What does a circularly polarized electromagnetic plane wave look like in a co-rotating reference frame?

Question

For a circularly polarized plane wave, the $\mathbf{E}$ and $\mathbf{B}$ vectors rotate in a particular direction. For concreteness, say the electric and magnetic fields are given by: \begin{align} \mathbf{E} & = \frac{E_0}{\sqrt{2}}\left(\hat{i} +i\,\hat{j}\right)e^{ikz - \omega t} \text{ and}\\ \mathbf{B} & = i \frac{\mathbf{E}}{c}. \end{align} Now, if I enter a frame rotating with angular frequency $\omega$ that, if $\mathbf{E}$ and $\mathbf{B}$ were rigid physical vectors, would render them stationary, what do I see?

I know that this is an exercise in electromagnetism in non-inertial frames, and so the domain of general relativity. Beyond that, I have not had time to look into it, and am asking out of curiosity if the problem has already been solved.

Answer 1

Difficult but fascinating idea. I try to partly answer the classical EM question (note definite photon z spin say +1 corresponds to right hand circular polarized light wave). Assume we have a square (!) cross sectioned cylindrical beam of finite width. To get an angular momentum is tricky and the outer surfaces are very important in the momentum integrals. Anyway the EM field is not like a spinning solid pencil. The outer field parts do not in any sense circulate around the central ray. True, each parallel ray independently has its local E and B field rotating. So it is like a lot of separate spinners over a transverse plane. The outer spinners just spin in place they don’t circulate around the central ray. Choosing a frame rotating about a central ray may give E B static along that ray but further out from the center ray the E B fields would not be static in direction but would be rotating more and more as we move away from axis. It would get quite tricky using rotating system in special relativistic flat spacetime, similar to GR.

But it is clear the E B fields cannot be fixed in such a rotating frame except right on the central ray you choose for axis of rotation.

Perhaps for a very thin beam it may be approximately possible. Experimentally we would probably need ultra low radio frequency to be able to rotate an observer/measuring device but then the wavelength would be long and diffraction would make difficult the narrow beam requirement. It is a very tricky question imo.

Answer 2

This is an answer to the original, before editing, question the title asked:

What does a circularly polarized photon look like in a co-rotating reference frame?

First of all photons are not electromagnetic light. Classical electromagnetic waves emerge from the quantum mechanical superposition of the wavefunctions of a large number of photons.

This illustration explains how the photons, which can only have spin +1 or -1 to their direction of momentum, build up a polarized beam

Left and right circular polarization and their associate angular momenta

You ask:

Now, if I enter a frame rotating with angular frequency ω that, if E and B were rigid physical vectors, would render them stationary, what do I see?

The "stationary" would be linearly polarized light at the instantaneous framework. The "seeing" may be achieved by rotating a linear polarisation filter at that frequency,( although I do not know it is feasible in the lab). Anyway one does not need general relativity to have non inertial frames.

As there exist no circularly polarized photons your question can be answered only for light.

Answer 3

The instantaneous electric field at any point in time from a monochromatic light source is given by:

$\mathbf{ \epsilon } (\mathbf{r}, t) = $Re$[ \mathbf{E}(\mathbf{r})$ exp$( i \omega t)] $

where $\omega$ is the optical frequency and $\mathbf{r}$ picks out some location in 3D space. $\mathbf{E}$ is the complex electric field vector.

Setting $\mathbf{E} = [1, i]$ as in your example we find:

$\mathbf{ \epsilon } (\mathbf{r}, t) = $ [cos($\omega t$), sin($\omega t$) ].

We can move to a frame rotating at the optical frequency by applying the rotation matrix (with time-dependent angle):

$R = [ \cos(\omega t), -\sin(\omega t) \\ \sin(\omega t), \cos(\omega t)] $

So we get

$R \epsilon = [1, 0] $

This makes perfect sense. The circularly polarised wave was described by an electric field vector of constant length but with orientation going in a circle over time. In the rotating frame we don't see this rotation and are left with a vector of constant length and direction.

Note that in the rotating frame the circular polarised wave does not become the same as a linearly polarised wave. A linearly polarised wave with $\mathbf{E} = [1, 0]$ gives rise to $\mathbf{ \epsilon }(t) = [\cos(\omega t), 0]$. This is not the same as $\mathbf{ \epsilon }(t) = [1, 0]$. The overall magnitude of the instantaneous electric field for a linearly polarised wave has zeros, the circular field does not. The move to a rotating frame does not change this. Similarly starting with a linearly polarised field a move to a rotating frame does not make it become a circularly polarised field (it still has zeros).