Planck's Radiation Law

Prerequisites: Statistical Mechanics. Full path (4 pages)

• Intuitive
• Concrete
• Abstract
• Why is it interesting?

Intuitive

Every object glows. A cold object glows mostly in the infrared, which we cannot see. Heat a piece of iron and it starts to glow dark red, then orange, then white. The hotter the object, the more light it gives off in total, and the more of that light comes at higher frequencies. Physicists study this using an idealised "black body", an object that absorbs all light falling on it. The light it emits then depends only on its temperature. A small hole in the wall of a hot oven is a good black body.

Planck's radiation law tells us how much light such a body emits at each frequency, for a given temperature. Classical physics cannot explain this curve. If we treat the light inside an oven as a collection of vibrations of many different frequencies and give each vibration the same average share of the heat energy, then, since there are more and more possible vibrations at higher frequencies, we find that an oven should radiate an infinite amount of energy at high frequencies. This failure is called the ultraviolet catastrophe.

In 1900 Max Planck found the formula that matches the measurements. To derive it, he had to assume that light can only exchange energy with matter in chunks whose size is the frequency times a new constant, now called Planck's constant. At high frequencies a single chunk costs more energy than the thermal motion typically provides, so these vibrations are rarely excited and the catastrophe goes away. This assumption of energy quanta was the first step towards quantum mechanics.

The law describes the light of stars quite well, and the cosmic microwave background follows it with very high precision.

• See Thermodynamics and Quanta in Plancks Work by Klein

Concrete

The spectral energy density of black-body radiation at temperature $T$, that is, the energy per unit volume and per unit frequency interval in a cavity in thermal equilibrium, is

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

with Planck's constant $h$, Boltzmann's constant $k_B$ and the speed of light $c$. The power emitted per unit area of a black surface, per unit frequency and per unit solid angle, is the spectral radiance $B_\nu = \frac{c}{4\pi}u = \frac{2h\nu^3}{c^2}\frac{1}{e^{h\nu/k_BT} - 1}$.

Derivation. The formula is the product of two factors.

1. The number of electromagnetic modes per unit volume with frequency between $\nu$ and $\nu + d\nu$. In a large box the allowed wave vectors $\vec k$ form a lattice with one point per volume $(2\pi)^3/V$ of $k$-space. Counting the points in a spherical shell, using $k = 2\pi\nu/c$, and multiplying by $2$ for the two polarizations gives $\frac{8\pi\nu^2}{c^3}\,d\nu$.
2. The mean thermal energy of one mode. Planck's assumption is that a mode of frequency $\nu$ can only have the energies $E_n = n h\nu$, $n = 0, 1, 2, \dots$ With the Boltzmann weights $e^{-\beta E_n}$, $\beta = 1/k_BT$, the partition function is a geometric series,

$$Z = \sum_{n=0}^\infty e^{-n\beta h\nu} = \frac{1}{1 - e^{-\beta h\nu}}, \qquad \langle E\rangle = -\frac{\partial\ln Z}{\partial\beta} = \frac{h\nu}{e^{\beta h\nu} - 1}.$$

Multiplying the two factors gives $u(\nu,T)$. (In quantum mechanics each mode is a harmonic oscillator with energies $(n + \frac{1}{2})h\nu$. The zero-point energy is the same at every temperature and is dropped here.)

Limits. For $h\nu \ll k_BT$ we expand the exponential and get $\langle E\rangle \approx k_BT$, the classical equipartition value. Then $u \approx 8\pi\nu^2 k_BT/c^3$ (the Rayleigh-Jeans law), whose integral over all frequencies diverges. For $h\nu \gg k_BT$ we get $u \approx \frac{8\pi h\nu^3}{c^3}e^{-h\nu/k_BT}$ (Wien's law): the high-frequency modes are exponentially suppressed, and the total energy is finite.

