---
title: Blackbody Thermodynamics and Radiation Pressure
module: Bosonic Systems
moduleNumber: 8
lessonNumber: 3
order: 803
summary: >
  Integrating the Planck spectrum over all frequencies gives the total energy
  density proportional to the fourth power of temperature — the Stefan-Boltzmann
  law — and the isotropy of a relativistic gas fixes the radiation pressure at
  one third of the energy density. From the free energy follow the entropy and
  heat capacity, both proportional to T cubed, and the adiabatic law for
  radiation. The results govern the pressure inside stars and the cooling of the
  cosmic microwave background as the universe expands.
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; §9.14–9.16"
  - book: Pathria & Beale
    ref: "Ch. 7 — Ideal Bose Systems; §7.3"
draft: false
---

The [Planck spectrum](/statistical-mechanics/bose-systems/the-photon-gas-and-plancks-radiation-law)
fixes the energy in each frequency interval. Integrating over the spectrum
collapses the radiation field to a handful of thermodynamic functions of $T$ and
$V$ alone, because the photon gas has no independent particle number to carry.
The energy density, pressure, entropy, and heat capacity all follow one power
law in temperature, and their exponents differ only by the counting. These are
the quantities a star's interior and the expanding universe are built on.

## The Stefan-Boltzmann law

The total energy density integrates the Planck density over all frequencies. With
$x = hf/k_B T$,

$$
u = \int_0^\infty u(f)\,\d f
= \frac{8\pi h}{c^3}\int_0^\infty \frac{f^3\,\d f}{e^{hf/k_B T} - 1}
= \frac{8\pi h}{c^3}\left(\frac{k_B T}{h}\right)^4
\int_0^\infty \frac{x^3\,\d x}{e^x - 1}.
$$

The dimensionless integral is a standard Bose integral,

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

so the energy density is proportional to $T^4$.[^zeta]

> **Theorem (Stefan-Boltzmann law).** The equilibrium energy density of the
> photon gas is
> $$
> u = a\,T^4, \qquad
> a = \frac{8\pi^5 k_B^4}{15\,c^3 h^3} = 7.566\times 10^{-16}\ \mathrm{J\,m^{-3}\,K^{-4}}.
> $$
> The power radiated per unit area from a blackbody surface is $I = \tfrac{1}{4}cu
> = \sigma T^4$, with the Stefan-Boltzmann constant
> $$
> \sigma = \frac{a c}{4} = \frac{2\pi^5 k_B^4}{15\,c^2 h^3}
> = 5.670\times 10^{-8}\ \mathrm{W\,m^{-2}\,K^{-4}}.
> $$

The factor $\tfrac{1}{4}c$ relating the interior energy density to the emitted
flux comes from averaging the outward photon flux over the hemisphere of
directions leaving a small hole in the cavity: a factor $c$ for the photon speed,
a factor $\tfrac{1}{2}$ for outgoing directions, and a factor $\tfrac{1}{2}$ for
the mean projection $\langle\cos\theta\rangle$ onto the surface normal.[^stefan]

$$
% caption: The total energy density is the area under the Planck curve; each doubling of temperature multiplies that area sixteenfold, the fourth-power Stefan-Boltzmann scaling.
\begin{tikzpicture}[scale=1.0, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[->, black] (0,0) -- (6.8,0) node[right] {frequency};
\draw[->, black] (0,0) -- (0,3.6) node[above] {u(f)};
% cool curve + fill
\draw[acc, thick, smooth] plot coordinates {(0,0)(0.6,0.42)(1.2,0.92)(1.7,1.1)(2.3,0.9)(3.0,0.55)(3.8,0.3)(4.6,0.15)(5.4,0.07)};
\fill[acc!12] (0,0) -- plot coordinates {(0,0)(0.6,0.42)(1.2,0.92)(1.7,1.1)(2.3,0.9)(3.0,0.55)(3.8,0.3)(4.6,0.15)(5.4,0.07)} -- (5.4,0) -- cycle;
\node[text=acc, anchor=west] at (1.9,0.55) {area at T};
% hot curve
\draw[black, thick, densely dashed, smooth] plot coordinates {(0,0)(0.8,0.8)(1.7,2.2)(2.7,3.0)(3.4,2.75)(4.3,1.9)(5.2,1.1)(6.0,0.62)(6.6,0.4)};
\node[text=black, anchor=south] at (3.1,3.05) {area at 2T is 16 times};
\end{tikzpicture}
$$

## Radiation pressure

Photons carry momentum $p = \varepsilon/c$. A photon striking a wall at angle
$\theta$ to the normal and reflecting transfers momentum $2p\cos\theta$ along the
normal. Averaging the momentum flux over an isotropic distribution of directions
gives a pressure equal to one third of the energy density — the universal result
for an ultrarelativistic gas, where $PV = \tfrac{1}{3}U$ because
$\varepsilon = cp$ makes $p\,\partial\varepsilon/\partial p = \varepsilon$.

