【微積分】ラプラシアンの極座標表示

微積分

 下記記事にて、直交座標系から極座標系への座標変換した際のヤコビ行列を導出し、座標変換後の偏微分の表式を求めた。

 今回はその結果を利用して、ラプラシアンの極座標表示を導出する。

 かなり骨が折れる計算量になるが、詳細計算も示すので不明な場合は参照しつつ、自分で手を動かして計算してみてほしい。


二次元系の場合

 二次元系の場合、ラプラシアンは

\begin{align}
\triangle =\frac{\partial^{2}}{\partial x^{2}}+\frac{\partial^{2}}{\partial y^{2}} \tag{1}\label{2lap}
\end{align}

で与えられる。

 また二次元直交座標系\((x,y)\)から二次元極座標系\((r,\theta)\)に座標変換、すなわち

\begin{align}
\begin{cases}
x=r\cos\theta \\
y=r\sin\theta
\end{cases} \tag{2}\label{2xyzrtz-1-2}
\end{align}

と座標変換したとき、変数変換後の変数による偏微分はそれぞれ

\begin{align}
&\frac{\partial}{\partial r}=\cos\theta\frac{\partial}{\partial x}+\sin\theta\frac{\partial}{\partial y} \tag{3}\label{2pr}\\
&\frac{\partial}{\partial \theta}=-r\sin\theta\frac{\partial}{\partial x}+r\cos\theta\frac{\partial}{\partial y} \tag{4}\label{2pt}
\end{align}

となる。

 

 よって(\ref{2pr})と(\ref{2pt})を\(\displaystyle{\frac{\partial}{\partial x},\frac{\partial}{\partial y}}\)の連立方程式として解き、その結果を2乗して(\ref{2lap})に代入すれば良い。

 実際に計算すると

\begin{align}
&\frac{\partial}{\partial x}=\cos\theta\frac{\partial}{\partial r}-\frac{\sin\theta}{r}\frac{\partial}{\partial \theta} \tag{5}\label{2px}\\
&\frac{\partial}{\partial y}=\sin\theta\frac{\partial}{\partial r}+\frac{\cos\theta}{r}\frac{\partial}{\partial \theta} \tag{6}\label{2py}
\end{align}

 (\ref{2pt})の両辺を\(r\)で割り、右辺の形を(\ref{2pr})に揃える。

\begin{align}
&\frac{\partial}{\partial r}=\cos\theta\frac{\partial}{\partial x}+\sin\theta\frac{\partial}{\partial y} \tag{a1}\label{2pr-2}\\
&\frac{1}{r}\frac{\partial}{\partial \theta}=-\sin\theta\frac{\partial}{\partial x}+\cos\theta\frac{\partial}{\partial y} \tag{a2}\label{2pt-2}
\end{align}

 まず\(\displaystyle{\frac{\partial}{\partial x}}\)を求める。(\ref{2pr-2})の両辺に\(\cos\theta\)、(\ref{2pt-2})の両辺に\(\sin\theta\)を掛けて

\begin{align}
&\cos\theta\frac{\partial}{\partial r}=\cos^{2}\theta\frac{\partial}{\partial x}+\sin\theta\cos\theta\frac{\partial}{\partial y} \tag{a3}\label{2pr-3}\\
&\frac{\sin\theta}{r}\frac{\partial}{\partial \theta}=-\sin^{2}\theta\frac{\partial}{\partial x}+\sin\theta\cos\theta\frac{\partial}{\partial y} \tag{a4}\label{2pt-3}
\end{align}

とし、(\ref{2pr-3})\(-\)(\ref{2pt-3})を計算して整理すると

\begin{align}
\cos\theta\frac{\partial}{\partial r}-\frac{\sin\theta}{r}\frac{\partial}{\partial \theta}&=\cos^{2}\theta\frac{\partial}{\partial x}+\sin\theta\cos\theta\frac{\partial}{\partial y}-\left(-\sin^{2}\theta\frac{\partial}{\partial x}+\sin\theta\cos\theta\frac{\partial}{\partial y}\right)\notag \\
&=(\cos^{2}\theta+\sin^{2}\theta)\frac{\partial}{\partial x}=\frac{\partial}{\partial x}
\end{align}

となり、(\ref{2px})が求められる。

 続いて\(\displaystyle{\frac{\partial}{\partial y}}\)を求める。(\ref{2pr-2})の両辺に\(\sin\theta\)、(\ref{2pt-2})の両辺に\(\cos\theta\)を掛けて

\begin{align}
&\sin\theta\frac{\partial}{\partial r}=\sin\theta\cos\theta\frac{\partial}{\partial x}+\sin^{2}\theta\frac{\partial}{\partial y} \tag{a5}\label{2pr-4}\\
&\frac{\cos\theta}{r}\frac{\partial}{\partial \theta}=-\sin\theta\cos\theta\frac{\partial}{\partial x}+\cos^{2}\theta\frac{\partial}{\partial y} \tag{a6}\label{2pt-4}
\end{align}

とし、両辺を足し上げて整理すると

\begin{align}
\sin\theta\frac{\partial}{\partial r}+\frac{\cos\theta}{r}\frac{\partial}{\partial \theta}&=\sin\theta\cos\theta\frac{\partial}{\partial x}+\sin^{2}\theta\frac{\partial}{\partial y}-\sin\theta\cos\theta\frac{\partial}{\partial x}+\cos^{2}\theta\frac{\partial}{\partial y} \notag \\
&=(\cos^{2}\theta+\sin^{2}\theta)\frac{\partial}{\partial y}=\frac{\partial}{\partial y}
\end{align}

