Part 4 · Probability and statistical physics

4.2 The Boltzmann distribution

Source: Postorino, week 4 slides, p. 16

Objectives

Picture. $N \sim 1 0^{23}$ particles sharing a fixed energy: one way of distributing them over the states is overwhelmingly more likely than all the others.
Theorem. That configuration is $\tilde{n}_{k} = N e^{- β E_{k}} / Z$ , with $Z = \sum_{k} e^{- β E_{k}}$ and $β = 1/ K_{B} T$ .
Recipe. Population ratios $e^{- Δ E / K_{B} T}$ ; mean energy $\overset{ˉ}{E} = - \partial ln Z / \partial β$ .
Vocabulary. Microcanonical ensemble, configuration, partition function, Boltzmann factor.

The setup

An isolated system of $N$ particles, identical and distinguishable, with states of energy $E_{1}, \dots, E_{k}, \dots$ available to them. The states can be discrete, quasi-continuous or continuous.

Take a “snapshot” of the system: $n_{1}$ particles are in state $E_{1}$ , $n_{2}$ in $E_{2}$ , and so on. Then $E_{tot} = k \sum states n_{k} E_{k}, N = k \sum states n_{k} .$ If the particles don’t interact this is an ideal gas, and $E_{tot} = const$ .

Definition 4.2.1Microcanonical ensembleslides w4, p. 16

$E_{tot}$ and $N$ fixed, with identical, distinguishable, non-interacting particles. One state of the whole system is a tuple $(n_{1}, \dots, n_{k}, \dots)$ , a configuration.

Even obeying the two sums, the variety is huge. But for very large $N$ (about $1 0^{23}$ ) one configuration $(\tilde{n}_{1}, \dots, \tilde{n}_{k}, \dots)$ is far more probable than all the others: the equilibrium configuration.

The distribution

If $P (k)$ is the probability that state $k$ is occupied, the number of particles in it is $n_{k} = P (k) N$ .

Definition 4.2.2Boltzmann distributionEssentialslides w4, p. 16

The most probable configuration is $\tilde{n}_{k} = N \frac{e ^{- β E_{k}}}{Z}, Z = k \sum e^{- β E_{k}}, β = \frac{1}{K _{B} T} .$ $Z$ is the partition function: it makes the probabilities $P (k) = e^{- β E_{k}} / Z$ add up to 1. $e^{- β E_{k}}$ is the Boltzmann factor.

Where it comes frombeyond the lectures

The slide says “it turns out”. The derivation:

Count the arrangements. The particles are distinguishable, so the number of ways of putting $n_{1}$ of them in state 1, $n_{2}$ in state 2, … is $W = \frac{N !}{n _{1} ! n _{2} ! \dots n _{k} ! \dots} .$ Each arrangement is equally likely, so the most probable configuration maximises $W$ , or $ln W$ .
Stirling’s formula. For large $n$ , $ln n! \approx n ln n - n$ . Then $ln W \approx N ln N - \sum_{k} n_{k} ln n_{k}$ .
Maximise under the two constraints, with Lagrange multipliers $a$ for $\sum n_{k} = N$ and $β$ for $\sum n_{k} E_{k} = E_{tot}$ : $\frac{\partial}{\partial n _{k}} [ln W - a j \sum n_{j} - β j \sum n_{j} E_{j}] = - ln n_{k} - 1 - a - β E_{k} = 0 \Rightarrow n_{k} = e^{- 1 - a} e^{- β E_{k}} .$
Normalise. $\sum n_{k} = N$ fixes $e^{- 1 - a} = N / Z$ .
Identify $β$ . For an ideal gas the mean kinetic energy comes out as $\frac{3}{2 β}$ ; matching it to $\frac{3}{2} K_{B} T$ (Definition 3.3.1) gives $β = 1/ K_{B} T$ .

The maximum is extremely sharp: configurations that differ noticeably from $\tilde{n}_{k}$ have a negligible share of all arrangements when $N \sim 1 0^{23}$ , so the system is practically always found in it.

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Figure 4.2.1

N

oscillators share a fixed number of energy quanta. Each step, a random oscillator hands one quantum to another random one: the total energy never changes. Start from the even split (every oscillator with the same energy) and press Run: the histogram of how many oscillators have

0, 1, 2, \dots

quanta settles onto the

Boltzmann

exponential, and stays there. No one designed it: it is simply the most probable configuration.

Where it works. The distribution works very well in the limit $\tilde{n}_{k} ≪ 1$ : very many states, most of them empty, a few with one particle. When states hold many particles, the indistinguishability of identical quantum particles matters, and other statistics replace it (Fermi–Dirac, Bose–Einstein) beyond the lectures.

