Blackbody Radiation & Planck's Law

Every warm body glows. Classical physics — Maxwell's waves plus statistical mechanics — makes a clean, parameter-free prediction for the glow's spectrum, and the prediction is not just wrong but infinitely wrong. Planck's 1900 repair, restricting each radiation mode to energies $E_n = nh\nu$, is the first appearance of $h$ in physics and the opening shot of the quantum revolution. A cat on a radiator has always known thermal physics matters; here we learn exactly how much.

Learning Objectives

After this lesson you will be able to:

1. Explain why the cavity radiator realizes an ideal blackbody, and state the Stefan–Boltzmann and Wien displacement laws with correct constants.
2. Derive the Rayleigh–Jeans law from standing-wave mode counting plus equipartition, and explain the ultraviolet catastrophe.
3. Derive Planck's law from the hypothesis $E_n = nh\nu$ via Boltzmann-weighted geometric sums, and recover both limits.
4. Obtain the Stefan–Boltzmann constant and Wien's constant by integrating and maximizing Planck's law.
5. Compute blackbody spectra numerically and apply them to real sources (the Sun, the CMB).
6. State Kirchhoff's law of thermal radiation and use it to explain why the blackbody spectrum depends on temperature alone.

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Intuition

Heat a poker: first infrared, then dull red, then orange, then white. Two facts are universal — hotter bodies radiate more (total power grows steeply with $T$) and radiate bluer (the peak shifts up in frequency). For an ideal absorber the entire spectrum depends on temperature alone, not on material — so it probes something fundamental about radiation itself.

The classical picture is seductive. The field in a hot cavity is a collection of standing-wave modes — exactly the modes of P.1.1 — and equipartition assigns each mode average energy $k_BT$. But a cavity supports ever more modes at higher frequency, without bound, so classical physics predicts infinite energy density, diverging in the ultraviolet. Planck's fix: a mode of frequency $\nu$ holds energy only in lumps of $h\nu$. High-frequency modes, whose lump dwarfs $k_BT$, freeze out — the spectrum bends over and the total is finite.

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Theory

Light as an electromagnetic wave

Maxwell's equations in vacuum make each field component obey the wave equation of P.1.1, $\nabla^2E = c^{-2}\,\partial_t^2E$ with $c = 1/\sqrt{\mu_0\varepsilon_0} = 2.998\times10^8\ \mathrm{m\,s^{-1}}$ and $\lambda\nu = c$. Light is transverse, with two independent polarizations per propagation direction — remember that factor of 2. The spectrum runs from radio ($\nu\sim10^6$ Hz) through visible ($4$–$7.5\times10^{14}$ Hz, 750–400 nm) to gamma rays. Accelerating charges radiate; thermally agitated charges in matter emit a continuous spectrum: thermal radiation.

Blackbodies and the cavity radiator

The spectral radiancy $R_T(\nu)\,d\nu$ is the power emitted per unit area in $[\nu,\nu+d\nu]$. Real surfaces emit a fraction $\epsilon\le1$ (emissivity) of the ideal maximum; a blackbody is the ideal $\epsilon=1$ absorber-and-emitter. The lab realization is a cavity with a small hole: entering radiation rattles around and is absorbed before escaping, so the hole absorbs everything — it is a blackbody — and its leakage samples the equilibrium energy density $u(\nu,T)$ inside. A kinetic-theory angle average relates the two: $R_T(\nu) = \tfrac{c}{4}\,u(\nu,T)$.

Kirchhoff's law: why the spectrum is universal

Why should the glow depend on temperature alone and not on the material? Kirchhoff (1859) answered before anyone knew the spectrum. Let the spectral irradiance $I_\nu$ be the power per unit area and unit frequency falling on a surface, and its spectral absorptance $a_\nu$ ($0 \le a_\nu \le 1$) the fraction it absorbs. Place any body inside a closed cavity whose walls are held at temperature $T$, and wait. At equilibrium the body must emit, in every frequency band, exactly what it absorbs in that band: $R_\nu = a_\nu\,I_\nu$. The balance has to hold band by band, not just in total. Otherwise a filter passing only one band would let the body warm or cool spontaneously against an equal-temperature cavity, a violation of the second law. (The same argument works direction by direction and polarization by polarization.) The irradiance inside the cavity is the same whatever body sits in it, so Kirchhoff's law follows: the ratio $R_\nu/a_\nu$ is a universal function of $\nu$ and $T$, identical for every material. A blackbody has $a_\nu = 1$, so its emission is that universal function, $R^{\text{bb}}_\nu(T) = I_\nu(T) = \tfrac{c}{4}\,u(\nu,T)$. Two corollaries:

• Good absorbers are good emitters: a surface's emissivity at each frequency equals its absorptance, $\epsilon_\nu = a_\nu$. Polished silver, which absorbs little, glows feebly when red hot; soot glows brightly.
• Every real body radiates $\epsilon_\nu$ times the blackbody spectrum at its temperature, so finding $u(\nu,T)$ once describes thermal radiation from everything. That universal function is what Rayleigh, Jeans and Planck set out to derive.

The empirical laws

Two facts were nailed down before 1900. Stefan–Boltzmann (1879/84): the total radiancy is

$$R_T = \sigma T^4, \qquad \sigma = 5.670\times10^{-8}\ \mathrm{W\,m^{-2}\,K^{-4}} .$$

Wien displacement (1893): the peak of the wavelength spectrum obeys

$$\lambda_{max}\,T = 2.898\times10^{-3}\ \mathrm{m\,K} .$$

A 5800 K surface (the Sun) peaks near 500 nm, mid-visible; your 310 K body peaks near 9.4 μm.

Rayleigh–Jeans: counting modes, feeding them equipartition

Take a cubic cavity of side $a$ ($V = a^3$) with conducting walls, so fields vanish at the walls — the 3D standing waves of P.1.1, $E \propto \sin(n_x\pi x/a)\sin(n_y\pi y/a)\sin(n_z\pi z/a)$ with $k_i = n_i\pi/a$, $n_i = 1,2,3,\dots$ The dispersion relation $|\mathbf k| = 2\pi\nu/c$ gives

$$n_x^2+n_y^2+n_z^2 = \left(\frac{2a\nu}{c}\right)^{\!2} \equiv r^2 .$$

Modes are lattice points in the positive octant of $\mathbf n$-space (sign flips give the same standing wave). Counting points below frequency $\nu$ as the octant volume of a sphere of radius $r$, times 2 for polarization:

$$N(<\nu) = \underbrace{\tfrac18}_{\text{octant}}\cdot\tfrac43\pi\left(\frac{2a\nu}{c}\right)^{\!3}\cdot\underbrace{2}_{\text{polarizations}} = \frac{8\pi V\nu^3}{3c^3} \quad\Longrightarrow\quad \frac{1}{V}\frac{dN}{d\nu} = \frac{8\pi\nu^2}{c^3} .$$

Classically each mode is a harmonic degree of freedom in equilibrium, and equipartition (Boltzmann-weighting over a continuum of energies, $\langle E\rangle = \int_0^\infty E e^{-E/k_BT}dE \big/ \int_0^\infty e^{-E/k_BT}dE$) assigns it $\langle E\rangle = k_BT$ independent of $\nu$, with $k_B = 1.381\times10^{-23}\ \mathrm{J\,K^{-1}}$. Modes × energy-per-mode is the Rayleigh–Jeans law:

$$u(\nu,T) = \frac{8\pi\nu^2}{c^3}\,k_BT .$$

It fits experiment at low frequency — and the total energy diverges: $\int_0^\infty u\,d\nu \propto \int_0^\infty \nu^2\,d\nu = \infty$. Every oven should contain infinite energy, mostly beyond the ultraviolet: the ultraviolet catastrophe. Nothing here is sloppy — the mode counting survives into quantum theory unchanged, and equipartition is a theorem. Classical physics itself is wrong.

Planck's hypothesis and the quantized average energy

Planck (1900) kept the mode counting and changed the energy assignment: a mode of frequency $\nu$ may only hold

$$E_n = n\,h\nu, \qquad n = 0,1,2,\dots, \qquad h = 6.626\times10^{-34}\ \mathrm{J\,s} .$$

Boltzmann factors still weight each allowed energy, but integrals become sums. With $x \equiv h\nu/k_BT$ and $q \equiv e^{-x}$, both sums are geometric:

$$Z = \sum_{n=0}^{\infty} e^{-nx} = \frac{1}{1-e^{-x}}, \qquad \sum_{n=0}^{\infty} n\,e^{-nx} = -\frac{dZ}{dx} = \frac{e^{-x}}{(1-e^{-x})^2},$$$$\langle E\rangle = \frac{\sum_n nh\nu\,e^{-nx}}{\sum_n e^{-nx}} = h\nu\,\frac{e^{-x}}{1-e^{-x}} = \frac{h\nu}{e^{h\nu/k_BT}-1} .$$

