球座標における微分演算子/メモ の履歴(No.5)
更新目次†
演習:偏微分の計算†
解答†
(1)
より、 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)
より、
、 、 、
&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)
より、
、 、 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†
&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†
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、 から 平面に垂線を下ろした点と原点との距離を と書き、
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、 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 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†
http://irobutsu.a.la9.jp/PhysTips/Lap.html
「第1の方法:変分法を使え。」において†
に対する の変化量はそれぞれ、 であるから、第一感で
&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
&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 の係数 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。
「第2の方法:ちゃんと基底ベクトルも微分しろ。」において†
&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、 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、 自体が 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、
の項から
の項から がもう1つ
の項から
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†
&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 );