ベクトル解析12 極座標 勾配、発散、ラプラシアン

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ベクトル解析12 極座標 勾配、発散、ラプラシアン

 

極座標

基本事項の整理。直交座標との関係性。

 

\(x=r\sin\theta\cos\phi\)、\(y=r\sin\theta\sin\phi\)、\(z=r\cos\theta\)

 

上から分かる以下3つの式。

\(r=\sqrt{x^2+y^2+z^2}\)、\(\phi=\tan^{-1} \displaystyle\frac{y}{x}\)、\(\theta=\cos^{-1}\displaystyle\frac{z}{\sqrt{x^2+y^2+z^2}}\)

 

勾配

ここから、勾配の成分を一つ一つ計算していく。

\(\nabla=\displaystyle\frac{\partial}{\partial x}\boldsymbol{e_{x}}+\displaystyle\frac{\partial}{\partial y}\boldsymbol{e_{y}}+\displaystyle\frac{\partial}{\partial \phi}\boldsymbol{e_{\phi}}\) のそれぞれの成分を計算していく。

 

x成分

\(\displaystyle\frac{\partial}{\partial x}=\displaystyle\frac{\partial r}{\partial x}\displaystyle\frac{\partial}{\partial r}+\displaystyle\frac{\partial \theta}{\partial x}\displaystyle\frac{\partial}{\partial \theta}+\displaystyle\frac{\partial \phi}{\partial x}\displaystyle\frac{\partial}{\partial \phi}\)

 

ここで、以下の3式が成立する。

\(\displaystyle\frac{\partial r}{\partial x}=\displaystyle\frac{\partial}{\partial x}\sqrt{x^2+y^2+z^2}=\displaystyle\frac{x}{\sqrt{x^2+y^2+z^2}}=\displaystyle\frac{x}{r}=\sin\theta\cos\phi\)

 

\(\displaystyle\frac{\partial \theta}{\partial x}=\displaystyle\frac{\partial}{\partial x}\cos^{-1}\displaystyle\frac{z}{\sqrt{x^2+y^2+z^2}}=\displaystyle\frac{1}{r\sin\theta}\cdot\displaystyle\frac{r\cos\theta}{r^2}\cdot(r\sin\theta\cos\phi)=\displaystyle\frac{\cos\theta\cos\phi}{r}\)

 

\(\displaystyle\frac{\partial \phi}{\partial x}=-\displaystyle\frac{y}{x^2+y^2}=-\displaystyle\frac{r\sin\theta\sin\phi}{r^2\sin^2 \theta}=-\displaystyle\frac{\sin\phi}{r\sin\theta}\)

 

これらをもとの式に代入すると

\(\displaystyle\frac{\partial}{\partial x}=\sin\theta\cos\phi \displaystyle\frac{\partial}{\partial r}+\displaystyle\frac{1}{r}\cos\theta\cos\phi\displaystyle\frac{\partial}{\partial \theta}-\displaystyle\frac{\sin\phi}{r\sin\theta}\displaystyle\frac{\partial}{\partial \phi}\)

 

y成分

上と同様に計算する。

\(\displaystyle\frac{\partial}{\partial y}=\sin\theta\sin\phi \displaystyle\frac{\partial}{\partial r}+\displaystyle\frac{1}{r}\cos\theta\sin\phi\displaystyle\frac{\partial}{\partial \theta}+\displaystyle\frac{\cos\phi}{r\sin\theta}\displaystyle\frac{\partial}{\partial \phi}\)

 

z成分

上と同様に計算する。

\(\displaystyle\frac{\partial}{\partial z}=\cos\theta\displaystyle\frac{\partial}{\partial r}-\displaystyle\frac{\sin\theta}{r}\displaystyle\frac{\partial}{\partial \theta}\)

 

まとめ



\(\nabla=\displaystyle\frac{\partial}{\partial x}\boldsymbol{e_{x}}+\displaystyle\frac{\partial}{\partial y}\boldsymbol{e_{y}}+\displaystyle\frac{\partial}{\partial z}\boldsymbol{e_{z}}\)を計算していく。

 

ここで基底変換に対して以下の式が成立。


\(\boldsymbol{e_{x}}=\sin\theta\cos\phi\boldsymbol{e_{r}}+\cos\theta\cos\phi\boldsymbol{e_{\theta}}-\sin\phi\boldsymbol{e_{\phi}}\)

 

\(\boldsymbol{e_{y}}=\sin\theta\sin\phi\boldsymbol{e_{r}}+\cos\theta\sin\phi\boldsymbol{e_{\theta}}+\cos\phi\boldsymbol{e_{\phi}}\)


\(\boldsymbol{e_{z}}=\cos\theta\boldsymbol{e_{r}}-\sin\theta\boldsymbol{e_{\theta}}\)

 

これらを、上の式に代入すると

 

\(\nabla=\)

