Stefan-Boltzmann Law and Radiation Pressure

Point a thermometer at the Sun and, without ever touching it, you can read its surface temperature. Aim an infrared camera at a wall and it paints the temperature of every square centimetre. Both tricks rest on a single fact discovered in the nineteenth century: the power a hot body radiates climbs as the fourth power of its absolute temperature. Double the temperature and the glow brightens sixteenfold. That startling steepness is the Stefan–Boltzmann law, and it is nothing more than the thermodynamics of a box full of light. The physics is disarmingly clean. Radiation in equilibrium inside a cavity is a gas — a gas of photons: massless, uncharged, spin-1 bosons with two polarizations, moving at $c$ , carrying energy $\varepsilon = \hbar\omega$ and momentum $\varepsilon/c$ . Because the cavity walls create and destroy photons at will, their number is not fixed and the chemical potential is pinned at $\mu = 0$ . From those facts the energy density comes out proportional to $T^4$ , and — because photons are ultrarelativistic — the pressure is exactly one third of the energy density, $P = u/3$ . Both the energy stored in radiation and the pressure it exerts scale as $T^4$ . Remarkably, the $T^4$ power was known before quantum mechanics existed. Boltzmann derived it in 1884 from Maxwell's radiation pressure and the second law alone, fixing the exponent without any $\hbar$ . What thermodynamics could not supply was the constant $\sigma$ out front — that had to wait for Planck's quantum, which is where the companion article aq-qm-blackbody-radiation-planck-quantum takes over the microscopic story. Here we take that spectrum as given and extract the thermodynamics: the flux $\sigma T^4$ that runs every pyrometer, the pressure that props up massive stars against gravity, and the $PV^{4/3}$ adiabat that cooled the cosmic microwave background from a fireball to 2.7 K.

Key takeaways

Blackbody radiation is a photon gas: massless spin-1 bosons, two polarizations, $\mu = 0$ , dispersion $\varepsilon = \hbar c k$ .
Integrating the Planck spectrum over frequency gives the energy density $u = aT^4$ with $a = \pi^2 k_B^4/(15\hbar^3 c^3)$ — the Stefan–Boltzmann law.
The flux leaving a small hole is $j = (c/4)u = \sigma T^4$ , with $\sigma = ac/4 = 5.670\times10^{-8}\ \mathrm{W\,m^{-2}\,K^{-4}}$ .
Because photons are ultrarelativistic ( $\varepsilon = pc$ ), the equation of state is $P = u/3$ — one third, not the two thirds of a non-relativistic gas.
Boltzmann (1884) fixed the exponent 4 from $P = u/3$ and the second law alone, long before Planck supplied the constant.
Key idea: a photon gas has $\mu = 0$ and $\varepsilon = pc$ ; the linear dispersion turns the energy integral into $u = aT^4$ , and the relativistic kinematics forces $P = u/3$ , so both energy density and pressure grow as the fourth power of temperature.

The core insight

Treat radiation as a Bose gas of photons with $\mu = 0$ and $\varepsilon = \hbar c k$ ; the energy integral gives $u = aT^4$ , and because photons move at $c$ their momentum is $\varepsilon/c$ , so the pressure is exactly one third of the energy density, $P = u/3$ — both scaling as $T^4$ .

Where this lives today

The flux law $j = \varepsilon\sigma T^4$ (with emissivity $\varepsilon \le 1$ ) is the engine of every non-contact thermometer: the forehead scanner at a clinic, the infrared camera on a drone surveying a solar farm for hot cells, the pyrometer reading a furnace too hot to touch. The fourth-power sensitivity is what makes these instruments precise — a small temperature change is a large signal. At the other extreme of scale, radiation pressure sets the maximum steady luminosity of a star (the Eddington limit $L_{\rm Edd} = 4\pi GMm_pc/\sigma_T$ ), shapes the outflows from the most massive stars and accreting black holes, and — through the $P = u/3$ , $\gamma = 4/3$ equation of state — defined the expansion of the Universe during its first fifty thousand years. The CMB's near-perfect blackbody spectrum, cooled adiabatically to 2.725 K along the $VT^3 = \text{const}$ track, is a photon gas you can still measure directly with a radio dish.

