曲線座標系におけるベクトル関数のカール
定理
曲線座標系でのベクトル関数$\mathbf{F}=\mathbf{F}(q_{1},q_{2},q_{3})=F_{1}\hat{\mathbf{q}}_{1}+F_{2}\hat{\mathbf{q}}_{2}+F_{3}\hat{\mathbf{q}}_{3}$のカールは以下の通りだ。
$$ \begin{align*} \nabla \times \mathbf{F} &= \frac{\hat{\mathbf{q}}_{1}}{h_{2}h_{3}}\left( \dfrac{\partial (F_{3}h_{3})}{\partial q_{2}} - \dfrac{\partial (F_{2}h_{2})}{\partial q_{3}} \right) + \frac{\hat{\mathbf{q}}_{2}}{h_{1}h_{3}}\left( \dfrac{\partial (F_{1}h_{1})}{\partial q_{3}} - \dfrac{\partial (F_{3}h_{3})}{\partial q_{1}} \right) \\ &\quad+ \frac{\hat{\mathbf{q}}_{3}}{h_{1}h_{2}}\left( \dfrac{\partial (F_{2}h_{2})}{\partial q_{1}} - \dfrac{\partial (F_{1}h_{1})}{\partial q_{2}} \right) \\ &= \frac{1}{h_{1}h_{2}h_{3}} \begin{vmatrix} h_{1}\hat{\mathbf{q}}_{1} & h_{2}\hat{\mathbf{q}}_{2} & h_{3}\hat{\mathbf{q}}_{3} \\[0.5em] \dfrac{\partial }{\partial q_{1}} & \dfrac{\partial }{\partial q_{2}} & \dfrac{\partial }{\partial q_{3}} \\[1em] F_{1}h_{1} & F_{2}h_{2} & F_{3}h_{3} \end{vmatrix} \end{align*} $$
$h_{i}$はスケールファクターだ。
公式
直交座標系:
$$ h_{1}=h_{2}=h_{3}=1 $$
$$ \nabla \times \mathbf{F} = \left( \dfrac{\partial F_{z}}{\partial y}-\dfrac{\partial F_{y}}{\partial z} \right)\hat{\mathbf{x}}+ \left( \dfrac{\partial F_{x}}{\partial z}-\dfrac{\partial F_{z}}{\partial x} \right)\hat{\mathbf{y}}+ \left( \dfrac{\partial F_{y}}{\partial x}-\dfrac{\partial F_{x}}{\partial y} \right)\hat{\mathbf{z}} $$
円筒座標系:
$$ h_{1}=1,\quad h_{2}=\rho,\quad h_{3}=1 $$
$$ \nabla \times \mathbf{F} = \left(\frac{1}{\rho}\dfrac{\partial F_{z}}{\partial \phi} - \dfrac{\partial F_{\phi}}{\partial z} \right)\boldsymbol{\hat \rho} + \left(\dfrac{\partial F_{\rho}}{\partial z} - \dfrac{\partial F_{z}}{\partial \rho} \right)\boldsymbol{\hat \phi} + \frac{1}{\rho}\left(\dfrac{\partial (\rho F_{\phi})}{\partial \rho} - \dfrac{\partial F_{\rho}}{\partial \phi} \right)\mathbf{\hat{\mathbf{z}}} $$
球座標系:
$$ h_{1}=1,\quad h_{2}=r\quad, h_{3}=r\sin\theta $$
$$ \nabla \times \mathbf{F} = \frac{1}{r\sin\theta}\left(\dfrac{\partial (F_{\phi} \sin\theta)}{\partial \theta} - \dfrac{\partial F_{\theta}}{\partial \phi} \right)\mathbf{\hat r} + \frac{1}{r}\left(\frac{1}{\sin\theta}\dfrac{\partial F_{r}}{\partial \phi} - \dfrac{\partial (r F_{\phi})}{\partial r} \right)\boldsymbol{\hat \theta} + \frac{1}{r}\left(\dfrac{\partial (r F_{\theta})}{\partial r} - \dfrac{\partial F_{r}}{\partial \theta} \right)\boldsymbol{\hat \phi} $$
