球座標における微分演算子/メモ の変更点

更新


* 目次 [#n498e6ef]

#contents

* 演習:偏微分の計算 [#j884fda6]

** 解答 [#f8689b48]

(1)

&math(r^2=x^2+y^2+z^2); より、&math(2r\frac{\PD r}{\PD x}=2x); などが得られて、

 &math(
\begin{cases}
\displaystyle\frac{\PD r}{\PD x}=\frac{x}{r}=\sin\theta\cos\phi\\[4mm]
\displaystyle\frac{\PD r}{\PD y}=\frac{y}{r}=\sin\theta\sin\phi\\[4mm]
\displaystyle\frac{\PD r}{\PD z}=\frac{z}{r}=\cos\theta\\
\end{cases}
);

(2)

&math(\tan^2\theta=\frac{x^2+y^2}{z^2}); より、
&math(\frac{1}{\cos^2\theta}\frac{\PD \theta}{\PD y}=\frac{2x}{z^2});

&math(\frac{\not\!2\tan\theta}{\cos^2\theta}\frac{\PD \theta}{\PD x}=\frac{\not\! 2x}{z^2});、
&math(\frac{\not\!2\tan\theta}{\cos^2\theta}\frac{\PD \theta}{\PD y}=\frac{\not\! 2y}{z^2});、
&math(\frac{\not\!2\tan\theta}{\cos^2\theta}\frac{\PD \theta}{\PD z}=-\not\!2\frac{x^2+y^2}{z^3});、

 &math(
\begin{cases}
\displaystyle\frac{\PD \theta}{\PD x}=\frac{r\sin\theta\cos\phi}{r^2\cos^2\theta}\frac{\cos^2\theta}{\tan\theta}=\frac{1}{r}\cos\theta\cos\phi\\[4mm]
\displaystyle\frac{\PD \theta}{\PD y}=\frac{r\sin\theta\sin\phi}{r^2\cos^2\theta}\frac{\cos^2\theta}{\tan\theta}=\frac{1}{r}\cos\theta\sin\phi\\[4mm]
\displaystyle\frac{\PD \theta}{\PD z}=-\frac{r^2\sin^2\theta}{r^3\cos^3\theta}\frac{\cos^2\theta}{\tan\theta}=-\frac{1}{r}\sin\theta
\end{cases}
);

(3)

&math(\tan\phi=\frac{y}{x}); より、

&math(\frac{1}{\cos^2\phi}\frac{\PD\phi}{\PD x}=-\frac{y}{x^2});、
&math(\frac{1}{\cos^2\phi}\frac{\PD\phi}{\PD y}=\frac{1}{x});、
&math(\frac{1}{\cos^2\phi}\frac{\PD\phi}{\PD z}=0); であるから、

 &math(
\begin{cases}
\displaystyle\frac{\PD \phi}{\PD x}=-\frac{r\sin\theta\sin\phi}{r^2\sin^2\theta\cos^2\phi}\cos^2\phi=-\frac{\sin\phi}{r\sin\theta}\\[4mm]
\displaystyle\frac{\PD \phi}{\PD y}=\frac{1}{r\sin\theta\cos\phi}\cos^2\phi=\frac{\cos\phi}{r\sin\theta}\\[4mm]
\displaystyle\frac{\PD \phi}{\PD z}=0
\end{cases}
);

* 球座標のラプラシアン [#o621468a]

&math(
\frac{\PD^2}{\PD x^2}
&=\Big(\sin\theta\cos\phi \frac{\PD}{\PD r}
+\frac{1}{r}\cos\theta\cos\phi \frac{\PD}{\PD \theta}
-\frac{\sin\phi}{r\sin\theta} \frac{\PD}{\PD \phi}\Big)^2\\

&=\sin\theta\cos\phi \frac{\PD}{\PD r}\Big(\sin\theta\cos\phi \frac{\PD}{\PD r}
+\frac{1}{r}\cos\theta\cos\phi \frac{\PD}{\PD \theta}
-\frac{\sin\phi}{r\sin\theta} \frac{\PD}{\PD \phi}\Big)\\
&\ \ \ +\frac{1}{r}\cos\theta\cos\phi \frac{\PD}{\PD \theta}
\Big(\sin\theta\cos\phi \frac{\PD}{\PD r}
+\frac{1}{r}\cos\theta\cos\phi \frac{\PD}{\PD \theta}
-\frac{\sin\phi}{r\sin\theta} \frac{\PD}{\PD \phi}\Big)\\
&\ \ \ -\frac{\sin\phi}{r\sin\theta} \frac{\PD}{\PD \phi}
\Big(\sin\theta\cos\phi \frac{\PD}{\PD r}
+\frac{1}{r}\cos\theta\cos\phi \frac{\PD}{\PD \theta}
-\frac{\sin\phi}{r\sin\theta} \frac{\PD}{\PD \phi}\Big)\\

