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리에나르-비케르트 전위의 시간 도함수 📂전자기학

리에나르-비케르트 전위의 시간 도함수

개요

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리에나르-비케르트 전위의 시간에 대한 도함수는 다음과 같다.

∂V∂t=qc4πϵ01(c−⋅v)2(c2−c2∂t∂tr−v2+⋅a)∂tr∂t∂A∂t=qc4πϵ01(c−⋅v)3[(c+⋅v)(a/c−v)+cv(c2−v2+⋅a)] \begin{align*} \frac{ \partial V}{ \partial t} &= \frac{qc}{4\pi \epsilon_{0}} \frac{1}{(\cR c -\bcR \cdot \mathbf{v})^{2}} \left( c^{2} -c^{2}\frac{ \partial t}{ \partial t_{r}}-v^{2}+\bcR\cdot \mathbf{a} \right)\frac{ \partial t_{r}}{ \partial t } \\ \frac{ \partial \mathbf{A}}{ \partial t } &= \frac{qc}{4\pi \epsilon_{0}} \frac{1}{(\cR c -\bcR \cdot \mathbf{v})^{3}}\left[ (\cR c +\bcR\cdot \mathbf{v})(\cR\mathbf{a}/c-\mathbf{v})+ \frac{\cR}{c}\mathbf{v}\left( c^{2} -v^{2}+\bcR\cdot \mathbf{a}\right) \right] \end{align*}

보조정리

지연 시각의 시간 미분은 아래와 같다.

∂tr∂t=cc−⋅v=c⋅u \frac{ \partial t_{r}}{ \partial t}=\frac{\cR c}{\cR c-\bcR\cdot \mathbf{v}}=\frac{ \cR c}{\bcR \cdot \mathbf{u}}

이 때 u=c−v\mathbf{u}=c\crH-\mathbf{v}이다.

증명

지연 시각의 정의에 따라

c(t−tr)= c(t-t_{r})=\cR

이므로 양변을 제곱하면

c2(t−tr)2=2=⋅ c^{2}(t-t_{r})^{2}=\cR ^{2}=\bcR \cdot \bcR

양변을 tt로 미분하면

2c2(t−tr)(1−∂tr∂t)=2∂∂t⋅(1) 2c^{2}(t-t_{r}) \left( 1-\frac{ \partial t_{r}}{ \partial t } \right)=2\frac{ \partial \bcR}{ \partial t }\cdot \bcR \tag{1}

이때 =r−w(tr)\bcR=\mathbf{r}-\mathbf{w}(t_{r})이므로

∂∂t=−∂w∂t=−∂w∂tr∂tr∂t=−v∂tr∂t \frac{ \partial \bcR}{ \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)(1)에 대입하고 정리하면

c(1−∂tr∂t)=−⋅v∂tr∂t  ⟹  c−c∂tr∂t=−⋅v∂tr∂t  ⟹  ∂tr∂t=cc−⋅v=c⋅u \begin{align*} && c\cR \left( 1-\frac{ \partial t_{r}}{ \partial t } \right) &= -\bcR\cdot \mathbf{v}\frac{ \partial t_{r}}{ \partial t } \\[1em] \implies && c\cR -c\cR \frac{ \partial t_{r}}{ \partial t} &= -\bcR\cdot\mathbf{v}\frac{ \partial t_{r} }{ \partial t } \\[1em] \implies && \frac{ \partial t_{r}}{ \partial t } &= \frac{c\cR}{c\cR -\bcR\cdot \mathbf{v}} \\[1em] && &= \frac{c\cR}{\bcR \cdot \mathbf{u}} \end{align*}

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증명

리에나르-비케르트 전위

지연시각 trt_{r}에서 속도 v\mathbf{v}로 움직이는 점전하 qq에 대한 전위는 다음과 같다.

V(r,t)=14πϵ0qc(c−⋅v)A(r,t)=μ04πqcv(c−⋅v)=vc2V(r,t) \begin{align*} V(\mathbf{r}, t) &= \frac{1}{4\pi \epsilon_{0}} \frac{qc}{ (\cR c -\bcR\cdot \mathbf{v})} \\ \mathbf{A}(\mathbf{r}, t) &= \frac{\mu_{0}}{4 \pi}\frac{qc \mathbf{v} }{(\cR c - \bcR\cdot \mathbf{v} )}=\frac{\mathbf{v}}{c^2}V(\mathbf{r}, t) \end{align*}

이때 =r−w(tr)\bcR=\mathbf{r} -\mathbf{w}(t_{r})는 지연위치에서 관찰점까지의 벡터, w(tr)\mathbf{w}(t_{r})은 지연시각에서의 점전하의 위치인 지연 위치이다.