となり、(\ref{2py})が求められる。

となるため、これを2乗して整理すると

\begin{align}
\frac{\partial^{2}}{\partial x^{2}}&=\cos^{2}\theta\frac{\partial^{2}}{\partial r^{2}}+\frac{\sin\theta\cos\theta}{r}\left(\frac{1}{r}\frac{\partial}{\partial \theta}-\frac{\partial}{\partial r}\frac{\partial}{\partial \theta}\right) \notag \\
&\quad+\frac{\sin\theta}{r}\left(\sin\theta\frac{\partial}{\partial r}-\cos\theta\frac{\partial}{\partial \theta}\frac{\partial}{\partial r}\right)+\frac{\sin\theta}{r^{2}}\left(\cos\theta\frac{\partial}{\partial \theta}+\sin\theta\frac{\partial^{2}}{\partial \theta^{2}}\right) \tag{7}\label{2px2}\\
\frac{\partial^{2}}{\partial y^{2}}&=\sin^{2}\theta\frac{\partial^{2}}{\partial r^{2}}-\frac{\sin\theta\cos\theta}{r}\left(\frac{1}{r}\frac{\partial}{\partial \theta}-\frac{\partial}{\partial r}\frac{\partial}{\partial \theta}\right) \notag \\
&\quad+\frac{\cos\theta}{r}\left(\cos\theta\frac{\partial}{\partial r}+\sin\theta\frac{\partial}{\partial \theta}\frac{\partial}{\partial r}\right)-\frac{\cos\theta}{r^{2}}\left(\sin\theta\frac{\partial}{\partial \theta}-\cos\theta\frac{\partial^{2}}{\partial \theta^{2}}\right)\tag{8}\label{2py2}
\end{align}

 まず\(\displaystyle{\frac{\partial^{2}}{\partial x^{2}}}\)は(\ref{2px})の両辺を2乗して整理すれば

\begin{align}
\frac{\partial^{2}}{\partial x^{2}}&=\left(\cos\theta\frac{\partial}{\partial r}-\frac{\sin\theta}{r}\frac{\partial}{\partial \theta}\right)\left(\cos\theta\frac{\partial}{\partial r}-\frac{\sin\theta}{r}\frac{\partial}{\partial \theta}\right) \notag \\
&=\cos\theta\frac{\partial}{\partial r}\left(\cos\theta\frac{\partial}{\partial r}\right)-\cos\theta\frac{\partial}{\partial r}\left(\frac{\sin\theta}{r}\frac{\partial}{\partial \theta}\right)\notag \\
&\quad-\frac{\sin\theta}{r}\frac{\partial}{\partial \theta}\left(\cos\theta\frac{\partial}{\partial r}\right)+\frac{\sin\theta}{r}\frac{\partial}{\partial \theta}\left(\frac{\sin\theta}{r}\frac{\partial}{\partial \theta}\right) \notag \\
&=\cos^{2}\theta\frac{\partial^{2}}{\partial r^{2}}-\cos\theta\left(-\frac{\sin\theta}{r^{2}}\frac{\partial}{\partial \theta}+\frac{\sin\theta}{r}\frac{\partial}{\partial r}\frac{\partial}{\partial \theta}\right) \notag \\
&\quad-\frac{\sin\theta}{r}\frac{\partial}{\partial \theta}\left(-\sin\theta\frac{\partial}{\partial r}+\cos\theta\frac{\partial}{\partial \theta}\frac{\partial}{\partial r}\right)+\frac{\sin\theta}{r^{2}}\left(\cos\theta\frac{\partial}{\partial \theta}+\sin\theta\frac{\partial^{2}}{\partial \theta^{2}}\right) \notag \\
&=\cos^{2}\theta\frac{\partial^{2}}{\partial r^{2}}+\frac{\sin\theta\cos\theta}{r}\left(\frac{1}{r}\frac{\partial}{\partial \theta}-\frac{\partial}{\partial r}\frac{\partial}{\partial \theta}\right) \notag \\
&\quad+\frac{\sin\theta}{r}\left(\sin\theta\frac{\partial}{\partial r}-\cos\theta\frac{\partial}{\partial \theta}\frac{\partial}{\partial r}\right)+\frac{\sin\theta}{r^{2}}\left(\cos\theta\frac{\partial}{\partial \theta}+\sin\theta\frac{\partial^{2}}{\partial \theta^{2}}\right)
\end{align}

となり(\ref{2px2})が求められる。

 続いて\(\displaystyle{\frac{\partial^{2}}{\partial y^{2}}}\)は(\ref{2py})の両辺を2乗して整理すれば

\begin{align}
\frac{\partial^{2}}{\partial y^{2}}&=\left(\sin\theta\frac{\partial}{\partial r}+\frac{\cos\theta}{r}\frac{\partial}{\partial \theta}\right)\left(\sin\theta\frac{\partial}{\partial r}+\frac{\cos\theta}{r}\frac{\partial}{\partial \theta}\right) \notag \\
&=\sin\theta\frac{\partial}{\partial r}\left(\sin\theta\frac{\partial}{\partial r}\right)+\sin\theta\frac{\partial}{\partial r}\left(\frac{\cos\theta}{r}\frac{\partial}{\partial \theta}\right)\notag \\
&\quad+\frac{\cos\theta}{r}\frac{\partial}{\partial \theta}\left(\sin\theta\frac{\partial}{\partial r}\right)+\frac{\cos\theta}{r}\frac{\partial}{\partial \theta}\left(\frac{\cos\theta}{r}\frac{\partial}{\partial \theta}\right) \notag \\
&=\sin^{2}\theta\frac{\partial^{2}}{\partial r^{2}}+\sin\theta\left(-\frac{\cos\theta}{r^{2}}\frac{\partial}{\partial \theta}+\frac{\cos\theta}{r}\frac{\partial}{\partial r}\frac{\partial}{\partial \theta}\right) \notag \\
&\quad+\frac{\cos\theta}{r}\left(\cos\theta\frac{\partial}{\partial r}+\sin\theta\frac{\partial}{\partial \theta}\frac{\partial}{\partial r}\right)+\frac{\cos\theta}{r^{2}}\left(-\sin\theta\frac{\partial}{\partial \theta}+\cos\theta\frac{\partial^{2}}{\partial \theta^{2}}\right) \notag \\
&=\sin^{2}\theta\frac{\partial^{2}}{\partial r^{2}}-\frac{\sin\theta\cos\theta}{r}\left(\frac{1}{r}\frac{\partial}{\partial \theta}-\frac{\partial}{\partial r}\frac{\partial}{\partial \theta}\right) \notag \\
&\quad+\frac{\cos\theta}{r}\left(\cos\theta\frac{\partial}{\partial r}+\sin\theta\frac{\partial}{\partial \theta}\frac{\partial}{\partial r}\right)-\frac{\cos\theta}{r^{2}}\left(\sin\theta\frac{\partial}{\partial \theta}-\cos\theta\frac{\partial^{2}}{\partial \theta^{2}}\right)
\end{align}