&=\sin^2\theta\cos^2\phi \frac{\PD^2}{\PD r^2}
-\frac{\sin\theta\cos\theta\cos^2\phi}{r^2} \frac{\PD}{\PD \theta}
+\frac{\sin\theta\cos\theta\cos^2\phi}{r} \frac{\PD}{\PD r}\frac{\PD}{\PD \theta}
+\frac{\sin\phi\cos\phi}{r^2} \frac{\PD}{\PD \phi}
-\frac{\sin\phi\cos\phi}{r} \frac{\PD}{\PD r}\frac{\PD}{\PD \phi}\\
&\ \ \ 
+\frac{\cos^2\theta\cos^2\phi}{r} \frac{\PD}{\PD r}
+\frac{\sin\theta\cos\theta\cos^2\phi}{r} \frac{\PD}{\PD r}\frac{\PD}{\PD \theta}
+\frac{-\sin\theta\cos\theta\cos^2\phi}{r^2}\frac{\PD}{\PD \theta}
+\frac{\cos^2\theta\cos^2\phi}{r^2} \frac{\PD^2}{\PD \theta^2}\\
&\hspace{9cm}+\frac{\cos^2\theta\sin\phi\cos\phi}{r^2\sin^2\theta} \frac{\PD}{\PD \phi}
-\frac{\cos\theta\sin\phi\cos\phi}{r^2\sin\theta} \frac{\PD}{\PD \theta}\frac{\PD}{\PD \phi}
\\
&\ \ \ 
+\frac{\sin^2\phi}{r} \frac{\PD}{\PD r}
-\frac{\sin\phi\cos\phi}{r} \frac{\PD}{\PD r}\frac{\PD}{\PD \phi}
+\frac{\cos\theta\sin^2\phi}{r^2\sin\theta} \frac{\PD}{\PD \theta}
-\frac{\cos\theta\sin\phi\cos\phi}{r^2\sin\theta} \frac{\PD}{\PD \theta}\frac{\PD}{\PD \phi}
+\frac{\sin\phi\cos\phi}{r^2\sin^2\theta} \frac{\PD}{\PD \phi}
+\frac{\sin^2\phi}{r^2\sin^2\theta} \frac{\PD^2}{\PD \phi^2}\\

&=\sin^2\theta\cos^2\phi \frac{\PD^2}{\PD r^2}
-\frac{\sin\theta\cos\theta\cos^2\phi}{r^2} \frac{\PD}{\PD \theta}
+\frac{2\sin\theta\cos\theta\cos^2\phi}{r} \frac{\PD}{\PD r}\frac{\PD}{\PD \theta}
+\frac{\sin\phi\cos\phi}{r^2} \frac{\PD}{\PD \phi}
-\frac{2\sin\phi\cos\phi}{r} \frac{\PD}{\PD r}\frac{\PD}{\PD \phi}\\
&\ \ \ 
+\frac{\cos^2\theta\cos^2\phi}{r} \frac{\PD}{\PD r}
+\frac{-\sin\theta\cos\theta\cos^2\phi}{r^2}\frac{\PD}{\PD \theta}
+\frac{\cos^2\theta\cos^2\phi}{r^2} \frac{\PD^2}{\PD \theta^2}
+\frac{\cos^2\theta\sin\phi\cos\phi}{r^2\sin^2\theta} \frac{\PD}{\PD \phi}\\
&\ \ \ 
-\frac{2\cos\theta\sin\phi\cos\phi}{r^2\sin\theta} \frac{\PD}{\PD \theta}\frac{\PD}{\PD \phi}
+\frac{\sin^2\phi}{r} \frac{\PD}{\PD r}
+\frac{\cos\theta\sin^2\phi}{r^2\sin\theta} \frac{\PD}{\PD \theta}
+\frac{\sin\phi\cos\phi}{r^2\sin^2\theta} \frac{\PD}{\PD \phi}
+\frac{\sin^2\phi}{r^2\sin^2\theta} \frac{\PD^2}{\PD \phi^2}\\
);

&math(
\frac{\PD^2}{\PD y^2}
&=\Big(
\sin\theta\sin\phi \frac{\PD}{\PD r}
+\frac{1}{r}\cos\theta\sin\phi \frac{\PD}{\PD \theta}
+\frac{\cos\phi}{r\sin\theta} \frac{\PD}{\PD \phi}
\Big)^2\\
&=
\sin^2\theta\sin^2\phi \frac{\PD^2}{\PD r^2}
-\frac{\sin\theta\cos\theta\sin^2\phi}{r^2} \frac{\PD}{\PD \theta}
+\frac{2\sin\theta\cos\theta\sin^2\phi}{r} \frac{\PD}{\PD r}\frac{\PD}{\PD \theta}
-\frac{\sin\phi\cos\phi}{r^2} \frac{\PD}{\PD \phi}
+\frac{2\sin\phi\cos\phi}{r} \frac{\PD}{\PD r}\frac{\PD}{\PD \phi}\\
&\ \ \ 
+\frac{\cos^2\theta\sin^2\phi}{r} \frac{\PD}{\PD r}
+\frac{-\sin\theta\cos\theta\sin^2\phi}{r^2} \frac{\PD}{\PD \theta}
+\frac{\cos^2\theta\sin^2\phi}{r^2} \frac{\PD^2}{\PD \theta^2}
-\frac{\cos^2\theta\sin\phi\cos\phi}{r^2\sin^2\theta} \frac{\PD}{\PD \phi}
+\frac{2\cos\theta\sin\phi\cos\phi}{r^2\sin\theta} \frac{\PD}{\PD \theta}\frac{\PD}{\PD \phi}\\
&\ \ \ 
+\frac{\cos^2\phi}{r} \frac{\PD}{\PD r}
+\frac{\cos\theta\cos^2\phi}{r^2\sin\theta} \frac{\PD}{\PD \theta}
+\frac{-\sin\phi\cos\phi}{r^2\sin^2\theta} \frac{\PD}{\PD \phi}
+\frac{\cos^2\phi}{r^2\sin^2\theta} \frac{\PD^2}{\PD \phi^2}
);