  • Part 1. VV

스칼라 전위 VV를 tt로 미분하면,

∂V∂t=qc4πϵ0∂∂t(1c−⋅v)=qc4πϵ0∂∂tr(1c−⋅v)∂tr∂t \begin{align*} \frac{ \partial V}{ \partial t } &= \frac{qc}{4\pi \epsilon_{0}}\frac{\partial}{\partial t} \left( \frac{1}{\cR c -\bcR\cdot \mathbf{v}} \right) \\ &= \frac{qc}{4\pi \epsilon_{0}}\frac{\partial}{\partial t_{r}} \left(\frac{1}{\cR c -\bcR\cdot \mathbf{v}}\right) \frac{ \partial t_{r} }{ \partial t } \end{align*}

이 때 ddx(1f(x))=ddf(x)(1f(x))df(x)dx=−1(f(x))2f′(x)\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

∂V∂t=qc4πϵ0−1(c−⋅v)2∂(c−⋅v)∂tr∂tr∂t=qc4πϵ0−1(c−⋅v)2(c∂∂tr−∂∂tr⋅v−⋅∂v∂tr)∂tr∂t \begin{align*} \frac{ \partial V}{ \partial t }&=\frac{qc}{4\pi \epsilon_{0}} \frac{-1}{(\cR c -\bcR \cdot \mathbf{v})^{2}}\frac{ \partial (\cR c -\bcR \cdot \mathbf{v})}{ \partial t_{r} }\frac{ \partial t_{r}}{ \partial t } \\ &= \frac{qc}{4\pi \epsilon_{0}} \frac{-1}{(\cR c -\bcR \cdot \mathbf{v})^{2}} \left( c\frac{ \partial \cR}{ \partial t_{r}} - \frac{ \partial \bcR}{ \partial t_{r} }\cdot \mathbf{v}-\bcR\cdot \frac{ \partial \mathbf{v}}{ \partial t_{r} }\right)\frac{ \partial t_{r}}{ \partial t } \end{align*}

=r−w(tr)\bcR=\mathbf{r}-\mathbf{w}(t_{r})이므로

∂∂tr=−∂w(tr)∂tr=−v(tr) \frac{ \partial \bcR}{ \partial t_{r}}=-\frac{ \partial \mathbf{w}(t_{r})}{ \partial t_{r} }=-\mathbf{v}(t_{r})

∣∣==c(t−tr)| \bcR|=\cR=c(t-t_{r})이므로

∂∂tr=c∂t∂tr−c \frac{ \partial \cR}{ \partial t_{r}}=c\frac{ \partial t}{ \partial t_{r} }-c

그러므로

∂V∂t=qc4πϵ0−1(c−⋅v)2(−c2+c2∂t∂tr+v2−⋅a)∂tr∂t=qc4πϵ01(c−⋅v)2(c2−c2∂t∂tr−v2+⋅a)∂tr∂t \begin{align*} \frac{ \partial V}{ \partial t} &=\frac{qc}{4\pi \epsilon_{0}} \frac{-1}{(\cR c -\bcR \cdot \mathbf{v})^{2}} \left( -c^{2} +c^{2}\frac{ \partial t}{ \partial t_{r}}+v^{2}-\bcR\cdot \mathbf{a} \right)\frac{ \partial t_{r}}{ \partial t } \\ &= \frac{qc}{4\pi \epsilon_{0}} \frac{1}{(\cR c -\bcR \cdot \mathbf{v})^{2}} \left( c^{2} -c^{2}\frac{ \partial t}{ \partial t_{r}}-v^{2}+\bcR\cdot \mathbf{a} \right)\frac{ \partial t_{r}}{ \partial t } \end{align*}

