In short

A degree of freedom (DOF) is an independent way for a molecule to store energy. A helium atom has three — its three velocity components. A nitrogen molecule at room temperature has five — three translational plus two rotational (it cannot rotate meaningfully about its long axis because the moment of inertia is too small). A rigid three-atom molecule like water has six — three translational plus three rotational. At still higher temperatures, molecules also start vibrating, and each vibrational mode counts as two degrees of freedom (kinetic plus potential).

The equipartition theorem says: in thermal equilibrium at temperature T, each quadratic term in the energy (each DOF) carries an average energy

\langle E_\text{per DOF}\rangle = \tfrac{1}{2}k_B T.

Adding up all active DOF gives the molar internal energy U = \tfrac{f}{2}R T, from which the molar specific heats follow:

C_v = \frac{f}{2}R, \qquad C_p = C_v + R = \frac{f+2}{2}R, \qquad \gamma = \frac{C_p}{C_v} = \frac{f+2}{f}.

For the three textbook cases:

  • Monatomic gas (He, Ne, Ar): f = 3, so C_v = \tfrac{3}{2}R, C_p = \tfrac{5}{2}R, \gamma = 5/3 \approx 1.667.
  • Diatomic gas at moderate T (N₂, O₂, H₂): f = 5, so C_v = \tfrac{5}{2}R, C_p = \tfrac{7}{2}R, \gamma = 7/5 = 1.4.
  • Polyatomic rigid gas (H₂O vapour, CO₂ linear: special): f = 6, so C_v = 3R, C_p = 4R, \gamma = 4/3 \approx 1.333.

This is why water vapour absorbs more heat per molecule than nitrogen per molecule — it has more degrees of freedom to spread the energy across. It is also why γ of air (78% N₂, 21% O₂) is very nearly 7/5, a result that appears in every sound-speed and adiabatic-process calculation.

At very low temperatures the rotational and then the translational modes "freeze out" — quantum mechanics forbids partial excitation of a mode, and k_B T needs to exceed the mode's energy spacing for it to carry its equipartition share. The hydrogen-gas C_v data in 1900 showing rotational DOF disappearing below 80 K was one of the key experimental inputs to the quantum revolution.

Pour some helium into a thick rubber balloon at room temperature. Measure how much heat it takes to warm that gas by one kelvin at constant volume, per mole. The answer is \tfrac{3}{2}R \approx 12.5\,\text{J mol}^{-1}\,\text{K}^{-1}. Now do the same experiment with nitrogen: the answer is \tfrac{5}{2}R \approx 20.8\,\text{J mol}^{-1}\,\text{K}^{-1}. With water vapour at 200 °C, about 3R. With CO₂ at room temperature, about \tfrac{5}{2}R that rises toward 3R as the gas is heated. The numbers are different — but they are suspiciously simple multiples of R, and the multiplier looks like it is counting something.

The something is degrees of freedom: the independent ways a molecule can carry energy. A single helium atom can move in three independent directions (translate in x, y, z) — that is three degrees of freedom. A dumbbell-shaped nitrogen molecule can also rotate about two axes perpendicular to its bond — that is two more. At higher temperatures it can also vibrate, stretching and compressing the bond, which counts for two more (kinetic and potential). Every such independent energy-storing "address" is what a DOF is.

The equipartition theorem says every degree of freedom carries on average \tfrac{1}{2}k_B T of energy in thermal equilibrium. Add up the DOF, multiply by \tfrac{1}{2}k_B T per molecule (or \tfrac{1}{2}R T per mole), and you have the molar internal energy. Differentiate with respect to T and you have the molar specific heat at constant volume. That is why C_v = \tfrac{f}{2}R where f is the number of active DOF. One theorem explains all the textbook values.

This article builds that machinery from the start. You should have read the ideal gas law (so that PV = nRT is a given) and ideally kinetic theory — pressure and temperature (which shows that \tfrac{1}{2}m\langle v^2\rangle = \tfrac{3}{2}k_B T per molecule, the translational equipartition made explicit). Familiarity with specific heat and calorimetry is useful — this chapter tells you why the specific heats have the values you measured in an earlier experiment.

