Equipartition theorem — ½k_BT per quadratic degree of freedom: Exercises

Throughout, remember the core statement from the parent note: each quadratic energy term carries $\frac{1}{2} k_{B} T$ on average, where $k_{B}$ is Boltzmann's constant and $T$ is the absolute temperature (in kelvin). We take $k_{B} = 1.380649 \times 1 0^{- 23} J/K$ and the gas constant $R = N_{A} k_{B} = 8.314 J/(mol\cdotK)$ where $N_{A} = 6.022 \times 1 0^{23}$ is Avogadro's number (atoms per mole).

Level 1 — Recognition

L1.1 — Count the quadratic slots

State how many quadratic degrees of freedom (DOF) each energy expression below contains. A quadratic DOF is one additive term of the form $α q^{2}$ (a constant times a squared coordinate or momentum). In part (c), $μ$ is the reduced mass of the two bonded atoms — a single effective mass that lets us treat the vibration of the pair as one particle of mass $μ$ moving along the bond; you can read it as "the effective mass of the vibration." The symbol $\overset{r}{˙}$ means the rate of change of the bond length $r$ (its stretching speed), and $κ$ is the bond's spring constant.

(a) $E = \frac{1}{2} m v_{x}^{2} + \frac{1}{2} m v_{y}^{2} + \frac{1}{2} m v_{z}^{2}$ (b) $E = \frac{1}{2} I ω^{2}$ (c) $E = \frac{1}{2} μ \overset{r}{˙}^{2} + \frac{1}{2} κ r^{2}$ (d) $E = \frac{1}{2} m v_{x}^{2} + a x^{4}$ (the $x^{4}$ term is not quadratic)

Recall Solution

Count only additive terms of the shape $α q^{2}$ .

(a) three such terms ( $v_{x}^{2}, v_{y}^{2}, v_{z}^{2}$ ) → $== 3 ==$ quadratic DOF.
(b) one term → $== 1 ==$ quadratic DOF.
(c) two terms, one in $\overset{r}{˙}^{2}$ (kinetic) and one in $r^{2}$ (potential) → $== 2 ==$ quadratic DOF. This is the vibration double-count.
(d) the $v_{x}^{2}$ term is quadratic → $== 1 ==$ quadratic DOF. The $a x^{4}$ term is quartic (a coordinate raised to the fourth power), not of the form $α q^{2}$ , so it is not a quadratic DOF and does not contribute $\frac{1}{2} k_{B} T$ . (Exactly how much it does contribute is worked out in L3.2/L5.2 below, once the tools are built — do not assume it here.) So: $== 1 ==$ quadratic DOF plus one non-quadratic term.

L1.2 — State the mean energy

A system has exactly $f$ quadratic DOF at temperature $T$ . Write its mean total energy.

Recall Solution

$⟨ E ⟩ = f \times \frac{1}{2} k_{B} T = \frac{f}{2} k_{B} T .$ Nothing about mass, stiffness, or geometry enters — only the count $f$ and the temperature.

Level 2 — Application

L2.1 — Mean translational energy of one molecule

Compute the mean translational kinetic energy of a single air molecule at $T = 300 K$ . Give the number in joules.

Recall Solution

Translation has 3 quadratic DOF (motion along $x, y, z$ ), so $⟨ E_{trans} ⟩ = \frac{3}{2} k_{B} T = \frac{3}{2} (1.380649 \times 1 0^{- 23}) (300) .$ $⟨ E_{trans} ⟩ \approx 6.21 \times 1 0^{- 21} J .$

L2.2 — Heat capacities of a diatomic gas

For nitrogen ( $N_{2}$ ) at room temperature, translation and rotation are active but vibration is frozen. Find $C_{V}$ , $C_{P}$ , and $γ = C_{P} / C_{V}$ per mole.

Recall Solution

Active DOF: 3 translation + 2 rotation $= 5$ . So $U = \frac{5}{2} R T, C_{V} = (\frac{\partial U}{\partial T})_{V} = \frac{5}{2} R .$ For an ideal gas $C_{P} = C_{V} + R = \frac{7}{2} R$ (this " $+ R$ " comes from the work done expanding at constant pressure via $P V = N k T$ , see Ideal gas law PV=NkT). $C_{V} = \frac{5}{2} (8.314) \approx 20.79 J/(mol\cdotK), C_{P} = \frac{7}{2} (8.314) \approx 29.10 J/(mol\cdotK) .$ $γ = \frac{C _{P}}{C _{V}} = \frac{7/2}{5/2} = \frac{7}{5} = 1.4.$

L2.3 — Dulong–Petit prediction

A classical solid models each atom as sitting in a 3D harmonic well. Predict the molar heat capacity and compare to the Dulong–Petit law value.

