The Equipartition Theorem

Derivation

The Equipartition Theorem

D-237 Home PU-302 Threads energy · chance Depends on The Partition Function and F = -kT ln Z

Statement

For a classical system in thermal equilibrium at temperature \(T\), described by a Hamiltonian \(H\) in which one canonical coordinate or momentum \(x\) appears as a single quadratic term \(\tfrac{1}{2}\alpha x^2\) (with \(\alpha>0\)) that separates additively from the rest of \(H\), the mean thermal energy stored in that term is exactly \(\left\langle \tfrac{1}{2}\alpha x^2\right\rangle = \tfrac{1}{2}k_B T\), independent of \(\alpha\). Summing over all such quadratic degrees of freedom gives the equipartition theorem: each contributes \(\tfrac{1}{2}k_B T\).

Why it matters

Equipartition is the bridge between microscopic mechanics and macroscopic thermodynamics for classical systems. It immediately yields the heat capacities of ideal gases (\(C_V = \tfrac{3}{2}Nk_B\) for a monatomic gas, \(\tfrac{5}{2}Nk_B\) with rotations), the Dulong–Petit law \(C_V = 3Nk_B\) for solids, and the mean-square thermal fluctuation of any harmonically-bound coordinate. It tells you, before solving any dynamics, how much energy a mode carries once it is "switched on".

Equally important is where it fails: the ultraviolet catastrophe of blackbody radiation and the freezing-out of molecular vibrations are both the breakdown of equipartition, and historically they were the cracks through which quantum mechanics entered. Understanding the theorem is understanding precisely the classical limit that quantum statistics must reduce to.

Assumptions

Classical (continuous) phase space.If energy levels are discrete and spaced by more than \(k_B T\), the integral over \(x\) must become a sum and the Gaussian result no longer holds — the mode carries less than \(\tfrac{1}{2}k_B T\) and freezes out.

Canonical (Boltzmann) weighting.The result is the ensemble average with weight \(e^{-\beta H}\); without contact with a heat bath at fixed \(T\) (e.g. a microcanonical or driven system) the mean energy per mode is not pinned to \(\tfrac{1}{2}k_B T\).

The coordinate appears purely quadratically and separably, \(H = \tfrac{1}{2}\alpha x^2 + H'\) with \(H'\) independent of \(x\).If \(H\) contains \(x^4\), cross terms \(x\,y\), or a coordinate-dependent \(\alpha(x)\), the average is no longer \(\tfrac{1}{2}k_B T\); anharmonic and coupled modes deviate.

The Gaussian integral converges, requiring \(\alpha>0\) and integration over the full line \(x\in(-\infty,\infty)\).For \(\alpha\le 0\) (unstable mode) or a bounded/half-line domain the normalisation diverges or the symmetric cancellation fails, and \(\tfrac{1}{2}k_B T\) is not recovered.

Derivation

\[ H = \tfrac{1}{2}\alpha x^2 + H'(\text{other variables}) \]

Isolate the single quadratic degree of freedom; by assumption \(H'\) does not depend on \(x\). A

\[ \left\langle \tfrac{1}{2}\alpha x^2 \right\rangle = \frac{\displaystyle\int \tfrac{1}{2}\alpha x^2\, e^{-\beta H}\, d\Gamma}{\displaystyle\int e^{-\beta H}\, d\Gamma},\qquad \beta \equiv \frac{1}{k_B T} \]

Definition of a canonical ensemble average; \(d\Gamma\) is the full phase-space measure and \(Z=\int e^{-\beta H}d\Gamma\) is the partition function (prior result). A

\[ e^{-\beta H} = e^{-\beta\,\frac{1}{2}\alpha x^2}\; e^{-\beta H'} \]

The Boltzmann factor factorises because \(H\) is additively separable in \(x\). A

\[ \left\langle \tfrac{1}{2}\alpha x^2 \right\rangle = \frac{\displaystyle\int_{-\infty}^{\infty} \tfrac{1}{2}\alpha x^2\, e^{-\beta\frac{1}{2}\alpha x^2}\, dx \;\int e^{-\beta H'} d\Gamma'}{\displaystyle\int_{-\infty}^{\infty} e^{-\beta\frac{1}{2}\alpha x^2}\, dx \;\int e^{-\beta H'} d\Gamma'} \]

The \(H'\) integrals are identical in numerator and denominator and cancel; only the \(x\)-integral survives. A

\[ \left\langle \tfrac{1}{2}\alpha x^2 \right\rangle = \frac{\displaystyle\int_{-\infty}^{\infty} \tfrac{1}{2}\alpha x^2\, e^{-\beta\frac{1}{2}\alpha x^2}\, dx}{\displaystyle\int_{-\infty}^{\infty} e^{-\beta\frac{1}{2}\alpha x^2}\, dx} \]

