Photon Gas and Planck Radiation Law

Open a small hole in the wall of a furnace and the light that leaks out carries no fingerprint of what the furnace is made of — only its temperature. Iron, ceramic, or coal, heated to 1000 K, all glow the same dull red; at 6000 K all glow the same white as the Sun. That universality is the signature of a photon gas: the electromagnetic field in thermal equilibrium inside a cavity, a collection of field oscillators — one per standing-wave mode — each holding an integer number of photons of energy $\hbar\omega$ . We take one thing from quantum mechanics and add one thing of our own. From aq-qm-blackbody-radiation-planck-quantum we inherit the single-mode result: the mean number of photons in a mode is the Planck factor $\langle n\rangle = 1/(e^{\hbar\omega/k_BT}-1)$ , and the classical mode count $8\pi\nu^2/c^3$ survives quantization untouched — that article "keep[s] the classical mode count, quantize[s] the energy exchange," and its $\langle n\rangle$ "is the mean number of quanta in the mode." The microscopic reason photons are freely made and destroyed lives in aq-qm-spontaneous-emission-einstein-coefficients. What statistical mechanics owns is the step from one mode to the whole radiation field as a thermodynamic body: because photon number is not conserved, the free energy is minimized over $N$ , forcing $\mu=0$ ; integrating the Planck spectrum over the density of states then delivers the Stefan–Boltzmann $T^4$ energy density, a radiation pressure equal to one third of that density, and an entropy scaling as $T^3$ . The photon gas is the simplest nontrivial quantum gas and a recurring template — it shares its mathematics with lattice phonons, with relic particle backgrounds, and with any gas of massless number-non-conserving bosons. The cosmic microwave background is a real photon gas that has been expanding and cooling for 13.8 billion years while keeping its Planck shape.

Key takeaways

A photon gas is an ideal Bose gas with zero chemical potential, because photon number is not conserved — walls freely create and absorb photons, so $\mu=0$ at every temperature.
The mean occupation of a mode of frequency $\omega$ is the Planck factor $\langle n\rangle = 1/(e^{\hbar\omega/k_BT}-1)$ — the $\mu=0$ special case of the Bose–Einstein distribution.
Multiplying by the photon density of states (two polarizations, linear dispersion $\omega=ck$ ) gives Planck's spectral energy density, whose limits reproduce Rayleigh–Jeans and Wien.
Integrating over frequency yields Stefan–Boltzmann $U/V=aT^4$ with $a=\pi^2 k_B^4/15\hbar^3c^3$ , radiation pressure $P=U/3V$ , and photon number $N/V\propto T^3$ .
Thermodynamics closes: $S=\tfrac43aVT^3$ , $C_V=4aVT^3$ , and adiabats $PV^{4/3}=$ const that cool the CMB as $T\propto 1/a_{\rm scale}$ .
Key idea: set $\mu=0$ in the Bose–Einstein occupation because photons are not conserved; everything — Planck's law, $T^4$ , $P=u/3$ — then follows from counting field modes and integrating.

The core insight

Set $\mu=0$ in the Bose–Einstein occupation because photons are not conserved; everything — Planck's law, the $T^4$ energy density, and $P=u/3$ — then follows from counting field modes and integrating.

Where this lives today

The photon gas is not a historical relic — it is working infrastructure. The cosmic microwave background is the most precisely measured blackbody in nature: COBE/FIRAS found its spectrum Planckian to a few parts in $10^5$ at $T=2.725$ K, and the tiny anisotropies map the seeds of every galaxy. In astrophysics, the radiation pressure $P=aT^4/3$ sets the Eddington luminosity that caps how bright a star can shine before its own light blows it apart, and it supports the cores of the most massive stars against gravity. On the laboratory bench, because a cavity's spectral radiance depends only on $T$ , blackbody radiation is an absolute, material-independent thermometer — the basis of non-contact pyrometry, the calibration of infrared cameras, and, since the 2019 SI redefinition fixed $k_B$ and $h$ exactly, of primary temperature metrology itself.