  • Part 2. A\mathbf{A}

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

∂A∂t=∂∂t(vc2V(r,t))=1c2(∂v∂tV+v∂V∂t)=1c2[∂v∂tr∂tr∂tV+vqc4πϵ01(c−⋅v)2(c2−c2∂t∂tr−v2+⋅a)∂tr∂t]=1c2∂tr∂t[aqc4πϵ01c−⋅v+vqc4πϵ01(c−⋅v)2(c2−c2∂t∂tr−v2+⋅a)]=1c2∂tr∂tqc4πϵ01(c−⋅v)2[a(c+⋅v)+v(c2−c2∂t∂tr−v2+⋅a)] \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}{(\cR c -\bcR \cdot \mathbf{v})^{2}} \left( c^{2} -c^{2}\frac{ \partial t}{ \partial t_{r}}-v^{2}+\bcR\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}{\cR c- \bcR \cdot \mathbf{v}}+ \mathbf{v}\frac{qc}{4\pi \epsilon_{0}} \frac{1}{(\cR c -\bcR \cdot \mathbf{v})^{2}} \left( c^{2} -c^{2}\frac{ \partial t}{ \partial t_{r}}-v^{2}+\bcR\cdot \mathbf{a} \right)\right] \\[1em] &= \frac{1}{c^{2}}\frac{\partial t_{r}}{\partial t}\frac{qc}{4\pi \epsilon_{0}} \frac{1}{(\cR c -\bcR \cdot \mathbf{v})^{2}}\left[ \mathbf{a}(\cR c +\bcR\cdot \mathbf{v})+ \mathbf{v} \left( c^{2} -c^{2}\frac{ \partial t}{ \partial t_{r}}-v^{2}+\bcR\cdot \mathbf{a} \right)\right] \end{align*}

보조정리에 의해

∂A∂t=1c2cc−⋅vqc4πϵ01(c−⋅v)2[a(c+⋅v)+v(c2−c2c−⋅vc−v2+⋅a)]=cqc4πϵ01(c−⋅v)3[a(c+⋅v)+c2v−c2v(c−⋅vc)−v2v+(⋅a)v]=cqc4πϵ01(c−⋅v)3[(c+⋅v)(a−c2vc)+v(c2−v2+⋅a)]=qc4πϵ01(c−⋅v)3[(c+⋅v)(a/c−v)+cv(c2−v2+⋅a)] \begin{align*} \frac{ \partial \mathbf{A}}{ \partial t }&= \frac{1}{c^{2}}\frac{\cR c}{\cR c-\bcR\cdot \mathbf{v}}\frac{qc}{4\pi \epsilon_{0}} \frac{1}{(\cR c -\bcR \cdot \mathbf{v})^{2}}\left[ \mathbf{a}(\cR c +\bcR\cdot \mathbf{v})+ \mathbf{v} \left( c^{2} -c^{2}\frac{\cR c-\bcR\cdot \mathbf{v}}{\cR c}-v^{2}+\bcR\cdot \mathbf{a} \right)\right] \\[1em] &=\frac{\cR}{c} \frac{qc}{4\pi \epsilon_{0}} \frac{1}{(\cR c -\bcR \cdot \mathbf{v})^{3}}\left[ \mathbf{a}(\cR c +\bcR\cdot \mathbf{v})+ c^{2}\mathbf{v} -c^{2}\mathbf{v}\left( \frac{\cR c-\bcR\cdot \mathbf{v}}{\cR c} \right)-v^{2 }\mathbf{v}+(\bcR\cdot \mathbf{a}) \mathbf{v}\right] \\[1em] &=\frac{\cR}{c} \frac{qc}{4\pi \epsilon_{0}} \frac{1}{(\cR c -\bcR \cdot \mathbf{v})^{3}}\left[ (\cR c +\bcR\cdot \mathbf{v})(\mathbf{a}-\frac{c^{2}\mathbf{v}}{\cR c})+ \mathbf{v}\left( c^{2} -v^{2}+\bcR\cdot \mathbf{a}\right) \right] \\[1em] &= \frac{qc}{4\pi \epsilon_{0}} \frac{1}{(\cR c -\bcR \cdot \mathbf{v})^{3}}\left[ (\cR c +\bcR\cdot \mathbf{v})(\cR\mathbf{a}/c-\mathbf{v})+ \frac{\cR}{c}\mathbf{v}\left( c^{2} -v^{2}+\bcR\cdot \mathbf{a}\right) \right] \end{align*}

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