Recall Solution

Each atom: 3 kinetic + 3 potential quadratic terms $= 6$ DOF. $U = 6 \times \frac{1}{2} R T = 3 R T ⟹ C_{V} = 3 R \approx 24.94 J/(mol\cdotK) .$ This is the Dulong–Petit law, confirmed for many solids near room temperature.

Level 3 — Analysis

L3.1 — Why does stiffness cancel? (numerical demonstration)

Two 1-D oscillators sit at the same temperature $T = 300 K$ : oscillator A has spring constant $κ_{A} = 1 N/m$ , oscillator B has $κ_{B} = 100 N/m$ . For each, the mean potential energy is $⟨ \frac{1}{2} κ x^{2} ⟩ = \frac{1}{2} k_{B} T$ . Compute the mean squared displacement $⟨ x^{2} ⟩$ for each and confirm the mean potential energies are equal.

Figure — Equipartition theorem — ½k_BT per quadratic degree of freedom

The figure plots, for each oscillator, the probability density $P (x)$ — how likely the mass is to be found at displacement $x$ from equilibrium (the Boltzmann weight $\propto e^{- κ x^{2} /2 k_{B} T}$ , normalised so the area under each curve is 1). The horizontal axis is displacement $x$ ; the vertical axis is that probability density. The cyan curve is the soft spring (A), which lets the mass wander far — a wide bell. The amber curve is the stiff spring (B), which pins the mass near $x = 0$ — a narrow, tall bell. The stiff bell is $100 = 10 \times$ narrower. The point of the picture: although the two bells have very different widths, the stored energy $\frac{1}{2} κ ⟨ x^{2} ⟩$ (width $^{2} \times$ stiffness) comes out identical.

Recall Solution

Set $\frac{1}{2} κ ⟨ x^{2} ⟩ = \frac{1}{2} k_{B} T$ , so $⟨ x^{2} ⟩ = k_{B} T / κ$ . $⟨ x^{2} ⟩_{A} = \frac{( 1.380649 \times 1 0 ^{- 23} ) ( 300 )}{1} \approx 4.14 \times 1 0^{- 21} m^{2},$ $⟨ x^{2} ⟩_{B} = \frac{( 1.380649 \times 1 0 ^{- 23} ) ( 300 )}{100} \approx 4.14 \times 1 0^{- 23} m^{2} .$ The stiff spring (B) confines the mass to a 100× smaller $⟨ x^{2} ⟩$ — its Gaussian is narrower (look at the tall thin amber curve in the figure). But the product $\frac{1}{2} κ_{A} ⟨ x^{2} ⟩_{A} = \frac{1}{2} κ_{B} ⟨ x^{2} ⟩_{B} = \frac{1}{2} k_{B} T \approx 2.07 \times 1 0^{- 21} J$ is identical. The stiffness $κ$ narrows the spread by exactly the amount needed to keep the stored energy fixed — this is why $α$ (here $κ /2$ ) cancels in the general derivation.

L3.2 — A quartic potential

A particle moves in 1-D with energy $E = \frac{1}{2} m v^{2} + b x^{4}$ ( $b > 0$ ). Find its total mean energy at temperature $T$ using the generalised equipartition rule $⟨ α q^{n} ⟩ = \frac{1}{n} k_{B} T$ (this rule is proved from scratch in L5.2; here we simply apply it — it says: an energy term where a coordinate appears to the power $n$ carries $\frac{1}{n} k_{B} T$ on average).

Recall Solution

The kinetic term $\frac{1}{2} m v^{2}$ is quadratic ( $n = 2$ ) → contributes $\frac{1}{2} k_{B} T$ .
The potential term $b x^{4}$ has $n = 4$ → contributes $\frac{1}{4} k_{B} T$ . $⟨ E ⟩ = \frac{1}{2} k_{B} T + \frac{1}{4} k_{B} T = \frac{3}{4} k_{B} T .$ A quartic well is "stiffer at large $x$ " than a spring, so it stores less energy than a quadratic potential's $\frac{1}{2} k_{B} T$ — the generalised rule captures this cleanly.

Level 4 — Synthesis

L4.1 — Vibration turns on

Take a diatomic gas heated from room temperature to high temperature, where the bond vibration unfreezes. Find $C_{V}$ and $γ$ in the high- $T$ regime, and state by how much $C_{V}$ jumped from the room-temperature value.

Recall Solution

Room temp: 3 translation + 2 rotation = 5 DOF → $C_{V} = \frac{5}{2} R$ . High temp: vibration adds 2 DOF (KE + PE) → total $5 + 2 = 7$ DOF. $C_{V} = \frac{7}{2} R \approx 29.10 J/(mol\cdotK), C_{P} = \frac{9}{2} R, γ = \frac{9/2}{7/2} = \frac{9}{7} \approx 1.286.$ The jump is $Δ C_{V} = \frac{7}{2} R - \frac{5}{2} R = R \approx 8.314 J/(mol\cdotK)$ — one full $R$ , because vibration contributes two half- $R$ slots. See Quantum freezing of degrees of freedom for why this unfreezing is gradual, not a sharp step.

