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分離ベクトルの大きさの勾配 📂数理物理学

分離ベクトルの大きさの勾配

数式

分離ベクトル \bcRの大きさのnn乗、n\cR ^{n}の勾配は以下の通りだ。

∇(n)=nn−1 \nabla (\cR^n)=n\cR^{n-1}\crH

説明

多項関数の微分と同じやり方で計算した後に、単位ベクトル \crHを付けるだけだ。

分離ベクトルは =r−r′\bcR=\mathbf{r}-\mathbf{r}^{\prime}なので、(x,y,z)(x,y,z)と(x′,y′,z′)(x^{\prime},y^{\prime},z^{\prime})を変数に持つ。だから、微分するときはこれに注意が必要だ。上付き標数がある座標とない座標に対する勾配を以下のように表す。

∇f=∂f∂xx^+∂f∂yy^+∂f∂zz^∇′f=∂f∂x′x^+∂f∂y′y^+∂f∂z′z^ \begin{align*} \nabla f&= \dfrac{\partial f}{\partial x}\hat {\mathbf{x}} + \dfrac{\partial f}{\partial y} \hat{\mathbf{y}} + \dfrac{\partial f} {\partial z} \hat{\mathbf{z}} \\ \nabla^{\prime} f&= \dfrac{\partial f}{\partial x^{\prime}}\hat {\mathbf{x}} + \dfrac{\partial f}{\partial y^{\prime}} \hat{\mathbf{y}} + \dfrac{\partial f} {\partial z^{\prime}} \hat{\mathbf{z}} \end{align*}

直交座標系では、分離ベクトルは以下の通りだ。

=(x−x′)x^+(y−y′)y^+(z−z′)z^=(x−x′)2+(y−y′)2+(z−z′)2=(x−x′)x^+(y−y′)y^+(z−z′)z^(x−x′)2+(y−y′)2+(z−z′)2 \begin{align*} \bcR &= (x-x^{\prime})\hat {\mathbf{x}} + (y-y^{\prime})\hat{\mathbf{y}} + (z-z^{\prime})\hat{\mathbf{z}} \\ \cR &= \sqrt{ (x-x^{\prime})^{2} + (y-y^{\prime})^{2} + (z-z^{\prime})^{2} } \\ \crH &= \dfrac{ (x-x^{\prime})\hat {\mathbf{x}} + (y-y^{\prime})\hat{\mathbf{y}} + (z-z^{\prime})\hat{\mathbf{z}}}{\sqrt{ (x-x^{\prime})^{2} + (y-y^{\prime})^{2} + (z-z^{\prime})^{2} }} \end{align*}

n=2n=2、n=−1n=-1の場合の結果を先に見て、一般的な場合について証明する。式が長すぎる場合は、同じ部分を赤い角括弧[  ]{\color{red}[ \ \ ]}で示して省略した。

証明

∇2=2=2\nabla \cR^{2} = 2\bcR=2\cR\crH

∇(2)= ∂∂x[(x−x′)2+(y−y′)2+(z−z′)2]x^+∂∂y[  ]y^+∂∂z[  ]z^= 2(x−x′)x^+2(y−y′)y^+2(z−z′)z^= 2((x−x′)x^+(y−y′)y^+(z−z′)z^)= 2= 2 \begin{align*} \nabla(\cR ^{2}) =&\ \frac{\partial }{\partial x} {\color{red} \left[ (x-x^{\prime})^{2}+(y-y^{\prime})^{2}+(z-z^{\prime})^{2} \right]} \hat{\mathbf{x}} +\frac{\partial }{\partial y}{\color{red}[ \ \ ]}\hat{\mathbf{y}} +\frac{\partial }{\partial z}{\color{red}[ \ \ ]}\hat{\mathbf{z}} \\ =&\ 2(x-x^{\prime})\hat{\mathbf{x}}+2(y-y^{\prime})\hat{\mathbf{y}}+2(z-z^{\prime})\hat{\mathbf{z}} \\ =&\ 2 \left( (x-x^{\prime})\hat {\mathbf{x}} + (y-y^{\prime})\hat{\mathbf{y}} + (z-z^{\prime})\hat{\mathbf{z}} \right) \\ =&\ 2\bcR \\ =&\ 2\cR\crH \end{align*}

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∇1=−12\nabla \dfrac{1}{\cR} = -\dfrac{1}{\cR^{2}}\crH

