Statistical Physics II, Chapter 6. Based on lecture notes by Prof. Sang Hoon Lee, Gyeongsang National University.

Boltzmann statistics

An interactive companion to the lecture notes. Each section below follows the notes in order: from a system that is completely isolated, to one that trades energy with a reservoir, to the Boltzmann factor, the partition function, average values and fluctuations.

6.0Isolated vs. in contact with a reservoir

Last semester, we counted the multiplicity Ω(U, V, N) directly and got everything else from S = k_B ln Ω. For more complicated systems this is prohibitively difficult, so we trade generality for power: we study a system in thermal equilibrium with a reservoir at a specific temperature.

Isolated atom

Its energy is fixed. All microstates with that energy are equally probable — the fundamental assumption, like a fair coin toss.

Atom + reservoir

The atom can be found in any of its microstates, but some are more likely than others, depending on their energies. Only the composite (atom + reservoir) is isolated.

Each small well is one oscillator of the Einstein-solid reservoir, with its occupied level marked. Every quantum the system gains is taken from the reservoir, so P(n) ∝ Ω_R(U_total − nε).

Below, the “atom” is one quantum oscillator (energies 0, ε, 2ε, …) and the reservoir is an Einstein solid. Energy quanta hop at random between oscillators, which samples every microstate of the composite system with equal probability. Watch how often the atom (pink) holds 0, 1, 2, … quanta.

Live: one atom exchanging quanta with a reservoir

Reservoir size 80 oscillators

Quanta per oscillator 1.0

simulated exact count Ω_R(q−n)

When isolated, the atom keeps whatever energy it had; its histogram collapses to a single bar.

6.1The Boltzmann factor

If the atom is in state s, the reservoir has Ω_R(s) accessible states, all equally probable. When the atom sits lower, more energy is left for the reservoir, so Ω_R is larger. Hence

P(s₂)P(s₁) = Ω_R(s₂)Ω_R(s₁) = e^{[S_R(s₂)−S_R(s₁)]/k_B} = e^{−[E(s₂)−E(s₁)]/k_BT}

using dS_R = (dU_R + P dV_R − μ dN_R)/T with dU_R = −dE; the P dV term is ~10⁻²⁵ J and there is no particle exchange in this chapter. Is the exponential exact? Only when the reservoir is truly large. Grow it and see.

Exact counting vs. the Boltzmann factor

exact, ∝ Ω_R(q−n) (1/Z) e^−nε/k_BT

Reservoir oscillators 10

Quanta per oscillator 1.0

Reservoir temperature from 1/T = ∂S_R/∂U_R, which for an Einstein solid gives ε/k_BT = ln(1 + N/q). With a handful of oscillators the curve misses; by a few thousand it is indistinguishable.

Each ratio is a ratio of simple exponentials, so every state carries a Boltzmann factor e^−E(s)/k_BT. Normalizing gives

P(s) = 1Z e^−E(s)/k_BT

“the most useful formula in all of statistical mechanics. Memorize it.”

Thermal excitation of hydrogen in a stellar atmosphere

For hydrogen, E(s₁) = −13.6 eV and E(s₂) = −3.4 eV. The n = 2 level holds 4 independent states (neglecting spin), n = 3 holds 9: a level can be degenerate. Only atoms in n = 2 absorb the visible Balmer lines (656, 486, 434 nm).

Hydrogen populations at temperature T

Temperature 5800 K

Bars use a logarithmic scale: each tick is a factor of 10 in population relative to the ground state. Ratios only; the full hydrogen partition function needs more care.

6.1The partition function

Normalization Σ_s P(s) = 1 fixes the constant:

Z = Σ_s e^−E(s)/k_BT = sum of all Boltzmann factors

“Zustandssumme”, the sum over states. It is a function because it depends on T, and it roughly counts how many states are accessible, each weighted by its probability. With the ground state at 0, Z → 1 as T → 0⁺ and Z ≫ 1 at high T. Shifting every energy by E₀ multiplies Z by e^−E₀/k_BT, an uninteresting factor that cancels in every probability.

Explore Z for a few systems

k_BT / ε 1.00

Shift all energies by E₀ / ε 0.0

Bars: Boltzmann factor of each level, g·e^−E/k_BT; the curve is e^−E/k_BT. Move E₀: the bars and Z rescale, the probabilities on the right do not.

For a degenerate level, the probability of the level is g(E)·e^−E/k_BT/Z = e^−F/k_BT/Z with F = E − TS and S = k_B ln g (Problem 6.2). Try the hydrogen-like preset at high temperature: the many excited states win on sheer numbers.

6.11Lithium nuclei and negative temperature

A ⁷Li nucleus has four spin states m = −3/2, −1/2, 1/2, 3/2 with E = −mμB, μ = 1.03 × 10⁻⁷ eV/T. In the Purcell–Pound experiment, B = 0.63 T and T = 300 K, so μB/k_BT ≈ 2.5 × 10⁻⁶ and every probability is within a millionth of 1/4. If the field is reversed suddenly, the populations have no time to change, and they now match a Boltzmann distribution with T = −300 K.

Spin populations, shown as deviation from 1/4

Field magnitude |B| 0.63 T

Lattice temperature 300 K

Lower the temperature to make the deviations large enough to see by eye, then reverse the field: the highest-energy state becomes the most populated.

6.2Average values and fluctuations

Often we only need the average of some property. Writing β ≡ 1/k_BT,

X̄ = Σ_s X(s) P(s) = 1Z Σ_s X(s) e^−βE(s)

Start with the five-atom toy model from the notes. Move atoms between levels.

Five-atom toy model (Problem 6.17)

σ_E² = mean of the squares − square of the mean, “제곱의 평균 − 평균의 제곱”.

Now let the canonical ensemble do the averaging. The derivatives of Z do the work (Problems 6.16 and 6.18):

Ē = −∂ ln Z∂β, E² = 1Z∂²Z∂β², σ_E = k_BT √(C/k_B)

Canonical ensemble: averages, fluctuations, heat capacity

Ē σ_E C / k_B

k_BT / ε 2.00

The two columns of numbers agree: summing over states, differentiating ln Z, and reading off the heat capacity are three routes to the same physics. Sampling shows the ensemble that these averages describe. For N independent copies, U = NĒ.

σ_E = k_BT√(C/k_B) is a first taste of the fluctuation–dissipation theorem: internal fluctuations are tied to the response to an external change.