Part 5 · The old quantum theory

5.2 Rayleigh–Jeans and Planck

Source: Postorino, week 4 slides, pp. 19–21

Objectives

Recipe. Mean energy from the partition function: $\overset{ε}{ˉ} = - \frac{\partial}{\partial β} ln (sum or integral of e^{- βε})$ .
Theorem. Continuous energies give $\overset{ε}{ˉ} = K_{B} T$ and the Rayleigh–Jeans law, which diverges at short wavelengths. Energies in steps $ε_{0} = h ν$ give $\overset{ε}{ˉ} = \frac{h ν}{e ^{h ν / K_{B} T} - 1}$ and Planck’s law.
Picture. A mode whose energy step is much larger than $K_{B} T$ is frozen out: that is what kills the ultraviolet catastrophe.
Vocabulary. Equipartition, ultraviolet catastrophe, Planck’s constant.

The classical mean energy

Classically the energy is continuous, $0 \leq ε < \infty$ . With Boltzmann’s probabilities (Proposition 5.1.11) the mean energy of an oscillator is $\overline{ε (T, λ)} = \frac{\int _{0}^{\infty} ε e ^{- βε} d ε}{\int _{0}^{\infty} e ^{- βε} d ε} .$

Theorem 5.2.1Classical mean energyslides w4, p. 19

$\overset{ε}{ˉ} = \frac{1}{β} = K_{B} T .$

Proof

The trick: the numerator is minus the $β$ -derivative of the denominator, and $f^{'} / f = (ln f)^{'}$ . So $\overset{ε}{ˉ} = - \frac{\partial}{\partial β} ln \int_{0}^{\infty} e^{- βε} d ε = - \frac{\partial}{\partial β} ln [- \frac{1}{β} e^{- βε}_{0}^{\infty}] = - \frac{\partial}{\partial β} ln [- \frac{1}{β} (0 - 1)] = - \frac{\partial}{\partial β} ln \frac{1}{β} = \frac{\partial}{\partial β} ln β = \frac{1}{β} .$ The three facts used, boxed on the slide: $\frac{d}{d x} ln [f (x)] = \frac{f ^{'} ( x )}{f ( x )}, \int e^{c x} d x = \frac{1}{c} e^{c x}, \frac{d}{d x} e^{f (x)} = f^{'} (x) e^{f (x)} .$

Key observation 5.2.2Equipartitionslides w4, p. 19

Without calculation, the equipartition theorem gives the same: $\frac{1}{2} K_{B} T$ for each quadratic term in the energy. An oscillator’s energy $\frac{p ^{2}}{2 m} + \frac{1}{2} K x^{2}$ has two quadratic terms, kinetic and potential, so $\overset{ε}{ˉ} = K_{B} T$ .

(The two terms spelled out are not on the slide; compare Exercise 1.3.5, where the two terms have equal averages.)

The ultraviolet catastrophe

Theorem 5.2.3Rayleigh–Jeans lawslides w4, p. 19

$ρ (T, λ) = n (λ) \overline{ε (T, λ)} = \frac{8 π}{λ ^{4}} K_{B} T .$

It works for $λ \to \infty$ : it matches the long-wavelength tail of the data.
It does not work for $λ \to 0$ : it diverges, the ultraviolet catastrophe.

RemarkWhy it divergesbeyond the lectures

Every mode gets the same $K_{B} T$ , however short its wavelength, and there are infinitely many short-wavelength modes (Theorem 5.1.10). The total energy $\int_{0}^{\infty} ρ d λ \propto \int_{0}^{\infty} d λ / λ^{4}$ is infinite.

The slide plots measured intensities at 5000, 4000 and 3000 K, each going to zero at both ends, against the classical theory at 5000 K: it agrees at long wavelengths but shoots off the top below about 500 nm instead of turning down.

Loading the demo…

Figure 5.2.1 The measured blackbody spectrum (Planck’s law,

violet

) against the classical

Rayleigh-Jeans

law at the same temperature. Drag the temperature. The two agree at long wavelengths; at short ones Rayleigh–Jeans runs away while the data turn down to zero.

