Equipartition, and the low-temperature failure that demanded quanta

Derivation · the atom

Equipartition, and the low-temperature failure that demanded quanta

D-020 Home PU-203 Threads energy · matter Depends on D-013 · D-010 verified

Statement

Each quadratic degree of freedom contributes ½k_BT to the mean energy of a classical system — and the observed failure of this rule at low temperature is direct evidence for energy quantisation.

Why it matters

This one line predicts the heat capacity of every classical system by counting: three translational terms for a gas atom, more for rotation and vibration, six per atom for a solid. It works beautifully at room temperature and then, on cooling, it breaks in a way classical physics cannot repair. That breakage was recognised as fundamental before it was explained, and it is one of the cleanest fingerprints of quantisation in all of thermodynamics.

Assumptions

The energy is quadratic in the variable considered.True for kinetic terms and harmonic potentials; false for anharmonic potentials and for relativistic kinetic energy, where the counting changes.

The variable is continuous.This is what lets step 2 replace a sum by an integral — and it is the assumption that fails, whenever the level spacing exceeds k_BT.

The variable ranges over the whole real line.Restricting the range changes the Gaussian integral, and with it the ½ prefactor.

Derivation

⟨ε⟩ = [ ∫ ax² e^−βax² dx ] ⁄ [ ∫ e^−βax² dx ]

Boltzmann average (D-013) for an energy term ε = ax², where x is a velocity component or a displacement. A

⟨ε⟩ = −∂/∂β · ln ∫ e^−βax² dx = −∂/∂β · ln ( π ⁄ βa )^1/2

Write the numerator as a β-derivative of the denominator; the Gaussian integral is then evaluated once and differentiated, avoiding the moment integral. B

= 1 ⁄ (2β) = ½k_BT

The constant a cancels — this is why the result is universal: spring stiffness and particle mass make no difference whatsoever. B

Z = Σ_n e^{−β(n+½)ℏω} = e^−βℏω/2 ⁄ ( 1 − e^−βℏω ) ⇒ ⟨E⟩ = ℏω⁄2 + ℏω ⁄ ( e^ℏω/k_BT − 1 )

Redo the average as a discrete sum over E_n = (n+½)ℏω (D-010) instead of an integral — the rigorous version, of which step 3 is only the high-temperature limit. C

Result

⟨ε⟩ = ½k_BT per quadratic degree of freedom

Reading. Monatomic gas: three translational terms give U = 3⁄2 Nk_BT, C_V = 3⁄2 Nk_B. A diatomic gas adds two rotational terms → 5⁄2 Nk_B; adding the vibrational mode (kinetic and potential, two more terms) would give 7⁄2 Nk_B. A solid with three vibrational modes per atom, each counted twice, gives C = 3Nk_B — the Dulong–Petit law.

Units check. k_B carries energy per kelvin, so k_BT is an energy; the ½ is dimensionless, and a has cancelled, leaving ⟨ε⟩ as a pure energy per degree of freedom. Correct.

Limiting cases

High temperature (k_BT ≫ ℏω): the exponential in step 4 expands and ⟨E⟩ → k_BT, recovering equipartition exactly. The classical rule is a limit, not a wrong theory.
Low temperature (k_BT ≪ ℏω): the second term is exponentially suppressed and the mode contributes essentially nothing — it is frozen out.
Measured diatomic heat capacities sit at 5⁄2 Nk_B at room temperature, not 7⁄2; solids fall below 3Nk_B on cooling, tending to zero as T³.

Breaks when

Level spacing is comparable to or larger than k_BT — the whole point above.
The energy is not quadratic — an ultra-relativistic gas has ε = pc and gives k_BT per momentum component, not ½.
Applied to photons or phonons, where the mode count is not fixed — the naive application gives the ultraviolet catastrophe.

Failure modes

Counting a vibrational mode once. It contributes twice, kinetic and potential — this halving error is extremely common.
Assuming a degree of freedom that exists must be active. Existence and excitation are different questions, and the difference is ℏω⁄k_BT.
Presenting equipartition as simply wrong. It is exactly right in its regime, and knowing the regime is the physics.

Worked number

For CO (from D-010), ℏω = 0.27 eV and k_BT = 0.026 eV at 300 K, so ℏω⁄k_BT ≈ 10.4 and the vibrational contribution to C_V is suppressed by roughly e^−10.4 ≈ 3 × 10⁻⁵. The mode exists; it is simply never excited. Rotational spacings are far smaller (~10⁻⁴ eV), so rotations are fully active — which is exactly why the observed value is 5⁄2 Nk_B and not 7⁄2. Heat the gas above about 3,000 K and the vibrational contribution appears, as observed.