The physical picture

Picture the cavity field as an infinite orchestra of independent oscillators, one per standing-wave mode. At temperature $T$ each oscillator is thermally excited to a mean energy $\hbar\omega\,\langle n\rangle$ . The low-frequency instruments ( $\hbar\omega\ll k_BT$ ) play loudly and classically, carrying $\sim k_BT$ each; the high-frequency instruments ( $\hbar\omega\gg k_BT$ ) are frozen silent, because exciting even one photon there costs more than the thermal budget $k_BT$ . The balance point $\hbar\omega\sim k_BT$ is the Planck peak — and the frozen high-frequency modes are exactly what save the total energy from the classical ultraviolet catastrophe. The chemical potential vanishes because the cavity can always lower its free energy by adjusting $N$ . Equilibrium with the walls fixes $T$ and $V$ but lets the photon number float; photons cost no rest energy to create, so $N$ self-adjusts until $\mu=(\partial F/\partial N)_{T,V}=0$ . This is the one structural fact that separates the photon gas from a conserved-number atomic Bose gas — and the reason ordinary blackbody photons never Bose-condense. The hero figure below is the Planck spectrum at three temperatures. Read three things off it: the low-frequency $\omega^2$ rise (Rayleigh–Jeans), the exponential Wien cutoff at high frequency, and the peak sliding rightward and the whole area (the total energy density) swelling as $T^4$ — a modest temperature increase makes the furnace vastly brighter and bluer at once.

Figure

Planck spectrum at three temperatures — the peak slides and the area grows as T⁴

The second picture is the mode-counting and pressure story. Each allowed wavevector $\mathbf{k}$ occupies a cell of volume $(2\pi/L)^3$ in $\mathbf{k}$ -space; a thin shell of radius $k$ holds $\propto k^2\,dk$ modes, and with two transverse polarizations and $\omega=ck$ this becomes the density of states $g(\omega)\propto\omega^2$ . The pressure factor $1/3$ is pure geometry: photons carry momentum $p=\varepsilon/c$ , and a wall feels the flux averaged over directions, $\langle\cos^2\theta\rangle=1/3$ .

Figure

Mode counting in k-space and the 1/3 pressure factor from angular averaging

The core equations

The starting point is the one thing statistical mechanics supplies that quantum mechanics does not fix on its own: the value of the chemical potential. Because the cavity walls create and destroy photons at no rest-energy cost, the equilibrium state minimizes the free energy over the photon number itself.

\mu = \left(\frac{\partial F}{\partial N}\right)_{T,V}=0 \qquad (1)

where: $\mu$ — chemical potential of the photon gas; $F$ — Helmholtz free energy; $N$ — total photon number, itself a free variable. The vanishing is exact and temperature-independent — it is the defining property that makes the photon gas the cleanest of all quantum gases, and the reason no fugacity $z=e^{\beta\mu}\neq1$ ever appears. With $\mu=0$ , the mean occupation of a mode is the Planck factor. This single-mode result is owned by the quantum-mechanics track (aq-qm-blackbody-radiation-planck-quantum); we quote it and build on it rather than re-deriving it.

\langle n(\omega)\rangle=\frac{1}{e^{\hbar\omega/k_BT}-1} \qquad (2)

where: $\langle n\rangle$ — mean photon number in a mode of frequency $\omega$ ; $\hbar\omega$ — photon energy. At low frequency $\langle n\rangle\approx k_BT/\hbar\omega\gg1$ (many photons per mode, classical); at high frequency $\langle n\rangle\approx e^{-\hbar\omega/k_BT}\ll1$ (mode almost empty). To turn one mode into a spectrum we need the density of modes. Counting wavevectors in $\mathbf{k}$ -space, including the factor 2 for the two transverse polarizations and the linear dispersion $\omega=ck$ , gives a density of states rising as $\omega^2$ .

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

where: $g(\omega)$ — number of field modes per unit angular frequency; $V$ — cavity volume; $c$ — speed of light. The $\omega^2$ growth (not the $\sqrt{\varepsilon}$ of massive particles) reflects the linear photon dispersion; it is the same $8\pi\nu^2/c^3$ "pure wave geometry" the QM owner keeps intact. Multiplying energy per photon, occupation, and mode density gives Planck's spectral energy density — the curve plotted above.

u(\omega,T)=\frac{\hbar}{\pi^2 c^3}\,\frac{\omega^3}{e^{\hbar\omega/k_BT}-1} \qquad (4)

where: $u(\omega,T)$ — energy per unit volume per unit angular frequency. The $\omega^3$ numerator is mode-count $\times$ photon-energy; the Bose denominator is the quantized occupation. This is Planck's 1900 law, the historical birthplace of the quantum. Integrating $u(\omega,T)$ over all frequencies collapses the spectrum to a single power law — the Stefan–Boltzmann result. The temperature dependence factors out completely, leaving a pure number (a Bose integral) times $T^4$ .

