In the Feynman Lectures, Volume I, 28-1, Feynman promulgates this expression for the electric field due to a moving charge. My notation differs from the original.

$$ \mathfrak{E}=-\frac{q}{4\pi\varepsilon_{o}}\left(\frac{\hat{\rho}}{\rho^{2}}+\frac{\rho}{c}\frac{d}{dt}\left[\frac{\hat{\rho}}{\rho^{2}}\right]+\frac{1}{c^{2}}\frac{d^{2}\hat{\rho}}{dt^{2}}\right). $$

Here $\rho$ represents the retarded position of the source; and $\hat{\rho}$ is the corresponding unit vector point away from the source.

He later asserts that

Of the terms appearing in [this equation], the first one evidently goes inversely as the square of the distance, and the second is only a correction for delay, so it is easy to show that both of them vary inversely as the square of the distance.

The first term is simply Coulomb's inverse square law:

$$ \mathfrak{E}_{1}=-\frac{q}{\rho^{2}4\pi\varepsilon_{o}}\hat{\rho}. $$

Representing the delay $\frac{\rho}{c}=\Delta t$, the second term can be written as

$$ \mathfrak{E}_{2}=\frac{d}{dt}\left[-\frac{q}{\rho^{2}4\pi\varepsilon_{o}}\hat{\rho}\right]\Delta t. $$

And it is indeed the time derivative of the first term (multiplied by the delay):

$$ \mathfrak{E}_{2}=\frac{d\mathfrak{E}_{1}}{dt}\Delta t. $$

The third term is not dependent on $\rho$:

$$ \mathfrak{E}_{3}=-\frac{q}{4\pi\varepsilon_{o}c^{2}}\frac{d^{2}\hat{\rho}}{dt^{2}}. $$

Carrying out the differentiation in the second term gives

$$ \mathfrak{E}_{2}=-\frac{q}{4\pi\varepsilon_{o}c}\rho\left(\frac{1}{\rho^{2}}\frac{d\hat{\rho}}{dt}-\frac{d\rho}{dt}\frac{2}{\rho^{3}}\hat{\rho}\right) $$

$$ =-\frac{q}{4\pi\varepsilon_{o}c}\left(\frac{1}{\rho}\frac{d\hat{\rho}}{dt}-\frac{d\rho}{dt}\frac{2}{\rho^{2}}\hat{\rho}\right). $$

The first term in the parentheses varies as the inverse of $\rho$. The second term does vary as the inverse squared.

I contend Feynman is incorrect. Is he?

I'm adding this graphic to illustrate the answer I received. I was, in fact, wrong. Diagram illustrating that I was wrong

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    $\begingroup$ Why was this question down-voted? It is clearly stated, with supporting reason, and direct access to the source text. I have submitted several errata to the maintainers of the Feynman Lectures. It would not be the first time I found an error. $\endgroup$ Dec 19, 2017 at 1:07
  • $\begingroup$ Was Feynman correct? Most likely... $\endgroup$
    – Natsfan
    Dec 20, 2017 at 2:13

1 Answer 1


I would suggest that you reconsider the quantity $d\hat{\rho}/dt$, which itself varies inversely with $\rho$.

Now, $\hat{\rho}$ is the unit vector pointing from the field point $P$ at which $\mathfrak{E}$ is measured to the charge $q$. At this point, let's introduce a new variable that Feynman evidently didn't want to introduce: the velocity $\vec{v}$ of the charge. You'll surely agree that this is independent of the field point $P$.

It also helps to think of a few specific scenarios, and consider how the result would change. So imagine that $\vec{v}$ is perpendicular to $\hat\rho$ and has magnitude $10\mathrm{m/s}$. If $\rho$ is just a few meters, you expect $d\hat{\rho}/dt$ to be quite large. On the other hand, if $\rho$ is several light years, you expect $d\hat{\rho}/dt$ to be extremely small. So you might begin to believe that it varies inversely with $\rho$.

But to be a little more specific, the only possible change in $\hat{\rho}$ is a rotation, and can therefore be described by an angular velocity $\vec{\omega}$. This angular velocity is just the angular velocity of the charge as seen from the field point. A standard formula gives the angular velocity as \begin{equation} \vec{\omega} = \frac{\hat{\rho} \times \vec{v}} {\rho}, \end{equation} so we have \begin{equation} \frac{d \hat{\rho}} {dt} = \vec{\omega} \times \hat{\rho} = \frac{(\hat{\rho} \times \vec{v}) \times \hat{\rho}} {\rho}. \end{equation} (Note that the cross-product is not associative, so the grouping by parentheses is crucial. In particular, the numerator of that final expression is not trivially zero.) This does indeed vary inversely with $\rho$, which means that your term $\frac{1}{\rho} \frac{d \hat{\rho}} {dt}$ varies inversely with the square of $\rho$, just as Feynman claimed.

  • $\begingroup$ I don't understand the bit about the angular velocity. But I think you're on to something regarding the relationship between $\frac{d\hat{\rho}}{dt}$ and $\rho$. Note that the unit vector $\hat{\rho}$ is the position vector of the source charge relative to the point of observstion (I think I had that backwards), so its value doesn't change with the angular velocity of the source charge. $\endgroup$ Dec 19, 2017 at 4:02
  • $\begingroup$ I think you’re misunderstanding this vector. Going back to the original text, he says “$\mathbf{e}_{r’}$ is the unit vector in the direction from the point $P$ where $\mathbf{E}$ is measured.” It’s always pointing from the field point $P$ to the charge. It certainly will change with the velocity of the charge; angular velocity is just one way of calculating it. And I would argue it is the simplest and most standard way. $\endgroup$
    – Mike
    Dec 19, 2017 at 4:12
  • $\begingroup$ I did have the sense of $\hat{\rho}$ wrong in the original post. I now see what you meant about angular velocity. I originally thought you were talking about the angular velocity about some distant point. $\endgroup$ Dec 19, 2017 at 4:28
  • $\begingroup$ I could feel embarrassed for asking that question. I might have figured it out had I kept reading. But your answer is superior to any I would have come up with. Sometimes being wrong is a boon. $\endgroup$ Dec 19, 2017 at 23:14
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    $\begingroup$ Glad to be of help. If you're not wrong frequently, you're not challenging yourself, and you're not getting better. So it sounds like you're going about things the right way. $\endgroup$
    – Mike
    Dec 20, 2017 at 1:45

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