Why the sky is blue and sunsets are red

The daytime sky is blue and the setting Sun is red, and one law explains both. Air throws short waves of light sideways far more readily than long ones. Look away from the Sun and you see the light that was thrown aside, which is mostly blue. Look at the Sun through a long stretch of air and you see what was left behind, which is mostly red.

What is being claimed

When light meets particles much smaller than its wavelength, some of it is scattered off in all directions. How much depends steeply on the colour. In 1871 Lord Rayleigh put the rule this way: the ratio of the amplitudes of the scattered and incident light “varies inversely as the square of the wave-length, and the intensity of the lights themselves as the inverse fourth power.” So scattered intensity goes as 1/λ⁴, where λ is the wavelength. Our arithmetic: blue light at 450 nm has a wavelength about two-thirds that of red at 700 nm, and (700/450)⁴ is about 5.9, so the blue is scattered roughly six times as strongly.

Why it is worth knowing

Two popular answers are wrong. The sky is not blue because it reflects the sea: the colour is made in the air itself, by scattering. Nor does it need water vapour or dust. Rayleigh first supposed the scatterers were fine particles of foreign matter, as in the experiments of Tyndall with “precipitated clouds” that he credits at the start of his paper. In 1899 he returned to the question and concluded: “even in the absence of foreign particles we should still have a blue sky.” The molecules of the air are enough. Large particles do not follow the law at all, which is why clouds are white: Rayleigh's condition is particles “very small compared with any of the wave-lengths”.

The fourth power from units alone

Rayleigh got the law from dimensions, before doing any detailed theory. The ratio of scattered to incident amplitude is a pure number. It can depend only on the particle's volume, the distance to the observer, the wavelength, the speed of light and two densities. The densities carry the only unit of mass, so they can enter only as a ratio, a plain number. The speed of light carries the only unit of time, so it cannot enter at all. What remains is volume, distance and wavelength. The scattered wave must grow with the particle's volume and fall off with distance, and a volume divided by a distance is a length squared. To make a pure number, divide by the wavelength squared. Intensity is amplitude squared, which gives the fourth power.

He also checked it against the sky. He compared blue light from near the zenith with sunlight passed through white paper, in about thirty comparisons from the Fraunhofer line C in the red to beyond F in the blue. With both sets scaled to 25 at C, theory gave 25, 40, 63 and 80 at the lines C, D, b₁ and F; he observed 25, 41, 71 and 90. The sky, he wrote, was “even bluer than theory makes it”, and he suspected slightly yellow paper or yellowed sunlight.

The same law reddens the Sun

Light scattered sideways has left the direct beam. Rayleigh wrote the loss as dI = −kIλ⁻⁴dx: over each short stretch of path x, a share of the light proportional to that stretch goes, and that fraction grows as 1/λ⁴. It is “a law altogether similar to that of absorption, and showing how the light tends to become yellow and finally red as the thickness of the medium increases.” He drew white light after paths in the ratio 1, 2, 4, 8, 16 and 32, and remarked “how little of the violet light remains when the red is still in nearly its original force.” At sunrise and sunset the sunlight crosses far more air than at noon, so the same removal that fills the sky with blue strips the blue out of the Sun.

He gave a number. At the line A in the far red and a line R in the ultraviolet the wavelengths are “7617 and 3108” (in tenth-metres, the unit now called the Ångström; our conversion is 761.7 and 310.8 nm), and “the ratio of the fourth powers is about 36 : 1”. So whatever fraction of A gets through, R gets that fraction to the 36th power. His example: if 0.9 of A gets through, “only ·018 of R” does. Our arithmetic gives 0.9³⁶ ≈ 0.023, close to but not quite his figure; either way, almost none.

Air alone is enough

In 1899 Rayleigh worked out how clear air made only of molecules would be. Using Maxwell's estimate of 19 × 10¹⁸ molecules per cubic centimetre, he found that light at standard pressure would be dimmed by a factor of e, about 2.7, over 83 kilometres. Starlight measured by Bouguer and others loses about a fifth of its light crossing the whole atmosphere, which is the same as 8.3 km of air at standard pressure; at that rate real air lets through only about a third as much over 83 km as molecules alone would, since it also carries suspended matter. So the molecules “would suffice to give us a blue sky, not so very greatly darker than that actually enjoyed.” Turned around, the argument sets a lower limit on how many molecules there are, about 7 × 10¹⁸ per cubic centimetre. How clear the air is sets a floor on how many molecules there are.

Why blue and not violet, if violet scatters more? These sources do not settle it. The usual answer, which is general knowledge and not from them, is that sunlight carries less violet than blue and our eyes are less sensitive to it.

In short

Air molecules scatter light as 1/λ⁴, so blue is thrown aside about six times as readily as red. The thrown-aside light is the blue sky. The light that stays in the beam loses its short waves first, and over the long path of sunset what reaches you is yellow, then red.