Time Derivatives of the Lienard-Wiechert Potentials
Overview
The time derivative of the Liénard-Wiechert potential is as follows.
$$ \begin{align*} \frac{ \partial V}{ \partial t} &= \frac{qc}{4\pi \epsilon_{0}} \frac{1}{(\cR c -\bcR \cdot \mathbf{v})^{2}} \left( c^{2} -c^{2}\frac{ \partial t}{ \partial t_{r}}-v^{2}+\bcR\cdot \mathbf{a} \right)\frac{ \partial t_{r}}{ \partial t } \\ \frac{ \partial \mathbf{A}}{ \partial t } &= \frac{qc}{4\pi \epsilon_{0}} \frac{1}{(\cR c -\bcR \cdot \mathbf{v})^{3}}\left[ (\cR c +\bcR\cdot \mathbf{v})(\cR\mathbf{a}/c-\mathbf{v})+ \frac{\cR}{c}\mathbf{v}\left( c^{2} -v^{2}+\bcR\cdot \mathbf{a}\right) \right] \end{align*} $$
Lemma
The time derivative of the retarded time is as follows.
$$ \frac{ \partial t_{r}}{ \partial t}=\frac{\cR c}{\cR c-\bcR\cdot \mathbf{v}}=\frac{ \cR c}{\bcR \cdot \mathbf{u}} $$
At this time, it is $\mathbf{u}=c\crH-\mathbf{v}$.
Proof
According to the definition of retarded time,
$$ c(t-t_{r})=\cR $$
Therefore, squaring both sides,
$$ c^{2}(t-t_{r})^{2}=\cR ^{2}=\bcR \cdot \bcR $$
Differentiating both sides by $t$,
$$ 2c^{2}(t-t_{r}) \left( 1-\frac{ \partial t_{r}}{ \partial t } \right)=2\frac{ \partial \bcR}{ \partial t }\cdot \bcR \tag{1} $$
At this time, since it is $\bcR=\mathbf{r}-\mathbf{w}(t_{r})$,
$$ \frac{ \partial \bcR}{ \partial t}=-\frac{ \partial \mathbf{w}}{ \partial t}=-\frac{ \partial \mathbf{w}}{ \partial t_{r}}\frac{ \partial t_{r}}{ \partial t}=-\mathbf{v}\frac{ \partial t_{r} }{ \partial t} $$
It is. Substituting this into $(1)$ and rearranging,
$$ \begin{align*} && c\cR \left( 1-\frac{ \partial t_{r}}{ \partial t } \right) &= -\bcR\cdot \mathbf{v}\frac{ \partial t_{r}}{ \partial t } \\[1em] \implies && c\cR -c\cR \frac{ \partial t_{r}}{ \partial t} &= -\bcR\cdot\mathbf{v}\frac{ \partial t_{r} }{ \partial t } \\[1em] \implies && \frac{ \partial t_{r}}{ \partial t } &= \frac{c\cR}{c\cR -\bcR\cdot \mathbf{v}} \\[1em] && &= \frac{c\cR}{\bcR \cdot \mathbf{u}} \end{align*} $$
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Proof
Retarded time The potential for a point charge moving at speed $t_{r}$ from $\mathbf{v}$ to $q$ is as follows.
$$ \begin{align*} V(\mathbf{r}, t) &= \frac{1}{4\pi \epsilon_{0}} \frac{qc}{ (\cR c -\bcR\cdot \mathbf{v})} \\ \mathbf{A}(\mathbf{r}, t) &= \frac{\mu_{0}}{4 \pi}\frac{qc \mathbf{v} }{(\cR c - \bcR\cdot \mathbf{v} )}=\frac{\mathbf{v}}{c^2}V(\mathbf{r}, t) \end{align*} $$
At this time, $\bcR=\mathbf{r} -\mathbf{w}(t_{r})$ is the vector from the retarded position to the observation point, and $\mathbf{w}(t_{r})$ is the retarded position, which is the position of the point charge at the retarded time.
- Part 1. $V$
Differentiating the scalar potential $V$ by $t$,
$$ \begin{align*} \frac{ \partial V}{ \partial t } &= \frac{qc}{4\pi \epsilon_{0}}\frac{\partial}{\partial t} \left( \frac{1}{\cR c -\bcR\cdot \mathbf{v}} \right) \\ &= \frac{qc}{4\pi \epsilon_{0}}\frac{\partial}{\partial t_{r}} \left(\frac{1}{\cR c -\bcR\cdot \mathbf{v}}\right) \frac{ \partial t_{r} }{ \partial t } \end{align*} $$
At this time, since it is $\dfrac{ d }{ dx }\left( \dfrac{1}{f(x)} \right)=\dfrac{ d }{ d f(x) }\left(\dfrac{1}{f(x)} \right)\dfrac{ d f(x)}{ d x }=\dfrac{-1}{\left(f(x)\right)^{2}}f^{\prime}(x)$,
$$ \begin{align*} \frac{ \partial V}{ \partial t }&=\frac{qc}{4\pi \epsilon_{0}} \frac{-1}{(\cR c -\bcR \cdot \mathbf{v})^{2}}\frac{ \partial (\cR c -\bcR \cdot \mathbf{v})}{ \partial t_{r} }\frac{ \partial t_{r}}{ \partial t } \\ &= \frac{qc}{4\pi \epsilon_{0}} \frac{-1}{(\cR c -\bcR \cdot \mathbf{v})^{2}} \left( c\frac{ \partial \cR}{ \partial t_{r}} - \frac{ \partial \bcR}{ \partial t_{r} }\cdot \mathbf{v}-\bcR\cdot \frac{ \partial \mathbf{v}}{ \partial t_{r} }\right)\frac{ \partial t_{r}}{ \partial t } \end{align*} $$
Since it is $\bcR=\mathbf{r}-\mathbf{w}(t_{r})$,
$$ \frac{ \partial \bcR}{ \partial t_{r}}=-\frac{ \partial \mathbf{w}(t_{r})}{ \partial t_{r} }=-\mathbf{v}(t_{r}) $$
Since it is $| \bcR|=\cR=c(t-t_{r})$,
$$ \frac{ \partial \cR}{ \partial t_{r}}=c\frac{ \partial t}{ \partial t_{r} }-c $$
Therefore,
$$ \begin{align*} \frac{ \partial V}{ \partial t} &=\frac{qc}{4\pi \epsilon_{0}} \frac{-1}{(\cR c -\bcR \cdot \mathbf{v})^{2}} \left( -c^{2} +c^{2}\frac{ \partial t}{ \partial t_{r}}+v^{2}-\bcR\cdot \mathbf{a} \right)\frac{ \partial t_{r}}{ \partial t } \\ &= \frac{qc}{4\pi \epsilon_{0}} \frac{1}{(\cR c -\bcR \cdot \mathbf{v})^{2}} \left( c^{2} -c^{2}\frac{ \partial t}{ \partial t_{r}}-v^{2}+\bcR\cdot \mathbf{a} \right)\frac{ \partial t_{r}}{ \partial t } \end{align*} $$
- Part 2. $\mathbf{A}$
Using the lemma and $(a)$, it can be easily obtained.
