Equipartition Theorem

Heat a gas and its energy rises — but by exactly how much, and where does the energy go? The equipartition theorem gives a startlingly simple answer for any classical system in thermal equilibrium: every independent way the energy can be stored as a quadratic term contributes precisely $\tfrac12 k_BT$ to the average energy, no more and no less. A momentum component $p^2/2m$ , a spring displacement $\tfrac12\kappa x^2$ , a rotational term $L^2/2I$ — each is worth $\tfrac12 k_BT$ , and the coefficient (mass, stiffness, moment of inertia) cancels out completely. Thermal energy is shared democratically among quadratic degrees of freedom. The proof is a one-line Gaussian integral. Because the Boltzmann factor $e^{-\beta H}$ factorizes over independent quadratic terms, the average of any single term $\langle aq^2\rangle$ reduces to a ratio of Gaussian integrals whose value is always $\tfrac12 k_BT$ . There is a sweeping generalization — $\langle x_i\,\partial H/\partial x_j\rangle=\delta_{ij}k_BT$ for any phase-space variable — that recovers the quadratic case, connects to the virial theorem, and even handles non-quadratic Hamiltonians like the ultrarelativistic gas. Equipartition was the engine of nineteenth-century kinetic theory: it explained the heat capacities of monatomic gases ( $\tfrac32 R$ ), diatomic gases ( $\tfrac52 R$ to $\tfrac72 R$ ), and the Dulong-Petit law for solids ( $3R$ ). But its most famous role was in failing. Equipartition predicts a temperature-independent heat capacity and an infinite energy for blackbody radiation — the ultraviolet catastrophe. The observed freeze-out of molecular vibrations at low temperature and the finite blackbody spectrum were among the clearest signals that classical physics was incomplete, driving the birth of quantum mechanics. Equipartition is thus both a triumph and a signpost: it works perfectly when $k_BT$ exceeds every quantum level spacing, and its breakdown maps exactly where quantization begins.

Key takeaways

In classical equilibrium at temperature $T$ , every quadratic term in the Hamiltonian contributes exactly $\tfrac12 k_BT$ to the average energy — the coefficient (mass, spring constant, moment of inertia) cancels.
The proof is a Gaussian integral: $\langle aq^2\rangle=1/(2\beta)=\tfrac12 k_BT$ , independent of $a$ .
The generalized form $\langle x_i\,\partial H/\partial x_j\rangle=\delta_{ij}k_BT$ holds for any phase-space variable and leads to the virial theorem.
Heat capacities follow directly: $\tfrac32 R$ (monatomic), $\tfrac52 R$ (diatomic with rotation), $\tfrac72 R$ (with vibration), and $3R$ (Dulong-Petit for solids).
Equipartition fails whenever $k_BT$ falls below a quantum level spacing — vibrations and rotations "freeze out," and blackbody radiation avoids the ultraviolet catastrophe.
Key idea: because the Boltzmann weight of a quadratic term is a Gaussian whose average energy is fixed by temperature alone, each quadratic degree of freedom carries $\tfrac12 k_BT$ regardless of its stiffness.

The core insight

For a Hamiltonian built from independent quadratic terms, the Boltzmann factor factorizes and each Gaussian integral gives $\langle aq^2\rangle=\tfrac12 k_BT$ — the coefficient $a$ cancels because rescaling $q\to q/\sqrt a$ removes $a$ from both the energy and the normalization. Thermal energy is $\tfrac12 k_BT$ per quadratic degree of freedom.

Where this lives today

Equipartition remains a daily working tool wherever thermal energy is quantitative. In single-molecule biophysics and nanotechnology, it calibrates force probes: for a particle of stiffness $\kappa$ in an optical or magnetic trap, $\tfrac12\kappa\langle x^2\rangle=\tfrac12 k_BT$ turns the measured positional variance directly into the trap stiffness — the standard calibration of optical tweezers and atomic-force microscopes. It fixes the mean kinetic energy of Brownian particles independent of mass, underlying thermometry from velocity fluctuations. In molecular-dynamics simulations, the equipartition energy is how thermostats define and check temperature. And its failures still teach: the temperature staircase of molecular heat capacities is a direct readout of which quantum modes are active, and the deviation of solids from Dulong-Petit below the Debye temperature is a textbook quantum effect. Equipartition is the classical baseline against which every quantum suppression is measured.

