この記事では、デカルト座標における3次元ラプラシアン
\begin{equation}
\Delta_\mathrm{C} = \frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2}
\end{equation}
の極座標表示を導出します。
方針
デカルト座標系$(x,y,z)$と極座標系$(r,\theta,\phi)$は、次の関係式で結ばれます。
\begin{eqnarray}
x &=& r\sin\theta\cos\phi \\[10pt]
y &=& r\sin\theta\sin\phi \\[10pt]
z &=& r\cos\theta
\end{eqnarray}
$(x,y,z)$はそれぞれ$(r,\theta,\phi)$の関数なので、以下の微分の連鎖律が成り立ちます。
\begin{eqnarray}
\frac{\partial}{\partial x} &=& \frac{\partial r}{\partial x}\frac{\partial}{\partial r}+\frac{\partial \theta}{\partial x}\frac{\partial}{\partial\theta}+\frac{\partial \phi}{\partial x}\frac{\partial}{\partial\phi} \\[10pt]
\frac{\partial}{\partial y} &=& \frac{\partial r}{\partial y}\frac{\partial}{\partial r}+\frac{\partial \theta}{\partial y}\frac{\partial}{\partial\theta}+\frac{\partial \phi}{\partial y}\frac{\partial}{\partial\phi} \\[10pt]
\frac{\partial}{\partial z} &=& \frac{\partial r}{\partial z}\frac{\partial}{\partial r}+\frac{\partial \theta}{\partial z}\frac{\partial}{\partial\theta}+\frac{\partial \phi}{\partial z}\frac{\partial}{\partial\phi}
\end{eqnarray}
これら9つの係数を$r,\theta,\phi$で表すことができれば、$\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z}$の極座標表示が得られるので、それぞれを2乗して足し合わせることで3次元ラプラシアンの極座標表示を計算できます。
一階微分の導出
$\frac{\partial r}{\partial x},\frac{\partial r}{\partial y},\frac{\partial r}{\partial z}$の導出
$r$と$x,y,z$は、以下の関係式で結ばれます。
\begin{equation}
r^2 = x^2 + y^2 + z^2
\end{equation}
この両辺を$x$で偏微分すると
\begin{eqnarray}
2r\frac{\partial r}{\partial x} &=& 2x \nonumber\\[10pt]
\therefore \frac{\partial r}{\partial x} &=& \frac{x}{r} = \frac{r\sin\theta\cos\phi}{r} = \sin\theta\cos\phi
\end{eqnarray}
が得られます。$y,z$についても同様に
\begin{eqnarray}
2r\frac{\partial r}{\partial y} &=& 2y \nonumber\\[10pt]
\therefore \frac{\partial r}{\partial y} &=& \frac{y}{r} = \frac{r\sin\theta\sin\phi}{r} = \sin\theta\sin\phi
\end{eqnarray}
\begin{eqnarray}
2r\frac{\partial r}{\partial z} &=& 2z \nonumber\\[10pt]
\therefore \frac{\partial r}{\partial z} &=& \frac{z}{r} = \frac{r\cos\theta}{r} = \cos\theta \hspace{35pt}
\end{eqnarray}
となります。
$\frac{\partial\theta}{\partial x},\frac{\partial\theta}{\partial y},\frac{\partial\theta}{\partial z}$の導出
$\theta$と$x,y,z$は、以下の関係式で結ばれます。
\begin{equation}
\tan^2\theta = \frac{x^2+y^2}{z^2}
\end{equation}
この両辺を$x$で偏微分すると
\begin{gather}
2\tan\theta\frac{1}{\cos^2\theta}\frac{\partial\theta}{\partial x} = \frac{2x}{z^2} \nonumber\\[10pt]
\therefore \frac{\partial\theta}{\partial x} = \frac{x}{z^2}\frac{\cos^2\theta}{\tan\theta} = \frac{r\sin\theta\cos\phi\cos^2\theta}{r^2\cos^2\theta\tan\theta} = \frac{\cos\theta\cos\phi}{r}