The physical picture

Picture the cavity as an orchestra of standing electromagnetic waves. Heating the walls excites more modes and pushes the population toward higher frequencies. Two effects compound: the number of modes available near frequency $\omega$ grows as $\omega^2$ , and the band of thermally accessible frequencies (those with $\hbar\omega \lesssim k_BT$ ) widens in proportion to $T$ . Multiply the energy per mode ( $\sim k_BT$ ), the density of modes, and the width of the active band, and the total energy density climbs as $T \times T \times T \times T = T^4$ . The extra powers beyond a naive equipartition $T$ are the price of counting how many modes wake up as the box heats. The hero picture below shows why the pressure is one third of the energy density — the single most important number distinguishing radiation from ordinary gas.

Figure

Why radiation pressure is one third of the energy density — photons carry momentum ε∕c and strike the wall at every angle

Radiation pressure is the momentum flux of photons bouncing off a wall. Each photon of energy $\varepsilon$ delivers momentum $\varepsilon/c$ , but only the component along the surface normal, $\propto \cos\theta$ , pushes; and the arrival rate is itself $\propto \cos\theta$ . Averaging $\cos^2\theta$ over the incoming hemisphere gives exactly $1/3$ . This is the same geometric factor that appears in the kinetic theory of gases — but there the momentum is $p = mv$ , whereas here $p = \varepsilon/c$ . That swap is the whole story: a non-relativistic gas gives $P = \tfrac23 u$ , radiation gives $P = \tfrac13 u$ . Because a photon costs no rest energy to make, equilibrium chooses the photon number rather than letting you set it. Minimizing the free energy over $N$ forces $\mu = 0$ , so a cavity at temperature $T$ holds a fixed, $T$ -dependent number $N \propto VT^3$ — you cannot dial it independently. The payoff figure shows the two headline scalings side by side.

Figure

The two fourth-power laws: energy density u = aT⁴ and pressure P = u∕3 both climb steeply with temperature

Finally, unlike a solid whose phonon heat capacity saturates at the Dulong–Petit plateau, the photon $C_V \propto T^3$ never levels off: there is no upper cutoff on photon frequencies, so heating always wakes new modes. The photon gas is the Debye model — treated in aq-sm-debye-model-phonons — with its cutoff pushed to infinity.

The core equations

The starting point is the mean occupation of a single radiation mode, which the companion QM article aq-qm-blackbody-radiation-planck-quantum derives from Planck's quantization of the energy exchange. We take it as given and build the thermodynamics on top.

\langle n(\mathbf{k},\lambda)\rangle = \frac{1}{e^{\beta\hbar\omega_{\mathbf{k}}} - 1}, \qquad \omega_{\mathbf{k}} = c|\mathbf{k}|, \qquad \mu = 0 \qquad (1)

where: $\langle n\rangle$ — mean number of photons in mode $(\mathbf{k},\lambda)$ ; $\beta = 1/(k_BT)$ ; $\lambda$ — polarization index, two values; $\mu$ — chemical potential, zero because photon number is unconstrained. Read physically: this is the ordinary Bose factor with $\mu$ deleted; the deletion is not an approximation but a thermodynamic identity for a gas whose particles can vanish. Multiplying the energy per mode by the number of modes and their occupation, and expressing the mode count with two polarizations, gives Planck's spectral energy density — again, the object aq-qm-blackbody-radiation-planck-quantum owns and states as $u(\nu,T) = \frac{8\pi h\nu^3}{c^3}\frac{1}{e^{h\nu/k_BT}-1}$ ; in angular frequency it reads:

u_\omega\,\mathrm{d}\omega = \frac{\hbar}{\pi^2 c^3}\,\frac{\omega^3}{e^{\hbar\omega/k_BT} - 1}\,\mathrm{d}\omega \qquad (2)

where: $u_\omega$ — energy per unit volume per unit angular frequency; the $\omega^3$ is the mode density $\omega^2$ times the energy $\hbar\omega$ ; the denominator is the Bose factor. This spectrum peaks at $\hbar\omega_{\max} \approx 2.82\,k_BT$ (Wien's displacement law) and its low- and high-frequency wings recover Rayleigh–Jeans and Wien. Summing the spectrum over all frequencies collapses it to a single number that carries all the temperature dependence — the Stefan–Boltzmann law. This is where the thermodynamics of the photon gas begins.