Consequences. With $x = h\nu/k_BT$ and $\int_0^\infty \frac{x^3\,dx}{e^x - 1} = \frac{\pi^4}{15}$ the total energy density is

$$u(T) = \int_0^\infty u(\nu,T)\,d\nu = \frac{8\pi^5 k_B^4}{15 h^3 c^3}\,T^4,$$

and the power radiated per unit area of a black surface is $\frac{c}{4}u(T) = \sigma T^4$ with the Stefan-Boltzmann constant $\sigma = \frac{2\pi^5 k_B^4}{15h^3c^2} \approx 5.67\times 10^{-8}\,\text{W}\,\text{m}^{-2}\,\text{K}^{-4}$. The maximum of the spectrum per unit wavelength sits at $\lambda_{\max} T \approx 2.90\times 10^{-3}\,\text{m}\,\text{K}$ (Wien's displacement law). For the surface of the Sun, $T \approx 5800\,\text{K}$, this gives $\lambda_{\max} \approx 500\,\text{nm}$, in the visible range. Note that the maximum per unit frequency, at $h\nu_{\max} \approx 2.82\,k_BT$, does not correspond to $c/\lambda_{\max}$, because $u_\nu\,d\nu$ and $u_\lambda\,d\lambda$ are densities with respect to different variables.

• A better presentation of Planck's Radiation Law by Marr and Wilkin

Abstract

In modern language, black-body radiation is the thermal equilibrium state of the free quantized electromagnetic field, a gas of massless bosons with two helicity states. We set $\hbar = c = k_B = 1$. The photon number is not conserved, since photons are absorbed and emitted by the walls, so in the grand canonical ensemble the chemical potential vanishes, $\mu = 0$. The field is a collection of independent harmonic oscillators labelled by $\vec k$ and the helicity $\lambda = \pm 1$, with $\omega = |\vec k|$. The partition function factorises,

$$\ln Z = -2V\int\frac{d^3k}{(2\pi)^3}\,\ln\left(1 - e^{-\beta|\vec k|}\right),$$

and the occupation number of each mode is the Bose-Einstein distribution at $\mu = 0$,

$$\langle n_{\vec k,\lambda}\rangle = \frac{1}{e^{\beta\omega} - 1}.$$

Planck's law is the energy density per unit frequency, $2\cdot\frac{4\pi\omega^2}{(2\pi)^3}\,\omega\,\langle n\rangle = \frac{\omega^3}{\pi^2}\frac{1}{e^{\beta\omega}-1}$, which agrees with $u(\nu,T)$ after $\omega = 2\pi\nu$ and $u_\nu = 2\pi u_\omega$. The thermodynamics of the photon gas follows from $\ln Z$:

$$P = \frac{\pi^2}{45}T^4, \qquad u = 3P = \frac{\pi^2}{15}T^4, \qquad s = \frac{u + P}{T} = \frac{4\pi^2}{45}T^3, \qquad n = \frac{2\zeta(3)}{\pi^2}T^3.$$

The relation $u = 3P$ expresses the vanishing trace of the energy-momentum tensor of the free Maxwell field, $T^\mu{}_\mu = 0$, which holds for the classical field because of its scale (conformal) invariance.

The thermal state singles out a rest frame, given by the four-velocity $U^\mu$ of the heat bath, so it breaks Lorentz invariance. Its energy-momentum tensor has the perfect-fluid form $T^{\mu\nu} = (u + P)U^\mu U^\nu + P\eta^{\mu\nu}$ (signature $(-,+,+,+)$). The occupation numbers are $\langle n\rangle = 1/(e^{-\beta\,U_\mu k^\mu} - 1)$, where $-U_\mu k^\mu$ is the photon energy measured in the rest frame of the bath, so an observer moving relative to the bath sees, in each direction, a Planck spectrum with a direction-dependent temperature. In an expanding Friedmann-Robertson-Walker universe, free-streaming photons redshift as $\omega \propto 1/a$, and a Planck distribution stays a Planck distribution with $T \propto 1/a$. This is why the cosmic microwave background still has a black-body spectrum long after photons stopped interacting with matter.

In the algebraic formulation of quantum field theory, thermal equilibrium states are characterised by the KMS condition with respect to time translations. For the free field this is equivalent to Euclidean time being periodic with period $\beta$, which gives the bosonic Matsubara frequencies $\omega_n = 2\pi n/\beta$ in the imaginary-time formalism of finite-temperature field theory. Summing over $\omega_n$ reproduces the factor $1/(e^{\beta\omega} - 1)$.

Why is it interesting?

Planck's Radiation Law was a crucial step in the historical development of Quantum Mechanics.

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1. Calculus
2. Statistics
3. Entropy
4. Statistical Mechanics