\end{gather}
となります。$y$についても同様に
\begin{gather}
2\tan\theta\frac{1}{\cos^2\theta}\frac{\partial\theta}{\partial y} = \frac{2y}{z^2} \nonumber\\[10pt]
\therefore \frac{\partial\theta}{\partial y} = \frac{y}{z^2}\frac{\cos^2\theta}{\tan\theta} = \frac{r\sin\theta\sin\phi\cos^2\theta}{r^2\cos^2\theta\tan\theta} = \frac{\cos\theta\sin\phi}{r}
\end{gather}
となります。$z$については、両辺の$\ln$をとって$z$で偏微分すると
\begin{gather}
2\ln(\tan\theta) = \ln(x^2+y^2)-\ln(z^2) \nonumber\\[10pt]
2\frac{1}{\tan\theta}\frac{1}{\cos^2\theta}\frac{\partial\theta}{\partial z} = -\frac{2}{z} \nonumber\\[10pt]
\therefore \frac{\partial\theta}{\partial z} = -\frac{\tan\theta\cos^2\theta}{z} = -\frac{\tan\theta\cos^2\theta}{r\cos\theta} = -\frac{\sin\theta}{r}
\end{gather}
が得られます。
$\frac{\partial\phi}{\partial x},\frac{\partial\phi}{\partial y},\frac{\partial\phi}{\partial z}$の導出
$\phi$と$x,y,z$は、以下の関係式で結ばれます。
\begin{equation}
\tan\phi = \frac{y}{x}
\end{equation}
この両辺を$x$で偏微分すると
\begin{gather}
\frac{1}{\cos^2\phi}\frac{\partial\phi}{\partial x} = -\frac{y}{x^2} \nonumber\\[10pt]
\therefore \frac{\partial\phi}{\partial x} = -\frac{y\cos^2\phi}{x^2} = -\frac{r\sin\theta\sin\phi\cos^2\phi}{r^2\sin^2\theta\cos^2\phi} = -\frac{\sin\phi}{r\sin\theta}
\end{gather}
となります。$y$についても同様に
\begin{gather}
\frac{1}{\cos^2\phi}\frac{\partial\phi}{\partial y} = \frac{1}{x} \nonumber\\[10pt]
\therefore \frac{\partial\phi}{\partial y} = \frac{\cos^2\phi}{x} = \frac{\cos^2\phi}{r\sin\theta\cos\phi} = \frac{\cos\phi}{r\sin\theta} \hspace{55pt}
\end{gather}
となります。$z$については、$\phi$は$z$に依存しないため
\begin{equation}
\frac{\partial\phi}{\partial z} = 0
\end{equation}
です。
以上をまとめると、$\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z}$の極座標表示は以下のようになります。
\begin{eqnarray}
\frac{\partial}{\partial x} &=& \sin\theta\cos\phi\frac{\partial}{\partial r}+\frac{1}{r}\cos\theta\cos\phi\frac{\partial}{\partial\theta}-\frac{1}{r}\frac{\sin\phi}{\sin\theta}\frac{\partial}{\partial\phi} \\[10pt]
\frac{\partial}{\partial y} &=& \sin\theta\sin\phi\frac{\partial}{\partial r}+\frac{1}{r}\cos\theta\sin\phi\frac{\partial}{\partial\theta}+\frac{1}{r}\frac{\cos\phi}{\sin\theta}\frac{\partial}{\partial\phi} \\[10pt]
\frac{\partial}{\partial z} &=& \cos\theta\frac{\partial}{\partial r}-\frac{1}{r}\sin\theta\frac{\partial}{\partial\theta}
\end{eqnarray}
二階微分の導出
次に、得られた$\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z}$の2乗を計算していきます。
$\frac{\partial^2}{\partial x^2}$の導出
\begin{eqnarray}
