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Rayleigh scattering is elastic scattering of electromagnetic radiation by particles much smaller than the incident wavelength. In the stationary particle approximation, the scattered electromagnetic radiation keeps the incident frequency: its direction changes, rather than its color. The incident electric field induces an oscillating electric dipole moment, which produces electric dipole radiation. This is a useful example of how a microscopic response can explain a familiar large-scale sight, why the sky is blue. ^[ref].

The smallness condition for Rayleigh scattering by a small sphere of radius $a$ is $ka\ll 1$, where the wavenumber is $k=2\pi /\lambda$ and $\lambda$ is the surrounding medium's wavelength. For ordinary materials with modest refractive index, the incident electric field then varies little across the particle, making the electric-dipole approximation for radiation useful. For very large refractive index, also check the internal size $|m|ka\ll 1$ before applying the simple electric dipole model. Once these conditions fail, use Mie scattering for a sphere instead of extrapolating the Rayleigh scattering wavelength dependence.

Historically, the 1871 paper ^[ref] connected the color and polarization of an electromagnetic wave of scattered skylight with particles smaller than the wavelength. Its dimensional analysis anticipates scattered light intensity proportional to particle volume squared and to ${\lambda }^{-4}$. The original mechanical picture is historical; the modern explanation uses induced electric dipole radiation.

For a fixed sphere and fixed refractive index, both the Rayleigh scattering cross-section and scattering efficiency are proportional to ${\lambda }^{-4}$. Halving wavelength multiplies either by $16$, as long as the shorter wavelength still satisfies $ka\ll 1$. If optical dispersion changes $m$, the material factor changes too; the pure fourth-power comparison then needs correction.

For equal incident light intensities at blue wavelength $450\mathrm{nm}$ and red wavelength $650\mathrm{nm}$, the same toy sphere givesBlue light therefore produces about $4.35$ times the scattered power. This compares the particle's response at two wavelengths; it does not assert that an actual sky has this ratio of observed blue and red light intensities.

In a clear terrestrial atmosphere, Rayleigh scattering by air molecules redirects some sunlight toward an observer looking away from the Sun. Shorter wavelengths are preferentially scattered, enriching the scattered light in blue compared with the incoming sunlight. Along the longer atmospheric path to the low Sun, more short-wavelength light is removed from the direct beam, helping produce red sunsets. ^[ref].

Violet light has an even shorter wavelength than blue light, so the Rayleigh scattering wavelength dependence alone cannot predict the perceived sky color. The incoming solar optical spectrum and the eye's differing response to wavelengths both matter; human vision is less sensitive to violet than to blue. ^[ref]. Multiple scattering and other atmospheric particles also affect the observed sky. Cloud droplets are generally too large for this simple Rayleigh scattering model: Mie scattering describes spherical droplets, with much weaker visible wavelength selectivity, while repeated scattering helps make clouds appear white. ^[ref].