\(\biggl(\sin\theta\cos\phi \displaystyle\frac{\partial}{\partial r}+\displaystyle\frac{\cos\theta\cos\phi}{r}\displaystyle\frac{\partial}{\partial \theta}-\displaystyle\frac{\sin\phi}{r\sin\theta}\displaystyle\frac{\partial}{\partial \phi}\biggr)(\sin\theta\cos\phi\boldsymbol{e_{r}}+\cos\theta\cos\phi\boldsymbol{e_{\theta}}-\sin\phi\boldsymbol{e_{\phi}})\)

\(+\biggl(\sin\theta\sin\phi \displaystyle\frac{\partial}{\partial r}+\displaystyle\frac{\cos\theta\sin\phi}{r}\displaystyle\frac{\partial}{\partial \theta}+\displaystyle\frac{\cos\phi}{r\sin\theta}\displaystyle\frac{\partial}{\partial \phi}\biggr)(\sin\theta\sin\phi\boldsymbol{e_{r}}+\cos\theta\sin\phi\boldsymbol{e_{\theta}}+\cos\phi\boldsymbol{e_{\phi}})\)

\(+\biggl(\cos\theta\displaystyle\frac{\partial}{\partial r}-\displaystyle\frac{\sin\theta}{r}\displaystyle\frac{\partial}{\partial \theta}\biggr)(\cos\theta\boldsymbol{e_{r}}-\sin\theta\boldsymbol{e_{\theta}})\)


\(=\displaystyle\frac{\partial}{\partial r}\boldsymbol{e_{r}}+\displaystyle\frac{1}{r}\displaystyle\frac{\partial}{\partial \theta}\boldsymbol{e_{\theta}}+\displaystyle\frac{1}{r\sin\theta}\displaystyle\frac{\partial}{\partial\phi}\boldsymbol{e_{\phi}}\)

 

結果

 

\[\nabla f=\displaystyle\frac{\partial f}{\partial r}\boldsymbol{e_{r}}+\displaystyle\frac{1}{r}\displaystyle\frac{\partial f}{\partial \theta}\boldsymbol{e_{\theta}}+\displaystyle\frac{1}{r\sin\theta}\displaystyle\frac{\partial f}{\partial\phi}\boldsymbol{e_{\phi}}\]

 

発散

 

\(\boldsymbol{A}=A_{r}\boldsymbol{e_{r}}+A_{\theta}\boldsymbol{e_{\theta}}+A_{\phi}\boldsymbol{e_{\phi}}\)

 

ここで発散は

\(\nabla\cdot\boldsymbol{A}=\displaystyle\frac{\partial A_{x}}{\partial x}+\displaystyle\frac{\partial A_{y}}{\partial y}+\displaystyle\frac{\partial A_{z}}{\partial z}\)

 

右辺第一項

 

\(\displaystyle\frac{\partial A_{x}}{\partial x}=\displaystyle\frac{\partial r}{\partial x}\displaystyle\frac{\partial A_{x}}{\partial r}+\displaystyle\frac{\partial \theta}{\partial x}
\displaystyle\frac{\partial A_{x}}{\partial \theta}+\displaystyle\frac{\partial \phi}{\partial x}\displaystyle\frac{\partial A_{x}}{\partial \phi}\)


\(=
\sin\theta\cos\phi\displaystyle\frac{\partial A_{x}}{\partial r}+\displaystyle\frac{1}{r}\cos\theta\cos\phi\displaystyle\frac{\partial A_{x}}{\partial\theta}-\displaystyle\frac{\sin\phi}{r\sin\theta}\displaystyle\frac{\partial A_{x}}{\partial \theta}\)

 

\(=\sin\theta\cos\phi\theta\biggl(\sin\theta\cos\phi\displaystyle\frac{\partial A_{r}}{\partial r}+\cos\theta\cos\phi\displaystyle\frac{A_{\theta}}{\partial r}-\sin\phi\displaystyle\frac{\partial A_{\phi}}{\partial r}\biggr)\)


\(+\displaystyle\frac{1}{r}\cos\theta\cos\phi\biggl(\cos\theta\cos\phi A_{r}+\sin\theta\cos\phi\displaystyle\frac{\partial A_{r}}{\partial \theta}-\sin\theta\cos\phi A_{\theta}+\cos\theta\cos\phi\displaystyle\frac{A_{\theta}}{\partial \theta}-\sin\phi\displaystyle\frac{\partial A_{\phi}}{\partial \theta}\biggr)\)


\(-\displaystyle\frac{\sin\phi}{r\sin\theta}\biggl(-\sin\theta\sin\phi A_{r}+\sin\theta\cos\phi\displaystyle\frac{\partial A_{r}}{\partial \phi}-\cos\theta\sin\phi A_{\theta}+\cos\theta\cos\phi\displaystyle\frac{A_{\theta}}{\partial \phi}-\cos\phi A_{\phi}-\sin\phi\displaystyle\frac{\partial A_{\phi}}{\partial \phi}\biggr)\)

 


右辺第二項

第一項と同様に計算する。

 

\(\displaystyle\frac{\partial A_{y}}{\partial y}=\sin\theta\sin\phi\biggl(\displaystyle\frac{\partial A_{r}}{\partial r}\sin\theta\sin\phi+\displaystyle\frac{\partial A_{\theta}}{\partial r}\cos\theta\sin\phi+\displaystyle\frac{\partial A_{\phi}}{\partial r}\cos\phi\biggr)\)

