Why is the sky blue, and sunsets golden?

The color of our sky is caused by the interplay of blue-light-scattering by air molecules, and white-light-scattering by water drops and dust...

Blue wavelengths are generally scattered down toward the earth. This makes the sky appear blue wherever it is daytime (and the sun is high in the sky). At sunset, however, the opposite occurs.



	Glowing beam of the "bat signal," as imagined in a 1950's comic book cover. In reality, rays of light are invisible in the vacuum of space. In earth's atmosphere, however, the rays of our sun are faintly visible, spanning across our planet... appearing as a blue sky.

		Leonardo da Vinci had observed that a very fine water spray produced light scattering, but for many centuries only confusing and misleading ideas abounded.

		The English experimentalist John Tyndall (1820-1893) demonstrated that the scattering from particles small compared to the wavelength of light depends on the wavelength, with blue being much more strongly scattered than red.

Why are sunsets red?

Far on the horizon, to the West, it is daylight. The distant clouds and dust create a diffuse white glow (Mie scattering) with all wavelengths of light. As the daylight travels east toward your eye, the blue hues are scattered toward to ground (Rayleigh scattering). Sun rays have to traverse long distances through the dense parts of the atmosphere until they reach the eye of the observer with the sun close to or even below the horizon. The traversed layer of the atmosphere is more than 30 times longer compared to the sun standing directly above the observer.The violets and blues are lost (they make the sky blue to the West of you). The light you see is missing the violets and blues... leaving you with various shades of yellows, reds, aven purples.

Small molecules make the sky blue

Although gases appear totally clear, even the purest substances, are found to scatter light when carefully examined.

It remained for english physicist John William Strutt Lord Rayleigh (1842-1919) to explain that scattering particles were not necessary, since even the purest of substances have fluctuations in their refractive index, which can scatter light. Every gas particle (mostly nitrogen and oxygen molecules) "scatters" light. When a particle is hit by a light wave, it creates a new wave which propagates in all directions. He also showed that the intensity of the scattered light I_S is related to that of the incident light I₀ by the inverse fourth power of the wavelength l:

If we take the intensity of scattered violet light at the 400nm limit of visibility to be 100, then red light at 700 nm is scattered only at an intensity of 10.7. The terms "Rayleigh scattering" and "Tyndall blue" are often applied to the scattered blue. A dark background, such as the dark of outer space, is required for an intense blue to be perceived.

In a gas, liquid, or glass the atoms and molecules are evenly distributed on a macroscopic scale, yet at the atomic level there is considerable nonrandomness. As one example, individual molecules, as well as small clusters in a gas or a liquid coming together in collision for a brief instant before dispersing again, will act as light scattering particles much as do particles of dust. In a glass there will be similar density and refi-active-index variations, both from the imperfect mixing of the various ingredients as well as from the frozen-in liquid fluctuations. Even in what might be thought of as a perfectly ordered single crystal, there usually will be a variety of point defects (impurity atoms, vacancies, and clusters of these) and line and plane defects (dislocations, low-angle grain boundaries, and the like) as well as density fluctuations from the thermal vibrations of the atoms or molecules, all of which scatter light.

Rayleigh scattering involves single light-emitting regions that are very small compared to the wavelength of light; they absorb light photons and re-ernit them as single wavelets. Since light is a transverse oscillation, the scattered light is polarized as indicated in Fig. 32; exactly perpendicular to the beam the scattered light is completely polarized in a direction perpendicular to the incident beam, while in other directions, such as at the angle 0 shown in this figure, there is an additional component of the polarization parallel to the incident-beam direction of cos' 0. The combination (I + cos' 0) gives the total light-scattering intensity distribution.