Reduced to a single one-dimensional ratio of Gaussian integrals. A

\[ \int_{-\infty}^{\infty} e^{-a x^2}\, dx = \sqrt{\frac{\pi}{a}},\qquad \int_{-\infty}^{\infty} x^2 e^{-a x^2}\, dx = -\frac{\partial}{\partial a}\sqrt{\frac{\pi}{a}} = \frac{1}{2a}\sqrt{\frac{\pi}{a}} \]

Standard Gaussian moment via differentiation under the integral sign with respect to the parameter \(a\); here \(a = \tfrac{1}{2}\beta\alpha\). B

\[ \frac{\displaystyle\int x^2 e^{-a x^2} dx}{\displaystyle\int e^{-a x^2} dx} = \frac{\tfrac{1}{2a}\sqrt{\pi/a}}{\sqrt{\pi/a}} = \frac{1}{2a} \]

Divide the two moments; the normalisation factor \(\sqrt{\pi/a}\) cancels exactly. A

\[ \left\langle x^2 \right\rangle = \frac{1}{2a} = \frac{1}{\beta\alpha} = \frac{k_B T}{\alpha} \]

Substitute \(a = \tfrac{1}{2}\beta\alpha\) and \(\beta = 1/k_B T\). A

\[ \left\langle \tfrac{1}{2}\alpha x^2 \right\rangle = \tfrac{1}{2}\alpha \cdot \frac{k_B T}{\alpha} = \tfrac{1}{2}k_B T \]

Multiply through by \(\tfrac{1}{2}\alpha\); the coupling constant \(\alpha\) cancels identically. A

\[ \langle H \rangle = \sum_{i=1}^{N_q} \left\langle \tfrac{1}{2}\alpha_i x_i^2 \right\rangle = \frac{N_q}{2}\,k_B T \]

If \(H\) is a sum of \(N_q\) independent quadratic terms, the averages add (linearity of the ensemble average), giving the full equipartition statement. B

Result

\[ \boxed{\;\left\langle \tfrac{1}{2}\alpha x^2 \right\rangle = \tfrac{1}{2}k_B T\;}\qquad \langle H\rangle = \tfrac{1}{2}N_q\, k_B T \]

Reading. Every independent coordinate or momentum that enters the Hamiltonian as a pure square carries, on average, an energy \(\tfrac{1}{2}k_B T\) — and remarkably the value is universal: it does not depend on the stiffness \(\alpha\), the mass, or the frequency of the mode, only on temperature. A stiff spring and a floppy one store the same thermal energy; the stiff one simply achieves it with a smaller mean-square displacement, since \(\langle x^2\rangle = k_B T/\alpha\).

Units check. \(k_B T\) has units of \(\mathrm{J\,K^{-1}}\times\mathrm{K}=\mathrm{J}\), an energy, as required. Consistency of \(\langle x^2\rangle = k_B T/\alpha\): if \(x\) is a length, \(\tfrac{1}{2}\alpha x^2\) is an energy so \([\alpha]=\mathrm{J\,m^{-2}}=\mathrm{N\,m^{-1}}\); then \([k_B T/\alpha]=\mathrm{J}/(\mathrm{J\,m^{-2}})=\mathrm{m^2}\), correctly a squared length.

Limiting cases

Monatomic ideal gas. \(H=\tfrac{1}{2m}(p_x^2+p_y^2+p_z^2)\) has 3 quadratic momenta per atom, giving \(\langle E\rangle = \tfrac{3}{2}k_B T\) and \(C_V=\tfrac{3}{2}Nk_B\).
Classical harmonic oscillator (1D). \(H=\tfrac{p^2}{2m}+\tfrac{1}{2}m\omega^2 q^2\) has 2 quadratic terms, so \(\langle E\rangle = k_B T\) — kinetic and potential each get \(\tfrac{1}{2}k_B T\). This gives Dulong–Petit \(C_V=3Nk_B\) for a solid (3 oscillators per atom).
High-temperature limit of any quantum mode. When \(k_B T \gg \hbar\omega\) (or the level spacing), the quantum average \(\langle E\rangle = \hbar\omega/(e^{\beta\hbar\omega}-1)\to k_B T\), recovering equipartition as the leading term.
Diatomic gas at room temperature. 3 translational \(+\) 2 rotational quadratic terms are active (vibration frozen), giving \(\langle E\rangle=\tfrac{5}{2}k_B T\) and \(\gamma=C_P/C_V=7/5\).