u(T) = \int_0^\infty u_\omega\,\mathrm{d}\omega = aT^4, \qquad a = \frac{\pi^2 k_B^4}{15\,\hbar^3 c^3} \qquad (3)

where: $u$ — total electromagnetic energy density; $a = 7.566\times10^{-16}\ \mathrm{J\,m^{-3}\,K^{-4}}$ — the radiation constant; $k_B$ — Boltzmann constant. The whole $T^4$ climbs out of the frequency integral: three powers of $T$ from the mode band-width scaling, one from the energy per mode. The energy escaping a small hole in the cavity is the isotropic flux projected onto the outward normal, giving the geometric factor $c/4$ :

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

where: $j$ — energy radiated per unit area per unit time from a black surface; $\sigma$ — the Stefan–Boltzmann constant; $c/4$ — the fraction of isotropic energy density crossing a unit area outward per unit time. As the QM owner memorably puts it, $\sigma$ "is $h$ , $k_B$ , and $c$ in a trench coat" — not a new constant of nature, but a combination. The equation of state closes the thermodynamics. Because photons are ultrarelativistic, the pressure is one third of the energy density:

P = \frac{1}{3}u = \frac{1}{3}aT^4 \qquad (5)

where: $P$ — pressure of the isotropic radiation. The factor $1/3$ is the $\langle\cos^2\theta\rangle$ hemisphere average combined with $p = \varepsilon/c$ ; contrast the non-relativistic $P = \tfrac23 u$ . Everything else follows from the free energy of the $\mu = 0$ boson gas, $F = -\tfrac13 aVT^4$ , giving $S = \tfrac43 aVT^3$ and $C_V = 4aVT^3$ .

Derivation AStefan-Boltzmann law from the Planck spectrum

We want the total energy density at temperature $T$ : the integral of the Planck spectrum over all frequencies. The strategy is to render the temperature dependence explicit with a dimensionless substitution, so that everything reduces to a pure number — the Bose integral $\int x^3/(e^x-1)\,\mathrm{d}x$ — times $T^4$ . The one subtlety is recognizing that this integral is a standard $\Gamma\zeta$ form; watch for the common slip of dropping the factor $\Gamma(4) = 6$ . The constant $a$ that emerges is exactly what Boltzmann's thermodynamic argument could not supply.

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

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

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

Why: this pulls all four factors of $T$ out front — three from $\omega^3\,\mathrm{d}\omega \to (k_BT/\hbar)^4$ minus the one already visible — and leaves 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: $\int_0^\infty x^{s-1}/(e^x-1)\,\mathrm{d}x = \Gamma(s)\zeta(s)$ by expanding $1/(e^x-1) = \sum_{n\ge1}e^{-nx}$ and integrating term by term; with $s=4$ , $\Gamma(4)=6$ and $\zeta(4)=\pi^4/90$ . This is the identical integral the QM owner evaluates for the same photon gas.

Step 4. Collect:

u = \frac{\pi^2 k_B^4}{15\hbar^3 c^3}\,T^4 \equiv aT^4.

Why: identifying the prefactor as the radiation constant $a$ completes the law. Thermodynamics fixed the power 4; the quantum integral fixed the number $a$ . *Verified numerically in verify_stefan-boltzmann-radiation-pressure.py: $a = 7.566\times10^{-16}\ \mathrm{J\,m^{-3}\,K^{-4}}$ and $\sigma = ac/4 = 5.670\times10^{-8}\ \mathrm{W\,m^{-2}\,K^{-4}}$ , reproduced from $k_B,\hbar,c$ to CODATA precision; the Bose integral confirmed $= \pi^4/15$ .*

Go deeper · Derivation B · 1 minThe equation of state P equals u over three

Derivation BThe equation of state P equals u over three

We show directly that radiation pressure is one third of the energy density — the fact that most sharply separates a photon gas from an ordinary one. The clean route is the grand potential $\Omega$ of the $\mu = 0$ boson gas: integrating $\Omega$ by parts converts the logarithm into the energy integral and produces $\Omega = -U/3$ exactly, whence $P = -\Omega/V = u/3$ . A second, purely kinematic route via the box-scaling $\varepsilon \propto V^{-1/3}$ gives the same $1/3$ and makes clear that the number is set by the linear dispersion, not by any detail of the spectrum.

Step 1. Write the grand potential of the photon gas ( $\mu = 0$ , two polarizations):

\Omega = k_BT\,\frac{V}{\pi^2 c^3}\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 mode sum to a frequency integral introduces the $\omega^2$ density of states.