&math(
\frac{\PD^2}{\PD z^2}
&=\Big(
\cos\theta \frac{\PD}{\PD r}
-\frac{1}{r}\sin\theta \frac{\PD}{\PD \theta}
\Big)^2\\
&=
\cos^2\theta \frac{\PD^2}{\PD r^2}
+\frac{\sin\theta\cos\theta}{r^2} \frac{\PD}{\PD \theta}
-\frac{2\sin\theta\cos\theta}{r} \frac{\PD}{\PD r}\frac{\PD}{\PD \theta}
\\
&\ \ \ 
+\frac{\sin^2\theta}{r} \frac{\PD}{\PD r}
+\frac{\sin\theta\cos\theta}{r^2} \frac{\PD}{\PD \theta}
+\frac{\sin^2\theta}{r^2} \frac{\PD^2}{\PD \theta^2}
);

足せばいい(本気?)。

&math(
&\frac{\PD^2}{\PD x^2}+\frac{\PD^2}{\PD y^2}+\frac{\PD^2}{\PD z^2}\\
&=
\sin^2\theta\cos^2\phi \frac{\PD^2}{\PD r^2}
-\frac{\sin\theta\cos\theta\cos^2\phi}{r^2} \frac{\PD}{\PD \theta}
+\frac{2\sin\theta\cos\theta\cos^2\phi}{r} \frac{\PD}{\PD r}\frac{\PD}{\PD \theta}
+\frac{\sin\phi\cos\phi}{r^2} \frac{\PD}{\PD \phi}
-\frac{2\sin\phi\cos\phi}{r} \frac{\PD}{\PD r}\frac{\PD}{\PD \phi}\\
&\ \ \ 
+\frac{\cos^2\theta\cos^2\phi}{r} \frac{\PD}{\PD r}
+\frac{-\sin\theta\cos\theta\cos^2\phi}{r^2}\frac{\PD}{\PD \theta}
+\frac{\cos^2\theta\cos^2\phi}{r^2} \frac{\PD^2}{\PD \theta^2}
+\frac{\cos^2\theta\sin\phi\cos\phi}{r^2\sin^2\theta} \frac{\PD}{\PD \phi}\\
&\ \ \ 
-\frac{2\cos\theta\sin\phi\cos\phi}{r^2\sin\theta} \frac{\PD}{\PD \theta}\frac{\PD}{\PD \phi}
+\frac{\sin^2\phi}{r} \frac{\PD}{\PD r}
+\frac{\cos\theta\sin^2\phi}{r^2\sin\theta} \frac{\PD}{\PD \theta}
+\frac{\sin\phi\cos\phi}{r^2\sin^2\theta} \frac{\PD}{\PD \phi}
+\frac{\sin^2\phi}{r^2\sin^2\theta} \frac{\PD^2}{\PD \phi^2}\\

&\ \ \ 
+\sin^2\theta\sin^2\phi \frac{\PD^2}{\PD r^2}
-\frac{\sin\theta\cos\theta\sin^2\phi}{r^2} \frac{\PD}{\PD \theta}
+\frac{2\sin\theta\cos\theta\sin^2\phi}{r} \frac{\PD}{\PD r}\frac{\PD}{\PD \theta}
-\frac{\sin\phi\cos\phi}{r^2} \frac{\PD}{\PD \phi}
+\frac{2\sin\phi\cos\phi}{r} \frac{\PD}{\PD r}\frac{\PD}{\PD \phi}\\
&\ \ \ 
+\frac{\cos^2\theta\sin^2\phi}{r} \frac{\PD}{\PD r}
+\frac{-\sin\theta\cos\theta\sin^2\phi}{r^2} \frac{\PD}{\PD \theta}
+\frac{\cos^2\theta\sin^2\phi}{r^2} \frac{\PD^2}{\PD \theta^2}
-\frac{\cos^2\theta\sin\phi\cos\phi}{r^2\sin^2\theta} \frac{\PD}{\PD \phi}
+\frac{2\cos\theta\sin\phi\cos\phi}{r^2\sin\theta} \frac{\PD}{\PD \theta}\frac{\PD}{\PD \phi}\\
&\ \ \ 
+\frac{\cos^2\phi}{r} \frac{\PD}{\PD r}
+\frac{\cos\theta\cos^2\phi}{r^2\sin\theta} \frac{\PD}{\PD \theta}
+\frac{-\sin\phi\cos\phi}{r^2\sin^2\theta} \frac{\PD}{\PD \phi}
+\frac{\cos^2\phi}{r^2\sin^2\theta} \frac{\PD^2}{\PD \phi^2}\\