What is a degree of freedom

Start with the simplest case: a single point particle in a box.

To describe the position of a point particle you need three numbers: its coordinates x, y, z. To describe its velocity you need three more: v_x, v_y, v_z. The kinetic energy is

E_\text{kin} = \tfrac{1}{2}m v_x^2 + \tfrac{1}{2}m v_y^2 + \tfrac{1}{2}m v_z^2. \tag{1}

Three independent quadratic terms. Each of the three velocity components is an independent way to store kinetic energy — three translational degrees of freedom.

A DOF is formally an independent quadratic term in the energy expression. The quadratic nature matters — the equipartition theorem is a theorem about quadratic terms, not about anything-the-molecule-can-do.

Now consider a rigid dumbbell — two point masses joined by a stiff rod. The centre of mass still has three translational DOF (kinetic energy like (1)). But the dumbbell can also rotate about two perpendicular axes through its centre of mass. Rotation about each axis carries kinetic energy \tfrac{1}{2}I\omega^2 — another quadratic term per axis. A rigid dumbbell therefore has

E = \underbrace{\tfrac{1}{2}m v_x^2 + \tfrac{1}{2}m v_y^2 + \tfrac{1}{2}m v_z^2}_{\text{3 translational}} + \underbrace{\tfrac{1}{2}I_1\omega_1^2 + \tfrac{1}{2}I_2\omega_2^2}_{\text{2 rotational}} \tag{2}

— five degrees of freedom.

Degrees of freedom for mono-, di-, and tri-atomic molecules Three sketches: a single atom with three translation arrows; a diatomic dumbbell with three translation arrows and two rotation arrows about axes perpendicular to the bond; and a non-linear tri-atomic molecule with three translation arrows and three rotation arrows. Monatomic (He): f = 3 3 translational Diatomic (N₂): f = 5 3 translational + 2 rotational Polyatomic (H₂O): f = 6 3 translational + 3 rotational
Molecular degree-of-freedom count for a monatomic atom (3 translational), a linear diatomic molecule (3 translational + 2 rotational; the third rotation about the bond axis does not count classically — the moment of inertia is negligibly small), and a non-linear polyatomic molecule such as water (3 translational + 3 rotational). Vibrational DOF are not shown; they become active only at higher temperatures.

Why only two rotational DOF for a linear diatomic instead of three? A mathematical dumbbell has three rotation axes: the two perpendicular to the bond (which do carry rotational kinetic energy) and the one along the bond. Rotating about the bond axis does not actually move the atoms if you treat them as point masses, so the moment of inertia is zero and no energy is stored. Even for real molecules the "along the bond" moment of inertia is tiny (\sim 10^{-48}\,\text{kg m}^2, small by molecular standards), and quantum mechanics freezes that DOF out at all practical temperatures. Two rotational DOF is therefore the right count for any linear molecule (diatomic, or CO₂ which is linear).

Non-linear molecules like water or ammonia have three distinct moments of inertia all of roughly the same size, so all three rotational DOF are active at ordinary temperatures.

A molecule can also vibrate. In a diatomic molecule, the bond can stretch and compress; that is one vibrational mode. In a non-linear triatomic molecule like water, there are three vibrational modes (symmetric stretch, antisymmetric stretch, bending). Each vibrational mode has two quadratic terms in its energy — the kinetic energy \tfrac{1}{2}\mu\dot{q}^2 and the potential energy \tfrac{1}{2}k q^2 of the spring-like bond, where q is the displacement and \mu is the reduced mass. Both are quadratic in some coordinate, so each vibrational mode contributes 2 DOF.

We will see that at room temperature most vibrational modes are "frozen" by quantum mechanics (their energy spacings are larger than k_B T), so they do not contribute to the specific heat. They start turning on at temperatures around 1000 K and above — which is why C_v of diatomic gases rises from \tfrac{5}{2}R toward \tfrac{7}{2}R at high temperature.

The equipartition theorem

Statement. For any classical system in thermal equilibrium at temperature T, the average energy associated with each quadratic term in the total energy expression (Hamiltonian) is \tfrac{1}{2}k_B T.