Multiplying by the unchanged mode density gives Planck's law:

$$\boxed{\,u(\nu,T) = \frac{8\pi\nu^2}{c^3}\,\frac{h\nu}{e^{h\nu/k_BT}-1}\,}$$

Low frequency ($h\nu\ll k_BT$): $e^x - 1 \approx x$, so $\langle E\rangle \to k_BT$ — Rayleigh–Jeans recovered exactly where it worked. High frequency ($h\nu\gg k_BT$): $\langle E\rangle \approx h\nu\,e^{-h\nu/k_BT}$, the exponential Wien form $u \propto \nu^3 e^{-h\nu/k_BT}$. Modes whose quantum exceeds the thermal budget are almost never excited: discreteness starves the ultraviolet.

The demo plots the same physics per unit wavelength, which moves the peak (the caution under Wien displacement below explains why), with the Rayleigh–Jeans curve as an optional overlay:

Caution. Planck quantized the energies of the cavity modes (equivalently, of the material oscillators exchanging energy with them) — he did not claim free light is made of particles. Radiation in flight was still, for Planck, a classical wave. The bolder step — the photon — is Einstein's (1905), next course: P.3.1 Light as Particles.

Stefan–Boltzmann from Planck

Integrate over all frequencies with $x = h\nu/k_BT$, $d\nu = (k_BT/h)\,dx$:

$$u_{tot}(T) = \frac{8\pi h}{c^3}\left(\frac{k_BT}{h}\right)^{\!4}\int_0^\infty\!\frac{x^3}{e^x-1}\,dx = \frac{8\pi^5k_B^4}{15\,h^3c^3}\,T^4,$$

using $\int_0^\infty x^3/(e^x-1)\,dx = \pi^4/15$. With $R_T = (c/4)\,u_{tot}$:

$$R_T = \sigma T^4, \qquad \sigma = \frac{2\pi^5k_B^4}{15\,h^3c^2} = 5.670\times10^{-8}\ \mathrm{W\,m^{-2}\,K^{-4}} .$$

The empirical $T^4$ law now follows from first principles. Historically the logic ran in reverse: Planck fitted his law to the data and extracted the first accurate values of both $h$ and $k_B$.

Wien displacement from Planck

Maximize $u \propto x^3/(e^x-1)$ over $x = h\nu/k_BT$: setting the derivative to zero, $3x^2(e^x-1) = x^3e^x$, i.e.

$$x = 3\left(1 - e^{-x}\right),$$

a transcendental equation with nonzero root $x \approx 2.821$. Hence $\nu_{max}/T = 2.821\,k_B/h = 5.88\times10^{10}\ \mathrm{Hz\,K^{-1}}$.

Caution. The peak of the wavelength distribution is not at $c/\nu_{max}$. Since $u(\lambda)|d\lambda| = u(\nu)|d\nu|$ with Jacobian $c/\lambda^2$, the $\lambda$-form is $u(\lambda,T) = (8\pi hc/\lambda^5)[e^{hc/\lambda k_BT}-1]^{-1}$, whose maximum solves $x = 5(1-e^{-x})$, root $x \approx 4.965$ — giving the familiar $\lambda_{max}T = hc/(4.965\,k_B) = 2.898\times10^{-3}\ \mathrm{m\,K}$. Same spectrum, different binning; always say which variable you maximized over.

Looking ahead: a cavity mode is a harmonic oscillator

The ladder $E_n = nh\nu$ is no accident: each electromagnetic mode is mathematically a harmonic oscillator, whose exact quantum levels are $E_n = (n+\tfrac12)h\nu$ — the ladder returns, with a zero-point offset, in P.5.3 The Harmonic Oscillator. The same quantized-mode structure, engineered in microwave resonators coupled to artificial atoms, is circuit quantum electrodynamics — the physics of superconducting quantum processors (Term 4.4).

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Worked Examples

Example 1 — The Sun as a blackbody

Solar surface: $T \approx 5800$ K, radius $R_\odot = 6.96\times10^{8}$ m.