Plot C_V(T) and overlay the classical limit →

Discussion

The arresting feature of the result is what is missing from it. The mean energy ⟨ε⟩ = ½k_BT retains no trace of the constant a — not the stiffness of a spring, not the mass of a molecule, not the moment of inertia. The reason is visible if you substitute u = (βa)^1/2x in both integrals of step 1: the factor a and the factor β always appear glued together in the combination βax², so a single rescaling of the dummy variable expels a from the problem entirely and leaves the answer a pure function of β. Physically, a stiff mode and a soft mode reach the same thermal energy at the same temperature; the stiff one simply sits with a smaller amplitude. Temperature, not the microscopic detail, is the shared currency of thermal contact, and it is spent equally on every quadratic term.

That "equally" is the whole content of the theorem, and it is why the result is structured as a flat ½k_BT per term rather than something mode-dependent. The ½ is the fingerprint of a Gaussian phase-space measure: a quadratic energy makes the Boltzmann weight e^−βax² a Gaussian, whose logarithm carries exactly one factor of ½lnβ, and −∂/∂β of that is 1/(2β). Change the power law and you change the fraction in a completely predictable way, which is the generalised equipartition theorem: for an energy term proportional to |x|^s over the whole real line, ⟨ε⟩ = k_BT/s. Quadratic (s = 2) gives ½; linear (s = 1), as for an ultra-relativistic particle, gives a full k_BT.

Seen this way, equipartition and its collapse are a single organising principle running through classical and quantum statistical physics. The same "½k_BT per quadratic slot" underlies the Dulong–Petit law for solids, the thermal (Johnson–Nyquist) voltage noise of a resistor — where ⟨½C V²⟩ = ½k_BT fixes the mean-square fluctuation of a capacitor's voltage — and Einstein's 1905 account of Brownian motion, where ⟨½m v²⟩ = ½k_BT per component ties a visible jiggle to k_B. Its most famous failure, the ultraviolet catastrophe, is not a different disease: assigning k_BT to each electromagnetic mode is exactly equipartition applied to an unbounded set of oscillators, and it diverges for precisely the reason a diatomic vibration freezes — the classical average ignores the discrete ladder E_n = (n+½)ħω. Whenever the level spacing ħω becomes comparable to k_BT, the continuous integral of step 2 is no longer a legitimate stand-in for the sum, and the mode either freezes out (matter) or is exponentially cut off at high frequency (radiation). One inequality, ħω ≳ k_BT, governs both.

Common misconceptions. The rule is ½k_BT per quadratic term in the energy, not per particle and not loosely "per degree of freedom": a one-dimensional oscillator has a kinetic term and a potential term, so it carries k_BT, and forgetting the potential half is the single most common error. A second confusion is treating a frozen mode as absent — the vibrational mode of room-temperature CO physically exists, it is simply never excited, and existence versus excitation is settled entirely by the ratio ħω/k_BT. A third is calling equipartition "wrong". It is exactly correct in its regime — the high-temperature limit — and the quantum result reduces to it precisely there; knowing the regime is the physics.

Worked examples

Example 1 — Heat capacity of an Einstein solid below its Einstein temperature. Model copper as 3N independent quantum oscillators of a single frequency ω, with Einstein temperature θ_E = ħω/k_B = 240 K. Find the molar heat capacity at T = 80 K and compare with the classical Dulong–Petit value.

⟨E⟩ = 3N [ ½ħω + ħω / (e^ħω/k_BT − 1) ]

Three oscillators per atom, each with the quantum-oscillator mean energy from the derivation. The ½ħω zero-point term is constant in T and will drop out of the heat capacity.

C_V = (∂⟨E⟩/∂T) = 3Nk_B x² e^x / (e^x − 1)², x ≡ ħω/k_BT = θ_E/T

Differentiate the second term; the chain rule brings down x²e^x/(e^x−1)². This is the Einstein heat-capacity function. As T → ∞ (x → 0) it → 3Nk_B, recovering equipartition.

x = θ_E/T = 240 / 80 = 3.00

The level spacing is three times the thermal energy, so we expect substantial freeze-out.

e^3.00 = 20.09; C_V = 3R · (3.00)² · 20.09 / (19.09)² = 3R · (180.8 / 364.4) = 3R · 0.496

Per mole, 3Nk_B = 3R. The bracket 0.496 is the fraction of the classical value that survives.

