Statistical Mechanics
Equipartition Theorem
Each quadratic degree of freedom contributes ½kT to the mean energy of a system in thermal equilibrium
In classical thermal equilibrium, every quadratic degree of freedom carries ½kT of mean energy. Three translational DOFs give Cv = 3R/2; adding two rotational gives Cv = 5R/2 = 20.8 J/(mol·K).
- StatementEach quadratic DOF → ½kT mean energy
- Monatomic gas Cv(3/2)R ≈ 12.5 J/(mol·K)
- Diatomic gas Cv(5/2)R ≈ 20.8 J/(mol·K)
- Atomic solid Cv (Dulong-Petit)3R ≈ 24.9 J/(mol·K)
- Fails whenkT ≪ quantum level spacing
- N₂ rotational T_rot≈ 2.9 K
Interactive visualization
A 2D gas of particles with a Maxwell-Boltzmann velocity histogram. Switch rotational and vibrational DOFs on and watch the heat capacity climb from 3R/2 to 5R/2 to 7R/2.
Watch the 60-second explainer
A condensed visual walkthrough — narrated, captioned, under a minute.
The theorem
For a classical system in thermal equilibrium at temperature T, each independent quadratic term in the Hamiltonian contributes ½kT to the mean energy.
Examples of quadratic terms:
- Translational kinetic energy in 3D: ½m·vₓ² + ½m·v_y² + ½m·v_z². Three quadratic DOFs.
- Rotational kinetic energy: ½I·ωₐ². One DOF per nontrivial principal axis.
- Harmonic oscillator: ½m·v² + ½k·x². Two quadratic DOFs (kinetic + potential).
Total mean energy: ⟨E⟩ = (f/2)·kT per particle, where f is the count of active quadratic DOFs.
Two-line proof
For a single quadratic coordinate q with energy E(q) = ½a·q², the canonical mean energy is:
⟨E⟩ = ∫(½a·q²)·exp(−β·½a·q²) dq / ∫exp(−β·½a·q²) dq
= (1/2β) · [Gaussian integral identity]
= ½kT
The coefficient a cancels — any quadratic DOF contributes the same ½kT, independent of the spring constant or moment of inertia. The DOFs are independent (cross terms vanish in the equilibrium average) so we just sum them.
Heat capacities by molecular structure
| System | Quadratic DOFs | U per particle | Cv (per mole) | Cv (numeric) |
|---|---|---|---|---|
| Monatomic ideal gas (He, Ne, Ar) | 3 translational | (3/2)kT | (3/2)R | 12.47 J/(mol·K) |
| Diatomic gas, no vib (N₂, O₂ at 300 K) | 3 trans + 2 rot | (5/2)kT | (5/2)R | 20.79 J/(mol·K) |
| Diatomic gas, full vib (high T) | 3 trans + 2 rot + 2 vib | (7/2)kT | (7/2)R | 29.10 J/(mol·K) |
| Linear triatomic (CO₂) | 3 trans + 2 rot + 4 vib | (13/2)kT | (13/2)R | 54.0 J/(mol·K) (asymptotic) |
| Non-linear polyatomic (H₂O, CH₄) | 3 trans + 3 rot + many vib | variable | (3 + N_vib)R | asymptotic Dulong-Petit-like |
| Atomic solid (Cu, Al, Au at high T) | 3 kinetic + 3 elastic | 3kT | 3R | 24.94 J/(mol·K) Dulong-Petit |
| Photon gas (electromagnetic radiation) | — | aT⁴ Stefan-Boltzmann | — | Equipartition fails (linear E-p) |
Diatomic data check: Cv(N₂, 300 K) measured = 20.76 J/(mol·K), within 0.1% of 5R/2. At 2000 K it climbs toward 29 J/(mol·K) as vibrations activate. Below 200 K rotational levels start freezing; below 20 K hydrogen heat capacity drops below 3R/2 even though equipartition would predict 5R/2.
Worked example — diatomic N₂ at room temperature
For 1 mole of N₂ at 300 K:
Active DOFs: 3 translational + 2 rotational = 5
U = (5/2) · R · T = 2.5 · 8.314 · 300 = 6235 J/mol
Cv = (5/2) · R = 20.79 J/(mol·K)
Cp = Cv + R = (7/2) · R = 29.10 J/(mol·K)
γ = Cp/Cv = 7/5 = 1.40 (measured 1.401 — exact)
Measured γ for air is 1.400 at room temperature — equipartition nails it. Heat 1 mol of N₂ from 300 K to 310 K and you must add 20.79 × 10 = 208 J. Exactly what equipartition predicts.