The physical picture

Put a single harmonic mode — a mass on a spring — in contact with a thermal bath. Its energy fluctuates as the bath pushes it around, but on average the bath deposits exactly $\tfrac12 k_BT$ in the kinetic term and $\tfrac12 k_BT$ in the potential term, for $k_BT$ total. The stiffness of the spring is irrelevant. A stiff spring stores the same average energy as a soft one, because a stiff spring simply explores a smaller range of displacement; the two effects cancel exactly in the Gaussian average. This is "energy democracy": every accessible quadratic mode gets an equal share. The payoff is a rule of thumb for heat capacity: count the active quadratic degrees of freedom, multiply by $\tfrac12 k_B$ (per molecule) or $\tfrac12 R$ (per mole). But the picture also contains its own limits. Quantum mechanics breaks the democracy — a mode whose excitation quantum $\hbar\omega$ exceeds $k_BT$ is effectively "disenfranchised," stuck in its ground state and contributing nothing. This is why the heat capacity of a real gas rises in steps as the temperature climbs, activating translation, then rotation, then vibration one regime at a time.

Figure

A quadratic energy term is a parabola; the Boltzmann factor makes a Gaussian cloud of displacements. A stiff parabola (narrow cloud) and a soft one (wide cloud) both store exactly ½kT on average — the coefficient cancels.

The second figure is the punchline: the heat-capacity staircase of a diatomic gas like hydrogen. At very low temperature only translation is active ( $C_V=\tfrac32 R$ ); as $T$ rises, rotation switches on ( $\tfrac52 R$ ); at high $T$ , vibration too ( $\tfrac72 R$ ). Each plateau is equipartition counting the active quadratic modes; each step is a quantum mode thawing as $k_BT$ crosses its level spacing. The frozen-out region at low $T$ is precisely where equipartition fails and quantum mechanics rules.

Figure

Punchline: the molar heat capacity of a diatomic gas climbs in steps — 3/2 R (translation) → 5/2 R (rotation activates) → 7/2 R (vibration activates). Each plateau is equipartition; each step is a quantum mode thawing. The low-T region is frozen out.

The core equations

The theorem in its classic form assigns each quadratic term of the Hamiltonian an average energy of $\tfrac12 k_BT$ . This is the statement to remember for counting heat capacities.

\left\langle \tfrac12 a_i q_i^2\right\rangle = \tfrac12 k_BT \qquad (1)

where: $a_i$ — the coefficient of the $i$ -th quadratic term (a mass for kinetic energy, a spring constant for potential, a moment of inertia for rotation); $q_i$ — the coordinate or momentum entering quadratically; $\langle\cdot\rangle$ — the canonical thermal average. The coefficient $a_i$ cancels; only $T$ survives. Summing over all $f$ quadratic terms gives the total average energy and, by differentiation, the heat capacity.

\langle H\rangle = \frac f2 k_BT,\qquad C_V=\frac{\partial\langle H\rangle}{\partial T}=\frac f2 k_B\ (\text{per molecule})=\frac f2 R\ (\text{per mole}) \qquad (2)

where: $f$ — the number of independent quadratic degrees of freedom; $R=N_A k_B=8.314\,\mathrm{J\,mol^{-1}K^{-1}}$ . Reading it: the molar heat capacity is just half the gas constant per active quadratic mode. The deep version replaces "quadratic" with a general phase-space identity, valid for any Hamiltonian with confining boundaries.

\left\langle x_i\,\frac{\partial H}{\partial x_j}\right\rangle = \delta_{ij}\,k_BT \qquad (3)

where: $x_i,x_j$ — any canonical variables (coordinates or momenta); $\delta_{ij}$ — the Kronecker delta. Setting $x_i=x_j=q$ and $H=aq^2$ recovers equation (1); summing over coordinates and forces gives the virial theorem. For a non-quadratic energy the theorem still applies through Euler's theorem on homogeneous functions, giving a different coefficient. The ultrarelativistic gas is the canonical example.

E=c|\mathbf p|\ \Rightarrow\ \langle E\rangle=3k_BT \qquad (\text{not } \tfrac32 k_BT) \qquad (4)

where: $E=c|\mathbf p|$ is homogeneous of degree $1$ in momentum, so $\sum_j p_j\,\partial H/\partial p_j=H$ ; applying equation (3) to each of the three components gives $\langle H\rangle=3k_BT$ . Linear energies give $k_BT$ per direction; quadratic ones give $\tfrac12 k_BT$ .