\frac{\partial^2}{\partial x^2} &=& \bigg(\sin\theta\cos\phi\frac{\partial}{\partial r}+\frac{1}{r}\cos\theta\cos\phi\frac{\partial}{\partial\theta}-\frac{1}{r}\frac{\sin\phi}{\sin\theta}\frac{\partial}{\partial\phi}\bigg)^2 \nonumber\\
&=& \sin\theta\cos\phi\frac{\partial}{\partial r}\bigg(\sin\theta\cos\phi\frac{\partial}{\partial r}+\frac{1}{r}\cos\theta\cos\phi\frac{\partial}{\partial\theta}-\frac{1}{r}\frac{\sin\phi}{\sin\theta}\frac{\partial}{\partial\phi}\bigg) \nonumber\\
&&+ \frac{1}{r}\cos\theta\cos\phi\frac{\partial}{\partial\theta}\bigg(\sin\theta\cos\phi\frac{\partial}{\partial r}+\frac{1}{r}\cos\theta\cos\phi\frac{\partial}{\partial\theta}-\frac{1}{r}\frac{\sin\phi}{\sin\theta}\frac{\partial}{\partial\phi}\bigg) \nonumber\\
&&- \frac{1}{r}\frac{\sin\phi}{\sin\theta}\frac{\partial}{\partial\phi}\bigg(\sin\theta\cos\phi\frac{\partial}{\partial r}+\frac{1}{r}\cos\theta\cos\phi\frac{\partial}{\partial\theta}-\frac{1}{r}\frac{\sin\phi}{\sin\theta}\frac{\partial}{\partial\phi}\bigg)
\end{eqnarray}
\begin{eqnarray}
第1項の微分部分 && = \frac{\partial}{\partial r}\bigg(\sin\theta\cos\phi\frac{\partial}{\partial r}+\frac{1}{r}\cos\theta\cos\phi\frac{\partial}{\partial\theta}-\frac{1}{r}\frac{\sin\phi}{\sin\theta}\frac{\partial}{\partial\phi}\bigg) \nonumber\\
&& \quad = \sin\theta\cos\phi\frac{\partial^2}{\partial r^2} \nonumber\\
&& \qquad – \frac{1}{r^2}\cos\theta\cos\phi\frac{\partial}{\partial\theta}+\frac{1}{r}\cos\theta\cos\phi\frac{\partial^2}{\partial r\partial\theta} \nonumber\\
&& \qquad + \frac{1}{r^2}\frac{\sin\phi}{\sin\theta}\frac{\partial}{\partial\phi}-\frac{1}{r}\frac{\sin\phi}{\sin\theta}\frac{\partial^2}{\partial r\partial\phi}
\end{eqnarray}
\begin{eqnarray}
第2項の微分部分 &&= \frac{\partial}{\partial\theta}\bigg(\sin\theta\cos\phi\frac{\partial}{\partial r}+\frac{1}{r}\cos\theta\cos\phi\frac{\partial}{\partial\theta}-\frac{1}{r}\frac{\sin\phi}{\sin\theta}\frac{\partial}{\partial\phi}\bigg) \nonumber\\
&& \quad = \cos\theta\cos\phi\frac{\partial}{\partial r}+\sin\theta\cos\phi\frac{\partial^2}{\partial\theta\partial r} \nonumber\\
&& \qquad – \frac{1}{r}\sin\theta\cos\phi\frac{\partial}{\partial\theta}+\frac{1}{r}\cos\theta\cos\phi\frac{\partial^2}{\partial\theta^2} \nonumber\\
&& \qquad + \frac{1}{r}\frac{\cos\theta\sin\phi}{\sin^2\theta}\frac{\partial}{\partial\phi}-\frac{1}{r}\frac{\sin\phi}{\sin\theta}\frac{\partial^2}{\partial\theta\partial\phi}
\end{eqnarray}
\begin{eqnarray}
第3項の微分部分 &&= \frac{\partial}{\partial\phi}\bigg(\sin\theta\cos\phi\frac{\partial}{\partial r}+\frac{1}{r}\cos\theta\cos\phi\frac{\partial}{\partial\theta}-\frac{1}{r}\frac{\sin\phi}{\sin\theta}\frac{\partial}{\partial\phi}\bigg) \nonumber\\