&\ \ \ 
+\cos^2\theta \frac{\PD^2}{\PD r^2}
+\frac{\sin\theta\cos\theta}{r^2} \frac{\PD}{\PD \theta}
-\frac{2\sin\theta\cos\theta}{r} \frac{\PD}{\PD r}\frac{\PD}{\PD \theta}
\\
&\ \ \ 
-\frac{\sin^2\theta}{r} \frac{\PD}{\PD r}-
-\frac{\sin\theta\cos\theta}{r^2} \frac{\PD}{\PD \theta}
-\frac{\sin^2\theta}{r^2} \frac{\PD^2}{\PD \theta^2}\\
);
&math(
&=
(\sin^2\theta\cos^2\phi+\sin^2\theta\sin^2\phi+\cos^2\theta)\frac{\PD^2}{\PD r^2}
\\&\ \ \ 
+\Big(\frac{\cos^2\theta\cos^2\phi}{r}+\frac{\sin^2\phi}{r}+\frac{\cos^2\theta\sin^2\phi}{r}+\frac{\cos^2\phi}{r} +\frac{\sin^2\theta}{r}\Big) \frac{\PD}{\PD r}
\\&\ \ \ 
+\Big(-\cancel{\frac{\sin\theta\cos\theta\cos^2\phi}{r^2}}+\cancel{\frac{-\sin\theta\cos\theta\cos^2\phi}{r^2}}+\frac{\cos\theta\sin^2\phi}{r^2\sin\theta}-\cancel{\frac{\sin\theta\cos\theta\sin^2\phi}{r^2}}\\
&\hspace{4cm}+\cancel{\frac{-\sin\theta\cos\theta\sin^2\phi}{r^2}}+\frac{\cos\theta\cos^2\phi}{r^2\sin\theta}
+\cancel{\frac{\sin\theta\cos\theta}{r^2}}+\cancel{\frac{\sin\theta\cos\theta}{r^2}}\Big) \frac{\PD}{\PD \theta}
\\&\ \ \ 
+\Big(\frac{\cos^2\theta\cos^2\phi}{r^2}+\frac{\cos^2\theta\sin^2\phi}{r^2}+\frac{\sin^2\theta}{r^2}\Big) \frac{\PD^2}{\PD \theta^2}
\\&\ \ \ 
+\Big(\cancel{\frac{\sin\phi\cos\phi}{r^2}}+\cancel{\frac{\cos^2\theta\sin\phi\cos\phi}{r^2\sin^2\theta}}+\cancel{\frac{\sin\phi\cos\phi}{r^2\sin^2\theta}}-\cancel{\frac{\sin\phi\cos\phi}{r^2}}-\cancel{\frac{\cos^2\theta\sin\phi\cos\phi}{r^2\sin^2\theta}}-\cancel{\frac{\sin\phi\cos\phi}{r^2\sin^2\theta}}\Big) \frac{\PD}{\PD \phi}
\\&\ \ \ 
+\Big(\frac{\sin^2\phi}{r^2\sin^2\theta}+\frac{\cos^2\phi}{r^2\sin^2\theta}\Big) \frac{\PD^2}{\PD \phi^2}
\\&\ \ \ 
+\Big(\cancel{\frac{2\sin\theta\cos\theta\cos^2\phi}{r}}+\cancel{\frac{2\sin\theta\cos\theta\sin^2\phi}{r}}-\cancel{\frac{2\sin\theta\cos\theta}{r}}\Big) \frac{\PD}{\PD r}\frac{\PD}{\PD \theta}
\\&\ \ \ 
+\Big(-\cancel{\frac{2\sin\phi\cos\phi}{r}}+\cancel{\frac{2\sin\phi\cos\phi}{r}}\Big) \frac{\PD}{\PD r}\frac{\PD}{\PD \phi}
\\&\ \ \ 
+\Big(-\cancel{\frac{2\cos\theta\sin\phi\cos\phi}{r^2\sin\theta}}+\cancel{\frac{2\cos\theta\sin\phi\cos\phi}{r^2\sin\theta}}\Big)\frac{\PD}{\PD \theta}\frac{\PD}{\PD \phi}\\

&=
\frac{\PD^2}{\PD r^2}+\frac{2}{r} \frac{\PD}{\PD r}+\frac{\cos\theta}{r^2\sin\theta} \frac{\PD}{\PD \theta}
+\frac{1}{r^2}\frac{\PD^2}{\PD \theta^2}+\frac{1}{r^2\sin^2\theta} \frac{\PD^2}{\PD \phi^2}
\\
&=
\frac{\PD^2}{\PD r^2}+\frac{2}{r} \frac{\PD}{\PD r}
+\frac{1}{r^2}\underbrace{\bigg[\frac{1}{\sin\theta} \frac{\PD}{\PD \theta}
\Big(\sin\theta\frac{\PD}{\PD \theta}\Big)+\frac{1}{\sin^2\theta} \frac{\PD^2}{\PD \phi^2}\bigg]}_{=\,\Lambda}
);

