Liénard-Wiechert Potentials
Overview1
The Liénard-Wiechert potentials for a point charge $q$ moving at a speed $\mathbf{v}$ at the retarded time $t_{r}$ are as follows.
$$ \begin{align*} V(\mathbf{r}, t) &= \frac{1}{4\pi \epsilon_{0}} \frac{qc}{ (\cR c -\bcR\cdot \mathbf{v})} \\ \mathbf{A}(\mathbf{r}, t) &= \frac{\mu_{0}}{4 \pi}\frac{qc \mathbf{v} }{(\cR c - \bcR\cdot \mathbf{v} )}=\frac{\mathbf{v}}{c^2}V(\mathbf{r}, t) \end{align*} $$
Here, $\bcR=\mathbf{r} -\mathbf{w}(t_{r})$ is the vector from the retarded position to the observation point, and $\mathbf{w}(t_{r})$ is the retarded position where the point charge is located at the retarded time.
Explanation
The French physicist Liénard and the German physicist Wiechert independently derived these equations in 1898 and 1900, respectively.
It might seem simple to calculate the potential for a moving point charge, but it’s actually not. When the charge/current density only changes in place, at least its position was fixed. But now, it’s not. In addition to considering the retarded time, one must also consider the retarded position. Especially now, the retarded time is no longer constant with respect to time. Without careful consideration, one can arrive at incorrect results like the following.
$$ \begin{align*} V(\mathbf{r}, t) &= \frac{1}{4\pi \epsilon_{0}} \int \frac{ \rho (\mathbf{r}^{\prime},t_{r}) }{\cR} d\tau^{\prime} \\ &= \frac{1}{4\pi \epsilon_{0}} \int \frac{ q\delta \big( \mathbf{r}^{\prime}-\mathbf{w}(t_{r}) \big) }{\cR} d\tau^{\prime} \\ &= \frac{1}{4 \pi \epsilon_{0}}\frac{q}{c} \end{align*} $$
Derivation
Consider calculating the potential and electromagnetic field for a changing charge/current density at a fixed position. Since it’s not moving, the retarded time is independent of time changes and only affected by the position. But now, assuming the point charge moves, the retarded time is affected by both position and time. Thus, it’s a variable, not a constant with respect to time, so we denote it not as $t_{r}$ but as $t^{\prime}$.
$$ V(\mathbf{r}, t) = \frac{1}{4\pi \epsilon_{0}} \int \frac{\rho (\mathbf{r}^{\prime},t^{\prime})}{ | \mathbf{r} -\mathbf{r}^{\prime} | }d\tau^{\prime} $$
Since the point charge exists precisely at one location, the charge density can be represented as follows.
$$ \rho (\mathbf{r}^{\prime},t)=q \delta \big( \mathbf{r}^{\prime} - \mathbf{w}(t^{\prime}) \big) $$
$\delta$ is the Dirac delta function. Therefore, the potential is
$$ \begin{equation} V(\mathbf{r}, t) = \frac{1}{4\pi \epsilon_{0}} \int \frac{q \delta \big( \mathbf{r}^{\prime} - \mathbf{w}(t^{\prime}) \big)}{ | \mathbf{r} -\mathbf{r}^{\prime} | }d\tau^{\prime} \end{equation} $$
The retarded time at the retarded position $\mathbf{w}(t^{\prime})$ can be obtained by subtracting the time it takes to go from the retarded position to the observation point from the current time $t$. Thus, the retarded time at $\mathbf{w}(t^{\prime})$ is
$$ t-\frac{ | \mathbf{r} -\mathbf{w}(t^{\prime}) | }{c} $$
To eliminate $t^{\prime}$ in favor of $(1)$, the delta function is used. By definition of the delta function, $1={\displaystyle \int }\delta \left( t^{\prime}-\big(t-\frac{ | \mathbf{r} -\mathbf{w}(t^{\prime}) | }{c} \big) \right) dt^{\prime}$, so multiplying it does not affect the result. Therefore,
$$ \begin{align*} V(\mathbf{r}, t) &= \frac{q}{4\pi \epsilon_{0}} \int \frac{ \delta \big( \mathbf{r}^{\prime} - \mathbf{w}(t^{\prime}) \big)}{ | \mathbf{r} -\mathbf{r}^{\prime} | }d\tau^{\prime} \int \delta \left( t^{\prime}-t+\frac{ | \mathbf{r} -\mathbf{w}(t^{\prime}) | }{c} \right) dt^{\prime} \\ &= \frac{q}{4\pi \epsilon_{0}} \int \int \frac{ \delta \big( \mathbf{r}^{\prime} - \mathbf{w}(t^{\prime}) \big)}{ | \mathbf{r} -\mathbf{r}^{\prime} | }\delta \left( t^{\prime}-t+\frac{ | \mathbf{r} -\mathbf{w}(t^{\prime}) | }{c} \right) d\tau^{\prime}dt^{\prime} \end{align*} $$