In symbols: if the total energy is

E = \sum_i \alpha_i q_i^2

for some coordinates q_i (which might be positions, velocities, angular velocities, or anything else) and positive coefficients \alpha_i, then \langle \alpha_i q_i^2\rangle = \tfrac{1}{2}k_B T for each i.

Why it is true — the sketch. The theorem follows from the Boltzmann distribution: in thermal equilibrium the probability of finding the system with coordinate q_i in the interval [q_i, q_i + dq_i] is proportional to e^{-E/(k_B T)}\,dq_i. For a single quadratic term E_i = \alpha_i q_i^2, the average value of E_i is

\langle E_i\rangle = \frac{\int_{-\infty}^{\infty}\alpha_i q_i^2\, e^{-\alpha_i q_i^2/(k_B T)}\,dq_i}{\int_{-\infty}^{\infty} e^{-\alpha_i q_i^2/(k_B T)}\,dq_i}.

Change variables u = q_i\sqrt{\alpha_i/(k_B T)}; the ratio becomes

\langle E_i\rangle = k_B T\,\frac{\int u^2 e^{-u^2}\,du}{\int e^{-u^2}\,du} = k_B T \cdot \frac{1}{2} = \tfrac{1}{2}k_B T. \tag{3}

Why: both Gaussian integrals are standard. \int e^{-u^2}\,du = \sqrt{\pi} and \int u^2 e^{-u^2}\,du = \sqrt{\pi}/2; their ratio is 1/2. The coefficient \alpha_i dropped out — which is the central result. Every quadratic term, regardless of its prefactor, carries the same average energy \tfrac{1}{2}k_B T.

So for a gas of molecules each with f quadratic DOF, the total internal energy of N molecules is

U = N f \cdot \tfrac{1}{2}k_B T = \tfrac{f}{2} N k_B T = \tfrac{f}{2}nRT, \tag{4}

using N k_B = nR.

Consequences — molar specific heats

The molar specific heat at constant volume is

C_v = \left(\frac{\partial U}{\partial T}\right)_V = \frac{f}{2}R. \tag{5}

At constant volume, no work is done, so all heat added goes into internal energy. The specific heat follows directly.

The molar specific heat at constant pressure uses Mayer's relation C_p = C_v + R (derived for an ideal gas in the ideal gas laws article, going-deeper section):

C_p = \frac{f}{2}R + R = \frac{f+2}{2}R. \tag{6}

The adiabatic index

\gamma = \frac{C_p}{C_v} = \frac{f+2}{f}. \tag{7}

Three textbook cases:

Gas type f C_v C_p \gamma = C_p/C_v
Monatomic (He, Ne, Ar) 3 \tfrac{3}{2}R \tfrac{5}{2}R 5/3 \approx 1.667
Diatomic (N₂, O₂, H₂) at 300 K 5 \tfrac{5}{2}R \tfrac{7}{2}R 7/5 = 1.4
Non-linear polyatomic (H₂O, NH₃, CH₄) 6 3R 4R 4/3 \approx 1.333
The adiabatic index $\gamma = (f+2)/f$ plotted against the number of active degrees of freedom. $\gamma$ starts at 5/3 for a monatomic gas and decreases toward 1 as more modes become active. The four marked points are the textbook values; measurements of $\gamma$ for real gases at different temperatures trace out this curve as vibrational modes progressively activate.

These values are borne out by experiment:

Every one of these matches comes directly from counting DOF and applying equipartition.

Interactive: count DOF and read off γ

Interactive: γ as a function of degrees of freedom A plot of γ = (f+2)/f as f varies from 1 to 10. A draggable red point lets you pick f; readouts display γ, C_v/R, and C_p/R for that value of f. number of degrees of freedom f γ = (f+2)/f 0 1 2 1 3 5 7 9 γ = 1 (many-DOF limit) drag red point for f
Drag the red point to vary the number of degrees of freedom $f$ from 1 to 10. Read off $\gamma = (f+2)/f$, $C_v/R$, and $C_p/R$ live. At $f = 3$ (monatomic), $\gamma = 5/3$; at $f = 5$ (diatomic at room temp), $\gamma = 7/5$; at $f = 6$ (non-linear polyatomic), $\gamma = 4/3$. As $f$ grows, $\gamma$ approaches 1 — the high-DOF limit where specific-heat differences between isobaric and isochoric paths become negligible.

