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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?

Soluty
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    The first term of $\mathbf{E}$ is not a vector. – SuperCiocia Jun 28 '20 at 03:31
  • Thank you! I've rectified it. – Soluty Jun 28 '20 at 12:15
  • Isn't $u$ a vector? So isn't the first term of $E$ a vector? – J Thomas Jun 29 '20 at 17:32
  • @verdelite can you try and clarify what about this question and existing answer you're looking to clarify? The question you pose in the bounty appears only tangentially related to the present question and it's not clear to me why you attached it as a bounty instead of asking as a separate question (possibly referencing this question). – Richard Myers Sep 14 '21 at 18:57
  • @RichardMyers The OP asked how Gauss's law (it is part of Maxwell's equations) could be derived from the Lienard-Wiechert fields. I'd like to see that too so I started the bounty. It is better if somebody can show all Maxwell's equations being derived from the Lienard-Wiechert fields. I know it is more than just Gauss's law, but I think it is justified to tie the bounty to the original post. – verdelite Sep 14 '21 at 23:16
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    @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 [...]"). – Richard Myers Sep 15 '21 at 02:56
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    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. – bolbteppa Sep 15 '21 at 22:38
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    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... – bolbteppa Sep 15 '21 at 22:42

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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

Soham Patil
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  • Is it should be $\mathbf {A}=\mathbf {v} \phi /c^2$ on the last line? – Alex Trounev Sep 17 '21 at 04:04
  • Thanks i changed it – Soham Patil Sep 17 '21 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? – Alex Trounev Sep 17 '21 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? – verdelite Sep 18 '21 at 00:44
  • @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. – Soham Patil Sep 18 '21 at 03:56
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They are derived from Maxwell's equations, so they satisfy Maxwell's equations, but taking vector derivatives is very complicated with retardation.

Jerrold Franklin
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  • ...so they satisfy Maxwell's equations... Proof ??? If $,\texttt A \boldsymbol \implies \texttt B,$ then we are not sure that $,\texttt B \boldsymbol \implies \texttt A$. – Frobenius Sep 14 '21 at 10:32
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    @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. – Richard Myers Sep 14 '21 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. – Frobenius Sep 14 '21 at 19:01
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    @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. – Richard Myers Sep 15 '21 at 02:59
  • @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). – Frobenius Sep 15 '21 at 03:40