--- title: The Photon Gas and Planck's Radiation Law module: Bosonic Systems moduleNumber: 8 lessonNumber: 2 order: 802 summary: > Electromagnetic radiation in equilibrium with cavity walls is a gas of non-conserved bosons, and non-conservation forces the chemical potential to zero. Counting standing-wave modes with two polarizations and weighting each by the Bose occupation gives the Planck spectral energy density. Its low-frequency tail reproduces the classical Rayleigh-Jeans law and the ultraviolet catastrophe; the Bose factor cuts the divergence off at high frequency and the peak obeys Wien's displacement law. topics: [Bosonic Systems] sources: - book: Schroeder ref: "Ch. 7 — Quantum Statistics; §7.4 Blackbody Radiation" - book: Reif ref: "Ch. 9 — Quantum Statistics of Ideal Gases; §9.13–9.15" - book: Pathria & Beale ref: "Ch. 7 — Ideal Bose Systems; §7.2–7.3" - book: Kardar (Statistical Physics of Particles) ref: "Ch. 7 — Quantum Statistical Mechanics; §7.5" draft: false --- Thermal radiation in a cavity held at temperature $T$ is a gas of photons. The walls emit and absorb photons continuously, so their number is not fixed by any conservation law but set by the condition of thermal equilibrium. The general [framework for ideal quantum gases](/statistical-mechanics/quantum-statistics/ideal-quantum-gases-general-framework) converts a sum over occupied single-particle modes into an integral weighted by the Bose-Einstein occupation; for photons that framework specializes cleanly because two features fix the inputs at once — the dispersion is linear, $\varepsilon = \hbar\omega = c\hbar k$, and the chemical potential is zero. The output is the Planck law, the first quantitative success of the quantum hypothesis and the spectrum of every blackbody from a furnace to the cosmic microwave background. ## Non-conservation and the vanishing chemical potential Photons are spin-$1$ massless bosons. A cavity wall at temperature $T$ absorbs a photon into an excited atomic state and re-emits one, and nothing constrains the total count $N$: the equilibrium value of $N$ is whatever minimizes the free energy. For a system in contact with a heat bath at fixed $T$ and $V$, the equilibrium state minimizes the Helmholtz free energy $F(T,V,N)$ over every variable free to adjust. If $N$ is free, the minimum condition is $$ \left(\frac{\partial F}{\partial N}\right)_{T,V} = 0. $$ The left-hand side is the definition of the chemical potential, $\mu = (\partial F/\partial N)_{T,V}$, so equilibrium of a non-conserved species forces $$ \mu = 0. $$ Every occupation number then reduces to the Bose-Einstein form evaluated at $\mu = 0$, $$ \langle n(\varepsilon)\rangle = \frac{1}{e^{\beta\varepsilon} - 1}, \qquad \beta = \frac{1}{k_B T}, $$ with no fugacity to determine.[^photon-mu] The same argument applies to any gas of non-conserved quasiparticles, and the next lesson applies it verbatim to [phonons](/statistical-mechanics/bose-systems/phonons-and-the-debye-model). > **Definition (Photon gas).** The equilibrium electromagnetic field in a cavity > of volume $V$ at temperature $T$, treated as an ideal gas of photons: bosons of > energy $\varepsilon = \hbar\omega$, momentum $\vec p = \hbar\vec k$ with > $\omega = c\lvert\vec k\rvert$, two transverse polarizations, and chemical > potential $\mu = 0$ because photon number is not conserved. ## Counting the cavity modes The allowed field configurations in a cubical cavity of side $L$ are standing waves that vanish on the walls. Each Cartesian component of the wavevector is quantized, $$ k_i = \frac{\pi n_i}{L}, \qquad n_i = 1, 2, 3, \dots, $$ so the allowed modes are the points of a cubic lattice in the positive octant of $\vec k$-space, spaced $\pi/L$ apart. A single standing wave carries a definite number