Law of Equipartition of Energy

Leading up to the law

The kinetic energy of a single molecule is

$$\text{K.E.} = \dfrac{1}{2}mv_{x}^{2} + \dfrac{1}{2}mv_{y}^{2} + \dfrac{1}{2}mv_{z}^{2}$$

so for a gas at temperature $T$ the average kinetic energy per molecule is

$$\overline{\text{K.E.}} = \dfrac{1}{2}m\overline{v_{x}^{2}} + \dfrac{1}{2}m\overline{v_{y}^{2}} + \dfrac{1}{2}m\overline{v_{z}^{2}}$$

But the mean energy per molecule is $\dfrac{3}{2}k_{B}T$, and since there is no preferred direction $x$, $y$ or $z$,

$$\dfrac{1}{2}m\overline{v_{x}^{2}} = \dfrac{1}{2}m\overline{v_{y}^{2}} = \dfrac{1}{2}m\overline{v_{z}^{2}} = \dfrac{1}{2}k_{B}T$$

Thus the mean energy associated with every component of translational kinetic energy — each of which is quadratic in a velocity component — is $\dfrac{1}{2}k_{B}T$, and the total translational contribution is $\dfrac{3}{2}k_{B}T$.

Degrees of freedom

Degrees of freedom of a system are defined as the total number of coordinates or independent quantities required to describe the position and configuration of the system completely.

A molecule free to move anywhere in three-dimensional space needs three coordinates and has three components of velocity, so it has 3 degrees of freedom. Restricted to a plane it needs only $x$ and $y$, with velocity components $v_{x}$ and $v_{y}$, so it has 2 degrees of freedom. Restricted to a straight line it needs only $x$ and $v_{x}$, so it has 1 degree of freedom.

Degrees of freedom depend on the atomicity of the molecule, on the geometric structure or shape of the molecule, and on the temperature — vibrational modes are excited only at sufficiently high temperatures.

Monatomic molecules

A monatomic gas such as helium contains single He atoms. Each has only 3 translational degrees of freedom, and therefore an average energy $3 \times \dfrac{1}{2}k_{B}T = \dfrac{3}{2}k_{B}T$.

Diatomic molecules

Consider an O$_2$ or N$_2$ molecule with its two atoms lying along the $x$-axis.

• It has 3 translational degrees of freedom.

• In addition it can rotate about the $z$-axis and about the $y$-axis. Rotation about the $x$-axis is not a rotation in any meaningful sense, since it does not involve a change in the positions of the two atoms. In general a diatomic molecule can rotate about its centre of mass in two directions perpendicular to its molecular axis, so it possesses 2 additional rotational degrees of freedom.

Diagram in words: a dumbbell-shaped molecule with its two atoms on the $x$-axis, one on each side of the centre of mass, and two curved arrows showing rotation about the $z$-axis (vertical, through the centre of mass) and about the $y$-axis (horizontal, perpendicular to the molecular axis).

If $I_{z}$ and $I_{y}$ are the moments of inertia about the $z$ and $y$ axes and $\omega_{z}$, $\omega_{y}$ the corresponding angular speeds, the rotational kinetic energies are $\dfrac{1}{2}I_{z}\omega_{z}^{2}$ and $\dfrac{1}{2}I_{y}\omega_{y}^{2}$. The total energy of a rigid diatomic molecule is then

$$E = E_{\text{translational}} + E_{\text{rotational}} = \dfrac{1}{2}mv_{x}^{2} + \dfrac{1}{2}mv_{y}^{2} + \dfrac{1}{2}mv_{z}^{2} + \dfrac{1}{2}I_{z}\omega_{z}^{2} + \dfrac{1}{2}I_{y}\omega_{y}^{2}$$

Every term is quadratic, and each contributes $\dfrac{1}{2}k_{B}T$.

Vibrational degrees of freedom

The above assumes implicitly that the rotating molecule is a rigid rotator. Real molecules contain covalent bonds between the atoms and can therefore also vibrate about their mean positions like a one-dimensional harmonic oscillator. In diatomic molecules such as O$_2$, N$_2$ and CO the atoms can oscillate along the internuclear axis only. Then

$$E = E_{\text{translational}} + E_{\text{rotational}} + E_{\text{vibrational}}$$

$$E_{\text{vibrational}} = \dfrac{1}{2}mu^{2} + \dfrac{1}{2}kr^{2}$$

where $u$ is the velocity of vibration of the atoms, $r$ is the separation between the oscillating atoms and $k$ is related to the force constant. These two terms are quadratic in velocity and in position respectively, and each contributes $\dfrac{1}{2}k_{B}T$. Hence each vibrational mode contributes $2 \times \dfrac{1}{2}k_{B}T = k_{B}T$ to the internal energy.

Statement of the law

For a gas in thermal equilibrium at a temperature $T$, the average energy per molecule associated with each quadratic term is $\dfrac{1}{2}k_{B}T$.

For a non-rigid diatomic gas in thermal equilibrium at temperature $T$, this gives: mean kinetic energy of translation along three directions $= 3 \times \dfrac{1}{2}k_{B}T$; mean kinetic energy of rotation about two perpendicular axes $= 2 \times \dfrac{1}{2}k_{B}T$; and total vibrational energy $= 2 \times \dfrac{1}{2}k_{B}T$, corresponding to the kinetic and potential terms.

The law of equipartition of energy is valid at high temperatures and not at extremely low temperatures, where quantum effects become important.

Degrees of freedom and energy, summarised

Standard results

• Average energy per quadratic term $= \dfrac{1}{2}k_{B}T$.

• Each vibrational mode contributes $k_{B}T$, because it has both a kinetic and a potential quadratic term.

• Energy per molecule with $f$ quadratic terms $= \dfrac{f}{2}k_{B}T$; internal energy of one mole $= \dfrac{f}{2}RT$.