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Tuorui "v1ncent19" Peng

En voyage dans l'espace de Hilbert.

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Mathematics & Statistics≈ 2 min readEnglish

Liénard–Wiechert Potential

In dealing with electro-magnetic field of moving charged particle qq at r⃗0(t)\vec{r}_0(t) with velocity v⃗0(t)\vec{v}_0(t):

ρ(r⃗,t)=qδ(r⃗−r⃗0(t)),j⃗(r⃗,t)=ρ(r⃗,t)v⃗0(t)=qv⃗0(t)δ(r⃗−r⃗0(t))\begin{align} \rho (\vec{r},t)=q\delta (\vec{r}-\vec{r}_0(t)),\quad \vec{j}(\vec{r},t)=\rho (\vec{r},t)\vec{v}_0(t)=q\vec{v}_0(t)\delta (\vec{r}-\vec{r}_0(t)) \end{align}

4-potential given with Lorentz gauge:

ϕ(r⃗,t)=∫ d3x′ϕ(r⃗′,t−Rc)4πε0R=∫ d3x′qδ(r⃗′−r⃗0(t−∣r⃗−r⃗′∣c))4πε0∣r⃗−r⃗′∣A⃗(r⃗,t)=∫ d3x′μ0j⃗(r⃗′,t−Rc)4πR=∫ d3x′μ0qv⃗0(t−∣r⃗−r⃗′∣c)δ(r⃗′−r⃗0(t−∣r⃗−r⃗′∣c))4π∣r⃗−r⃗′∣\begin{align} \phi (\vec{r},t) =&\int \,\mathrm{d}^3x' \dfrac{\phi (\vec{r}',t-\dfrac{R}{c})}{4\pi \varepsilon _0R}=\int \,\mathrm{d}^3x' \dfrac{q\delta \left(\vec{r}'-\vec{r}_0(t-\frac{|\vec{r}-\vec{r}'|}{c})\right)}{4\pi \varepsilon _0|\vec{r}-\vec{r}'|}\\ \vec{A}(\vec{r},t)=&\int \,\mathrm{d}^3x' \dfrac{\mu _0\vec{j}(\vec{r}',t-\dfrac{R}{c})}{4\pi R}=\int \,\mathrm{d}^3x' \dfrac{\mu _0q\vec{v}_0(t-\frac{|\vec{r}-\vec{r}'|}{c})\delta \left( \vec{r}'-\vec{r}_0(t-\frac{|\vec{r}-\vec{r}'|}{c}) \right)}{4\pi |\vec{r}-\vec{r}'|} \end{align}

Note that only when r⃗′\vec{r}' s.t. r′⃗−r⃗0(t−∣r⃗−r⃗′∣c)=0\vec{r'}-\vec{r}_0(t-\frac{|\vec{r}-\vec{r}'|}{c})=0, the formula in integral is non-zero. We denote the solution of it as (r⃗∗,t∗)(\vec{r}^*,t^*), i.e.:

{r⃗∗−r⃗0(t−∣r⃗−r⃗∗∣c)=0t∗=t−∣r⃗−r⃗∗∣c⇒{r⃗∗=r⃗0(t∗)v⃗∗:=v⃗0(t∗)∣r⃗−r⃗∗∣=c(t−t∗)\begin{align} \begin{cases} \vec{r}^*-\vec{r}_0(t-\frac{|\vec{r}-\vec{r}^*|}{c})=0\\ t^*=t-\frac{|\vec{r}-\vec{r}^*|}{c} \end{cases}\Rightarrow \begin{cases} \vec{r}^*=\vec{r}_0(t^*)\\ \vec{v}^*:=\vec{v}_0(t^*)\\ |\vec{r}-\vec{r}^*|=c(t-t^*) \end{cases} \end{align}

in this way:

ϕ(r⃗,t)=q4πε0∣r⃗−r⃗∗∣∫ d3x′δ(r⃗′−r⃗0(t−∣r⃗−r⃗′∣c))A⃗(r⃗,t)=μ0qv⃗∗4π∣r⃗−r⃗∗∣∫ d3x′δ(r⃗′−r⃗0(t−∣r⃗−r⃗′∣c))\begin{align} \phi (\vec{r},t)=\dfrac{q}{4\pi\varepsilon _0|\vec{r}-\vec{r}^*|}\int \,\mathrm{d}^3x' \delta \left( \vec{r}'-\vec{r}_0(t-\dfrac{|\vec{r}-\vec{r}'|}{c}) \right)\\ \vec{A} (\vec{r},t)=\dfrac{\mu _0q\vec{v}^*}{4\pi|\vec{r}-\vec{r}^*|}\int \,\mathrm{d}^3x' \delta \left( \vec{r}'-\vec{r}_0(t-\dfrac{|\vec{r}-\vec{r}'|}{c}) \right) \end{align}

with coordinate transformation r⃗′↦r⃗∗′=r⃗′−r⃗0(t−∣r⃗−r⃗′∣c)\vec{r}'\mapsto \vec{r}^{*\prime}=\vec{r}'-\vec{r}_0(t-\frac{|\vec{r}-\vec{r}'|}{c}):

∥∂r⃗∗′∂r⃗′∥∣r⃗′=r⃗∗=∥∂∂xjxi′−x0i(t−∣r⃗−r⃗′∣c)∥∣r⃗′=r⃗∗=1−(r⃗−r⃗∗)⋅v⃗∗∣r⃗−r⃗∗∣c∫ d3x′δ(r⃗′−r⃗0(t−∣r⃗−r⃗′∣c))=∫ d3x∗′δ(r⃗∗′)∥∂r⃗′∂r⃗∗′∥=1/∥∂r⃗∗′∂r⃗′∥∣r⃗′=r⃗∗\begin{align} \left\Vert \dfrac{\partial^{} \vec{r}^{*\prime}}{\partial \vec{r}^\prime} \right\Vert\Big|_{\vec{r}'=\vec{r}^*}=&\left\Vert \dfrac{\partial^{} }{\partial x_j^{}} x_i'-x_{0i}(t-\dfrac{|\vec{r}-\vec{r}'|}{c}) \right\Vert\Big|_{\vec{r}'=\vec{r}^*}=1-\dfrac{(\vec{r}-\vec{r}^*)\cdot \vec{v}^*}{|\vec{r}-\vec{r}^*|c}\\ \int \,\mathrm{d}^3x' \delta \left( \vec{r}'-\vec{r}_0(t-\dfrac{|\vec{r}-\vec{r}'|}{c}) \right)=&\int \,\mathrm{d}^3x^{*\prime}\delta (\vec{r}^{*\prime})\left\Vert \dfrac{\partial^{} \vec{r}'}{\partial \vec{r}^{*\prime}} \right\Vert\\ =&1\Bigg/ \left\Vert \dfrac{\partial^{} \vec{r}^{*\prime}}{\partial \vec{r}^\prime} \right\Vert\Big|_{\vec{r}'=\vec{r}^*} \end{align}

then we get Liénard–Wiechert Potential:

ϕ(r⃗,t)=q4πε0∣r⃗−r⃗∗∣11−(r⃗−r⃗∗)⋅v⃗∗∣r⃗−r⃗∗∣cA⃗(r⃗,t)=μ0qv⃗∗4π∣r⃗−r⃗∗∣11−(r⃗−r⃗∗)⋅v⃗∗∣r⃗−r⃗∗∣c(r⃗∗,v⃗∗,t∗):{r⃗∗−r⃗0(t−∣r⃗−r⃗∗∣c)=0t∗=t−∣r⃗−r⃗∗∣cv⃗∗:=v⃗0(t∗)\begin{align} \phi (\vec{r},t)=&\dfrac{q}{4\pi\varepsilon _0|\vec{r}-\vec{r}^*|}\dfrac{1}{1-\frac{(\vec{r}-\vec{r}^*)\cdot \vec{v}^*}{|\vec{r}-\vec{r}^*|c}}\\ \vec{A} (\vec{r},t)=&\dfrac{\mu _0q\vec{v}^*}{4\pi|\vec{r}-\vec{r}^*|}\dfrac{1}{1-\frac{(\vec{r}-\vec{r}^*)\cdot \vec{v}^*}{|\vec{r}-\vec{r}^*|c}}\\ (\vec{r}^*,\vec{v}^*,t^*):&\begin{cases} \vec{r}^*-\vec{r}_0(t-\frac{|\vec{r}-\vec{r}^*|}{c})=0\\ t^*=t-\frac{|\vec{r}-\vec{r}^*|}{c}\\ \vec{v}^*:=\vec{v}_0(t^*) \end{cases} \end{align}