&& \quad = -\sin\theta\sin\phi\frac{\partial}{\partial r}+\sin\theta\cos\phi\frac{\partial^2}{\partial\phi\partial r} \nonumber\\
&& \qquad – \frac{1}{r}\cos\theta\sin\phi\frac{\partial}{\partial\theta}+\frac{1}{r}\cos\theta\cos\phi\frac{\partial^2}{\partial\phi\partial\theta} \nonumber\\
&& \qquad – \frac{1}{r}\frac{\cos\phi}{\sin\theta}\frac{\partial}{\partial\phi}-\frac{1}{r}\frac{\sin\phi}{\sin\theta}\frac{\partial^2}{\partial\phi^2}
\end{eqnarray}
$\frac{\partial^2}{\partial y^2}$の導出
\begin{eqnarray}
\frac{\partial^2}{\partial y^2} &=& \bigg(\sin\theta\sin\phi\frac{\partial}{\partial r}+\frac{1}{r}\cos\theta\sin\phi\frac{\partial}{\partial\theta}+\frac{1}{r}\frac{\cos\phi}{\sin\theta}\frac{\partial}{\partial\phi}\bigg)^2 \nonumber\\
&=& \sin\theta\sin\phi\frac{\partial}{\partial r}\bigg(\sin\theta\sin\phi\frac{\partial}{\partial r}+\frac{1}{r}\cos\theta\sin\phi\frac{\partial}{\partial\theta}+\frac{1}{r}\frac{\cos\phi}{\sin\theta}\frac{\partial}{\partial\phi}\bigg) \nonumber\\
&& + \frac{1}{r}\cos\theta\sin\phi\frac{\partial}{\partial\theta}\bigg(\sin\theta\sin\phi\frac{\partial}{\partial r}+\frac{1}{r}\cos\theta\sin\phi\frac{\partial}{\partial\theta}+\frac{1}{r}\frac{\cos\phi}{\sin\theta}\frac{\partial}{\partial\phi}\bigg) \nonumber\\
&& + \frac{1}{r}\frac{\cos\phi}{\sin\theta}\frac{\partial}{\partial\phi}\bigg(\sin\theta\sin\phi\frac{\partial}{\partial r}+\frac{1}{r}\cos\theta\sin\phi\frac{\partial}{\partial\theta}+\frac{1}{r}\frac{\cos\phi}{\sin\theta}\frac{\partial}{\partial\phi}\bigg)
\end{eqnarray}
\begin{eqnarray}
第1項の微分部分 &&= \frac{\partial}{\partial r}\bigg(\sin\theta\sin\phi\frac{\partial}{\partial r}+\frac{1}{r}\cos\theta\sin\phi\frac{\partial}{\partial\theta}+\frac{1}{r}\frac{\cos\phi}{\sin\theta}\frac{\partial}{\partial\phi}\bigg) \nonumber\\
&& \quad = \sin\theta\sin\phi\frac{\partial^2}{\partial r^2} \nonumber\\
&& \qquad – \frac{1}{r^2}\cos\theta\sin\phi\frac{\partial}{\partial\theta}+\frac{1}{r}\cos\theta\sin\phi\frac{\partial^2}{\partial r\partial\theta} \nonumber\\
&& \qquad – \frac{1}{r^2}\frac{\cos\phi}{\sin\theta}\frac{\partial}{\partial\phi}+\frac{1}{r}\frac{\cos\phi}{\sin\theta}\frac{\partial^2}{\partial r\partial\phi}
\end{eqnarray}
\begin{eqnarray}
第2項の微分部分 &&= \frac{\partial}{\partial\theta}\bigg(\sin\theta\sin\phi\frac{\partial}{\partial r}+\frac{1}{r}\cos\theta\sin\phi\frac{\partial}{\partial\theta}+\frac{1}{r}\frac{\cos\phi}{\sin\theta}\frac{\partial}{\partial\phi}\bigg) \nonumber\\
&& \quad = \cos\theta\sin\phi\frac{\partial}{\partial r}+\sin\theta\sin\phi\frac{\partial^2}{\partial\theta\partial r} \nonumber\\
