On chapter 27 of "The Feynman Lectures on Physics Volume II: Mainly Electromagnetism and Matter", Richard Feynman says:

Finally, in order to really convince you that this theory is obviously nuts, we will take one more example — an example in which an electric charge and a magnet are at rest near each other — both sitting quite still. Suppose we take the example of a point charge sitting near the center of a bar magnet, as shown in Fig. 27-6 Everything is at rest, so the energy is not changing with time. Also, E and B are quite static. But the Poynting vector says that there is a flow of energy, because there is an E X B that is not zero. If you look at the energy flow, you find that it just circulates around and around. There isn't any change in the energy anywhere — everything which flows into one volume flows out again It is like incompressible water flowing around. So there is a circulation of energy in this so-called static condition. How absurd it gets!

Perhaps it isn't so terribly puzzling, though, when you remember that what we called a "static" magnet is really a circulating permanent current. In a permanent magnet the electrons are spinning permanently inside. So maybe a circulation of the energy outside isn't so queer after all.

I was wondering whether it would be possible to calculate the Poynting vector of an electron, considering its static electric field and the magnetic field associated to its magnetic dipole moment, caused by its intrinsic property of spin.

In case it is possible to do such calculation, what would be the interpretation of the Poynting vector?

Fig. 27-6. A charge and a magnet produce a Poynting vector that circulates in closed loops.

  • $\begingroup$ I've been meaning to ask this question for a while actually. Btw don't you just love the name... Poynting vector... heh $\endgroup$
    – Skyler
    Commented Nov 18, 2014 at 23:42
  • $\begingroup$ Your question may be fully answered in this paper: arxiv.org/abs/physics/0208072 $\endgroup$
    – Realist753
    Commented Dec 21, 2015 at 17:35
  • $\begingroup$ Wow! Thanks @Realist753 for bringing this article to my attention! It is really interesting to me. $\endgroup$
    – roy
    Commented Jan 19, 2016 at 15:11

1 Answer 1


To start with the electric field of the electron is as far as possible to measure symmetric, the electric dipole moment is very small. In this article

The electron's EDM must be collinear with the direction of the electron's magnetic moment (spin). Within the standard model of elementary particle physics, such a dipole is predicted to be non-zero but very small, at most 10^−38 e·cm,] where e stands for the elementary charge.

So we have the electric field parallel to the magnetic moment direction .

The magnetic field from a magnetic dipole treated classically :

As seen in the geometry of a current loop, this torque tends to line up the magnetic moment with the magnetic field B

So we have E parallel to B . The Poynting vector is ExB. So at the elementary particle level, we have a very small electric dipole ( Delta (E) differentiated from a uniform E ) parallel to a magnetic field given by the magnetic dipole.

Statistical fluctuations and the Heisenberg uncertainty principle will ensure that deviations from parallelism will have both positive and negative signs so, together with the infinitesimally small electric dipole moment, it will be a very very small transient effect. It will probably turn out to be zero if the calculations are done.

  • $\begingroup$ Thanks for your answer @annav. Nevertheless, I believe that the electric field due to an electron has, at least, two terms in a multipole expansion: the monopole term and the dipole term. I mean that the total electric field of an electron should be the superposition of the dipolar field you considered and a Coulombic field due to its negative charge. $\bar{\mathbf{E}}={1\over4\pi\varepsilon_0} {q_e\over r^2} \,\hat{\mathbf{r}} + \frac {3\; (\bar{\mathbf{p}}\cdot\hat{\mathbf{r}})}{4 \pi \varepsilon_0 r^3} \hat{\mathbf{r}}-\frac {\bar{\mathbf{p}}}{4 \pi \varepsilon_0 r^3}$. $\endgroup$
    – roy
    Commented Nov 19, 2014 at 14:45
  • $\begingroup$ And, since the electron electric dipole moment is quite small compared to its electric charge per cm ($10^{−38}$ e·cm vs $1$ e·cm), then it is quite reasonable that the electric field of an electron would be predominantly monopolar. $\endgroup$
    – roy
    Commented Nov 19, 2014 at 14:54
  • $\begingroup$ well that is what I mean with a small dipole contribution and spherically symmetric distribution of the electric field $\endgroup$
    – anna v
    Commented Nov 19, 2014 at 15:46
  • $\begingroup$ Oh, OK. Now I understand. But then I don't know why you say that the electric field should be parallel to the magnetic field if E has a monopole term plus a dipole term while B has only a dipole term. $\endgroup$
    – roy
    Commented Nov 19, 2014 at 19:24
  • $\begingroup$ There exists an experimental limit on the electric dipole moment. I have given a link that says that the electric dipole lines up with the magnetic moment in the case of a dipole, thus any asymmetry induced by the dipole will be parallel to B, and thus zero Poynting vector. I argue that the monopole part can only transiently contribute to an ExB , because it is spherically symmetric it will again average to zero. $\endgroup$
    – anna v
    Commented Nov 19, 2014 at 20:29

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