of half-wavelengths along each axis. $$ % caption: A standing electromagnetic mode of the cavity vanishes at both walls and fits a whole number of half-wavelengths across the width L; higher modes pack more nodes into the same box. \begin{tikzpicture}[scale=1.0, font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \draw[black, thick] (0,0) -- (0,2.4); \draw[black, thick] (6,0) -- (6,2.4); \draw[black, dashed] (0,1.2) -- (6,1.2); % n=1 half wave \draw[acc, thick, smooth] plot[domain=0:6, samples=80] (\x,{1.2+0.85*sin(180*\x/6)}); % n=3 three half waves \draw[black, thick, densely dashed, smooth] plot[domain=0:6, samples=120] (\x,{1.2+0.55*sin(3*180*\x/6)}); \node[text=acc, anchor=south] at (3,2.1) {lowest mode}; \node[text=black, anchor=north] at (1.0,0.5) {third mode}; \node[black, anchor=north] at (3,-0.05) {cavity width L}; \end{tikzpicture} $$ The number of modes with wavevector magnitude up to $k$ is the volume of the octant of a sphere of radius $n = kL/\pi$ in the integer lattice, one lattice point per unit cell, multiplied by two for the polarizations: $$ \mathcal{N}(k) = 2 \cdot \frac{1}{8} \cdot \frac{4}{3}\pi\left(\frac{kL}{\pi}\right)^3 = \frac{V k^3}{3\pi^2}. $$ Differentiating gives the mode density in $k$, and converting through $\omega = ck$ and $\omega = 2\pi f$ gives the density in frequency, $$ \frac{\d\mathcal{N}}{\d k} = \frac{V k^2}{\pi^2}, \qquad \frac{\d\mathcal{N}}{\d f} = \frac{8\pi V}{c^3}\, f^2. $$ The factor $f^2$ is the geometric growth of the number of modes with frequency, the same factor that drives the classical divergence below. $$ % caption: Allowed wavevectors are lattice points in the positive octant of k-space; the modes up to magnitude k fill one-eighth of a sphere, and their count grows as k cubed. \begin{tikzpicture}[scale=1.0, font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \draw[->, black] (0,0) -- (3.4,0) node[right] {$n_x$}; \draw[->, black] (0,0) -- (0,3.4) node[above] {$n_y$}; \draw[->, black] (0,0) -- (-1.7,-1.3) node[left] {$n_z$}; % lattice points \foreach \i in {0,1,2,3} \foreach \j in {0,1,2,3} \fill[acc] (\i*0.7,\j*0.7) circle (1.3pt); % quarter arc (octant boundary in plane) \draw[black, thick] (2.1,0) arc (0:90:2.1); \draw[black, dashed] (0,0) -- (2.1,0); \draw[black, dashed] (0,0) -- (0,2.1); \node[text=black, anchor=south west] at (1.5,1.5) {radius n}; \node[text=acc, anchor=west] at (2.4,2.6) {allowed modes}; \end{tikzpicture} $$ ## The Planck spectral density Each mode of frequency $f$ is an independent harmonic oscillator whose quantized excitations are photons. The mean energy stored in a mode, measured from its ground state, is the occupation times the quantum $hf$, $$ \bar\varepsilon(f) = \langle n\rangle\, hf = \frac{hf}{e^{hf/k_B T} - 1}. $$ The spectral energy density — energy per unit volume per unit frequency — is the mode density per volume times the mean energy per mode: $$ u(f) = \frac{1}{V}\,\frac{\d\mathcal{N}}{\d f}\,\bar\varepsilon(f) = \frac{8\pi f^2}{c^3}\,\frac{hf}{e^{hf/k_B T} - 1}. $$ > **Theorem (Planck radiation law).** The energy per unit volume of cavity > radiation with frequency between $f$ and $f + \d f$ is > $$ > u(f)\,\d f = \frac{8\pi h}{c^3}\,\frac{f^3\,\d f}{e^{hf/k_B T} - 1}. > $$ > In terms of wavelength, using $u(f)\,\d f = -u(\lambda)\,\d\lambda$ and > $f = c/\lambda$, > $$ > u(\lambda)\,\d\lambda = \frac{8\pi hc}{\lambda^5}\, > \frac{\d\lambda}{e^{hc/\lambda k_B T} - 1}. > $$ The two forms describe the same radiation field but peak at different arguments, because the Jacobian $\d f/\d\lambda = -c/\lambda^2$ reshapes the curve.[^planck] ## Classical limits and the ultraviolet catastrophe The mean energy per mode carries the whole quantum content. Expanding it in the two limits recovers the classical law and its failure. - **Rayleigh-Jeans limit ($hf \ll k_B T$).