&& \qquad – \frac{1}{r}\sin\theta\sin\phi\frac{\partial}{\partial\theta}+\frac{1}{r}\cos\theta\sin\phi\frac{\partial^2}{\partial\theta^2} \nonumber\\
&& \qquad – \frac{1}{r}\frac{\cos\theta\cos\phi}{\sin^2\theta}\frac{\partial}{\partial\phi}+\frac{1}{r}\frac{\cos\phi}{\sin\theta}\frac{\partial^2}{\partial\theta\partial\phi}
\end{eqnarray}
\begin{eqnarray}
第3項の微分部分 &&= \frac{\partial}{\partial\phi}\bigg(\sin\theta\sin\phi\frac{\partial}{\partial r}+\frac{1}{r}\cos\theta\sin\phi\frac{\partial}{\partial\theta}+\frac{1}{r}\frac{\cos\phi}{\sin\theta}\frac{\partial}{\partial\phi}\bigg) \nonumber\\
&& \quad = \sin\theta\cos\phi\frac{\partial}{\partial r}+\sin\theta\sin\phi\frac{\partial^2}{\partial\phi\partial r} \nonumber\\
&& \qquad + \frac{1}{r}\cos\theta\cos\phi\frac{\partial}{\partial\theta}+\frac{1}{r}\cos\theta\sin\phi\frac{\partial^2}{\partial\phi\partial\theta} \nonumber\\
&& \qquad – \frac{1}{r}\frac{\sin\phi}{\sin\theta}\frac{\partial}{\partial\phi}+\frac{1}{r}\frac{\cos\phi}{\sin\theta}\frac{\partial^2}{\partial\phi^2}
\end{eqnarray}
$\frac{\partial^2}{\partial z^2}$の導出
\begin{eqnarray}
\frac{\partial^2}{\partial z^2} &=& \bigg(\cos\theta\frac{\partial}{\partial r}-\frac{1}{r}\sin\theta\frac{\partial}{\partial\theta}\bigg)^2 \nonumber\\
&=& \cos\theta\frac{\partial}{\partial r}\bigg(\cos\theta\frac{\partial}{\partial r}-\frac{1}{r}\sin\theta\frac{\partial}{\partial\theta}\bigg) \nonumber\\
&& – \frac{1}{r}\sin\theta\frac{\partial}{\partial\theta}\bigg(\cos\theta\frac{\partial}{\partial r}-\frac{1}{r}\sin\theta\frac{\partial}{\partial\theta}\bigg)
\end{eqnarray}
\begin{eqnarray}
第1項の微分部分 &&= \frac{\partial}{\partial r}\bigg(\cos\theta\frac{\partial}{\partial r}-\frac{1}{r}\sin\theta\frac{\partial}{\partial\theta}\bigg) \nonumber\\
&& \quad = \cos\theta\frac{\partial^2}{\partial r^2} \nonumber\\
&& \qquad + \frac{1}{r^2}\sin\theta\frac{\partial}{\partial\theta}-\frac{1}{r}\sin\theta\frac{\partial^2}{\partial r\partial\theta}
\end{eqnarray}
\begin{eqnarray}
第2項の微分部分 &&= \frac{\partial}{\partial\theta}\bigg(\cos\theta\frac{\partial}{\partial r}-\frac{1}{r}\sin\theta\frac{\partial}{\partial\theta}\bigg) \nonumber\\
&& \quad = -\sin\theta\frac{\partial}{\partial r}+\cos\theta\frac{\partial^2}{\partial\theta\partial r} \nonumber\\
&& \qquad – \frac{1}{r}\cos\theta\frac{\partial}{\partial\theta}-\frac{1}{r}\sin\theta\frac{\partial^2}{\partial\theta^2}
\end{eqnarray}
まとめ
最後に、ここまで計算した$\frac{\partial^2}{\partial x^2},\frac{\partial^2}{\partial y^2},\frac{\partial^2}{\partial z^2}$すべて足し上げます。