\frac{U}{V}=a\,T^4,\qquad a=\frac{\pi^2 k_B^4}{15\hbar^3 c^3}\approx 7.566\times10^{-16}\ \mathrm{J\,m^{-3}\,K^{-4}} \qquad (5)

where: $U$ — total internal energy; $a$ — radiation constant, $a=4\sigma/c$ . Four powers of $T$ : three from phase space ( $\int\omega^2\,d\omega$ ) and one from the energy $\hbar\omega\sim k_BT$ of each active mode. The equation of state is what most sharply separates radiation from an ordinary gas. Because photons are ultrarelativistic ( $\varepsilon=cp$ ), the pressure is one third of the energy density, not two thirds.

P=\frac{1}{3}\frac{U}{V}=\frac{a}{3}T^4 \qquad (6)

where: $P$ — radiation pressure. The $1/3$ is the $\langle\cos^2\theta\rangle$ hemisphere average combined with $p=\varepsilon/c$ ; contrast the non-relativistic $P=\tfrac23u$ . At fixed $T$ this pressure is independent of volume — unlike $PV=Nk_BT$ . The radiant power leaving a black surface follows from the energy density by a geometric factor $c/4$ , giving the familiar Stefan–Boltzmann constant.

j^\star=\sigma T^4,\qquad \sigma=\frac{\pi^2 k_B^4}{60\hbar^3 c^2}=5.670\times10^{-8}\ \mathrm{W\,m^{-2}\,K^{-4}} \qquad (7)

where: $j^\star$ — radiant exitance (power per unit area of a black surface); $\sigma=ac/4$ . This constant is, in the QM owner's phrase, just " $h$ , $k_B$ , and $c$ in a trench coat" — no new constant of nature. Finally the caloric quantities. Differentiating and integrating the energy gives an entropy and heat capacity that both scale as $T^3$ , and a photon number that also scales as $T^3$ .

S=\frac{4}{3}\,a\,V\,T^3,\qquad C_V=4aVT^3=3S,\qquad \frac{N}{V}=\frac{2\zeta(3)}{\pi^2}\left(\frac{k_BT}{\hbar c}\right)^3 \qquad (8)

where: $S$ — entropy; $C_V$ — heat capacity at constant volume; $\zeta(3)\approx1.20206$ . As $T\to0$ every one of these vanishes, so the third law is manifest — there is nothing left to excite in an empty cavity at absolute zero.

Derivation AFrom mode counting to Planck's law

The goal is to assemble Planck's spectral energy density from three ingredients and then check that it reduces to the two known classical limits. We combine the QM-owned single-mode occupation $\langle n\rangle$ (at $\mu=0$ ) with the photon density of states, forming the energy in a frequency shell as (number of modes) $\times$ (energy per photon) $\times$ (mean occupation). The subtlety to watch is that the density of states here grows as $\omega^2$ because the dispersion is linear — a massive gas with $\varepsilon\propto p^2$ would give $g\propto\sqrt\varepsilon$ instead, and a completely different spectrum. Success is recovering Rayleigh–Jeans at low $\omega$ and Wien at high $\omega$ from one formula.

Step 1. Write the energy in modes with frequency in $[\omega,\omega+\mathrm{d}\omega]$ : $\mathrm{d}U=\hbar\omega\,\langle n(\omega)\rangle\,g(\omega)\,\mathrm{d}\omega$ . Why: each mode carries energy $\hbar\omega$ times its mean occupation, and there are $g(\omega)\,\mathrm{d}\omega$ modes in the shell.

Step 2. Insert the Planck occupation (2) and the density of states (3): $\mathrm{d}U=\hbar\omega\cdot\dfrac{1}{e^{\hbar\omega/k_BT}-1}\cdot\dfrac{V\omega^2}{\pi^2c^3}\,\mathrm{d}\omega$ . Why: the occupation is QM-owned (aq-qm-blackbody-radiation-planck-quantum, "the mean number of quanta in the mode"); the mode density is the standing-wave count with two polarizations.