C_V = 3 · 8.314 · 0.496 = 12.4 J mol⁻¹ K⁻¹ vs 3R = 24.9 J mol⁻¹ K⁻¹

The Dulong–Petit value 3R is the high-temperature ceiling.

C_V(80 K) ≈ 12.4 J mol⁻¹ K⁻¹ ≈ ½ (3R)

Answer. At T = 80 K ≈ θ_E/3, roughly half of each oscillator's classical heat capacity has frozen out, giving C_V ≈ 12.4 J mol⁻¹ K⁻¹ against the classical 24.9. Cool the copper further and C_V collapses toward zero; the Einstein model gives an exponential fall, and the more accurate Debye model the observed T³.

Example 2 — A vibration that is not frozen: iodine gas at room temperature. The derivation showed CO's stiff bond (ħω ≈ 0.27 eV) is frozen at 300 K. Iodine, I₂, has a much softer bond, vibrational wavenumber ν̃ = 214 cm⁻¹. Find the vibrational contribution to its molar C_V at 300 K and the total C_V.

x = ħω/k_BT = ν̃ / (k_BT/hc)

Work in wavenumbers. At 300 K the thermal energy in the same units is k_BT/hc = 208.5 cm⁻¹.

x = 214 / 208.5 = 1.026

Level spacing and thermal energy are comparable — the borderline case, where neither the classical integral nor total freeze-out is a good approximation.

C_V,vib = R x² e^x / (e^x − 1)²

One vibrational mode = two quadratic terms, so its classical ceiling is R (not ½R). This single Einstein function captures the kinetic and potential halves together.

e^1.026 = 2.790; C_V,vib = R · (1.026)² · 2.790 / (1.790)² = R · (2.937 / 3.204) = 0.917 R

About 92% of the classical vibrational heat capacity is active.

C_V = C_trans + C_rot + C_vib = (3/2)R + R + 0.917R = 3.417 R

Translation (3 terms) + rotation of a linear molecule (2 terms = R) + the partially active vibration.

C_V,vib ≈ 0.92R ≈ 7.6 J mol⁻¹ K⁻¹, C_V ≈ 3.42R ≈ 28.4 J mol⁻¹ K⁻¹

Answer. Because iodine's bond is soft, ħω ≈ k_BT at room temperature and the vibration is nearly fully active, contributing 7.6 J mol⁻¹ K⁻¹. The molecule sits close to the classical (7/2)R that CO only reaches above ~3000 K — a direct, measurable demonstration that freeze-out is controlled by ħω/k_BT and nothing else.

