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The equipartition theorem

Use the equipartition theorem and molecular degrees of freedom to predict internal energy and heat capacity.

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  • University Physics Year 1
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Translation, rotation and vibration store energy differently, but an independent quadratic term has a simple classical equilibrium average. The equipartition theorem connects that average to internal energy, heat capacity and fluctuation size. Its useful rule counts energy terms, with explicit limits on when the classical model applies.

At a glance

  • Prerequisites: kinetic energy, degrees of freedom, ideal gas internal energy.
  • Outcomes: apply equipartition to translational, rotational, and vibrational modes.
  • Key result: each quadratic term contributes (1/2)kB T per particle.
  • Common trap: counting one vibrational mode as (1/2)kBT instead of kBT.

Identify the quadratic terms

Classical equilibrium system with Hamiltonian terms quadratic in coordinates/momenta:

  • Translational: pₓ²/2m, py²/2m, pz²/2m.
  • Rotational (rigid rotor): quadratic angular momentum terms.
  • Vibrational mode: kinetic + potential quadratic pair.

Assumption:

  • Use classical equilibrium and independent quadratic variables ranging over a sufficiently unrestricted domain. Quantum level spacing must be small compared with kBT for the classical approximation; a constrained or non-quadratic term cannot simply be counted with this rule.

Physical interpretation:

  • Equipartition is a statement about energy sharing at equilibrium: if the energy contains a sum of independent quadratic pieces, thermal fluctuations spread energy equally among them.
  • “Quadratic term” is the key phrase. The theorem does not say “each physical mode always gets the same energy”; it says each quadratic contribution to the energy gets 1/2 kBT (when that contribution is active).
  • Counting the active quadratic terms f is the bookkeeping step (see UY1: counting the number of modes).

Connect the Gaussian weight to mean energy

Derivation spine (why 1/2 kBT appears):

  • In the canonical (fixed-T) distribution, a quadratic variable x has a Gaussian weight e^(-ax²/(kBT)).
  • Gaussian integrals give ⟨x²⟩ ∝ kBT/a, so the mean energy stored in ax² scales linearly with T. For a positive quadratic coefficient a and x ranging over the real line, the Gaussian second moment is
⟨x²⟩ = (∫_(-∞)^∞x²e^(-ax²/(kBT)) dx)/(∫_(-∞)^∞e^(-ax²/(kBT)) dx) = kBT/2a.

Multiplying by a gives ⟨ax²⟩ = kBT/2. For momentum, a = 1/(2m); for a harmonic displacement, a = κ/2. The coefficient changes the width of the distribution, while the mean quadratic energy is fixed by temperature.

Equipartition theorem (classical):

⟨each quadratic term⟩ = (1/2)kB T.

If the thermal energy of a molecule is represented by f active quadratic terms,

⟨Eₜₕₑᵣₘₐₗ⟩ = (f/2)kB T, Uₜₕₑᵣₘₐₗ = (f/2)NkB T = (f/2)nRT.

Thus, for constant f, the molar heat capacity at constant volume is

CV = (1/n)((∂ U)/(∂ T))V = (f/2)R.

A temperature-independent energy offset can be added to U without changing CV. In a changing activation regime, use the actual U(T) or measured capacity and differentiate or integrate; do not insert a temperature-dependent effective f into a fixed-f formula and assume CV = fR/2 still follows.

For a classical harmonic coordinate with restoring constant κ, E = p²/2m + (1/2)κ x², there are two quadratic terms. Each averages kBT/2, so ⟨E⟩ = kBT, ⟨x²⟩ = kBT/κ. A stiffer oscillator has smaller displacement fluctuations at the same temperature, while its mean classical thermal energy stays unchanged. This is why the coefficient of a quadratic term does not determine its mean energy.

Examples:

  • Monatomic ideal gas: f = 3 (translation only) ⇒ CV = (3/2)R.
  • Diatomic with classical rotation but frozen vibration: f = 5 (3 translational + 2 rotational) ⇒ CV = (5/2)R.
  • If one vibrational mode activates, add 2 quadratic terms ⇒ f = 7, so CV = (7/2)R.

Units check:

[CV] = J/(mol K).

From (f/2)R, units are inherited from R.

Physical sanity check:

  • As temperature rises, extra modes may become thermally active, so CV can rise smoothly between approximate plateaus. A fixed integer f describes a limiting regime, not the activation transition itself.

Limit checks:

  • Equipartition is a high-temperature / classical limit: it works when the mode’s quantum energy spacing is small compared with kBT.
  • When kBT is too small, modes “freeze out” and contribute little to heat capacity (see UY1: crisis for equipartition theorem).
Pitfalls (equipartition)
  • Count quadratic energy terms. A classical harmonic vibration has two, giving a mean energy of kBT.
  • Two rotational terms contribute for a linear molecule in its classical rigid-rotor regime. Check that temperature is high enough for those rotations.
  • Interactions and constraints can also invalidate a simple ideal-gas term count.

