Why is the daytime sky blue and the setting sun red — and can that colour be derived from scratch, without ever assuming the answer?
▶ Launch the interactive simulationModel an electron bound in an air molecule (natural frequency ω₀ in the deep UV) as a driven damped oscillator ẍ+γẋ+ω₀²x=(e/mₑ)E₀cos(ωt); its steady amplitude x₀(ω)=(eE₀/mₑ)/√((ω₀²−ω²)²+(γω)²) contains NO exponent. The wiggling charge radiates by the Larmor formula ⟨P⟩=e²ω⁴x₀²/(12πε₀c³). Generate the scattered-power curve ⟨P⟩(ω), measure its local log–log slope by finite difference, and extrapolate to ω→0 to read off the exponent p in ⟨P⟩∝λ⁻ᵖ — a recovery in which the number 4 never appears. Repeat across binding stiffness ω₀, damping γ and intensity E₀ (universality), and unbind the electron (ω₀=0, Thomson) as a falsification control. An RK4 lock-in of the actual equation of motion confirms the generator's amplitude. ?world=rayleigh.
Rayleigh's law ⟨P⟩∝ω⁴∝λ⁻⁴ — the exponent p=4.000 recovered (to ~1e-4) as the ω→0 limit of the scattered-power slope, and shown to be UNIVERSAL: independent of the binding stiffness ω₀, the damping γ and the intensity E₀, exactly as the law demands. Blue (450 nm) therefore scatters (700/450)⁴≈5.9× more than red — the sky's blue, and the reddened setting sun whose long air path scatters the blue away. The result is decisive because binding is what makes it happen: unbinding the electron (ω₀=0, Thomson scattering) collapses the slope to 0 — a free electron scatters every colour equally (grey), which is why a free-electron plasma does not blue the sky. The same machinery hands back a second textbook constant, the wavelength-independent Thomson cross-section σ_T=(8π/3)r_e²=6.6525e-29 m² (and r_e=2.818e-15 m), to ~1e-11. Distinct from ?world=compton (the inelastic photon–electron energy shift): Rayleigh is elastic dipole re-radiation whose wavelength SCALING makes colour. The scattering branch of the optics arc, after Snell, Brewster and the rainbow.