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曲線座標系におけるベクトル関数のカール 📂数理物理学

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F=F(q1,q2,q3)=F1q^1+F2q^2+F3q^3\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}カールは以下の通りだ。

×F=q^1h2h3((F3h3)q2(F2h2)q3)+q^2h1h3((F1h1)q3(F3h3)q1)+q^3h1h2((F2h2)q1(F1h1)q2)=1h1h2h3h1q^1h2q^2h3q^3q1q2q3F1h1F2h2F3h3 \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*}

hih_{i}スケールファクターだ。

公式

  • 直交座標系:

    h1=h2=h3=1 h_{1}=h_{2}=h_{3}=1

    ×F=(FzyFyz)x^+(FxzFzx)y^+(FyxFxy)z^ \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:

    h1=1,h2=ρ,h3=1 h_{1}=1,\quad h_{2}=\rho,\quad h_{3}=1

    ×F=(1ρFzϕFϕz)ρ^+(FρzFzρ)ϕ^+1ρ((ρFϕ)ρFρϕ)z^ \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:

    h1=1,h2=r,h3=rsinθ h_{1}=1,\quad h_{2}=r\quad, h_{3}=r\sin\theta

    ×F=1rsinθ((Fϕsinθ)θFθϕ)r^+1r(1sinθFrϕ(rFϕ)r)θ^+1r((rFθ)rFrθ)ϕ^ \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(q1,q2,q3)(q_{1}, q_{2}, q_{3})としよう。

F=F1q^1+F2q^2+F3q^3 \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、

×F=×(F1q^1+F2q^2+F3q^3)=i=13×(Fiq^i) \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}

曲線座標系でのグラディエント

f=1h1fq1q^1+1h2fq2q^2+1h3fq3q^3=i=131hifqiq^i \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。

qi=m=131hmqiqmq^m=1hiq^i \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}

    hiqi=q^i \begin{equation} \implies h_{i}\nabla q_{i} = \hat{\mathbf{q}}_{i} \end{equation}

デル演算子を含む乗算公式

×(fA)=(f)×A+f(×A) \nabla \times (f \mathbf{A}) = (\nabla f) \times \mathbf{A} + f (\nabla \times \mathbf{A})

(2)(2)(1)(1)に代入し、上の乗算公式を適用すると、以下を得る。

×(Fiq^i)=×(Fihiqi)=[(Fihi)]×qi+Fihi×(qi)=[(Fihi)]×qi \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0\mathbf{0}であるため成立する。式に再度(2)(2)を代入し、グラディエントを展開してみると、

[(Fihi)]×qi=[m=131hm(Fihi)qmq^m]×1hiq^i=1hj(Fihi)qjq^j×1hiq^i+1hk(Fihi)qkq^k×1hiq^i=1hihk(Fihi)qkq^j1hihj(Fihi)qjq^k \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q^i×q^j=q^k\hat{\mathbf{q}}_{i} \times \hat{\mathbf{q}}_{j} = \hat{\mathbf{q}}_{k}とした。これを(1)(1)に代入すると、

×F=(1h1h3(F1h1)q3q^21h1h2(F1h1)q2q^3)+(1h2h1(F2h2)q1q^31h2h3(F2h2)q3q^1)+(1h3h2(F3h3)q2q^11h3h1(F3h3)q1q^2)=h1q^1h1h2h3((F3h3)q2(F2h2)q3)+h2q^2h1h2h3((F1h1)q3(F3h3)q1)+h3q^3h1h2h3((F2h2)q1(F1h1)q2)=1h1h2h3h1q^1h2q^2h3q^3q1q2q3F1h1F2h2F3h3 \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

×F\nabla \times \mathbf{F}の最初の成分を求めるために、q1q_{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平面を突き抜ける方向を意味する。それではdσ=h2h3dq2dq3q^1d\boldsymbol{\sigma} = h_{2}h_{3} dq_{2}dq_{3}\hat{\mathbf{q}}_{1}だから、

×Fdσ(×F)h2h3dq2dq3q^1 \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}

ストークスの定理

S(×v)da=Pvdl \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。

(×F)h2h3dq2dq3q^1=Fdr (\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、×F\nabla \times \mathbf{F}の最初の成分は、以下を計算して得られる。

(×F)1=(×F)q^1=1h2h3dq2dq3Fdr (\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1\textcircled{1}~4\textcircled{4}に分けて考えよう。

Fdr=1Fdr+2Fdr+3Fdr+4Fdr \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1\textcircled{1}に対する積分を求めると、

1FdrF(q2,q3)h2(q2,q3)dq2q^2=F2h2dq2 \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2\textcircled{2}は、

2FdrF(q2+dq2,q3)h3(q2+dq2,q3)dq3q^3=F3(q2+dq2,q3)h3(q2+dq2,q3)dq3[F3h3+(F3h3)q2dq2]dq3 \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。

テイラーの定理

f(x+dx)f(x)+f(x)dx f(x+dx) \approx f(x) + f^{\prime}(x)dx

同様に、残りの積分を計算すると、

3Fdr=3FdrF(q2,q3+dq3)h2(q2,q3+dq3)dq2q^2=F2(q2,q3+dq3)h2(q2,q3+dq3)dq2[F2h2+(F2h2)q3dq3]dq2 \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*}

4Fdr=4FdrF(q2,q3)h3(q2,q3)dq3q^3=F3h3dq3 \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、

(×F)1=1h2h3dq2dq3Fdr=1h2h3dq2dq3[(F3h3)q2dq2dq3(F2h2)q3dq3dq2]=1h2h3[(F3h3)q2(F2h2)q3] \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