となり(\ref{2py2})が求められる。

となる。

 よって(\ref{2px2})と(\ref{2py2})を(\ref{2lap})に代入して整理すると

\begin{align}
\boxed{\triangle=\frac{\partial^{2}}{\partial r^{2}}+\frac{1}{r}\frac{\partial}{\partial r}+\frac{1}{r^{2}}\frac{\partial^{2}}{\partial \theta^{2}}}
\end{align}

 (\ref{2px2})と(\ref{2py2})の右辺の第2項同士が相殺されるため、残りの項を足し上げればよい。

\begin{align}
\triangle&=\frac{\partial^{2}}{\partial x^{2}}+\frac{\partial^{2}}{\partial y^{2}} \notag \\
&=\cos^{2}\theta\frac{\partial^{2}}{\partial r^{2}}+\frac{\sin\theta}{r}\left(\sin\theta\frac{\partial}{\partial r}-\cos\theta\frac{\partial}{\partial \theta}\frac{\partial}{\partial r}\right)+\frac{\sin\theta}{r^{2}}\left(\cos\theta\frac{\partial}{\partial \theta}+\sin\theta\frac{\partial^{2}}{\partial \theta^{2}}\right) \notag \\
&+\sin^{2}\theta\frac{\partial^{2}}{\partial r^{2}}+\frac{\cos\theta}{r}\left(\cos\theta\frac{\partial}{\partial r}+\sin\theta\frac{\partial}{\partial \theta}\frac{\partial}{\partial r}\right)-\frac{\cos\theta}{r^{2}}\left(\sin\theta\frac{\partial}{\partial \theta}-\cos\theta\frac{\partial^{2}}{\partial \theta^{2}}\right) \notag \\
&=(\sin^{2}\theta+\cos^{2}\theta)\frac{\partial^{2}}{\partial r^{2}}+\frac{\sin^{2}\theta+\cos^{2}\theta}{r}\frac{\partial}{\partial r}+\frac{\sin^{2}\theta+\cos^{2}\theta}{r^{2}}\frac{\partial^{2}}{\partial \theta^{2}} \notag \\
&=\frac{\partial^{2}}{\partial r^{2}}+\frac{1}{r}\frac{\partial}{\partial r}+\frac{1}{r^{2}}\frac{\partial^{2}}{\partial \theta^{2}}
\end{align}

となる。

三次元系の場合

 三次元系の場合、ラプラシアンは

\begin{align}
\triangle =\frac{\partial^{2}}{\partial x^{2}}+\frac{\partial^{2}}{\partial y^{2}}+\frac{\partial^{2}}{\partial z^{2}} \tag{9}\label{3lap}
\end{align}

で与えられる。

 また三次元直交座標系\((x,y,z)\)から三次元極座標系\((r,\theta,\phi)\)に座標変換、すなわち

\begin{align}
\begin{cases}
x=r\sin\theta\cos\phi \\
y=r\sin\theta\sin\phi \\
z=r\cos\theta
\end{cases} \tag{10}\label{3xyzrtp-1-2}
\end{align}

と座標変換したとき、変数変換後の変数による偏微分はそれぞれ

\begin{align}
&\frac{\partial}{\partial r}=\sin\theta\cos\phi\frac{\partial}{\partial x}+\sin\theta\sin\phi\frac{\partial}{\partial y}+\cos\theta\frac{\partial}{\partial z} \tag{11}\label{3pr}\\
&\frac{\partial}{\partial \theta}=r\cos\theta\cos\phi\frac{\partial}{\partial x}+r\cos\theta\sin\phi\frac{\partial}{\partial y}-r\sin\theta\frac{\partial}{\partial z} \tag{12}\label{3pt}\\
&\frac{\partial}{\partial \phi}=-r\sin\theta\sin\phi\frac{\partial}{\partial x}+r\sin\theta\cos\phi\frac{\partial}{\partial y}\tag{13}\label{3pp}
\end{align}

となる。

 

 よって(\ref{3pr})~(\ref{3pp})を\(\displaystyle{\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z}}\)の連立方程式として解き、その結果を2乗して(\ref{3lap})に代入すれば良い。

 実際に計算すると

\begin{align}
&\frac{\partial}{\partial x}=\sin\theta\cos\phi\frac{\partial}{\partial r}+\frac{\cos\theta\cos\phi}{r}\frac{\partial}{\partial \theta}-\frac{\sin\phi}{r\sin\theta}\frac{\partial}{\partial \phi} \tag{14}\label{3px}\\
&\frac{\partial}{\partial y}=\sin\theta\sin\phi\frac{\partial}{\partial r}+\frac{\cos\theta\sin\phi}{r}\frac{\partial}{\partial \theta}+\frac{\cos\phi}{r\sin\theta}\frac{\partial}{\partial \phi} \tag{15}\label{3py}\\
&\frac{\partial}{\partial z}=\cos\theta\frac{\partial}{\partial r}-\frac{\sin\theta}{r}\frac{\partial}{\partial \theta} \tag{16}\label{3pz}
\end{align}

 (\ref{3pt})と(\ref{3pp})の両辺を\(r\)で割り、右辺の形を(\ref{3pr})に揃える。

\begin{align}
&\frac{\partial}{\partial r}=\sin\theta\cos\phi\frac{\partial}{\partial x}+\sin\theta\sin\phi\frac{\partial}{\partial y}+\cos\theta\frac{\partial}{\partial z} \tag{b1}\label{3pr-2}\\
&\frac{1}{r}\frac{\partial}{\partial \theta}=\cos\theta\cos\phi\frac{\partial}{\partial x}+\cos\theta\sin\phi\frac{\partial}{\partial y}-\sin\theta\frac{\partial}{\partial z} \tag{b2}\label{3pt-2}\\
&\frac{1}{r}\frac{\partial}{\partial \phi}=-\sin\theta\sin\phi\frac{\partial}{\partial x}+\sin\theta\cos\phi\frac{\partial}{\partial y}\tag{b3}\label{3pp-2}
\end{align}