Derivation AEquipartition from the Gaussian integral

The goal is to prove that any single quadratic term in the Hamiltonian has thermal average energy exactly $\tfrac12 k_BT$ , independent of its coefficient, so that a Hamiltonian of $f$ quadratic terms has average energy $\tfrac f2 k_BT$ . The strategy is to isolate one quadratic term, use the fact that the Boltzmann factor of a sum factorizes, and evaluate the resulting one-dimensional Gaussian integrals directly — showing the coefficient cancels between numerator and denominator.

Step 1. Write $H=a q^2+H'$ where $H'$ is independent of $q$ . The canonical average of the quadratic term is $\langle aq^2\rangle=\dfrac{\int aq^2 e^{-\beta aq^2}\,dq}{\int e^{-\beta aq^2}\,dq}$ , because the $H'$ part cancels between numerator and denominator (it factors out of both).

Step 2. Evaluate the denominator with the standard Gaussian integral $\int_{-\infty}^{\infty}e^{-\beta aq^2}\,dq=\sqrt{\pi/\beta a}$ . Its dependence on $\beta$ is $\beta^{-1/2}$ ; the coefficient $a$ appears only inside the square root.

Step 3. Obtain the numerator by differentiating the denominator with respect to the parameter $\beta$ : $\int aq^2 e^{-\beta aq^2}\,dq=-\dfrac{\partial}{\partial\beta}\int e^{-\beta aq^2}\,dq=-\dfrac{\partial}{\partial\beta}\sqrt{\pi/\beta a}=\dfrac{1}{2\beta}\sqrt{\pi/\beta a}$ . Why the coefficient cancels. The substitution $q\to q/\sqrt a$ removes $a$ from the exponent entirely, appearing only as an overall $1/\sqrt a$ in both integrals. Equivalently, rescaling the energy and the normalization the same way leaves the average untouched — the mean energy is fixed by temperature alone.

Step 4. Divide: $\langle aq^2\rangle=\dfrac{(1/2\beta)\sqrt{\pi/\beta a}}{\sqrt{\pi/\beta a}}=\dfrac{1}{2\beta}=\tfrac12 k_BT$ , equation (1). For $f$ independent quadratic terms the energies simply add, giving $\langle H\rangle=\tfrac f2 k_BT$ and $C_V=\tfrac f2 k_B$ . Verified: the Gaussian-integral result $\langle aq^2\rangle=1/(2\beta)=\tfrac12 k_BT$ (coefficient cancels), and the harmonic-oscillator total $\langle H\rangle=k_BT$ , are confirmed symbolically in verify_aq-sm-equipartition-theorem.py.

Go deeper · Derivation B · 1 minThe generalized theorem by integration by parts, and the ultrarelativistic gas

Derivation BThe generalized theorem by integration by parts, and the ultrarelativistic gas

The goal is to establish the general phase-space identity $\langle x_i\,\partial H/\partial x_j\rangle=\delta_{ij}k_BT$ , valid for any confining Hamiltonian, and to use it to obtain the ultrarelativistic result $\langle E\rangle=3k_BT$ for $E=c|\mathbf p|$ . The strategy is integration by parts on the canonical average, using that the Boltzmann factor vanishes at the phase-space boundaries, followed by Euler's theorem on homogeneous functions.

Step 1. Write the average $\langle x_i\,\partial H/\partial x_j\rangle=\dfrac1Z\int x_i\dfrac{\partial H}{\partial x_j}e^{-\beta H}\,d\Gamma$ , where $d\Gamma$ is the phase-space measure and $Z$ the partition function. Note $\dfrac{\partial H}{\partial x_j}e^{-\beta H}=-\dfrac{1}{\beta}\dfrac{\partial}{\partial x_j}e^{-\beta H}$ .

Step 2. Integrate by parts in $x_j$ : $-\dfrac{1}{\beta Z}\int x_i\dfrac{\partial}{\partial x_j}e^{-\beta H}\,d\Gamma=\dfrac{1}{\beta Z}\int\dfrac{\partial x_i}{\partial x_j}e^{-\beta H}\,d\Gamma$ . The boundary term vanishes because $e^{-\beta H}\to0$ as $|x_j|\to\infty$ (the energy confines the system).

