2.2 Temperature and equipartition

The pressure derivation of 2.1 ended on a striking identity: the mean translational kinetic energy of a molecule is $\langle\varepsilon\rangle = \tfrac32 k_B T$ . Temperature, the macroscopic quantity, is a measure of microscopic kinetic energy. This lesson generalises that statement into the equipartition theorem and uses it to read off the heat capacity of a gas.

Temperature as mean energy

A monatomic molecule stores energy only in its translation, $\varepsilon = \tfrac12 m(v_x^2 + v_y^2 + v_z^2)$ — three independent quadratic terms, one per direction. With $\langle\varepsilon\rangle = \tfrac32 k_B T$ , each direction carries $\tfrac12 k_B T$ on average. That the share is the same for every direction, and for every molecule regardless of mass, is the first instance of a general rule.

The equipartition theorem

Every quadratic term in a system’s energy carries, on average, $\tfrac12 k_B T$ in thermal equilibrium. The translational $\tfrac12 m v_x^2$ , a rotational $\tfrac12 I\omega^2$ , the kinetic and potential terms $\tfrac12 m\dot q^2$ and $\tfrac12\kappa q^2$ of a vibration — each is one such term, and each gets the same $\tfrac12 k_B T$ .

▶ ½α⟨q²⟩ = ½kᴮT — the equipartition integral Derivation

For a coordinate $q$ entering the energy as $\tfrac12\alpha q^2$ , the Boltzmann-weighted mean is

\Big\langle \tfrac12\alpha q^2 \Big\rangle \;=\; \frac{\displaystyle\int_{-\infty}^{\infty} \tfrac12\alpha q^2\, e^{-\beta\alpha q^2/2}\, dq}{\displaystyle\int_{-\infty}^{\infty} e^{-\beta\alpha q^2/2}\, dq}, \qquad \beta = \frac{1}{k_B T}.

The denominator is the Gaussian integral $\sqrt{2\pi/(\beta\alpha)}$ . The numerator follows by differentiating the denominator with respect to $\beta$ :

\int_{-\infty}^{\infty} q^2\, e^{-\beta\alpha q^2/2}\, dq \;=\; -\frac{2}{\alpha}\frac{d}{d\beta}\sqrt{\frac{2\pi}{\beta\alpha}} \;=\; \frac{1}{\beta}\sqrt{\frac{2\pi}{\beta\alpha}}.

The ratio is $\tfrac12\cdot\tfrac1\beta = \tfrac12 k_B T$ , independent of the stiffness $\alpha$ . ✓

The independence from $\alpha$ is the heart of the theorem: a stiffer coordinate has a narrower Boltzmann distribution (smaller $\langle q^2\rangle$ ), but the mean energy $\tfrac12\alpha\langle q^2\rangle$ is unchanged. That is what makes equipartition a clean accounting tool — count the quadratic terms, multiply by $\tfrac12 k_B T$ .

spring α1.00

temperature k_BT1.00

σ_q = √(k_BT/α)1.000

∫ ½αq² p(q) dq0.5000

½ k_BT0.5000

Spring constant α = 1.00 Temperature k_BT = 1.00

Vary α: the parabola widens or narrows, and the Gaussian narrows or widens in compensation. The integral of energy × density is *unchanged* at fixed T; it always equals ½k_BT. This is equipartition — each quadratic degree of freedom contributes exactly ½k_BT regardless of stiffness.

Vary $\alpha$ at fixed $T$ : the parabola widens or narrows, the Gaussian narrows or widens in compensation, and the area under the integrand is invariant — always $\tfrac12 k_B T$ .

Heat capacity and the ratio of specific heats

Counting the active quadratic terms — the degrees of freedom $d$ — gives the internal energy $U = \tfrac{d}{2} N k_B T$ , the molar heat capacity at constant volume $c_v = \tfrac{d}{2} R$ , and (with Mayer’s relation $c_p - c_v = R$ ) the ratio of specific heats $\gamma = c_p/c_v = (d+2)/d$ :

| Molecule | Active DOF at room $T$ | $U/N$ | $c_v/R$ | $\gamma$ | |---|---|---|---|---| | Monatomic (He, Ar) | 3 (translation) | $\tfrac32 k_B T$ | $3/2$ | $5/3 \approx 1.67$ | | Diatomic (N₂, O₂) | 5 (3 trans. + 2 rot.) | $\tfrac52 k_B T$ | $5/2$ | $7/5 = 1.4$ | | Triatomic (CO₂, H₂O) | 6+ | $\ge 3 k_B T$ | $\ge 3$ | $\le 4/3$ |

The vibrational degrees of freedom of N₂ and O₂ are frozen out at room temperature: the energy quantum of the molecular vibration is much larger than $k_B T \approx 25\,\text{meV}$ , so the oscillator sits in its ground state and contributes nothing to the heat capacity. This is the first sign that classical equipartition is incomplete — it overcounts the energy of stiff oscillators, and the correction is quantum: a mode whose energy quantum exceeds $k_B T$ is frozen in its ground state and drops out of the count. Classically, though, equipartition is exact, and it fixes $\gamma$ for a gas from nothing but a count of its molecular degrees of freedom.