Problems

Bookkeeping the quadratic terms. Using equipartition alone, predict the molar C_V of (a) argon gas, (b) nitrogen gas at 300 K, (c) nitrogen with its vibration fully active, and (d) crystalline diamond. Give each as a multiple of R and in J mol⁻¹ K⁻¹.
Solution
(a) Monatomic: 3 translational terms → C_V = (3/2)R = 12.5 J mol⁻¹ K⁻¹. (b) Diatomic with rotation active, vibration frozen: 3 translational + 2 rotational = 5 half-terms → (5/2)R = 20.8. (c) Add the vibrational mode (2 more terms, kinetic + potential) → (7/2)R = 29.1. (d) A solid has 3 vibrational modes per atom, each with 2 quadratic terms → 3R = 24.9 (Dulong–Petit). Diamond in fact sits far below this at 300 K because its θ_E is very high (~1300 K), a foreshadowing of Problem 4.
The generalised theorem and a relativistic gas. For an energy term ε = A|x|^s with x over the whole real line, prove ⟨ε⟩ = k_BT/s. Then find U and C_V for an ultra-relativistic monatomic gas (ε = pc for each of three momentum components), and compare with the ordinary non-relativistic gas.
Solution
Following step 2 of the derivation, ⟨ε⟩ = −∂/∂β ln ∫ e^{−βA|x|^s} dx. Substitute u = β^1/sx, so dx = β^−1/sdu and the integral becomes β^−1/s ∫ e^{−A|u|^s} du, a constant times β^−1/s. Then ln(...) = −(1/s)lnβ + const, and −∂/∂β gives 1/(sβ) = k_BT/s. For quadratic s = 2 this is ½k_BT; for the linear relativistic term s = 1 it is a full k_BT per component. Three components give U = 3Nk_BT and C_V = 3Nk_B = 3R = 24.9 J mol⁻¹ K⁻¹ — exactly double the non-relativistic (3/2)R. (This is why the interior of a hot, radiation-pressure-dominated star has an adiabatic index near 4/3 rather than 5/3.)
Oxygen: nearly frozen, then thawing. The vibrational wavenumber of O₂ is 1580 cm⁻¹. Compute the fraction of the classical vibrational heat capacity active at (a) 300 K and (b) 2000 K. Take k_BT/hc = 208.5 cm⁻¹ at 300 K (and scale linearly with T).
Solution
Fraction active = x²e^x/(e^x−1)² with x = 1580/(k_BT/hc). (a) At 300 K, x = 1580/208.5 = 7.58; e^7.58 = 1957, so fraction = (57.4)(1957)/(1956)² = 1.123×10⁵/3.826×10⁶ = 0.029. Only ~3% active — essentially frozen, so O₂ shows the expected (5/2)R at room temperature. (b) At 2000 K, k_BT/hc = 208.5×(2000/300) = 1390 cm⁻¹, x = 1580/1390 = 1.137; e^1.137 = 3.117, fraction = (1.293)(3.117)/(2.117)² = 4.030/4.482 = 0.90. About 90% active. The vibration thaws over roughly a thousand kelvin, not abruptly.
Where is a mode "half active"? Define a vibration as half active when C_V,vib = R/2. Solve the transcendental condition for x = ħω/k_BT numerically, then find the temperature at which O₂ (1580 cm⁻¹) reaches this point. Note 1 cm⁻¹ ↔ 1.4388 K.
Solution
Require x²e^x/(e^x−1)² = ½. Iterate: at x = 2.90, value = 0.518; at x = 3.00, value = 0.496; interpolating gives x ≈ 2.98. So a mode is half-frozen when the thermal energy is about one third of the level spacing — freeze-out is well underway before k_BT even reaches ħω. For O₂, θ_vib = ħω/k_B = 1580 × 1.4388 = 2273 K, so T_½ = θ_vib/2.98 = 763 K. Below ~760 K the O₂ vibration contributes less than half of R; the value 0.90R at 2000 K found in Problem 3 is consistent with this.
The ultraviolet catastrophe as equipartition, and Planck's cure. The number of electromagnetic modes per unit volume with angular frequency in [ω, ω+dω] is dn = (ω²/π²c³) dω (two polarisations included). (a) Assign each mode the classical ⟨E⟩ = k_BT and write the spectral energy density u(ω); show it diverges. (b) Replace ⟨E⟩ by the quantum-oscillator value from the derivation (drop the zero point) and obtain the Planck law; show it recovers part (a) when ħω ≪ k_BT and is exponentially cut off when ħω ≫ k_BT. (c) Integrate to recover the Stefan–Boltzmann T⁴ law.
Solution
(a) With ⟨E⟩ = k_BT per mode, u(ω) = (ω²/π²c³)·k_BT — the Rayleigh–Jeans law. It grows without bound as ω → ∞ and ∫u dω diverges: the ultraviolet catastrophe. This is precisely equipartition applied to infinitely many oscillators, exactly the assumption (continuous energy) flagged in the derivation. (b) The quantised mode has ⟨E⟩ = ħω/(e^ħω/k_BT−1), so
u(ω) = ħω³ / [ π²c³ (e^ħω/k_BT − 1) ]
the Planck law. For ħω ≪ k_BT, e^x−1 ≈ x = ħω/k_BT, and u → ω²k_BT/π²c³, the Rayleigh–Jeans result — equipartition is the low-frequency limit. For ħω ≫ k_BT the denominator ≈ e^ħω/k_BT and u is exponentially suppressed — the same freeze-out that silences a stiff vibration silences the high-frequency modes, curing the catastrophe. (c) Total energy density U/V = ∫₀^∞ u dω. Substitute y = ħω/k_BT: U/V = (k_BT)⁴/(π²c³ħ³) ∫₀^∞ y³/(e^y−1) dy. The integral is π⁴/15, giving U/V = (π²k_B⁴/15ħ³c³) T⁴ — the Stefan–Boltzmann law, with the T⁴ dependence emerging entirely from the quantum cutoff that classical equipartition lacked.