Peak: $\lambda_{max} = 2.898\times10^{-3}/5800 = 5.00\times10^{-7}$ m $= 500$ nm — mid-visible, matching the evolution of eyes. Flux: $R_T = \sigma T^4 = 5.670\times10^{-8}\times(5800)^4 = 6.42\times10^{7}\ \mathrm{W\,m^{-2}}$. Luminosity: $L = 4\pi R_\odot^2R_T = 4\pi(6.96\times10^8)^2\times6.42\times10^7 = 3.9\times10^{26}$ W — the accepted solar output, from two constants and a temperature.

Example 2 — Why high-frequency modes freeze out

At $T = 300$ K, $k_BT = 4.14\times10^{-21}$ J $= 0.0259$ eV.

Radio mode, $\nu = 1$ GHz: $h\nu = 6.63\times10^{-25}$ J, so $x = 1.6\times10^{-4}$ and $\langle E\rangle = h\nu/(e^x-1) \approx k_BT(1-x/2) = 0.99992\,k_BT$ — equipartition holds to 0.01%; Rayleigh–Jeans is excellent here.

Visible mode, $\nu = 5\times10^{14}$ Hz: $h\nu = 3.31\times10^{-19}$ J $= 2.07$ eV, so $x = 80$ and $\langle E\rangle \approx h\nu\,e^{-80} \approx 6\times10^{-54}$ J — below equipartition by a factor $\sim10^{33}$. The mode exists, but the bath almost never scrapes together its first quantum. That suppression is the resolution of the catastrophe — and why a warm room is dark.

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Hands-on (Python)

import numpy as np
import matplotlib.pyplot as plt
from scipy.integrate import quad
from scipy.optimize import brentq

h, c, kB = 6.626e-34, 2.998e8, 1.381e-23        # J s, m/s, J/K

def planck(nu, T):
    """Spectral energy density u(nu,T) in J m^-3 Hz^-1."""
    return (8 * np.pi * h * nu**3 / c**3) / np.expm1(h * nu / (kB * T))

def rayleigh_jeans(nu, T):
    return 8 * np.pi * nu**2 * kB * T / c**3

def wien_approx(nu, T):
    return (8 * np.pi * h * nu**3 / c**3) * np.exp(-h * nu / (kB * T))

# --- 1. Planck vs Rayleigh-Jeans vs Wien at T = 5800 K -------------------
T = 5800.0
nu = np.linspace(1e12, 2.5e15, 2000)
plt.plot(nu, planck(nu, T), label="Planck")
plt.plot(nu, rayleigh_jeans(nu, T), "--", label="Rayleigh–Jeans (diverges)")
plt.plot(nu, wien_approx(nu, T), ":", label="Wien approximation")
plt.ylim(0, 1.8 * planck(nu, T).max()); plt.xlabel(r"$\nu$ (Hz)")
plt.ylabel(r"$u(\nu,T)$"); plt.legend(); plt.title("T = 5800 K"); plt.show()
# Expected: RJ hugs Planck at low nu then blows up; Wien matches the high-nu tail.

# --- 2. Verify the T^4 law and extract sigma ------------------------------
for T in [300.0, 1000.0, 3000.0, 5800.0]:
    nu_max = 100 * kB * T / h              # integrand is dead beyond x = 100
    u_tot, _ = quad(planck, 1.0, nu_max, args=(T,))
    print(f"T = {T:6.0f} K   sigma = {(c/4) * u_tot / T**4:.4e}")
# Expected: sigma = 5.676e-08 W m^-2 K^-4 for EVERY T -> R_T = sigma T^4.
# (CODATA: 5.670e-08; the 0.1% offset is rounding in our 4-digit h, kB.)

# --- 3. Wien displacement via brentq, applied to the CMB ------------------
x_nu  = brentq(lambda x: x - 3 * (1 - np.exp(-x)), 0.1, 10)
x_lam = brentq(lambda x: x - 5 * (1 - np.exp(-x)), 0.1, 10)
print(x_nu, x_lam)                                   # 2.8214...  4.9651...
print(f"Wien constant = {h*c/(x_lam*kB):.4e} m K")   # 2.898e-03  (verified)

T_cmb = 2.725                                        # cosmic microwave background
print(f"CMB peak = {x_nu * kB * T_cmb / h / 1e9:.1f} GHz")   # ~160.2 GHz

The CMB — the 2.725 K afterglow of the Big Bang — is the most perfect blackbody ever measured; COBE found deviations from Planck's law below 50 parts per million.

