本稿では、
\begin{eqnarray}
x &=& r \sin \theta \cos \phi \\
y &=& r \sin \theta \sin \phi \\
z &=& r \cos \theta \tag{1}
\end{eqnarray}で表される3次元の球座標系$(r, \theta, \phi)$の回転について述べます。
ここでは、ベクトルの外積のように導きます。
ベクトルの発散 - 数式で独楽する
極座標 - 数式で独楽する
球座標系のナブラとベクトルはそれぞれ
\begin{eqnarray}
\nabla &=& \boldsymbol{e}_r \, \frac{\partial}{\partial r} + \boldsymbol{e}_\theta \, \frac{1}{r} \frac{\partial}{\partial \theta} + \boldsymbol{e}_\phi \, \frac{1}{r \sin \theta} \frac{\partial}{\partial \phi} \tag{2} \\
\boldsymbol{A} &=& A_r \,\boldsymbol{e}_r + A_\theta \, \boldsymbol{e}_\theta + A_\phi \, \boldsymbol{e}_\phi \tag{3}
\end{eqnarray}で表します。
これより、直接、外積を求めるようにしていきます。ただし、単位ベクトルの偏微分に注意して計算していきます。
3次元球座標系の単位ベクトルの偏微分 - 数式で独楽する
\begin{eqnarray}
\nabla \times \boldsymbol{A}&=& \left( \boldsymbol{e}_r \, \frac{\partial}{\partial r} + \boldsymbol{e}_\theta \, \frac{1}{r} \frac{\partial}{\partial \theta} + \boldsymbol{e}_\phi \, \frac{1}{r \sin \theta} \frac{\partial}{\partial \phi} \right) \times (A_r \, \boldsymbol{e}_r + A_\theta \, \boldsymbol{e}_\theta + A_\phi \, \boldsymbol{e}_\phi \\
&=& \left( \frac{1}{r} \frac{\partial A_\phi}{\partial \theta} + \frac{\cos \theta \, A_\phi}{r \sin \theta} - \frac{1}{r \sin \theta} \frac{\partial A_\theta}{\partial \phi} \right) \, \boldsymbol{e}_r \\
&& + \left( \frac{1}{r \sin \theta} \frac{\partial A_r}{\partial \phi} - \frac{\partial A_\theta}{\partial r} - \frac{A_r}{r} \right) \, \boldsymbol{e}_\theta + \left( \frac{\partial A_\theta}{\partial r} + \frac{A_\theta}{r} - \frac{1}{r} \frac{\partial A_r}{\partial \theta} \right) \, \boldsymbol{e}_\phi
\end{eqnarray}整理すると、
\begin{eqnarray}
\nabla \times \boldsymbol{A} &=& \left( \frac{1}{r \sin \theta} \frac{\partial}{\partial \theta} (\sin \theta \, A_\phi) - \frac{1}{r \sin \theta} \frac{\partial A_\theta}{\partial \phi} \right) \, \boldsymbol{e}_r \\
&& + \left( \frac{1}{r \sin \theta} \frac{\partial A_r}{\partial \phi} - \frac{1}{r} \frac{\partial}{\partial r} (r A_\phi) \right) \, \boldsymbol{e}_\theta + \left( \frac{\partial}{\partial r} (r A_\theta) - \frac{1}{r} \frac{\partial A_r}{\partial \theta} \right) \, \boldsymbol{e}_\phi
\end{eqnarray}を得ます。
省略した計算は次の通りです。
\begin{eqnarray}
\boldsymbol{e}_r \times \frac{\partial}{\partial r} (A_r \, \boldsymbol{e}_r) &=& \boldsymbol{e}_r \times \frac{\partial A_r}{\partial r} \, \boldsymbol{e}_r + \boldsymbol{e}_r \times A_r \, \frac{\partial \boldsymbol{e}_r}{\partial r} = 0 \\
\boldsymbol{e}_r \times \frac{\partial}{\partial r} (A_\theta \, \boldsymbol{e}_\theta) &=& \boldsymbol{e}_r \times \frac{\partial A_\theta}{\partial r} \, \boldsymbol{e}_\theta + \boldsymbol{e}_r \times A_\theta \, \frac{\partial \boldsymbol{e}_\theta}{\partial r} = \frac{\partial A_\theta}{\partial r} \, \boldsymbol{e}_\phi \\
