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I have just learned that sunlight traveling through the earth's atmosphere picks up a net polarization by collisions with molecules ($O_2$, $N_2$, etc.) that the photons encounter. One would think that those molecules would have to possess some degree of common alignment in order to produce light that possesses nonrandom polarization. Since those molecules are randomly positioned in the atmosphere, how can a net polarization of the transmitted light be produced?

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    $\begingroup$ Is that linear polarisation or circular? $\endgroup$ Aug 8 '20 at 5:17
  • $\begingroup$ it's the polarization type which is revealed by polaroid filters. $\endgroup$ Aug 8 '20 at 5:28
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    $\begingroup$ similar physics.stackexchange.com/questions/341442/… $\endgroup$
    – anna v
    Aug 8 '20 at 6:28
  • $\begingroup$ I never thought about it.The same thing can be asked for glass and water , two of the most famous refractive materials.This questions even working of polarimeter and practically everything except polarizers. $\endgroup$
    – Protein
    Aug 8 '20 at 7:31
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    $\begingroup$ Rob Jeffries has written a relevant answer here. $\endgroup$
    – PM 2Ring
    Aug 8 '20 at 18:18
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One would think that those molecules would have to possess some degree of common alignment in order to produce light that possesses nonrandom polarization.

Molecules are quantum mechanical entities and light interacting with individual molecules should be thought of in photons. Nevertheless, the classical elecromagnetic light with its description by the maxwell equations emerges from the quantum framework correctly, so it is better for collective detection, as is the bulk polarization, to think in terms of classical light as the answer by David explains.

So the differences in polarization come from the fact that the rays from the sun are unidirectional, the perpendicular to the ray defines a plane for the E field , and depending on the scattering angle different components will scatter differently.

Look at the classical formulation of Raleigh scattering

Rayleigh scattering results from the electric polarizability of the particles. The oscillating electric field of a light wave acts on the charges within a particle, causing them to move at the same frequency. The particle, therefore, becomes a small radiating dipole whose radiation we see as scattered light. The particles may be individual atoms or molecules; it can occur when light travels through transparent solids and liquids, but is most prominently seen in gases.

The same results will come out by thinking photons and molecules, in a much more complicated mathematically way.

The "common alignment" you seek to produce polarization comes from the directionality of the light, and the stratification of levels in the atmosphere.

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I think your question is really based on isotropy and anisotropy: How can an isotropic medium produce an anisotropc effect? The answer is that the direction of the sunlight provides a preferred direction.

Given the fact that the atmosphere is isotropic, polarization (if it exists) of scattered sunlight must have radial symmetry around the axis defined by the direction of the sunlight.

The scattered light is polarized, due to the facts that A) the E field of the sunlight is always perpendicular to its direction of propagation, and B) light is scattered by very small particles primarily at a right angle to the E field of the light, with polarization aligned with the E field. This is a classical effect, as pointed out by @Ruslan. See the section on "Polarization by Scattering in this paper.

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  • $\begingroup$ Did you mean -isotrop- instead of -isometr-? I only find pages about fitness when I google for "isometric". $\endgroup$
    – Ruslan
    Aug 8 '20 at 19:23
  • $\begingroup$ Thank you for catching that! I was distracted- when I wrote it. Edited the answer. $\endgroup$
    – S. McGrew
    Aug 8 '20 at 19:46
  • $\begingroup$ Now to the point you're making in the last sentence: dipole radiation (and thus Rayleigh-scattered radiation) being polarized is a classical phenomenon, not a quantum effect. $\endgroup$
    – Ruslan
    Aug 8 '20 at 19:50
  • $\begingroup$ Thanks again; I've corrected the answer. $\endgroup$
    – S. McGrew
    Aug 8 '20 at 22:43
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Light passing through a fluid can be polarized for two reasons. We call this interaction, where different polarizations of light scatter differently "birefringence." First, if the components of the fluid have some preferred direction, then the light will be polarized in the corresponding orientation (e.g. chiral molecules). Second, if the fluid itself is arranged into some structure (e.g. stratification).

Let's consider the first scenario by addressing the case of chiral molecules. A molecule is called "chiral" if it does not possess mirror symmetry. In this way, non-chiral molecules possess a handedness, related by a mirror flip, left handed molecules being the counterparts to right handed particles. Light similarly has a handedness: we may think of light as coming in left and right handed circular polarizations. Left handed light will interact differently with a chiral molecule than right handed light. It is not hard to convince yourself that the net-chirality of randomly oriented chiral molecules does not cancel out. Consider, for example, the direction you need to spin a bolt when tightening it, and then flip the bolt. Does the direction you need to spin the bolt to tighten it change? Steve Mould has a nice video on this subject https://www.youtube.com/watch?v=SKhcan8pk2w

The other scenario is bouncing off a surface. Similar to the two circular polarizations of light, we can think of light as being composed of two linear polarizations. The direction of the two polarizations are orthogonal to the light trajectory. Therefore, when bouncing off a surface at an oblique angle, only one polarization vector will be tangential to the surface, while the other will point into or out of the surface. The two linear polarizations of light bouncing off a flat surface will scatter differently, leading to birefringence.

In the atmosphere, O2 and N2 are mirror symmetric, i.e. they are not chiral. Perhaps there are chiral molecules in the atmosphere, but they likely only appear in trace amounts. The more significant source of birefringence in the atmosphere is stratification. Because the atmosphere gets less dense at higher altitudes, the density gradient effectively acts like a surface, interacting differently with the two linear polarizations of light. If you have a pair of polarized sunglasses, you can view this effect for yourself by looking at the sky on a sunny day and tilting your head. You should notice the sky getting brighter and darker depending on the angle at which you tilt your polarized sunglasses.

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    $\begingroup$ You'd see the polarization even in homogeneous atmosphere (e.g. constant density up to some altitude). I.e. stratification is not the definitive reason for polarizedness of the scattered light: it's the Rayleigh scattering, which is anisotropic relative to $\vec E$, that polarizes scattered light. Indeed, at sunset you can even determine direction to the Sun (obscured by e.g. buildings) by noting the orientation of the Haidinger's brush in the zenith—it will be consistent with the Rayleigh phase function. $\endgroup$
    – Ruslan
    Aug 8 '20 at 19:18

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