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I know that the scattering of light decreases as inversely proportional to the 4th power of wavelength. But what happens at the atomic level?

Does the photon get absorbed and re-emitted?

Does the photon undergo Compton effect?

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From memory: there are two effects that cause the sky to be blue.

  1. At the molecular level: the response of Lorentz oscillators at frequencies below the resonances in the UV; this explains normal dispersion, the index of refraction increasing at shorter wavelengths.

  2. Fluctuations in the number density (and consequently of the refractive index) on the length scale of a wavelength — these are larger for shorter wavelengths.

Both effects cause more scattering at short wavelengths. Elastic scattering.

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    $\begingroup$ You might want to identify (1) as dipole or Rayleigh scattering by individual air molecules $\endgroup$ – Paul Young Jan 7 at 15:36
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    $\begingroup$ This paper, users.df.uba.ar/bragas/Web%20roberto/Papers/…, discusses the effect of fluctuation. Indeed a homogeneous gas would not scatter. The scattering is caused by fluctuation, however, the resulting effect is precisely that the air nevertheless behaves as a sum of independent scatterers ! $\endgroup$ – my2cts Jan 7 at 16:02
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This is a good question as there is some confusion over the description of reflection and scattering. Next time please formulate 1 question at a time.

Q1 At atomic/molecular level the photon wave function mixes with one or more electronic excitations which may alter its momentum.

Q2 The photon does not get absorbed and reemitted. Rayleigh scattering is a parametric, elastic process.

Q3 The photon does not undergo a Compton effect. We are talking about optical photons here and excitation energies are significantly higher. It may undergo a Raman process (inelastic) in which the rotational state of the molecule is altered.

Quote from wikipedia: "Rayleigh scattering , named after the British physicist Lord Rayleigh [ ] is the predominantly elastic scattering of light or other electromagnetic radiation by particles much smaller than the wavelength of the radiation. Rayleigh scattering does not change the state of material and is, hence, a parametric process. The particles may be individual atoms or molecules. It [ ] is most prominently seen in gases. 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."

Note that what was the last sentence unfortunately seems to contradict the second one ! Dipole radiation from an atom implies deexcitation, which does not occur. I just added one sentence the wiki article: "This radiation is integral part of the photon and no excitation or deexcitation occurs.".

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  • $\begingroup$ I challenge the anonymous complainer to bravely come forward and share his or her own views explicitly. $\endgroup$ – my2cts Jan 7 at 16:04
  • $\begingroup$ I gave you +1 because I agree with you - I'm not the anonymous complainer. $\endgroup$ – Cinaed Simson Jul 4 at 8:15
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The major factor of different scattering is the ratio of wavelength to the size of particles which are working as microscopic scattering mirrors.

In a sparse particle medium, the longer the wavelength, the easier it is to transmit and not be scattered. The shorter the wavelength, the easier it is to scatter and the difficult it is to transmit.

Given the size of mirrors, the shorter the wavelength, the stronger the scattered light, the weeker the transmitted light.

In sky, the mirrors are the air molecules. Red light gets more transmission, blue light gets more scattered.

So the sky is blue, the sun is red and yellow.

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