Step 2. Integrate by parts with $u = \ln(1-e^{-\beta\hbar\omega})$ and $\mathrm{d}v = \omega^2\,\mathrm{d}\omega$ (so $v = \omega^3/3$ ). The boundary term vanishes at both ends, leaving

\Omega = -\frac{V}{3}\,\frac{\hbar}{\pi^2 c^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 $U = uV$ from Equation (3); the $1/3$ comes entirely from the $\omega^3/3$ antiderivative of $\omega^2$ .

Step 3. For a homogeneous system $\Omega = -PV$ , so

P = -\frac{\Omega}{V} = \frac{1}{3}\frac{U}{V} = \frac{1}{3}u.

Why: combining $\Omega = -U/3$ with $\Omega = -PV$ gives the ultrarelativistic equation of state.

Step 4. Confirm kinematically. In a box of side $L = V^{1/3}$ each mode has $\varepsilon_{\mathbf{k}} \propto k \propto V^{-1/3}$ , so

P = -\sum_{\rm modes}\langle n\rangle\frac{\partial\varepsilon}{\partial V} = \frac{1}{3V}\sum_{\rm modes}\langle n\rangle\varepsilon = \frac{U}{3V}.

Why: $\partial\varepsilon/\partial V = -\varepsilon/3V$ for $\varepsilon \propto V^{-1/3}$ ; a non-relativistic gas has $\varepsilon \propto k^2 \propto V^{-2/3}$ , giving $-2\varepsilon/3V$ and hence $P = \tfrac23 u$ . The exponent in the dispersion is precisely the numerator of the pressure fraction. *Verified in verify_stefan-boltzmann-radiation-pressure.py: the integration-by-parts identity $\int\omega^2\ln(1-e^{-\omega})\,\mathrm{d}\omega = -\tfrac13\int\omega^3/(e^{\omega}-1)\,\mathrm{d}\omega$ confirmed numerically, and the box-scaling coefficient $s/3$ gives $1/3$ for $s=1$ (photon) and $2/3$ for $s=2$ (non-relativistic).*

Go deeper · Derivation C · 1 minBoltzmann's thermodynamic route to the fourth power

Derivation CBoltzmann's thermodynamic route to the fourth power

Here is the historical gem: the $T^4$ law without a single quantum. In 1884 Boltzmann derived it from Maxwell's result $P = u/3$ and the second law alone. The strategy uses one thermodynamic identity relating $(\partial U/\partial V)_T$ to the pressure, feeds in the two radiation facts ( $U = uV$ and $P = u/3$ ), and integrates the resulting ordinary differential equation. The power 4 falls out entirely from the equation of state; the constant $a$ stays undetermined — exactly the gap the quantum integral above fills.

Step 1. Start from the exact thermodynamic identity (a consequence of $\mathrm{d}U = T\,\mathrm{d}S - P\,\mathrm{d}V$ and a Maxwell relation):

\left(\frac{\partial U}{\partial V}\right)_T = T\left(\frac{\partial P}{\partial T}\right)_V - P.

Why: this holds for any simple substance; it converts a statement about energy into a statement about how pressure varies with temperature.

Step 2. Insert the radiation properties $U = uV$ (so $(\partial U/\partial V)_T = u$ ) and $P = u/3$ with $u = u(T)$ only:

u = \frac{T}{3}\frac{\mathrm{d}u}{\mathrm{d}T} - \frac{1}{3}u.

Why: the energy density depends only on temperature, so the partial derivatives become ordinary derivatives of $u(T)$ .

Step 3. Rearrange and integrate:

\frac{4}{3}u = \frac{T}{3}\frac{\mathrm{d}u}{\mathrm{d}T} \;\Rightarrow\; \frac{\mathrm{d}u}{u} = 4\,\frac{\mathrm{d}T}{T} \;\Rightarrow\; u = aT^4.

Why: the factor 4 in the exponent is literally the $4/3$ over $1/3$ from the equation of state. Change $P = u/3$ to $P = u/2$ and the exponent would become 5 — the power is a fingerprint of ultrarelativistic kinematics. *Verified in verify_stefan-boltzmann-radiation-pressure.py: the exponent 4 is forced by $P = u/3$ and the identity; quantum statistics enters only in the constant $a$ .*

Limiting cases

The single law $u = aT^4$ controls a wide range of regimes through its derived quantities.

