Chapter 20

Thermodynamics & Statistical Mechanics

College
At a glance
Core ideaEntropy counts microstates, S = kB ln Ω; the partition function Z yields all thermodynamics.
Key termBoltzmann distribution — a state's probability ∝ e−E/k_BT.
You can…Solve the two-level system and derive the ideal gas law from counting.
Watch outThe second law is overwhelming probability, not a new force — the micro-laws stay reversible.
Theory

From microstates to macroscopic law

Chapter 14 treated temperature and entropy phenomenologically. Statistical mechanics derives them from counting. A macroscopic state (fixed energy, volume, particle number) corresponds to an enormous number of possible microscopic arrangements, or microstates, Ω. Boltzmann's foundational relation defines entropy directly from that count:

S = kB ln Ω kB = 1.381×10⁻²³ J·K⁻¹

For a system in thermal contact with a much larger reservoir at temperature T, the probability of finding it in a microstate of energy Ei follows the Boltzmann distribution Pi ∝ e−Eᵢ/k_BT. Normalizing introduces the central object of the theory, the partition function:

Z = Σᵢ e−Eᵢ/k_BT

Every thermodynamic quantity follows from Z: mean energy ⟨E⟩ = −∂lnZ/∂β (with β = 1/kBT), Helmholtz free energy F = −kBT lnZ, and entropy S = (⟨E⟩ − F)/T.

Explanation

Why the second law is a matter of overwhelming probability

Chapter 14 called entropy "the deepest asymmetry in physics" and noted disordered states vastly outnumber ordered ones. Statistical mechanics makes this exact: entropy S = kBlnΩ increases in a spontaneous process simply because the system evolves toward whichever macrostate has overwhelmingly more microstates. A gas doesn't spread to fill a room because of some active "force of disorder" — it spreads because there are astronomically more ways for its molecules to be spread throughout the room than crammed in one corner, and random molecular motion will, essentially with certainty, sample the far more numerous "spread out" arrangements. The second law is not fundamental dynamics — Newton's (or Schrödinger's) equations are reversible — it is a statement about probability at the scale of 10²³ particles, where "overwhelmingly likely" is functionally indistinguishable from "certain".

Concrete example. For an ideal monatomic gas, the classical partition function of N particles in volume V gives exactly pV = NkBT — Chapter 14's ideal gas law falls straight out of counting microstates, with no separate postulate needed.
Practical

Worked example — the two-level system

Consider N independent particles, each with two possible energy states: 0 and ε. Find the partition function, mean energy, and high/low temperature limits.

  1. Single-particle partition function: z = e−0/k_BT + e−ε/k_BT = 1 + e−ε/k_BT.
  2. For N independent, distinguishable particles: Z = zN.
  3. Mean energy of one particle: ⟨E⟩ = −∂lnz/∂β = ε e−βε/(1+e−βε), with β=1/kBT. Total: ⟨Etot⟩ = N⟨E⟩.
  4. Low-T limit (kBT ≪ ε): e−βε → 0, so ⟨E⟩ → 0 — nearly every particle sits in the ground state, as expected near absolute zero.
  5. High-T limit (kBT ≫ ε): e−βε → 1, so ⟨E⟩ → ε/2 — particles are equally likely in either state, the maximum-entropy configuration.

This toy model is the backbone of paramagnetism (spins up/down in a field) and of Einstein's early model of specific heat in solids.

Q&A
Why is entropy additive for two independent subsystems, given S = k_B ln Ω?

If subsystem A has ΩA microstates and subsystem B independently has ΩB, the combined system has ΩAΩB microstates (every combination). Then S = kBln(ΩAΩB) = kBlnΩA + kBlnΩB = SA + SB — the logarithm converts multiplicative counting into additive entropy.

What is the relationship between the partition function Z and the Helmholtz free energy F?

F = −kBT lnZ. Since Z encodes every accessible microstate weighted by its Boltzmann factor, its logarithm directly yields the free energy, from which pressure, entropy, and chemical potential all follow by differentiation.

Why does the two-level system's mean energy approach ε/2, not ε, at high temperature?

At high T both states become equally probable (probability ½ each), so the average energy is the simple average of the two levels: (0 + ε)/2 = ε/2. It can never exceed this because no state has energy above ε.

How does this chapter's statistical view resolve the paradox that Newton's laws are time-reversible but the second law is not?

Reversing every particle's velocity in a gas is a perfectly valid solution of the (reversible) microscopic equations — but it corresponds to one absurdly special microstate among the astronomical number available. Generic initial conditions overwhelmingly evolve toward higher-Ω macrostates. Irreversibility is a property of typical, macroscopic observation, not a violation of the reversible microscopic laws.

Concept mind map

How the ideas connect

Every key idea in this chapter, branching from the core concept — use it to see the whole picture at a glance.

MicrostatesS = k*ln(W)Boltzmann factorPartition functionSecond lawTemperatureStatistical Mechanics
Infographic

The key facts, visualised

S = k*ln(W)
Entropy from number of microstates W
k
Boltzmann constant, 1.38e-23 J/K
exp(-E/kT)
Boltzmann factor: relative state probability
Z
Partition function, sum over states
Solved examples

Worked problems, step by step

Follow each solution line by line, then try to reproduce it on paper before moving on.

Example 1A two-level system has energies 0 and E. Write the partition function Z.

  1. Sum over both states with Boltzmann factors
  2. Z = exp(0) + exp(-E/kT)

Example 2For that two-level system, find the probability of the upper state.

  1. P(upper) = exp(-E/kT) / Z
  2. Z = 1 + exp(-E/kT)
Practice problem set

Now you try

Work each one out first, then tap to reveal the worked answer.

1What is a microstate?
A specific detailed arrangement of all particles consistent with the macroscopic state.
2How does entropy relate to microstates?
S = k*ln(W); more accessible microstates means higher entropy.
3Why is the second law statistical rather than absolute?
High-entropy states are overwhelmingly more probable, so systems almost always move toward them.
4What does the Boltzmann factor tell you?
The relative probability of a state falls exponentially with its energy over kT.
5As T rises, what happens to the upper state probability?
It increases toward equal population of the levels.
6What role does the partition function Z play?
It normalizes probabilities and generates thermodynamic quantities like energy and entropy.