** The denominator linearizes, $e^{hf/k_B T} - 1 \approx hf/k_B T$, so $\bar\varepsilon \to k_B T$ and $$ u(f) \to \frac{8\pi f^2}{c^3}\,k_B T. $$ This is exactly equipartition — $k_B T$ per electromagnetic mode — and it is the classical prediction. It grows without bound as $f^2$, so the total energy density $\int_0^\infty u(f)\,\d f$ diverges. That divergence is the **ultraviolet catastrophe**: classical physics assigns $k_B T$ to arbitrarily high-frequency modes, of which there are infinitely many. - **Wien limit ($hf \gg k_B T$).** The $-1$ is negligible against the exponential, so $$ u(f) \to \frac{8\pi h f^3}{c^3}\, e^{-hf/k_B T}. $$ The exponential suppression reflects the cost of exciting even one photon in a mode whose quantum $hf$ exceeds the thermal energy: such modes are frozen out, their mean occupation exponentially small. The quantization $\bar\varepsilon = hf/(e^{hf/k_B T}-1)$ interpolates between the two. At low frequency it returns $k_B T$; past $hf \approx k_B T$ it collapses toward zero and tames the $f^2$ growth of the mode count. The finite peak is the balance of a rising mode density against a falling occupation. $$ % caption: The mean energy per mode is the flat classical value kT at low frequency and falls off exponentially once the photon energy exceeds the thermal energy, cutting the mode-counting divergence. \begin{tikzpicture}[scale=1.0, font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \draw[->, black] (0,0) -- (6.4,0) node[right] {frequency}; \draw[->, black] (0,0) -- (0,3.2) node[above] {energy per mode}; \draw[black, dashed] (0,2.6) -- (6.0,2.6); \node[black, anchor=east] at (-0.05,2.6) {$k_B T$}; % classical flat \draw[black, thick, dashed] (0,2.6) -- (6.0,2.6); \node[text=black, anchor=south] at (4.6,2.65) {classical}; % quantum: kT then decays \draw[acc, thick, smooth] plot coordinates {(0,2.55)(0.7,2.5)(1.4,2.38)(2.1,2.12)(2.7,1.72)(3.3,1.25)(3.9,0.82)(4.6,0.45)(5.3,0.22)(6.0,0.1)}; \node[text=acc, anchor=west] at (2.9,1.55) {quantum}; \end{tikzpicture} $$ $$ % caption: The Planck energy density with its two classical asymptotes: the Rayleigh-Jeans law rising as frequency squared and diverging, and the Wien exponential tail. The true spectrum tracks each in its regime and peaks between them. \begin{tikzpicture}[scale=1.0, font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \draw[->, black] (0,0) -- (6.8,0) node[right] {frequency}; \draw[->, black] (0,0) -- (0,3.6) node[above] {u(f)}; % Rayleigh-Jeans f^2 \draw[black, thick, dashed, smooth] plot coordinates {(0,0)(0.8,0.16)(1.6,0.64)(2.4,1.44)(3.0,2.25)(3.5,3.06)(3.75,3.5)}; \node[text=black, anchor=south west] at (2.9,3.0) {Rayleigh-Jeans}; % Wien tail \draw[black, thick, dash dot, smooth] plot coordinates {(2.3,3.4)(2.9,2.4)(3.6,1.5)(4.3,0.95)(5.0,0.58)(5.7,0.34)(6.4,0.2)}; \node[text=black, anchor=west] at (4.7,1.15) {Wien}; % Planck \draw[acc, very thick, smooth] plot coordinates {(0,0)(0.6,0.5)(1.2,1.3)(1.8,2.05)(2.35,2.5)(2.9,2.4)(3.6,1.5)(4.3,0.95)(5.0,0.58)(5.7,0.34)(6.4,0.2)}; \node[text=acc, anchor=south] at (1.9,2.15) {Planck}; \end{tikzpicture} $$ ## Wien's displacement law The spectrum peaks at a frequency proportional to temperature. Writing $x = hf/k_B T$, the frequency density is $u \propto x^3/(e^x - 1)$, and setting its derivative to zero gives the transcendental condition $$ 3\left(1 - e^{-x}\right) = x, $$ with root $x_{\max} \approx 2.821$. The peak frequency is therefore $$ f_{\max} = 2.821\,\frac{k_B T}{h} \propto T. $$ Repeating the calculation for the wavelength density $u(\lambda) \propto \lambda^{-5}/(e^{hc/\lambda k_B T} - 1)$ gives the condition $5(1 - e^{-y}) = y$ with $y = hc/\lambda k_B T$ and root $y_{\max} \approx 4.965$, so $$ \lambda_{\max}\, T = \frac{hc}{4.965\,k_B} = 2.898\times 10^{-3}\ \mathrm{m\,K}. $$ The two peaks disagree, $x_{\max} \neq y_{\max}$, precisely because of the Jacobian between $f$ and $\lambda$; the peak of the frequency curve and the peak of the wavelength curve are different physical features of the same field.