 さらに(\ref{3pr-2})の両辺に\(\sin\theta\)、(\ref{3pt-2})の両辺に\(\cos\theta\)を掛けて

\begin{align}
&\sin\theta\frac{\partial}{\partial r}=\sin^{2}\theta\cos\phi\frac{\partial}{\partial x}+\sin^{2}\theta\sin\phi\frac{\partial}{\partial y}+\sin\theta\cos\theta\frac{\partial}{\partial z} \tag{b4}\label{3pr-3}\\
&\frac{\cos\theta}{r}\frac{\partial}{\partial \theta}=\cos^{2}\theta\cos\phi\frac{\partial}{\partial x}+\cos^{2}\theta\sin\phi\frac{\partial}{\partial y}-\sin\theta\cos\theta\frac{\partial}{\partial z} \tag{b5}\label{3pt-3}
\end{align}

とし、両辺を足し上げて整理すると

\begin{align}
&\sin\theta\frac{\partial}{\partial r}+\frac{\cos\theta}{r}\frac{\partial}{\partial \theta}\notag \\
=&(\sin^{2}\theta+\cos^{2}\theta)\cos\phi\frac{\partial}{\partial x}+(\sin^{2}\theta+\cos^{2}\theta)\sin\phi\frac{\partial}{\partial y} \notag \\
=&\cos\phi\frac{\partial}{\partial x}+\sin\phi\frac{\partial}{\partial y} \tag{b6}\label{rpt}
\end{align}

となるため、(\ref{rpt})の最左辺と最右辺の両辺に\(\sin\theta\)を掛ければ

\begin{align}
\sin\theta\left(\sin\theta\frac{\partial}{\partial r}+\frac{\cos\theta}{r}\frac{\partial}{\partial \theta}\right)=\sin\theta\cos\phi\frac{\partial}{\partial x}+\sin\theta\sin\phi\frac{\partial}{\partial y} \tag{b7}\label{rpt2}
\end{align}

となり、(\ref{3pp-2})と(\ref{rpt2})で\(\displaystyle{\frac{\partial}{\partial x},\frac{\partial}{\partial y}}\)の連立方程式になる。

 まず\(\displaystyle{\frac{\partial}{\partial x}}\)を求める。(\ref{3pp-2})の両辺に\(\sin\phi\)、(\ref{rpt2})の両辺に\(\cos\phi\)を掛けて

\begin{align}
&\frac{\sin\phi}{r}\frac{\partial}{\partial \phi}=-\sin\theta\sin^{2}\phi\frac{\partial}{\partial x}+\sin\theta\sin\phi\cos\phi\frac{\partial}{\partial y}\tag{b8}\label{rtpxy1}\\
&\sin\theta\cos\phi\left(\sin\theta\frac{\partial}{\partial r}+\frac{\cos\theta}{r}\frac{\partial}{\partial \theta}\right)=\sin\theta\cos^{2}\phi\frac{\partial}{\partial x}+\sin\theta\sin\phi\cos\phi\frac{\partial}{\partial y}\tag{b9}\label{rtpxy2}
\end{align}

とし、(\ref{rtpxy2})\(-\)(\ref{rtpxy1})を計算して整理すると

\begin{align}
&\sin\theta\cos\phi\left(\sin\theta\frac{\partial}{\partial r}+\frac{\cos\theta}{r}\frac{\partial}{\partial \theta}\right)-\frac{\sin\phi}{r}\frac{\partial}{\partial \phi} \notag \\
=&\sin\theta\cos^{2}\phi\frac{\partial}{\partial x}-\left(-\sin\theta\sin^{2}\phi\frac{\partial}{\partial x}\right) \notag \\
=&\sin\theta(\sin^{2}\phi+\cos^{2}\phi)\frac{\partial}{\partial x}=\sin\theta\frac{\partial}{\partial x} \tag{b10}\label{rtpxx}
\end{align}

となるため、(\ref{rtpxx})の最左辺と最右辺を\(\sin\theta\)で割って

\begin{align}
\frac{\partial}{\partial x}&=\cos\phi\left(\sin\theta\frac{\partial}{\partial r}+\frac{\cos\theta}{r}\frac{\partial}{\partial \theta}\right)-\frac{\sin\phi}{r\sin\theta}\frac{\partial}{\partial \phi} \notag \\
&=\sin\theta\cos\phi\frac{\partial}{\partial r}+\frac{\cos\theta\cos\phi}{r}\frac{\partial}{\partial \theta}-\frac{\sin\phi}{r\sin\theta}\frac{\partial}{\partial \phi}
\end{align}

となり、(\ref{3px})が求められる。

 続いて\(\displaystyle{\frac{\partial}{\partial y}}\)を求める。(\ref{3pp-2})の両辺に\(\cos\phi\)、(\ref{rpt2})の両辺に\(\sin\phi\)を掛けて

\begin{align}
&\frac{\cos\phi}{r}\frac{\partial}{\partial \phi}=-\sin\theta\sin\phi\cos\phi\frac{\partial}{\partial x}+\sin\theta\cos^{2}\phi\frac{\partial}{\partial y}\tag{b11}\label{rtpxy3}\\
&\sin\theta\sin\phi\left(\sin\theta\frac{\partial}{\partial r}+\frac{\cos\theta}{r}\frac{\partial}{\partial \theta}\right)=\sin\theta\sin\phi\cos\phi\frac{\partial}{\partial x}+\sin\theta\sin^{2}\phi\frac{\partial}{\partial y}\tag{b12}\label{rtpxy4}
\end{align}