Step 3. Divide by $V$ to get the spectral energy density:

u(\omega,T)=\frac{1}{V}\frac{\mathrm{d}U}{\mathrm{d}\omega}=\frac{\hbar\omega^3}{\pi^2 c^3}\,\frac{1}{e^{\hbar\omega/k_BT}-1}.

Why: this is Planck's law (4). The peak sits where $\partial u/\partial\omega=0$ , i.e. $x=3(1-e^{-x})$ , giving $x_{\max}=2.8214$ — Wien's displacement.

Step 4. Low-frequency limit ( $\hbar\omega\ll k_BT$ ): expand $e^{\hbar\omega/k_BT}-1\approx\hbar\omega/k_BT$ , so $u\to\omega^2k_BT/\pi^2c^3$ . Why: this is Rayleigh–Jeans — classical equipartition, $\hbar$ -free, and it diverges on integration. The ultraviolet catastrophe is real; quantization cures it by freezing the high-frequency modes.

Step 5. High-frequency limit ( $\hbar\omega\gg k_BT$ ): the $-1$ is negligible, so $u\to(\hbar\omega^3/\pi^2c^3)e^{-\hbar\omega/k_BT}$ . Why: exponential suppression makes the frequency integral converge — Wien's empirical law, recovered when $\langle n\rangle\approx e^{-\beta\hbar\omega}$ . *Verified: the Wien peak $x_{\max}=2.8214$ and both classical limits are checked numerically in verify_photon-gas-planck-radiation.py.*

Go deeper · Derivation B · 1 minStefan-Boltzmann law by frequency integration

Derivation BStefan-Boltzmann law by frequency integration

The goal is to integrate the Planck spectrum over all frequencies and obtain the total energy density $U/V=aT^4$ , with the radiation constant $a$ expressed in fundamental constants. The strategy is a dimensionless substitution $x=\hbar\omega/k_BT$ that pulls every factor of $T$ out front, leaving a pure number — the Bose integral $\int_0^\infty x^3/(e^x-1)\,\mathrm{d}x$ . The one slip to avoid is dropping the $\Gamma(4)=6$ : the integral is a standard $\Gamma\zeta$ form, and the same integral is evaluated by the QM owner for exactly this photon gas, so our value must match theirs.

Step 1. Integrate the spectral density over all frequencies: $\dfrac{U}{V}=\displaystyle\int_0^\infty\frac{\hbar\omega^3}{\pi^2c^3}\frac{\mathrm{d}\omega}{e^{\hbar\omega/k_BT}-1}$ . Why: total energy density is the spectrum summed over its whole support; photons have no frequency ceiling, so the range is $[0,\infty)$ .

Step 2. Substitute $x=\hbar\omega/k_BT$ , so $\mathrm{d}\omega=(k_BT/\hbar)\,\mathrm{d}x$ :

\frac{U}{V}=\frac{(k_BT)^4}{\pi^2c^3\hbar^3}\int_0^\infty\frac{x^3\,\mathrm{d}x}{e^x-1}.

Why: four powers of $T$ come out — three from $\omega^3\,\mathrm{d}\omega\to(k_BT/\hbar)^4$ and the visible one — leaving a pure number.

Step 3. Evaluate the standard Bose integral:

\int_0^\infty\frac{x^3\,\mathrm{d}x}{e^x-1}=\Gamma(4)\,\zeta(4)=6\cdot\frac{\pi^4}{90}=\frac{\pi^4}{15}.

Why: expanding $1/(e^x-1)=\sum_{n\ge1}e^{-nx}$ and integrating term by term gives $\Gamma(s)\zeta(s)$ at $s=4$ ; $\Gamma(4)=6$ , $\zeta(4)=\pi^4/90$ . This is the identical integral aq-qm-blackbody-radiation-planck-quantum evaluates for the same gas — the seam check.

