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\appendix

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\chapter{本文中の各式の導出}
\markright{本文中の各式の導出}


\section{(\ref{hadou4})の導出}

式(\ref{kiso3_motion-2})
\begin{equation}
  \bm{v}_{1}=\frac{a^{2}\rho_{1}}{\omega\rho_{0}}\bm{k}+\frac{1}{\mu\rho_{0}\omega}\bm{B}_{0}\times(\bm{k}\times\bm{B}_{1})
\end{equation}
から始める. ここに(\ref{kiso2_mass-2})と(\ref{kiso4_induction-2})を代入して $\rho_{1}, \bm{B}_{1}$ を消去すると,
\begin{subequations}
  \begin{align}
    \bm{v}_{1}&=\frac{a^{2}}{\mu\rho_{0}}\frac{\rho_{0}}{\omega}(\bm{k}\cdot\bm{v}_{1})\bm{k}+\frac{1}{\mu\rho_{0}\omega}\bm{B}_{0}\times\left[\bm{k}\times\left\{\frac{1}{\omega}\bm{k}\times(\bm{B}_{0}\times\bm{v}_{1})\right\}\right] \notag \\
    &=\frac{a^{2}}{\omega^{2}}(\bm{k}\cdot\bm{v}_{1})\bm{k}+\frac{1}{\mu\rho_{0}\omega^{2}}\bm{B}_{0}\times[\bm{k}\times\{\bm{k}\times(\bm{B}_{0}\times\bm{v}_{1})\}] \tag{A.2}
  \end{align}
\end{subequations}
となる. これより, 式(\ref{hadou3})
\begin{equation}
  \omega^{2}\bm{v}_{1}=a^{2}(\bm{k}\cdot\bm{v}_{1})\bm{k}-\frac{1}{\mu\rho_{0}}\bm{B}_{0}\times[\bm{k}\times\{\bm{k}\times(\bm{v}_{1}\times\bm{B}_{0})\}]　\label{hadou3_A}
\end{equation}
を得る. 次にこの式の右辺第二項に対してベクトル解析の公式
\begin{equation}
  \bm{A}\times(\bm{B}\times\bm{C})=\bm{B}(\bm{A}\cdot\bm{C})-\bm{C}(\bm{A}\cdot\bm{B})
\end{equation}
を用いると,
\begin{subequations}
  \begin{align}
    \bm{B}_{0}\times[\bm{k}\times\{\bm{k}\times(\bm{v}_{1}\times\bm{B}_{0})\}]&=\bm{k}[\bm{B}_{0}\cdot\{\bm{k}\times(\bm{v}_{1}\times\bm{B}_{0})\}]-\{\bm{k}\times(\bm{v}_{1}\times\bm{B}_{0})\}(\bm{B}_{0}\cdot\bm{k}) \notag \\
    &=\bm{k}\{(\bm{B}_{0}\cdot\bm{v}_{1})(\bm{k}\cdot\bm{B}_{0})-\bm{B}^{2}_{0}(\bm{k}\cdot\bm{v}_{1})\} \notag \\
    &\,\,\,\,\,\,\,\,\,\,-\{\bm{v}_{1}(\bm{k}\cdot\bm{B}_{0})\}(\bm{B}_{0}\cdot\bm{k})+\{\bm{B}_{0}(\bm{k}\cdot\bm{v}_{1})\}(\bm{B}_{0}\cdot\bm{k}) \notag \\
    &=(\bm{B}_{0}\cdot\bm{k})(\bm{B}_{0}\cdot\bm{v}_{1})\bm{k}-\bm{B}^{2}_{0}\bm{k}(\bm{k}\cdot\bm{v}_{1}) \notag \\
    &\,\,\,\,\,\,\,\,\,\,-(\bm{B}_{0}\cdot\bm{k})^{2}\bm{v}_{1}+\bm{B}_{0}(\bm{B}_{0}\cdot\bm{k})(\bm{k}\cdot\bm{v}_{1}) \tag{A.5}
  \end{align}
\end{subequations}
となる. これを(\ref{hadou3_A})に代入すると,
\begin{subequations}
  \begin{align}
    \omega^{2}\bm{v}_{1}=a^{2}(\bm{k}\cdot\bm{v}_{1})&\bm{k}-\frac{1}{\mu\rho_{0}}(\bm{B}_{0}\cdot\bm{k})(\bm{B}_{0}\cdot\bm{v}_{1})\bm{k}+\frac{1}{\mu\rho_{0}}\bm{B}^{2}_{0}\bm{k}(\bm{k}\cdot\bm{v}_{1}) \notag \\
    &+\frac{1}{\mu\rho_{0}}(\bm{B}_{0}\cdot\bm{k})^{2}\bm{v}_{1}-\frac{1}{\mu\rho_{0}}\bm{B}_{0}(\bm{B}_{0}\cdot\bm{k})(\bm{k}\cdot\bm{v}_{1}) \tag{A.6}
  \end{align}
\end{subequations}
となり, これを整理すると,
\begin{subequations}
  \begin{align}
  \left\{\frac{1}{\mu\rho_{0}}(\bm{B}_{0}\cdot\bm{k})^{2}-\omega^{2}\right\}&\bm{v}_{1}+\left\{\left(\frac{\bm{B}^{2}_{0}}{\mu\rho_{0}}+a^{2}\right)\bm{k}-\frac{1}{\mu\rho_{0}}\bm{B}_{0}(\bm{B}_{0}\cdot\bm{k})\right\}(\bm{k}\cdot\bm{v}_{1}) \notag \\
  &-\frac{1}{\mu\rho_{0}}(\bm{B}_{0}\cdot\bm{k})(\bm{B}_{0}\cdot\bm{v}_{1})\bm{k}=0 \tag{A.7}
  \end{align}
\end{subequations}
が得られ, (\ref{hadou4})が導出された.