Breaks when

Quantum freeze-out (\(k_B T \lesssim \hbar\omega\)). When the thermal energy is small compared with the level spacing, the phase-space integral must be replaced by a Boltzmann sum. The mode is stuck in its ground state, contributes far less than \(\tfrac{1}{2}k_B T\), and its heat capacity vanishes exponentially, \(C\sim (\hbar\omega/k_BT)^2 e^{-\hbar\omega/k_BT}\). This is why molecular vibrations and the blackbody UV modes drop out — averting the ultraviolet catastrophe.
Anharmonic / non-quadratic Hamiltonians. If a coordinate enters as \(bx^4\) instead of \(\tfrac{1}{2}\alpha x^2\), Gaussian integration no longer applies and \(\langle bx^4\rangle = \tfrac{1}{4}k_B T\) — a general homogeneous term \(x^n\) gives \(\tfrac{1}{n}k_B T\) (a generalised equipartition), not \(\tfrac{1}{2}k_B T\).
Coupled or non-separable modes. Cross terms such as \(\lambda x y\) or a position-dependent stiffness \(\alpha(x)\) spoil the factorisation in Step 3; the \(H'\) integral no longer cancels and the per-mode energy shifts.
Divergent or bounded phase space. An unstable mode (\(\alpha<0\)), a hard-wall constraint, or relativistic dispersion \(E=|\mathbf{p}|c\) (linear, not quadratic, in \(p\)) all break the assumptions; e.g. an ultrarelativistic gas has \(\langle E\rangle = 3k_B T\) per particle, not \(\tfrac{3}{2}k_B T\).

Failure modes

Counting atoms or molecules instead of quadratic terms. Equipartition assigns \(\tfrac{1}{2}k_B T\) per quadratic degree of freedom in \(H\), not per particle. A 1D oscillator has 2, not 1.
Forgetting the potential energy. Students count the 3 kinetic terms of a gas but omit the potential terms of a solid, getting \(\tfrac{3}{2}k_B T\) instead of \(3\times k_B T\) and missing Dulong–Petit.
Applying it to vibrations at room temperature. Assuming diatomic vibrational modes contribute \(k_B T\) at 300 K gives \(C_V=\tfrac{7}{2}Nk_B\); they are frozen, so the correct answer is \(\tfrac{5}{2}Nk_B\).
Thinking the answer depends on \(\alpha\), \(\omega\), or \(m\). The whole point is that these cancel; only the mean-square amplitude \(\langle x^2\rangle=k_BT/\alpha\) carries them.
Using \(\tfrac{1}{2}k_B T\) for a linear or quartic term. Only pure quadratic terms give \(\tfrac{1}{2}\); \(x^4\) gives \(\tfrac{1}{4}k_BT\), \(|p|c\) gives \(k_BT\).
Confusing \(k_B\) and \(R\). Per particle use \(k_B\); per mole use \(R=N_A k_B\). Mixing them mis-scales the energy by \(6\times10^{23}\).

Discussion

The deep reason equipartition is so clean is that the result comes entirely from the shape of the energy in phase space, not its scale. A quadratic well is scale-free: rescaling \(x\to x/\sqrt{\beta\alpha}\) removes both \(\alpha\) and \(\beta\) from the integrand's shape and pushes them entirely into the Jacobian, which cancels between numerator and denominator. What survives is a pure number, \(\tfrac{1}{2}\), times \(k_BT\). This is why the theorem is a statement about counting quadratic directions in phase space rather than about any dynamical detail.

A cleaner route uses the partition function directly (the prior result). For one quadratic mode \(Z_x \propto \int e^{-\beta\alpha x^2/2}dx \propto \beta^{-1/2}\), so \(\langle \tfrac{1}{2}\alpha x^2\rangle = -\partial_\beta \ln Z_x = -\partial_\beta(-\tfrac{1}{2}\ln\beta + \text{const}) = \tfrac{1}{2\beta}=\tfrac{1}{2}k_BT\). Each quadratic factor of \(Z\) contributes a \(\beta^{-1/2}\), hence a \(\tfrac{1}{2}k_BT\) to \(\langle E\rangle\) — the same counting, phrased through the free energy \(F=-k_BT\ln Z\).