Step 4. Collect: $\dfrac{U}{V}=\dfrac{\pi^2 k_B^4}{15\hbar^3c^3}\,T^4\equiv aT^4$ , and the emissive power is $j^\star=\tfrac{c}{4}\,aT^4=\sigma T^4$ . Why: identifying the prefactor as $a$ finishes Stefan–Boltzmann; the $c/4$ is the geometric conversion from energy density to one-sided flux. *Verified: $a=7.566\times10^{-16}$ and $\sigma=ac/4=5.670\times10^{-8}$ reproduced from $k_B,\hbar,c$ to CODATA precision, and $\int x^3/(e^x-1)\,\mathrm{d}x=\pi^4/15$ confirmed, in verify_photon-gas-planck-radiation.py.*

Go deeper · Derivation C · 1 minThermodynamics: the 1/3 pressure, entropy, and adiabats

Derivation CThermodynamics: the 1/3 pressure, entropy, and adiabats

The goal is to derive the equation of state $P=U/3V$ , the entropy $S=\tfrac43aVT^3$ , and the adiabatic law $PV^{4/3}=$ const from the grand potential of the massless $\mu=0$ Bose gas. The clean route is $\Omega=k_BT\int g\ln(1-e^{-\beta\hbar\omega})\,\mathrm{d}\omega$ ; integrating by parts converts the logarithm into the energy integral and produces $\Omega=-U/3$ exactly, whence $P=-\Omega/V=u/3$ . The number $1/3$ is set entirely by the linear dispersion — the same $\omega^2\to\omega^3/3$ antiderivative that a massive gas would replace by a different power. Success is a closed thermodynamics with a manifest third law.

Step 1. Write the grand potential of the photon gas ( $\mu=0$ , two polarizations): $\Omega=k_BT\dfrac{V}{\pi^2c^3}\displaystyle\int_0^\infty\omega^2\ln\!\big(1-e^{-\beta\hbar\omega}\big)\,\mathrm{d}\omega$ . Why: for bosons $\Omega=k_BT\sum_{\rm modes}\ln(1-e^{-\beta\hbar\omega})$ ; converting the sum to a frequency integral introduces the $\omega^2$ density of states.

Step 2. Integrate by parts with $\mathrm{d}v=\omega^2\,\mathrm{d}\omega$ (so $v=\omega^3/3$ ); the boundary term vanishes at both ends, leaving

\Omega=-\frac{1}{3}\,\frac{V\hbar}{\pi^2c^3}\int_0^\infty\frac{\omega^3\,\mathrm{d}\omega}{e^{\beta\hbar\omega}-1}=-\frac{1}{3}U.

Why: the surviving integral is exactly the total energy $U$ ; the $1/3$ comes entirely from the $\omega^3/3$ antiderivative of $\omega^2$ .

Step 3. For a homogeneous system $\Omega=-PV$ , so $P=-\Omega/V=\tfrac13(U/V)=\tfrac13u$ . Why: this is the ultrarelativistic equation of state; a non-relativistic gas ( $\varepsilon\propto p^2$ ) would give $\Omega=-\tfrac23U$ and $P=\tfrac23u$ instead.

Step 4. Entropy from $S=-(\partial\Omega/\partial T)_V$ with $\Omega=-aVT^4/3$ : $S=\tfrac43aVT^3$ , and $C_V=T(\partial S/\partial T)_V=4aVT^3=3S$ . Why: differentiating the $T^4$ potential; as $T\to0$ , $S\to0$ , so the third law holds automatically.

Step 5. Adiabat: $S=$ const gives $VT^3=$ const, and with $P\propto T^4$ eliminating $T$ yields $PV^{4/3}=$ const. Why: the photon gas has adiabatic index $\gamma=4/3$ ; this governs CMB cooling, $T\propto V^{-1/3}\propto a_{\rm scale}^{-1}$ . *Verified: $\langle\cos^2\theta\rangle=1/3$ , the symbolic $U/PV=3$ for photons versus $3/2$ for a massive gas, and $C_V=3S$ with $S\to0$ are all checked in verify_photon-gas-planck-radiation.py.*

Limiting cases

The single Planck spectrum interpolates between two classical laws and obeys one scaling relation.

