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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.

In Rayleigh scattering, a small bound system develops an induced electric dipole moment $\mathbf{p}(t)=\mathrm{Re}[\alpha (\omega ){\mathbf{E}}_0{e}^{-i\omega t}]$. Well below its binding resonances, the electric polarizability is approximately its static value $\alpha (0)$, independent of frequency. The mean radiated power and incident light intensity areDividing gives the total scattering cross sectionThis Rayleigh scattering wavelength dependence comes from the two time derivatives of an oscillating electric dipole moment, whose radiated field is proportional to ${\omega }^2{p}_0$. It requires the same scatterer with approximately constant low-frequency electric polarizability; resonances and strong optical dispersion invalidate the simple fourth-power law.

Expand the cloud's potential energy about a stable equilibrium: $U(z)=U(0)+U^{\prime }(0)z+U^{\prime \prime }(0)z^2/2+\cdots$. Equilibrium gives $U^{\prime }(0)=0$, stability gives $U^{\prime \prime }(0)=k>0$, and reflection symmetry also removes odd terms. For small displacements, $U-U(0)\simeq k{z}^2/2$ and $F=-dU/dz\simeq -kz$.

Including the electric and damping forces, Newton's second law is $m\ddot{z}=qE(t)-b\dot{z}-kz$. Dividing by $m$ gives the Lorentz oscillator modelThe specified convention uses $2\gamma$ as the damping coefficient per unit mass.

For the steady response, represent the field by $E_0{e}^{i\omega t}$ and the displacement by $\tilde{z}{e}^{i\omega t}$, taking real parts for physical quantities. Substitution givesDefine the nonnegative amplitude $z_0=|\tilde{z}|$ and $\delta =atan2(2\gamma \omega ,{\omega }_0^2-{\omega }^2)$. ThenFor positive $q$, $\varphi =-\delta$. For electrons $q<0$, the phase of a negatively charged driven oscillator is $\varphi =\pi -\delta$: the low-frequency displacement is opposite to the field, rather than in phase with it. A signed amplitude could absorb this extra $\pi$, but the phase convention must then be stated. The induced dipole $p=qz$ has phase $-\delta$ for either charge sign. Transient free oscillations have been omitted because the question asks for steady state.

For the electron cloud, $q/m=-e/m_e$ andThe visible range corresponds to $\omega =2\pi c/\lambda$ between $2.35\times 10^{15}$ and $4.71\times 10^{15}\mathrm{rad}{\mathrm{s}}^{-1}$, below this resonance. The resonant wavelength is about $180\mathrm{nm}$, in the ultraviolet.

The acceleration amplitude is ${\omega }^2{z}_0$. At fixed incident field amplitude, the scattered intensity is therefore proportional toWell below resonance and outside strong damping corrections, $z_0$ is nearly frequency-independent and this becomes the Rayleigh scattering wavelength dependence $I_{\mathrm{sc}}\propto {\omega }^4\propto {\lambda }^{-4}$. The shortest visible wavelengths are consequently scattered much more strongly into directions away from the direct solar beam than the longest ones, producing a blue sky. The simple fourth-power approximation gives a factor 16 between 400 and 800 nm at equal incident intensity. Retaining the undamped denominator with the given resonance raises that factor to about 22.7, since the upper end is not extremely far below resonance. The exact ratio needs the unspecified damping and incident solar spectrum; the stronger scattering of the blue end is the robust conclusion here.

The model describes frequency-dependent scattering, not the detailed perceived color after solar-spectrum weighting and human visual response. It also explains why treating the electrons as free at visible frequencies would miss the bound-charge fourth-power limit.

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].