∇1=∂∂x[(x−x′)2+(y−y′)2+(z−z′)2]−12x^+∂∂y[  ]−12y^+∂∂z[  ]−12z^=−122(x−x′)[(x−x′)2+(y−y′)2+(z−z′)2]32x^−122(y−y′)[  ]32y^−122(z−z′)[  ]32=−1[(x−x′)2+(y−y′)2+(z−z′)2][(x−x′)[  ]12x^+(y−y′)[  ]12y^+(z−z′)[  ]12]=−12[(x−x′)[(x−x′)2+(y−y′)2+(z−z′)2]12x^+(y−y′)[  ]12y^+(z−z′)[  ]12z^]=−12(x−x′)x^+(y−y′)y^+(z−z′)z^(x−x′)2+(y−y′)2+(z−z′)2=−12 \begin{align*} \nabla \dfrac{1}{\cR} &= \dfrac{\partial }{\partial x} {\color{red} \left[ (x-x^{\prime})^{2}+(y-y^{\prime})^{2}+(z-z^{\prime})^{2} \right]}^{-\frac{1}{2}} \hat{\mathbf{x}} +\dfrac{\partial }{\partial y}{\color{red}[ \ \ ]}^{-\frac{1}{2}} \hat{\mathbf{y}} +\dfrac{\partial }{\partial z}{\color{red}[ \ \ ]}^{-\frac{1}{2}} \hat{\mathbf{z}} \\ &= -\frac{1}{2}\dfrac{2(x-x^{\prime})}{ {\color{red} \left[(x-x^{\prime})^{2}+(y-y^{\prime})^{2}+(z-z^{\prime})^{2} \right]}^{\frac{3}{2}} }\hat{\mathbf{x}} - \frac{1}{2}\dfrac{2(y-y^{\prime})}{ {\color{red}[ \ \ ]}^{\frac{3}{2}} } \hat{\mathbf{y}} -\frac{1}{2}\dfrac{2(z-z^{\prime})}{ {\color{red}[ \ \ ]}^{\frac{3}{2}} } \\ &= -\dfrac{1}{ {\color{red} \left[(x-x^{\prime})^{2}+(y-y^{\prime})^{2}+(z-z^{\prime})^{2} \right]} } \left[ \dfrac{(x-x^{\prime})}{ {\color{red}[ \ \ ]}^{\frac{1}{2}} } \hat{\mathbf{x}} + \dfrac{(y-y^{\prime})}{ {\color{red}[ \ \ ]}^{\frac{1}{2}} } \hat{\mathbf{y}} + \dfrac{(z-z^{\prime})}{ {\color{red}[ \ \ ]}^{\frac{1}{2}} } \right] \\ &= -\dfrac{1}{ {\color{red}\cR^{2}} } \left[ \dfrac{(x-x^{\prime})}{ {\color{red} \left[ (x-x^{\prime})^{2}+(y-y^{\prime})^{2}+(z-z^{\prime})^{2} \right]} ^{\frac{1}{2}}} \hat{\mathbf{x}} + \dfrac{(y-y^{\prime})}{ {\color{red}[ \ \ ]}^{\frac{1}{2}}} \hat{\mathbf{y}} +\dfrac{(z-z^{\prime})}{ {\color{red}[ \ \ ]}^{\frac{1}{2}}} \hat{\mathbf{z}} \right] \\ &= -\dfrac{1}{\cR^{2}} \dfrac{ (x-x^{\prime})\hat{\mathbf{x}} + (y-y^{\prime})\hat{\mathbf{y}} +(z-z^{\prime})\hat{\mathbf{z}}}{\sqrt{(x-x^{\prime})^{2}+(y-y^{\prime})^{2}+(z-z^{\prime})^{2}}} \\ &= -\dfrac{1}{\cR^{2}}\crH \end{align*}

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∇(n)=nn−1\nabla (\cR^n)=n\cR^{n-1}\crH

∇(n)=∂∂x(n)x^+∂∂y(n)y^+∂∂z(n)z^=∂∂(n)∂∂xx^+∂∂(n)∂∂yy^+∂∂(n)∂∂zz^ \begin{align*} \nabla (\cR^n) &= \frac{\partial}{\partial x}(\cR^n)\hat{\mathbf{x}}+ \frac{\partial}{\partial y}(\cR^n)\hat{\mathbf{y}}+\frac{\partial}{\partial z}(\cR^n)\hat{\mathbf{z}} \\ &= \frac{\partial}{\partial \cR}(\cR^n)\frac{\partial \cR}{\partial x}\hat{\mathbf{x}}+ \frac{\partial}{\partial \cR}(\cR^n)\frac{\partial \cR}{\partial y}\hat{\mathbf{y}}+\frac{\partial}{\partial \cR}(\cR^n)\frac{\partial \cR}{\partial z}\hat{\mathbf{z}} \end{align*}

2番目の等号は連鎖律によって成り立つ。この時、以下の式が成り立つ。

∂∂x=∂∂x[(x−x′)2+(y−y′)2+(z−z′)2)]12=12[2(x−x′)][(x−x′)2+(y−y′)2+(z−z′)2]−12=x−x′ \begin{align*} \frac{\partial \cR}{\partial x} &=\frac{\partial }{\partial x}[(x-x^{\prime})^{2} + (y-y^{\prime})^{2} +(z-z^{\prime})^{2})]^{\frac{1}{2}} \\ &= \frac{1}{2}[2(x-x^{\prime})][(x-x^{\prime})^{2}+(y-y^{\prime})^{2}+(z-z^{\prime})^{2}]^{-\frac{1}{2}} \\ &= \frac{x-x^{\prime}}{\cR} \end{align*}

同様に∂∂y=y−y′\dfrac{\partial \cR}{\partial y}= \dfrac {y-y^{\prime}}{\cR}、∂∂z=z−z′\dfrac{\partial \cR}{\partial z} = \dfrac {z-z^{\prime}}{\cR}だ。従って、まとめると以下の通りだ。

∇(n)=∂∂(n)x−x′x^+∂∂(n)y−y′y^+∂∂(n)z−z′z^=∂∂(n)(x−x′x^+y−y′y^+z−z′z^)=nn−1 \begin{align*} \nabla (\cR^n) &= \frac{\partial}{\partial \cR}(\cR^n)\frac{x-x^{\prime}}{\cR}\hat{\mathbf{x}}+ \frac{\partial}{\partial \cR}(\cR^n)\frac{y-y^{\prime}}{\cR}\hat{\mathbf{y}}+\frac{\partial}{\partial \cR}(\cR^n)\frac{z-z^{\prime}}{\cR}\hat{\mathbf{z}} \\ &= \frac{\partial}{\partial \cR}(\cR^n) \left( \frac{x-x^{\prime}}{\cR}\hat{\mathbf{x}}+ \frac{y-y^{\prime}}{\cR}\hat{\mathbf{y}}+\frac{z-z^{\prime}}{\cR}\hat{\mathbf{z}} \right) \\ &= n\cR^{n-1}\crH \end{align*}

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