とし、両辺を足し上げて整理すると

\begin{align}
&\frac{\cos\phi}{r}\frac{\partial}{\partial \phi}+\sin\theta\sin\phi\left(\sin\theta\frac{\partial}{\partial r}+\frac{\cos\theta}{r}\frac{\partial}{\partial \theta}\right)\notag \\
=&\sin\theta\cos^{2}\phi\frac{\partial}{\partial y}+\sin\theta\sin^{2}\phi\frac{\partial}{\partial y} \notag \\
=&\sin\theta(\sin^{2}\phi+\cos^{2}\phi)\frac{\partial}{\partial y}=\sin\theta\frac{\partial}{\partial y} \tag{b13}\label{rtpyy}
\end{align}

となるため、(\ref{rtpyy})の最左辺と最右辺を\(\sin\theta\)で割って

\begin{align}
\frac{\partial}{\partial y}&=\frac{\cos\phi}{r\sin\theta}\frac{\partial}{\partial \phi}+\sin\phi\left(\sin\theta\frac{\partial}{\partial r}+\frac{\cos\theta}{r}\frac{\partial}{\partial \theta}\right) \notag \\
&=\sin\theta\sin\phi\frac{\partial}{\partial r}+\frac{\cos\theta\sin\phi}{r}\frac{\partial}{\partial \theta}+\frac{\cos\phi}{r\sin\theta}\frac{\partial}{\partial \phi}
\end{align}

となり、(\ref{3py})が求められる。\\

 最後に\(\displaystyle{\frac{\partial}{\partial z}}\)を求める。(\ref{3pr-2})の両辺に\(\cos\theta\)、(\ref{3pt-2})の両辺に\(\sin\theta\)を掛けて

\begin{align}
&\cos\theta\frac{\partial}{\partial r}=\sin\theta\cos\theta\cos\phi\frac{\partial}{\partial x}+\sin\theta\cos\theta\sin\phi\frac{\partial}{\partial y}+\cos^{2}\theta\frac{\partial}{\partial z} \tag{b14}\label{3pr-2-2}\\
&\frac{\sin\theta}{r}\frac{\partial}{\partial \theta}=\sin\theta\cos\theta\cos\phi\frac{\partial}{\partial x}+\sin\theta\cos\theta\sin\phi\frac{\partial}{\partial y}-\sin^{2}\theta\frac{\partial}{\partial z} \tag{b15}\label{3pt-2-2}
\end{align}

とし(\ref{3pr-2-2})\(-\)(\ref{3pt-2-2})を計算して整理すると

\begin{align}
\cos\theta\frac{\partial}{\partial r}-\frac{\sin\theta}{r}\frac{\partial}{\partial \theta}&=\cos^{2}\theta\frac{\partial}{\partial z}-\left(-\sin^{2}\theta\frac{\partial}{\partial z}\right) \notag \\
&=(\sin^{2}\theta+\cos^{2}\theta)\frac{\partial}{\partial z}=\frac{\partial}{\partial z}
\end{align}

となり、(\ref{3pz})が求められる。

となるため、これを2乗して足し上げるわけだが、ただ闇雲にそれぞれ2乗して足し上げると半端ない計算量になる。

 そこで、以下のように少し工夫して2乗の和を計算する。

(i) 各項の2乗和

 まず(\ref{3px})~(\ref{3pz})の右辺の第1項の2乗同士、第2項の2乗同士、第3項の2乗同士をそれぞれ足し上げ、最後にこれらの総和を計算する。

\begin{align}
&\text{第1項の2乗和}:\left(\sin\theta\cos\phi\frac{\partial}{\partial r}\right)^{2}+\left(\sin\theta\sin\phi\frac{\partial}{\partial r}\right)^{2}+\left(\cos\theta\frac{\partial}{\partial r}\right)^{2}=\frac{\partial^{2}}{\partial r^{2}} \tag{17}\label{2-1sum}\\
&\text{第2項の2乗和}:\left(\frac{\cos\theta\cos\phi}{r}\frac{\partial}{\partial \theta}\right)^{2}+\left(\frac{\cos\theta\sin\phi}{r}\frac{\partial}{\partial \theta}\right)^{2}+\left(-\frac{\sin\theta}{r}\frac{\partial}{\partial \theta}\right)^{2}=\frac{1}{r^{2}}\frac{\partial^{2}}{\partial \theta^{2}} \tag{18}\label{2-2sum}\\
&\text{第3項の2乗和}:\left(-\frac{\sin\phi}{r\sin\theta}\frac{\partial}{\partial \phi}\right)^{2}+\left(\frac{\cos\phi}{r\sin\theta}\frac{\partial}{\partial \phi}\right)^{2}=\frac{1}{r^{2}\sin^{2}\theta}\frac{\partial^{2}}{\partial \phi^{2}} \tag{19}\label{2-3sum}
\end{align}

\begin{align}
&\left(\sin\theta\cos\phi\frac{\partial}{\partial r}\right)^{2}+\left(\sin\theta\sin\phi\frac{\partial}{\partial r}\right)^{2}+\left(\cos\theta\frac{\partial}{\partial r}\right)^{2} \notag \\
=&\sin^{2}\theta\cos^{2}\phi\frac{\partial^{2}}{\partial r^{2}}+\sin^{2}\theta\sin^{2}\phi\frac{\partial^{2}}{\partial r^{2}}+\cos^{2}\theta\frac{\partial^{2}}{\partial r^{2}} \notag \\
=&(\sin^{2}\phi+\cos^{2}\phi)\sin^{2}\theta\frac{\partial^{2}}{\partial r^{2}}+\cos^{2}\theta\frac{\partial^{2}}{\partial r^{2}} \notag \\
=&(\sin^{2}\theta+\cos^{2}\theta)\frac{\partial^{2}}{\partial r^{2}}=\frac{\partial^{2}}{\partial r^{2}}
\end{align}