Why H₂O absorbs more heat than N₂ — the specific-heat puzzle solved

One of the cleanest applications of equipartition is explaining why water vapour has a higher heat capacity (per mole) than nitrogen.

At 400 K:

Water carries 20% more energy per mole per kelvin than nitrogen, simply because the water molecule has three rotational DOF instead of two. You are reading a direct consequence of molecular geometry in every steam table.

(The commonly cited specific heat of liquid water at 4.18 J/g/K is a separate and larger number because liquid water has additional modes: intermolecular hydrogen-bond vibrations and librations. The equipartition count applies cleanly to gases; for liquids the picture gets more complicated and is outside this article's scope.)

Worked examples

Example 1: Molar specific heat of helium versus nitrogen

A 0.10 mol sample of helium gas and a separate 0.10 mol sample of nitrogen gas each sit in rigid 1-litre containers at 300 K. Each is heated by adding 50 J of heat at constant volume. Find the final temperature of each gas.

Comparing helium and nitrogen heat responses Two rigid 1 L containers, each with 0.1 mol of gas at 300 K. One labelled He receives 50 J, warming to 340 K. The other labelled N2 receives 50 J, warming to 324 K. He (monatomic, f = 3) 0.1 mol, 300 K +50 J ΔT = 40 K → 340 K N₂ (diatomic, f = 5) 0.1 mol, 300 K +50 J ΔT = 24 K → 324 K
Same number of moles, same heat added, but nitrogen warms up less because it has more degrees of freedom to spread the energy across.

Step 1. Count DOF. Helium is monatomic, f = 3. Nitrogen at 300 K is diatomic with translational + rotational modes active, f = 5.

Step 2. Compute C_v for each.

C_{v,\text{He}} = \tfrac{3}{2} R = \tfrac{3}{2} \times 8.314 = 12.47\,\text{J mol}^{-1}\,\text{K}^{-1}.
C_{v,\text{N}_2} = \tfrac{5}{2} R = \tfrac{5}{2} \times 8.314 = 20.79\,\text{J mol}^{-1}\,\text{K}^{-1}.

Why: each DOF contributes \tfrac{1}{2}R to the molar specific heat by equipartition. Count the DOF, divide by two, multiply by R.

Step 3. Heat equation at constant volume: Q = n C_v \Delta T. Solve for \Delta T.

For He: \Delta T = Q/(n C_v) = 50/(0.10 \times 12.47) = 40.1 K. Final T = 340 K.

For N₂: \Delta T = 50/(0.10 \times 20.79) = 24.1 K. Final T = 324 K.

Why: same heat input, same amount of gas. The difference is entirely in how many DOF are sharing the energy. More DOF means each one gets a smaller share, so the translational temperature rises less.

Step 4. Sanity check with total energy. For He: U = \tfrac{3}{2}nRT, so \Delta U = \tfrac{3}{2}(0.10)(8.314)(40.1) = 50.0 J. ✓ For N₂: \Delta U = \tfrac{5}{2}(0.10)(8.314)(24.1) = 50.1 J. ✓

Result: the same 50 J of heat raises helium by 40 K but nitrogen by only 24 K — a factor of 5/3 difference, exactly the ratio of their DOF counts.

What this shows: the equipartition theorem gives you every molecular gas's specific heat for free, just from geometry. The factor \tfrac{5}{3} between He and N₂ is not an empirical constant — it is the ratio of their degree-of-freedom counts.

Example 2: Sound speed in air versus helium

The speed of sound in an ideal gas is v_s = \sqrt{\gamma RT/M} where \gamma is the adiabatic index and M is the molar mass. Use equipartition to predict and compare the sound speed in dry air at 20 °C with the sound speed in helium at the same temperature.

Air: treat as 78% N₂ + 21% O₂ + 1% Ar; take effective molar mass M_\text{air} = 28.96 g/mol. Helium: M_\text{He} = 4.003 g/mol.

