# Rotational partition function

The rotational partition function relates the rotational degrees of freedom to the rotational part of the energy.

## Definition

The total partition function $Z$ of a system of identical particles can be divided into molecular partition functions $\zeta$. Under the assumption that that all energy levels $E_j$ can be partitioned into its contributions from different degrees of freedom (weakly coupled degrees of freedom)

$E_j = \sum_i E_j^i=E_j^{trans}+E_j^{rot}+E_j^{vib}+E_j^{e}$

and the number of degenerate states are given as products of the single contributions

$g_j =\prod_i g_j^i =g_j^{trans}g_j^{rot}g_j^{vib}g_j^{e},$

where "trans", "rot", "vib" and "e" denotes translational, rotational and vibrational contributions as well as electron excitation, the molecular partition functions

$\zeta=\sum_j g_j e^{-E_j/k_B T}$

can be written as a product itself

$\zeta =\prod_i \zeta^i =\zeta^{trans}\zeta^{rot}\zeta^{vib}\zeta^{e}.$

Rotational energies are quantized. For a diatomic molecule like CO or HCl, the allowed rotational energies are

$E_j^{rot}=\frac{\bold{J}^2}{2I}=\frac{j(j+1)\hbar^2}{2I}=j(j+1)\epsilon.$

Now $g_j$ for each level $j$ is $2j+1$, so the rotational partition function is therefore

$\zeta^{rot}=\sum_{j=0}^{\infty}g_je^{-E_j/kT}=\sum_{j=0}^{\infty}(2j+1)e^{-j(j+1)\epsilon/kT}.$

If the difference between energy levels $E_j^{rot}$ is very small compared to $k_B T$, then the sum can be approximated by an integral and is found to be proportional to the temperature. For more complex polyatomic molecules, the situation is more difficult.

For the CO molecule at $T=300K$, the (unit less) contribution $\zeta^{rot}$ to $\zeta$ turns out to be in the range of $10^2$.

The rotational energy can now be computed by taking the derivative of $\zeta^{rot}$ with respect to temperature $T$. This can of course also be done by using the equipartition theorem, which says that in an equilibrium the energies of all degrees of freedom are proportional to the temperature.