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PU-203 / L23 · lecture

Equipartition and Where It Fails

~7 min read · Tier toggle changes depth (T1 plain · T2 standard · T3 rigour)

BEFORE YOU START

You should be comfortable with the Boltzmann distribution and the definition of the canonical partition function. Familiarity with quadratic terms in Hamiltonians is assumed.

Why this matters

Equipartition is the simplest bridge between microscopic motion and macroscopic heat capacity, yet its failure is the first crack in classical physics.

The law of equipartition of energy states that every quadratic degree of freedom in the Hamiltonian contributes an average energy of \(\frac{1}{2}k_B T\). For a classical ideal gas, this predicts a constant molar heat capacity of \(\frac{3}{2}R\) for monatomic gases and \(\frac{5}{2}R\) for diatomic gases (ignoring vibration). While this works well at high temperatures, it fails spectacularly at low temperatures or for high-frequency modes, predicting heat capacities that are independent of temperature. This failure is not a minor error; it is the direct evidence that quantum mechanics is required to describe thermal systems correctly. Understanding where equipartition breaks down is the gateway to understanding quantum statistics and the specific heat anomalies observed in solids and molecules.

The Classical Intuition

In classical mechanics, the energy of a system is a continuous function of its coordinates and momenta. If the Hamiltonian \(H\) can be written as a sum of independent quadratic terms, such as \(H = \sum_i a_i x_i^2 + \sum_j b_j p_j^2\), the probability of finding the system in a particular state is proportional to \(e^{-H/k_B T}\). Because the exponential of a quadratic term is a Gaussian, the average value of each quadratic term \(a_i x_i^2\) or \(b_j p_j^2\) is identical, regardless of the coefficients \(a_i\) or \(b_j\). This is the essence of equipartition: the "stiffness" of the spring or the mass of the particle does not matter; only the number of independent quadratic terms matters. Each term acts as a reservoir that shares the thermal energy equally. This leads to the prediction that the internal energy \(U\) is simply \(N f \frac{1}{2} k_B T\), where \(f\) is the number of quadratic degrees of freedom per particle. Consequently, the heat capacity \(C_V = \left(\frac{\partial U}{\partial T}\right)_V\) becomes a constant, independent of temperature. This is a powerful and simple result, but it relies entirely on the assumption that the energy levels are continuous and that the thermal energy \(k_B T\) is large compared to the spacing between energy levels.

Mathematical Statement of the Law

For a system in the canonical ensemble, the average energy associated with a degree of freedom \(q\) that appears quadratically in the Hamiltonian as \(\frac{1}{2} m \dot{q}^2\) or \(\frac{1}{2} k q^2\) is derived from the partition function. The result, derived in the embedded derivation below, is that the mean energy per quadratic degree of freedom is \(\frac{1}{2} k_B T\).

\[ \langle \epsilon \rangle = \frac{1}{2} k_B T \quad \text{per quadratic degree of freedom} \]

If a molecule has \(f\) such degrees of freedom, the total internal energy per mole is \(U = N_A f \frac{1}{2} k_B T = \frac{f}{2} R T\). The molar heat capacity at constant volume is then \(C_V = \frac{f}{2} R\). For a monatomic ideal gas, \(f=3\) (three translational degrees of freedom), so \(C_V = \frac{3}{2} R\). For a diatomic gas at moderate temperatures, we add two rotational degrees of freedom (rotation about the axis of symmetry is negligible due to small moment of inertia), giving \(f=5\) and \(C_V = \frac{5}{2} R\). If vibrational modes are active, they contribute two quadratic terms (kinetic and potential energy of the oscillator), adding \(k_B T\) to the energy and \(R\) to the heat capacity, resulting in \(C_V = \frac{7}{2} R\).

The Quantum Failure: Freezing Out

The failure of equipartition arises because quantum mechanics discretizes energy levels. A harmonic oscillator, for example, has energy levels \(E_n = \hbar \omega (n + \frac{1}{2})\). The average energy is not \(\frac{1}{2} k_B T\) but rather \(\langle E \rangle = \frac{\hbar \omega}{2} + \frac{\hbar \omega}{e^{\hbar \omega / k_B T} - 1}\). At high temperatures, where \(k_B T \gg \hbar \omega\), the exponential term can be expanded, and \(\langle E \rangle \approx k_B T\), recovering the classical result (including the zero-point energy which is often ignored in classical thermodynamics). However, at low temperatures, where \(k_B T \ll \hbar \omega\), the exponential term becomes very large, and \(\langle E \rangle \approx \frac{\hbar \omega}{2}\). The energy becomes nearly constant, meaning the heat capacity \(C_V = \frac{\partial \langle E \rangle}{\partial T}\) approaches zero. This phenomenon is known as "freezing out." The vibrational modes of a diatomic molecule, for instance, have characteristic temperatures \(\Theta_{\text{vib}} = \hbar \omega / k_B\) often in the thousands of Kelvin. At room temperature, \(k_B T\) is much smaller than \(\hbar \omega\), so the vibrational modes are frozen out and do not contribute to the heat capacity. This explains why the experimental \(C_V\) for diatomic gases is \(\frac{5}{2} R\) rather than \(\frac{7}{2} R\) at standard conditions.

