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Will two freely rotatable linear polarizers (placed in sequence and at some angular offset less than, say, 45 degrees) eventually align if you shine (plenty of) unpolarized light at the first one?

If so, will the second polarizer align itself with the first one, or will both of them start to rotate towards alignment (although perhaps not necessarily equally quick)?

Edit. There are over 400 views and five answers now (one at +1, the others at 0), which are all highly appreciated. However, I thought it might be useful to provide a bit of my initial intuition with regards to the first part of the question.

I am assuming that the light is polarized vertically after passing the first polarizer and that the second polarizer (pictured) is slightly offset.

To represent the vertical polarization I give a single photon some vertical extent (possibly in terms of probability amplitude). This the segment A-B. I am not at all sure that such representation is allowed or misguided.

enter image description here

Now the observation is that a collision (absorption or reflection) at B will produce more angular momentum (of the polarizer) than a collision at A.

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The intuition provided after "Edit" is plainly wrong, as explained here. One can therefore safely disregard everything after "Edit". –  Glen The Udderboat Mar 23 '13 at 7:14

5 Answers 5

I believe this question can be addressed using conservation of angular momentum. If the incident light waves are linearly polarized, then their angular momentum is zero, therefore the polarizer won't experience a torque. If, however, the incident light carries elliptical polarization then rotation will occur to conserve angular momentum. The resulting beam will be linearly polarized and therefore carry no angular momentum. Therefore the second polarizer won't rotate. Just to be clear, the beam requires a net angular momentum for a net torque to exist. If you're using some generic light source (the sun, a lamp, etc.) the net angular momentum will be zero so there won't be an effect. What you could do (and I now want to try if I can find a big enough laser in lab) is circularly polarize the beam and then observe the torque on a linear polarizer.

I'm sure we can also think of this in terms of photons. The photon spin eigenstates correspond to left and right handed polarization, so a circularly polarized beam of light is a pure spin state. Hence, if the photons coming out of the polarizer are in a state of zero angular momentum (linear polarization) then conservation of angular momentum requires the polarizer feel a torque.

Please take this is a grain of salt. I'm still working my way through undergrad EM and QM.

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What you are trying to ask is:

Does radiation with a well defined polarization, that passes through a polarizer that is not completely alligned with that radiation polarization, does interact with the atoms in a way that makes the structure to turn.

I cannot say for sure, but I do think that yes. Although you should not consider any kind of friction, or sufficiently intense radiation (none of the case seem realistic to me).

But academically... The radiation does interact with the atoms: as electromagnetic field is created in the material because of the radiation. The atoms/structure in a polarizer, can be more easily polarized along a certain direction, but the structure is not completely rigid (nor perfect).

So, if a radiation tries to polarize the material in a direction slightly different from that preferential direction, the overall structure will begin to try to be polarized along that direction even if it cannot. This will lead, eventually, to a lot of very small "changes" that would the structure to turn in order to ease (energy minimization) its polarization.

As I said. I cannot tell this for sure...

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Interesting. I would certainly like somebody to firm this up or to show this wrong. –  Glen The Udderboat Mar 23 '13 at 7:28

Only objects that interacts differently with the two circular polarizations of light will begin to rotate. This will happen, in general, even if the incoming light is linearly polarized or unpolarized. This effect has been seen. See references in this Wikipedia article.

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Thanks! There are 20 references in the Wikipedia entry. Could you point out which one(s) are most relevant? –  Glen The Udderboat Mar 23 '13 at 7:31

I believe that the confusion here is that you are thinking of light in analogy to water. While this is helpful for some things, it is not applicable here. Light moves in straight lines until it is absorbed or reflected by something. Light that is blocked by a polarized lens is absorbed by the lens and becomes heat. Light that is not absorbed passes straight through. If you have, say, a nozzle with two wafers with slits cut across them blocking the exit, and you try and force water through it, the water that does not pass both slits will become trapped and begin to increase the pressure. When both slits are aligned so that the maximum amount of water passes through, the pressure will be at a minimum, and that's why with waterfall toys the little gadgets that block the water move around. Since we are assuming unpolarized light from a single bright source, all of the lateral (in the plane of the polarizers) components will cancel, and the only net pressure on the polarizers will be into the plane. There is nothing there to create a rotational component, so the polarizers will not align.

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I was actually thinking more about analogies with 1) misaligned coins eventually turning a slot machine or 2) jumping ropes eventually turning a fence. :) Anyway, your explanation involves heat, which to my taste is a bit too macroscopic for a rigorous explanation. (I'm thinking: Heat could be hiding some directional component.) –  Glen The Udderboat Mar 23 '13 at 7:25

I don't see a source for the angular momentum required for them to move. So no, unless something is unstated.

Does the manufacture of the polarization cause the polarized material to have a density dependent on angle?

It might be that in zero gravity that the polarizers align because of their polarization etching completely independent of the light falling on them

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