恐らくもっと簡単に求める方法もあるはず。

** 別解 [#p2feb906]

2次元の極座標表示を利用すると少し楽らしい。~
https://twitter.com/Paul_Painleve/status/995843040244711424

2次元の極座標:

&math(
\begin{cases}
x&=r\cos\theta\\
y&=r\sin\theta
\end{cases}
);   
&math(
\begin{cases}
r^2=x^2+y^2\\
\tan^2\theta=y^2/x^2
\end{cases}
);

より、

&math(
2r\frac{\partial r}{\partial x}=2x=2r\cos\theta,\ \ 
2r\frac{\partial r}{\partial y}=2y=2r\sin\theta,);

&math(
\frac{1}{\cos^2\theta}\frac{\partial\theta}{\partial x}=-\frac{y}{x^2}=-\frac{r\sin\theta}{r^2\cos^2\theta},\ \ 
\frac{1}{\cos^2\theta}\frac{\partial\theta}{\partial y}=\frac{1}{x}=\frac{1}{r\cos\theta}
); 

などを使って、

 &math(
\frac{\partial}{\partial x}
&=\left(\frac{\partial r}{\partial x}\right)\frac{\partial}{\partial r}
+\left(\frac{\partial \theta}{\partial x}\right)\frac{\partial}{\partial \theta}
=\cos\theta\frac{\partial}{\partial r}-\frac{\sin\theta}{r}\frac{\partial}{\partial \theta}\\
);~
 &math(
\frac{\partial}{\partial y}
&=\left(\frac{\partial r}{\partial y}\right)\frac{\partial}{\partial r}
+\left(\frac{\partial \theta}{\partial y}\right)\frac{\partial}{\partial \theta}
=\sin\theta\frac{\partial}{\partial r}+\frac{\cos\theta}{r}\frac{\partial}{\partial \theta}\\
);

 &math(
\frac{\partial^2}{\partial x^2}f
&=\left(\cos\theta\frac{\partial}{\partial r}-\frac{\sin\theta}{r}\frac{\partial}{\partial \theta}\right)
  \left(\cos\theta\frac{\partial f}{\partial r}-\frac{\sin\theta}{r}\frac{\partial f}{\partial \theta}\right)\\
&=\cos^2\theta\frac{\partial^2 f}{\partial r^2}
-\cos\theta\sin\theta\left(-\frac{1}{r^2}\frac{\partial f}{\partial \theta}+\frac{1}{r}\frac{\partial^2f}{\partial r\partial\theta}\right)\\
&\hspace{1cm}-\frac{\sin\theta}{r}\left(-\sin\theta\frac{\partial f}{\partial r}+\cos\theta\frac{\partial^2 f}{\partial\theta\partial r}\right)
+\frac{\sin\theta}{r^2}\left(\cos\theta\frac{\partial f}{\partial \theta}+\sin\theta\frac{\partial^2f}{\partial\theta^2}\right)\\
&=\left[\cos^2\theta\frac{\partial^2 }{\partial r^2}+\frac{\sin^2\theta}{r}\frac{\partial}{\partial r}
-2r\cos\theta\sin\theta\frac{\partial}{\partial r}\left(\frac{1}{r}\frac{\partial}{\partial\theta}\right)
+\frac{\sin^2\theta}{r^2}\frac{\partial^2}{\partial\theta^2}
\right]f);~
 &math(
\frac{\partial^2}{\partial y^2}f
&=\left(\sin\theta\frac{\partial}{\partial r}+\frac{\cos\theta}{r}\frac{\partial}{\partial \theta}\right)
  \left(\sin\theta\frac{\partial f}{\partial r}+\frac{\cos\theta}{r}\frac{\partial f}{\partial \theta}\right)\\
&=\sin^2\theta\frac{\partial^2 f}{\partial r^2}
+\sin\theta\cos\theta\left(-\frac{1}{r^2}\frac{\partial f}{\partial \theta}+\frac{1}{r}\frac{\partial^2f}{\partial r\partial\theta}\right)\\
&\hspace{1cm}+\frac{\cos\theta}{r}\left(\cos\theta\frac{\partial f}{\partial r}-\sin\theta\frac{\partial^2 f}{\partial\theta\partial r}\right)
+\frac{\cos\theta}{r^2}\left(-\sin\theta\frac{\partial f}{\partial \theta}+\cos\theta\frac{\partial^2f}{\partial\theta^2}\right)\\
&=\left[\sin^2\theta\frac{\partial^2 }{\partial r^2}+\frac{\cos^2\theta}{r}\frac{\partial}{\partial r}
+2r\cos\theta\sin\theta\frac{\partial}{\partial r}\left(\frac{1}{r}\frac{\partial}{\partial\theta}\right)
+\frac{\cos^2\theta}{r^2}\frac{\partial^2}{\partial\theta^2}
\right]f\\
);