\begin{align}
&\left(\frac{\cos\theta\cos\phi}{r}\frac{\partial}{\partial \theta}\right)^{2}+\left(\frac{\cos\theta\sin\phi}{r}\frac{\partial}{\partial \theta}\right)^{2}+\left(-\frac{\sin\theta}{r}\frac{\partial}{\partial \theta}\right)^{2}\notag \\
=&\frac{\sin^{2}\phi+\cos^{2}\phi}{r^{2}}\left(\cos\theta\frac{\partial}{\partial \theta}\right)^{2}+\frac{1}{r^{2}}\left(\sin\theta\frac{\partial}{\partial \theta}\right)^{2} \notag \\
=&\frac{\cos\theta}{r^{2}}\frac{\partial}{\partial \theta}\left(\cos\theta\frac{\partial}{\partial \theta}\right)+\frac{\sin\theta}{r^{2}}\frac{\partial}{\partial \theta}\left(\sin\theta\frac{\partial}{\partial \theta}\right) \notag \\
=&\frac{\cos\theta}{r^{2}}\left(-\sin\theta\frac{\partial}{\partial \theta}+\cos\theta\frac{\partial^{2}}{\partial \theta^{2}}\right)+\frac{\sin\theta}{r^{2}}\left(\cos\theta\frac{\partial}{\partial \theta}+\sin\theta\frac{\partial^{2}}{\partial \theta^{2}}\right) \notag \\
=&\frac{\sin^{2}\theta+\cos^{2}\theta}{r^{2}}\frac{\partial^{2}}{\partial \theta^{2}}=\frac{1}{r^{2}}\frac{\partial^{2}}{\partial \theta^{2}}
\end{align}

\begin{align}
&\left(-\frac{\sin\phi}{r\sin\theta}\frac{\partial}{\partial \phi}\right)^{2}+\left(\frac{\cos\phi}{r\sin\theta}\frac{\partial}{\partial \phi}\right)^{2}\notag \\
=&\frac{\sin\phi}{r^{2}\sin^{2}\theta}\frac{\partial}{\partial \phi}\left(\sin\phi\frac{\partial}{\partial \phi}\right)+\frac{\cos\phi}{r^{2}\sin^{2}\theta}\frac{\partial}{\partial \phi}\left(\cos\phi\frac{\partial}{\partial \phi}\right) \notag \\
=&\frac{\sin\phi}{r^{2}\sin^{2}\theta}\left(\cos\phi\frac{\partial}{\partial \phi}+\sin\phi\frac{\partial^{2}}{\partial \phi^{2}}\right)+\frac{\cos\phi}{r^{2}\sin^{2}\theta}\left(-\sin\phi\frac{\partial}{\partial \phi}+\cos\phi\frac{\partial^{2}}{\partial \phi^{2}}\right) \notag \\
=&\frac{\sin^{2}\phi+\cos^{2}\phi}{r^{2}\sin^{2}\theta}\frac{\partial^{2}}{\partial \phi^{2}}=\frac{1}{r^{2}\sin^{2}\theta}\frac{\partial^{2}}{\partial \phi^{2}}
\end{align}

よってこれらの総和\(D_{1}\)は

\begin{align}
D_{1}=\frac{\partial^{2}}{\partial r^{2}}+\frac{1}{r^{2}}\frac{\partial^{2}}{\partial \theta^{2}}+\frac{1}{r^{2}\sin^{2}\theta}\frac{\partial^{2}}{\partial \phi^{2}} \tag{20}\label{d1}
\end{align}

となる。

(ii) 第1項と第2項の積の和

 次に(\ref{3px})~(\ref{3pz})の右辺の第1項と第2項の積の和を計算する。

\begin{align}
&\frac{\partial}{\partial x}:\sin\theta\cos\phi\frac{\partial}{\partial r}\left(\frac{\cos\theta\cos\phi}{r}\frac{\partial}{\partial \theta}\right)+\frac{\cos\theta\cos\phi}{r}\frac{\partial}{\partial \theta}\left(\sin\theta\cos\phi\frac{\partial}{\partial r}\right) \tag{21}\label{x1-2seki}\\
&\frac{\partial}{\partial y}:\sin\theta\sin\phi\frac{\partial}{\partial r}\left(\frac{\cos\theta\sin\phi}{r}\frac{\partial}{\partial \theta}\right)+\frac{\cos\theta\sin\phi}{r}\frac{\partial}{\partial \theta}\left(\sin\theta\sin\phi\frac{\partial}{\partial r}\right) \tag{22}\label{y1-2seki}\\
&\frac{\partial}{\partial z}:\cos\theta\frac{\partial}{\partial r}\left(-\frac{\sin\theta}{r}\frac{\partial}{\partial \theta}\right)-\frac{\sin\theta}{r}\frac{\partial}{\partial \theta}\left(\cos\theta\frac{\partial}{\partial r}\right) \tag{23}\label{z1-2seki}
\end{align}

 ここで(\ref{x1-2seki})と(\ref{y1-2seki})の第1項同士を足し上げると

\begin{align}
&\sin\theta\cos\phi\frac{\partial}{\partial r}\left(\frac{\cos\theta\cos\phi}{r}\frac{\partial}{\partial \theta}\right)+\sin\theta\sin\phi\frac{\partial}{\partial r}\left(\frac{\cos\theta\sin\phi}{r}\frac{\partial}{\partial \theta}\right)\notag \\
=&\sin\theta\cos\theta\left(-\frac{1}{r^{2}}\frac{\partial}{\partial \theta}+\frac{1}{r}\frac{\partial}{\partial r}\frac{\partial}{\partial \theta}\right) \tag{24}\label{xy1-2-1sum}
\end{align}

\begin{align}
&\sin\theta\cos\phi\frac{\partial}{\partial r}\left(\frac{\cos\theta\cos\phi}{r}\frac{\partial}{\partial \theta}\right)+\sin\theta\sin\phi\frac{\partial}{\partial r}\left(\frac{\cos\theta\sin\phi}{r}\frac{\partial}{\partial \theta}\right)\notag \\
=&\sin\theta\cos\theta(\cos^{2}\phi+\sin^{2}\phi)\frac{\partial}{\partial r}\left(\frac{1}{r}\frac{\partial}{\partial \theta}\right) \notag \\
=&\sin\theta\cos\theta\frac{\partial}{\partial r}\left(\frac{1}{r}\frac{\partial}{\partial \theta}\right) \notag \\
=&\sin\theta\cos\theta\left(-\frac{1}{r^{2}}\frac{\partial}{\partial \theta}+\frac{1}{r}\frac{\partial}{\partial r}\frac{\partial}{\partial \theta}\right)
\end{align}