\boldsymbol{e}_r \times \frac{\partial}{\partial r} (A_\phi \, \boldsymbol{e}_\phi) &=& \boldsymbol{e}_r \times \frac{\partial A_\phi}{\partial r} \, \boldsymbol{e}_\phi + \boldsymbol{e}_r \times A_\phi \, \frac{\partial \boldsymbol{e}_\phi}{\partial r} = - \frac{\partial A_\phi}{\partial r} \, \boldsymbol{e}_\theta \\
\boldsymbol{e}_\theta \times \frac{1}{r} \frac{\partial}{\partial \theta} (A_r \, \boldsymbol{e}_r) &=& \boldsymbol{e}_\theta \times \frac{1}{r} \frac{\partial A_r}{\partial \theta} \, \boldsymbol{e}_r + \boldsymbol{e}_\theta \times \frac{1}{r} \, A_r \, \frac{\partial \boldsymbol{e}_r}{\partial \theta} = - \frac{1}{r} \frac{\partial A_r}{\partial \theta} \, \boldsymbol{e}_\phi \\
\boldsymbol{e}_\theta \times \frac{1}{r} \frac{\partial}{\partial \theta} (A_\theta \, \boldsymbol{e}_\theta) &=& \boldsymbol{e}_\theta \times \frac{1}{r} \frac{\partial A_\theta}{\partial \theta} \, \boldsymbol{e}_\theta + \boldsymbol{e}_\theta \times \frac{1}{r} \, A_\theta \, \frac{\partial \boldsymbol{e}_\theta}{\partial \theta} = \frac{A_\theta}{r} \, \boldsymbol{e}_\phi \\
\boldsymbol{e}_\theta \times \frac{1}{r} \frac{\partial}{\partial \theta} (A_\phi \, \boldsymbol{e}_\phi) &=& \boldsymbol{e}_\theta \times \frac{1}{r} \frac{\partial A_\phi}{\partial \theta} \, \boldsymbol{e}_\phi + \boldsymbol{e}_\theta \times \frac{1}{r} \, A_\phi \, \frac{\partial \boldsymbol{e}_\phi}{\partial \theta} = \frac{1}{r} \frac{\partial A_\phi}{\partial \theta} \, \boldsymbol{e}_r \\
\boldsymbol{e}_\phi \times \frac{1}{r \sin \theta} \frac{\partial}{\partial \phi} (A_r \, \boldsymbol{e}_r) &=& \boldsymbol{e}_\phi \times \frac{1}{r \sin \theta} \frac{\partial A_r}{\partial \phi} \, \boldsymbol{e}_r + \boldsymbol{e}_\phi \times \frac{1}{r \sin \theta} \, A_r \, \frac{\partial \boldsymbol{e}_r}{\partial \phi} = \frac{1}{r \sin \theta} \frac{\partial A_r}{\partial \phi} \, \boldsymbol{e}_\theta \\
\boldsymbol{e}_\phi \times \frac{1}{r \sin \theta} \frac{\partial}{\partial \phi} (A_\theta \, \boldsymbol{e}_\theta) &=& \boldsymbol{e}_\phi \times \frac{1}{r \sin \theta} \frac{\partial A_\theta}{\partial \phi} \, \boldsymbol{e}_\theta + \boldsymbol{e}_\phi \times \frac{1}{r \sin \theta} \, A_\theta \, \frac{\partial \boldsymbol{e}_\theta}{\partial \phi} = - \frac{1}{r \sin \theta} \frac{\partial A_\theta}{\partial \phi} \, \boldsymbol{e}_r \\
\boldsymbol{e}_\phi \times \frac{1}{r \sin \theta} \frac{\partial}{\partial \phi} (A_\phi \, \boldsymbol{e}_\phi) &=& \boldsymbol{e}_\phi \times \frac{1}{r \sin \theta} \frac{\partial A_\phi}{\partial \phi} \, \boldsymbol{e}_\phi + \boldsymbol{e}_\phi \times \frac{1}{r \sin \theta} \, A_\phi \, \frac{\partial \boldsymbol{e}_\phi}{\partial \phi} \\
&=& - \frac{A_\phi}{r} \, \boldsymbol{e}_\theta + \frac{\cos \theta \, A_\phi}{r \sin \theta} \boldsymbol{e}_r
\end{eqnarray}
