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Molecules in clean air are much smaller than visible wavelengths. Their induced electric dipole moments undergo Rayleigh scattering. Away from molecular resonances, the induced dipole amplitude is approximately $p=\alpha E$, with only weak wavelength variation in the polarizability $\alpha$. The radiation-zone electric field is proportional to $\ddot{p}$, so the scattered intensity satisfiesThis is the Rayleigh scattering wavelength dependence. For equal incident intensities, $450\mathrm{nm}$ light scatters about $(650/450{)}^4=4.35$ times as strongly as $650\mathrm{nm}$ light. Thus light scattered out of the direct solar beam and into our line of sight is enriched in blue wavelengths.

The dipole mechanism also determines the Rayleigh scattering angular pattern. For unpolarized light, its angular factor is $1+{\cos }^2\vartheta$; near $90^{\circ }$ scattering, the scattered light has strong linear polarization. The sky's perceived colour depends additionally on the solar spectrum, atmospheric transmission and the eye's spectral response, so the ${\lambda }^{-4}$ law alone does not predict a violet sky. Over a long near-horizontal path the direct beam loses more short-wavelength light, explaining red sunsets. Large aerosol particles instead produce Mie scattering, which is less strongly wavelength-selective and can make haze look whitish.

For incident unpolarized light and the isotropic electric dipole response of Rayleigh scattering by a small sphere, let $\theta$ be the angle between the incident beam's propagation direction and the outgoing direction. The differential scattering cross-section and its relative angular pattern arewhere $d\Omega$ is an element of solid angle. Averaging over the two transverse directions of incident polarization of an electromagnetic wave gives this pattern. Forward scattering at $0^{\circ }$ and backward scattering at $180^{\circ }$ are equally strong: $F=1$. Sideways scattering at $90^{\circ }$ is weakest, but nonzero: $F=1/2$. ^[ref].

The normalization is consistent becauseThus integrating $d\sigma /d\Omega$ gives the total Rayleigh scattering cross-section $\sigma$. The plotted $F$ has maximum one; it is not itself normalized to unit solid angle integral.

For a single electric dipole driven by incident linear polarization, the angular distribution of electric dipole radiation instead has a ${\sin }^2\alpha$ factor, where $\alpha$ is measured from the electric dipole moment direction, parallel to the incident electric field. That direction is perpendicular to the incident beam. Confusing $\alpha$ with the beam angle $\theta$ would incorrectly predict zero forward scattering. Averaging the dipole patterns over incident unpolarized light yields the displayed $F(\theta )$.