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I attempted to grasp the retarded potentials by staring at them and wanted to know if my thoughts seem to work out.

Equation taken from wikipedia (replaced $t_r$ with its definition): $$ \mathrm\varphi (\mathbf r , t) = \frac{1}{4\pi\epsilon_0}\int \frac{\rho (\mathbf r' , t-\frac{|\mathbf r - \mathbf r'|}{c})}{|\mathbf r - \mathbf r'|}\, \mathrm{d}^3\mathbf r'\\ \mathbf A (\mathbf r , t) = \frac{\mu_0}{4\pi}\int \frac{\mathbf J (\mathbf r' , t-\frac{|\mathbf r - \mathbf r'|}{c})}{|\mathbf r - \mathbf r'|}\, \mathrm{d}^3\mathbf r' $$

Each charge emits at all times a spherical shell of information propagating straight outward (my interpretation of $(\mathbf r' , t-\frac{|\mathbf r - \mathbf r'|}{c})$). Each point on the shell remembers the quantity of electric charge that emitted it (used for $\rho$), the distance traveled since emission (used for $|\mathbf r - \mathbf r'|$), the direction it is going (used for flow from $\mathbf r'$ or something; not sure yet) and the velocity the charge that emitted it had (used for $\mathbf{J} = \rho \cdot \mathbf v$). Using this information it can compute the scalar and vector potentials for (the benefit of) each electric charge the shell bumps into (I think this would be when the point on the shell is at $(\mathbf r, t)$; summing the (algebraically massaged) information from all shells that have been emitted that intersects that point would then give the potential).

The next step if this thought model is correct would be to attempt to take spatial and time derivatives to get the electric and magnetic fields, which in turn can be used to compute the Lorentz force on the charges that gets visited by the shells. (Perhaps the time derivative can be alleviated by each point on the shell remembering the latest state it had, the spatial derivative would require it to be able to query its surroundings, or some combination of that)

It seems similar to Huygens principle so I think I am on the right track.

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  • $\begingroup$ Each charge emits at all times a spherical shell... Non-accelerating charges don't emit anything. $\endgroup$
    – G. Smith
    Feb 19 '20 at 7:51
  • $\begingroup$ Each point on the shell remembers the quantity of electric charge that emitted it, the distance traveled since emission, the direction it is going and the velocity the charge that emitted it had. I’m fairly sure that none of these four quantities can be back-computed from the potential at one point, so I disagree that you are on the right track. $\endgroup$
    – G. Smith
    Feb 19 '20 at 7:55
  • $\begingroup$ I would not back-compute; I would try to make something like a toy simulation/game of life/... and see if these rules would simulate electromagnetism. $\endgroup$
    – Emil
    Feb 19 '20 at 17:47
  • $\begingroup$ I wonder If this isn't the content of Green's functions? I don't really remember them. $\endgroup$
    – Emil
    Feb 19 '20 at 18:00
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Your thoughts are essentially right, with this restriction: you should not think that what is emitted is anything physical, that actually will act on a body at arrival: these are potentials, after all, not fields. The resulting formulae for the fields are known as the Liénard-Wiechert formulae, and they are not really transparent as they are usually given. But they do separate into 2 parts, one due to taking the gradient of the potential, and thus going as $1/r^2$, and the other obtained by translating the gradient into a time-derivative: these typically go as $1/r$ and are proportional to acceleration. Staring long and hard at their derivation may give some insight.

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  • $\begingroup$ That does seem like it could be a better place to stare. $\endgroup$
    – Emil
    Mar 25 '20 at 7:03

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