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I would like to estimate the effect that Earth's seasonal tilt (with respect to the sun) has on local variations in polarization of sunlight.

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If the atmosphere were treated as an interface with index of refraction $n_\text{atm}$ (occuring either discretely at some arbitrary boundary based on density etc, or smoothly increasing over differential layers of atmosphere), then the transition from $\mu_0, \epsilon_0 \rightarrow \mu_\text{atm}, \epsilon_\text{atm}$ at some angle of incidence $\theta_i$ would selectively transmit and reflect different polarizations of sunlight.

I would think that dipole scattering would then preserve the plane of polarization for light transmitted into the atmosphere, so that a difference in proportion of perpindicular- vs. parallel-polarized light at the upper atmosphere would be preserved as the light propagated/scattered towards the surface of the Earth. From the Figure, this would lead an observer in, say, South Africa ($\theta_i\approx 45^\text{o}$) to measure different amounts of s-pol or p-pol light than an observe in India ($\theta_i\approx 0^\text{o}$)


The two questions, compactly:

  1. Can the atmosphere be treated as an interface that preferentially transmits/reflects different polarizations of sunlight for a given $\theta_i$?
  2. Would this effect cause observable differences in fractions of polarization types measured by an observer at the surface?
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Well, you can model the earth's atmosphere, like you have said, as an interface but the refractive index of the interface will be a time dependent one, since the concentration of gases in the atmosphere is not static; it is dynamic process, continuously changing the refractive index of the atmospheric media. As a result, the so-called preferential treatment dealt by the earth's atmosphere to the transmitted polarisation of the sunlight is not something absolute; rather it changes from time to time. I am not sure but maybe this change has a diurnal variation or even a seasonal variation. As part of the second question, the answer is yes. Strictly speaking, observers on different parts of the earth will see different kind of polarised light. So the sunset in Greenland, India and Africa have different characteristics; they are not unique but distinguishable.

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  • $\begingroup$ For your answer to (2), is the location-dependent sunset due to the effect I asked about or are there other physical mechanisms you're aware of? $\endgroup$
    – forky40
    Commented Jun 26, 2017 at 17:53
  • $\begingroup$ @forky40 To some extent, the location dependent sunset effect is due to the refraction of the sunlight by the earth's atmosphere. But that certainly contributes to all the locations in different proportions. For example the refraction experienced by the sunlight incident in the tropics is greater than received at the equator. This is due to the peculiar oblate spheroid shape of the earth. Another point that can be taken into consideration is that the composition of the atmosphere at different locations vary. So the scattering that the sunlight is subjected to at different locations (contd) $\endgroup$ Commented Jun 26, 2017 at 18:27
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    $\begingroup$ Is not the same. This contributes significantly to the location dependent sunset effects that I have already described . However there maybe a few factors that I have not mentioned . $\endgroup$ Commented Jun 26, 2017 at 18:30
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That mechanism causes no significant polarization. The formula for specular reflection ought not to exclude the refractive index of the medium above the elevation of the hypothetical reflective sphere. When the two indices of refraction are nearly equal (I expect atmospheric density doesn't have vertical discontinuities below the ionosphere), the resulting reflection is the sum of squares inversely proportional to small differences in refractive index. As the sum is made into a continuous integral, the increase of summed elements only rises linearly, while the element size decreases quadratically. The sum tends to zero in a continuous atmosphere gradient.

Skylight, though, IS polarized, by a different effect. The Rayleigh scattering that makes the sky blue, creates scattered light that propogates only from the direct sun's ray in a direction perpendicular to the E-field, and that E-field (in a ray from the sun) has no component oriented toward the sun. So if you point at a patch of blue sky, expect that the light coming from that patch has polarization mainly perpendicular to the patch-to-sun direction. This will not be the case for multiply scattered light (like the dim gray glow under a cloud).

So, while the visible solar disk is not polarized, the blue sky surrounding it IS.

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  • $\begingroup$ To check my understanding, (in the first paragraph) you're saying that if the atmosphere very smoothly transitions from vacuum to index of refraction $n_\text{atm} \neq 1$ then the reflection coefficient tends to zero? I couldn't follow the math; is there a resource you'd recommend for more detail on the calculation you're describing? $\endgroup$
    – forky40
    Commented Jun 27, 2017 at 2:15

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