Do the electric field and magnetic field derived from the Lienard-Wiechert potentials satisfy Gauss's law?

Question

I've already got the electric fields and magnetic fields derived from the Lienard-Wiechert potentials:

$${\bf E}=\frac{q}{4\pi\epsilon_0}\frac{R}{(\bf R\cdot u)^3}[(c^2-v^2){\bf u}+\bf R\times(u\times a)]$$

$${\bf B}=\frac{\bf R}{cR}\times\bf E$$

where ${\bf R=r-r'}$ and ${\bf u}=\frac{c\bf R}{R}-\bf v$.

I wonder if they satisfy Maxwell's equations, I've tried to derive Gauss's law, but in vain. So do they? Or is there something wrong in my derivation?

Answer 1

I will take you back to Maxwell's equations in Lorenz Gauge. $$\vec \nabla^2 \varphi-\frac{1}{c^2} \frac{\partial^2\varphi}{\partial t^2}=-\frac{\rho}{\varepsilon_0} $$ $$\vec \nabla^2 \bf{A}-\frac{1}{c^2} \frac{\partial^2\bf A}{\partial t^2}=\mu_0\bf J$$ also $$\bf E=-\vec \nabla \varphi-\frac{\partial \bf A}{\partial t}$$ $$\bf B=\vec \nabla \times \bf A$$ Note that $\bf E$ and $\bf B$ satisfy all Maxwell's Equations.Solution to these can be given as $$\varphi = \frac{1}{4\pi \varepsilon_0}\int \frac{\rho(\bf r',t_r')}{\vert \bf r -\bf r' \vert}d^3\bf r'$$ $$\bf{A}=\frac{\mu_0}{4 \pi} \int \frac{\bf J (\bf r',t_r')}{\vert \bf r- \bf r' \vert}d^3\bf r'$$

Using $\rho =q\delta^3(\bf r-r_0)$ and $ \bf J$ $=$ $q\bf v \delta^3(\bf r- \bf r_0)$ Remember that both $\bf r_0$ and $\bf v$ are functions of retarted time $t_r'$ to simplify we add another delta term $\delta (t' -t_r')$ we get that. $$\varphi=\frac{q}{4 \pi \varepsilon_0}\int \frac{\delta(t'-t_r')}{\vert \bf r - \bf r_0 (t_r') \vert}dt'$$ $$\bf A=\frac {q\mu_0}{4 \pi} \int \bf v \frac{\delta (t'-t_r')}{\vert \bf r-\bf r_0 \vert}dt'$$ Now using $\vert \bf R \vert'=\frac{\bf R}{\vert \bf R \vert} \cdot \bf v$ and $\delta (t'-t_r')=\frac{\delta (t'-t_r)}{ \frac{\partial }{\partial t'}(t'-t_r') \vert _{t'=t_r} } $ we get our potential $$\varphi = \frac{q}{4 \pi \varepsilon_0} \frac{1}{(1-\bf \hat R \cdot \frac{\bf v}{c})\vert \bf R \vert}$$ $$\bf A = \frac{\bf v}{c^2} \varphi$$

So we just derived our Potential starting from the assumption that Maxwell's Equations are valid which includes Gauss's Law

Answer 2

They are derived from Maxwell's equations, so they satisfy Maxwell's equations, but taking vector derivatives is very complicated with retardation.