움직이는 점전하가 만드는 전기장

움직이는 점전하가 만드는 전기장


🚧 이 포스트는 아직 이관 작업이 완료되지 않았습니다 🚧

5D6FB6AA3.png

움직이는 점전하가 만드는 전자기장 $$ \mathbf{E}(\mathbf{r}, t) = \frac{q}{4\pi\epsilon_0} \frac{\eta} {( \boldsymbol{\eta}\cdot \mathbf{u} )^3 } \left[(c^2-v^2)\mathbf{u} +\boldsymbol{\eta}\times (\mathbf{u} \times \mathbf{a} ) \right] $$

$$ \mathbf{B} (\mathbf{ r}, t) =\frac{1}{c} \hat{\boldsymbol{\eta}}\times \mathbf{ E } (\mathbf{ r}, t) $$

움직이는 점전하가 만드는 전기장, 자기장은 리에나르-비케르트 전위를 사용하여 구할 수 있다. $$ V(\mathbf{r}, t)= \frac{1}{4\pi \epsilon_0} \frac{qc}{ (\eta c -\boldsymbol{\eta}\cdot \mathbf{v})} ,\quad \mathbf{A}(\mathbf{r}, t) = \frac{ \mathbf{v} } {c^2} V(\mathbf{r}, t) $$ 또한 전자기장은 아래의 식으로 구할 수 있다. $$ \mathbf{ E} = -\nabla V -\frac{\partial \mathbf{ A} }{\partial t},\quad \mathbf{B}=\nabla \times \mathbf{A} $$ 움직이는 점전하가 만드는 전기장 $\nabla V$와 $\dfrac{\partial \mathbf{A}}{\partial t}$를 차례로 구해보자. 우선 $\nabla V$를 계산해보면 $$ \nabla V=\frac{qc}{4 \pi \epsilon_0} \nabla \dfrac{1}{(\eta c - \boldsymbol{\eta} \cdot \mathbf{v}) } $$ 이때 $\dfrac{ d}{dx} \left( \dfrac{1}{f(x)} \right)=\dfrac{-1}{ [f(x)]^2} f^{\prime}(x)$이므로 $$ \nabla V = \frac{qc}{4\pi\epsilon_0} \frac{-1}{ (\eta c -\boldsymbol{\eta} \cdot \mathbf{v} )^2}\nabla (\eta c -\boldsymbol{\eta} \cdot \mathbf{v} )\tag{1} $$ 따라서 $\nabla \eta$와 $\nabla (\boldsymbol{\eta} \cdot \mathbf{v})$를 계산하면 된다. 우선 $\nabla ( \boldsymbol{\eta} \cdot \mathbf{v} )$부터 계산해보면 곱셈규칙 2에 의해서 $$ \nabla (\boldsymbol{\eta} \cdot \mathbf{v}) = (\boldsymbol{\eta} \cdot \nabla ) \mathbf{v}+(\mathbf{v} \cdot \nabla ) \boldsymbol{\eta} + \boldsymbol{\eta}\times (\nabla \times \mathbf{v}) +\mathbf{v}\times (\nabla \times \boldsymbol{\eta}) \tag{2} $$ 첫 번째 항 $(\boldsymbol{\eta} \cdot \nabla ) \mathbf{v}(t_r)$ $$ \begin{align*} (\boldsymbol{\eta} \cdot \nabla) \mathbf{v} &= \left( \eta_{x} \frac{\partial}{\partial x} + \eta _{y} \frac{\partial}{\partial y}+\eta_{z}\frac{\partial }{\partial z} \right)\mathbf{v} \\ &= \eta_{x}\frac{d \mathbf{v} }{dt_r}\frac{\partial t_r}{\partial x} +\eta_{y}\frac{d \mathbf{v} }{dt_r}\frac{\partial t_r}{\partial y} + \eta_{z}\frac{d \mathbf{v} }{dt_r}\frac{\partial t_r}{\partial z} \\ &= \frac{d \mathbf{v} }{ dt_r} \left( \eta_{x}\frac{\partial t_r}{\partial x} + \eta_{y}\frac{\partial t_r}{\partial y}+\eta_{z}\frac{\partial t_r}{\partial z}\right) \\ &= \mathbf{a} ( \boldsymbol{\eta} \cdot \nabla t_r) \end{align*} $$ 이때 $\mathbf{a}$는 지연시각일 때 입자(점전하)의 가속도이다.