Physic Labs

Classical statistical mechanics

The equipartition theorem

At classical equilibrium, each independent quadratic degree of freedom in the Hamiltonian contributes kBT/2 to the mean energy.

At classical equilibrium, each independent quadratic degree of freedom in the Hamiltonian contributes kBT/2 to the mean energy.

⟨x∂H∂x⟩=kBT,Hx=ax2⇒⟨Hx⟩=12kBT\left\langle x\frac{\partial H}{\partial x}\right\rangle=k_B T, \quad H_x=ax^2\Rightarrow\langle H_x\rangle=\frac12k_BT

Definition: Quantities and meaning

The result applies to continuous variables in classical equilibrium and independent quadratic energy terms; each momentum component p²/(2m), for example, contributes kBT/2. A classical harmonic oscillator has one kinetic and one potential term, totaling kBT. At low temperature quantum effects freeze modes, so the classical theorem cannot be extrapolated to all temperatures.

Đóng góp của các bậc tự do

Quantitative relation

⟨Etranslation⟩=32kBT,⟨Eoscillator⟩=kBT\langle E_{\rm translation}\rangle=\frac32k_BT, \quad \langle E_{\rm oscillator}\rangle=k_BT

Example: Worked example

A monatomic ideal gas has 3 translational degrees of freedom. What is the mean kinetic energy per particle from equipartition?

Solution

The three quadratic momentum components contribute 3(kBT/2)=3kBT/2.

Example: Example: molar heat capacity of a monatomic gas

A mole of monatomic ideal gas has three quadratic translational kinetic terms. Equipartition gives U=3RT/2, so C_V=(∂U/∂T)_V=3R/2≈12.5 J·mol⁻¹·K⁻¹. This assumes a dilute classical gas; at sufficiently low temperature, discrete quantum translational states are not fully excited and the heat capacity falls below the classical prediction. Ludwig Boltzmann helped establish the statistical foundations behind this classical result.

The theorem follows from an equilibrium integral: for a coordinate x with energy term ax², the Boltzmann weight is proportional to e^(−βax²), and averaging gives ⟨ax²⟩=1/(2β)=kBT/2. Each independent quadratic coordinate contributes this amount, but constraints or nonquadratic terms alter the result. For example, a quartic potential x⁴ is not a quadratic degree of freedom and does not contribute exactly kBT/2. A model with f independent quadratic terms therefore has mean internal energy fkBT/2 and classical molar heat capacity fR/2. This provides a useful high-temperature benchmark for oscillator models. The theorem fails when discrete energy levels matter or the system has not reached thermal equilibrium. Quantum contributions emerge gradually as temperature rises. This explains the Dulong–Petit law at high temperatures, while quantum mechanics is needed for the low-temperature decline.

Quick check

What is the mean contribution of one independent quadratic energy term at classical equilibrium?

A classical harmonic oscillator has two independent quadratic terms. What is its total mean energy?

References

  1. Charles Kittel and Herbert Kroemer (1980). Thermal Physics
  2. L. D. Landau and E. M. Lifshitz (1980). Statistical Physics