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We know that a free electron (an electron which isn't subject to accelerated motion due to external fields) can't emit a photon/cannot radiate. I tried to explain it without getting STR involved: if the electron isn't accelerating, then it means it's moving with a constant velocity $\vec{v}$. Since it's constant, there exists an inertial frame where the electron is at rest in a point $\vec{r}'$. In that frame, its field is that of a stationary charge, i.e. $$\vec{E}(\vec{r})=\frac{e}{4\pi\epsilon_0}\frac{\vec{r}-\vec{r}'}{|\vec{r}-\vec{r}'|^{3/2}},$$ and since $\vec{B}=0$ in that frame, it follows that $\vec{S}=\frac{1}{\mu_0}\vec{E}\times\vec{B}=0$, and so there is no radiation occuring, i.e. no photons are emitted. Now, I don't know enough QFT to confirm my speculations, so my question is this: does this intuitive explanation hold for virtual photons, or photons which are in any way not associated with radiation which we can detect at some macroscopic distances from the electron? If so, are they somewhat classicaly predicted by the fact that, even though there is no radiation, there still exists an electric field around the electron? For that matter, is my initial reasoning even correct in the context of relativity?

Note: I'm aware of other reasons why a classical free electron can't radiate photons, e.g. from Lienard-Wiechert potentials, conservation of energy and momentum etc. The question is focused around the given intuitive explanation, and the validity of it.

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  • $\begingroup$ I don't understand why you assume that a free electron means no acceleration: an electron can be free in the universe but there might always exist a reference frame in which it accelerates (just take a rotating reference frame). However, as a side note, emissions are caused by interactions, so if no other interactions is present in the universe (gravitational, weak, strong, electromagnetic) an electron will remain an electron forever. $\endgroup$ – gented Jul 10 '17 at 17:34
  • $\begingroup$ That's true, but I limited myself to observing an electron in an inertial reference frame (one in which, for example, distant stars don't rotate around the point, but all move with constant velocities). The key underlying point of my argument is that the electron stands still in an inertial reference frame, which means that in other inertial frames it's only permitted to move at constant velocities. $\endgroup$ – Soba noodles Jul 10 '17 at 17:41
  • $\begingroup$ Sure, I agree, but I was objecting the identification of your assumption with the word "free": an electron can be in an inertial frame and be free/not free according to the cases. $\endgroup$ – gented Jul 10 '17 at 17:44
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In the end it turns out that this proof makes sense. In classical electrodynamics we define radiation as a plane wave with a dispersion relation $\omega=c|\vec{k}|$, and if the energy transfer of the wave is zero (by virtue of $\vec{S}=0$), it follows that $\omega=0$ and $\vec{k}=0$, thus there is no photon.

Of course, in QED things get more complicated already at the 1L level with virtual off shell photons, at which point the dispersion relation breaks down and the proof doesn't hold anymore. However, the fact that the photons are virtual means they are unphysical in the sense of measuring them as ingoing and outgoing states. Instead, they contribute to the mass renormalisation of the electron, but that's another story.

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