なので、

 &math(
\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}
&=\frac{\partial^2 }{\partial r^2}+\frac{1}{r}\frac{\partial}{\partial r}+\frac{1}{r^2}\frac{\partial^2}{\partial\theta^2}\\
&=\frac{1}{r}\frac{\partial}{\partial r}\left(r\frac{\partial}{\partial r}\right)+\frac{1}{r^2}\frac{\partial^2}{\partial\theta^2}
);

を得る。これ自体有用な式なのだけれど、球座標系の計算にどう使うかというと、
&math(\bm r); から &math(x\mathrm-y); 平面に垂線を下ろした点と原点との距離を
&math(\rho=r\sin\theta); と書き、

 &math((x,y,z)\to(\rho,\phi,z)\to(r,\theta,\phi));

の2段階の変数変換を考える。1段目は、

 &math(
\begin{cases}
x=\rho\cos\phi\\
y=\rho\sin\phi\\
z=z
\end{cases}
);

であるから、上記を参考に、

 &math(
\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}
&=\frac{\partial^2 }{\partial \rho^2}+\frac{1}{\rho}\frac{\partial}{\partial \rho}+\frac{1}{\rho^2}\frac{\partial^2}{\partial\phi^2}\\
);

を得る。一方、2段目は、

 &math(
\begin{cases}
\rho=r\sin\theta\\
\phi=\phi\\
z=r\cos\theta
\end{cases}
);

であるから、上記を参考に、

 &math(
\frac{\partial^2}{\partial z^2}+\frac{\partial^2}{\partial \rho^2}
&=\frac{\partial^2 }{\partial r^2}+\frac{1}{r}\frac{\partial}{\partial r}+\frac{1}{r^2}\frac{\partial^2}{\partial\theta^2}\\
);

辺々加えると、&math(\partial^2/\partial\rho^2); の項は打消し合って、

 &math(
\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2}
&=\frac{\partial^2 }{\partial r^2}+\frac{1}{r}\frac{\partial}{\partial r}
+\frac{1}{r^2}\frac{\partial^2}{\partial\theta^2}
+\frac{1}{\rho}\frac{\partial}{\partial \rho}
+\frac{1}{\rho^2}\frac{\partial^2}{\partial\phi^2}
);

2次元の &math(\partial/\partial y); の計算結果から、

 &math(
\frac{\partial}{\partial \rho}
=\sin\theta\frac{\partial}{\partial r}+\frac{\cos\theta}{r}\frac{\partial}{\partial \theta}\\
);

なので、

 &math(
\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2}
&=\frac{\partial^2 }{\partial r^2}+\frac{1}{r}\frac{\partial}{\partial r}
+\frac{1}{r^2}\frac{\partial^2}{\partial\theta^2}
+\frac{1}{r\sin\theta}\left(\sin\theta\frac{\partial}{\partial r}+\frac{\cos\theta}{r}\frac{\partial}{\partial \theta}\right)
+\frac{1}{r^2\sin^2\theta}\frac{\partial^2}{\partial\phi^2}\\
&=\frac{\partial^2 }{\partial r^2}+\frac{2}{r}\frac{\partial}{\partial r}
+\frac{1}{r^2}\frac{\partial^2}{\partial\theta^2}
+\frac{1}{r^2}\frac{\cos\theta}{\sin\theta}\frac{\partial}{\partial \theta}
+\frac{1}{r^2\sin^2\theta}\frac{\partial^2}{\partial\phi^2}\\
&=\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial}{\partial r}\right)
+\frac{1}{r^2\sin\theta}\frac{\partial}{\partial \theta}\left(
\sin\theta\frac{\partial}{\partial\theta}\right)
+\frac{1}{r^2\sin^2\theta}\frac{\partial^2}{\partial\phi^2}\\
);

として、上で得たのと同じ結果が得られる。

これはこれで大変だけれど、完全に力ずくでやるより見通しが良い。

** こちらもとても参考になる [#d755cebe]

http://irobutsu.a.la9.jp/PhysTips/Lap.html

*** 「第1の方法:変分法を使え。」において [#ya83c81d]

&math(dr, d\theta, d\phi); に対する &math(\bm r); の変化量はそれぞれ、
&math(dr, r\,d\theta, r\sin\theta\,d\phi); で、互いに直交するから、
これらの方向に &math(x',y',z'); 軸を取るのと比べれば、第一感では

 &math(
\nabla^2
&\overset{??}{=}\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}\\
);

などとなりそうなところ、実際には

 &math(
\nabla^2&=\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial}{\partial r}\right)
+\frac{1}{r^2}\frac{1}{\sin\theta}\frac{\partial}{\partial \theta}\left(\sin\theta\frac{\partial}{\partial \theta}\right)
+\frac{1}{r^2\sin^2\theta}\frac{\partial^2}{\partial \phi^2}\\
&\hspace{5mm}\uparrow\hspace{11mm}\uparrow\hspace{21mm}\uparrow\hspace{16mm}\uparrow
);