Where equipartition breaks
Real molecules don't have a continuous energy spectrum — they have quantized levels. A degree of freedom contributes ½kT to ⟨E⟩ only when kT is much greater than the level spacing. As T drops below the characteristic temperature of a DOF, that DOF "freezes out" and stops contributing.
| DOF | Characteristic energy ℏω or ΔE | Characteristic T | Behavior at 300 K |
|---|---|---|---|
| Translation (free particle) | ~h²/(8mL²) ~ 10⁻³⁹ J | ~10⁻¹⁶ K | Fully classical |
| Rotation, H₂ | ℏ²/(2I) ≈ 7.5 meV | ~88 K | Mostly active |
| Rotation, N₂ | ≈ 0.25 meV | ~2.9 K | Fully classical |
| Vibration, H₂ | ≈ 0.55 eV (ℏω) | ~6300 K | Frozen out |
| Vibration, N₂ | ≈ 0.29 eV | ~3400 K | Frozen out |
| Vibration, Cl₂ | ≈ 70 meV | ~800 K | Partially active |
| Lattice vibration in Cu | Debye T ≈ 343 K | ~343 K | Approaching Dulong-Petit |
| Lattice vibration in diamond | Debye T ≈ 2230 K | ~2230 K | Cv ≪ 3R at 300 K |
The freeze-out problem for blackbody radiation (continuous EM modes with no upper cutoff) was the "ultraviolet catastrophe" that triggered the quantum revolution: equipartition predicted infinite energy density, Planck's quantization rescued the spectrum.
JavaScript — heat-capacity arithmetic
const R = 8.314; // J/(mol·K)
const k_B = 1.380649e-23;
// Classical equipartition heat capacity
function Cv_equipartition(f) { return (f / 2) * R; }
console.log(`Monatomic gas Cv: ${Cv_equipartition(3).toFixed(2)} J/(mol·K)`); // 12.47
console.log(`Diatomic gas Cv (5 DOF): ${Cv_equipartition(5).toFixed(2)}`); // 20.79
console.log(`Diatomic gas Cv (7 DOF): ${Cv_equipartition(7).toFixed(2)}`); // 29.10
console.log(`Atomic solid Cv (6 DOF): ${Cv_equipartition(6).toFixed(2)}`); // 24.94
// Einstein heat capacity (one mode, quantum)
function Cv_Einstein(T_Einstein, T) {
const x = T_Einstein / T;
return R * x * x * Math.exp(x) / Math.pow(Math.exp(x) - 1, 2);
}
// N₂ vibrational T ≈ 3400 K
const T_vib_N2 = 3400;
console.log(`N₂ vib Cv at 300 K: ${Cv_Einstein(T_vib_N2, 300).toExponential(2)}`); // ~0 (frozen)
console.log(`N₂ vib Cv at 3000 K: ${Cv_Einstein(T_vib_N2, 3000).toFixed(2)}`); // approaches R
// Total Cv for diatomic gas with quantum vibrations
function Cv_diatomic(T, T_vib) {
return Cv_equipartition(5) + Cv_Einstein(T_vib, T);
}
console.log(`N₂ total Cv at 300 K: ${Cv_diatomic(300, 3400).toFixed(2)}`); // 20.79
console.log(`N₂ total Cv at 2000 K: ${Cv_diatomic(2000, 3400).toFixed(2)}`); // ~26
// Mean speed from equipartition
function meanSpeed(T, m) {
return Math.sqrt(8 * k_B * T / (Math.PI * m));
}
// N₂ molecule m ≈ 4.65e-26 kg at 300 K
console.log(`N₂ mean speed at 300 K: ${meanSpeed(300, 4.65e-26).toFixed(0)} m/s`); // ~476
// Adiabatic index γ
function gamma_from_f(f) { return (f + 2) / f; }
console.log(`γ(monatomic) = ${gamma_from_f(3).toFixed(3)}`); // 1.667
console.log(`γ(diatomic) = ${gamma_from_f(5).toFixed(3)}`); // 1.400
Where equipartition shows up
- Gas thermodynamics. The textbook predictions Cv = (3/2)R, Cp = (5/2)R, γ = 5/3 for monatomic gases all come from equipartition.
- Brownian motion. A colloidal particle of any size has ⟨½m·v²⟩ = (3/2)kT — the thermal speed scales as 1/√m. This is how Perrin measured Avogadro's number in 1908.
- Trapped particles. A bead in a harmonic optical trap with stiffness κ has ⟨½κ·x²⟩ = ½kT, so position variance is ⟨x²⟩ = kT/κ. Used directly to calibrate optical tweezers.
- Plasma physics. Electron and ion temperatures in plasmas use the same ½kT-per-DOF accounting; magnetic moments give two perpendicular DOFs producing the perpendicular vs parallel temperature distinction in tokamaks.
- Specific heats of solids. Dulong-Petit (Cv = 3R per mole of atoms) holds above the Debye temperature; deviations below it motivated Einstein's 1907 quantum model and Debye's 1912 phonon refinement.
- Acoustic noise. Johnson-Nyquist thermal noise in resistors: ⟨V²⟩ = 4kT·R·Δf — equipartition for the electromagnetic modes of the circuit.
- Astrophysics. Stellar interiors use equipartition to relate temperature to ion kinetic energy; departures help diagnose degenerate matter in white dwarfs and neutron stars.
