In the introductory books we find the equation of electric field, by a point charge, as $$ \mathbf{E} = \frac{q}{4\pi\epsilon_0}~\frac{\hat r}{r^2}$$

In Feynman Lectures on Physics we find some paragraphs like this

Maxwell noted that the equations or the laws that had been discovered up to this time were mutually inconsistent when he tried to put them all together, and in order for the whole system to be consistent, he had to add another term to his equations. With this new term there came an amazing prediction, which was that a part of the electric and magnetic fields would fall off much more slowly with the distance than the inverse square, namely, inversely as the first power of the distance!

You can find it here. And therefore, he attempts to teach electromagnetic radiation theory with some very little assumptions and the only pre-requisite he assumes is classical electromagnetism theory (those four Maxwell's equations). And then he comes up with this equation for electric field

What is the formula for the electric and magnetic field produced by one individual charge? It turns out that this is very complicated, and it takes a great deal of study and sophistication to appreciate it. But that is not the point. We write down the law now only to impress the reader with the beauty of nature, so to speak, i.e., that it is possible to summarize all the fundamental knowledge on one page, with notations that he is now familiar with. This law for the fields of an individual charge is complete and accurate, so far as we know (except for quantum mechanics) but it looks rather complicated. We shall not study all the pieces now; we only write it down to give an impression, to show that it can be written, and so that we can see ahead of time roughly what it looks like. $$\mathbf{E} = \frac{-q}{4\pi\epsilon_0} \left[ \frac{\hat r}{r^2} + \frac{r}{c}~\frac{d}{dt}~(\frac{\hat r}{r^2})+\frac{1}{c^2}~\frac{d^2}{dt^2}~\hat r\right]~~~~~~~~~~~(2) $$

It's quite hard to understand all those new terms (terms except that $\frac{\hat r}{r^2}$). He tries to explain these new terms from elementary point of view (I may be wrong in saying elementary) but his explanation seems more complicated than the equation itself.

For example, when uses this equation $$\mathbf{E}=\frac{-q}{4\pi\epsilon_0 c^2}\, \frac{d^2 \hat r}{dt^2}$$

make arguments like this

If the charged object is moving in a very small motion and it is laterally displaced by the distance $x(t)$, then the angle that the unit vector $\hat r$ is displaced is $x/r$, and since r is practically constant, the x-component of $\frac{d^2 \hat r}{dt^2}$ is simply the acceleration of $x$ itself at an earlier time divided by $r$, and so finally we get the law we want, which is $$E_x(t)=\frac{q}{4\pi\epsilon_0 c^2r}\,a_x \Bigl(t-\frac{r}{c}\Bigr).~~~~~~~(3)$$

I really can't understand why he divided by $r$ in the equation above and how that time delay has come in. His simplicity is very complicated. So, due to all this I searched for books on web and found Classical Electromagnetic Radiation but I couldn't find the equation (2) in the book, I found a similar equation which contained the retarded factor but was not of the form of equation (2).

So, I request you to please explain me the equation (2) from introductory point of view or guide me somewhere. Please also show me how can we derive equation (3).

Thank you.

  • $\begingroup$ My recollection is the problem with Coulomb's law was it did not take into account that the electric field at a distance from the charge does not change instantaneously when the source (charged object) of the field moves, that the change radiates out at the speed of light $\endgroup$
    – Bob D
    Dec 20, 2019 at 12:03
  • $\begingroup$ @BobD Yes. But that should only add $r/c$ that is the time taken for the influence to reach to $r$. $\endgroup$
    – user240696
    Dec 20, 2019 at 12:21

1 Answer 1


To clarify what is happening note that Eq. (2) in your question is incorrect and should be written as Feynman wrote it with "primes", that is $$\mathbf{E} = \frac{-q}{4\pi\epsilon_0} \left[ \frac{\hat r'}{r'^2} + \frac{r'}{c}~\frac{d}{dt}~(\frac{\hat r'}{r'^2})+\frac{1}{c^2}~\frac{d^2}{dt^2}~\hat r'\right]~~~~~~~~~~~(2) $$ where $r'=r(t-r/c)$ is the distance between the moving charge to the fixed observation point as a function of the retarded time, and in your notation $\hat r'$ is the unit vector along that line of sight.

Then the first term is just Coulomb's law taken into account that even the static field propagates with a finite speed for it is also referenced to the retarded time but whatever it is if $r$ (and $r'$) is large relative to the actual displacement of the charge over which we are observing then it decays as $1/r^2$. The second term has a time derivative and it decays at least as fast as $1/r^3$ because $\frac{d}{dt} \frac {1}{r^2}=-\frac{2}{r^3}\frac{dr}{dt}$, and when multiplied by $r$ it will also decay at least as fast as $1/r^2$. The more interesting piece is the last term, here you have the $2^{nd}$ time derivative of the unit line of sight vector, so it is an acceleration of a unit vector. Here Feynman simplifies the situation by assuming that this acceleration is lateral, i.e., perpendicular to the line of sight.

In a Cartesian coordinate system $(x,y,z)$ with $\hat r=(x/r, y/z, z/r)$ where $x/r$, $y/r$, $z/r$ are direction cosines, pick an axis, say, $x$ along which the charge is moving while $r$ is essentially constant. Then $a_x$ is the acceleration of the lateral displacement itself $x$ and if you take retardation into account you get $a_x(t')$ divided by $r$. Now you see that the last term is proportional to $1/r$ while the first and second terms are proportional to $1/r^2$, and if $r$ is much larger than the displacement of the charge during the observation interval then the third term, the so-called radiation term, dominates.

While an accelerating charge radiates in other directions besides perpendicular to its motion the radiation is maximum in that plane and reaching zero along the direction of motion. This is exactly how linear antennas radiate: nothing along the wire, maximum radiation in the plane perpendicular to the wire and in-between a smooth transition. Antennas radiate because the charges forming their current are subjected to acceleration, as a minimum they must stop at the end of the wire.

  • $\begingroup$ $r(t-r/c)$ has units $m\cdot s$ which is not the distance. $\endgroup$
    – user240696
    Dec 20, 2019 at 14:27
  • $\begingroup$ $r(t)$ is the line of sight distance at time $t$, $r(t-r/c)$ is the LoS distance at time $t-r/c$ $\endgroup$
    – hyportnex
    Dec 20, 2019 at 14:29
  • $\begingroup$ Oh I’m sorry for misunderstanding it. $\endgroup$
    – user240696
    Dec 20, 2019 at 14:46
  • $\begingroup$ Please explain this while 𝑟 is essentially constant and this Then $𝑎_𝑥$ is the acceleration of the lateral displacement itself 𝑥 and if you take retardation into account you get $𝑎_𝑥(𝑡′)$ divided by 𝑟. I understand that $a_x$ is the acceleration of charge but why $r$ is essentially a constant and why $\frac{a_x (t')}{r}$? $\endgroup$
    – user240696
    Dec 20, 2019 at 15:05
  • $\begingroup$ Draw a line, call it the $z$ axis along which at $t=0$ the charge and the observation point are. Perpendicular to that is the $x$ axis along which the charge accelerates. For a short time interval the sideways displacement is much smaller than a large $r$, so during the motion the latter is essentially constant. $\endgroup$
    – hyportnex
    Dec 20, 2019 at 15:12

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