University Physics IV · Limits of Classical Physics · 1.5

Heat Capacities & Frozen Degrees of Freedom

Equipartition's plateaus — 3R for a solid, 5/2 R for a diatomic gas — are real measurements that die on cooling. This lesson shows what kills them: a mode whose quantum ħω is several times k_BT cannot take a first step, so it stops contributing to dU/dT long before the atoms stop moving.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Equipartition is a theorem about a continuous variable: integrate a Boltzmann factor over a quadratic coordinate that runs over all the reals and each such coordinate returns exactly ½k_BT, whatever its stiffness. That is why it predicts plateaus with no temperature in them — 3R per mole for a monatomic solid, 5/2 R for a diatomic gas — and why those plateaus hold to a couple of percent at 300 K. The premise fails when the coordinate's energy comes in steps of ħω: the integral becomes a Boltzmann-weighted sum, the mean energy is Planck's ħω/(e^(ħω/k_BT) − 1), and one dimensionless ratio x = θ/T tells the whole story.

For x ≪ 1 the sum looks continuous and equipartition returns; for x ≫ 1 the mode is stuck in its ground state and adds nothing to dU/dT. It has not vanished and the atoms have not stopped — zero-point motion survives to absolute zero — it simply cannot buy the next quantum. Einstein made this concrete with 3N oscillators at one frequency, the first model to send C to zero at T = 0; Debye replaced the single frequency with the crystal's sound-wave spectrum and recovered the measured T³ tail, because long-wavelength modes have ω → 0 and never freeze.

The cost is that both assume a spectrum rather than derive one, so θ_E and θ_D are fitted numbers that drift with temperature.

Simple definition: A degree of freedom is frozen out when the quantum it must absorb, ħω, is far larger than the thermal energy k_BT available, so it stores almost no energy and contributes almost nothing to the heat capacity.
Example: Nitrogen's stretch has θ_vib = 3.37 × 10³ K, so at 300 K x = 11.2 and the Einstein term R x² e^x/(e^x − 1)² contributes 0.014 J mol⁻¹ K⁻¹ of the 8.31 the mode would hold classically; C_V, m reads 5/2 R = 20.8, not 7/2 R = 29.1 J mol⁻¹ K⁻¹.

Equipartition and its plateausC_V, m = (f/2)R monatomic gas 3/2 R diatomic gas 5/2 R solid 3R

A plateau with no temperature in it, so any measured slope in C_V is already outside the theorem.

f counts quadratic energy terms, kinetic and potential alike; R = 8.314 J mol⁻¹ K⁻¹, so 3R = 24.94 J mol⁻¹ K⁻¹

Mean energy of a quantised mode⟨E⟩ = ½ħω + ħω / (e^(ħω/k_BT) − 1)

Planck's cavity result moved onto matter; expand it for ħω ≪ k_BT and k_BT per mode comes back.

ħω is the level spacing in J; the zero-point ½ħω is a constant, so it cancels out of C = dU/dT

The ratio that decidesθ = ħω/k_B x = θ/T

One number per mode: x ≪ 1 classical, x ≫ 1 frozen, and the crossover spans about a decade in T.

θᵣₒₜ = ħ²/2Ik_B, θ_vib = ħω_vib/k_B, in K. H₂: 85.4 K and 6.33 × 10³ K. N₂: 2.88 K and 3.37 × 10³ K.

Einstein solidC_V, m = 3R x² e^x / (e^x − 1)², with x = θ_E/T

Correct at both ends — 3R as x → 0, zero as x → ∞ — but the approach to zero is exponential, and no measured solid falls that fast.

θ_E = ħω_E/k_B. Einstein's 1907 diamond fit took θ_E ≈ 1.32 × 10³ K; copper fits with θ_E ≈ 0.75 θ_D ≈ 257 K.

Debye T³ lawC_V, m → (12π⁴/5) R (T/θ_D)³ = 1944 (T/θ_D)³ J mol⁻¹ K⁻¹

A power law with both exponent and coefficient fixed, so once θ_D is known a log–log plot of C against T tests Debye with no free parameter.

θ_D = ħω_D/k_B; valid for T ≲ θ_D/50. Copper 343 K, lead 105 K, diamond 2.23 × 10³ K.

The Debye cutoff, from countingω_D = vₛ (6π² n)¹⁄³

The cutoff is forced by the crystal owning exactly 3N modes, so θ_D follows from density and sound speed.

n = N/V in m⁻³; vₛ is the Debye average of one longitudinal and two transverse branches, 3/vₛ³ = 1/v_L³ + 2/v_T³ — not an arithmetic mean.