Internal energy from the active term count

Worked example 1

Monatomic and diatomic energy

Problem

Calculate the thermal internal energy of 2.0 mol of classical monatomic ideal gas at 400 K. Compare it with a diatomic gas of the same amount and temperature, with translation and rotation active but vibration frozen. Use the equipartition energy reference and R = 8.314 J mol⁻¹K⁻¹.
Show full solution

  1. Count monatomic contributions

    Method

    Use f = 3 in U = fnRT/2.

    Reason

    The point-particle monatomic model has three independent translational kinetic-energy terms.

    Working

    Answer

    U = (3/2)nRT = (3/2)(2.0)(8.314)(400) ≈ 9.98 × 10³ J.

  2. Include the active rotations

    Working

    Answer

    The monatomic energy is 9.98 × 10³ J (about 10 kJ). The diatomic count is f = 3 + 2 = 5:

    U = (5/2)nRT ≈ 1.66 × 10⁴ J.

    Thus the diatomic energy is about 17 kJ, a factor 5/3 larger for this active-mode regime. Frozen vibration contributes no temperature-dependent equipartition term.

Practice set

Check your understanding 1

An ideal gas has six active quadratic terms throughout a classical temperature range. For n moles at temperature T, give its thermal internal energy and molar CV in the equipartition model.

Show hint
Use U = fnRT/2, then differentiate and divide by n for a molar capacity.
Show answer
U = 3nRT in J and CV = (1/n)(∂ U/∂ T)V = 3R in J mol⁻¹K⁻¹. The whole sample’s capacity is 3nR, not 3R.

Check your understanding 2

Why does one thermally active harmonic vibrational mode contribute mean energy kBT rather than kBT/2?

Show hint

Write its kinetic and potential energy terms separately.

Show answer

Its energy contains one quadratic kinetic term and one quadratic potential term. Each contributes kBT/2, so their sum is kBT per molecule.

Check your understanding 3

Compare the molar heat-capacity plateaus for an ideal gas with five active quadratic terms and with seven active terms. What is the ratio of the latter to the former?

Show hint
The common factor R/2 cancels.

Show answer

C_((V, 7))/C_((V, 5)) = (7R/2)/(5R/2) = 7/5 = 1.4. This compares the limiting plateaus, not an abrupt jump during mode activation.

Relate energy to displacement fluctuations

Guided practice 2

Connect energy sharing to displacement fluctuations

About 8 min

Problem

A one-dimensional harmonic coordinate has energy E = p²/(2m) + κ x²/2 and zero mean displacement. It is in classical thermal equilibrium at 300 K with κ = 0.0400 N m⁻¹. Quantum freezing is negligible. Use kB = 1.38 × 10⁻²³ J K⁻¹. Find mean potential energy divided by kBT, rms displacement in nanometres and mean total thermal energy divided by kBT. Explain how stiffness can change displacement without changing the equilibrium energy per term.

Work through the model, then compare your reasoning

Unit: dimensionless
Unit: nm
Unit: dimensionless

Hints

Hint 1

Check the model

Write the two energy terms before counting. What displacement variance gives the required mean potential energy?
Show solution step by step

  1. Use each energy term

    Method

    Assign kBT/2 to each independent quadratic contribution.

    Reason

    One coordinate supplies a kinetic and a potential term.

    Working

    Answer

    ⟨Eₚₒₜₑₙₜᵢₐₗ⟩ = kBT/2 and ⟨x²⟩ = kBT/κ = 1.035 × 10⁻¹⁹ m², so xᵣₘₛ ≈ 0.322 nm.

  2. Interpret the coefficient

    Method

    Add the two energy averages and compare stiffness changes.

    Reason

    The probability distribution adjusts its width as the quadratic coefficient changes.

    Working

    Answer

    ⟨E⟩ = kBT, so the requested ratios are 1/2 and 1. At the same temperature, quadrupling κ halves rms displacement and leaves each mean classical energy term unchanged.

Infer temperature from equal fluctuations

Check your understanding 4

Two classical one-dimensional harmonic systems with the same mass have restoring constants κA = 0.0400 N m⁻¹ and κB = 0.160 N m⁻¹. They have the same measured mean squared displacement about equilibrium. A is at 300 K. Infer B’s temperature, the ratio of their mean thermal energies and the ratio of rms velocities. A solution declares equal fluctuation size means equal temperature. State why its inference fails and the classical assumption needed.

Show answer

⟨x²⟩ = kBT/κ, so equal displacement variance requires TB/κB = TA/κA. Thus TB = 1200 K. Each one-dimensional oscillator has mean thermal energy kBT, giving energy ratio 4. With the same mass, ⟨v²⟩ = kBT/m, so rms velocity ratio is 2. Equal displacement alone fixes temperature only when the restoring constants are also the same. Both systems must satisfy the stated classical equilibrium harmonic model; a quantum-frozen mode would require another relation.

What to keep

Count independent quadratic energy terms in a valid classical equilibrium model. A harmonic coordinate contributes kinetic and potential terms; the coefficient changes fluctuation size rather than energy per term. Temperature-independent energy offsets do not change heat capacity.

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