분리벡터의 회전
curl of separation vector 1r^2
공식
$$ \nabla \times \dfrac{\hat{\boldsymbol{\eta}} }{\eta ^2} =0 $$
설명
이 식이 특별한 의미를 가지는 것은 아니다. 자기장의 발산을 구하는 과정에서 나오는데 계산이 간단하지 않아 따로 설명한다.
증명
$\boldsymbol{\eta}=(x-x^{\prime})\hat{\mathbf{x}} + (y-y^{\prime})\hat{\mathbf{y}} + (z-z^{\prime})\hat{\mathbf{z}}$를 분리벡터라고 하면 다음과 같다.
$$ | \boldsymbol{\eta} |=\eta=\sqrt{(x-x^{\prime})^2+(y-y^{\prime})^2 + (z-z^{\prime})^2} $$
$$ \hat{ \boldsymbol{\eta}}=\dfrac{ \boldsymbol{\eta} } { \eta}=\dfrac{(x-x^{\prime})\hat{\mathbf{x}} + (y-y^{\prime})\hat{\mathbf{y}} + (z-z^{\prime})\hat{\mathbf{z}}}{\sqrt{(x-x^{\prime})^2+(y-y^{\prime})^2 + (z-z^{\prime})^2}} $$
$$ \dfrac{\hat{\boldsymbol{\eta}}}{\eta^2}=\dfrac{1}{(x-x^{\prime})^2+(y-y^{\prime})^2 + (z-z^{\prime})^2}\dfrac{(x-x^{\prime})\hat{\mathbf{x}} + (y-y^{\prime})\hat{\mathbf{y}} + (z-z^{\prime})\hat{\mathbf{z}}}{\sqrt{(x-x^{\prime})^2+(y-y^{\prime})^2 + (z-z^{\prime})^2}} $$
계산의 편의를 위해
$$ \dfrac{\hat{\boldsymbol{\eta}} }{\eta ^2} = H_{x} \hat{\mathbf{x}} + H_{y} \hat{\mathbf{y}} + H_{z}\hat{\mathbf{z}} $$
라고 하자. 그러면
$$ \begin{align*} \nabla \times \dfrac{ \hat{\boldsymbol{\eta}}}{\eta^2} &= \begin{vmatrix} \hat{\mathbf{x}} & \hat{\mathbf{y}} & \hat{\mathbf{z}} \\ \dfrac{\partial }{\partial x} & \dfrac{\partial }{\partial y} & \dfrac{\partial }{\partial z} \\ H_{x} & H_{y} & H_{z} \end{vmatrix} \\ &= \left( \dfrac{ \partial}{\partial y} H_{z} -\dfrac{\partial }{\partial z}H_{y} \right) \hat{\mathbf{x}} + \left( \dfrac{ \partial}{\partial z} H_{x} -\dfrac{\partial }{\partial x}H_{z} \right) \hat{\mathbf{y}} +\left( \dfrac{ \partial}{\partial x} H_{y} -\dfrac{\partial }{\partial y}H_{x} \right) \hat{\mathbf{z}} \end{align*} $$
$\hat{\mathbf{x}}$항만 먼저 계산해보자.
$$ \begin{align*} & \dfrac{\partial }{\partial y} H_{z} - \dfrac{\partial}{\partial z}H_{y} \\ =&\ \dfrac{\partial }{\partial y} \left[ \left[ (x-x^{\prime})^2+(y-y^{\prime})^2+(z-z^{\prime})^2 \right]^{-\frac{3}{2}}(z-z^{\prime}) \right] \\ &- \dfrac{\partial }{\partial z} \left[ \left[ (x-x^{\prime})^2+(y-y^{\prime})^2+(z-z^{\prime})^2 \right]^{-\frac{3}{2}}(y-y^{\prime}) \right] \\ =&\ -\dfrac{3}{2}\left[ \left[ (x-x^{\prime})^2+(y-y^{\prime})^2+(z-z^{\prime})^2 \right]^{-\frac{5}{2}}(z-z^{\prime}) \right]\cdot 2(y-y^{\prime}) \\ & +\dfrac{3}{2}\left[ \left[ (x-x^{\prime})^2+(y-y^{\prime})^2+(z-z^{\prime})^2 \right]^{-\frac{5}{2}}(y-y^{\prime}) \right]\cdot 2(z-z^{\prime}) \\ =&\ -3 \left[ (x-x^{\prime})^2+(y-y^{\prime})^2+(z-z^{\prime})^2 \right]^{-\frac{5}{2}}(z-z^{\prime})(y-y^{\prime}) \\ & +3 \left[ (x-x^{\prime})^2+(y-y^{\prime})^2+(z-z^{\prime})^2 \right]^{-\frac{5}{2}}(y-y^{\prime})(z-z^{\prime}) \\ =&\ 0 \end{align*} $$
같은 방식으로 계산하면 $\hat{\mathbf{y}}$항과 $\hat{\mathbf{z}}$항 역시 $0$이 된다.
$$ \therefore \nabla \times \dfrac{ \hat{\boldsymbol{\eta}}}{\eta^2} = \mathbf{0} $$
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