$$ \begin{align*} \frac{ \partial \mathbf{A}}{ \partial t } &= \frac{ \partial }{ \partial t }\left( \frac{ \mathbf{v}}{c^{2}}V(\mathbf{r},t) \right) \\[1em] &= \frac{1}{c^{2}} \left( \frac{ \partial \mathbf{v}}{ \partial t }V +\mathbf{v}\frac{ \partial V }{ \partial t} \right) \\[1em] &= \frac{1}{c^{2}} \left[ \frac{ \partial \mathbf{v}}{ \partial t_{r} } \frac{ \partial t_{r}}{ \partial t }V + \mathbf{v}\frac{qc}{4\pi \epsilon_{0}} \frac{1}{(\cR c -\bcR \cdot \mathbf{v})^{2}} \left( c^{2} -c^{2}\frac{ \partial t}{ \partial t_{r}}-v^{2}+\bcR\cdot \mathbf{a} \right)\frac{ \partial t_{r}}{ \partial t }\right] \\[1em] &= \frac{1}{c^{2}}\frac{\partial t_{r}}{\partial t}\left[ \mathbf{a}\frac{qc}{4\pi \epsilon_{0}} \frac{1}{\cR c- \bcR \cdot \mathbf{v}}+ \mathbf{v}\frac{qc}{4\pi \epsilon_{0}} \frac{1}{(\cR c -\bcR \cdot \mathbf{v})^{2}} \left( c^{2} -c^{2}\frac{ \partial t}{ \partial t_{r}}-v^{2}+\bcR\cdot \mathbf{a} \right)\right] \\[1em] &= \frac{1}{c^{2}}\frac{\partial t_{r}}{\partial t}\frac{qc}{4\pi \epsilon_{0}} \frac{1}{(\cR c -\bcR \cdot \mathbf{v})^{2}}\left[ \mathbf{a}(\cR c +\bcR\cdot \mathbf{v})+ \mathbf{v} \left( c^{2} -c^{2}\frac{ \partial t}{ \partial t_{r}}-v^{2}+\bcR\cdot \mathbf{a} \right)\right] \end{align*} $$
According to the lemma,
$$ \begin{align*} \frac{ \partial \mathbf{A}}{ \partial t }&= \frac{1}{c^{2}}\frac{\cR c}{\cR c-\bcR\cdot \mathbf{v}}\frac{qc}{4\pi \epsilon_{0}} \frac{1}{(\cR c -\bcR \cdot \mathbf{v})^{2}}\left[ \mathbf{a}(\cR c +\bcR\cdot \mathbf{v})+ \mathbf{v} \left( c^{2} -c^{2}\frac{\cR c-\bcR\cdot \mathbf{v}}{\cR c}-v^{2}+\bcR\cdot \mathbf{a} \right)\right] \\[1em] &=\frac{\cR}{c} \frac{qc}{4\pi \epsilon_{0}} \frac{1}{(\cR c -\bcR \cdot \mathbf{v})^{3}}\left[ \mathbf{a}(\cR c +\bcR\cdot \mathbf{v})+ c^{2}\mathbf{v} -c^{2}\mathbf{v}\left( \frac{\cR c-\bcR\cdot \mathbf{v}}{\cR c} \right)-v^{2 }\mathbf{v}+(\bcR\cdot \mathbf{a}) \mathbf{v}\right] \\[1em] &=\frac{\cR}{c} \frac{qc}{4\pi \epsilon_{0}} \frac{1}{(\cR c -\bcR \cdot \mathbf{v})^{3}}\left[ (\cR c +\bcR\cdot \mathbf{v})(\mathbf{a}-\frac{c^{2}\mathbf{v}}{\cR c})+ \mathbf{v}\left( c^{2} -v^{2}+\bcR\cdot \mathbf{a}\right) \right] \\[1em] &= \frac{qc}{4\pi \epsilon_{0}} \frac{1}{(\cR c -\bcR \cdot \mathbf{v})^{3}}\left[ (\cR c +\bcR\cdot \mathbf{v})(\cR\mathbf{a}/c-\mathbf{v})+ \frac{\cR}{c}\mathbf{v}\left( c^{2} -v^{2}+\bcR\cdot \mathbf{a}\right) \right] \end{align*} $$
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