# Rotational contribution towards heat capacity of an ideal gas made up of arbitrary shaped particles

Consider the particles of an ideal gas to be arbitrarily shaped, but all particles have the same shape. The Hamiltonian has a translation term and a rotational term. The translation contribution towards the heat capacity is $$\frac{3}{2}k_B$$. The rotational part of the Hamiltonian $$H$$ is $$\frac{1}{2}I_i \omega^2$$ where $$I_i$$ is the instantaneous axis of rotation. This may be rewritten as $$H=\frac{L_i^2}{2I_i}$$. A similar expression was found for the diatomic case, and borrowing the details from there, I have my energy $$\epsilon = \frac{\hbar^2 l (l+1)}{2I_i}$$ and my partition function $$Z = \Sigma_{l=0}^{\infty}(2l+1)e^{-\beta\epsilon}$$. Defining $$\Theta_R=\frac{\hbar^2}{2Ik_B}$$ and for $$T>>\Theta_R$$, we should recover the classical expression. Problem is that I'm getting $$C_{rot}=k_B$$ whereas my arbitrary shaped body has 3 rotational degrees of freedom, so I should be getting $$C_{rot}=\frac{3}{2}k_B$$. What have I done wrong above? Assume $$N=1$$ throughout.

Another question which I think is relevant and so I'm including it here: Doesn't my instantaneous axis keep changing as my particles keep colliding? How do I account for this?

• This may not be the issue - but you are assuming a single instantaneous axis, whilst for an arbitrary body $\vec{L}$ and $\vec{\omega}$ don't need to be aligned so there are at least two axes in the problem. – jacob1729 May 22 '19 at 22:26