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I've done plenty of Googling on the topic, but I'm still not completely getting it. The way my E&M professor's PowerPoint explained it, you find the polarization of a wave by imagining inserting a plane into the wave at a right angle and see how the plane moves. That's the exact same analogy often used to describe the curl of a vector field, but polarization can't just be the curl of the EM field, because the units don't match -- the curl of the E field has units of $\frac{\text{kg}}{\text{s}^2 \text{C}}$ (and the curl of the B field is of course just zero), while polarization has units of $\frac{\text{C}}{\text{m}^2}$.

Wikipedia says, "Polarization (also polarisation) is a property applying to transverse waves that specifies the geometrical orientation of the oscillations." Several other websites explicitly point out that the direction of the polarization vector is the direction of the E field, and also make it sound like it's all just about the direction. But if that's the case, then why does it have units of $\frac{\text{C}}{\text{m}^2}$ instead of just being a dimensionless unit vector? And, for that matter, why do we even need another vector to specify orientation when we already know that the E and B fields are orthogonal to each other and the direction of wave propagation is orthogonal to them both?

I did find this page which has some helpful pictures. It sort of seems like it's about how the individual photons are all lined up, which makes sense with the idea of polarized light in experiments demonstrating Bell's inequality in QM (there are several youtube videos about that that I watched a while ago) -- in that context, they talked about the probabilities of individual photons being measured at a particular orientation, both before and after passing through the polarization filter used in the experiment. That makes sense conceptually, but it still doesn't tell me why polarization isn't a dimensionless quantity if it's just about direction/orientation.

FYI: I did see this question, but the single answer doesn't answer my question about why it's not dimensionless if it's only an indication of orientation.

This article did give me some insight as to what the units mean, since it sounds like it's not just about the direction of the E field, but how that E field affects the spacial distribution of charged particles in the region. That almost answers my question, but not entirely -- what about light that's just been through a polarization filter but isn't, on average, interacting with any charged particles? What do the units mean then? Is it just a description of how that light would affect the distribution of charge if there were a bunch of charged particles around? But even then, wouldn't that depend on the initial distribution of the charges?

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You appear to have confused two different meanings of the word "polarization." The polarization of an EM wave is the orientation and aspect ratio of the ellipse traced out by the $E$ field in a plane perpendicular to the propagation direction over a cycle. This information is dimensionless. E.g. I can specify an EM plane wave in a vacuum as, say, a "400 THz wave traveling left with amplitude $500\;\mathrm V\,\mathrm m^{-1}$ and polarized vertically" (the only thing missing is phase). I need dimensions for amplitude and frequency. The direction of travel and the polarization are dimensionless (and so is phase). The polarization of a wave is also not defined by a curl. It is, again, simply the shape (but not size) of the ellipse traced out by the $E$ vector at any single point over time. (The curl finds the amount that $E$ rotates around a point in space, which is entirely not what we want.)

The other use of "polarization" is in the study of EM inside bulk materials. A body of matter may, one way or another, develop microscopic charge separations that result in macroscopic electric fields. Such a material has been electrically polarized. We often treat the amount and direction of charge separation at each point in the material as a vector field. This field is called the polarization field and has units of $\mathrm C\,\mathrm m^{-2}.$ You integrate it over a volume of matter and it tells you the net dipole moment (in $\mathrm C\,\mathrm m$) due to charge separation over that volume. This is a completely different concept to the polarization of a wave. It just uses the same word.

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  • $\begingroup$ Thanks for clarification! Follow-up question: Is there any relationship between the two types of polarization? Like, does the direction of the E field that's causing polarization of a dielectric affect the dielectric's polarization? E.g. If I were to pass light through a polarizing filter onto a dielectric, would there be any difference in the dielectric's polarization than if I shined unpolarized light on it? $\endgroup$ Commented Feb 7, 2023 at 22:50
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    $\begingroup$ @MikaylaEckelCifrese Yes, sure. Since the polarization of a wave gives the direction of the $E$ field, and the polarization of a dielectric is usually parallel to the applied field ($P\propto E$), shining polarized light of different orientations onto a dielectric will generally produce different patterns of polarization in the material. (When the produced $P$ is not parallel to $E,$ that is the origin of birefringence.) $\endgroup$
    – HTNW
    Commented Feb 9, 2023 at 18:41

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