Case	Result	Meaning
Low frequency (Rayleigh–Jeans)	$u_\omega \approx \dfrac{k_BT\,\omega^2}{\pi^2 c^3}\ (\hbar\omega \ll k_BT)$	Classical equipartition per mode; integrated to $\infty$ it diverges — the ultraviolet catastrophe the Planck factor cures.
High frequency (Wien tail)	$u_\omega \approx \dfrac{\hbar\omega^3}{\pi^2 c^3}e^{-\hbar\omega/k_BT}\ (\hbar\omega \gg k_BT)$	Exponentially frozen; the spectrum peaks at $\hbar\omega_{\max} \approx 2.82\,k_BT$ (Wien's displacement).
Equation of state	$P = \tfrac13 u = \tfrac13 aT^4$	Ultrarelativistic; at fixed $T$ the pressure is independent of volume, unlike $PV = Nk_BT$ .
Non-relativistic contrast	$P = \tfrac23 u$	Same $\langle\cos^2\theta\rangle$ geometry, but $\varepsilon \propto p^2$ instead of $\varepsilon = pc$ — the source of the $2/3$ vs $1/3$ .
Adiabatic expansion	$VT^3 = \text{const}, \quad PV^{4/3} = \text{const}$	Polytropic index $\gamma = 4/3$ ; describes the redshift cooling $T \propto 1/a(t)$ of the CMB.

Common trapRadiation is just an ordinary gas, so P = ⅔u and I can tune the photon number

Both halves of this thought are wrong, and both are common. First: photons move at $c$ , so their momentum is $p = \varepsilon/c$ , not $mv$ ; the correct kinematic relation is $P = \tfrac13 u$ , not the non-relativistic $\tfrac23 u$ . The factor is not a convention — it is what makes $\gamma = 4/3$ and drives radiation-dominated cosmology. Second: you cannot independently specify the photon number. Because photon number is not conserved, equilibrium enforces $\mu = 0$ , which determines $N \propto VT^3$ . This is also why equilibrium photons in a simple cavity never Bose-condense — there is no $\mu \to \varepsilon_0$ approach to trigger it, unlike the atomic gas of aq-sm-bose-einstein-condensation-ideal-gas.

Worked examples

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

We first check that the constant $\sigma$ really is $k_B,\hbar,c$ in combination, then use it to read the Sun's surface temperature backwards from its brightness.

Step 1. Assemble the radiation constant from fundamental constants: $a = \dfrac{\pi^2 k_B^4}{15\hbar^3 c^3}$ . Plugging $k_B = 1.381\times10^{-23}$ , $\hbar = 1.055\times10^{-34}$ , $c = 2.998\times10^8$ gives $a = \boxed{7.566\times10^{-16}\ \mathrm{J\,m^{-3}\,K^{-4}}}$ .

Step 2. The flux constant is $\sigma = ac/4 = (7.566\times10^{-16})(2.998\times10^8)/4 = \boxed{5.670\times10^{-8}\ \mathrm{W\,m^{-2}\,K^{-4}}}$ — the tabulated value, recovered with no fitting.

Step 3. The Sun's photosphere is $T = 5778$ K. Its emitted flux is $j = \sigma T^4 = (5.670\times10^{-8})(5778)^4$ . Now $(5778)^4 = 1.115\times10^{15}$ , so $j = \boxed{6.3\times10^{7}\ \mathrm{W\,m^{-2}}}$ — 63 megawatts off every square metre of the solar surface.

Step 4. The steepness is the point: a sunspot at $4000$ K radiates $(4000/5778)^4 = 0.23$ as much per unit area — which is why sunspots look dark against the photosphere despite being white-hot. The fourth power turns a 30 % temperature drop into a fourfold dimming.

2 · Radiation pressure inside the Sun and in the cosmic microwave background (exam level)

Now we compare radiation pressure to gas pressure where it matters — a stellar core — and compute the energy density of the oldest light in the Universe.

Step 1. At the Sun's centre $T \approx 1.5\times10^7$ K. The radiation pressure is $P_{\rm rad} = \tfrac13 aT^4 = \tfrac13(7.566\times10^{-16})(1.5\times10^7)^4$ . With $(1.5\times10^7)^4 = 5.06\times10^{28}$ , this is $P_{\rm rad} = \boxed{1.3\times10^{13}\ \mathrm{Pa}}$ — about $10^8$ atmospheres of pressure from light alone.