The generalised version, sometimes called the equipartition or virial theorem in statistical mechanics, states \(\langle x_i\,\partial H/\partial x_j\rangle = \delta_{ij}k_BT\). For \(H=\tfrac{1}{2}\alpha x^2\) this reproduces our result, but it also governs the pressure of interacting gases (the virial equation of state) and the mean kinetic energy of stars in a self-gravitating cluster. Equipartition is thus one face of a far broader identity relating the ensemble average of \(x_i\partial_j H\) to temperature.

The theorem's failure is more instructive than its success. Rayleigh and Jeans applied equipartition to the electromagnetic field — infinitely many quadratic modes, each demanding \(k_BT\) — and obtained a spectral density diverging as \(\nu^2\): the ultraviolet catastrophe. Planck's resolution was precisely to deny the classical phase-space integral, quantising each mode so that when \(h\nu\gg k_BT\) the Gaussian is replaced by a nearly one-term Boltzmann sum and the mode freezes. Equipartition is therefore exactly the classical limit that any correct quantum theory must recover as \(h\to 0\) or \(T\to\infty\); its domain of validity is the boundary line \(k_BT\gtrsim\hbar\omega\).

Common misconceptions. Equipartition does not say every particle has energy \(\tfrac{3}{2}k_BT\) — that is the mean; individual energies are Maxwell–Boltzmann distributed with large spread. Nor does it say energy is shared equally between particles at any instant; it is an ensemble time-average per quadratic mode. And it is not a quantum result gone classical by accident — it is intrinsically classical and simply fails quantum-mechanically at low temperature.

Worked examples

Example A — RMS speed of nitrogen at room temperature. A \(\mathrm{N_2}\) molecule (\(m = 4.65\times10^{-26}\ \mathrm{kg}\)) in air at \(T=300\ \mathrm{K}\). Find the root-mean-square translational speed. A

\[ \left\langle \tfrac{1}{2}m v^2\right\rangle = \tfrac{3}{2}k_B T \;\Rightarrow\; \langle v^2\rangle = \frac{3k_B T}{m} \]

Three quadratic momentum terms, each \(\tfrac{1}{2}k_BT\); symbols first. A

\[ v_{\rm rms}=\sqrt{\frac{3(1.381\times10^{-23}\,\mathrm{J\,K^{-1}})(300\,\mathrm{K})}{4.65\times10^{-26}\,\mathrm{kg}}} \]

Insert numbers with units; \(k_B=1.381\times10^{-23}\ \mathrm{J\,K^{-1}}\). A

\[ v_{\rm rms} \approx 517\ \mathrm{m\,s^{-1}} \]

Reading. Air molecules move at roughly the speed of sound scale, a few hundred metres per second — consistent with sound speed \(\sim 340\ \mathrm{m\,s^{-1}}\), which is a fraction of \(v_{\rm rms}\).

Units check. \(\sqrt{\mathrm{J/kg}}=\sqrt{\mathrm{kg\,m^2 s^{-2}/kg}}=\mathrm{m\,s^{-1}}\). Correct.

Example B — thermal jitter of an AFM cantilever. A micro-cantilever behaves as a 1D harmonic mode with stiffness \(k=0.10\ \mathrm{N\,m^{-1}}\) at \(T=300\ \mathrm{K}\). Find the RMS thermal deflection \(\sqrt{\langle x^2\rangle}\), the noise floor of the instrument. A

\[ \left\langle \tfrac{1}{2}k x^2\right\rangle = \tfrac{1}{2}k_B T \;\Rightarrow\; \langle x^2\rangle = \frac{k_B T}{k} \]

The potential term is a single quadratic degree of freedom; \(\alpha\to k\). A

\[ \sqrt{\langle x^2\rangle}=\sqrt{\frac{(1.381\times10^{-23}\,\mathrm{J\,K^{-1}})(300\,\mathrm{K})}{0.10\,\mathrm{N\,m^{-1}}}} \]

Substitute numbers and units. A

\[ \sqrt{\langle x^2\rangle} \approx 6.4\times10^{-10}\ \mathrm{m}\;\approx\;0.64\ \mathrm{nm} \]

Reading. Even a soft cantilever jitters by sub-nanometre amounts purely from thermal energy; this sets the fundamental detection limit for atomic-force microscopy, independent of the cantilever's resonance frequency.

Units check. \(\sqrt{\mathrm{J}/(\mathrm{N\,m^{-1})}}=\sqrt{\mathrm{N\,m}/(\mathrm{N\,m^{-1}})}=\sqrt{\mathrm{m^2}}=\mathrm{m}\). Correct.

