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

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?

• The first term of $\mathbf{E}$ is not a vector. Jun 28, 2020 at 3:31
• Thank you! I've rectified it. Jun 28, 2020 at 12:15
• @verdelite Then I think you will be left disappointed. The Lienard-Wiechert potentials are solutions to Maxwell's equations. It is not possible to derive a set of equations from a solution thereto. The most you can do is show that the LW potentials satisfy Maxwell's equations. It is this latter point OP appears to be asking about ("I wonder if they satisfy Maxwell's equations [...]"). Sep 15, 2021 at 2:56
• Lienard-Wiechert is the solution for a single point charge, Maxwell's equations obviously apply to more general systems, the question people should be asking is about 'deriving' Maxwell's equations from the Jefimenko solutions, which reduce to e.g. Coulomb in the static cases, and is equivalent to Lienard-Wiechert in the case of a single charge. These complicated solutions makes it obvious one should be thinking of the equations rather than their solutions, unlike the way EM is usually taught e.g. beginning via Coulomb's Law. Sep 15, 2021 at 22:38
• See this for the Jefimenko discussion, it's really as simple as realizing the general electric and magnetic fields satisfy the inhomogenous wave equation (the inhomogeneous terms are given in the link) and the retarded solution of the wave equation for these inhomogeneous terms give you the Jefimenkos solutions, which immediately satisfy all of Maxwell's equations by construction... Sep 15, 2021 at 22:42

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

• Is it should be $\mathbf {A}=\mathbf {v} \phi /c^2$ on the last line? Sep 17, 2021 at 4:04
• Thanks i changed it Sep 17, 2021 at 13:13
• Also, please, could you correct $\bf J=q\mu_0/4 \pi\int ...$ to $\bf A=q\mu_0/4 \pi\int ...$ in the line before? Sep 17, 2021 at 15:00
• It seems you are doing the forward derivation, from Maxwell's Equations to the Lienard-Wiechert fields. Could we do the backward derivation, namely, from the Lienard-Wiechert fields to Gauss's Law? Sep 18, 2021 at 0:44
• @verdelite sure you can, generally in a textbook this is how liembert potentials are derived. But in case you are said to start with the potentials and you have to check if they follow maxwell's equations you can plug those in the maxwell,s equation and check if they are valid(Note the potentials you were given need not follow Lorenz gauze so you can either check if they do or plug them in general maxwell equation). The fact that the potentials are derived from maxwell equations they definitely do follow maxwell equation. Sep 18, 2021 at 3:56

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

• ...so they satisfy Maxwell's equations... Proof ??? If $\,\texttt A \boldsymbol \implies \texttt B\,$ then we are not sure that $\,\texttt B \boldsymbol \implies \texttt A$. Sep 14, 2021 at 10:32
• @Frobenius If you look at the dates, this is an answer to the original question, when it was originally posted. It does, in fact, answer OP's question. The boundary is more recent and poses a different question. Sep 14, 2021 at 18:54
• @Richard Myers : I apologize, but I don't understand. Looking in the edits of the post I don't see a different original question. Sep 14, 2021 at 19:01
• @Frobenius OP's question was "[do] they satisfy Maxwell's equations [...]" to which the answer is "yes," as reported in this answer. The only more one could ask for is an explicit demonstration that they indeed satisfy Maxwell's equations. Sep 15, 2021 at 2:59
• @Richard Myers : OK, thank you. I think that a first test would be if the field $\left(\mathbf E,\mathbf B\right)$ produced by a uniformly moving charge satisfies the Maxwell equations. As field I mean that derived from the Lienard-Wiechert potentials which depends on the present position and time, see in my answer here Magnetic field due to a single moving charge, equations (01a) & (01b). Sep 15, 2021 at 3:40