Speed of sound (vertical, m/s) versus temperature (horizontal, kelvin) for air (red, $\gamma = 7/5$) and helium (black, $\gamma = 5/3$). At 293 K, sound in helium is about three times faster than in air — hence the squeaky voice after breathing helium.

Step 1. Compute \gamma for each gas.

Air is mostly diatomic (N₂ and O₂ are both diatomic; Ar is only 1% monatomic). So \gamma_\text{air} \approx 7/5 = 1.4.

Helium is monatomic, so \gamma_\text{He} = 5/3.

Step 2. Speed of sound in air.

v_\text{air} = \sqrt{\frac{\gamma RT}{M}} = \sqrt{\frac{1.4 \times 8.314 \times 293}{0.02896}} = \sqrt{117{,}800} = 343\,\text{m/s}.

Why: 20 °C = 293 K (absolute temperature matters). The formula follows from treating a sound wave as an adiabatic compression-rarefaction — which in turn uses PV^\gamma = \text{const}, which comes from equipartition via \gamma.

Step 3. Speed of sound in helium.

v_\text{He} = \sqrt{\frac{1.667 \times 8.314 \times 293}{0.004003}} = \sqrt{1{,}015{,}000} = 1007\,\text{m/s}.

Why: two factors push He higher than air: \gamma is larger (1.667 vs 1.4) and M is much smaller (4 vs 29 g/mol). The ratio is \sqrt{(1.667/1.4)(29/4)} = \sqrt{1.19 \times 7.25} = \sqrt{8.63} = 2.94, matching the factor of \approx 3 between He and air.

Step 4. Connect to the "helium voice" demonstration. The resonant frequencies of your vocal tract are f \propto v_s/L, where L is the tract length. Swap air for helium in your lungs and the sound speed jumps by a factor of 2.94 — all resonant frequencies shift upward by the same factor, giving the characteristic Donald Duck voice. Importantly, your vocal cord vibration frequency (the fundamental pitch) does not change — that is set by muscle tension, not by the gas. What changes is the formant frequencies (the resonance peaks of the tract), which is why the timbre is altered even though the fundamental pitch stays close to normal.

Result: v_\text{air} = 343 m/s and v_\text{He} = 1007 m/s at 20 °C. The ratio is about 2.94, set by both the \gamma ratio and the mass ratio.

What this shows: equipartition fixes \gamma for each gas, which in turn determines acoustic properties. Every acoustic calculation — from loudspeaker design to ultrasound imaging to the pitch of a Diwali-rocket whistle — rests on these DOF counts.

Example 3: CO₂ versus H₂O — why CO₂ has lower C_v

CO₂ is a linear molecule: O=C=O, atoms on a line. H₂O is non-linear (bent, 104.5° angle). Both are triatomic. Predict the molar specific heat at constant volume C_v for each at moderate temperature (around 400 K, before significant vibration excitation), and compare to the experimental values C_v(\text{CO}_2) = 28.5\,\text{J mol}^{-1}\,\text{K}^{-1} and C_v(\text{H}_2\text{O}) = 25.3\,\text{J mol}^{-1}\,\text{K}^{-1} at 400 K.

Geometry of CO2 and H2O molecules Left: CO2 drawn as a linear O=C=O molecule, with two rotational axes perpendicular to the bond. Right: water drawn as a bent H-O-H molecule with three distinct rotational axes. CO₂ (linear) O C O f_rot = 2 (rotation about bond axis has I ≈ 0) f_total = 3 + 2 = 5 H₂O (bent) O H H f_rot = 3 (three distinct I's) f_total = 3 + 3 = 6
Molecular geometry determines degree-of-freedom count. CO₂ is linear, so one rotational axis contributes no energy and $f = 5$. H₂O is bent, so all three rotational axes are active and $f = 6$.

Step 1. Count DOF for each molecule ignoring vibration.

CO₂ (linear): 3 translational + 2 rotational (rotation about the O=C=O axis has essentially zero moment of inertia) = 5 DOF.

H₂O (bent, non-linear): 3 translational + 3 rotational = 6 DOF.

