# Rayleigh scattering

Rayleigh scattering is the cause of the aerial perspective
Rayleigh scattering is what causes the sky to turn blue during the day and the sun to turn red when it sets or rises.

The Rayleigh scattering [ reɪlɪ- ], named after John William Strutt, 3rd Baron Rayleigh , denotes the (mainly) elastic scattering of electromagnetic waves to particles having a diameter small compared to the wavelength is so for example in the scattering of light at small Molecules. In the case of scattering in the earth's atmosphere by molecular oxygen and nitrogen, the inelastic component due to rotational Raman scattering is typically also counted as Rayleigh scattering, since this only causes a shift in the wavenumber of the photon by less than 50 cm −1 . The cross section of this contribution has the same wavelength dependence as the elastic component. ${\ displaystyle \ lambda}$

The scattering cross section of Rayleigh scattering is proportional to the fourth power of the frequency of the electromagnetic wave. This applies not only to independently scattering particles, i.e. at particle distances greater than the coherence length of the radiation, but also at higher particle concentrations for the scattering of inhomogeneities in the refractive index through a statistical arrangement of the particles, for example in gases or glasses . Blue light has a higher frequency than red and is therefore more strongly scattered. ${\ displaystyle \ sigma}$${\ displaystyle f}$${\ displaystyle f}$

The waxing crescent moon also appears reddish when it is only a few degrees above the horizon. The moonlight only reaches the observer after a long passage of over 200 kilometers through the earth's atmosphere.

The frequency-dependent scattering of sunlight on the particles of the earth's atmosphere, which varies in strength, causes the blue of the sky during the day, and the dawn and the evening red during twilight . Standing just above the horizon, the moon also appears reddish.

Rayleigh-scattered light is polarized , particularly strong at scattering angles of 90 °. The left picture is taken without a polarization filter , the right one with a reverse
polarization filter.

Rayleigh scattering occurs because the incident light excites the electrons of a molecule and induces a dipole moment that oscillates in the same way as the incident electromagnetic radiation . The induced dipole moment now acts like a Hertzian dipole and emits light that has the same wavelength as the incident light. ${\ displaystyle {\ vec {p}} _ {\ text {ind}} = \ alpha {\ vec {E}}}$

## Cross section

The cross-section of Rayleigh scattering for a single particle results from the oscillator model . In the limiting case of low frequencies (compared to the natural frequency, ) the following applies: ${\ displaystyle \ sigma}$${\ displaystyle \ omega \ ll \ omega _ {0}}$

${\ displaystyle \ sigma (\ omega) \ approx \ sigma _ {\ mathrm {Th}} {\ frac {\ omega ^ {4}} {\ omega _ {0} ^ {4}}}}$

where is the Thomson cross section . The angular distribution and polarization is that of a dipole in the direction of the incident wave. ${\ displaystyle \ sigma _ {\ mathrm {Th}} = 0 {,} 665 \ cdot 10 ^ {- 24} \, \ mathrm {cm} ^ {2}}$

Milky opal is red-orange in transmission because density inhomogeneities scatter the blue light laterally.

If there are several particles in the coherence volume , the scattered waves interfere . With many particles per coherence volume, spatial fluctuations in the refractive index act as scattering centers. For a sphere with diameter and refractive index in a medium, the scattering cross-section is: ${\ displaystyle d \ ll {\ tfrac {\ lambda} {2 \ pi}} = {\ tfrac {1} {k}}}$${\ displaystyle n_ {2}}$${\ displaystyle n_ {1}}$

${\ displaystyle \ sigma = {\ frac {8 \ pi d ^ {6} k ^ {4}} {3}} \ left ({\ frac {\ left ({\ frac {n_ {2}} {n_ { 1}}} \ right) ^ {2} -1} {\ left ({\ frac {n_ {2}} {n_ {1}}} \ right) ^ {2} +1}} \ right) ^ { 2}.}$

## The blue or the red of the sky

Power distribution of scattered sunlight

Rayleigh scattering explains why the sky appears blue. The wavelength of blue light,, is around 450 nm, that of red light around 650 nm. Thus, the ratio of the effective cross sections follows: ${\ displaystyle \ lambda _ {\ mathrm {blue}}}$

{\ displaystyle {\ begin {aligned} {\ frac {\ sigma _ {\ text {blue}}} {\ sigma _ {\ text {red}}}} & = {\ frac {\ frac {1} {\ lambda _ {\ text {blue}} ^ {4}}} {\ frac {1} {\ lambda _ {\ text {red}} ^ {4}}}} = \ left ({\ frac {650 \, {\ text {nm}}} {450 \, {\ text {nm}}}} \ right) ^ {4} \ approx 4 {,} 4 \ end {aligned}}}

In the picture, the radiated power distribution of the sun, approximated by Planck's law of radiation from a surface temperature of 5777 K, is shown in red. The spectral maximum is then in green light (500 nm wavelength). The spectral maximum of daylight , however, is u. a. due to the scattering effect described here at 550 nm. The power distribution of the scattered light (blue curve) results from multiplication by ω 4 . Accordingly, the maximum moves far into the UV range. In fact, however, it is in the near UV, since molecular absorptions are added at shorter wavelengths.