Planck’s hypothesis

Definition 5.2.4Planck’s hypothesisEssentialslides w4, p. 20

Energy is a discrete variable. It still ranges over $0 \leq ε < \infty$ , but only in steps: $ε_{n} = n ε_{0}, n = 0, 1, 2, \dots, ε_{0} = ε (λ) .$ The step $ε_{0}$ may depend on the wavelength. Comparison with the data fixes it as $ε_{0} = h ν, h = 6.62 \times 1 0^{- 34} J \cdot s (Planck’s constant) .$

Theorem 5.2.5Planck’s mean energyslides w4, p. 20

$\overline{ε (T, λ)} = \frac{ε _{0}}{e ^{β ε_{0}} - 1} .$

Proof

The same trick as in the continuous case, with sums: $\overset{ε}{ˉ} = \frac{\sum _{n = 0}^{\infty} n ε _{0} e ^{- β n ε_{0}}}{\sum _{n = 0}^{\infty} e ^{- β n ε_{0}}} = - \frac{\partial}{\partial β} ln n = 0 \sum \infty e^{- β n ε_{0}} = - \frac{\partial}{\partial β} ln n = 0 \sum \infty [e^{- β ε_{0}}]^{n} .$ But $1/∣ e^{β ε_{0}} ∣ < 1$ , so this is a geometric series: $\sum_{n = 0}^{\infty} x^{n} = \frac{1}{1 - x}$ for $∣ x ∣ < 1$ . Hence $\overset{ε}{ˉ} = - \frac{\partial}{\partial β} ln \frac{1}{1 - e ^{- β ε_{0}}} = \frac{\partial}{\partial β} ln [1 - e^{- β ε_{0}}] = \frac{ε _{0} e ^{- β ε_{0}}}{1 - e ^{- β ε_{0}}} = \frac{ε _{0}}{e ^{β ε_{0}} - 1},$ the last step multiplying top and bottom by $e^{β ε_{0}}$ .

Key observation 5.2.6Why discreteness cures the catastrophebeyond the lectures

The mean energy depends on how the step $ε_{0}$ compares with $K_{B} T$ :

$ε_{0} ≪ K_{B} T$ (long wavelengths): $e^{β ε_{0}} - 1 \approx β ε_{0}$ and $\overset{ε}{ˉ} \approx K_{B} T$ . The steps are too fine to notice: the classical result comes back.
$ε_{0} ≫ K_{B} T$ (short wavelengths): $\overset{ε}{ˉ} \approx ε_{0} e^{- ε_{0} / K_{B} T} \to 0$ . Even the first step costs far more than the thermal energy available, so the mode is almost never excited: it is frozen out.

Loading the demo…

Figure 5.2.2 The mean energy of one oscillator, in units of

K_{B} T

, against its energy step

ε_{0} / K_{B} T

: classical (

1

, always) and Planck’s (

violet

). Drag the step. The ladder on the right shows the levels

n ε_{0}

with bars for their Boltzmann populations: when the step is large, only

n = 0

is occupied.

Planck’s law

Theorem 5.2.7Planck’s lawslides w4, p. 21

With $ε_{0} = h ν = h c / λ$ , $ρ (T, λ) = \frac{8 π}{λ ^{4}} \frac{h c / λ}{e ^{β h c / λ} - 1} = \frac{8 π}{λ ^{5}} \frac{h c}{e ^{β h c / λ} - 1} .$ It works very well, and the ultraviolet catastrophe is gone.

Proposition 5.2.8The two limitsslides w4, p. 21

$λ \to \infty$ : $β h c / λ \to 0$ and $e^{x} \approx 1 + x$ , so $ρ \approx \frac{8 π}{λ ^{5}} \frac{h c}{1 + β h c / λ - 1} = \frac{8 π}{λ ^{5}} \frac{λ}{β} = \frac{8 π}{λ ^{4}} K_{B} T :$ Rayleigh–Jeans, OK.
$λ \to 0$ : $e^{β h c / λ} ≫ 1$ , so $ρ \approx \frac{8 π}{λ ^{5}} h c e^{- β h c / λ} \to 0 :$ the exponential beats the power $λ^{- 5}$ , OK with experiment.

Change of variable $λ \to ν$ . With $λ ν = c$ :

$n (λ) d λ = \frac{8 π}{λ ^{4}} d λ$ is the number of modes per volume with wavelength between $λ$ and $λ + d λ$ ;
$n (ν) d ν$ is the number of modes per volume with frequency between $ν$ and $ν + d ν$ .

Proposition 5.2.9Modes per unit frequencyslides w4, p. 21

They count the same modes, so $n (λ) d λ = n (ν) d ν$ : $n (ν) = \frac{\partial λ}{\partial ν} n (λ) = \frac{c}{ν ^{2}} \frac{8 π ν ^{4}}{c ^{4}} = \frac{8 π ν ^{2}}{c ^{3}} .$ The absolute value is there because $λ$ decreases as $ν$ increases, while numbers of modes are positive.

