Q:

An intuitive reason why (classicaly) a free electron can't emit a photon

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.

A:

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.