[^wien] > **Theorem (Wien displacement law).** The wavelength at which the blackbody > spectrum is brightest is inversely proportional to temperature, > $\lambda_{\max} = b/T$ with $b = 2.898\ \mathrm{mm\,K}$. Hotter bodies radiate > at shorter wavelength: a stove element glows red, the Sun's photosphere at > $5800\ \mathrm{K}$ peaks in the visible, and the $2.725\ \mathrm{K}$ cosmic > background peaks in the microwave. $$ % caption: Blackbody spectra at three temperatures. Raising the temperature lifts the whole curve and slides its peak to higher frequency in fixed proportion to T, the displacement law. \begin{tikzpicture}[scale=1.0, font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \draw[->, black] (0,0) -- (7.0,0) node[right] {frequency}; \draw[->, black] (0,0) -- (0,3.8) node[above] {u(f)}; % cool \draw[black, thick, densely dotted, smooth] plot coordinates {(0,0)(0.5,0.28)(1.0,0.62)(1.4,0.78)(1.9,0.62)(2.5,0.38)(3.2,0.2)(4.0,0.09)(4.8,0.04)}; \node[text=black, anchor=west] at (2.4,0.42) {cooler}; % mid \draw[acc, thick, smooth] plot coordinates {(0,0)(0.7,0.5)(1.4,1.2)(2.1,1.62)(2.6,1.48)(3.3,1.0)(4.1,0.58)(4.9,0.32)(5.6,0.17)}; \node[text=acc, anchor=south] at (2.2,1.68) {warmer}; % hot \draw[black, thick, dashed, smooth] plot coordinates {(0,0)(0.9,0.7)(1.8,1.9)(2.9,2.85)(3.5,2.72)(4.3,2.0)(5.1,1.25)(5.9,0.72)(6.6,0.4)}; \node[text=black, anchor=south] at (3.2,2.9) {hottest}; % peak locus \draw[black, dashed] (1.4,0.78) -- (2.6,1.48) -- (3.5,2.72); \node[black, anchor=west] at (3.55,2.5) {peak locus}; \end{tikzpicture} $$ > **Worked example.** The Sun's photosphere radiates as a blackbody near > $T = 5800\ \mathrm{K}$. The brightest wavelength is > $$ > \lambda_{\max} = \frac{2.898\times 10^{-3}\ \mathrm{m\,K}}{5800\ \mathrm{K}} > \approx 5.0\times 10^{-7}\ \mathrm{m} = 500\ \mathrm{nm}, > $$ > in the middle of the visible band, so the Sun peaks in green-yellow. A stove > element at $800\ \mathrm{K}$ peaks at $\lambda_{\max} \approx 3.6\ \mu\mathrm{m}$ > in the infrared, with only the short-wavelength tail of its spectrum reaching > the red end of the visible, which is why it glows dull red rather than white. ## Summary - Photon number is not conserved; equilibrium at fixed $T,V$ minimizes $F$ over $N$, giving $\mu = (\partial F/\partial N)_{T,V} = 0$ and Bose occupation $\langle n\rangle = 1/(e^{hf/k_B T} - 1)$. - Standing-wave modes in a cavity number $\d\mathcal{N}/\d f = 8\pi V f^2/c^3$ after counting two polarizations; the $f^2$ growth is the geometric origin of the classical divergence. - The Planck density $u(f) = (8\pi h f^3/c^3)/(e^{hf/k_B T} - 1)$ returns the Rayleigh-Jeans law $8\pi f^2 k_B T/c^3$ for $hf \ll k_B T$ and the Wien tail for $hf \gg k_B T$; the Bose factor cuts off the ultraviolet catastrophe. - The peak obeys Wien's displacement law: $f_{\max} \propto T$, and $\lambda_{\max} T = 2.898\ \mathrm{mm\,K}$, the frequency and wavelength peaks differing by the Jacobian between them. [^photon-mu]: Schroeder, §7.4, derives $\mu = 0$ for photons from the equilibrium condition on $F$; Pathria & Beale, §7.3, obtain the same result by noting the photon gas has no fugacity in its grand partition function. [^planck]: Reif, §9.13–9.14, and Pathria & Beale, §7.3, give both the frequency and wavelength forms and the mode count with two polarizations. [^wien]: Schroeder, §7.4, works the peak conditions; the CODATA value of the Wien displacement constant is $b = 2.897\,771\times 10^{-3}\ \mathrm{m\,K}$. See NIST, .