})
  \begin{Bmatrix}
    0\\  0\\  H(\boldsymbol{r})
  \end{Bmatrix}
  \\
  =& \nabla \times \left\{  \frac{1}{ \epsilon(\boldsymbol{r}) } 
  \begin{vmatrix}
    \boldsymbol{i}&  \boldsymbol{j}&  \boldsymbol{k}\\
    \frac{\partial}{\partial x}&  \frac{\partial}{\partial y}&  \frac{\partial}{\partial z}\\
    0&  0&  H(\boldsymbol{r})
  \end{vmatrix}
   \right\}
    -\frac{\omega^2}{c^2} \mu(\boldsymbol{r})
  \begin{Bmatrix}
    0\\ 0\\ H(\boldsymbol{r})
  \end{Bmatrix}
  \\
  =& \nabla \times \left\{  \frac{1}{ \epsilon(\boldsymbol{r}) } 
  \begin{Bmatrix}
    \frac{\partial H(\boldsymbol{r})}{\partial y}\\
    -\frac{\partial H(\boldsymbol{r})}{\partial x}\\
    0
  \end{Bmatrix}
   \right\}
    -\frac{\omega^2}{c^2} \mu(\boldsymbol{r})
  \begin{Bmatrix}
    0\\ 0\\ H(\boldsymbol{r})
  \end{Bmatrix}
  \\
  =& \begin{vmatrix}
       \boldsymbol{i}&  \boldsymbol{j}&  \boldsymbol{k}\\
       \frac{\partial}{\partial x}&  \frac{\partial}{\partial y}&  \frac{\partial}{\partial z}\\
       \frac{1}{\epsilon(\boldsymbol{r})} \frac{\partial H(\boldsymbol{r})}{\partial y}&  -\frac{1}{\epsilon(\boldsymbol{r})} \frac{\partial H(\boldsymbol{r})}{\partial x}&  0
     \end{vmatrix}
     -\frac{\omega^2}{c^2} \mu(\boldsymbol{r})
     \begin{Bmatrix}
       0\\ 0\\ H(\boldsymbol{r})
     \end{Bmatrix}
  \\
  =& \begin{Bmatrix}
       0\\
       0\\
       -\frac{\partial}{\partial x} \left[ \frac{1}{\epsilon(\boldsymbol{r})} \frac{\partial H(\boldsymbol{r})}{\partial x} \right]  -\frac{\partial}{\partial y} \left[ \frac{1}{\epsilon(\boldsymbol{r})} \frac{\partial H(\boldsymbol{r})}{\partial y} \right]
     \end{Bmatrix}
     -\frac{\omega^2}{c^2} \mu(\boldsymbol{r})
     \begin{Bmatrix}
       0\\ 0\\ H(\boldsymbol{r})
     \end{Bmatrix}
  \\
  =& \boldsymbol{0}
\end{align}

 これより$z$成分において,次式が成り立つ.

\begin{align}
  \frac{\partial}{\partial x} \left[ \frac{1}{\epsilon(\boldsymbol{r})} \frac{\partial H(\boldsymbol{r})}{\partial x} \right]
  +\frac{\partial}{\partial y} \left[ \frac{1}{\epsilon(\boldsymbol{r})} \frac{\partial H(\boldsymbol{r})}{\partial y} \right]
  +\frac{\omega^2}{c^2} \mu(\boldsymbol{r}) H(\boldsymbol{r})
  =  0
\end{align}
\begin{align}
  \therefore \nabla \cdot \left[  \frac{1}{\epsilon(\boldsymbol{r})} \nabla H(\boldsymbol{r})  \right]  +\frac{\omega^2}{c^2} H(\boldsymbol{r})  =  0
  %\label{HelmholtzHz}
\end{align}

 この式をTransverse Electric(TE)偏光,またはH偏光などという.なお,この式はHelmholtz方程式の形をしている.

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