> **Theorem (Radiation pressure).** The pressure of the photon gas is one third
> of its energy density,
> $$
> P = \frac{u}{3} = \frac{a}{3}\,T^4,
> $$
> independent of volume at fixed temperature. Contrast the nonrelativistic ideal
> gas, where $PV = \tfrac{2}{3}U$; the factor $\tfrac{1}{3}$ rather than
> $\tfrac{2}{3}$ is the signature of the linear photon dispersion.

$$
% caption: A photon reflecting from a cavity wall reverses the normal component of its momentum; summing the momentum transfer over an isotropic photon flux gives a pressure equal to one third of the energy density.
\begin{tikzpicture}[scale=1.0, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
% wall
\fill[black] (5.0,-0.3) rectangle (5.6,3.0);
\draw[black, thick] (5.0,-0.3) -- (5.0,3.0);
\node[black, anchor=south, rotate=90] at (5.35,1.35) {wall};
% incoming photon
\draw[acc, very thick, ->] (0.6,2.4) -- (4.8,1.6);
\node[text=acc, anchor=south] at (2.0,2.25) {photon in};
% outgoing photon
\draw[black, very thick, dashed, ->] (4.8,1.6) -- (0.9,0.7);
\node[text=black, anchor=north] at (2.2,0.95) {photon out};
% normal
\draw[black, dashed] (5.0,1.6) -- (3.6,1.6);
\node[black, anchor=east] at (3.6,1.6) {normal};
\end{tikzpicture}
$$

## Free energy, entropy, and heat capacity

Because $\mu = 0$, the grand potential equals the Helmholtz free energy, and both
equal $-PV$:

$$
F = -PV = -\frac{1}{3}\,a V T^4.
$$

Every other function follows by differentiation. The entropy is

$$
S = -\left(\frac{\partial F}{\partial T}\right)_V = \frac{4}{3}\,a V T^3,
$$

and the internal energy reconstructs correctly, $U = F + TS = -\tfrac{1}{3}aVT^4
+ \tfrac{4}{3}aVT^4 = aVT^4$. The heat capacity at constant volume is

$$
C_V = \left(\frac{\partial U}{\partial T}\right)_V = 4 a V T^3 = 3S.
$$

Both the entropy and the heat capacity vanish as $T^3$, so the photon gas obeys
the third law: $S \to 0$ and $C_V \to 0$ as $T \to 0$. Unlike a material gas,
there is no residual entropy and no additive constant, because the number of
photons is itself a function of temperature.

$$
% caption: The photon-gas entropy and heat capacity both rise as temperature cubed from zero, with the heat capacity three times the entropy at every temperature.
\begin{tikzpicture}[scale=1.0, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[->, black] (0,0) -- (6.0,0) node[right] {T};
\draw[->, black] (0,0) -- (0,3.4) node[above] {};
% C_V = 3S, steeper
\draw[black, thick, densely dashed, smooth] plot coordinates {(0,0)(1,0.16)(2,0.72)(3,1.6)(3.6,2.4)(4.05,3.1)};
\node[text=black, anchor=west] at (3.2,2.5) {heat capacity};
% S, one third
\draw[acc, thick, smooth] plot coordinates {(0,0)(1,0.055)(2,0.24)(3,0.53)(4,1.05)(4.7,1.7)(5.2,2.25)};
\node[text=acc, anchor=west] at (4.3,1.35) {entropy};
\node[black, anchor=north] at (3,-0.15) {both proportional to T cubed};
\end{tikzpicture}
$$

## The adiabatic law and cosmological cooling

A reversible adiabatic expansion holds the entropy fixed. Since $S \propto V
T^3$, constant entropy requires

$$
V T^3 = \text{const}, \qquad T \propto V^{-1/3},
$$

and combined with $P \propto T^4$ this gives the adiabatic relation

$$
P V^{4/3} = \text{const},
$$

so the photon gas has adiabatic index $\gamma = \tfrac{4}{3}$, distinct from the
$\tfrac{5}{3}$ of a monatomic ideal gas. The universe expands adiabatically to
good approximation, its volume scaling as the cube of the scale factor,
$V \propto R^3$, so $T \propto R^{-1}$: the blackbody radiation that filled the
early universe cools in inverse proportion to expansion, and a Planck spectrum
stays a Planck spectrum at a lower temperature.[^cosmo]

## Photon number and the cosmic microwave background

The photon number density integrates the mode density times the occupation,

$$
n = \frac{N}{V} = \frac{8\pi}{c^3}\int_0^\infty \frac{f^2\,\d f}{e^{hf/k_B T} - 1}
= 8\pi\left(\frac{k_B T}{hc}\right)^3\int_0^\infty \frac{x^2\,\d x}{e^x - 1}
= 16\pi\,\zeta(3)\left(\frac{k_B T}{hc}\right)^3,
$$

using $\int_0^\infty x^2/(e^x-1)\,\d x = 2\zeta(3) = 2.404$. Numerically
$n \approx 2.03\times 10^7\, T^3\ \mathrm{m^{-3}}$ with $T$ in kelvin. Dividing the
energy density by the number density gives the mean energy per photon,

$$
\frac{U}{N} = \frac{a T^4}{n} = \frac{\pi^4}{30\,\zeta(3)}\,k_B T \approx 2.70\,k_B T,
$$

somewhat below the peak-frequency quantum $2.82\,k_B T$ because the broad
low-frequency tail pulls the mean down.