のように余計な因子が紛れ込むのだが、上記のリンク先ではラプラシアンを
のように余計な因子が紛れ込むのだが、上記のリンク先ではラプラシアンが

 &math(
\nabla^2
&=\frac{1}{r^2\sin\theta}\frac{\partial}{\partial r}\left(r^2\sin\theta\frac{\partial}{\partial r}\right)
+\frac{1}{r^2\sin\theta}\frac{1}{r}\frac{\partial}{\partial \theta}\left(r^2\sin\theta\frac{1}{r}\frac{\partial}{\partial \theta}\right)
+\frac{1}{r^2\sin\theta}\frac{1}{r\sin\theta}\frac{\partial}{\partial \phi}\left(r^2\sin\theta\frac{1}{r\sin\theta}\frac{\partial}{\partial \phi}\right)\\
);

の形に導き、1回微分した後に体積素 &math((r^2\sin\theta)dr\,d\theta\,d\phi); の係数 &math(r^2\sin\theta); を掛け、2回目の微分後に同じ値で割る形になっていることを示している。
の形に導かれる。

「1回目の微分」をした後に体積素 &math((r^2\sin\theta)dr\,d\theta\,d\phi); の係数 &math(r^2\sin\theta); を掛け、「2回目の微分」をした後に同じ値で割る形になっている。

これは覚えやすい。

*** 「第2の方法:ちゃんと基底ベクトルも微分しろ。」において [#aa4b3355]

 &math(
\bm \nabla=\bm e_r\frac{\partial}{\partial r}+\bm e_\theta\frac{1}{r}\frac{\partial}{\partial \theta}+\bm e_\phi\frac{1}{r\sin\theta}\frac{\partial}{\partial \phi}
);

から

 &math(
\nabla^2&=\bm\nabla\cdot\bm\nabla\\
&=\bigg(\bm e_r\frac{\partial}{\partial r}+\bm e_\theta\frac{1}{r}\frac{\partial}{\partial \theta}+\bm e_\phi\frac{1}{r\sin\theta}\frac{\partial}{\partial \phi}\bigg)\cdot
\bigg(\bm e_r\frac{\partial}{\partial r}+\bm e_\theta\frac{1}{r}\frac{\partial}{\partial \theta}+\bm e_\phi\frac{1}{r\sin\theta}\frac{\partial}{\partial \phi}\bigg)
);

のように計算する際、&math(\bm e_r,\bm e_\theta,\bm e_\phi); は直交するため一見すると

 &math(
\nabla^2
&\overset{??}{=}\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}\\
);

となりそうなところ、上記リンク先では

 &math(
\begin{pmatrix}
\frac{\partial}{\partial r} \\
\frac{\partial}{\partial \theta} \\
\frac{\partial}{\partial \phi}
\end{pmatrix}
\begin{pmatrix}
\bm e_r &
\bm e_\theta &
\bm e_\phi
\end{pmatrix}=\begin{pmatrix}
\bm 0 & \bm 0 & \bm 0 \\
\bm e_\theta & -\bm e_r & \bm 0 \\
(\sin\theta)\bm e_\phi & (\cos\theta)\bm e_\phi & -\bm e_r
\end{pmatrix}
);

のように、&math(\bm e_r,\bm e_\theta,\bm e_\phi); 自体が &math(\theta,\phi); の関数であることを考慮しなければならないことを指摘し、

 &math(\bm e_\theta\frac{\partial\bm e_r}{\partial \theta}); の項から &math(\frac{1}{r}\frac{\partial}{\partial r});

 &math(\bm e_\phi\frac{\partial\bm e_r}{\partial \theta}); の項から &math(\frac{1}{r}\frac{\partial}{\partial r}); がもう1つ

 &math(\bm e_\phi\frac{\partial\bm e_\theta}{\partial \theta}); の項から &math(\frac{\cos\theta}{r^2\sin\theta}\frac{\partial}{\partial\theta});

がそれぞれ出ることにより、正しいラプラシアンが得られることを示している。

* 球座標の角運動量演算子 [#nad66919]

&math(
\hat l_x&=-i\hbar\Big(y\frac{\PD}{\PD z}-z\frac{\PD}{\PD y}\Big)\\
&=-i\hbar\bigg[r\sin\theta\sin\phi\Big(\cancel{\cos\theta \frac{\PD}{\PD r}}-\frac{1}{r}\sin\theta \frac{\PD}{\PD \theta}\Big)
-r\cos\theta\Big(\cancel{\sin\theta\sin\phi \frac{\PD}{\PD r}}
+\frac{1}{r}\cos\theta\sin\phi \frac{\PD}{\PD \theta}
+\frac{\cos\phi}{r\sin\theta} \frac{\PD}{\PD \phi}\Big)\bigg]\\
&=i\hbar\Big(\sin\phi\frac{\PD}{\PD\theta}+\frac{\cos\phi}{\tan\theta}\frac{\PD}{\PD\phi}\Big)
);

