Given an approximation of the spectrum of extraterrestrial sunlight by blackboby radiation at 5800K, between 360 and 800nm, is there a simple formula or algorithm that can be used to make an approximation of the spectrum of the diffuse blue sky at one point, at noon, on earth (with fixed humidity, pressure and temperature and zero pollutants, if relevant). And what would be the parameters to use, if any, to change the position of observation on the planet, i.e longitude and latitude ?


2 Answers 2


If you rule out the presence of dust and aerosol particles in the atmosphere then the spectrum of scattered sunlight is governed by Raleigh scattering.

The Rayleigh scattering cross-section goes as $\sigma \simeq \lambda^{-4}$ and thus the amount of scattered light will go as $$ I(\lambda) \simeq I_0 \lambda^{-4} \exp(-a/\lambda^4),$$ where $I_0$ is the incident solar spectrum, with the exponential term accounting for light scattered out of the beam. Generally speaking, the exponential term is $\sim 1$.

Therefore a simple algorithm to determine the spectrum of blue sky would be to take the intrinsic direct solar spectrum (after passage through the atmosphere at zenith) and multiply it by $\lambda^{-4}$.

The likelihood of significant Mie scattering from particles in the atmosphere, which has a more wavelength-independent cross-section, would dilute the blue-ing effect of Raleigh scattering.

A second order of precision might be to use a combination of Rayleigh and Mie scattering of the form $$ I(\lambda) = I_0 \lambda^{-4} \exp(-a/\lambda^4) \exp(-b/\lambda^p),$$ where the first term accounts for Rayleigh scattering and the second term accounts for Mie scattering. According to Zagury (2012), appropriate coefficients for the daytime sky at noon are $a=0.005$ ($\mu$m)$^4$, $p=1.3$ and $b=0.45$ ($\mu$m)$^{1.3}$.

Note that the above considers just the form of the spectrum. To calculate an accurate intensity would require a far more sophisticated radiative transfer treatment in a realistic atmosphere with density and composition gradients and perhaps the influence of multiple scattering events and absorption by water and ozone.


Have a look Not simple, even to get the energy distribution. In addition note that at 100C


Essentially all of the radiation from the human body and its ordinary surroundings is in the infrared portion of the electromagnetic spectrum, which ranges from about 1000 to 1,000,000 on this scale.

The energy coming from the black body of the sun is reradiated as a black body from the earth, with frequencies in the infrared. The formula connects them but does not explain the colors of the sky.

The colors we see in the sky are the result of scatters of the visible part of the sun's spectrum.:


Blue comes from Rayleigh scattering, for example. The sun's black body radiation is the source, but the observed colors depend on scattering amplitudes, given the arriving frequencies.

  • $\begingroup$ Thanks, but I already know how to calculate the blackboby curve. However I don't know how to calculate the rest. $\endgroup$ Oct 6, 2018 at 12:44
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    $\begingroup$ well, you need the crossections which depend on quantum dynamical calculations, not one simple formula, that is what I am saying. It is not a simple formula business which you should understand if you read the links. $\endgroup$
    – anna v
    Oct 6, 2018 at 14:24

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