 

\(+\displaystyle\frac{\cos\theta\sin\phi}{r}\biggl(A_{r}\cos\theta\sin\phi+\sin\theta\sin\phi\displaystyle\frac{\partial A_{r}}{\partial \theta}-A_{\theta}\sin\theta\sin\phi+\displaystyle\frac{\partial A_{\theta}}{\partial \theta}\cos\theta\sin\phi+\displaystyle\frac{\partial A_{\phi}}{\partial \theta}\cos\phi\biggr)\)

 

\(+\displaystyle\frac{\cos\phi}{r\sin\theta}\biggl(A_{r}\sin\theta\cos\phi+\displaystyle\frac{\partial A_{r}}{\partial \theta}\sin\theta\sin\phi+A_{\theta}\cos\theta\cos\phi+\displaystyle\frac{\partial A_{\theta}}{\partial \phi}\cos\theta\sin\phi-A_{\phi}\sin\phi+\displaystyle\frac{\partial A_{\phi}}{\partial \phi}\cos\phi\biggr)\)

 

右辺第三項

第一項と同様に計算する。

 

\(\displaystyle\frac{\partial A_{z}}{\partial z}=\cos\theta\biggl(\displaystyle\frac{\partial A_{r}}{\partial r}\cos\theta-\displaystyle\frac{\partial A_{\theta}}{\partial r}\sin\theta\biggr)\)

 

\(-\displaystyle\frac{\sin\theta}{r}\biggl(-A_{r}\sin\theta+\displaystyle\frac{\partial A_{r}}{\partial\theta}\cos\theta-A_{\theta}\cos\theta-\displaystyle\frac{\partial A_{\theta}}{\partial\theta}\sin\theta\biggr)\)

 

まとめ


これらの結果を、発散の式に代入して整理すると

 

\[\nabla\cdot\boldsymbol{A}=\displaystyle\frac{\partial A_{r}}{\partial r}+\displaystyle\frac{2}{r}E_{r}+\displaystyle\frac{\cos\theta}{r\sin\theta}E_{\theta}+\displaystyle\frac{1}{r}\displaystyle\frac{\partial A_{\theta}}{\partial\theta}+\displaystyle\frac{1}{r\sin\theta}\displaystyle\frac{\partial A_{\phi}}{\partial \phi}\]
\[=\displaystyle\frac{1}{r^2}\displaystyle\frac{\partial}{\partial r}(r^2 A_{r})+\displaystyle\frac{1}{r\sin\theta}\displaystyle\frac{\partial}{\partial \theta}(E_{\theta}\sin\theta)+\displaystyle\frac{1}{r\sin\theta}\displaystyle\frac{\partial A_{\phi}}{\partial \phi}\]

 

ラプラシアン


\(\Delta f=\nabla\cdot(\nabla f)=\nabla\cdot\biggl(\displaystyle\frac{\partial f}{\partial r}\boldsymbol{e_{r}}+\displaystyle\frac{1}{r}\displaystyle\frac{\partial f}{\partial \theta}\boldsymbol{e_{\theta}}+\displaystyle\frac{1}{r\sin\theta}\displaystyle\frac{\partial f}{\partial \phi}\boldsymbol{e_{\phi}}\biggr)\)


ただし、勾配の式を利用した。ここで、

 

\(A_{r}=\displaystyle\frac{\partial f}{\partial r}\)、\(A_{\theta}=\displaystyle\frac{1}{r}\displaystyle\frac{\partial f}{\partial \theta}\)、\(A_{z}=\displaystyle\frac{1}{r\sin\theta}\displaystyle\frac{\partial f}{\partial \phi}\)として、上の発散の式を適用すると

 

\(\Delta f\)\(=\displaystyle\frac{1}{r^2}\displaystyle\frac{\partial}{\partial r}\biggl(r^2\displaystyle\frac{\partial f}{\partial r}\biggr)+\displaystyle\frac{1}{r\sin\theta}\displaystyle\frac{\partial}{\partial \theta}\biggl(\displaystyle\frac{1}{r}\displaystyle\frac{\partial f}{\partial \theta}\sin\theta\biggr)+\displaystyle\frac{1}{r\sin\theta}\displaystyle\frac{\partial}{\partial \phi}\biggl(\displaystyle\frac{1}{r\sin\theta}\displaystyle\frac{\partial f}{\partial \phi}\biggr)\)


\(=\displaystyle\frac{1}{r^2}\displaystyle\frac{\partial}{\partial r}\biggl(r^2\displaystyle\frac{\partial f}{\partial r}\biggr)+\displaystyle\frac{1}{r^2\sin\theta}\displaystyle\frac{\partial}{\partial \theta}\biggl(\sin\theta\displaystyle\frac{\partial f}{\partial \theta}\biggr)+\displaystyle\frac{1}{r^2\sin^2 \theta}\displaystyle\frac{\partial^2 f}{\partial \phi^2}\)

 

 

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