Step 3. Use $\partial x_i/\partial x_j=\delta_{ij}$ and normalize by $Z$ : $\langle x_i\,\partial H/\partial x_j\rangle=\delta_{ij}/\beta=\delta_{ij}k_BT$ , equation (3). Setting $H=aq^2$ , $x_i=x_j=q$ recovers $\langle q(2aq)\rangle=k_BT$ , i.e. $\langle aq^2\rangle=\tfrac12 k_BT$ — consistent with the previous card. Why quadratic gives $\tfrac12$ but linear gives $1$ . For a homogeneous energy of degree $n$ in a variable, $x\,\partial H/\partial x = nH$ for that term, so $\langle H_{\text{term}}\rangle=k_BT/n$ . Quadratic ( $n=2$ ) gives $\tfrac12 k_BT$ ; linear ( $n=1$ ) gives $k_BT$ .

Step 4. Ultrarelativistic gas: $H=c|\mathbf p|$ is homogeneous of degree $1$ in $\mathbf p$ , so by Euler's theorem $\sum_j p_j\,\partial H/\partial p_j=H$ . Taking the average and applying equation (3) to each of the three momentum components, $\langle H\rangle=\sum_j\langle p_j\,\partial H/\partial p_j\rangle=3k_BT$ . A photon-like gas carries $3k_BT$ per particle, not $\tfrac32 k_BT$ . Verified: the integration-by-parts identity, the recovery of the quadratic case, and the Euler-homogeneity result $\sum_j p_j\partial_j H=H$ for $E=c|\mathbf p|$ (giving $\langle E\rangle=3k_BT$ ) are all confirmed symbolically in the verify script.

Limiting cases

Equipartition is the high-temperature, classical limit of more complete results; its breakdown is as informative as its validity.

Case	Result	Meaning
Classical quadratic mode	$\langle E_i\rangle=\tfrac12 k_BT$	Full equipartition; each kinetic or harmonic term contributes equally.
Quantum oscillator, $k_BT\gg\hbar\omega$	$\langle E\rangle\to k_BT$	Recovers equipartition (two quadratic terms) at high $T$ .
Quantum oscillator, $k_BT\ll\hbar\omega$	$\langle E\rangle\approx\hbar\omega\,e^{-\beta\hbar\omega}\to0$	Mode freezes out; contribution exponentially suppressed. Equipartition fails.
Ultrarelativistic gas, $E=c\|\mathbf p\|$	$\langle E\rangle=3k_BT$	Linear (degree-1) energy gives $k_BT$ per direction, not $\tfrac12 k_BT$ .
Blackbody radiation (classical)	energy density diverges	Assigning $k_BT$ per mode gives the ultraviolet catastrophe.

Common trapCommon pitfall

Equipartition counts only degrees of freedom that appear quadratically, with unbounded range, and are thermally active — and it is purely classical. Three traps follow. First, do not count all $3N$ atomic coordinates and blindly assign $\tfrac12 k_BT$ each; a mode contributes only if $k_BT$ exceeds its quantum level spacing, so vibrations and (at very low $T$ ) rotations are frozen out and contribute nothing. Second, the $\tfrac12 k_BT$ rule is for quadratic terms only — a linear energy like $c|\mathbf p|$ gives $k_BT$ per direction, and a quartic term $bq^4$ gives $\tfrac14 k_BT$ . Third, a vibrational mode is two quadratic terms (kinetic plus potential), so it contributes $k_BT$ , not $\tfrac12 k_BT$ , when active. Getting the count wrong is the single most common equipartition error.

Worked examples

Example 1 — Heat capacity of a monatomic ideal gas from counting quadratic terms (enumerate, apply, boundary number). Consider argon, a monatomic gas, at room temperature.

Step 1. Enumerate the quadratic terms. A point atom has only translational kinetic energy, $H=\sum_i \mathbf p_i^2/2m$ , which is $3N$ quadratic momentum terms ( $p_x,p_y,p_z$ per atom) and no potential or rotational structure. So $f=3$ per atom.

Step 2. Apply equipartition: $\langle H\rangle=\tfrac f2 Nk_BT=\tfrac32 Nk_BT$ , so the molar heat capacity is $C_V=\tfrac32 R$ .

Step 3. Boundary number: $C_V=\tfrac32(8.314)=\mathbf{12.47\,J\,mol^{-1}K^{-1}}$ . The measured $C_V$ of argon at room temperature is $\approx12.5\,\mathrm{J\,mol^{-1}K^{-1}}$ — an essentially exact match, and $\gamma=C_P/C_V=(\tfrac52 R)/(\tfrac32 R)=5/3\approx1.667$ agrees with noble gases.

