In dealing with electro-magnetic field of moving charged particle \(q\) at \(\vec{r}_0(t)\) with velocity \(\vec{v}_0(t)\): \(\begin{align} \rho (\vec{r},t)=q\delta (\vec{r}-\vec{r}_0(t)),\quad \vec{j}(\vec{r},t)=\rho (\vec{r},t)\vec{v}_0(t)=q\vec{v}_0(t)\delta (\vec{r}-\vec{r}_0(t)) \end{align}\)
4-potential given with Lorentz gauge: \(\begin{align} \phi (\vec{r},t) =&\int \,\mathrm{d}^3x' \dfrac{\phi (\vec{r}',t-\dfrac{R}{c})}{4\pi \varepsilon _0R}=\int \,\mathrm{d}^3x' \dfrac{q\delta \left(\vec{r}'-\vec{r}_0(t-\frac{|\vec{r}-\vec{r}'|}{c})\right)}{4\pi \varepsilon _0|\vec{r}-\vec{r}'|}\\ \vec{A}(\vec{r},t)=&\int \,\mathrm{d}^3x' \dfrac{\mu _0\vec{j}(\vec{r}',t-\dfrac{R}{c})}{4\pi R}=\int \,\mathrm{d}^3x' \dfrac{\mu _0q\vec{v}_0(t-\frac{|\vec{r}-\vec{r}'|}{c})\delta \left( \vec{r}'-\vec{r}_0(t-\frac{|\vec{r}-\vec{r}'|}{c}) \right)}{4\pi |\vec{r}-\vec{r}'|} \end{align}\)
Note that only when \(\vec{r}'\) s.t. \(\vec{r'}-\vec{r}_0(t-\frac{|\vec{r}-\vec{r}'|}{c})=0\), the formula in integral is non-zero. We denote the solution of it as \((\vec{r}^*,t^*)\), i.e.: \(\begin{align} \begin{cases} \vec{r}^*-\vec{r}_0(t-\frac{|\vec{r}-\vec{r}^*|}{c})=0\\ t^*=t-\frac{|\vec{r}-\vec{r}^*|}{c} \end{cases}\Rightarrow \begin{cases} \vec{r}^*=\vec{r}_0(t^*)\\ \vec{v}^*:=\vec{v}_0(t^*)\\ |\vec{r}-\vec{r}^*|=c(t-t^*) \end{cases} \end{align}\)
in this way: \(\begin{align} \phi (\vec{r},t)=\dfrac{q}{4\pi\varepsilon _0|\vec{r}-\vec{r}^*|}\int \,\mathrm{d}^3x' \delta \left( \vec{r}'-\vec{r}_0(t-\dfrac{|\vec{r}-\vec{r}'|}{c}) \right)\\ \vec{A} (\vec{r},t)=\dfrac{\mu _0q\vec{v}^*}{4\pi|\vec{r}-\vec{r}^*|}\int \,\mathrm{d}^3x' \delta \left( \vec{r}'-\vec{r}_0(t-\dfrac{|\vec{r}-\vec{r}'|}{c}) \right) \end{align}\)
with coordinate transformation \(\vec{r}'\mapsto \vec{r}^{*\prime}=\vec{r}'-\vec{r}_0(t-\frac{|\vec{r}-\vec{r}'|}{c})\): \(\begin{align} \left\Vert \dfrac{\partial^{} \vec{r}^{*\prime}}{\partial \vec{r}^\prime} \right\Vert\Big|_{\vec{r}'=\vec{r}^*}=&\left\Vert \dfrac{\partial^{} }{\partial x_j^{}} x_i'-x_{0i}(t-\dfrac{|\vec{r}-\vec{r}'|}{c}) \right\Vert\Big|_{\vec{r}'=\vec{r}^*}=1-\dfrac{(\vec{r}-\vec{r}^*)\cdot \vec{v}^*}{|\vec{r}-\vec{r}^*|c}\\ \int \,\mathrm{d}^3x' \delta \left( \vec{r}'-\vec{r}_0(t-\dfrac{|\vec{r}-\vec{r}'|}{c}) \right)=&\int \,\mathrm{d}^3x^{*\prime}\delta (\vec{r}^{*\prime})\left\Vert \dfrac{\partial^{} \vec{r}'}{\partial \vec{r}^{*\prime}} \right\Vert\\ =&1\Bigg/ \left\Vert \dfrac{\partial^{} \vec{r}^{*\prime}}{\partial \vec{r}^\prime} \right\Vert\Big|_{\vec{r}'=\vec{r}^*} \end{align}\)
then we get Liénard–Wiechert Potential: \(\begin{align} \phi (\vec{r},t)=&\dfrac{q}{4\pi\varepsilon _0|\vec{r}-\vec{r}^*|}\dfrac{1}{1-\frac{(\vec{r}-\vec{r}^*)\cdot \vec{v}^*}{|\vec{r}-\vec{r}^*|c}}\\ \vec{A} (\vec{r},t)=&\dfrac{\mu _0q\vec{v}^*}{4\pi|\vec{r}-\vec{r}^*|}\dfrac{1}{1-\frac{(\vec{r}-\vec{r}^*)\cdot \vec{v}^*}{|\vec{r}-\vec{r}^*|c}}\\ (\vec{r}^*,\vec{v}^*,t^*):&\begin{cases} \vec{r}^*-\vec{r}_0(t-\frac{|\vec{r}-\vec{r}^*|}{c})=0\\ t^*=t-\frac{|\vec{r}-\vec{r}^*|}{c}\\ \vec{v}^*:=\vec{v}_0(t^*) \end{cases} \end{align}\)