Step 2. The gas (ideal) pressure there is $\sim 2\times10^{16}$ Pa, so radiation contributes a percent-level fraction in the Sun — but the $T^4$ scaling means that in a star ten times more massive, with a hotter core, radiation pressure dominates and sets the Eddington luminosity ceiling.

Step 3. For the cosmic microwave background, $T = 2.725$ K. The energy density is $u = aT^4 = (7.566\times10^{-16})(2.725)^4$ . With $(2.725)^4 = 55.1$ , $u = \boxed{4.2\times10^{-14}\ \mathrm{J\,m^{-3}}}$ .

Step 4. The corresponding radiation pressure is $P = u/3 = 1.4\times10^{-14}$ Pa — utterly negligible today, but in the hot early Universe (with $u \propto T^4$ and $T$ thousands of times larger) it was the dominant term in the cosmic equation of state, and its $\gamma = 4/3$ adiabat governed the expansion during the radiation-dominated era.

Origins · 1 minradiation pressure and how this was discovered

radiation pressure

momentum flux of light · Maxwell 1873 → Lebedev/Nichols–Hull 1901

That light carries momentum, and hence pressure, was predicted by Maxwell from electromagnetic theory in 1873 — the pressure equals the energy density over $c$ for a directed beam, and $u/3$ for isotropic radiation. It was measured only in 1901, by Pyotr Lebedev in Moscow and independently by Ernest Nichols and Gordon Hull in the United States, using delicate torsion balances against the pressure of a light beam. The same $P = u/3$ propels solar sails today and, scaled up by $T^4$ , holds massive stars against their own gravity.

History

1873

— Maxwell predicts radiation pressure from electromagnetic theory: an isotropic field exerts P = u/3.

1879

— Stefan infers the T⁴ law empirically from measurements (including Tyndall's platinum-filament data).

1884

— Boltzmann derives the T⁴ law thermodynamically from P = u/3 and the second law, fixing the exponent.

1900

— Planck supplies the constant a via the quantum hypothesis, completing the microscopic picture.

1901

— Lebedev, and Nichols and Hull, measure radiation pressure directly with torsion balances.

1990

— COBE/FIRAS confirms the CMB as a 2.725 K blackbody, the most perfect ever measured.

Going further

References. Pathria & Beale, Statistical Mechanics, 3rd ed., Ch. 7 (The Theory of the Photon Gas) · Kardar, Statistical Physics of Particles, Ch. 7 · Reif, Fundamentals of Statistical and Thermal Physics, Ch. 9 · Landau & Lifshitz, Statistical Physics Part 1, §63 (Black-body radiation) · Sethna, Statistical Mechanics: Entropy, Order Parameters, and Complexity, Ch. 7.

Provenance. Equations and constants standard in Pathria & Beale Ch. 7, Kardar Ch. 7, Landau–Lifshitz §63. A verifier should recompute: (i) the Bose integral $\int_0^\infty x^3/(e^x-1)\,\mathrm{d}x = \Gamma(4)\zeta(4) = \pi^4/15$ ; (ii) $a = \pi^2 k_B^4/(15\hbar^3 c^3) = 7.566\times10^{-16}$ and $\sigma = ac/4 = 5.670\times10^{-8}$ (CODATA); (iii) the integration-by-parts identity $\Omega = -U/3$ yielding $P = u/3$ ; (iv) the Wien peak $\hbar\omega_{\max} \approx 2.82\,k_BT$ . All confirmed in verify_stefan-boltzmann-radiation-pressure.py. CMB values use $T = 2.725$ K.

Builds on: Photon Gas and Planck Radiation Law · Density of States for Free Particles · Bose-Einstein Condensation in the Ideal Gas · aq-qm-blackbody-radiation-planck-quantum. Leads to: Thermodynamics of the BEC Transition · Degeneracy Pressure of Fermions.