導出
方法1 1
曲線座標系を$(q_{1}, q_{2}, q_{3})$としよう。
$$ \mathbf{F} = F_{1}\hat{\mathbf{q}}_{1} + F_{2}\hat{\mathbf{q}}_{2} + F_{3}\hat{\mathbf{q}}_{3} $$
カールは線形性を持っているので、
$$ \begin{equation} \nabla \times \mathbf{F} = \nabla \times \left( F_{1}\hat{\mathbf{q}}_{1} + F_{2}\hat{\mathbf{q}}_{2} + F_{3}\hat{\mathbf{q}}_{3} \right) = \sum_{i=1}^{3} \nabla \times (F_{i}\hat{\mathbf{q}}_{i}) \end{equation} $$
$$ \nabla f= \frac{1}{h_{1}}\frac{ \partial f }{ \partial q_{1} } \hat{\mathbf{q}}_{1} + \frac{1}{h_{2}}\frac{ \partial f }{ \partial q _{2}}\hat{\mathbf{q}}_{2}+\frac{1}{h_{3}}\frac{ \partial f }{ \partial q_{3} } \hat{\mathbf{q}}_{3}=\sum \limits _{i=1} ^{3}\frac{1}{h_{i}}\frac{ \partial f}{ \partial q_{i}}\hat{\mathbf{q}}_{i} $$
グラディエントの公式により、以下を得る。
$$ \nabla q_{i} = \sum \limits _{m=1} ^{3}\frac{1}{h_{m}}\frac{ \partial q_{i}}{ \partial q_{m}}\hat{\mathbf{q}}_{m} = \frac{1}{h_{i}}\hat{\mathbf{q}}_{i} $$
$$ \begin{equation} \implies h_{i}\nabla q_{i} = \hat{\mathbf{q}}_{i} \end{equation} $$
$$ \nabla \times (f \mathbf{A}) = (\nabla f) \times \mathbf{A} + f (\nabla \times \mathbf{A}) $$
$(2)$を$(1)$に代入し、上の乗算公式を適用すると、以下を得る。
$$ \begin{align*} \nabla \times (F_{i}\hat{\mathbf{q}}_{i}) &= \nabla \times (F_{i}h_{i}\nabla q_{i}) \\ &= [\nabla (F_{i}h_{i})] \times \nabla q_{i} + F_{i}h_{i} \nabla \times (\nabla q_{i}) \\ &= [\nabla (F_{i}h_{i})] \times \nabla q_{i} \end{align*} $$
最後の等号はグラディエントのカールは$\mathbf{0}$であるため成立する。式に再度$(2)$を代入し、グラディエントを展開してみると、
$$ \begin{align*} [\nabla (F_{i}h_{i})] \times \nabla q_{i} &= \left[ \sum_{m=1}^{3} \dfrac{1}{h_{m}}\dfrac{\partial (F_{i}h_{i})}{\partial q_{m}} \hat{\mathbf{q}}_{m} \right] \times \frac{1}{h_{i}}\hat{\mathbf{q}}_{i} \\ &= \dfrac{1}{h_{j}}\dfrac{\partial (F_{i}h_{i})}{\partial q_{j}} \hat{\mathbf{q}}_{j} \times \frac{1}{h_{i}}\hat{\mathbf{q}}_{i} + \dfrac{1}{h_{k}}\dfrac{\partial (F_{i}h_{i})}{\partial q_{k}} \hat{\mathbf{q}}_{k} \times \frac{1}{h_{i}}\hat{\mathbf{q}}_{i} \\ &= \dfrac{1}{h_{i}h_{k}}\dfrac{\partial (F_{i}h_{i})}{\partial q_{k}} \hat{\mathbf{q}}_{j} - \dfrac{1}{h_{i}h_{j}}\dfrac{\partial (F_{i}h_{i})}{\partial q_{j}} \hat{\mathbf{q}}_{k} \end{align*} $$
ここで、インデックスを$\hat{\mathbf{q}}_{i} \times \hat{\mathbf{q}}_{j} = \hat{\mathbf{q}}_{k}$とした。これを$(1)$に代入すると、