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Exercises

E1 (easy). Your body: $T = 310$ K, area $A = 1.7\ \mathrm{m^2}$, $\epsilon \approx 1$ in the infrared. Find $\lambda_{max}$ and the total power you radiate.

Solution

$\lambda_{max} = 2.898\times10^{-3}/310 = 9.35\ \mu$m (far infrared — thermal cameras look here). $P = \sigma T^4A = 5.670\times10^{-8}\times(310)^4\times1.7 \approx 890$ W. You also absorb $\sigma T_{env}^4A \approx 690$ W from a 293 K room; the net $\sim$150–200 W is your metabolic scale.

E2 (easy). Expand Planck's $\langle E\rangle$ for $h\nu \ll k_BT$ one order beyond equipartition and show $\langle E\rangle \approx k_BT - \tfrac12h\nu$.

Solution

With $x \ll 1$: $e^x-1 = x(1 + x/2 + O(x^2))$, so $\langle E\rangle = \frac{h\nu}{x}(1+x/2)^{-1} \approx k_BT(1 - x/2) = k_BT - \tfrac12h\nu$. Quantization always lowers the average below equipartition, more at higher frequency.

E3 (medium). Reproduce the geometric-sum derivation of $\langle E\rangle = h\nu/(e^{h\nu/k_BT}-1)$ without looking, evaluating both sums explicitly.

Solution

Let $x = h\nu/k_BT$, $q = e^{-x} < 1$. Partition sum $Z = \sum_{n\ge0}q^n = 1/(1-q)$; energy sum $\sum_n nh\nu\,q^n = h\nu\,q\,\frac{d}{dq}\frac{1}{1-q} = h\nu\,\frac{q}{(1-q)^2}$. Ratio: $\langle E\rangle = h\nu\frac{q}{1-q} = \frac{h\nu}{e^x-1}$. Limits: $k_BT$ as $x\to0$; $h\nu e^{-x}$ as $x\to\infty$.

E4 (medium). Compute the CMB's total energy density $u_{tot} = \frac{4\sigma}{c}T^4$ at $T = 2.725$ K, in $\mathrm{J\,m^{-3}}$ and $\mathrm{eV\,cm^{-3}}$.

Solution

$a \equiv 4\sigma/c = 7.566\times10^{-16}\ \mathrm{J\,m^{-3}K^{-4}}$ and $T^4 = 55.1$, so $u_{tot} = 4.17\times10^{-14}\ \mathrm{J\,m^{-3}} = 2.60\times10^{5}\ \mathrm{eV\,m^{-3}} = 0.26\ \mathrm{eV\,cm^{-3}}$. Every cubic centimeter of the universe holds about a quarter electron-volt of primordial light.

E5 (hard). Derive the $\lambda$-form of Planck's law from the $\nu$-form, maximize it to get $x = 5(1-e^{-x})$, and compute Wien's constant. Then explain in one sentence why $\lambda_{max}\nu_{max} \ne c$.

Solution

Equal energy in corresponding bins: $u(\lambda) = u(c/\lambda)\,|d\nu/d\lambda| = u(c/\lambda)\,c/\lambda^2 = \frac{8\pi hc}{\lambda^5}\frac{1}{e^{hc/\lambda k_BT}-1}$. With $x = hc/\lambda k_BT$, maximizing $x^5/(e^x-1)$ gives $5x^4(e^x-1) = x^5e^x$, i.e. $x = 5(1-e^{-x})$, root $x \approx 4.965$; then $\lambda_{max}T = \frac{hc}{4.965\,k_B} = \frac{(6.626\times10^{-34})(2.998\times10^{8})}{4.965\times1.381\times10^{-23}} = 2.898\times10^{-3}\ \mathrm{m\,K}$. ✓ The two peaks differ because maximizing a density depends on the variable it is a density in — the Jacobian $c/\lambda^2$ reweights the spectrum, so $x_\lambda = 4.965 \ne x_\nu = 2.821$.

E6 (easy). Stefan's $T^4$ law (1879) began with data published by Tyndall: a platinum wire at 1200 °C radiated about 11.7 times as much as at 525 °C. (a) Show that a $T^4$ law fits, and what it would have predicted had the temperatures been used in Celsius. (b) Platinum is far from black. What does the agreement silently assume, by Kirchhoff's law?