Case	Result	Meaning
Rayleigh–Jeans ( $\hbar\omega\ll k_BT$ )	$u\to \dfrac{\omega^2}{\pi^2 c^3}\,k_BT$	Classical equipartition, $k_BT$ per mode; $\int\omega^2\,\mathrm{d}\omega$ diverges — the ultraviolet catastrophe.
Wien ( $\hbar\omega\gg k_BT$ )	$u\to \dfrac{\hbar\omega^3}{\pi^2 c^3}\,e^{-\hbar\omega/k_BT}$	High modes frozen; exponential cutoff makes the total energy finite.
Wien displacement	$\hbar\omega_{\max}=2.821\,k_BT$ ; $\lambda_{\max}T=2.898\times10^{-3}$ m K	Spectral peak shifts linearly with $T$ ; solves $x=3(1-e^{-x})$ in $\omega$ , $x=5(1-e^{-x})$ in $\lambda$ .
Equation of state	$P=\tfrac13 u=\tfrac13 aT^4$	Ultrarelativistic; at fixed $T$ , pressure is independent of $V$ .
Non-relativistic contrast	$P=\tfrac23 u$	Same $\langle\cos^2\theta\rangle$ geometry, but $\varepsilon\propto p^2$ — the source of $2/3$ versus $1/3$ .

Common trapPhotons behave like an ordinary gas, so I can fix their number and use P = ⅔u

Both halves of this are wrong, and both are common. First, you cannot independently specify the photon number: because photon number is not conserved, equilibrium enforces $\mu=0$ , which determines $N\propto VT^3$ — there is no fugacity $z=e^{\beta\mu}\neq1$ and no Bose–Einstein condensation of ordinary cavity photons (unlike the atomic gas of aq-sm-bose-einstein-condensation-ideal-gas, where $\mu\to\varepsilon_0$ triggers condensation). Second, photons are ultrarelativistic ( $p=\varepsilon/c$ ), so the pressure is $P=\tfrac13u$ , not the non-relativistic $\tfrac23u$ ; the factor is what gives the adiabatic index $\gamma=4/3$ and drives radiation-dominated cosmology.

Worked examples

1 · The Stefan-Boltzmann constant and the Sun's surface flux (confidence builder)

Step 1. Assemble $\sigma$ from fundamental constants: $\sigma=\dfrac{\pi^2 k_B^4}{60\hbar^3 c^2}$ . Plugging $k_B=1.381\times10^{-23}$ , $\hbar=1.055\times10^{-34}$ , $c=2.998\times10^8$ gives $\sigma=\mathbf{5.670\times10^{-8}\ W\,m^{-2}\,K^{-4}}$ — matching the measured Stefan–Boltzmann constant, with no adjustable parameter.

Step 2. The Sun's effective temperature is $T_{\rm eff}\approx5772$ K. The surface radiant exitance is $j^\star=\sigma T^4=(5.670\times10^{-8})(5772)^4$ .

Step 3. Evaluate $(5772)^4=1.110\times10^{15}\ \mathrm{K^4}$ , so $j^\star=(5.670\times10^{-8})(1.110\times10^{15})=\mathbf{6.29\times10^{7}\ W\,m^{-2}}$ .

Step 4. Sanity check: multiply by the solar surface area $4\pi R_\odot^2=6.09\times10^{18}\ \mathrm{m^2}$ to get the luminosity $L=j^\star\cdot4\pi R_\odot^2\approx3.8\times10^{26}$ W — the measured solar luminosity. The furnace formula, fed only $k_B,\hbar,c$ and the surface temperature, reproduces the Sun's power output.

2 · Radiation pressure and photon number in the cosmic microwave background (exam level)

Step 1. The CMB is a photon gas at $T=2.725$ K. Its energy density is $u=aT^4$ with $a=7.566\times10^{-16}$ : $u=(7.566\times10^{-16})(2.725)^4=(7.566\times10^{-16})(55.13)=\mathbf{4.17\times10^{-14}\ J\,m^{-3}}$ .

Step 2. The radiation pressure is one third of this: $P=u/3=\mathbf{1.39\times10^{-14}\ Pa}$ — utterly negligible today, but in the radiation-dominated early universe (when $T$ was $\sim10^4$ times larger, so $u$ was $\sim10^{16}$ times larger) it drove the expansion.

Step 3. The photon number density is $\dfrac{N}{V}=\dfrac{2\zeta(3)}{\pi^2}\left(\dfrac{k_BT}{\hbar c}\right)^3$ . Compute $\dfrac{k_BT}{\hbar c}=\dfrac{(1.381\times10^{-23})(2.725)}{(1.055\times10^{-34})(2.998\times10^8)}=1.19\times10^{3}\ \mathrm{m^{-1}}$ , cube it to $1.68\times10^{9}\ \mathrm{m^{-3}}$ .