となり、さらに(\ref{x1-2seki})と(\ref{y1-2seki})の第2項同士を足し上げると

\begin{align}
&\frac{\cos\theta\cos\phi}{r}\frac{\partial}{\partial \theta}\left(\sin\theta\cos\phi\frac{\partial}{\partial r}\right)+\frac{\cos\theta\sin\phi}{r}\frac{\partial}{\partial \theta}\left(\sin\theta\sin\phi\frac{\partial}{\partial r}\right) \notag \\
=&\frac{\cos\theta}{r}\left(\cos\theta\frac{\partial}{\partial r}+\sin\theta\frac{\partial}{\partial \theta}\frac{\partial}{\partial r}\right)\tag{25}\label{xy1-2-2sum}
\end{align}

\begin{align}
&\frac{\cos\theta\cos\phi}{r}\frac{\partial}{\partial \theta}\left(\sin\theta\cos\phi\frac{\partial}{\partial r}\right)+\frac{\cos\theta\sin\phi}{r}\frac{\partial}{\partial \theta}\left(\sin\theta\sin\phi\frac{\partial}{\partial r}\right) \notag \\
=&\frac{\cos\theta}{r}(\cos^{2}\phi+\sin^{2}\phi)\frac{\partial}{\partial \theta}\left(\sin\theta\frac{\partial}{\partial r}\right) \notag \\
=&\frac{\cos\theta}{r}\frac{\partial}{\partial \theta}\left(\sin\theta\frac{\partial}{\partial r}\right) \notag \\
=&\frac{\cos\theta}{r}\left(\cos\theta\frac{\partial}{\partial r}+\sin\theta\frac{\partial}{\partial \theta}\frac{\partial}{\partial r}\right)
\end{align}

となる。

 さらに(\ref{z1-2seki})は

\begin{align}
&\cos\theta\frac{\partial}{\partial r}\left(-\frac{\sin\theta}{r}\frac{\partial}{\partial \theta}\right)-\frac{\sin\theta}{r}\frac{\partial}{\partial \theta}\left(\cos\theta\frac{\partial}{\partial r}\right) \notag \\
=&-\sin\theta\cos\theta\left(-\frac{1}{r^{2}}\frac{\partial}{\partial \theta}+\frac{1}{r}\frac{\partial}{\partial r}\frac{\partial}{\partial \theta}\right)-\frac{\sin\theta}{r}\left(-\sin\theta\frac{\partial}{\partial r}+\cos\theta\frac{\partial}{\partial \theta}\frac{\partial}{\partial r}\right) \tag{26}\label{z1-2seki2}
\end{align}

となる。

 よってこれらの総和\(D_{2}\)を計算すると、(\ref{xy1-2-1sum})と(\ref{z1-2seki2})の第1項が相殺され、残りは(\ref{xy1-2-2sum})と(\ref{z1-2seki2})の第2項となり

\begin{align}
D_{2}=\frac{1}{r}\frac{\partial}{\partial r} \tag{27}\label{d2}
\end{align}

\begin{align}
D_{2}&=\frac{\cos\theta}{r}\left(\cos\theta\frac{\partial}{\partial r}+\sin\theta\frac{\partial}{\partial \theta}\frac{\partial}{\partial r}\right)-\frac{\sin\theta}{r}\left(-\sin\theta\frac{\partial}{\partial r}+\cos\theta\frac{\partial}{\partial \theta}\frac{\partial}{\partial r}\right) \notag \\
&=\frac{\cos^{2}\theta+\sin^{2}\theta}{r}\frac{\partial}{\partial r}+\frac{\sin\theta\cos\theta}{r}\frac{\partial}{\partial \theta}\frac{\partial}{\partial r}-\frac{\sin\theta\cos\theta}{r}\frac{\partial}{\partial \theta}\frac{\partial}{\partial r} \notag \\
&=\frac{1}{r}\frac{\partial}{\partial r}
\end{align}

となる。

(iii) 第2項と第3項の積の和

 続いて(\ref{3px})、(\ref{3py})の右辺の第2項と第3項の積の和を計算する。

\begin{align}
&\frac{\partial}{\partial x}:\frac{\cos\theta\cos\phi}{r}\frac{\partial}{\partial \theta}\left(-\frac{\sin\phi}{r\sin\theta}\frac{\partial}{\partial \phi}\right)-\frac{\sin\phi}{r\sin\theta}\frac{\partial}{\partial \phi}\left(\frac{\cos\theta\cos\phi}{r}\frac{\partial}{\partial \theta}\right) \tag{28}\label{x2-3seki}\\
&\frac{\partial}{\partial y}:\frac{\cos\theta\sin\phi}{r}\frac{\partial}{\partial \theta}\left(\frac{\cos\phi}{r\sin\theta}\frac{\partial}{\partial \phi}\right)+\frac{\cos\phi}{r\sin\theta}\frac{\partial}{\partial \phi}\left(\frac{\cos\theta\sin\phi}{r}\frac{\partial}{\partial \theta}\right) \tag{29}\label{y2-3seki}
\end{align}

 (\ref{x2-3seki})と(\ref{y2-3seki})の第1項同士が相殺されて0になる。

 よって、第2項同士の和を求めればよい。これを\(D_{3}\)とすれば

\begin{align}
D_{3}=\frac{\cos\theta}{r^{2}\sin\theta}\frac{\partial}{\partial \theta} \tag{30}\label{d3}
\end{align}