The plateaus equipartition earns

Equipartition says each quadratic term in a particle's energy holds ½k_BT in thermal equilibrium — and it is a theorem, not a guess: integrate e^{(−αq²/k_BT)} over all real q and the mean of αq² comes out ½k_BT whatever α is. Count the terms and you have the heat capacity. A monatomic gas has three kinetic terms, so C_V, m = 3/2 R = 12.47 J mol⁻¹ K⁻¹, which argon obeys to four figures. A diatomic gas adds two rotational terms: 5/2 R = 20.79, against 20.81 for N₂ at 298 K. An atom in a solid has three kinetic and three potential terms, giving Dulong–Petit's 3R = 24.94; the measured Cₚ, m at 298 K — a few per cent above C_V for a solid, 0.7 J mol⁻¹ K⁻¹ in copper's case — is 24.44 for copper, 24.20 for aluminium and 25.35 for silver. That is a serious record, and it contains no temperature at all.

Where the theorem breaks: the integral becomes a sum

The derivation assumed q ranges continuously. Quantise the mode — Eₙ = nħω — and the integral becomes a Boltzmann-weighted sum of a geometric series, giving ⟨E⟩ = ħω/(e^x − 1) with x = ħω/k_BT, the same expression Planck wrote for a cavity mode. Expand for small x: e^x − 1 = x + x²/2 + …, so ⟨E⟩ = k_BT(1 − x/2 + …). Equipartition is the x → 0 limit, exactly as Rayleigh–Jeans was. The energy corrections arrive early — at x = 0.5 the mode holds 0.771 k_BT, already 23% short; at x = 2 it holds 0.313 k_BT; at x = 10 it holds 4.5 × 10⁻⁴ k_BT — but the heat capacity is more patient, because the −x/2 term is the constant −½ħω and dU/dT removes it. Differentiating gives C = k_B x²e^x/(e^x − 1)² = k_B(1 − x²/12 + …): 0.979 k_B at x = 0.5, 0.724 k_B at x = 2, 0.496 k_B at x = 3, and only 4.5 × 10⁻³ k_B at x = 10. So a mode stays almost fully classical until its quantum is about the size of k_BT, loses half its heat capacity near x = 3, and beyond that dies as x²e^(−x) — because what C measures is the mode's willingness to take the next quantum when T rises, and there is no next quantum on offer.

One number per mode: θ = ħω/k_B

Turn the level spacing into a temperature and you can read the answer off before any algebra. For rotation θᵣₒₜ = ħ²/2Ik_B; for vibration θ_vib = ħω/k_B, which is 1.439 K for every cm⁻¹ of measured wavenumber. Hydrogen is the outlier twice over: its tiny moment of inertia puts θᵣₒₜ at 85.4 K, high enough that the rotational step is visible in an ordinary cryostat, and its stiff light bond puts θ_vib at 6.33 × 10³ K, so at 300 K one vibrational quantum costs 21 k_BT and the mode is dead. Nitrogen has θᵣₒₜ = 2.88 K and θ_vib = 3.37 × 10³ K, which is why every common diatomic gas sits flat at 5/2 R across the whole laboratory range. Chlorine is the exception that proves the rule: θ_vib = 806 K gives x = 2.70 at 298 K, and its measured C_V, m is 25.6, not 20.8.

Einstein's solid: right shape, wrong tail

Einstein modelled a solid as 3N independent oscillators all at one frequency ω_E. Each holds ħω_E/(e^x − 1), so differentiating gives C_V, m = 3R x²e^x/(e^x − 1)² with x = θ_E/T. As T → ∞ the bracket → 1 and Dulong–Petit returns; as T → 0 it goes to zero, which no classical model had ever managed and which the third law demands. Fitted with θ_E ≈ 1320 K, the curve tracked the anomalously low heat capacity of diamond that had puzzled everyone since Weber's measurements. But the low-temperature form is 3R x²e^(−x), an exponential, and measured solids do not fall exponentially. Take copper with θ_E = 0.75 θ_D ≈ 257 K at T = 20 K: x = 12.9 and the model gives 0.011 J mol⁻¹ K⁻¹ against a measured 0.46 — forty times too small, and the gap widens with every kelvin down.