So, per unit frequency beyond the lectures, $ρ (T, ν) = \frac{8 π ν ^{2}}{c ^{3}} \frac{h ν}{e ^{h ν / K_{B} T} - 1} .$

Theorem 5.2.10Stefan–Boltzmann and Wien from Planckbeyond the lectures

With $x = h c / λ K_{B} T$ :

$R = \frac{c}{4} \int_{0}^{\infty} ρ d λ = \frac{2 π K _{B}^{4} T ^{4}}{h ^{3} c ^{2}} \int_{0}^{\infty} \frac{x ^{3}}{e ^{x} - 1} d x$ , and the integral is $π^{4} /15$ , so $σ = \frac{2 π ^{5} K _{B}^{4}}{15 h ^{3} c ^{2}} = 5.67 \times 1 0^{- 8}$ W m⁻² K⁻⁴.
$d ρ / d λ = 0$ gives $5 (1 - e^{- x}) = x$ , solved by $x = 4.965$ : $λ_{max} T = h c / (4.965 K_{B}) = 2.898 \times 1 0^{- 3}$ m·K.

Proofbeyond the lectures

For Stefan–Boltzmann substitute $λ = h c / (x K_{B} T)$ , $d λ = - \frac{h c}{x ^{2} K _{B} T} d x$ : $\frac{8 πh c}{λ ^{5}} d λ$ becomes $8 πh c (\frac{x K _{B} T}{h c})^{5} \frac{h c}{x ^{2} K _{B} T} d x = 8 π \frac{( K _{B} T ) ^{4}}{( h c ) ^{3}} x^{3} d x$ ; times $c /4$ gives the prefactor. For Wien, $ρ \propto λ^{- 5} (e^{x} - 1)^{- 1}$ with $x \propto 1/ λ$ ; setting the $λ$ -derivative to zero gives $- 5 (e^{x} - 1) + x e^{x} = 0$ , i.e. $5 (1 - e^{- x}) = x$ .

A constant measured with thermometers is computed from $h$ , $c$ and $K_{B}$ .

Exercises

Exercise 5.2.1beyond the lectures

Show that Planck’s mean energy tends to $K_{B} T$ when $ε_{0} ≪ K_{B} T$ , and find the next correction.

Solution

With $x = β ε_{0} ≪ 1$ , $e^{x} - 1 = x + x^{2} /2 + \dots$ , so $\overset{ε}{ˉ} = \frac{ε _{0}}{x ( 1 + x /2 )} \approx K_{B} T (1 - x /2) = K_{B} T - ε_{0} /2$ .

Exercise 5.2.2beyond the lectures

Why does the Rayleigh–Jeans law fail at short wavelengths but not at long ones? One sentence each, classically and with Planck.

Solution

Classically every mode gets $K_{B} T$ , and the number of modes grows as $λ^{- 4}$ , so the energy diverges at short $λ$ . With Planck, a mode of step $h c / λ ≫ K_{B} T$ is almost never excited, so short-wavelength modes carry exponentially little energy; at long $λ$ the steps are tiny and the classical result holds.

Exercise 5.2.3beyond the lectures

At 300 K, what is the mean energy of a mode at 10 µm and at 500 nm, compared with $K_{B} T$ ?

Solution

$K_{B} T = 0.0259$ eV. At 10 µm, $h ν = 0.124$ eV, $x = 4.79$ , $\overset{ε}{ˉ} / K_{B} T = x / (e^{x} - 1) = 0.040$ . At 500 nm, $h ν = 2.48$ eV, $x = 96$ , $\overset{ε}{ˉ} / K_{B} T = 96 e^{- 96} \approx 1 0^{- 40}$ : completely frozen out. That is why a room-temperature body does not glow.

Exercise 5.2.4beyond the lectures

Check that $σ = \frac{2 π ^{5} K _{B}^{4}}{15 h ^{3} c ^{2}}$ gives $5.67 \times 1 0^{- 8}$ .

Solution

$K_{B}^{4} = (1.381 \times 1 0^{- 23})^{4} = 3.64 \times 1 0^{- 92}$ ; $h^{3} = 2.91 \times 1 0^{- 100}$ ; $c^{2} = 8.99 \times 1 0^{16}$ ; $2 π^{5} /15 = 40.8$ . So $σ = 40.8 \times 3.64 \times 1 0^{- 92} / (2.91 \times 1 0^{- 100} \times 8.99 \times 1 0^{16}) = 5.67 \times 1 0^{- 8}$ W m⁻² K⁻⁴.