Key observation 4.2.3High energies are sparsely populatedslides w4, p. 16

At fixed $β$ (fixed $T$ ), the larger $E_{k}$ , the smaller $\tilde{n}_{k}$ : high-energy states are sparsely populated, exponentially so on the scale $K_{B} T$ .
Lowering $T$ makes $β$ larger, and the distribution shifts from high to low energies. Raising $T$ spreads it to higher energies.

WarningSlide slip

The slide says that lowering $T$ makes $β$ smaller. Since $β = 1/ K_{B} T$ , lowering $T$ makes $β$ larger. The conclusion it draws, the shift towards low energies, is right.

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Figure 4.2.2 Populations of equally spaced levels (

E_{k} = k ε

, as for Planck’s oscillator, §5.2 Rayleigh–Jeans and Planck) at temperature

T

. Drag the temperature: when

K_{B} T ≪ ε

everything sits in the ground state; as

T

grows the populations spread up the ladder, always falling exponentially. The readouts give

Z

and the mean energy.

Two consequences

Proposition 4.2.4Population ratiobeyond the lectures

$\frac{n ~ _{j}}{n ~ _{i}} = e^{- (E_{j} - E_{i}) / K_{B} T} .$

Proofbeyond the lectures

Divide the two Boltzmann expressions: $N$ and $Z$ cancel.

Hydrogen’s first excited level is 10.2 eV up, so at room temperature, $K_{B} T = 0.025$ eV, the ratio is $e^{- 408}$ : atoms sit in their ground state (Key observation 1.1.5).

Proposition 4.2.5Mean energy from the partition functionbeyond the lectures

$\overset{ˉ}{E} = \frac{\sum _{k} E _{k} e ^{- β E_{k}}}{Z} = - \frac{\partial ln Z}{\partial β} .$

Proofbeyond the lectures

$\frac{\partial Z}{\partial β} = - \sum_{k} E_{k} e^{- β E_{k}}$ , and $\frac{\partial l n Z}{\partial β} = \frac{1}{Z} \frac{\partial Z}{\partial β}$ .

Planck’s derivation uses exactly this, with $E_{k} = k ε_{0}$ (Theorem 5.2.5).

Exercises

Exercise 4.2.1beyond the lectures

A two-level system has energies 0 and $Δ = 0.1$ eV. Find the fraction in the upper level at 300 K and at 3000 K.

Solution

$P (up) = \frac{e ^{- Δ/ K_{B} T}}{1 + e ^{- Δ/ K_{B} T}}$ . At 300 K, $Δ/ K_{B} T = 3.87$ and $P = 0.020$ . At 3000 K, $Δ/ K_{B} T = 0.387$ and $P = 0.40$ . It never exceeds 1/2, however hot.

Exercise 4.2.2beyond the lectures

For the same system, find the mean energy $\overset{ˉ}{E} (T)$ and its limits.

Solution

$Z = 1 + e^{- β Δ}$ , $\overset{ˉ}{E} = - \partial_{β} ln Z = \frac{Δ}{e ^{β Δ} + 1}$ . As $T \to 0$ , $\overset{ˉ}{E} \to 0$ ; as $T \to \infty$ , $\overset{ˉ}{E} \to Δ/2$ .

Exercise 4.2.3beyond the lectures

For equally spaced levels $E_{k} = k ε$ , $k = 0, 1, 2, \dots$ , compute $Z$ and the fraction in the ground state.

Solution

$Z = \sum_{k} e^{- k βε} = \frac{1}{1 - e ^{- βε}}$ (a geometric series). $P (0) = 1/ Z = 1 - e^{- βε}$ : close to 1 when $K_{B} T ≪ ε$ , close to $βε$ (small) when $K_{B} T ≫ ε$ .

Exercise 4.2.4beyond the lectures

Four distinguishable particles share three quanta, on levels $E_{k} = k ε$ . Count the arrangements $W$ of the configurations $(n_{0}, n_{1}, n_{2}, n_{3}) = (2, 1, 1, 0)$ and $(3, 0, 0, 1)$ , and say which is more likely.

Solution

Both have total energy $3 ε$ . $(2, 1, 1, 0)$ : $W = \frac{4 !}{2 ! 1 ! 1 !} = 12$ . $(3, 0, 0, 1)$ : $W = \frac{4 !}{3 ! 1 !} = 4$ . The spread-out configuration is three times as likely. With $1 0^{23}$ particles such ratios become astronomically large, which is why one configuration wins.