Debye: the modes that never freeze

Einstein's error was independence. A crystal's normal modes are collective sound waves, and their frequencies run from ω_D all the way down to ω ≈ 2πvₛ/L, essentially zero. Debye counted them the way the cavity was counted — g(ω) = 3Vω²/2π²vₛ³ — and truncated at ω_D so the total is exactly 3N, which fixes ω_D = vₛ(6π²n)¹⁄³. Now the argument is pure counting. However cold it gets, the modes with ħω < k_BT are still classical and still take k_B each; their number is proportional to T³ because g ∝ ω². So C ∝ T³, and the exact coefficient is (12π⁴/5)R = 1944 J mol⁻¹ K⁻¹ divided by θ_D³. For copper, n = 8.47 × 10²⁸ m⁻³ and vₛ = 2.61 × 10³ m s⁻¹ give θ_D = 341 K, which is the calorimetric 343 K to better than 1%.

What both models cost

Neither model derives the vibrational spectrum; each assumes one. Einstein assumes a single frequency, which no crystal has. Debye assumes ω = vₛ k all the way to a sharp cutoff, which is right only for long wavelengths — real spectra bend over at the zone boundary and carry optical branches and van Hove peaks that neither shape contains. Both still fit, because C is an integral over g(ω) and integrals forgive detail. The tell is that θ_D is not a constant of the material: fit it separately at each temperature and copper's value drops roughly ten per cent below its 0 K figure near 30 K before recovering. A genuine constant would not drift. Measuring g(ω) takes inelastic neutron scattering, not calorimetry. And in a metal below about 4 K the conduction electrons take over — plot C/T against T² and the intercept is γ, the slope A.

Change one variable at a time

Make the relationship visible.

Interactive model

Debye temperature θ_D343 K

Einstein ratio θ_E / θ_D0.75

Temperature T20 K

Keep θ_D at 343 K and drag T down to 5 K: the Debye reading still holds near 6 mJ mol⁻¹ K⁻¹ while the Einstein curve has fallen off the bottom of the scale. Now sweep θ_E/θ_D — the curve slides sideways but never straightens into the slope-3 line. No choice of frequency repairs that.

θ_E / T12.86

EINSTEIN C_V0.011 J mol⁻¹ K⁻¹

T³ LAW C_V, VALID T < θ_D/50, CAPPED AT 3R0.385 J mol⁻¹ K⁻¹

QUANTA PER MODE ⟨n⟩0.000

Live interpretationθ_E / T: 12.86. EINSTEIN C_V: 0.011 J mol⁻¹ K⁻¹. T³ LAW C_V, VALID T < θ_D/50, CAPPED AT 3R: 0.385 J mol⁻¹ K⁻¹. QUANTA PER MODE ⟨n⟩: 0.000

Catch the common trap

Explain before calculating.

Hydrogen gas at 300 K has C_V, m ≈ 5/2 R = 20.8 J mol⁻¹ K⁻¹, not the 7/2 R = 29.1 that equipartition gives once the H–H stretch is counted. Which statement explains the missing R?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA 250 g block of copper (M = 63.55 g mol⁻¹) sits at 300 K. Predict its heat capacity from Dulong–Petit, compare with the measured specific heat capacity 0.385 J g⁻¹ K⁻¹, and say whether the classical plateau is justified given copper's Debye temperature θ_D = 343 K.

Amount of substance: n = 250 g ÷ 63.55 g mol⁻¹ = 3.934 mol.
Dulong–Petit assigns six quadratic terms per atom — three kinetic, three potential — so C_V, m = 3R = 3 × 8.314 = 24.94 J mol⁻¹ K⁻¹, and the block should have C = 3.934 × 24.94 = 98.1 J K⁻¹.
Measured: C = 0.385 × 250 = 96.3 J K⁻¹, or per mole 0.385 × 63.55 = 24.47 J mol⁻¹ K⁻¹ — 98.1% of 3R. That figure is at constant pressure, and C_V for a solid runs about 3% lower still.
Is the plateau earned? The stiffest mode has x = θ_D/T = 343/300 = 1.14, and every other mode has a smaller x, so most of the spectrum is already classical. The shortfall is the quantum correction that has not quite vanished.

AnswerC ≈ 98 J K⁻¹ predicted against a measured 96 J K⁻¹ at constant pressure. Dulong–Petit is 2% high on Cₚ and rather more on C_V, because at T = 0.87 θ_D the stiffest modes are still slightly frozen.