$$ \begin{align*} \nabla \times \mathbf{F} &= \left( \dfrac{1}{h_{1}h_{3}}\dfrac{\partial (F_{1}h_{1})}{\partial q_{3}} \hat{\mathbf{q}}_{2} - \dfrac{1}{h_{1}h_{2}}\dfrac{\partial (F_{1}h_{1})}{\partial q_{2}} \hat{\mathbf{q}}_{3} \right) + \left( \dfrac{1}{h_{2}h_{1}}\dfrac{\partial (F_{2}h_{2})}{\partial q_{1}} \hat{\mathbf{q}}_{3} - \dfrac{1}{h_{2}h_{3}}\dfrac{\partial (F_{2}h_{2})}{\partial q_{3}} \hat{\mathbf{q}}_{1} \right) \\ &\quad+ \left( \dfrac{1}{h_{3}h_{2}}\dfrac{\partial (F_{3}h_{3})}{\partial q_{2}} \hat{\mathbf{q}}_{1} - \dfrac{1}{h_{3}h_{1}}\dfrac{\partial (F_{3}h_{3})}{\partial q_{1}} \hat{\mathbf{q}}_{2} \right) \\ &= \frac{h_{1}\hat{\mathbf{q}}_{1}}{h_{1}h_{2}h_{3}}\left( \dfrac{\partial (F_{3}h_{3})}{\partial q_{2}} - \dfrac{\partial (F_{2}h_{2})}{\partial q_{3}} \right) + \frac{h_{2}\hat{\mathbf{q}}_{2}}{h_{1}h_{2}h_{3}}\left( \dfrac{\partial (F_{1}h_{1})}{\partial q_{3}} - \dfrac{\partial (F_{3}h_{3})}{\partial q_{1}} \right) \\ &\quad+ \frac{h_{3}\hat{\mathbf{q}}_{3}}{h_{1}h_{2}h_{3}}\left( \dfrac{\partial (F_{2}h_{2})}{\partial q_{1}} - \dfrac{\partial (F_{1}h_{1})}{\partial q_{2}} \right) \\ &= \frac{1}{h_{1}h_{2}h_{3}} \begin{vmatrix} h_{1}\hat{\mathbf{q}}_{1} & h_{2}\hat{\mathbf{q}}_{2} & h_{3}\hat{\mathbf{q}}_{3} \\[0.5em] \dfrac{\partial }{\partial q_{1}} & \dfrac{\partial }{\partial q_{2}} & \dfrac{\partial }{\partial q_{3}} \\[1em] F_{1}h_{1} & F_{2}h_{2} & F_{3}h_{3} \end{vmatrix} \end{align*} $$
■
方法2 2
$\nabla \times \mathbf{F}$の最初の成分を求めるために、$q_{1}$の座標が定数である曲面上の閉じた曲面を考えよう。
ここで、$\odot$は平面を突き抜ける方向を意味する。それでは$d\boldsymbol{\sigma} = h_{2}h_{3} dq_{2}dq_{3}\hat{\mathbf{q}}_{1}$だから、
$$ \int \nabla \times \mathbf{F} \cdot d\boldsymbol{\sigma} \approx (\nabla \times \mathbf{F})h_{2}h_{3}dq_{2}dq_{3} \cdot \hat{\mathbf{q}}_{1} $$
$$ \int_{\mathcal{S}} (\nabla \times \mathbf{v} )\cdot d\mathbf{a} = \oint_{\mathcal{P}} \mathbf{v} \cdot d\mathbf{l} $$
ストークスの定理により、以下を得る。
$$ (\nabla \times \mathbf{F})h_{2}h_{3}dq_{2}dq_{3} \cdot \hat{\mathbf{q}}_{1} = \oint \mathbf{F} \cdot d \mathbf{r} $$
したがって、$\nabla \times \mathbf{F}$の最初の成分は、以下を計算して得られる。
$$ (\nabla \times \mathbf{F})_{1} = (\nabla \times \mathbf{F}) \cdot \hat{\mathbf{q}}_{1} = \frac{1}{h_{2}h_{3}dq_{2}dq_{3}}\oint \mathbf{F} \cdot d \mathbf{r} $$
閉曲線積分を$\textcircled{1}$~$\textcircled{4}$に分けて考えよう。