Step 2. Predict C_v from C_v = \tfrac{f}{2}R.

C_v(\text{CO}_2)^\text{rigid} = \tfrac{5}{2} R = 20.8\,\text{J mol}^{-1}\,\text{K}^{-1}.

C_v(\text{H}_2\text{O})^\text{rigid} = 3R = 24.9\,\text{J mol}^{-1}\,\text{K}^{-1}.

Step 3. Compare to experiment.

CO₂ measured: 28.5 J/mol/K. Difference from rigid prediction: 28.5 - 20.8 = 7.7 J/mol/K — suggests about 7.7/(R) = 0.93 \approx 1 additional degree-of-freedom pair (a vibration is 2 DOF, so this is about half a vibration excited). CO₂ has four vibrational modes; at 400 K one of them (the 667 cm⁻¹ bending mode, energy = h c \bar\nu = 1.32\times 10^{-20} J, compared to k_B T = 5.52\times 10^{-21} J) has a non-negligible thermal population, adding roughly its fair share to C_v. The partial contribution matches the 7.7 J/mol/K excess reasonably well.

H₂O measured: 25.3 J/mol/K. Difference: 25.3 - 24.9 = 0.4 J/mol/K — the vibrations of H₂O have much higher frequencies (1595 cm⁻¹, 3657 cm⁻¹, 3756 cm⁻¹), corresponding to energies 1.98 h c / k_B \approx 2800 K expressed as equivalent temperatures, so at 400 K they are essentially unexcited. The rigid-body equipartition prediction nails H₂O within 2%.

Why: the reason CO₂'s measured C_v exceeds the rigid prediction while H₂O's does not is straight quantum mechanics. Low-frequency modes unfreeze at lower temperatures than high-frequency modes; CO₂'s bending mode is softest and comes on first.

Result: CO₂ rigid C_v = 20.8, measured 28.5 — partially activated vibrations. H₂O rigid C_v = 24.9, measured 25.3 — essentially pure rigid-body, in excellent agreement with equipartition.

What this shows: DOF counting plus a check of the vibrational-mode frequencies gives you the full specific-heat story for any gas. The equipartition theorem is exact in the classical limit; corrections come only from quantum freezing-out of individual modes.

Common confusions

If you came here to use equipartition to predict specific heats, you have what you need. What follows is the quantum derivation of why modes freeze out, the Einstein and Debye theories of solid specific heats, and the Maxwell-Boltzmann speed distribution that underpins the whole enterprise.

Why modes freeze out — the quantum correction

The classical equipartition derivation assumed the coordinate q (or the mode amplitude) is continuous. Quantum mechanically, each harmonic-oscillator mode has quantised energies E_n = (n + \tfrac{1}{2}) h\nu with n = 0, 1, 2, \ldots. The average energy of such a mode in thermal equilibrium is no longer k_B T but

\langle E\rangle = \frac{h\nu}{e^{h\nu/(k_B T)} - 1} + \tfrac{1}{2}h\nu. \tag{8}

In the high-temperature limit (h\nu \ll k_B T), expand the exponential: e^{h\nu/(k_B T)} - 1 \approx h\nu/(k_B T), giving \langle E\rangle \approx k_B T + \tfrac{1}{2}h\nu \approx k_B T. So equipartition is recovered at high T.

In the low-temperature limit (h\nu \gg k_B T), the exponential factor is huge: \langle E\rangle \approx h\nu\, e^{-h\nu/(k_B T)} + \tfrac{1}{2}h\nu \to \tfrac{1}{2}h\nu. The mode sits in its ground state, thermally inaccessible. The specific-heat contribution of this mode goes to zero.

The crossover temperature for each mode is \Theta = h\nu/k_B — the characteristic temperature of the mode. For N₂ vibration (\nu = 2360 cm⁻¹), \Theta \approx 3400 K — which is why N₂ vibrations are essentially frozen at room temperature. For N₂ rotation (\Theta \approx 3 K), rotations are fully active at any practical gas temperature.

