Gradient of the Magnitude of Separation Vectors
📂Mathematical Physics Gradient of the Magnitude of Separation Vectors The square of the magnitude of the separation vector \bcR gradient of n \cR ^{n} n
∇ ( n ) = n n − 1
\nabla (\cR^n)=n\cR^{n-1}\crH
∇ ( n ) = n n − 1
Explanation It is calculated in the same way as the derivative of a polynomial function, and then just attach the unit vector \crH
Since the separation vector is = r − r ′ \bcR=\mathbf{r}-\mathbf{r}^{\prime} = r − r ′ ( x , y , z ) (x,y,z) ( x , y , z ) ( x ′ , y ′ , z ′ ) (x^{\prime},y^{\prime},z^{\prime}) ( x ′ , y ′ , z ′ )
∇ f = ∂ f ∂ x x ^ + ∂ f ∂ y y ^ + ∂ f ∂ z z ^ ∇ ′ 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*}
∇ f ∇ ′ f = ∂ x ∂ f x ^ + ∂ y ∂ f y ^ + ∂ z ∂ f z ^ = ∂ x ′ ∂ f x ^ + ∂ y ′ ∂ f y ^ + ∂ z ′ ∂ f z ^
In Cartesian coordinates, the separation vector is as follows.
= ( 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*}
= ( x − x ′ ) x ^ + ( y − y ′ ) y ^ + ( z − z ′ ) z ^ = ( x − x ′ ) 2 + ( y − y ′ ) 2 + ( z − z ′ ) 2 = ( x − x ′ ) 2 + ( y − y ′ ) 2 + ( z − z ′ ) 2 ( x − x ′ ) x ^ + ( y − y ′ ) y ^ + ( z − z ′ ) z ^
Let’s first examine the results for the case of n = 2 n=2 n = 2 n = − 1 n=-1 n = − 1 [ ] {\color{red}[ \ \ ]} [ ]
Proof ∇ 2 = 2 = 2 \nabla \cR^{2} = 2\bcR=2\cR\crH ∇ 2 = 2 = 2 ∇ ( 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*}
∇ ( 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
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∇ 1 = − 1 2 \nabla \dfrac{1}{\cR} = -\dfrac{1}{\cR^{2}}\crH ∇ 1 = − 2 1 ∇ 1 = ∂ ∂ x [ ( x − x ′ ) 2 + ( y − y ′ ) 2 + ( z − z ′ ) 2 ] − 1 2 x ^ + ∂ ∂ y [ ] − 1 2 y ^ + ∂ ∂ z [ ] − 1 2 z ^ = − 1 2 2 ( x − x ′ ) [ ( x − x ′ ) 2 + ( y − y ′ ) 2 + ( z − z ′ ) 2 ] 3 2 x ^ − 1 2 2 ( y − y ′ ) [ ] 3 2 y ^ − 1 2 2 ( z − z ′ ) [ ] 3 2 = − 1 [ ( x − x ′ ) 2 + ( y − y ′ ) 2 + ( z − z ′ ) 2 ] [ ( x − x ′ ) [ ] 1 2 x ^ + ( y − y ′ ) [ ] 1 2 y ^ + ( z − z ′ ) [ ] 1 2 ] = − 1 2 [ ( x − x ′ ) [ ( x − x ′ ) 2 + ( y − y ′ ) 2 + ( z − z ′ ) 2 ] 1 2 x ^ + ( y − y ′ ) [ ] 1 2 y ^ + ( z − z ′ ) [ ] 1 2 z ^ ] = − 1 2 ( x − x ′ ) x ^ + ( y − y ′ ) y ^ + ( z − z ′ ) z ^ ( x − x ′ ) 2 + ( y − y ′ ) 2 + ( z − z ′ ) 2 = − 1 2
