Bosons at Work: Photons, Phonons, and Blackbody Radiation

The photon gas: a boson gas with μ = 0

Electromagnetic radiation in a hot cavity is a gas of photons — spin-1 bosons. But photons are freely created and destroyed by the cavity walls, so their number is not conserved. The system minimizes its free energy with respect to N, which forces the chemical potential to zero: \mu = 0.

To count modes we need the photon density of states. Photons are massless with \varepsilon = \hbar\omega = \hbar c k, and each mode has two polarizations. Counting standing waves in a box of volume V gives:

g(\omega)\,d\omega = \frac{V\,\omega^{2}}{\pi^{2} c^{3}}\,d\omega

The number of electromagnetic modes between ω and ω + dω, including both polarizations. It grows as ω² — there are far more high-frequency modes than low-frequency ones.

Planck's law and the end of the ultraviolet catastrophe

Multiply the density of states by the mean energy per mode — the Bose occupation \bar n(\omega) with \mu = 0, times \hbar\omega — and you get the spectral energy density. This is Planck's radiation law.

u(\omega)\,d\omega = \frac{\hbar\,\omega^{3}}{\pi^{2} c^{3}}\,\frac{1}{e^{\hbar\omega / k_B T} - 1}\,d\omega

Planck's law: the energy density of blackbody radiation per unit angular frequency. The Bose factor cuts off the ω³ growth exponentially at high frequency.

The two limits tell the historical story. At low frequency (\hbar\omega \ll k_B T) the Bose factor gives k_B T per mode and we recover the classical Rayleigh-Jeans law — which, taken to all frequencies, would diverge (the ultraviolet catastrophe). At high frequency the exponential wins and the spectrum falls off (Wien's law). Planck's \hbar tames the catastrophe.

\hbar\omega \ll k_B T:\;\; u \to \frac{\omega^{2} k_B T}{\pi^{2} c^{3}} \;\text{(Rayleigh-Jeans)}; \qquad \hbar\omega \gg k_B T:\;\; u \to \frac{\hbar\omega^{3}}{\pi^{2} c^{3}}\,e^{-\hbar\omega / k_B T} \;\text{(Wien)}

The classical (Rayleigh-Jeans) and quantum-suppressed (Wien) limits of Planck's law. Only the full Bose expression bridges them without disaster.

Worked example: Stefan-Boltzmann and Wien

Two celebrated laws fall straight out of Planck's spectrum. Let us derive the total radiated energy density (Stefan-Boltzmann) explicitly.

Integrate the spectrum. The total energy density is \frac{U}{V} = \int_0^\infty u(\omega)\,d\omega = \frac{\hbar}{\pi^2 c^3}\int_0^\infty \frac{\omega^3\,d\omega}{e^{\hbar\omega/k_B T}-1}.
Substitute x = \hbar\omega/k_B T. The temperature factors pull out as T^4, leaving the pure number \int_0^\infty \frac{x^3\,dx}{e^x - 1} = \frac{\pi^4}{15} (a standard integral, tied to the gamma and Riemann-zeta functions).
Collect. The result is U/V = a\,T^4 with a = \pi^2 k_B^4 / (15\hbar^3 c^3). The power radiated per unit area is j = \sigma T^4 with \sigma = ac/4 \approx 5.67\times10^{-8}\ \text{W m}^{-2}\text{K}^{-4}.

\frac{U}{V} = a\,T^{4}, \quad a = \frac{\pi^{2} k_B^{4}}{15\,\hbar^{3} c^{3}}; \qquad \lambda_{\max}\,T = b \approx 2.898\times10^{-3}\ \text{m·K}

Left: the Stefan-Boltzmann law — total radiated energy grows as T⁴. Right: Wien's displacement law — the peak wavelength shifts inversely with temperature, which is why hotter objects glow bluer.

Phonons and the heat capacity of solids

A crystal's atoms vibrate collectively in normal modes; quantize each mode and its excitations are phonons — again bosons with \mu = 0. The mathematics mirrors the photon gas, with one crucial difference: a crystal of N atoms has a finite number of modes ($3N$), so there is a maximum frequency. The Debye model captures this with a cutoff.

C_V \xrightarrow{\;T \ll \Theta_D\;} \frac{12\pi^{4}}{5} N k_B \!\left(\frac{T}{\Theta_D}\right)^{3}, \qquad C_V \xrightarrow{\;T \gg \Theta_D\;} 3 N k_B

The Debye heat capacity: a T³ law at low temperature (the celebrated Debye result) rising to the classical Dulong-Petit value 3Nk_B at high temperature. Θ_D is the Debye temperature.