Thermodynamics · Upper Division

Statistical Mechanics

Statistical mechanics derives thermodynamics from first principles by counting microscopic states. Temperature, entropy, and pressure emerge from probability theory applied to systems with ~10²³ particles.

PrerequisitesLawsofthermodynamics(Ch.12)⋅Quantummechanics(Ch.20)forquantumstatistics⋅BasiLaws of thermodynamics (Ch. 12) \cdot Quantum mechanics (Ch. 20) for quantum statistics \cdot Basic calculus and probability
Learning Goals
  • StateBoltzmann′sentropyformulaS=kBlnΩanduseittocomparelikelihoodsofmacrosState Boltzmann's entropy formula S = k_{B} ln \Omega and use it to compare likelihoods of macrostates.
  • Derive the Boltzmann distribution from the maximum-entropy principle and define the partition function Z.
  • Extract mean energy and free energy from Z using the standard thermodynamic relations.
  • Apply the equipartition theorem to predict heat capacities of monatomic and diatomic gases.
  • Contrast Fermi-Dirac and Bose-Einstein statistics and describe their physical consequences.

S.1 Microstates and the Boltzmann Entropy

The central insight of statistical mechanics is that macroscopic thermodynamic quantities — temperature, pressure, entropy — correspond to averages over an enormous number of microscopic configurations (microstates). Each microstate is a complete specification of every particle's position and momentum.

Definition S.1 — Boltzmann Entropy
TheentropyofamacrostateisproportionaltothelogarithmofthenumberofmicrostatesΩhe entropy of a macrostate is proportional to the logarithm of the number of microstates \Omegaconsistent with it:S=kBlnΩ(kB=1.381×10−23J/KS = k_{B} ln \Omega \qquad (k_{B} = 1.381\times10^{-23} J/KThisiscarvedonBoltzmann′sgravestone.Itconnectsthemicroscopicworld(Ω,countingThis is carved on Boltzmann's gravestone. It connects the microscopic world (\Omega, counting tothemacroscopic(S,measurable).AsystemevolvestowardstatesofmaximumΩ—thesecoto the macroscopic (S, measurable). A system evolves toward states of maximum \Omega — the secod law is just the law of large numbers.
Example S.1 — Entropy of a Two-State System

N=100coins,eachshowingheads(H)ortails(T).Howmanywayscanexactly50beheadsN = 100 coins, each showing heads (H) or tails (T). How many ways can exactly 50 be heads? Compare to all-heads. What does this say about the second law?

Ω(50H):C(100,50)=100!/(50!50!)≈1029—anastronomicalnumberC(100,50) = 100!/(50!50!) \approx 10^{29} — an astronomical number
Ω(100H):C(100,100)=1C(100,100) = 1
Entropy difference:\DeltaS=kBln(1029/1)=kB×29×ln10≈10−21J/K\DeltaS = k_{B} ln(10^{29}/1) = k_{B} \times 29 \times ln10 \approx 10^{-21} J/K
Moral:Theall−headsstateisfantasticallyimprobable.For1023coins(molesofgasmoleculesThe all-heads state is fantastically improbable. For 10^{23} coins (moles of gas molecules, the overwhelmingly most probable macrostate is the uniform distribution. Deviations are essentially impossible — this is the second law.

S.2 The Boltzmann Distribution

Consider a small system in contact with a large thermal reservoir at temperature T. The probability that the system occupies a microstate with energy E is theBoltzmann distribution:

P(E)=(1/Z)e−E/kBT=(1/Z)e−\betaE(β≡1/kBT)P(E) = (1/Z) e^{-E/k_{BT}} = (1/Z) e^{-\betaE} \qquad (\beta \equiv 1/k_{BT})(S.1)

The partition function Z = Σ_i e^(−βEᵢ) (sum over all microstates) is the normalization factor and encodes all thermodynamic information. Once Z is known:

⟨E⟩=−∂(lnZ)/∂βF=−kBTlnZS=−\partialF/\partialT⟨E⟩ = -\partial(ln Z)/\partial\beta \qquad F = -k_{BT} ln Z \qquad S = -\partialF/\partialT(S.2)

The partition function is to statistical mechanics what the wavefunction is to quantum mechanics — everything follows from it.