• During the day, when the sun is high in the sky, light only travels a short distance through the atmosphere. Only a small amount of blue light is scattered in other directions. That is why the sun appears yellow. From soaring aircraft, the sun appears “whiter” because less blue light is scattered away.
• The sum of all scattered light makes the sky appear blue from all other directions. On the other hand, on the moon, where there is no dense atmosphere, the sky appears black even during the day.
• When the sun is low, the distance of sunlight through the earth's atmosphere is much longer. As a result, a large part of the high-frequency light components (blue) is scattered sideways, mostly light with long wavelengths remains and the color impression of the sun shifts towards red. This effect is reinforced by additional particles in the air (e.g. haze, aerosols, dust). However, the Chappuis absorption is responsible for the blue coloration of the sky at the zenith after sunset , which is hardly noticeable when the sun is higher.

## Strength of light attenuation due to Rayleigh scattering

In order to quantitatively calculate the strength of Rayleigh scattering, it must be taken into account that the elementary waves emanating from the molecules interfere within a coherence volume of the radiation , so that it is not intensities according to the above formula, but scatter amplitudes that have to be added. The particle density below which this effect may be neglected for sunlight is about 1 / μm³, seven orders of magnitude below the value relevant for the atmosphere. The mean density within a scattering volume element is irrelevant for the scattering, the density fluctuations are effective.

One result of the statistics is that the fluctuation amplitude of the number of particles according to the Poisson distribution only increases with the square root of the particle number. Since shorter wavelengths are scattered on finer structures from correspondingly fewer particles, this radiation 'sees' stronger fluctuation amplitudes than longer-wave radiation. With a fixed wavelength, the fluctuation amplitude depends on the square root of the particle density of the gas. The scattered intensity depends on the square of the fluctuation amplitude, i.e. linearly on the density. Overall, according to Paetzold (1952), the following applies to the attenuation of light in the atmosphere at perpendicular incidence :

${\ displaystyle k = {\ frac {\, 5 \ cdot 32 \, \ pi ^ {3} \, (n_ {0} -1) ^ {2}} {3 \ cdot \ ln 100 \ cdot \ lambda ^ {4}}} \, {\ frac {H} {N_ {0}}}}$

This contains the so-called extinction in astronomical size classes , the refractive index of the air under normal conditions, the effective thickness of the atmosphere (scale height, see barometric height formula ) and the Loschmidt constant (particle density of the air under normal conditions). The fact that the latter is in the denominator is only an apparent contradiction to what has been said above, because the term is proportional to the density and is the square of the numerator. ${\ displaystyle k}$${\ displaystyle n _ {\ mathrm {0}}}$${\ displaystyle H}$${\ displaystyle N_ {0}}$${\ displaystyle n_ {0} -1}$

The transmission , the ratio between the intensity transmitted by the scattering layer and the incident intensity, follows from the extinction : ${\ displaystyle T}$

${\ displaystyle T = 10 ^ {- 0 {,} 4 \, k}}$

This is the form of the Lambert-Beer law commonly used in astronomy . In practice, too

${\ displaystyle T = \ mathrm {e} ^ {- \ tau}}$

used, with as optical depth. The simple conversion applies: ${\ displaystyle \ tau}$

${\ displaystyle k = 1 {,} 086 \, \ tau}$
Transmission of the clear atmosphere at sea level as a function of wavelength and angle of incidence

In the case of an inclined incidence at a zenith angle , the effective layer thickness is approximately (with plane-parallel layering): ${\ displaystyle z}$

${\ displaystyle H (z) = {\ frac {H (0)} {\ cos (z)}}}$

According to Paetzold (1952) is as well . According to Stoecker (1997) is . Inserting supplies ${\ displaystyle N_ {0} = 2 {,} 70 \ cdot 10 ^ {25} ~ \ mathrm {m} ^ {- 3}}$${\ displaystyle H = 7990 ~ \ mathrm {m}}$${\ displaystyle n_ {0} = 1 {,} 000292}$

${\ displaystyle \ tau = {\ frac {0 {,} 0084 ~ \ mu \ mathrm {m} ^ {4}} {\ cos (z) \, \ lambda ^ {4}}}}$

In the visual (550 nm), about 90% of the light passes through the atmosphere at normal incidence, in the blue (440 nm) about 80%. At a shallow incidence at a zenith angle of 80 °, these proportions are only 60% and 25%. The already discussed reddening of the light due to Rayleigh scattering is thus clearly understandable.

In practice, the attenuation of light due to further scattering from aerosol and dust particles (see Mie scattering ) is significantly greater. This additional extinction is particularly strong after natural disasters such as volcanic eruptions . For example, after the eruption of the Pinatubo in 1991 on La Silla (one of the locations of the European Southern Observatory (ESO)), Grothues and Gochermann (1992) found a visual attenuation of 0.21 size classes (normal are 0.13 size classes) with perpendicular incidence of light ). The transmission was therefore reduced from 89% to 82%. In the blue the extinction coefficient had risen from 0.23 to 0.31 size classes, i.e. that is, the transmission had dropped from 81% to 75%.