Problems

(A) Heat capacity of argon. Compute the molar heat capacity at constant volume \(C_V\) of argon gas (monatomic) and the ratio \(\gamma=C_P/C_V\).
Solution
Three translational quadratic terms give \(\langle E\rangle=\tfrac{3}{2}k_BT\) per atom, so per mole \(U=\tfrac{3}{2}RT\) and \(C_V=\left(\partial U/\partial T\right)_V=\tfrac{3}{2}R=\tfrac{3}{2}(8.314)=12.5\ \mathrm{J\,mol^{-1}K^{-1}}\). Then \(C_P=C_V+R=\tfrac{5}{2}R\) and \(\gamma=C_P/C_V=\tfrac{5}{2}/\tfrac{3}{2}=5/3\approx1.67\), matching measured monatomic values.
(B) Diatomic energy budget. For an ideal diatomic gas at 300 K with vibration frozen, state \(\langle E\rangle\) per molecule and \(C_V\) per mole; then say what \(C_V\) becomes once the vibrational mode unfreezes at high \(T\).
Solution
Active modes: 3 translational \(+\) 2 rotational \(=5\) quadratic terms, so \(\langle E\rangle=\tfrac{5}{2}k_BT\) and \(C_V=\tfrac{5}{2}R=20.8\ \mathrm{J\,mol^{-1}K^{-1}}\). When vibration activates it adds 2 quadratic terms (kinetic \(+\) potential), giving \(7\) total: \(\langle E\rangle=\tfrac{7}{2}k_BT\) and \(C_V=\tfrac{7}{2}R=29.1\ \mathrm{J\,mol^{-1}K^{-1}}\).
(B) Mean-square charge on a capacitor. A capacitor \(C\) in a thermal environment stores energy \(\tfrac{Q^2}{2C}\). Find \(\langle Q^2\rangle\) and evaluate the RMS voltage noise \(\sqrt{\langle V^2\rangle}\) for \(C=1\ \mathrm{pF}\) at \(T=300\ \mathrm{K}\).
Solution
The energy is quadratic in \(Q\) with \(\alpha=1/C\), so \(\langle \tfrac{Q^2}{2C}\rangle=\tfrac{1}{2}k_BT\Rightarrow\langle Q^2\rangle=Ck_BT\). Voltage \(V=Q/C\), so \(\langle V^2\rangle=\langle Q^2\rangle/C^2=k_BT/C\). Numerically \(\sqrt{k_BT/C}=\sqrt{(1.381\times10^{-23})(300)/(1\times10^{-12})}=\sqrt{4.14\times10^{-9}}\approx 6.4\times10^{-5}\ \mathrm{V}=64\ \mu\mathrm{V}\). This is the famous \(kT/C\) noise of sampled-capacitor circuits.
(C) Quartic degree of freedom. A coordinate enters the Hamiltonian as \(H=bq^4\) (with \(b>0\)), the rest separable. Show its mean energy is \(\tfrac{1}{4}k_BT\), not \(\tfrac{1}{2}k_BT\).
Solution
Use the generalised result \(\langle q\,\partial H/\partial q\rangle=k_BT\) (equipartition/virial identity, provable by integration by parts of \(\int q\,\partial_q H\,e^{-\beta H}dq\)). Here \(q\,\partial H/\partial q=q(4bq^3)=4bq^4=4H\). So \(\langle 4H\rangle=k_BT\Rightarrow\langle H\rangle=\tfrac{1}{4}k_BT\). Equivalently, for \(H\propto q^n\) the mean energy is \(\tfrac{1}{n}k_BT\); \(n=2\) recovers \(\tfrac{1}{2}k_BT\), \(n=4\) gives \(\tfrac{1}{4}k_BT\).
(C) Where equipartition dies. A vibrational mode has \(\hbar\omega/k_B = 3390\ \mathrm{K}\) (like \(\mathrm{N_2}\)). Estimate its contribution to \(C_V\) at \(T=300\ \mathrm{K}\) using the quantum result, and compare with the equipartition prediction \(R\).
Solution
The quantum (Einstein) mode heat capacity is \(C_{\rm vib}=R\,x^2\dfrac{e^{x}}{(e^{x}-1)^2}\) with \(x=\hbar\omega/k_BT=3390/300=11.3\). Since \(e^{x}\gg1\), \(C_{\rm vib}\approx R\,x^2 e^{-x}=8.314\times(11.3)^2 e^{-11.3}=8.314\times128\times1.23\times10^{-5}\approx0.013\ \mathrm{J\,mol^{-1}K^{-1}}\). This is about \(0.16\%\) of the equipartition value \(R=8.314\); the mode is essentially frozen, so treating it classically overestimates its heat capacity by nearly three orders of magnitude — the failure that motivated quantum statistics.