Example 2 — The Dulong-Petit law and its quantum breakdown. Model a crystalline solid as $3N$ independent classical harmonic oscillators (each atom oscillates in three dimensions about its lattice site).

Step 1. Each one-dimensional oscillator has two quadratic terms (kinetic $p^2/2m$ and potential $\tfrac12 m\omega^2 x^2$ ), so it contributes $k_BT$ . With $3N$ oscillators, $\langle E\rangle=3Nk_BT$ .

Step 2. Molar heat capacity: $C_V=3Nk_B/N_{\text{mol}}=3R=3(8.314)=\mathbf{24.94\,J\,mol^{-1}K^{-1}}$ — the Dulong-Petit law, independent of the material.

Step 3. Boundary check: this matches copper, silver, and lead near and above room temperature. But below the Debye temperature the oscillator modes freeze out: for $\hbar\omega\gg k_BT$ the quantum energy is $\langle E\rangle\approx\hbar\omega\,e^{-\beta\hbar\omega}$ , exponentially small, and $C_V\to0$ as $C_V\propto T^3$ (Debye law). Numerically, at $\beta\hbar\omega=20$ the contribution is $\hbar\omega\,e^{-20}\approx4\times10^{-8}\,\hbar\omega$ — utterly frozen. Equipartition holds only for $T$ well above the characteristic quantum temperature.

Origins · 1 minequipartition theorem and how this was discovered

equipartition theorem

statistical-mechanics concept · Maxwell 1859, Boltzmann 1868–77

The equipartition theorem states that in classical thermal equilibrium each quadratic degree of freedom carries an average energy of $\tfrac12 k_BT$ . Maxwell first argued (1859) that translational and rotational energy should be shared equally among molecular degrees of freedom, and Boltzmann extended and formalized it. The theorem was the quantitative backbone of nineteenth-century kinetic theory, predicting the heat capacities of gases and the Dulong-Petit law for solids. Its dramatic failures — the frozen-out heat capacity of diatomic gases at low temperature and the ultraviolet catastrophe of blackbody radiation — were among the decisive clues that classical statistical mechanics was incomplete, and directly motivated Planck's and Einstein's quantum ideas.

History

1845

— Waterston's kinetic-theory manuscript, containing an early equipartition idea, is rejected and forgotten.

1859

— Maxwell argues that energy is shared equally among translational and rotational degrees of freedom.

1868

— Boltzmann generalizes equipartition to arbitrary quadratic degrees of freedom.

1900

— Rayleigh applies equipartition to blackbody radiation, exposing the ultraviolet catastrophe.

1907

— Einstein's quantum model of specific heat explains the low-temperature failure of equipartition in solids.

Going further

References. Pathria & Beale, Statistical Mechanics, 3rd ed., §3.7 (equipartition and virial theorems) · Reif, Fundamentals of Statistical and Thermal Physics, §7.5–7.7 · Kardar, Statistical Physics of Particles, §4 · Sethna, Statistical Mechanics: Entropy, Order Parameters, and Complexity, Ch. 3 & 6 · Landau & Lifshitz, Statistical Physics Part 1, §28.

Provenance. The result $\tfrac12 k_BT$ per quadratic term, the generalized theorem $\langle x_i\partial H/\partial x_j\rangle=\delta_{ij}k_BT$ , the heat capacities $\tfrac32 R,\tfrac52 R,\tfrac72 R$ , Dulong-Petit $3R$ , and the ultrarelativistic $3k_BT$ are standard (Pathria §3.7, Reif §7.5). A verifier should confirm: (i) $\langle aq^2\rangle=\int aq^2 e^{-\beta aq^2}dq/\int e^{-\beta aq^2}dq=1/2\beta=\tfrac12 k_BT$ ; (ii) the integration-by-parts proof of equation (3); (iii) $\sum_j p_j\partial H/\partial p_j=H$ for $E=c|\mathbf p|$ , giving $\langle E\rangle=3k_BT$ ; (iv) $C_V(\text{monatomic})=\tfrac32 R=12.47$ and Dulong-Petit $3R=24.94\,\mathrm{J\,mol^{-1}K^{-1}}$ ; (v) the high-T limit $\hbar\omega/(e^{\beta\hbar\omega}-1)\to k_BT$ and low-T freeze-out. All checks are in verify_aq-sm-equipartition-theorem.py. Constants: $k_B=1.380649\times10^{-23}\,\mathrm{J/K}$ , $R=8.314\,\mathrm{J\,mol^{-1}K^{-1}}$ .