\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*}
∇ 1 = ∂ x ∂ [ ( x − x ′ ) 2 + ( y − y ′ ) 2 + ( z − z ′ ) 2 ] − 2 1 x ^ + ∂ y ∂ [ ] − 2 1 y ^ + ∂ z ∂ [ ] − 2 1 z ^ = − 2 1 [ ( x − x ′ ) 2 + ( y − y ′ ) 2 + ( z − z ′ ) 2 ] 2 3 2 ( x − x ′ ) x ^ − 2 1 [ ] 2 3 2 ( y − y ′ ) y ^ − 2 1 [ ] 2 3 2 ( z − z ′ ) = − [ ( x − x ′ ) 2 + ( y − y ′ ) 2 + ( z − z ′ ) 2 ] 1 [ [ ] 2 1 ( x − x ′ ) x ^ + [ ] 2 1 ( y − y ′ ) y ^ + [ ] 2 1 ( z − z ′ ) ] = − 2 1 [ [ ( x − x ′ ) 2 + ( y − y ′ ) 2 + ( z − z ′ ) 2 ] 2 1 ( x − x ′ ) x ^ + [ ] 2 1 ( y − y ′ ) y ^ + [ ] 2 1 ( z − z ′ ) z ^ ] = − 2 1 ( x − x ′ ) 2 + ( y − y ′ ) 2 + ( z − z ′ ) 2 ( x − x ′ ) x ^ + ( y − y ′ ) y ^ + ( z − z ′ ) z ^ = − 2 1
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∇ ( n ) = n n − 1 \nabla (\cR^n)=n\cR^{n-1}\crH ∇ ( n ) = n n − 1 ∇ ( n ) = ∂ ∂ x ( n ) x ^ + ∂ ∂ y ( n ) y ^ + ∂ ∂ z ( n ) z ^ = ∂ ∂ ( n ) ∂ ∂ x x ^ + ∂ ∂ ( n ) ∂ ∂ y y ^ + ∂ ∂ ( n ) ∂ ∂ z z ^
\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*}
∇ ( n ) = ∂ x ∂ ( n ) x ^ + ∂ y ∂ ( n ) y ^ + ∂ z ∂ ( n ) z ^ = ∂ ∂ ( n ) ∂ x ∂ x ^ + ∂ ∂ ( n ) ∂ y ∂ y ^ + ∂ ∂ ( n ) ∂ z ∂ z ^
The second equality holds by the chain rule. At this time, the following equation holds.
∂ ∂ x = ∂ ∂ x [ ( x − x ′ ) 2 + ( y − y ′ ) 2 + ( z − z ′ ) 2 ) ] 1 2 = 1 2 [ 2 ( x − x ′ ) ] [ ( x − x ′ ) 2 + ( y − y ′ ) 2 + ( z − z ′ ) 2 ] − 1 2 = 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*}
∂ x ∂ = ∂ x ∂ [( x − x ′ ) 2 + ( y − y ′ ) 2 + ( z − z ′ ) 2 ) ] 2 1 = 2 1 [ 2 ( x − x ′ )] [( x − x ′ ) 2 + ( y − y ′ ) 2 + ( z − z ′ ) 2 ] − 2 1 = x − x ′
Similarly, ∂ ∂ y = y − y ′ \dfrac{\partial \cR}{\partial y}= \dfrac {y-y^{\prime}}{\cR} ∂ y ∂ = y − y ′ ∂ ∂ z = z − z ′ \dfrac{\partial \cR}{\partial z} = \dfrac {z-z^{\prime}}{\cR} ∂ z ∂ = z − z ′
∇ ( n ) = ∂ ∂ ( n ) x − x ′ x ^ + ∂ ∂ ( n ) y − y ′ y ^ + ∂ ∂ ( n ) z − z ′ z ^ = ∂ ∂ ( n ) ( x − x ′ x ^ + y − y ′ y ^ + z − z ′ z ^ ) = n n − 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*}
∇ ( n ) = ∂ ∂ ( n ) x − x ′ x ^ + ∂ ∂ ( n ) y − y ′ y ^ + ∂ ∂ ( n ) z − z ′ z ^ = ∂ ∂ ( n ) ( x − x ′ x ^ + y − y ′ y ^ + z − z ′ z ^ ) = n n − 1
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