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An answer I found online was: At high altitude where molecules are far apart, scattered photons can travel without interfering with each other, thus they fill the sky with blue light. But at low altitude, molecules are so close to each other that scattered photons destructively interfere with each other to cancel each other out, that's why we see air as transparent. However, if they perfectly cancel out, then wouldn't the energy be not conserved?

Another explanation was simply that the effect is only apparent in a very large column of air since the scattering effect is not that noticeable.

So why doesn't Rayleigh scattering happen at low altitude in air?

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  • $\begingroup$ Photons only interfere with themselves. $\endgroup$
    – my2cts
    Feb 28, 2020 at 7:00
  • $\begingroup$ I wrote about this on wordpress.com/post/rogerjbarlow.com/56 . The Rayleigh scattering occurs off fluctuations which are greater at low densities, i.e. high altitudes $\endgroup$
    – RogerJBarlow
    Feb 28, 2020 at 9:36
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    $\begingroup$ @RogerJBarlow write an answer. $\endgroup$
    – ProfRob
    Feb 28, 2020 at 10:53
  • $\begingroup$ @RogerJBarlow your link requires logging in or directs to the home page. $\endgroup$
    – Ruslan
    Feb 28, 2020 at 19:22
  • $\begingroup$ Sorry: try rogerjbarlow.com/2018/06/03/why-is-the-sky-blue $\endgroup$
    – RogerJBarlow
    Feb 29, 2020 at 10:02

2 Answers 2

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The assumption is wrong. Rayleigh-scattering does happen independent of altitude. That it happens at sea level, too, becomes clear when you look at the other bank of a large river mouth or a nearby coast over a bay: they loose much contrast and everything is blue-ish; the same effect happens in the mountains looking from one peak to the other.

In the sky there is no other light source, so that only scattered light arrives from there which makes it so nicely blue.

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    $\begingroup$ Beware that your example of ground-level effects is probably illustrating Mie scattering (off aerosols) instead of Rayleigh. $\endgroup$
    – BowlOfRed
    Feb 28, 2020 at 6:52
  • $\begingroup$ @BowlOfRed Yes-ish, near the surface mie scattering definitely plays a role as well. Mie generally turns down contrast (muss less wavelength dependent), while Rayleigh scatters blue much stronger (lambda^-4). Both happen and their relative effect depends highly on the local atmosphere. I was thinking of and hinting at clear days outside urban influence where distant objects appear not only less in contrast but mostly blue. $\endgroup$
    – planetmaker
    Feb 28, 2020 at 8:12
  • $\begingroup$ Rayleigh scattering is worse at lower altitudes because there is more air. $\endgroup$
    – Kevin Kostlan
    Feb 28, 2020 at 16:32
  • $\begingroup$ Of course. But speaking of ratios Rayleigh vs. Mie? $\endgroup$
    – planetmaker
    Feb 28, 2020 at 17:17
  • $\begingroup$ Aerosol profiles very much depend on weather. See e.g. this bunch of profiles. At the same time, for air molecules the barometric formula is a reasonable approximation, so molecular scattering doesn't depend on weather that much. $\endgroup$
    – Ruslan
    Feb 28, 2020 at 19:20
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There is no problem with energy, and an explanation in terms of waves is more intuitive. How else could one easily understand a wavelength dependence?

First there is dispersion, the increase of the index of refraction at short wavelength, which can be understood with the model of Lorentz oscillators.

In a gas there are fluctuations in the number density on short length scales. That means that the index of refraction of cubes with a size of $\lambda^3$ will fluctuate. This gives more scattering at shorter wavelengths, together the $\lambda^{-4}$ dependence.

I am not sure about the form of the pressure dependence, but I would agree with @planetmaker that your assumption is wrong. Fluctuations are relatively larger at low densities, but the lower the density the less scattering there will be. In the limiting case of the vacuum there is no blue sky.

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