리에나르-비케르트 전위의 시간 도함수 📂전자기학

리에나르-비케르트 전위의 시간 도함수

time derivative of lienard wiechert potentials

개요

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리에나르-비케르트 전위의 시간에 대한 도함수는 다음과 같다.

$$ \begin{align*} \frac{ \partial V}{ \partial t} &= \frac{qc}{4\pi \epsilon_{0}} \frac{1}{(\eta c -\boldsymbol{\eta} \cdot \mathbf{v})^{2}} \left( c^{2} -c^{2}\frac{ \partial t}{ \partial t_{r}}-v^{2}+\boldsymbol{\eta}\cdot \mathbf{a} \right)\frac{ \partial t_{r}}{ \partial t } \\ \frac{ \partial \mathbf{A}}{ \partial t } &= \frac{qc}{4\pi \epsilon_{0}} \frac{1}{(\eta c -\boldsymbol{\eta} \cdot \mathbf{v})^{3}}\left[ (\eta c +\boldsymbol{\eta}\cdot \mathbf{v})(\eta\mathbf{a}/c-\mathbf{v})+ \frac{\eta}{c}\mathbf{v}\left( c^{2} -v^{2}+\boldsymbol{\eta}\cdot \mathbf{a}\right) \right] \end{align*} $$

보조정리

지연 시각의 시간 미분은 아래와 같다.

$$ \frac{ \partial t_{r}}{ \partial t}=\frac{\eta c}{\eta c-\boldsymbol{\eta}\cdot \mathbf{v}}=\frac{ \eta c}{\boldsymbol{\eta} \cdot \mathbf{u}} $$

이 때 $\mathbf{u}=c\hat{\boldsymbol{\eta}}-\mathbf{v}$이다.

증명

지연 시각의 정의에 따라

$$ c(t-t_{r})=\eta $$

이므로 양변을 제곱하면

$$ c^{2}(t-t_{r})^{2}=\eta ^{2}=\boldsymbol{\eta} \cdot \boldsymbol{\eta} $$

양변을 $t$로 미분하면

$$ 2c^{2}(t-t_{r}) \left( 1-\frac{ \partial t_{r}}{ \partial t } \right)=2\frac{ \partial \boldsymbol{\eta}}{ \partial t }\cdot \boldsymbol{\eta} \tag{1} $$

이때 $\boldsymbol{\eta}=\mathbf{r}-\mathbf{w}(t_{r})$이므로

$$ \frac{ \partial \boldsymbol{\eta}}{ \partial t}=-\frac{ \partial \mathbf{w}}{ \partial t}=-\frac{ \partial \mathbf{w}}{ \partial t_{r}}\frac{ \partial t_{r}}{ \partial t}=-\mathbf{v}\frac{ \partial t_{r} }{ \partial t} $$

이다. 이를 $(1)$에 대입하고 정리하면

$$ \begin{align*} && c\eta \left( 1-\frac{ \partial t_{r}}{ \partial t } \right) &= -\boldsymbol{\eta}\cdot \mathbf{v}\frac{ \partial t_{r}}{ \partial t } \\[1em] \implies && c\eta -c\eta \frac{ \partial t_{r}}{ \partial t} &= -\boldsymbol{\eta}\cdot\mathbf{v}\frac{ \partial t_{r} }{ \partial t } \\[1em] \implies && \frac{ \partial t_{r}}{ \partial t } &= \frac{c\eta}{c\eta -\boldsymbol{\eta}\cdot \mathbf{v}} \\[1em] && &= \frac{c\eta}{\boldsymbol{\eta} \cdot \mathbf{u}} \end{align*} $$

증명

리에나르-비케르트 전위

지연시각 $t_r$에서 속도 $\mathbf{v}$로 움직이는 점전하 $q$에 대한 전위는 다음과 같다.

$$ \begin{align*} V(\mathbf{r}, t) &= \frac{1}{4\pi \epsilon_0} \frac{qc}{ (\eta c -\boldsymbol{\eta}\cdot \mathbf{v})} \\ \mathbf{A}(\mathbf{r}, t) &= \frac{\mu_0}{4 \pi}\frac{qc \mathbf{v} }{(\eta c - \boldsymbol{\eta}\cdot \mathbf{v} )}=\frac{\mathbf{v}}{c^2}V(\mathbf{r}, t) \end{align*} $$

이때 $\boldsymbol{\eta}=\mathbf{r} -\mathbf{w}(t_r)$는 지연위치에서 관찰점까지의 벡터, $\mathbf{w}(t_r)$은 지연시각에서의 점전하의 위치인 지연 위치이다.