$$ \oint \mathbf{F} \cdot d \mathbf{r} = \int_{\textcircled{1}} \mathbf{F} \cdot d \mathbf{r} + \int_{\textcircled{2}} \mathbf{F} \cdot d \mathbf{r} + \int_{\textcircled{3}} \mathbf{F} \cdot d \mathbf{r} + \int_{\textcircled{4}} \mathbf{F} \cdot d \mathbf{r} $$
経路$\textcircled{1}$に対する積分を求めると、
$$ \int_{\textcircled{1}} \mathbf{F} \cdot d \mathbf{r} \approx \mathbf{F}(q_{2},q_{3}) \cdot h_{2}(q_{2},q_{3}) dq_{2}\hat{\mathbf{q}}_{2} = F_{2}h_{2}dq_{2} $$
経路$\textcircled{2}$は、
$$ \begin{align*} \int_{\textcircled{2}} \mathbf{F} \cdot d \mathbf{r} &\approx \mathbf{F}(q_{2}+dq_{2},q_{3}) \cdot h_{3}(q_{2}+dq_{2},q_{3}) dq_{3}\hat{\mathbf{q}}_{3} \\ &= F_{3}(q_{2}+dq_{2},q_{3})h_{3}(q_{2}+dq_{2},q_{3})dq_{3} \\ &\approx \left[ F_{3}h_{3} + \dfrac{\partial (F_{3}h_{3})}{\partial q_{2}}dq_{2} \right]dq_{3} \end{align*} $$
最後の行では、テイラー近似を使用した。
$$ f(x+dx) \approx f(x) + f^{\prime}(x)dx $$
同様に、残りの積分を計算すると、
$$ \begin{align*} \int_{\textcircled{3}} \mathbf{F} \cdot d \mathbf{r} &= -\int_{-\textcircled{3}} \mathbf{F} \cdot d \mathbf{r} \\ &\approx -\mathbf{F}(q_{2},q_{3}+dq_{3}) \cdot h_{2}(q_{2},q_{3}+dq_{3}) dq_{2}\hat{\mathbf{q}}_{2} \\ &= -F_{2}(q_{2},q_{3}+dq_{3})h_{2}(q_{2},q_{3}+dq_{3})dq_{2} \\ &\approx -\left[ F_{2}h_{2} + \dfrac{\partial (F_{2}h_{2})}{\partial q_{3}}dq_{3} \right]dq_{2} \end{align*} $$
$$ \int_{\textcircled{4}} \mathbf{F} \cdot d \mathbf{r} = -\int_{-\textcircled{4}} \mathbf{F} \cdot d \mathbf{r} \approx \mathbf{F}(q_{2},q_{3}) \cdot h_{3}(q_{2},q_{3}) dq_{3}\hat{\mathbf{q}}_{3} = F_{3}h_{3}dq_{3} $$
したがって、
$$ \begin{align*} (\nabla \times \mathbf{F})_{1} &= \dfrac{1}{h_{2}h_{3}dq_{2}dq_{3}}\oint \mathbf{F} \cdot d \mathbf{r} \\ &= \dfrac{1}{h_{2}h_{3}dq_{2}dq_{3}} \left[ \dfrac{\partial (F_{3}h_{3})}{\partial q_{2}}dq_{2}dq_{3} - \dfrac{\partial (F_{2}h_{2})}{\partial q_{3}}dq_{3} dq_{2} \right]\\ &= \dfrac{1}{h_{2}h_{3}} \left[ \dfrac{\partial (F_{3}h_{3})}{\partial q_{2}} - \dfrac{\partial (F_{2}h_{2})}{\partial q_{3}} \right]\\ \end{align*} $$
二番目と三番目の成分も同じ方法で得る。
■
参照
https://math.libretexts.org/Bookshelves/Differential_Equations/Introduction_to_Partial_Differential_Equations_(Herman)/06%3A_Problems_in_Higher_Dimensions/6.09%3A_Curvilinear_Coordinates ↩︎
George B. Arfken, MATHEMATICAL METHODS FOR PHYSICISTS (6th Edition), p112 ↩︎