Example S.2 — Two-Level System

Asystemhastwoenergylevels:E=0andE=ε.Findthepartitionfunction,meanenergyA system has two energy levels: E = 0 and E = \varepsilon. Find the partition function, mean energy, and heat capacity.

Z:Z=e−β⋅0+e−βε=1+e(−βεZ = e^{-\beta\cdot0} + e^{-\beta\varepsilon} = 1 + e^(-\beta\varepsilon
ln Z:lnZ=ln(1+e(−βεln Z = ln(1 + e^(-\beta\varepsilon
⟨E⟩:⟨E⟩=−∂(lnZ)/∂β=εe−βε/(1+e−βε)=ε/(eβε+1E⟩ = -\partial(lnZ)/\partial\beta = \varepsilon e^{-\beta\varepsilon}/(1 + e^{-\beta\varepsilon}) = \varepsilon/(e^{\beta\varepsilon} + 1
Heat capacity:C=d⟨E⟩/dT=kB(βε)2eβε/(eβε+1)2C = d⟨E⟩/dT = k_{B} (\beta\varepsilon)^{2} e^{\beta\varepsilon}/(e^{\beta\varepsilon}+1)^{2}
Schottky anomaly:CpeaksatkBT≈0.42εthendecays.AthighT(kBT≫ε):⟨E⟩→ε/2(equalpopulation).AC peaks at k_{BT} \approx 0.42\varepsilon then decays. At high T (k_{BT} ≫ \varepsilon): ⟨E⟩ \to \varepsilon/2 (equal population). At low T: system freezes into ground state.

S.3 The Equipartition Theorem

Theorem S.1 — Equipartition Theorem
In thermal equilibrium at temperature T, every quadratic degree of freedom contributes \frac{1}{2}k_BT to the mean energy:⟨½mx˙2⟩=12kBTforeachtranslational,rotational,orvibrationalmodemẋ^{2}⟩ = \frac{1}{2}k_{BT} \qquad for each translational, rotational, or vibrational modeAmonatomicidealgashas3translationalDOF→⟨E⟩=3/2kBTperparticle.AdiatomicmoA monatomic ideal gas has 3 translational DOF \to ⟨E⟩ = 3/2 k_{BT} per particle. A diatomic moleculehas3translational+2rotational=5DOF→⟨E⟩=5/2kBT.ThisgivesCv=(f/2)Nlecule has 3 translational + 2 rotational = 5 DOF \to ⟨E⟩ = 5/2 k_{BT}. This gives Cv = (f/2)NkB,wherefisthenumberofactivequadraticmodesk_{B}, where f is the number of active quadratic modes.

The equipartition theorem breaks down at low temperatures where quantum effects freeze out modes with spacing ε ≫ k_BT. This is why the heat capacity of hydrogen drops from 5/2 Nk_B (room temperature, 5 modes) to 3/2 Nk_B (low temperature, only translation) — a purely quantum effect observed by Boltzmann himself but not understood until quantum mechanics.

S.4 Quantum Statistics

Identical quantum particles obey one of two statistics, depending on their spin:

Definition S.2 — Quantum Distribution Functions
Bosons(integerspin,e.g.,photons,4He):Bose−Einsteindistribution⟨nk⟩=1/(e(β(εk−μ))−integer spin, e.g., photons, ^{4}He): Bose-Einstein distribution \qquad ⟨n_{k}⟩ = 1/(e^(\beta(\varepsilon_k-\mu)) -1)Fermions(half−integerspin,e.g.,electrons,protons):Fermi−Diracdistribution⟨nk⟩=1/(e(β(εhalf-integer spin, e.g., electrons, protons): Fermi-Dirac distribution \qquad ⟨n_{k}⟩ = 1/(e^(\beta(\varepsilonk−μ))+1k-\mu)) + 1μisthechemicalpotential.AtT=0,theFermi−Diracdistributionisastepfunction—all\mu is the chemical potential. At T=0, the Fermi-Dirac distribution is a step function — all statesbelowtheFermienergyEFarefilled,allaboveareemptystates below the Fermi energy E_{F} are filled, all above are empty

