Suppose an electron is created somehow in the universe -- through particle-pair production or what have you. Suppose it is stationary so that electrodynamics (and specifically propagating electromagnetic waves) is unavailable for explanation. Certainly distance charges will feel Coulomb's force, but how quickly?

I'm wondering if, since electromagnetism is manifestly relativistic (Lorentz rather than Galilean invariant) we can use that to always move to a frame where the electron is travelling and then claim that there are in fact EM waves? I don't know.


2 Answers 2


The electrostatic field, by definition does not propagate. That is what the static part of electrostatic means.

The key to this is to recognize that the continuity equation forbids a single charge from being created alone. It is necessary that any charge creation be accompanied by the creation of the equal and opposite charge in the same location. In other words, what is created is a dipole, not just a charge.

Before the dipole is created there is no field, equivalently it is the field of the positive and negative charges canceling together in exactly the same location, or equivalently a dipole with 0 dipole moment. As the charges separate a small electric dipole is formed and as they continue to separate the dipole moment increases. The increasing dipole moment leads to standard dipole radiation which propagates outward at c as any other dipole radiation would.

At distances that are large compared to the charge separation then we are in the far field and the field is approximately a pure dipole field propagating at c.

At distances from one pole that are very small compared to the dipole charge separation the field is approximately a pure monopole field which is not propagating. Since the charge separation distance always increases at some $v<c$ and since the monopole approximation is only valid for distances $d<<vt<ct $ at any point where the monopole approximation is valid the dipole radiation has already passed.

So by paying attention to the continuity equation we find that the electrostatic field does not propagate, nor does it need to. What propagates is ordinary dipole radiation. The electrostatic field is only an approximation that becomes valid well after the dipole radiation has passed.


We cannot say that the charge would emit radiation simply due to it being in motion. For emission to occur, the charge must be accelerating - that is, moving with a nonuniform velocity. Charges moving at constant velocity do not emit EM waves. So, to your point about changing reference frames, the answer is no, we cannot change our frame to capture EM wave generation if the charge is not accelerating.

Now consider the non-accelerating charge, with a magnitude $Q$. It pops into existence in vacuum and sits still. Since it's charged, the electric field emanates radially outward:

$$\vec{E} = \frac{1}{4\pi\epsilon_0}\frac{Q}{r^2}\hat{r}$$

How quickly does this field reach some radial distance $L$? It's not EM radiation, so we cannot say $t=L/c$. It's more complicated, and rather contentious.

We can try to perform a calculation based on the Liénard-Weichert potential, which describes the motion of electric point charges in terms of their vector and scalar potentials in the Lorenz gauge. Upon computation, one finds that the field at distance $L$ is identical to that calculated if the electric field is assumed to move at infinite velocity [Measuring Propagation Speed of Coulomb Fields]. This clearly violates special relativity, as information is being conveyed across space at a speed exceeding that of light.

After a bit of digging, it seems that consensus has not been formed. Feynman had his own views on the matter, recognizing it as a paradox that can be explained away by considering that uniform motion lasts indefinitely (see third citation in [Measuring Propagation Speed of Coulomb Fields]).

I recommend reading through the 'static fields' section in the Wiki article [Speed of gravity]. The static field is presented as emanating all space instantaneously, and the relativistic violation is explained away by claiming the static field does not convey information (i.e., it is not a signal). To which I respond: the field imparts a force on a charged particle a distance $L$ from the source - certainty this constitutes information transfer.

It's a fascinating question that deserves more attention. It might be the case that this is an ill-posed question, which warrants an explanation in and of itself.


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