First solving the integral with respect to position, by the property of the delta function,
$$ V(\mathbf{r}, t) = \frac{q}{4\pi \epsilon_{0}} \int \frac{ 1}{ | \mathbf{r} -\mathbf{w}(t^{\prime}) | }\delta \left( t^{\prime}-t+\frac{ | \mathbf{r} -\mathbf{w}(t^{\prime}) | }{c} \right) dt^{\prime} $$
Also, by the property of the delta function[^2],
is obtained. Here, $t_{r}$ is the time when the particle that sends the message (electromagnetic wave) to the observing point $\mathbf{r}$ existed. In other words, it’s the retarded time with respect to $t$. Since we’re dealing with a moving ‘point charge’, there is only one point charge affecting any given position $\mathbf{r}$ at any time $t$, and thus there is only one retarded time. By substituting the above equation into the potential,
$$ \begin{align*} V(\mathbf{r}, t) &= \frac{q}{4\pi \epsilon_{0}} \int \frac{ 1}{ | \mathbf{r} -\mathbf{w}(t^{\prime}) | }\dfrac{\delta (t^{\prime}-t_{r}) }{1-\frac{\smallcrH\cdot \mathbf{v}}{c} } dt^{\prime} \\ &= \frac{q}{4\pi \epsilon_{0}} \frac{ 1}{ | \mathbf{r} -\mathbf{w}(t_{r}) | }\dfrac{1}{1-\frac{\smallcrH\cdot \mathbf{v}}{c} } \\ &= \frac{q}{4\pi \epsilon_{0}} \frac{ 1}{ \cR}\dfrac{1}{1-\frac{\smallcrH\cdot \mathbf{v}}{c} } \\ &= \frac{ 1}{ 4\pi \epsilon_{0} }\frac{qc}{(\cR c -\bcR\cdot \mathbf{v} )} \end{align*} $$ The same logic applies to the vector potential, leading to the result. The derivation is described in as much detail as possible, but the explanation is omitted since it’s almost the same. The current density is $\mathbf{J}(\mathbf{r}^{\prime},t^{\prime})=\rho (\mathbf{r}^{\prime}, t^{\prime})\mathbf{v}(t^{\prime})$, so
$$ \begin{align*} \mathbf{A} (\mathbf{r}, t) &=\frac{\mu_{0}}{4 \pi} \int \frac{ \mathbf{J}(\mathbf{r}^{\prime},t^{\prime})}{| \mathbf{r} -\mathbf{r}^{\prime}|}d\tau^{\prime} \\ &= \frac{\mu_{0}}{4 \pi} \int \frac{ \rho (\mathbf{r}^{\prime}, t^{\prime})\mathbf{v}(t^{\prime}) }{| \mathbf{r} -\mathbf{r}^{\prime}|}d\tau^{\prime} \\ &= \frac{q\mu_{0}}{4 \pi} \int \frac{ \delta \big( \mathbf{r}^{\prime} - \mathbf{w}(t^{\prime}) \big)\mathbf{v}(t^{\prime}) }{| \mathbf{r} -\mathbf{r}^{\prime}|}d\tau^{\prime} \\ &= \frac{q\mu_{0}}{4 \pi}\int \int \frac{ \delta \big( \mathbf{r}^{\prime} - \mathbf{w}(t^{\prime}) \big)\mathbf{v}(t^{\prime}) }{| \mathbf{r} -\mathbf{r}^{\prime}|} \delta \left( t^{\prime}-t+\frac{ | \mathbf{r} -\mathbf{w}(t^{\prime}) | }{c} \right)d\tau^{\prime} dt^{\prime} \\ &= \frac{q\mu_{0}}{4 \pi}\int \frac{ \mathbf{v}(t^{\prime}) }{| \mathbf{r} -\mathbf{w}(t^{\prime})|} \delta \left( t^{\prime}-t+\frac{ | \mathbf{r} -\mathbf{w}(t^{\prime}) | }{c} \right) dt^{\prime} \\ &= \frac{q\mu_{0}}{4 \pi}\int \frac{ \mathbf{v}(t^{\prime}) }{| \mathbf{r} -\mathbf{w}(t^{\prime})|} \dfrac{\delta (t^{\prime}-t_{r}) }{1-\frac{\smallcrH\cdot \mathbf{v}}{c} } dt^{\prime} \\ &= \frac{q\mu_{0}}{4 \pi} \frac{ \mathbf{v}(t_{r}) }{| \mathbf{r} -\mathbf{w}(t_{r})|} \dfrac{ 1}{1-\frac{\smallcrH\cdot \mathbf{v}}{c} } \\ &= \frac{q\mu_{0}}{4 \pi} \frac{ \mathbf{v}(t_{r}) }{\cR} \dfrac{ 1}{1-\frac{\smallcrH\cdot \mathbf{v}}{c} } \\ &= \frac{\mu_{0}}{4 \pi}\frac{qc \mathbf{v}}{(\cR c -\bcR\cdot \mathbf{v} )} \\ &= \frac{\mu_{0}\epsilon_{0}}{4 \pi\epsilon_{0}}\frac{qc \mathbf{v}}{(\cR c -\bcR\cdot \mathbf{v} )} \\ &= \frac{\mathbf{v} }{c^2} \frac{1}{4 \pi\epsilon_{0}}\frac{qc }{(\cR c -\bcR\cdot \mathbf{v} )} \\ &=\dfrac{\mathbf{v} } {c^2}V(\mathbf{r},t) \end{align*} $$
The second to last equality holds by virtue of $\mu_{0}\epsilon_{0}=\dfrac{1}{c^2}$.
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David J. Griffiths, 기초전자기학(Introduction to Electrodynamics, 김진승 역) (4th Edition1 2014), p488-492 ↩︎