# 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.
---
## 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](../c01-theoretical-mechanics/01-waves-and-the-wave-equation.md) —
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.
---
## Theory
### Light as an electromagnetic wave
Maxwell's equations in vacuum make each field component obey the wave equation of
[P.1.1](../c01-theoretical-mechanics/01-waves-and-the-wave-equation.md),
$\nabla^2E = c^{-2}\,\partial_t^2E$ 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:
```qm-demo blackbody-spectrum
```
> **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](../c03-wave-particle-duality/01-light-as-particles.md).
### 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](../c05-bound-states/03-harmonic-oscillator.md).
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).
---
## 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.
---
## Hands-on (Python)
```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.
---
## 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.
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.
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 $x\to\infty$.
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.
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 $x_\lambda = 4.965 \ne x_\nu = 2.821$.
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.
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.
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](../c03-wave-particle-duality/01-light-as-particles.md).
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.