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분리벡터의 회전 📂수리물리

분리벡터의 회전

공식

×2=0 \nabla \times \dfrac{\crH }{\cR ^2} = \mathbf{0}

설명

이 식이 특별한 의미를 가지는 것은 아니다. 자기장의 발산을 구하는 과정에서 나오는데 계산이 간단하지 않아 따로 설명한다.

증명

=(xx)x^+(yy)y^+(zz)z^\bcR=(x-x^{\prime})\hat{\mathbf{x}} + (y-y^{\prime})\hat{\mathbf{y}} + (z-z^{\prime})\hat{\mathbf{z}}를 분리벡터라고 하면 다음과 같다.

==(xx)2+(yy)2+(zz)2 | \bcR |=\cR=\sqrt{(x-x^{\prime})^2+(y-y^{\prime})^2 + (z-z^{\prime})^2}

==(xx)x^+(yy)y^+(zz)z^(xx)2+(yy)2+(zz)2 \crH = \dfrac{ \bcR } { \cR}=\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}}

2=1(xx)2+(yy)2+(zz)2(xx)x^+(yy)y^+(zz)z^(xx)2+(yy)2+(zz)2 \dfrac{\crH}{\cR^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

2=Axx^+Ayy^+Azz^ \dfrac{\crH }{\cR ^2} = A_{x} \hat{\mathbf{x}} + A_{y} \hat{\mathbf{y}} + A_{z}\hat{\mathbf{z}}

라고 하자. 그러면

×2=x^y^z^xyzAxAyAz=(yAzzAy)x^+(zAxxAz)y^+(xAyyAx)z^ \begin{align*} \nabla \times \dfrac{ \crH}{\cR^2} &= \begin{vmatrix} \hat{\mathbf{x}} & \hat{\mathbf{y}} & \hat{\mathbf{z}} \\ \dfrac{\partial }{\partial x} & \dfrac{\partial }{\partial y} & \dfrac{\partial }{\partial z} \\ A_{x} & A_{y} & A_{z} \end{vmatrix} \\ &= \left( \dfrac{ \partial}{\partial y} A_{z} -\dfrac{\partial }{\partial z}A_{y} \right) \hat{\mathbf{x}} + \left( \dfrac{ \partial}{\partial z} A_{x} -\dfrac{\partial }{\partial x}A_{z} \right) \hat{\mathbf{y}} +\left( \dfrac{ \partial}{\partial x} A_{y} -\dfrac{\partial }{\partial y}A_{x} \right) \hat{\mathbf{z}} \end{align*}

x^\hat{\mathbf{x}}항만 먼저 계산해보자.

yAzzAy= y[[(xx)2+(yy)2+(zz)2]32(zz)]z[[(xx)2+(yy)2+(zz)2]32(yy)]= 32[[(xx)2+(yy)2+(zz)2]52(zz)]2(yy)+32[[(xx)2+(yy)2+(zz)2]52(yy)]2(zz)= 3[(xx)2+(yy)2+(zz)2]52(zz)(yy)+3[(xx)2+(yy)2+(zz)2]52(yy)(zz)= 0 \begin{align*} & \dfrac{\partial }{\partial y} A_{z} - \dfrac{\partial}{\partial z}A_{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 y^\hat{\mathbf{y}}항과 z^\hat{\mathbf{z}}항 역시 00이 된다.

×2=0 \therefore \nabla \times \dfrac{ \crH}{\cR^2} = \mathbf{0}