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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?

Thanks in advance.

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  • $\begingroup$ I'm pretty sure the issue is that you seem to think the energy depends on angular momentum in the same way it did for a diatomic molecule, but I can't see how to do it properly right now. $\endgroup$ – jacob1729 May 22 '19 at 22:24
  • $\begingroup$ 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. $\endgroup$ – jacob1729 May 22 '19 at 22:26
  • $\begingroup$ Okay, I think I get it. $\endgroup$ – Alpha7200 May 23 '19 at 6:03

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