Hydrogen gas (H₂) has a notably high rotational \Theta \approx 85 K — so below that temperature, rotational DOF freeze out and C_v drops toward \tfrac{3}{2}R. This was measured around 1900 and was one of the signatures that pointed experimentally toward the quantum hypothesis.

Einstein's and Debye's models of solid specific heats

For solids, the Dulong-Petit rule C_v = 3R per mole breaks down at low temperatures: C_v \to 0 as T \to 0, in violation of equipartition. Einstein in 1907 proposed treating each atom in the solid as an independent 3D quantum harmonic oscillator of frequency \nu_E. Using (8), the molar specific heat becomes

C_v^\text{Einstein} = 3R\,\frac{(\Theta_E/T)^2\, e^{\Theta_E/T}}{(e^{\Theta_E/T} - 1)^2}, \tag{9}

which goes to 3R at high T (Dulong-Petit recovered) and to zero exponentially at low T (as e^{-\Theta_E/T}). This was the first quantum explanation of a solid's specific heat.

Debye in 1912 improved on Einstein by recognising that the vibrations of a solid are not all at one frequency but spread over a range of modes (phonons, essentially standing sound waves in the solid). The Debye model predicts C_v \propto T^3 at low T — a famous result that agrees beautifully with experiment.

The Maxwell-Boltzmann speed distribution

Equipartition gives the mean kinetic energy per molecule: \langle \tfrac{1}{2}mv^2\rangle = \tfrac{3}{2}k_B T. The full distribution is the Maxwell-Boltzmann distribution:

f(v)\,dv = 4\pi n \left(\frac{m}{2\pi k_B T}\right)^{3/2}\, v^2\, e^{-mv^2/(2k_B T)}\,dv, \tag{10}

where f(v)\,dv is the number density of molecules with speed between v and v+dv. Three characteristic speeds fall out of this:

  • Most probable speed (peak of the distribution): v_p = \sqrt{2k_B T/m}.
  • Mean speed: \langle v\rangle = \sqrt{8 k_B T/(\pi m)}.
  • Root-mean-square speed: v_\text{rms} = \sqrt{\langle v^2\rangle} = \sqrt{3k_B T/m}.

These satisfy v_p < \langle v\rangle < v_\text{rms}, with ratios 1 : 1.128 : 1.225.

For nitrogen at 300 K: v_p = 422 m/s, \langle v\rangle = 476 m/s, v_\text{rms} = 517 m/s. A nitrogen molecule at room temperature moves at roughly half a kilometre per second, covering its mean free path of \sim 60 nm in about 10^{-10} s. That is a lot of collisions per second.

Connection to the speed of sound

The adiabatic speed of sound in an ideal gas is

v_s = \sqrt{\gamma R T / M} = \sqrt{\gamma k_B T/m},

where m is the molecular mass. Compare to v_\text{rms} = \sqrt{3 k_B T/m}. The ratio is

\frac{v_s}{v_\text{rms}} = \sqrt{\frac{\gamma}{3}}.

For air (\gamma = 1.4): \sqrt{1.4/3} = 0.683. Sound speed is about 68% of the rms molecular speed. This makes physical sense: a sound wave propagates by molecular collisions, but not every molecule is moving in the wave direction — only a particular average matters.

Equipartition and the ultraviolet catastrophe revisited

The infamous UV catastrophe in black-body radiation (see the going-deeper section of the radiation article) was equipartition applied to the continuum of electromagnetic modes in a cavity. Each mode gets k_B T of energy (2 DOF per mode — electric and magnetic — times \tfrac{1}{2}k_B T each). But because the number of modes per unit wavelength grows as \lambda^{-4}, the total energy integrated over all wavelengths diverges. Planck's fix (quantised mode energies) froze out high-frequency modes by the same mechanism as (8), curing the catastrophe and giving birth to quantum mechanics.

Equipartition is therefore not merely a 19th-century gas-physics result. It is the classical limit of quantum thermodynamics — the answer you get when every mode has enough thermal energy to be in the classical regime. When a mode does not, quantum mechanics takes over and the specific heat drops. The transition temperature between the two regimes is set by the mode frequency; the physics of degrees-of-freedom counting is the physics of finding which modes are classical at a given T.

Where this leads next