  • Part 1. $V$

스칼라 전위 $V$를 $t$로 미분하면,

$$ \begin{align*} \frac{ \partial V}{ \partial t } &= \frac{qc}{4\pi \epsilon_{0}}\frac{\partial}{\partial t} \left( \frac{1}{\eta c -\boldsymbol{\eta}\cdot \mathbf{v}} \right) \\ &= \frac{qc}{4\pi \epsilon_{0}}\frac{\partial}{\partial t_{r}} \left(\frac{1}{\eta c -\boldsymbol{\eta}\cdot \mathbf{v}}\right) \frac{ \partial t_{r} }{ \partial t } \end{align*} $$

이 때 $\dfrac{ d }{ dx }\left( \dfrac{1}{f(x)} \right)=\dfrac{ d }{ d f(x) }\left(\dfrac{1}{f(x)} \right)\dfrac{ d f(x)}{ d x }=\dfrac{-1}{\left(f(x)\right)^{2}}f^{\prime}(x)$ 이므로

$$ \begin{align*} \frac{ \partial V}{ \partial t }&=\frac{qc}{4\pi \epsilon_{0}} \frac{-1}{(\eta c -\boldsymbol{\eta} \cdot \mathbf{v})^{2}}\frac{ \partial (\eta c -\boldsymbol{\eta} \cdot \mathbf{v})}{ \partial t_{r} }\frac{ \partial t_{r}}{ \partial t } \\ &= \frac{qc}{4\pi \epsilon_{0}} \frac{-1}{(\eta c -\boldsymbol{\eta} \cdot \mathbf{v})^{2}} \left( c\frac{ \partial \eta}{ \partial t_{r}} - \frac{ \partial \boldsymbol{\eta}}{ \partial t_{r} }\cdot \mathbf{v}-\boldsymbol{\eta}\cdot \frac{ \partial \mathbf{v}}{ \partial t_{r} }\right)\frac{ \partial t_{r}}{ \partial t } \end{align*} $$

$\boldsymbol{\eta}=\mathbf{r}-\mathbf{w}(t_{r})$이므로

$$ \frac{ \partial \boldsymbol{\eta}}{ \partial t_{r}}=-\frac{ \partial \mathbf{w}(t_{r})}{ \partial t_{r} }=-\mathbf{v}(t_{r}) $$

$| \boldsymbol{\eta}|=\eta=c(t-t_{r})$이므로

$$ \frac{ \partial \eta}{ \partial t_{r}}=c\frac{ \partial t}{ \partial t_{r} }-c $$

그러므로

$$ \begin{align*} \frac{ \partial V}{ \partial t} &=\frac{qc}{4\pi \epsilon_{0}} \frac{-1}{(\eta c -\boldsymbol{\eta} \cdot \mathbf{v})^{2}} \left( -c^{2} +c^{2}\frac{ \partial t}{ \partial t_{r}}+v^{2}-\boldsymbol{\eta}\cdot \mathbf{a} \right)\frac{ \partial t_{r}}{ \partial t } \\ &= \frac{qc}{4\pi \epsilon_{0}} \frac{1}{(\eta c -\boldsymbol{\eta} \cdot \mathbf{v})^{2}} \left( c^{2} -c^{2}\frac{ \partial t}{ \partial t_{r}}-v^{2}+\boldsymbol{\eta}\cdot \mathbf{a} \right)\frac{ \partial t_{r}}{ \partial t } \end{align*} $$

  • Part 2. $\mathbf{A}$

보조정리와 $(a)$를 이용해서 잘 정리하면 어려움 없이 얻을 수 있다.