Implications for Solids: The Dulong-Petit Law

The same principle applies to solids. In the classical model (Einstein or Debye model in the high-temperature limit), each atom in a solid lattice is treated as a 3D harmonic oscillator. There are \(6N\) quadratic terms (3 kinetic and 3 potential per atom), each contributing \(\tfrac12 k_BT\), so \(U = 3Nk_BT\). Equipartition predicts a molar heat capacity of \(C_V = 3R\), known as the Dulong-Petit law. This law holds well for many solids at room temperature because the characteristic Debye temperature \(\Theta_D\) is often comparable to or lower than room temperature. However, at low temperatures, the quantum nature of the lattice vibrations (phonons) becomes dominant. The heat capacity drops as \(T^3\) (Debye law) rather than remaining constant. This temperature dependence is a direct violation of the classical equipartition theorem and provides strong evidence for the quantization of vibrational energy in solids.

The derivation

◆ DERIVATION EMBED · D-020 · not re-derived here

Equipartition and Its Failure

Each quadratic degree of freedom contributes ½kT classically, but the partition function shows this dies once level spacings exceed kT.

⟨ε⟩ = ½kBT  per quadratic degree of freedom
Open full derivation →
Tier 3 · watch out

A subtle point: discreteness alone does not destroy equipartition. Equipartition holds whenever the level spacing is small compared with \(k_BT\), because then the sum over levels is well approximated by the classical integral. It fails for a given mode when \(k_BT\) drops to the order of the spacing (\(\hbar\omega\) for a vibration, \(\hbar^2/2I\) for a rotation): the Boltzmann factor then suppresses all excited levels and the mode stops absorbing energy. The zero-point energy \(\tfrac12\hbar\omega\) is part of \(U\) but, being independent of \(T\), contributes nothing to \(C_V\). C

Worked example

Problem. Consider a diatomic molecule with a vibrational frequency corresponding to a characteristic temperature \(\Theta_{\text{vib}} = 3000\ \mathrm{K}\). Calculate the molar heat capacity at constant volume \(C_V\) at \(T = 300\ \mathrm{K}\), assuming the vibrational mode is the only quantum-restricted degree of freedom and rotations are fully active.

  1. Identify the active degrees of freedom. Translational: 3 quadratic terms. Rotational: 2 quadratic terms (diatomic). Vibrational: 2 quadratic terms (kinetic + potential), but quantum restricted.
  2. Classical contribution from translation and rotation: \(C_{V,\text{classical}} = \frac{3}{2}R + \frac{2}{2}R = \frac{5}{2}R\).
  3. Quantum contribution from vibration. The average energy of a quantum harmonic oscillator is \(\langle E_{\text{vib}} \rangle = \frac{\hbar \omega}{2} + \frac{\hbar \omega}{e^{\Theta_{\text{vib}}/T} - 1}\). The heat capacity contribution is \(C_{V,\text{vib}} = R \left( \frac{\Theta_{\text{vib}}/T}{e^{\Theta_{\text{vib}}/T} - 1} \right)^2 e^{\Theta_{\text{vib}}/T}\).
  4. Calculate the exponent: \(\Theta_{\text{vib}}/T = 3000/300 = 10\).
  5. Evaluate the vibrational term: \(C_{V,\text{vib}} = R \left( \frac{10}{e^{10} - 1} \right)^2 e^{10}\). Since \(e^{10} \approx 22026\), the term is \(R \left( \frac{10}{22025} \right)^2 (22026) \approx R \frac{100}{22025^2} \cdot 22026 \approx R \frac{100}{22025} \approx 0.0045 R\). This is negligible compared to \(R\).
  6. Total \(C_V \approx \frac{5}{2}R + 0.0045 R \approx 2.5045 R\).
\[ C_V \approx 2.50 R \]

Units check. \(R\) is \(\mathrm{J/(mol \cdot K)}\), which is the correct unit for molar heat capacity. The vibrational contribution is small because \(T \ll \Theta_{\text{vib}}\), confirming the freezing out of the mode.

CHECK YOURSELF

Three quick tests

  1. Why does the heat capacity of a diatomic gas approach \(\frac{7}{2}R\) at very high temperatures?
    AnswerAt high temperatures, \(k_B T \gg \hbar \omega_{\text{vib}}\), so the vibrational modes are fully excited and contribute \(R\) to the heat capacity, adding to the \(\frac{5}{2}R\) from translation and rotation.
  2. What is the classical prediction for the molar heat capacity of a solid?
    Answer\(3R\), known as the Dulong-Petit law, arising from six quadratic terms per atom (three kinetic, three potential), each worth \(\tfrac12 k_BT\).
  3. Does the zero-point energy contribute to the heat capacity?
    AnswerNo, because the zero-point energy \(\frac{1}{2}\hbar \omega\) is independent of temperature, so its derivative with respect to temperature is zero.
PROBLEMS

Problem set for PU-203 →

Each derivation in this unit ends with five graded problems with full solutions.

NOTES

Lecture notes checked line by line by the editor; every number in the worked example was recomputed.

Derivations cited: D-020 · Threads: energy, matter, chance