Fermi-Dirac statistics explains why metals conduct electricity (electrons near E_F are mobile), why white dwarf stars don't collapse (electron degeneracy pressure), and why the specific heat of metals is linear in T (only electrons within ~k_BT of E_F contribute). Bose-Einstein statistics permits photon bunching (laser light) and Bose-Einstein condensation — all bosons collapsing into the ground state below a critical temperature.

Definition S.3 — Common Traps
  • Microstates and macrostates are different levels: entropy counts microscopic arrangements compatible with one macrostate.
  • Z is more than normalization: derivatives of ln Z generate energy, entropy, and free energy.
  • Equipartition has limits: quantumlevelspacingfreezesoutmodeswhenkBTistoosmallquantum level spacing freezes out modes when k_{BT} is too small
  • Fermions and bosons differ by occupancy rules: the plus/minus sign changes low-temperature behavior completely.
Exercises — S.1–S.4 Statistical Mechanics
1.
Using equipartition, derive the heat capacity Cv of a monatomic ideal gas. Express in terms of N, k_ number for one mole.
J/(mol·K)
Straightforward
2.Forthetwo−levelsystem(E=0,E=ε),plottheheatcapacityC(T)qualitatively.WhereistFor the two-level system (E=0, E=\varepsilon), plot the heat capacity C(T) qualitatively. Where is theSchottkypeak?WhathappensasT→0andT→∞he Schottky peak? What happens as T\to0 and T\to\infty?
Intermediate
3.DerivethePlanckblackbodydistributionfromtheBose−Einsteindistribution(μ=0).ShowhDerive the Planck blackbody distribution from the Bose-Einstein distribution (\mu=0). Show how this resolves the ultraviolet catastrophe.
Intermediate
4.CalculatetheFermienergyEFforafreeelectrongaswithdensityn.FindthetotalgrouCalculate the Fermi energy E_{F} for a free electron gas with density n. Find the total ground-state energy and the degeneracy pressure. Why is Cv of electrons linear in T rather than constant?
Challenging
Key Takeaways
  • S=kBlnΩ—entropycountsmicrostates;thesecondlawisjustprobabilitytheoryatscS = k_{B} ln \Omega — entropy counts microstates; the second law is just probability theory at scale.
  • Boltzmanndistribution:P(E)∝e−\betaE,β=1/kBT.Higherenergy→exponentiallylessproBoltzmann distribution: P(E) \propto e^{-\betaE}, \beta = 1/k_{BT}. Higher energy \to exponentially less probable.
  • PartitionfunctionZ=\sume−\betaEencodesallthermodynamics:⟨E⟩=−\partiallnZ/∂βPartition function Z = \sume^{-\betaE} encodes all thermodynamics: ⟨E⟩ = -\partiallnZ/\partial\beta.
  • Equipartition:12kBTperquadraticDOF.BreaksdownwhenkBT≪ε(quantumfreezeoutEquipartition: \frac{1}{2}k_{BT} per quadratic DOF. Breaks down when k_{BT} ≪ \varepsilon (quantum freezeout.
  • FermionsobeyFermi−Dirac:atT=0allstatesbelowEFfilled—explainsmetals,whitedwaFermions obey Fermi-Dirac: at T=0 all states below E_{F} filled — explains metals, white dwarfs.
  • BosonsobeyBose−Einstein:photons→blackbodyspectrum;atlowT→Bose−EinsteincondensaBosons obey Bose-Einstein: photons \to blackbody spectrum; at low T \to Bose-Einstein condensation.