Common mistakes
- Counting non-quadratic terms. A linear restoring force like gravity (U = m·g·z) is NOT quadratic. A column of air in gravity contributes ½kT (kinetic) + kT (potential, after integration with the barometric formula) — not ½kT for the potential. Be careful.
- Forgetting symmetry reductions. Linear molecules (CO₂, N₂) have only 2 rotational DOFs (rotation about the bond axis carries no moment of inertia). Don't naively use 3 rotational for linear molecules.
- Applying equipartition at low T. Below the characteristic T of a DOF, that DOF freezes. H₂ heat capacity at 50 K is below 3R/2 because even rotations are partially frozen.
- Using ½kT for relativistic systems. The theorem assumes E ∝ q². For photons (E ∝ |p|) the per-mode energy is kT, not ½kT, in the classical limit — and that limit itself fails as ℏω rises.
- Forgetting the factor 2 in solids. Atomic vibrations have BOTH kinetic and potential quadratic terms, giving 2 DOFs per spatial direction = 6 total → Cv = 3R. Don't just count 3.
- Treating Cv as constant. For real molecules Cv depends on T as DOFs activate or freeze. Use Einstein/Debye corrections for solids and quantum vibrational partition functions for molecules near their T_vib.
Frequently asked questions
What counts as a quadratic degree of freedom?
Any coordinate or momentum variable that enters the energy as a square: kinetic energy ½m·vₓ² (one DOF), ½I·ω² (rotational), ½k·x² (spring potential), and so on. Each such variable contributes ½kT to ⟨E⟩ in classical thermal equilibrium. Linear or higher-order terms don't follow the rule; only quadratic ones.
How does it predict heat capacities?
Mean energy U = (f/2)·N·k·T where f counts the active quadratic DOFs per particle. Then Cv = (∂U/∂T)_V = (f/2)·Nk, and per mole Cv = (f/2)R. Monatomic gas (3 translational): Cv = (3/2)R = 12.5 J/(mol·K). Diatomic gas (3 trans + 2 rot): Cv = (5/2)R = 20.8 J/(mol·K). Diatomic with vibrations active (3 trans + 2 rot + 2 vib): Cv = (7/2)R = 29.1 J/(mol·K). Atomic solid (3 kinetic + 3 elastic): Cv = 3R = 24.9 J/(mol·K), the Dulong-Petit law.
Why are diatomic gases 5/2 R at room temperature, not 7/2 R?
Vibration is frozen out. The vibrational quantum ħω of N₂ is about 0.29 eV — equivalent to T ≈ 3400 K. At 300 K we're far below the vibrational temperature, so the first excited vibrational state is hardly ever populated and contributes no heat capacity. Rotational quanta are much smaller (T_rot ≈ 2.9 K for N₂), so rotations are fully active at room T. Hence the measured Cv = 5/2 R for diatomic gases at room temperature.
Why does equipartition fail at low temperature?
Quantum mechanics. Real degrees of freedom have discrete energy levels. When the spacing ħω exceeds kT, the first excited state is exponentially suppressed and that DOF contributes nothing to Cv. The classical equipartition formula assumes a continuous quadratic energy, which is only valid when kT ≫ ΔE. As T drops past each characteristic temperature (T_vib, T_rot, etc.) the corresponding DOF "freezes out" and Cv decreases stepwise. This breakdown of the classical theorem was a major motivation for the quantum revolution.
How is equipartition derived?
Start from the canonical Boltzmann distribution P(q) ∝ exp(−β·E(q)) with quadratic E(q) = ½a·q². Compute the mean: ⟨E⟩ = ∫(½a·q²)·exp(−β·½a·q²)·dq / ∫exp(−β·½a·q²)·dq. Both integrals are Gaussians; the ratio is exactly 1/(2β) = ½kT, regardless of a. The result holds for every quadratic coordinate independently — hence "equi" partition.
Does the theorem hold for photons or relativistic particles?
Not directly. Photons have linear dispersion E = c·|p|, not quadratic; their mean energy per mode is kT only in the Rayleigh-Jeans limit (kT ≫ ħω), and the full Planck distribution corrects this. Relativistic particles with E = √(p²c² + m²c⁴) have a mixed dispersion; in the ultra-relativistic limit each mode gets kT (not ½kT). The theorem assumes purely quadratic energy.
Where does the Dulong-Petit law come from?
An atomic solid has each atom oscillating in 3 directions. Each oscillator has 1 kinetic and 1 potential quadratic DOF (½p²/m and ½kx²), so 2 quadratic DOFs per direction times 3 directions equals 6 DOFs per atom. Each contributes ½kT, so U = 3kT per atom and Cv = 3R ≈ 24.9 J/(mol·K). This 1819 Dulong-Petit empirical rule held for most solids above their Debye temperature; the low-T failure (Cv ∝ T³) was finally explained by Einstein and Debye using quantized phonons.