In this chapter, Feynman writes down the retarded Coulomb's law, $$\begin{equation} \label{Eq:II:21:1} \mathbf E=\frac{q}{4\pi\epsilon_0}\biggl[ \frac{\mathbf e_{r'}}{r'^2}+\frac{r'}{c}\,\frac{d}{dt}\biggl( \frac{\mathbf e_{r'}}{r'^2}\biggr)+\frac{1}{c^2}\,\frac{d^2}{dt^2}\,\mathbf e_{r'} \biggr] \end{equation}$$

and further explains about the various terms.

Disclaimer: All the text from here is, what I could understand, it might be egregious, so please bare with me.

  1. The logic behind the first term, as far as I could see, is to account for the following thing; When we try to find the $\mathbf E$ at a farther distance from the charge, the normal coulomb's law namely, $\mathbf E=\dfrac{q}{4\pi\epsilon_0}\dfrac{\mathbf e_{r'}}{r^2}$ doesn't hold good, as Coulomb's law gives a value for the field when given a position, neglecting the time taken for the field to travel to that point. So if the source charge is moving, when we try to measure $\mathbf{E}$ at time $t$, we plug into Coulomb's law, the distance from the charge at time t, and it will give a value, but that value in reality, takes some time to travel the distance $r\,$. So to correct this, we give coulomb's law the distance at a time $\,t-r'/c\,$ rather than $t$. Therefore the solution of the potentials and the fields are functions of $\,t-r/c\,$; $\,f(t-r/c)$ rather than $f(t)$.

  2. And later he explains that the second term, is correcting this retardation by making a linear approximation. But the whole point of giving Coulomb's law $r'$ was to account for the thing mentioned above? Why do we correct this again?

  3. And I couldn't understand the third term at all!

Can someone please point the places where I'm wrong?

  • $\begingroup$ Related: physics.stackexchange.com/q/227484 $\endgroup$ – G. Smith Oct 13 '19 at 3:56
  • $\begingroup$ I suggest that understanding the derivation of this equation from Maxwell’s equations is more important than trying to justify the three terms by hand-waving arguments. $\endgroup$ – G. Smith Oct 13 '19 at 4:00
  • $\begingroup$ @G.Smith Yes, I agree, but I really don't understand the role of the last two terms. In the answer that you've linked, it's specified that the last two terms are present to satisfy various frames of reference. In that sense, yeah I understand, but what do they do in a stationary frame of reference? $\endgroup$ – Aravindh Vasu Oct 13 '19 at 4:15
  • $\begingroup$ I am more familiar with the equivalent Liénard-Wiechert fields. The two terms there are seen to be a $1/r^2$ “static” field that depends only on velocity and a $1/r$ “radiative” field that depends on acceleration. I like this better than Feynman’s formula. I suppose Feynman’s third term must be the radiative field since it has a second time derivative. $\endgroup$ – G. Smith Oct 13 '19 at 4:44

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