Light emission paradox

Question

If we shake a electron back an forth we get light similarly but if go in the electrons frame the electrons is still and the rest is moving so what will we see it emits light or not

Answer 1

It is not enough to describe "shake an electron".

The electron is an elementary particle and follows the models of quantum mechanics.

An electron can be accelerated/decelerated and give off radiation by interacting with another particle or with a field. If in the laboratory, an electron radiates when it enters a region with a magnetic field, it will radiate when it is the magnetic field that moves, at the rest frame of the electron.

Take the Brehmssstrahlung Feynman diagrams for an electron decelerating in the electric field of a nucleus:

By construction of the feynman diagrams, the calculated crossections(probabilities of interaction happening) are Lorenz invariant so it does not matter if one thinks of the interaction in the center of mass of the electron.

When the acceleration/deceleration i.e. the interaction happens with a general electric or magnetic field, one has to to replace in the diagrams the nucleus with a model of the source of the field, in order to calculate the crossection, and again the frame makes no difference to the calculation by construction of the theory. The theory describes data very successfully.

Answer 2

Sitting on a shaked electron you will

1. feel an acceleration forth and back and
2. You will see not only the emitted electrons but also the electrons which shake the electron and get after this absorbed by the electron.

if go in the electrons frame the electrons is still and the rest is moving so what will we see it emits light...

Since you feel an acceleration you are not in a stationary frame of reference and not stand still. In any case, being on the electron or not, you will see the emitted light.

Answer 3

If we shake a electron back an forth we get light similarly but if go in the electrons frame the electrons is still and the rest is moving so what will we see it emits light or not

There is no inertial rest frame for an oscillating particle as it is nearly always accelerating [e.g., simple harmonic oscillator or SHO has $a \propto \cos{\left( \omega t \right)}$]. Thus, it would nearly always emit electromagnetic radiation, since the power emitted is proportional to the acceleration squared. This is given by the Larmor formula where: $$ P = \frac{ 2 }{ 3 } \frac{ e^{2} \ a^{2} }{ 4 \ \pi \ \varepsilon_{o} \ c^{3} } $$ where $e$ is the fundamental charge, $\varepsilon_{o}$ is the permittivity of free space, $c$ is the speed of light in vacuum, and $a$ is the magnitude of the acceleration.

For SHO the power would go as $P \propto \cos^{2}{\left( \omega t \right)}$, where $\omega$ is the angular frequency at which you "shake" the electron. Though there are very specific, infinitesimally short times when $P \rightarrow 0$ corresponding to $\omega t = \tfrac{n \ \pi}{2}$, where $n = 1, 2, ..., N$, one can safely say that a "shaking" electron would always emit electromagnetic radiation.

Answer 4

If we are talking about classical electromagnetism then an observer "riding with" the oscillating charge is in the near field of the electromagnetic field. The near field region is where the electric and magnetic fields self couple or have a self-impedance so that transmitted and self-reflected waves partially cancel out. For this reason the field near the oscillating charge or antenna with an oscillating current is approximately static. This happens in a region defined by the Fraunhofer distance $d~=~2D/\lambda$, where $D$ is the diameter or length of the antenna.

For a classical theory of the electron we can take the classical radius of the electron. This is $$ r_{cl}~=~\frac{e^2}{4\pi\epsilon_0mc^2}. $$ This can be derived by considering the energy of the electron as $mc^2~=~e\Phi$ for $\Phi$ the classical field. This is about $2.82\times 10^{-13}cm$. Hence for an electron emitting a IR photon with $\lambda~=~10^6m$ or a micron, we have then that $d~=~5.64\times 10^{-7}m$. This is close to half the wavelength of the emitted photon, which is about what we might expect. it also corresponds pretty well with a Nyquist criterion.

As Anna V writes the full theory is quantum mechanical or quantum electrodynamic. We also have to consider that an observer on the frame of the electron will not observe EM radiation, at least not classically, but the observer will also detect the fact they are not on an inertial frame. So the observer or probe would detect that though the fields are absent there must be propagating radiation in the far field region.