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Liénard-Wiechert Potentials 📂Electrodynamics

Liénard-Wiechert Potentials

Overview1

The Liénard-Wiechert potentials for a point charge qq moving at a speed v\mathbf{v} at the retarded time trt_{r} are as follows.

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*}

Here, =r−w(tr)\bcR=\mathbf{r} -\mathbf{w}(t_{r}) is the vector from the retarded position to the observation point, and w(tr)\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.

V(r,t)=14πϵ0∫ρ(r′,tr)dτ′=14πϵ0∫qδ(r′−w(tr))dτ′=14πϵ0qc \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 trt_{r} but as t′t^{\prime}.

V(r,t)=14πϵ0∫ρ(r′,t′)∣r−r′∣dτ′ 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.

ρ(r′,t)=qδ(r′−w(t′)) \rho (\mathbf{r}^{\prime},t)=q \delta \big( \mathbf{r}^{\prime} - \mathbf{w}(t^{\prime}) \big)

δ\delta is the Dirac delta function. Therefore, the potential is

V(r,t)=14πϵ0∫qδ(r′−w(t′))∣r−r′∣dτ′ \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 w(t′)\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 tt. Thus, the retarded time at w(t′)\mathbf{w}(t^{\prime}) is

t−∣r−w(t′)∣c t-\frac{ | \mathbf{r} -\mathbf{w}(t^{\prime}) | }{c}

To eliminate t′t^{\prime} in favor of (1)(1), the delta function is used. By definition of the delta function, 1=∫δ(t′−(t−∣r−w(t′)∣c))dt′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,

V(r,t)=q4πϵ0∫δ(r′−w(t′))∣r−r′∣dτ′∫δ(t′−t+∣r−w(t′)∣c)dt′=q4πϵ0∫∫δ(r′−w(t′))∣r−r′∣δ(t′−t+∣r−w(t′)∣c)dτ′dt′ \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(r,t)=q4πϵ0∫1∣r−w(t′)∣δ(t′−t+∣r−w(t′)∣c)dt′ 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],

δ(t′−t+∣r−w(t′)∣c)=δ(t′−tr)1−⋅vc \delta \left( t^{\prime}-t+\frac{ | \mathbf{r} -\mathbf{w}(t^{\prime}) | }{c} \right) =\dfrac{\delta (t^{\prime}-t_{r}) }{1-\frac{\smallacrH\cdot \mathbf{v}}{c} }

is obtained. Here, trt_{r} is the time when the particle that sends the message (electromagnetic wave) to the observing point r\mathbf{r} existed. In other words, it’s the retarded time with respect to tt. Since we’re dealing with a moving ‘point charge’, there is only one point charge affecting any given position r\mathbf{r} at any time tt, and thus there is only one retarded time. By substituting the above equation into the potential,

V(r,t)=q4πϵ0∫1∣r−w(t′)∣δ(t′−tr)1−⋅vcdt′=q4πϵ01∣r−w(tr)∣11−⋅vc=q4πϵ0111−⋅vc=14πϵ0qc(c−⋅v) \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 J(r′,t′)=ρ(r′,t′)v(t′)\mathbf{J}(\mathbf{r}^{\prime},t^{\prime})=\rho (\mathbf{r}^{\prime}, t^{\prime})\mathbf{v}(t^{\prime}), so

A(r,t)=μ04π∫J(r′,t′)∣r−r′∣dτ′=μ04π∫ρ(r′,t′)v(t′)∣r−r′∣dτ′=qμ04π∫δ(r′−w(t′))v(t′)∣r−r′∣dτ′=qμ04π∫∫δ(r′−w(t′))v(t′)∣r−r′∣δ(t′−t+∣r−w(t′)∣c)dτ′dt′=qμ04π∫v(t′)∣r−w(t′)∣δ(t′−t+∣r−w(t′)∣c)dt′=qμ04π∫v(t′)∣r−w(t′)∣δ(t′−tr)1−⋅vcdt′=qμ04πv(tr)∣r−w(tr)∣11−⋅vc=qμ04πv(tr)11−⋅vc=μ04πqcv(c−⋅v)=μ0ϵ04πϵ0qcv(c−⋅v)=vc214πϵ0qc(c−⋅v)=vc2V(r,t) \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 μ0ϵ0=1c2\mu_{0}\epsilon_{0}=\dfrac{1}{c^2}.

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  1. David J. Griffiths, 기초전자기학(Introduction to Electrodynamics, 김진승 역) (4th Edition1 2014), p488-492 ↩︎