\section{(\ref{eq-of-motion2})の導出}

運動方程式(\ref{kiso3_motion})
\begin{equation}
  \rho_{0}\frac{\partial\bm{v}_{1}}{\partial t}+\nabla p_{1}+\frac{1}{\mu}\bm{B}_{0}\times(\nabla\times\bm{B}_{1})=0 \label{eq-of-motion-app}
\end{equation}
から始める. 左辺第三項について,
\begin{subequations}
  \begin{align}
    \bm{B}_{0}\times(\nabla\times\bm{B}_{1})&=\nabla(\bm{B}_{0}\cdot\bm{B}_{1})-(\bm{B}_{1}\cdot\nabla)\bm{B}_{0}-(\bm{B}_{0}\cdot\nabla)\bm{B}_{1}-\bm{B}_{1}\times(\nabla\times\bm{B}_{0}) \notag \\
    &=\nabla(\bm{B}_{0}\cdot\bm{B}_{1})-(\bm{B}_{0}\cdot\nabla)\bm{B}_{1} \tag{A.9}
  \end{align}
\end{subequations}
である. したがって(\ref{eq-of-motion-app})は,
\begin{subequations}
  \begin{align}
    \rho_{0}\frac{\partial\bm{v}_{1}}{\partial t}&=-\nabla p_{1}-\frac{1}{\mu}\nabla(\bm{B}_{0}\cdot\bm{B}_{1})+\frac{1}{\mu}(\bm{B}_{0}\cdot\nabla)\bm{B}_{1} \notag \\
    &=-\nabla\left\{p_{1}+\frac{1}{\mu}(\bm{B}_{0}\cdot\bm{B}_{1})\right\}+\frac{1}{\mu}(\bm{B}_{0}\cdot\nabla)\bm{B}_{1} \tag{A.10}
  \end{align}
\end{subequations}
と書き換えることができ, (\ref{eq-of-motion2})が導出できた.


\section{(\ref{hadou-eq1})の導出}

$\bm{B}_{0}$ の方向が $z$ 方向であるから,
\begin{equation}
  (\bm{B}_{0}\cdot\nabla)=B_{0}\frac{\partial}{\partial z}
\end{equation}
となる. そのため式(\ref{hadou1})と(\ref{hadou2})はそれぞれ
\begin{equation}
  \frac{\partial\bm{v}_{1}}{\partial t}=\frac{B_{0}}{\mu\rho_{0}}\frac{\partial\bm{B}_{1}}{\partial z} \label{hadou1_A}
\end{equation}
\begin{equation}
  \frac{\partial\bm{B}_{1}}{\partial t}={B}_{0}\frac{\partial\bm{v}_{1}}{\partial z} \label{hadou2_A}
\end{equation}
と書き換えられる. 次に(\ref{hadou1_A})の $z$ 微分をとると,
\begin{equation}
  \frac{\partial^{2}\bm{v}_{1}}{\partial z \partial t}=\frac{B_{0}}{\mu\rho_{0}}\frac{\partial^{2}\bm{B}_{1}}{\partial z^{2}} \label{hadou1-1_A}
\end{equation}
となり, (\ref{hadou2_A})の時間微分をとると,
\begin{equation}
  \frac{\partial^{2}\bm{B}_{1}}{\partial t^{2}}=B_{0}\frac{\partial^{2}\bm{v}_{1}}{\partial t \partial z} \label{hadou2-1_A}
\end{equation}
となる. したがって, (\ref{hadou1-1_A})と(\ref{hadou2-1_A})より, 波動方程式
\begin{equation}
  \frac{\partial^{2}\bm{B}_{1}}{\partial t^{2}}=\frac{B^{2}_{0}}{\mu\rho_{0}}\frac{\partial^{2}\bm{B}_{1}}{\partial z^{2}}
\end{equation}
が導出できた.




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