&math(
\hat l_y&=-i\hbar\Big(z\frac{\PD}{\PD x}-x\frac{\PD}{\PD z}\Big)\\
&=-i\hbar\bigg[r\cos\theta\Big(\cancel{\sin\theta\cos\phi \frac{\PD}{\PD r}}
+\frac{1}{r}\cos\theta\cos\phi \frac{\PD}{\PD \theta}
-\frac{\sin\phi}{r\sin\theta} \frac{\PD}{\PD \phi}\Big)
-r\sin\theta\cos\phi\Big(\cancel{\cos\theta \frac{\PD}{\PD r}}
-\frac{1}{r}\sin\theta \frac{\PD}{\PD \theta}\Big)\bigg]\\
&=i\hbar\Big(-\cos\phi\frac{\PD}{\PD\theta}+\frac{\sin\phi}{\tan\theta}\frac{\PD}{\PD\phi}\Big)
);

&math(
\hat l_z&=-i\hbar\Big(x\frac{\PD}{\PD y}-y\frac{\PD}{\PD x}\Big)\\
&=-i\hbar\bigg[r\sin\theta\cos\phi\Big(\cancel{\sin\theta\sin\phi \frac{\PD}{\PD r}}
+\cancel{\frac{1}{r}\cos\theta\sin\phi \frac{\PD}{\PD \theta}}
+\frac{\cos\phi}{r\sin\theta} \frac{\PD}{\PD \phi}\Big)\\
&\hspace{1cm}-r\sin\theta\sin\phi\Big(\cancel{\sin\theta\cos\phi \frac{\PD}{\PD r}}
+\cancel{\frac{1}{r}\cos\theta\cos\phi \frac{\PD}{\PD \theta}}
-\frac{\sin\phi}{r\sin\theta} \frac{\PD}{\PD \phi}\Big)\bigg]\\
&=-i\hbar\frac{\PD}{\PD\phi}
);

&math(
\hat{\bm l}^2&=\hat l_x^2+\hat l_y^2+\hat l_z^2\\
&=
-\hbar^2\Big(\sin\phi\frac{\PD}{\PD\theta}+\frac{\cos\phi}{\tan\theta}\frac{\PD}{\PD\phi}\Big)^2
-\hbar^2\Big(-\cos\phi\frac{\PD}{\PD\theta}+\frac{\sin\phi}{\tan\theta}\frac{\PD}{\PD\phi}\Big)^2
-\hbar^2\frac{\PD^2}{\PD\phi^2}\\
&=
-\hbar^2\Big(
\sin^2\phi\frac{\PD^2}{\PD\theta^2}
-\cancel{\frac{\sin\phi\cos\phi}{\sin^2\theta}\frac{\PD}{\PD\phi}}
+\cancel{\frac{2\sin\phi\cos\phi}{\tan\theta}\frac{\PD}{\PD\theta}\frac{\PD}{\PD\phi}}
+\frac{\cos^2\phi}{\tan\theta}\frac{\PD}{\PD\theta}
+\cancel{\frac{-\sin\phi\cos\phi}{\tan^2\theta}\frac{\PD}{\PD\phi}}
+\frac{\cos^2\phi}{\tan^2\theta}\frac{\PD^2}{\PD\phi^2}
\Big)\\
&\ \ \ -\hbar^2\Big(
\cos^2\phi\frac{\PD^2}{\PD^2\theta}
+\cancel{\frac{\sin\phi\cos\phi}{\sin^2\theta}\frac{\PD}{\PD\phi}}
-\cancel{\frac{2\sin\phi\cos\phi}{\tan\theta}\frac{\PD}{\PD\theta}\frac{\PD}{\PD\phi}}
+\frac{\sin^2\phi}{\tan\theta}\frac{\PD}{\PD\theta}
+\cancel{\frac{\sin\phi\cos\phi}{\tan^2\theta}\frac{\PD}{\PD\phi}}
+\frac{\sin^2\phi}{\tan^2\theta}\frac{\PD^2}{\PD\phi^2}
\Big)\\
&\ \ \ -\hbar^2\frac{\PD^2}{\PD\phi^2}\\
&=-\hbar^2\Big[\frac{\PD^2}{\PD\theta^2}+\frac{1}{\tan\theta}\frac{\PD}{\PD\theta}+\Big(\frac{1}{\tan^2\theta}+1\Big)\frac{\PD^2}{\PD\phi^2}\Big]\\
&=-\hbar^2\Big(\frac{\PD^2}{\PD\theta^2}+\frac{\cos\theta}{\sin\theta}\frac{\PD}{\PD\theta}+\frac{1}{\sin^2\theta}\frac{\PD^2}{\PD\phi^2}\Big)\\
&=-\hbar^2\Big[\frac{1}{\sin\theta}\frac{\PD}{\PD\theta}\Big(\sin\theta\frac{\PD}{\PD\theta}\Big)+\frac{1}{\sin^2\theta}\frac{\PD^2}{\PD\phi^2}\Big]\\
&=-\hbar^2\hat\Lambda
);

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