MediumChlorine's stretching vibration has wavenumber 560 cm⁻¹, and 1 cm⁻¹ corresponds to θ = 1.439 K. Find θ_vib, predict C_V, m for Cl₂ gas at 298 K including the vibrational term, and compare with the measured Cₚ, m = 33.9 J mol⁻¹ K⁻¹. Take θᵣₒₜ = 0.35 K.

θ_vib = 1.439 K × 560 = 806 K, so x = θ_vib/T = 806/298 = 2.70. One vibrational quantum costs 2.7 k_BT — enough to matter, not enough to be free.
Translation and rotation are fully classical here, since θᵣₒₜ = 0.35 K is far below 298 K. Those five quadratic terms give 5/2 R = 20.79 J mol⁻¹ K⁻¹.
The vibration contributes the Einstein term R x²e^x/(e^x − 1)². With x = 2.70: e^x = 14.88, so (e^x − 1)² = 192.7 and x² = 7.29. The bracket is 7.29 × 14.88 ÷ 192.7 = 0.563, and the term is 0.563 × 8.314 = 4.68 J mol⁻¹ K⁻¹.
Total C_V, m = 20.79 + 4.68 = 25.47 J mol⁻¹ K⁻¹. Measured, using Cₚ − C_V = R for an ideal gas: 33.9 − 8.31 = 25.6 J mol⁻¹ K⁻¹.
Equipartition's two integer answers bracket the truth and miss it: 5/2 R = 20.8 is 19% low, 7/2 R = 29.1 is 14% high. Only the partial occupancy 0.563 R lands on the measurement.

Answerθ_vib = 806 K and C_V, m = 25.5 J mol⁻¹ K⁻¹, against 25.6 measured. The stretch is 56% switched on — a fraction, not an integer.

HardCopper has n = 8.47 × 10²⁸ atoms m⁻³ and a Debye-averaged sound speed vₛ = 2.61 × 10³ m s⁻¹ over one longitudinal and two transverse branches. (a) Find ω_D and θ_D. (b) Use the T³ law for the lattice heat capacity at 5.0 K. (c) The measured value there is 9.6 mJ mol⁻¹ K⁻¹ — account for the rest, given the electronic coefficient γ = 0.695 mJ mol⁻¹ K⁻².

Mode counting fixes the cutoff: ω_D = vₛ(6π²n)¹⁄³. Here 6π² = 59.22, so 6π²n = 5.02 × 10³⁰ m⁻³, whose cube root is 1.712 × 10¹⁰ m⁻¹. Then ω_D = 2.61 × 10³ × 1.712 × 10¹⁰ = 4.468 × 10¹³ rad s⁻¹.
θ_D = ħω_D/k_B: 1.055 × 10⁻³⁴ J s × 4.468 × 10¹³ s⁻¹ = 4.713 × 10⁻²¹ J, and dividing by 1.381 × 10⁻²³ J K⁻¹ gives 341 K — within 1% of the calorimetric 343 K, from density and sound speed alone.
T/θ_D = 5.0/341 = 0.01466, comfortably inside the T ≲ θ_D/50 = 6.8 K window where the T³ law holds. C_lattice = 1944 × (0.01466)³ = 1944 × 3.15 × 10⁻⁶ = 6.1 × 10⁻³ J mol⁻¹ K⁻¹, or 6.1 mJ mol⁻¹ K⁻¹.
The 3.5 mJ mol⁻¹ K⁻¹ shortfall is the conduction electrons: γT = 0.695 × 5.0 = 3.5 mJ mol⁻¹ K⁻¹, and 6.1 + 3.5 = 9.6 mJ mol⁻¹ K⁻¹ as measured. Plotting C/T against T² separates the two — intercept γ, slope A.
Scale check. Dulong–Petit predicts 24.94 J mol⁻¹ K⁻¹, so the whole measured value at 5 K is 2600 times smaller. An Einstein solid with θ_E = 0.75 θ_D = 256 K would give 3R x²e^(−x)/(1 − e^(−x))² at x = 51.2, which is 4 × 10⁻¹⁸ J mol⁻¹ K⁻¹ — fifteen orders of magnitude below the lattice term the calorimeter actually reads.

Answerω_D = 4.47 × 10¹³ rad s⁻¹ and θ_D = 341 K; the lattice gives 6.1 mJ mol⁻¹ K⁻¹ and the electrons 3.5, summing to the measured 9.6. Einstein's exponential misses by 10¹⁵.