**두 번째 항 $(\mathbf{v} \cdot \nabla ) \boldsymbol{\eta}$** $\boldsymbol{\eta}=\mathbf{r}-\mathbf{w}$이므로 $$ (\mathbf{v} \cdot \nabla ) \boldsymbol{\eta}=(\mathbf{v}\cdot \nabla)\mathbf{r} -(\mathbf{v}\cdot \nabla ) \mathbf{w} \tag{3} $$ 임의의 벡터 $\mathbf{A}=(A_{x}, A_{y}, A_{z})$에 대해서 $$ \begin{align*} (\mathbf{A} \cdot \nabla ) \mathbf{r} &= \left( A_{x}\frac{\partial}{\partial x}+A_{y}\frac{\partial}{\partial y}+A_{z}\frac{\partial}{\partial z} \right) ( x \hat{\mathbf{x}} + y \hat{\mathbf{y}} +z\hat{ \mathbf{z} }) \\ &= A_{x}\hat{\mathbf{x}}+A_{y} \hat{\mathbf{y}} + A_{z}\hat{\mathbf{z}} \\ &= \mathbf{A} \end{align*} $$ 이고 $$ \begin{align*} (\mathbf{A} \cdot \nabla ) \mathbf{w}(t_r) &= \left( A_{x}\frac{\partial}{\partial x}+A_{y}\frac{\partial}{\partial y}+A_{z}\frac{\partial}{\partial z} \right) \mathbf{w} \\ &= A_{x}\frac{\partial \mathbf{w} } {\partial x}+A_{y}\frac{\partial \mathbf{w} }{\partial y}+A_{z}\frac{\partial \mathbf{w} }{\partial z} \\ &= A_{x}\frac{\partial \mathbf{w} }{\partial t_r}\frac{\partial t_r} {\partial x}+A_{y}\frac{\partial \mathbf{w} }{\partial t_r}\frac{\partial t_r}{\partial y}+A_{z}\frac{\partial \mathbf{w} }{\partial t_r}\frac{\partial t_r}{\partial z} \\ &= A_{x} \mathbf{v}\frac{\partial t_r} {\partial x}+A_{y}\mathbf{v}\frac{\partial t_r}{\partial y}+A_{z}\mathbf{v}\frac{\partial t_r}{\partial z} \\ &= \mathbf{v} \left( A_{x} \frac{\partial t_r} {\partial x}+A_{y}\frac{\partial t_r}{\partial y}+A_{z}\frac{\partial t_r}{\partial z}\right) \\ &= \mathbf{v}(\mathbf{A} \cdot \nabla t_r) \end{align*} $$ 이므로 $(3)$은 $$ (\mathbf{v} \cdot \nabla ) \boldsymbol{\eta}=\mathbf{v} -\mathbf{v}(\mathbf{v}\cdot \nabla t_r ) $$ 이다.**세 번째 항 $\boldsymbol{\eta}\times (\nabla \times \mathbf{v})$** 우선 $$ \begin{align*} \nabla \times \mathbf{v} &= \left( \frac{\partial v_{z}}{\partial y} - \frac{\partial v_{y}}{\partial z} \right)\hat{\mathbf{x}} + \left( \frac{\partial v_{x}}{\partial z} - \frac{\partial v_{z}}{\partial x} \right)\hat{\mathbf{y}} + \left( \frac{\partial v_{y}}{\partial x} - \frac{\partial v_{x}}{\partial y} \right)\hat{\mathbf{z}} \\ &= \left( \frac{\partial v_{z}}{\partial t_r} \frac{t_r}{\partial y} - \frac{\partial v_{y}}{\partial t_r} \frac{t_r}{\partial z} \right)\hat{\mathbf{x}} + \left( \frac{\partial v_{x}}{\partial