Step 4. Multiply by $2\zeta(3)/\pi^2=2(1.202)/9.87=0.2437$ : $N/V=(0.2437)(1.68\times10^{9})=\mathbf{4.1\times10^{8}\ m^{-3}}\approx411\ \mathrm{cm^{-3}}$ . There are roughly 411 relic photons in every cubic centimetre of the universe — about a billion for every atom.

Origins · 1 minPhoton and how this was discovered

Photon

coined 1926 (G. N. Lewis) · quantum of the electromagnetic field

The word "photon" was coined by the chemist Gilbert N. Lewis in a 1926 letter to Nature, though the object long predates its name. Planck (1900) quantized the exchange of energy between cavity walls and field but resisted the idea that light itself is grainy; Einstein (1905) took the decisive step, treating radiation as a gas of energy quanta to explain the photoelectric effect, and (1917) derived the A and B coefficients that fix emission and absorption rates. The photon gas thus sits at the confluence of thermodynamics, electromagnetism, and quantum theory — the first system where "counting states" and "quantizing energy" had to be done together.

History

1859

— Kirchhoff states that cavity radiation depends only on temperature, defining the blackbody problem.

1879

— Josef Stefan infers the

T^4

law empirically from Tyndall's platinum-wire data.

1884

— Boltzmann derives the

T^4

law thermodynamically from

P=u/3

and a Carnot cycle on radiation.

1900

— Planck fits Rubens' far-infrared data and introduces the quantum hypothesis

E_n=nh\nu

1905

— Einstein proposes light quanta to explain the photoelectric effect.

1917

— Einstein derives the A and B coefficients, re-obtaining Planck's law from rate balance.

1924

— Bose re-derives Planck's law from photon statistics alone, with Einstein championing the paper.

1926

— Lewis coins the word "photon" for the light quantum.

1990

— COBE/FIRAS measures the cosmic microwave background as a 2.725 K Planck curve.

Going further

References. Pathria & Beale, Statistical Mechanics, 3rd ed., §7.3–7.4 (Black-body radiation) · Kardar, Statistical Physics of Particles, Ch. 7 · Reif, Fundamentals of Statistical and Thermal Physics, §9.13–9.15 · Landau & Lifshitz, Statistical Physics Part 1, §63 · Sethna, Statistical Mechanics: Entropy, Order Parameters, and Complexity, Ch. 7.

Provenance. Equations and limits cross-checked against Pathria & Beale §7.3–7.4, Kardar Ch. 7, Reif §9.13–9.15, and Landau–Lifshitz §63. The single-mode occupation and mode count are owned by aq-qm-blackbody-radiation-planck-quantum; the A/B emission mechanism by aq-qm-spontaneous-emission-einstein-coefficients. A verifier should recompute: (i) the density of states with polarization factor 2 and linear dispersion; (ii) the Bose integral $\int_0^\infty x^3/(e^x-1)\,\mathrm{d}x=\Gamma(4)\zeta(4)=\pi^4/15$ ; (iii) the constants $a=\pi^2k_B^4/15\hbar^3c^3\approx7.566\times10^{-16}$ and $\sigma=ac/4=5.670\times10^{-8}$ (CODATA); (iv) the Wien peak $x_{\max}=2.8214$ ; (v) the photon number density $N/V=2\zeta(3)(k_BT/\hbar c)^3/\pi^2$ .

Builds on: Quantum Statistics of Ideal Gases: Occupation Numbers · Density of States for Free Particles · Bose-Einstein Condensation in the Ideal Gas · Blackbody Radiation and the Planck Quantum (QM) · Spontaneous Emission and the Einstein Coefficients (QM). Leads to: Debye Model of Solids · Cosmic Microwave Background · Thermodynamics of the BEC Transition.

Why exactly is the chemical potential of the photon gas zero, and not just small?

It is exactly zero, and the reason is a conservation law that is absent. Ordinary particles are conserved, so their number $N$ is fixed and $\mu$ adjusts to enforce it. Photons are freely created and absorbed by the cavity walls, so $N$ is not fixed — it is a free variable that the system relaxes to minimize the free energy. The minimum of $F(N)$ at fixed $T,V$ occurs where $(\partial F/\partial N)_{T,V}=0$ , and that derivative is the chemical potential. So $\mu=0$ is exact and temperature-independent, not an approximation.