Builds on: Classical Ideal Gas · Maxwell-Boltzmann Velocity Distribution · Thermal de Broglie Wavelength Leads to: Generalized Equipartition Theorem · Virial Theorem in Statistical Mechanics · Classical Limit of Quantum Statistics

Why does the stiffness of a spring not affect its average energy?

Because a stiffer spring explores a smaller range of displacement, and the two effects cancel exactly. Mathematically, the substitution $q\to q/\sqrt a$ removes the coefficient $a$ from both the Boltzmann factor $e^{-\beta aq^2}$ and the energy $aq^2$ in the same way, so the ratio of integrals that gives $\langle aq^2\rangle$ is independent of $a$ . The mean energy of a quadratic mode is fixed by temperature alone: $\tfrac12 k_BT$ , whether the spring is stiff or soft.

Why is a diatomic gas 5/2 R and not 7/2 R at room temperature?

Because vibration is frozen out. A diatomic molecule has three translational and two rotational quadratic terms active at room temperature ( $f=5$ , giving $C_V=\tfrac52 R$ ), but its vibrational mode has a quantum $\hbar\omega$ far larger than $k_BT$ — for N $_2$ the vibrational temperature is about $3390\,\mathrm{K}$ — so the mode is stuck in its ground state and contributes nothing. Only at high temperature does vibration activate and $C_V$ climb toward $\tfrac72 R$ .

How does equipartition connect to the virial theorem?

Through the generalized identity $\langle x_i\,\partial H/\partial x_j\rangle=\delta_{ij}k_BT$ . Applying it to coordinates and forces, $\langle\sum_i\mathbf r_i\cdot\mathbf F_i\rangle=-3Nk_BT$ (plus a wall term that yields $PV$ ), which is the statistical virial theorem. Expanding the interparticle-force contribution generates the virial equation of state and the virial coefficients, linking microscopic potentials to deviations from ideal-gas behaviour. So equipartition and the virial theorem are two faces of the same phase-space identity.

Does equipartition hold in the microcanonical ensemble?

Yes, in the thermodynamic limit. It is most easily derived in the canonical ensemble via the Boltzmann factor, but it is an ensemble-equivalent result: the microcanonical and canonical averages of a quadratic term coincide to leading order in $1/N$ , both giving $\tfrac12 k_BT$ , with the microcanonical temperature defined by $1/T=\partial S/\partial E$ . For a macroscopic system the distinction is negligible.

What exactly goes wrong in the ultraviolet catastrophe?

Assigning $k_BT$ to every electromagnetic mode (each mode is two quadratic terms, like an oscillator) and summing over the infinitely many high-frequency modes gives an energy density that diverges: $\int g(\omega)k_BT\,d\omega\propto\int\omega^2 k_BT\,d\omega\to\infty$ . Experiment shows a finite spectrum that falls off at high frequency. The resolution is that a mode of frequency $\omega$ cannot be excited unless $k_BT\gtrsim\hbar\omega$ — quantization caps the high-frequency contribution, giving Planck's law. Equipartition's failure here was a direct route to the quantum.

Why does a linear energy give kT instead of ½kT?

Because the $\tfrac12 k_BT$ rule is specific to quadratic (degree-2) terms. For a term homogeneous of degree $n$ , the generalized theorem gives $\langle H_{\text{term}}\rangle=k_BT/n$ . Quadratic ( $n=2$ ) gives $\tfrac12 k_BT$ ; linear ( $n=1$ ), like the ultrarelativistic $c|\mathbf p|$ , gives $k_BT$ per component; quartic ( $n=4$ ) gives $\tfrac14 k_BT$ . The power of the variable in the energy sets the coefficient.

Can I use equipartition to calibrate a trapped-particle experiment?

Yes — it is the standard method. For a particle held in a harmonic trap of stiffness $\kappa$ , the potential energy is $\tfrac12\kappa x^2$ , a single quadratic term, so equipartition gives $\tfrac12\kappa\langle x^2\rangle=\tfrac12 k_BT$ . Measuring the variance $\langle x^2\rangle$ of the particle's position (from video tracking or a quadrant photodiode) and knowing $T$ yields the trap stiffness $\kappa=k_BT/\langle x^2\rangle$ directly. This is how optical tweezers and atomic-force cantilevers are calibrated.