Blackbody radiation and Planck's quantum hypothesis

Inside a cubical cavity of side L with conducting walls, each field component obeys the wave equation ∂²E/∂x² + ∂²E/∂y² + ∂²E/∂z² = (1/c²) ∂²E/∂t², and the walls allow only standing waves with wave vector k = (π/L)(nₓ, n_y, n_z) for whole numbers n. Since ω = ck, a mode's frequency is ν = (c/2L)√(nₓ² + n_y² + n_z²). Counting the modes below ν is counting lattice points inside one octant of a sphere of radius R = 2Lν/c: N(ν) = 2 × (1/8)(4π/3)R³ = 8πVν³/(3c³), where the 2 counts the two polarisations of a transverse wave. Differentiate and divide by V to get g(ν) = 8πν²/c³ modes per unit volume per unit frequency. The count is enormous: a 1 cm³ cavity holds about 2.3 × 10¹¹ modes in every 1 THz band near 500 THz, so a smooth density is safe, and the cavity's shape drops out once L ≫ λ.

Each mode is a harmonic oscillator: its energy passes back and forth between electric and magnetic fields as a spring's does between kinetic and potential. Classical statistical mechanics gives any oscillator in equilibrium a mean energy kT, half from each quadratic term, whatever its frequency. Multiply by the mode density and you have the Rayleigh–Jeans law, u_RJ = 8πν²kT/c³. It fits measured spectra at long wavelengths and contains no h at all. Its integral over frequency, though, grows as ν³ without bound. Stopping at 750 THz, the violet end of the visible, a cavity at 300 K would already hold 0.54 J m⁻³, about 9 × 10⁴ times the 6.1 × 10⁻⁶ J m⁻³ actually present, and every higher band adds more. A classical cavity could never reach equilibrium: its walls would keep pouring energy into ever-shorter waves. This is the ultraviolet catastrophe, and it is a failure of kT per mode, not of the counting.

Planck kept the mode count and changed one assumption: a mode of frequency ν can hold only Eₙ = nhν, with n = 0, 1, 2, … and h = 6.626 × 10⁻³⁴ J s. In equilibrium level n has Boltzmann weight e^(−nx), with x = hν/kT. The geometric series Z = Σ e^(−nx) = 1/(1 − e^(−x)) normalises the probabilities to Pₙ = (1 − e^(−x)) e^(−nx), and the mean energy is ⟨E⟩ = Σ nhν Pₙ, or more compactly −∂(ln Z)/∂β with β = 1/kT. Either route gives ⟨E⟩ = hν/(eˣ − 1). Now put numbers in. At 300 K a mode at 30 THz has x = 4.80: it sits in its ground state 99.2% of the time, its mean occupation is 0.0083 steps, and ⟨E⟩ = 0.040kT. Reaching even the first level costs 4.8kT, which the Boltzmann factor makes exponentially rare, so high-frequency modes freeze out instead of each claiming kT.

The correspondence check is an expansion. For x = hν/kT ≪ 1, eˣ − 1 = x + x²/2 + x³/6 + …, so ⟨E⟩ = kT(1 − x/2 + x²/12 − …): at x = 0.01 a mode holds 0.995kT, and at x = 0.1 still 0.951kT. When a step is a small fraction of kT the ladder cannot be told from a continuum, and Planck's law collapses onto Rayleigh–Jeans; letting h → 0 at fixed ν does the same formally. The crossover sits at ν = kT/h, which is 2.08 × 10¹⁰ Hz for every kelvin of temperature: 6.25 THz, a wavelength of 48 μm, at 300 K. A 2.45 GHz microwave mode at room temperature has x = 3.9 × 10⁻⁴ and is thoroughly classical; a 540 THz green-light mode has x ≈ 86 and is frozen solid. What separates the regimes is this ratio, not whether the light is ultraviolet and not the size of the cavity. In the opposite limit, x ≫ 1, ⟨E⟩ ≈ hν e^(−x), which is Wien's earlier approximation.

Mode density times mean energy is Planck's distribution, u(ν, T) = (8πhν³/c³)/(e^(hν/kT) − 1), in J m⁻³ Hz⁻¹. Written in x it becomes (8πk³T³/(h²c³)) x³/(eˣ − 1), so every temperature shares one shape and the peak sits at a fixed x. Setting d/dx[x³/(eˣ − 1)] = 0 gives 3(1 − e^(−x)) = x, which has no closed-form root; iterating x → 3(1 − e^(−x)) from x = 3 gives 2.851, 2.827, 2.822 and settles at 2.821. So hνₚₑₐₖ = 2.821kT, or νₚₑₐₖ = (5.879 × 10¹⁰ Hz K⁻¹) × T: double the temperature and the peak moves to twice the frequency while its height grows eightfold, as T³. Per unit wavelength the same algebra gives 5(1 − e^(−x)) = x, with root 4.965, and λₘₐₓ T = hc/(4.965k) = 2.898 × 10⁻³ m K, the Wien displacement law in its familiar form.

The total energy density is the area under the curve. Substituting ν = kTx/h, U/V = (8πk⁴T⁴/(h³c³)) ∫ from 0 to ∞ of x³/(eˣ − 1) dx, and the dimensionless integral is π⁴/15 = 6.494. So U/V = aT⁴ with a = 8π⁵k⁴/(15h³c³) = 7.566 × 10⁻¹⁶ J m⁻³ K⁻⁴. Radiation leaking from a small hole carries flux (c/4)(U/V), the 1/4 from averaging the isotropic flow over directions, so the exitance is M = σT⁴ with σ = 5.670 × 10⁻⁸ W m⁻² K⁻⁴. Dimensions: k⁴T⁴/(h³c²) is J⁴ ÷ (J³ s³ × m² s⁻²) = J s⁻¹ m⁻², a power per area. Limit: σ ∝ h⁻³, so h → 0 sends the total to infinity and the catastrophe returns. Data: the Sun's photosphere at 5772 K gives M = 6.29 × 10⁷ W m⁻², and multiplying by 4πR² with R = 6.96 × 10⁸ m gives 3.83 × 10²⁶ W, the measured solar luminosity.