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

It is exactly zero as a matter of thermodynamic principle, not approximation. Photon number is not conserved — cavity walls absorb and re-emit them freely — so at fixed $T$ and $V$ the system is free to adjust $N$ to minimize its free energy. Equilibrium is reached at $(\partial F/\partial N)_{T,V} = 0$ , and that derivative is the chemical potential. A nonzero $\mu$ would require an externally enforced number conservation, which a plain cavity does not have. (Engineered photon condensates in dye-filled microcavities create an effective conservation and can behave differently, but that is a special driven system, not equilibrium blackbody radiation.)

A photon gas and a monatomic ideal gas have the same energy density u. Which exerts more pressure?

The monatomic gas, by a factor of two: it has $P = \tfrac23 u$ while radiation has $P = \tfrac13 u$ . The difference is entirely kinematic. Both pressures come from the same geometric average $\langle\cos^2\theta\rangle = \tfrac13$ of momentum transfer over incidence angles, but the momentum itself differs: a non-relativistic particle carries $p = mv$ with $\varepsilon = p^2/2m$ , whereas a photon carries $p = \varepsilon/c$ . Tracking that difference through the momentum-flux integral produces the $2/3$ versus $1/3$ . It is the same watershed you meet for an ultrarelativistic Fermi gas in aq-sm-degeneracy-pressure-fermions.

Where does the factor c∕4 in the flux law j = (c∕4)u come from?

Photons stream isotropically at speed $c$ , but only those moving outward through the hole, and only the component of their velocity along the outward normal, carry energy across the surface. Integrating $c\cos\theta$ over the outgoing hemisphere and dividing by the full $4\pi$ solid angle gives $\int_{\rm hemi}\cos\theta\,\mathrm{d}\Omega/4\pi = 1/4$ . So the one-sided energy flux is $(c/4)u$ , and combining with $u = aT^4$ gives $j = \sigma T^4$ with $\sigma = ac/4$ . The $c/4$ is a pure geometry factor, identical to the effusion result in kinetic theory.

Boltzmann derived T⁴ without quantum mechanics — so what did Planck actually add?

Boltzmann's thermodynamic argument fixes the exponent 4 (from the equation of state $P = u/3$ and the second law) but leaves the constant $a$ completely undetermined — thermodynamics has no way to compute an absolute number of that kind. Planck's quantum hypothesis supplied the microscopic spectrum, and integrating it gives $a = \pi^2 k_B^4/(15\hbar^3 c^3)$ in terms of fundamental constants. So the two results are complementary: thermodynamics gives the shape of the law, quantum statistics gives its size. Historically Planck's constant $h$ was first extracted from fitting $\sigma$ to blackbody data.

Why does radiation pressure become important only in massive stars, if it scales the same T⁴ everywhere?

Because gas pressure and radiation pressure scale differently with temperature and density. Ideal-gas pressure is $P_{\rm gas} = nk_BT$ (linear in $T$ ), while radiation pressure is $P_{\rm rad} = \tfrac13 aT^4$ (fourth power). Their ratio $P_{\rm rad}/P_{\rm gas} \propto T^3/n$ grows steeply with temperature and falls with density. Massive stars have hotter, less dense cores, so this ratio can approach or exceed one, and radiation pressure comes to dominate the support against gravity — which is exactly what sets the Eddington luminosity limit and the upper mass cutoff for stable stars.

If the photon number is fixed by temperature, is it conserved during an adiabatic expansion of the CMB?

Yes, to an excellent approximation. In a reversible adiabatic expansion the entropy $S = \tfrac43 aVT^3$ is constant, so $VT^3 = \text{const}$ . But the photon number is $N \propto VT^3$ as well, so it too stays fixed — the photons already present simply redshift together as the volume grows and the temperature falls. This is why the CMB retains a blackbody spectrum as the Universe expands: no photons are created or destroyed, each just stretches, and a redshifted blackbody is still a (cooler) blackbody.

The photon C_V grows as T³ forever — doesn't that violate the third law or run away?

No. The third law requires $C_V \to 0$ as $T \to 0$ , and $C_V = 4aVT^3 \to 0$ obeys it beautifully — in fact it vanishes faster than a metal's linear electronic term. The absence of a high-temperature plateau is the honest physics: a solid has a finite number of vibrational modes ( $3N$ ) so its phonon heat capacity saturates at Dulong–Petit, but a cavity has infinitely many photon modes with no upper frequency cutoff, so heating forever finds new modes to populate. There is no runaway — $C_V \propto T^3$ is perfectly finite at any finite $T$ ; it simply never levels off.