\begin{align}
D_{3}&=\frac{\cos\phi}{r\sin\theta}\frac{\partial}{\partial \phi}\left(\frac{\cos\theta\sin\phi}{r}\frac{\partial}{\partial \theta}\right)-\frac{\sin\phi}{r\sin\theta}\frac{\partial}{\partial \phi}\left(\frac{\cos\theta\cos\phi}{r}\frac{\partial}{\partial \theta}\right) \notag \\
&=\frac{\cos\theta\cos\phi}{r^{2}\sin\theta}\frac{\partial}{\partial \phi}\left(\sin\phi\frac{\partial}{\partial \theta}\right)-\frac{\cos\theta\sin\phi}{r^{2}\sin\theta}\frac{\partial}{\partial \phi}\left(\cos\phi\frac{\partial}{\partial \theta}\right) \notag \\
&=\frac{\cos\theta\cos\phi}{r^{2}\sin\theta}\left(\cos\phi\frac{\partial}{\partial \theta}+\sin\phi\frac{\partial}{\partial \phi}\frac{\partial}{\partial \theta}\right)-\frac{\cos\theta\sin\phi}{r^{2}\sin\theta}\left(-\sin\phi\frac{\partial}{\partial \theta}+\cos\phi\frac{\partial}{\partial \phi}\frac{\partial}{\partial \theta}\right) \notag \\
&=\frac{\cos\theta(\cos^{2}\phi+\sin^{2}\phi)}{r^{2}\sin\theta}\frac{\partial}{\partial \theta}+\frac{\cos\theta\sin\phi\cos\phi}{r^{2}\sin\theta}\frac{\partial}{\partial \phi}\frac{\partial}{\partial \theta}-\frac{\cos\theta\sin\phi\cos\phi}{r^{2}\sin\theta}\frac{\partial}{\partial \phi}\frac{\partial}{\partial \theta} \notag \\
&=\frac{\cos\theta}{r^{2}\sin\theta}\frac{\partial}{\partial \theta}
\end{align}

となる。

(iv) 第1項と第3項の積の和

 最後に(\ref{3px})、(\ref{3py})の右辺の第1項と第3項の積の和を計算する。

\begin{align}
&\frac{\partial}{\partial x}:\sin\theta\cos\phi\frac{\partial}{\partial r}\left(-\frac{\sin\phi}{r\sin\theta}\frac{\partial}{\partial \phi}\right)-\frac{\sin\phi}{r\sin\theta}\frac{\partial}{\partial \phi}\left(\sin\theta\cos\phi\frac{\partial}{\partial r}\right) \tag{31}\label{x1-3seki}\\
&\frac{\partial}{\partial y}:\sin\theta\sin\phi\frac{\partial}{\partial r}\left(\frac{\cos\phi}{r\sin\theta}\frac{\partial}{\partial \phi}\right)+\frac{\cos\phi}{r\sin\theta}\frac{\partial}{\partial \phi}\left(\sin\theta\sin\phi\frac{\partial}{\partial r}\right) \tag{32}\label{y1-3seki}
\end{align}

 (\ref{x1-3seki})と(\ref{y1-3seki})の第1項同士が相殺されて0になる。

 よって、第2項同士の和を求めればよい。これを\(D_{4}\)とすれば

\begin{align}
D_{4}=\frac{1}{r}\frac{\partial}{\partial r} \tag{33}\label{d4}
\end{align}

\begin{align}
D_{4}&=-\frac{\sin\phi}{r\sin\theta}\frac{\partial}{\partial \phi}\left(\sin\theta\cos\phi\frac{\partial}{\partial r}\right)+\frac{\cos\phi}{r\sin\theta}\frac{\partial}{\partial \phi}\left(\sin\theta\sin\phi\frac{\partial}{\partial r}\right) \notag \\
&=-\frac{\sin\phi}{r}\frac{\partial}{\partial \phi}\left(\cos\phi\frac{\partial}{\partial r}\right)+\frac{\cos\phi}{r}\frac{\partial}{\partial \phi}\left(\sin\phi\frac{\partial}{\partial r}\right) \notag \\
&=-\frac{\sin\phi}{r}\left(-\sin\phi\frac{\partial}{\partial r}+\cos\phi\frac{\partial}{\partial \phi}\frac{\partial}{\partial r}\right)+\frac{\cos\phi}{r}\left(\cos\phi\frac{\partial}{\partial r}+\sin\phi\frac{\partial}{\partial \phi}\frac{\partial}{\partial r}\right) \notag \\
&=\frac{\sin^{2}\phi+\cos^{2}\phi}{r}\frac{\partial}{\partial r}-\frac{\sin\phi\cos\phi}{r}\frac{\partial}{\partial \phi}\frac{\partial}{\partial r}+\frac{\sin\phi\cos\phi}{r}\frac{\partial}{\partial \phi}\frac{\partial}{\partial r} \notag \\
&=\frac{1}{r}\frac{\partial}{\partial r}
\end{align}

となる。

総和を計算

 以上より、ラプラシアンは\(D_{1}\)~\(D_{4}\)の総和で表されるため、(\ref{d1})、(\ref{d2})、(\ref{d3})、(\ref{d4})より、

\begin{align}
\boxed{\triangle=\frac{\partial^{2}}{\partial r^{2}}+\frac{2}{r}\frac{\partial}{\partial r}+\frac{1}{r^{2}}\left(\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\sin\theta\frac{\partial}{\partial\theta}+\frac{1}{\sin^{2}\theta}\frac{\partial^{2}}{\partial\phi^{2}}\right)}
\end{align}

\begin{align}
\triangle&=D_{1}+D_{2}+D_{3}+D_{4} \notag \\
&=\frac{\partial^{2}}{\partial r^{2}}+\frac{1}{r^{2}}\frac{\partial^{2}}{\partial \theta^{2}}+\frac{1}{r^{2}\sin^{2}\theta}\frac{\partial^{2}}{\partial \phi^{2}}+\frac{1}{r}\frac{\partial}{\partial r}+\frac{\cos\theta}{r^{2}\sin\theta}\frac{\partial}{\partial \theta}+\frac{1}{r}\frac{\partial}{\partial r} \notag \\
&=\frac{\partial^{2}}{\partial r^{2}}+\frac{2}{r}\frac{\partial}{\partial r}+\frac{1}{r^{2}}\left(\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\sin\theta\frac{\partial}{\partial\theta}+\frac{1}{\sin^{2}\theta}\frac{\partial^{2}}{\partial\phi^{2}}\right)
\end{align}

と求められる。

終わりに

 これを一通りこなせれば、かなりの計算力がつくと思う。

 もちろん、詳細計算の部分は必ずしもこの通りに実施する必要はない。
 自分なりの計算手順を見つけて、最終的な結果が合っていればOKだ。

 次は寄り道から戻って多重積分について書いていこうと思う。

 

 END

 

 ※追記
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