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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.

Sean E. Lake
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    a coherent (IQ) receiver for a circularly polarized RF signal effectively rotates the instantaneous linear polarization at the rate of the carrier frequency – hyportnex Jan 14 '20 at 14:56
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    You would indeed see non-rotating fields, but there's nothing wrong with that, because in a noninertial frame Maxwell's equations don't hold anyway. – knzhou Sep 20 '20 at 01:13

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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.

blanci
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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

photspin

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.

anna v
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  • "As there exist no circularly polarized photons" That seems incorrect, since circularly polarized photons are photons of definite helicity, which is just spin measured along the direction or propagation. The different polarization states would correspond, in principle, to measuring spin along different axes than the propagation one, I think. – Sean E. Lake Jan 14 '20 at 10:29
  • Regardless, I have removed "photon" from the title, since it was irrelevant to the question as intended. – Sean E. Lake Jan 14 '20 at 10:34
  • So if I detect a single photon, I have not detected light? That does not seem consistent. – my2cts Jan 14 '20 at 10:51
  • @my2cts See this single photon double slit experiment http://www.sps.ch/en/articles/progresses/wave_particle_duality_of_light_for_the_classroom_13/ – anna v Jan 14 '20 at 11:17
  • @SeanE.Lake a photon is a zero mass quantum mechanical particle, and can be described only by its energy, and has the spin projected +1 to -1 to its direction of motion. It is the photon wavefuncion that has phases and the frequency, which affects the probability of finding the photon, at x,y,z,t. the picture above describes the case. the middle column are the photons orientation to get polarized light. also see http://www.sps.ch/en/articles/progresses/wave_particle_duality_of_light_for_the_classroom_13/ to gain some intution – anna v Jan 14 '20 at 11:23
  • None of the linked articles support your contentions. They seem, to me, to boil down to a mix of undecided questions about the epistemic status of the wave function and preferred terminology that springs from that. If there's a discussion to be had, please make a new question/answer for them, they're off topic here. – Sean E. Lake Jan 15 '20 at 03:07
  • I dislike this answer. The exact relation between classical electric fields and quantised photons might well have quite a lot of nuance and technical detail, but strong statements like "there exist no circularly polarized photons" are not useful to understanding this nuance, and are in direct contradiction to the standard language of quantum optics, where it is completely normal to refer to circularly polarised photons. Most especially in the case of polarisation entangled photons. eg: https://doi.org/10.1364/OE.20.019297, https://doi.org/10.1364/OPTICA.6.000034 – Dast Nov 05 '21 at 11:25
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    @Dast all the negative votes I get as a particle physicist are from quantum optics physicists who have used the terminology "photon" in contrast to particle physicists for solution in mediums of quantum equations where photons can have a mass. In elementary particle physics all that identifies the point particle photon is its energy and spin. – anna v Nov 06 '21 at 06:00
  • To clarify, I am not saying you are wrong. I am sure their are important subtlties that we quantum optics people skate over. [Our definition is very operational. Did it get through the polariser? Does the number resolving detector (g2) say it was a single photon pulse? If yes to both it was a circularly polarised photon.] My point was just that from your answer I still don't know what these subtleties are, and its clear that your terminology is different from what is (to me) ordinary, which leaves me in doubt about whether the technicalities are scientific or semantic. – Dast Nov 06 '21 at 17:08
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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).

Dast
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  • Are you sure this is accurate? $\mathbf{E}$ and $\mathbf{B}$ have a non-trival mixing from uniform boosts. This is at least a non-uniform boost. Worse, its an accelerating reference frame, so there may be some GR considerations. Thanks for trying, though. :) – Sean E. Lake Nov 05 '21 at 15:53