Solution

(a) In kelvin the ratio is $1473/798 = 1.846$, and $1.846^4 = 11.6$, within 1% of the measured 11.7. In Celsius, $(1200/525)^4 = 2.29^4 = 27$, wildly off: only an absolute temperature can enter a power law. (b) A real surface radiates $R = \epsilon\sigma T^4$, so the measured ratio is $(\epsilon_{1473}/\epsilon_{798})(1473/798)^4$. The fit assumes the platinum's emissivity was the same at both temperatures, which is not true of metals in general. The $T^4$ law was put on firm ground later: Boltzmann derived it from thermodynamics in 1884, and cavity measurements in the 1890s confirmed it for true blackbody radiation.

E7 (easy). Two stars have their spectral peaks (per unit wavelength) at 510 nm and 350 nm. (a) Find their surface temperatures. (b) Find the ratio of the power each radiates per unit area of surface. (c) Can you tell which star is more luminous?

Solution

(a) $T = b/\lambda_{max}$: $2.898\times10^{-3}/510\times10^{-9} = 5.7\times10^3$ K and $2.898\times10^{-3}/350\times10^{-9} = 8.3\times10^3$ K. The bluer star is hotter. (b) $R_T \propto T^4$: $(8280/5682)^4 = 1.457^4 = 4.5$. (c) No. Luminosity is $L = 4\pi R^2\sigma T^4$, and the radius is not known: a cool giant easily outshines a hot dwarf. Color gives temperature; it takes a distance or a size as well to get luminosity.

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Checkpoint

1. Why does a small hole in a cavity behave as a perfect blackbody?
2. Derive the mode density $8\pi\nu^2/c^3$, identifying where the octant factor $\tfrac18$ and the polarization factor 2 enter.
3. Which classical ingredient causes the ultraviolet catastrophe — mode counting or equipartition — and how does Planck's hypothesis repair it?
4. Show in two lines how Planck's law reduces to Rayleigh–Jeans for $h\nu\ll k_BT$ and to Wien's exponential for $h\nu\gg k_BT$.
5. What did Planck quantize — and what did he not quantize?
6. State Kirchhoff's law. Why does it make the blackbody spectrum a universal function of $\nu$ and $T$?

Answers

1. Any ray entering the hole reflects many times and is absorbed before escaping, so the hole absorbs essentially everything ($\epsilon = 1$); in equilibrium it must therefore also emit the ideal blackbody spectrum of the cavity interior.
2. Standing waves require $k_i = n_i\pi/a$ with $n_i > 0$; counting lattice points inside the sphere $|\mathbf n| = 2a\nu/c$ uses the positive octant only (factor $\tfrac18$), and each spatial mode has two transverse polarizations (factor 2): $N = 8\pi V\nu^3/3c^3$, whose $\nu$-derivative per volume is $8\pi\nu^2/c^3$.
3. Equipartition. The mode counting survives into QM; the error is giving every mode $k_BT$. With $E_n = nh\nu$, $\langle E\rangle = h\nu/(e^{h\nu/k_BT}-1) \to 0$ exponentially at high $\nu$, so the total energy integral converges.
4. $h\nu\ll k_BT$: $e^{h\nu/k_BT}-1 \approx h\nu/k_BT$, so $u \to (8\pi\nu^2/c^3)k_BT$. $h\nu\gg k_BT$: drop the $-1$, so $u \to (8\pi h\nu^3/c^3)e^{-h\nu/k_BT}$.
5. He quantized the allowed energies of cavity modes / material oscillators ($E_n = nh\nu$) — energy exchange with radiation. He did not quantize free light; the photon is Einstein's step, in P.3.1.
6. In equilibrium a body emits, band by band, what it absorbs: $R_\nu = a_\nu I_\nu$. The irradiance inside an isothermal cavity is the same whatever body sits there, so $R_\nu/a_\nu$ is the same for all materials. A blackbody ($a_\nu = 1$) emits exactly that universal function, and every other body emits a fraction $\epsilon_\nu = a_\nu$ of it.

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Further Reading

• [ER] Eisberg & Resnick, §1.1–1.6 — thermal radiation, cavity mode counting, and Planck's postulate; the definitive elementary treatment.
• [ER] Eisberg & Resnick, Appendix to Ch. 1 — the equipartition theorem and its breakdown.
• [Gri] Griffiths & Schroeter, §5.4 — the photon gas and the blackbody spectrum, revisited with proper quantum statistics.

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