t_r} \frac{t_r}{\partial z} - \frac{\partial v_{z}}{\partial t_r} \frac{t_r}{\partial x} \right)\hat{\mathbf{y}} + \left( \frac{\partial v_{y}}{\partial t_r} \frac{t_r}{\partial x} - \frac{\partial v_{x}}{\partial t_r} \frac{t_r}{\partial y} \right)\hat{\mathbf{z}} \\ &= \Big( a_{z} (\nabla t_r)_{y} - a_{y} (\nabla t_r)_{z} \Big)\hat{\mathbf{x}} + \Big( a_{x} (\nabla t_r)_{z} - a_{z} (\nabla t_r)_{x} \Big)\hat{\mathbf{y}} + \Big( a_{y} (\nabla t_r)_{x} - a_{x} (\nabla t_r)_{y} \Big)\hat{\mathbf{z}} \\ &= -\mathbf{a} \times \nabla t_r \end{align*} $$ 이다. 따라서 $$ \boldsymbol{\eta} \times ( \nabla \times \mathbf{v} )= -\boldsymbol{\eta}\times (\mathbf{a} \times \nabla t_r) $$ **네 번째 항 $\mathbf{v}\times (\nabla \times \boldsymbol{\eta})$** $\boldsymbol{\eta}=\mathbf{r}-\mathbf{w}$이고, $\nabla \times \mathbf{r}=0$이므로 $$ \nabla \times \boldsymbol{\eta} = \nabla \times \mathbf{r} -\nabla \times \mathbf{w}=-\nabla \times \mathbf{w} $$ 이때 바로 위에서 계산했던 $\nabla \times \mathbf{v} = -\mathbf{a} \times \nabla t_r$의 결과를 이용하면 $\nabla \times \mathbf{w} = -\mathbf{v} \times \nabla t_r$임을 알 수 있다. 따라서 $$ \mathbf{v}\times (\nabla \times \boldsymbol{\eta})=\mathbf{v} \times (- \nabla \times \mathbf{w})=\mathbf{v} \times (\mathbf{v} \times \nabla t_r) $$ 이제 위의 계산 결과들을 $(2)$에 대입하면 $$ \begin{align*} \nabla (\boldsymbol{\eta} \cdot \mathbf{v} ) &= \mathbf{a} ( \boldsymbol{\eta} \cdot \nabla t_r) + \mathbf{v} -\mathbf{v}(\mathbf{v} \cdot \nabla t_r) -\color{blue}{\boldsymbol{\eta}\times(\mathbf{a} \times \nabla t_r )} +\color{green}{\mathbf{v} \times (\mathbf{v} \times \nabla t_r)} \\ &= \mathbf{a} ( \boldsymbol{\eta} \cdot \nabla t_r) + \mathbf{v} -\mathbf{v}(\mathbf{v} \cdot \nabla t_r) -\color{blue}{\left[\mathbf{a} ( \boldsymbol{\eta} \cdot \nabla t_r) - \nabla t_r(\boldsymbol{\eta} \cdot \mathbf{a} ) \right]}+ \color{green}{\left[ \mathbf{v}(\mathbf{v} \cdot \nabla t_r ) -\nabla t_r (\mathbf{v} \cdot \mathbf{v}) \right] } \\ &=\mathbf{v} + \nabla t_r(\boldsymbol{\eta} \cdot \mathbf{a}) -\nabla t_r( \mathbf{v} \cdot \mathbf{v}) \\ &= \mathbf{v}+( \boldsymbol{\eta} \cdot \mathbf{a} -v^2) \nabla t_r \quad (4) \end{align*} $$ 두번째 등호는 BAC-CAB 공식에 의해 성립한다. 또한 $\eta=c(t-t_r)$이므로 $\nabla \eta = -c\nabla t_r$이다. 