$$ \begin{align*} \frac{ \partial \mathbf{A}}{ \partial t } &= \frac{ \partial }{ \partial t }\left( \frac{ \mathbf{v}}{c^{2}}V(\mathbf{r},t) \right) \\[1em] &= \frac{1}{c^{2}} \left( \frac{ \partial \mathbf{v}}{ \partial t }V +\mathbf{v}\frac{ \partial V }{ \partial t} \right) \\[1em] &= \frac{1}{c^{2}} \left[ \frac{ \partial \mathbf{v}}{ \partial t_{r} } \frac{ \partial t_{r}}{ \partial t }V + \mathbf{v}\frac{qc}{4\pi \epsilon_{0}} \frac{1}{(\eta c -\boldsymbol{\eta} \cdot \mathbf{v})^{2}} \left( c^{2} -c^{2}\frac{ \partial t}{ \partial t_{r}}-v^{2}+\boldsymbol{\eta}\cdot \mathbf{a} \right)\frac{ \partial t_{r}}{ \partial t }\right] \\[1em] &= \frac{1}{c^{2}}\frac{\partial t_{r}}{\partial t}\left[ \mathbf{a}\frac{qc}{4\pi \epsilon_{0}} \frac{1}{\eta c- \boldsymbol{\eta} \cdot \mathbf{v}}+ \mathbf{v}\frac{qc}{4\pi \epsilon_{0}} \frac{1}{(\eta c -\boldsymbol{\eta} \cdot \mathbf{v})^{2}} \left( c^{2} -c^{2}\frac{ \partial t}{ \partial t_{r}}-v^{2}+\boldsymbol{\eta}\cdot \mathbf{a} \right)\right] \\[1em] &= \frac{1}{c^{2}}\frac{\partial t_{r}}{\partial t}\frac{qc}{4\pi \epsilon_{0}} \frac{1}{(\eta c -\boldsymbol{\eta} \cdot \mathbf{v})^{2}}\left[ \mathbf{a}(\eta c +\boldsymbol{\eta}\cdot \mathbf{v})+ \mathbf{v} \left( c^{2} -c^{2}\frac{ \partial t}{ \partial t_{r}}-v^{2}+\boldsymbol{\eta}\cdot \mathbf{a} \right)\right] \end{align*} $$

보조정리에 의해

$$ \begin{align*} \frac{ \partial \mathbf{A}}{ \partial t }&= \frac{1}{c^{2}}\frac{\eta c}{\eta c-\boldsymbol{\eta}\cdot \mathbf{v}}\frac{qc}{4\pi \epsilon_{0}} \frac{1}{(\eta c -\boldsymbol{\eta} \cdot \mathbf{v})^{2}}\left[ \mathbf{a}(\eta c +\boldsymbol{\eta}\cdot \mathbf{v})+ \mathbf{v} \left( c^{2} -c^{2}\frac{\eta c-\boldsymbol{\eta}\cdot \mathbf{v}}{\eta c}-v^{2}+\boldsymbol{\eta}\cdot \mathbf{a} \right)\right] \\[1em] &=\frac{\eta}{c} \frac{qc}{4\pi \epsilon_{0}} \frac{1}{(\eta c -\boldsymbol{\eta} \cdot \mathbf{v})^{3}}\left[ \mathbf{a}(\eta c +\boldsymbol{\eta}\cdot \mathbf{v})+ c^{2}\mathbf{v} -c^{2}\mathbf{v}\left( \frac{\eta c-\boldsymbol{\eta}\cdot \mathbf{v}}{\eta c} \right)-v^{2 }\mathbf{v}+(\boldsymbol{\eta}\cdot \mathbf{a}) \mathbf{v}\right] \\[1em] &=\frac{\eta}{c} \frac{qc}{4\pi \epsilon_{0}} \frac{1}{(\eta c -\boldsymbol{\eta} \cdot \mathbf{v})^{3}}\left[ (\eta c +\boldsymbol{\eta}\cdot \mathbf{v})(\mathbf{a}-\frac{c^{2}\mathbf{v}}{\eta c})+ \mathbf{v}\left( c^{2} -v^{2}+\boldsymbol{\eta}\cdot \mathbf{a}\right) \right] \\[1em] &= \frac{qc}{4\pi \epsilon_{0}} \frac{1}{(\eta c -\boldsymbol{\eta} \cdot \mathbf{v})^{3}}\left[ (\eta c +\boldsymbol{\eta}\cdot \mathbf{v})(\eta\mathbf{a}/c-\mathbf{v})+ \frac{\eta}{c}\mathbf{v}\left( c^{2} -v^{2}+\boldsymbol{\eta}\cdot \mathbf{a}\right) \right] \end{align*} $$

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