If I open a hot oven, why does everything inside glow the same colour regardless of what it is made of?

Because the equilibrium radiation inside a cavity depends only on the temperature, never on the wall material — a result Kirchhoff proved in 1859. Any surface that absorbs strongly at a frequency must, in equilibrium, emit equally strongly there (otherwise it would heat up or cool relative to the field), so a good absorber is a good emitter and every object approaches the same universal blackbody spectrum. That universality is what makes the Planck curve a temperature standard: the colour tells you $T$ and nothing else.

The article says the mode count is owned by the quantum-mechanics track. What does statistical mechanics actually add?

Quantum mechanics (aq-qm-blackbody-radiation-planck-quantum) supplies two things per mode: the classical count of standing waves $8\pi\nu^2/c^3$ and the quantized mean occupation $\langle n\rangle=1/(e^{\hbar\omega/k_BT}-1)$ . What statistical mechanics adds is the step from one mode to the whole field as a thermodynamic body. That means fixing $\mu=0$ (the number is not conserved), integrating $\hbar\omega\langle n\rangle g(\omega)$ over all frequencies to get $U/V=aT^4$ , and deriving the pressure, entropy, and adiabats. The occupation is quantum; the thermodynamics of the ensemble of modes is ours.

Why is the radiation pressure

u/3

while an ordinary gas gives

2u/3

It comes down to the dispersion relation. For any gas, the grand potential of massless-or-massive bosons integrated by parts gives $\Omega=-U/(d/n)$ where the single-particle energy scales as $\varepsilon\propto p^n$ in $d$ dimensions. Photons have $\varepsilon=cp$ ( $n=1$ , $d=3$ ), giving $\Omega=-U/3$ and $P=u/3$ . A non-relativistic gas has $\varepsilon\propto p^2$ ( $n=2$ ), giving $\Omega=-2U/3$ and $P=2u/3$ . Equivalently, the wall feels the momentum flux $p=\varepsilon/c$ averaged over directions, $\langle\cos^2\theta\rangle=1/3$ . The exponent in the dispersion is literally the numerator of the pressure fraction.

Does the ultraviolet catastrophe mean classical physics is simply wrong here?

The Rayleigh–Jeans law is not garbage — it is the exact low-frequency limit of Planck's law and is still used as the "brightness temperature" formula in radio astronomy. What fails is extrapolating it to all frequencies: classical equipartition assigns $k_BT$ to every mode, and since there are infinitely many high-frequency modes, the total energy diverges. Quantization fixes this by making high-frequency modes expensive — exciting one costs $\hbar\omega\gg k_BT$ , so they stay empty and the integral converges. The catastrophe is real; it is the sharpest early evidence that energy exchange is quantized.

How can there be so many CMB photons — 411 per cubic centimetre — yet the radiation pressure be negligible?

Number density and pressure measure different things. The photon number density $\sim(k_BT/\hbar c)^3$ is large because the wavelengths are long (millimetre-scale) so many modes fit in a small volume. But each CMB photon carries a tiny energy $\sim k_BT\approx2.3\times10^{-4}$ eV, so the energy density $u=aT^4\approx4\times10^{-14}\ \mathrm{J\,m^{-3}}$ is minuscule, and the pressure $u/3$ with it. In the early universe the same $T^4$ scaling made both enormous — at recombination ( $T\sim3000$ K) the energy density was $\sim10^{12}$ times today's, and radiation pressure genuinely governed the dynamics.

Why does the Debye model of solids reuse the photon-gas mathematics?

Because lattice vibrations (phonons) are also number-non-conserving bosons with an approximately linear dispersion $\omega=c_s k$ . Replace the speed of light $c$ by the sound speed $c_s$ , use three polarizations (two transverse, one longitudinal) instead of two, and impose a high-frequency Debye cutoff (a crystal has a finite number of modes, unlike the field), and the identical mode-counting gives the Debye $T^3$ heat-capacity law at low temperature. The photon gas is the template; the Debye solid is the same calculation with a ceiling — which is why aq-sm-debye-model-phonons builds directly on this article.