이와 $(4)$를 $(1)$에 대입하면 $$ \begin{align*} \nabla V &= \frac{qc}{4\pi\epsilon_0} \frac{-1}{ (\eta c -\boldsymbol{\eta} \cdot \mathbf{v} )^2}\nabla (\eta c -\boldsymbol{\eta} \cdot \mathbf{v} ) \\ &= \frac{qc}{4\pi\epsilon_0} \frac{-1}{ (\eta c -\boldsymbol{\eta} \cdot \mathbf{v} )^2} \Big[ c\nabla \eta -\nabla (\boldsymbol{\eta} \cdot \mathbf{v} ) \Big] \\ &=\frac{qc}{4\pi\epsilon_0} \frac{-1}{ (\eta c -\boldsymbol{\eta} \cdot \mathbf{v} )^2} \Big[ -c^2\nabla t_r -\mathbf{v}-(\boldsymbol{\eta} \cdot \mathbf{a} -v^2)\nabla t_r \Big] \\ &=\frac{qc}{4\pi\epsilon_0} \frac{1}{ (\eta c -\boldsymbol{\eta} \cdot \mathbf{v} )^2} \Big[ \mathbf{v} + (c^2 -v^2+\boldsymbol{\eta} \cdot \mathbf{a} )\nabla t_r\Big] \tag{5} \end{align*} $$ 그리고 지연 시각의 기울기가 아래와 같다. $$ \nabla t_r = \frac {-\boldsymbol{\eta} }{\eta c -\boldsymbol{\eta} \cdot \mathbf{v} } $$ 이를 $(5)$에 대입하면 $$ \begin{align*} \nabla V &= \frac{qc}{4\pi\epsilon_0} \frac{1}{ (\eta c -\boldsymbol{\eta} \cdot \mathbf{v} )^2} \left[ \mathbf{v} + (c^2 -v^2+\boldsymbol{\eta} \cdot \mathbf{a} )\nabla t_r\right] \\ &= \frac{qc}{4\pi\epsilon_0} \frac{1}{ (\eta c -\boldsymbol{\eta} \cdot \mathbf{v} )^2} \left[ \mathbf{v} + (c^2 -v^2+\boldsymbol{\eta} \cdot \mathbf{a} ) \frac {-\boldsymbol{\eta} }{\eta c -\boldsymbol{\eta} \cdot \mathbf{v} } \right] \\ &= \frac{qc}{4\pi\epsilon_0} \frac{1}{ (\eta c -\boldsymbol{\eta} \cdot \mathbf{v} )^3} \Big[ (\eta c -\boldsymbol{\eta} \cdot \mathbf{v})\mathbf{v} - (c^2 -v^2+\boldsymbol{\eta} \cdot \mathbf{a} ) \boldsymbol{\eta} \Big] \end{align*} $$ 이제 $\frac{\partial \mathbf{A}}{\partial t}$를 구하면 끝이다. 계산 해보면 결과는 아래와 같다. $$ \frac{ \partial \mathbf{A}}{ \partial t }=\frac{qc}{4\pi \epsilon_{0}} \frac{1}{(\eta c -\boldsymbol{\eta} \cdot \mathbf{v})^{3}}\left[ (\eta c +\boldsymbol{\eta}\cdot \mathbf{v})(\eta\mathbf{a}/c-\mathbf{v})+ \frac{\eta}{c}\mathbf{v}\left( c^{2} -v^{2}+\boldsymbol{\eta}\cdot \mathbf{a}\right) \right] $$ $\mathbf{u}\equiv c \hat{\boldsymbol{\eta}}-\mathbf{v}$라고 두고 정리하면 $$ \begin{align*} \mathbf{E}(\mathbf{r},t) &= -\nabla V - \frac{ \partial \mathbf{A}}{ \partial t} \\ &= \frac{q}{4\pi\epsilon_{0}}\frac{\eta}{(\boldsymbol{\eta}\cdot \mathbf{u})^{3}}\left[ (c^{2}-v^{2})\mathbf{u} + \mathbf{u}(\boldsymbol{\eta}\cdot\mathbf{a})-\mathbf{a}(\boldsymbol{\eta}\cdot\mathbf{u}) \right] \\ &= \frac{q}{4\pi\epsilon_{0}}\frac{\eta}{(\boldsymbol{\eta}\cdot \mathbf{u})^{3}}\left[ (c^{2}-v^{2})\mathbf{u} + \boldsymbol{\eta} \times (\mathbf{u}\times \mathbf{a}) \right] \end{align*} $$

댓글