> **Worked example.** The cosmic microwave background is blackbody radiation at
> $T = 2.725\ \mathrm{K}$. The number density is
> $$
> n = 2.03\times 10^7 \times (2.725)^3\ \mathrm{m^{-3}}
> \approx 4.1\times 10^8\ \mathrm{m^{-3}} = 411\ \mathrm{cm^{-3}}.
> $$
> The energy density is $u = aT^4 = 7.566\times 10^{-16}\times(2.725)^4
> \approx 4.2\times 10^{-14}\ \mathrm{J\,m^{-3}}$, about $0.26\ \mathrm{eV\,cm^{-3}}$,
> and the mean photon energy is $2.70\,k_B T \approx 6.3\times 10^{-4}\ \mathrm{eV}$,
> in the microwave. The spectrum measured by COBE matches a Planck curve to
> better than one part in $10^4$, the most precise blackbody known.

$$
% caption: The cosmic microwave background spectrum is a Planck curve at 2.725 kelvin, peaking near a wavelength of a millimetre in the microwave band; measured points fall on the curve within the line width.
\begin{tikzpicture}[scale=1.0, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[->, black] (0,0) -- (6.8,0) node[right] {frequency};
\draw[->, black] (0,0) -- (0,3.4) node[above] {intensity};
\draw[acc, very thick, smooth] plot coordinates {(0,0)(0.6,0.5)(1.2,1.35)(1.9,2.4)(2.5,2.95)(3.0,3.05)(3.6,2.75)(4.3,2.05)(5.0,1.3)(5.7,0.72)(6.4,0.38)};
% data points on curve
\foreach \pt in {(0.9,0.9),(1.6,1.95),(2.5,2.95),(3.3,2.9),(4.3,2.05),(5.3,1.05)}
  \fill[black] \pt circle (1.6pt);
\node[text=acc, anchor=west] at (3.3,3.0) {Planck spectrum};
\node[text=black, anchor=west] at (0.7,0.7) {measurements};
\end{tikzpicture}
$$

## Summary

- The energy density integrates the Planck spectrum to $u = aT^4$ with
  $a = 8\pi^5 k_B^4/15c^3h^3$; the surface flux is $\sigma T^4$ with
  $\sigma = ac/4 = 5.670\times 10^{-8}\ \mathrm{W\,m^{-2}\,K^{-4}}$.
- Isotropy and the linear dispersion give $P = u/3 = \tfrac{1}{3}aT^4$,
  independent of volume, versus $\tfrac{2}{3}U/V$ for a nonrelativistic gas.
- From $F = -\tfrac{1}{3}aVT^4$ follow $S = \tfrac{4}{3}aVT^3$ and
  $C_V = 4aVT^3 = 3S$, both vanishing as $T^3$ in accord with the third law.
- Adiabatic expansion holds $VT^3$ fixed, so $T \propto V^{-1/3}$ and
  $PV^{4/3} = \text{const}$; the same relation cools the cosmic background as
  $T \propto R^{-1}$.
- The number density is $n = 16\pi\zeta(3)(k_B T/hc)^3 \approx 2.03\times 10^7\,T^3
  \ \mathrm{m^{-3}}$, giving $411\ \mathrm{cm^{-3}}$ for the $2.725\ \mathrm{K}$
  microwave background, with mean photon energy $2.70\,k_B T$.

[^stefan]: Schroeder, §7.4, derives the $\tfrac{1}{4}c$ effusion factor and the
Stefan-Boltzmann constant; the CODATA value is
$\sigma = 5.670\,374\times 10^{-8}\ \mathrm{W\,m^{-2}\,K^{-4}}$. See NIST,
<https://physics.nist.gov/cuu/Constants/>.
[^cosmo]: Reif, §9.15, gives the adiabatic law $PV^{4/3} = \text{const}$ for
radiation; Pathria & Beale, §7.3, connect it to the $T \propto R^{-1}$ cooling of
the cosmological photon gas.
[^zeta]: The **Riemann zeta function** is $\zeta(s)=\sum_{n=1}^{\infty} n^{-s}=1+2^{-s}+3^{-s}+\cdots$. It enters these gas integrals through the standard Bose result $\int_0^\infty \frac{x^{s-1}}{e^x-1}\,\d x=\Gamma(s)\,\zeta(s)$. Here $\zeta(4)=\pi^4/90$ (even arguments reduce to powers of $\pi$), while $\zeta(3)\approx 1.2021$ is Apéry's constant, with no closed form in terms of $\pi$.