L4.2 — Speed from energy

Using the equipartition result $\frac{3}{2} k_{B} T = \frac{1}{2} m ⟨ v^{2} ⟩$ , find the root-mean-square speed $v_{rms} = ⟨ v^{2} ⟩$ of a nitrogen molecule ( $m = 4.65 \times 1 0^{- 26} kg$ ) at $T = 300 K$ .

Recall Solution

Equate total translational KE to $\frac{3}{2} k_{B} T$ : $\frac{1}{2} m ⟨ v^{2} ⟩ = \frac{3}{2} k_{B} T ⟹ ⟨ v^{2} ⟩ = \frac{3 k _{B} T}{m} .$ $⟨ v^{2} ⟩ = \frac{3 ( 1.380649 \times 1 0 ^{- 23} ) ( 300 )}{4.65 \times 1 0 ^{- 26}} \approx 2.672 \times 1 0^{5} m^{2} / s^{2} .$ $v_{rms} = ⟨ v^{2} ⟩ \approx 517 m/s .$ This is the same $v_{rms}$ that appears in the Maxwell–Boltzmann speed distribution — equipartition fixes the width of that distribution.

Level 5 — Mastery

L5.1 — Derive equipartition for one quadratic term from scratch

Starting from the Boltzmann distribution (probability of a value $q$ is $\propto e^{- α q^{2} / k_{B} T}$ ), prove that $⟨ α q^{2} ⟩ = \frac{1}{2} k_{B} T$ without quoting the theorem. You may use the Gaussian integral $\int_{- \infty}^{\infty} e^{- a x^{2}} d x = π / a$ for $a > 0$ .

Recall Solution

Step 1 — the average as a weighted integral. Write $β \equiv 1/ k_{B} T$ . The Boltzmann-weighted average is $⟨ α q^{2} ⟩ = \frac{\int _{- \infty}^{\infty} α q ^{2} e ^{- β α q^{2}} d q}{\int _{- \infty}^{\infty} e ^{- β α q^{2}} d q} .$ Why: the numerator weights each $α q^{2}$ by its probability; the denominator normalises. All other coordinates cancel between top and bottom.

Step 2 — turn the numerator into a derivative. Define $Z (β) = \int_{- \infty}^{\infty} e^{- β α q^{2}} d q$ . Differentiating under the integral pulls down a factor $- α q^{2}$ : $\frac{\partial Z}{\partial β} = \int_{- \infty}^{\infty} (- α q^{2}) e^{- β α q^{2}} d q .$ Therefore the numerator equals $- Z^{'} (β)$ , and $⟨ α q^{2} ⟩ = - \frac{Z ^{'} ( β )}{Z ( β )} = - \frac{d l n Z}{d β} .$ Why this trick: it converts an awkward $q^{2}$ -weighted integral into a plain derivative of one Gaussian. (This is exactly the log-derivative move used with the Partition function Z.)

Step 3 — evaluate the Gaussian. With $a = β α$ : $Z (β) = \frac{π}{β α} = \frac{π}{α} β^{- 1/2} .$

Step 4 — take the log-derivative. $ln Z = constant in β \frac{1}{2} ln \frac{π}{α} - \frac{1}{2} ln β ⟹ - \frac{d l n Z}{d β} = \frac{1}{2 β} = \frac{1}{2} k_{B} T .$ The stiffness $α$ sat only in the additive constant $\frac{1}{2} ln (π / α)$ , which vanishes under $d / d β$ . Hence $⟨ α q^{2} ⟩ = \frac{1}{2} k_{B} T,$ independent of $α$ . $■$

L5.2 — Generalise to $E = α ∣ q ∣^{n}$

Repeat the argument for $E = α ∣ q ∣^{n}$ and show $⟨ E ⟩ = \frac{1}{n} k_{B} T$ . (Hint: substitute $u = β^{1/ n} q$ to extract the $β$ -dependence.)

Recall Solution

$Z (β) = \int_{- \infty}^{\infty} e^{- β α ∣ q ∣^{n}} d q .$ Substitute $u = β^{1/ n} q$ , so $q = β^{- 1/ n} u$ and $d q = β^{- 1/ n} d u$ : $Z (β) = β^{- 1/ n} constant in β \int_{- \infty}^{\infty} e^{- α ∣ u ∣^{n}} d u = C β^{- 1/ n} .$ Then $ln Z = ln C - \frac{1}{n} ln β$ , and $⟨ E ⟩ = - \frac{d l n Z}{d β} = \frac{1}{n β} = \frac{1}{n} k_{B} T .$ Setting $n = 2$ recovers the familiar $\frac{1}{2} k_{B} T$ ; $n = 4$ gives $\frac{1}{4} k_{B} T$ (matching L3.2). $■$

Related deep dives: Heat capacity of gases, Brownian motion.