$\frac{\partial^2}{\partial r^2}$の係数
\begin{equation}
\sin^2\theta\cos^2\phi+\sin^2\theta\cos^2\phi+\cos^2\phi = 1
\end{equation}
$\frac{\partial}{\partial r}$の係数
\begin{equation}
\frac{1}{r}\big(\cos^2\theta\cos^2\phi+\sin^2\phi+\cos^2\theta\sin^2\phi+\cos^2\phi+\sin^2\theta\big) = \frac{2}{r}
\end{equation}
$\frac{\partial^2}{\partial\theta^2}$の係数
\begin{equation}
\frac{1}{r^2}\big(\cos^2\theta\cos^2\phi+\cos^2\theta\sin^2\phi+\sin^2\theta\big) = \frac{1}{r^2}
\end{equation}
$\frac{\partial}{\partial\theta}$の係数
\begin{eqnarray}
& & \frac{1}{r^2}\bigg(-2\sin\theta\cos\theta\cos^2\phi+\frac{\cos\theta\sin^2\phi}{\sin\theta} \nonumber\\
& & -2\sin\theta\cos\theta\sin^2\phi+\frac{\cos\theta\cos^2\phi}{\sin\theta}+2\sin\theta\cos\theta\bigg) \nonumber\\
& & = \frac{1}{r^2}\frac{\cos\theta}{\sin\theta}
\end{eqnarray}
$\frac{\partial^2}{\partial\phi^2}$の係数
\begin{equation}
\frac{1}{r^2}\bigg(\frac{\sin^2\phi}{\sin^2\theta}+\frac{\cos^2\phi}{\sin^2\theta}\bigg) = \frac{1}{r^2}\frac{1}{\sin^2\theta}
\end{equation}
$\frac{\partial}{\partial\phi}$の係数
\begin{eqnarray}
& & \frac{1}{r^2}\bigg(\sin\phi\cos\phi+\frac{\cos^2\theta\sin\phi\cos\phi}{\sin^2\theta}+\frac{\sin\phi\cos\phi}{\sin^2\theta} \nonumber\\
& & -\sin\phi\cos\phi-\frac{\cos^2\theta\sin\phi\cos\phi}{\sin^2\theta}-\frac{\sin\phi\cos\phi}{\sin^2\theta}\bigg) \nonumber\\
& & = 0
\end{eqnarray}
$\frac{\partial^2}{\partial r\partial\theta}$の係数
\begin{equation}
\frac{1}{r}\big(2\sin\theta\cos\theta\cos^2\phi+2\sin\theta\cos\theta\sin^2\phi-2\sin\theta\cos\theta\big) = 0
\end{equation}
$\frac{\partial^2}{\partial\theta\partial\phi}$の係数
\begin{equation}
\frac{1}{r^2}\bigg(-\frac{2\cos\theta\sin\phi\cos\phi}{\sin\theta}+\frac{2\cos\theta\sin\phi\cos\phi}{\sin\theta}\bigg) = 0
\end{equation}
$\frac{\partial^2}{\partial\phi\partial r}$の係数
\begin{equation}
\frac{1}{r}\big(-2\sin\phi\cos\phi+2\sin\phi\cos\phi\big) = 0
\end{equation}
以上をまとめると、3次元ラプラシアンの極座標表示は以下のようになります。
\begin{eqnarray}
\Delta_\mathrm{P} &=& \frac{\partial^2}{\partial r^2}+\frac{2}{r}\frac{\partial}{\partial r}+\frac{1}{r^2}\bigg(\frac{\partial^2}{\partial\theta^2}+\frac{\cos\theta}{\sin\theta}\frac{\partial}{\partial\theta}+\frac{1}{\sin^2\theta}\frac{\partial^2}{\partial\phi^2}\bigg) \nonumber\\
&=& \frac{1}{r^2}\frac{\partial}{\partial r}\bigg(r^2\frac{\partial}{\partial r}\bigg)+\frac{1}{r^2}\bigg(\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\bigg(\sin\theta\frac{\partial}{\partial\theta}\bigg)+\frac{1}{\sin^2\theta}\frac{\partial^2}{\partial\phi^2}\bigg)
\end{eqnarray}