Moon during the day with bright blue sky light Virgin Galactic's Unity at high altitude

left: from this answer to Moon during the day with bright blue sky light right: from Why does Unity look transparent? below: from this answer

Virgin Galactic's Unity at low altitude note Unity is the bit in the middle.

I'm trying to estimate roughly how much of the sky's diffuse blue light is produced between us and this very high altitude object.


Answers to Where in the atmosphere is the blue light scattered? don't provide a quantitative answer to the question's title so I thought I would try to estimate the contribution to the blue light we see at sea level from each altitude.

I started with a simplistic approximation for the density profile using a scale height $H$ of 8.5 km. Using pressure as a proxy for density assumes constant temperature and composition so it's only a starting point.

$$\rho(h) \approx \exp(-h/H)$$

Next is the Rayleigh scattering. Wikipedia gives the following for a volume of dielectric spheres of index $n$, diameter $d$ and a distance $R$:

$$I_{0}{\frac {1+\cos ^{2}\theta }{2R^{2}}\left({\frac {2\pi }{\lambda }}\right)^{4}\left({\frac {n^{2}-1}{n^{2}+2}}\right)^{2}\left({\frac {d}{2}}\right)^{6}},$$

and for a certain region of volume $V$ of the atmosphere:

$$I_{0}{\frac {\pi ^{2}V^{2}\sigma _{\epsilon }^{2}}{2\lambda ^{4}R^{2}}}{\left(1+\cos ^{2}\theta \right)}$$

where $\sigma_{\epsilon}^{2}$ "represents the variance of the fluctuation in the dielectric constant $\epsilon$.

The Wikipedia article's reference #18 links to

McQuarrie, Donald A. (Donald Allan) (2000). Statistical mechanics. Sausalito, Calif.: University Science Books. pp. 62. ISBN 1891389157. OCLC 43370175.

and on page 63:

This is called Rayleigh scattering. The dielectric constant $\epsilon$ is related to the density by the so-called Clausius-Mossoti equation

$$\frac{\epsilon-1}{\epsilon+2} = A \rho$$

which is derived and discussed in most physical chemistry texts. The quantity A is a constant and $\rho$ is the density. We can see from this equation that fluctuations in $\rho$ lead to fluctuations in $\epsilon$, and hence to Rayeligh scattering by Eq. (3-56). If we calculate $\sigma_{\epsilon}^{2}$ n terms of $\sigma_{p}^{2}$ from Eq. (3-57) and use Eq. (3-54) for $\sigma_{p}^{2}$, we can find (see Problem 3-21)

$$\frac{I(\theta)}{I_0} = \frac{\pi k T}{18 \lambda^4} \kappa (\epsilon-1)^2(\epsilon+2)^2 V \frac{1+cos^2 \theta}{R^2}$$

where $\kappa$ is the isothermal compressibility...

One hopes to somehow express $(\epsilon-1)^2(\epsilon+2)^2$ in terms of $\rho$ using the relations above, but I don't know how to address the isothermal compressibility $\kappa$ as a function of height.

The $V/R^2$ cancels in the same way that a diffuse wall doesn't appear to get brighter nor darker as we walk away from it.

Question: How much of the sky's "blue" comes from each altitude; how can I get the variation of the atmosphere's dielectric constant and isothermal compressibility in order to estimate this?

  • $\begingroup$ Are you asking how the blue light accumulates with altitude? That is, how to set up an integral to define the final spectrum? $\endgroup$ Commented Jul 19, 2021 at 13:40
  • 1
    $\begingroup$ @honeste_vivere I can set up integrals but I need the goodies that go inside; "...how can I get the variation of the atmosphere's dielectric constant and isothermal compressibility in order to estimate this?" seems pretty clear? $\endgroup$
    – uhoh
    Commented Jul 19, 2021 at 13:43

1 Answer 1


Landau and Lifshitz, in their "Electrodynamics of Continuous Media", give the following expression for Rayleigh extinction coefficient $h$ (that's directly proportional to coefficient of scattering) in gases:

$$h=\frac{2\omega^4}{3\pi c^4}\frac{(n-1)^2}N, \tag{120.4}$$

where $\omega$ is frequency of light, $n$ is refractive index of the gas, and $N$ is number density of gas particles. Here the $n-1$ term is the approximately simplified $(n^2-1)/2$, the latter being the form you'll more often encounter in other discussions.

Thus, you can just use refractivity of air, which changes linearly with density, so you can insert your profile in it as follows (using $z$ here to avoid confusion with $h$ in $(120.4)$):

$$n(z)=(n_0-1) \exp(-z/H)+1.$$

Note though, that the trivial way of integration of inscattered light over altitudes will only give you first-order scattering, which in clear air will underestimate zenith brightness by about 30%.

To do actual calculation to multiple scattering orders you can use E. Bruneton's Precomputed Atmospheric Scattering paper and the updated implementation of the demo that I was using when working on my (still WIP) CalcMySky project.

Another option is to use Monte-Carlo simulation with e.g. libradtran's MYSTIC simulator. This one is not limited to any number of scattering orders, and generally is the production-ready package, unlike the demo mentioned above.

  • $\begingroup$ Thanks for your answer, it will give me plenty to read. Can I ask what is $h$? It seems to have units of inverse distance. $\endgroup$
    – uhoh
    Commented Jul 22, 2021 at 9:35
  • $\begingroup$ @uhoh as said above the equation, it's the coefficient of extinction, as in Lambert-Beer law: $I=I_0\exp(-h\ell)$. $\endgroup$
    – Ruslan
    Commented Jul 22, 2021 at 9:42
  • $\begingroup$ Okay it's the "h" that threw me off, the units are 1/height, got it. $\endgroup$
    – uhoh
    Commented Jul 22, 2021 at 9:50
  • $\begingroup$ @uhoh inverse distance along the ray, not necessarily to the zenith. $\endgroup$
    – Ruslan
    Commented Jul 22, 2021 at 9:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.