I understand that the derivation of the Maxwell-Boltzmann statistics is based on particles only having translational kinetic energy and that it is derived by maximizing the number of ways $\Omega_T$ a microstate can be achieved. $$\Omega_T = \frac{N!}{n_{T1}!n_{T2}!n_{T3}!...n_{Tn}!}$$ I also understand that rotational kinetic energy would only be present in the case of very high temperatures (especially for one-atomic particles). Thinking of that scenario, I would deduce that the number of ways a microstate can be achieved would be by multiplying $\Omega_T$ by the number of ways that a microstate can have in the rotational energy distribution $\Omega_R$. Thus: $$\Omega=\Omega_T \cdot \Omega_R = \frac{N!}{n_{T1}!n_{T2}!n_{T3}!...n_{Tn}!}\cdot \frac{N!}{n_{R1}!n_{R2}!n_{R3}!...n_{Rn}!}$$ However, when maximizing this by taking the derivative (after rewriting the formula in terms of basenumber $e$ and applying Stirling's Approximation), I'd have to deal with two variables, $n_T$ and $n_R$.

If this formula is correct, is it actually possible to derive the MB Statistics formula using this approach?


Yes! Rotational kinetic energy can absolutely be accounted-for with Maxwell-Boltzmann statistics.

For monoatomic gases, there are no rotational degrees of freedom, but for diatomic gases ($H_2$, $O_2$) and more complicated molecules, rotational and molecular vibrational modes contribute to the heat capacity. These modes are frozen out at low temperatures, but 'turn on' as you raise the temperature. See Hyperphysics for molecular hydrogen gas.

I am not sure that it is easy to do this calculation in your current approach, however. I think you will want to use the Canonical ensemble partition function

For low temperatures (just translational energy):

$Z = \frac{1}{N!} \left( \frac{1}{h^3} \int \int \exp \left[-\beta \frac{p^2}{2m} \right] d^3p d^3x \right)^N$

When you add rotational kinetic energy in, you just add those to the exponent:

$Z = \frac{1}{N!} \left( \frac{1}{h^3} \int \int \exp \left[ -\beta \left( \frac{\vec p^2}{2m} + \frac{I\vec \omega^2}{2} \right) \right] d^3p d^3x d^3\omega \right)^N$

Disclaimer: I have not double checked these equations.

  • $\begingroup$ Thanks for your explanation. You made me realise that my question is wrong, I wanted to specifically know if my approach is possible. I will edit it now. $\endgroup$ – Phy Apr 3 '20 at 15:44
  • $\begingroup$ @JohnnyGui: It might be unfair to change the question after someone else has already put so much effort into answer it. This user (a relatively new one), has correctly answered the question and has spent a lot of time helping you, would it not be possible to ask a new question on whether or not the method works? $\endgroup$ – user1271772 Apr 4 '20 at 0:35
  • $\begingroup$ Your specific approach is more challenging. I'm not sure, but I think you have to go a bit further back in the derivation to see how to add the rotational degrees of freedom (two per molecule for a diatomic gas). Is there a reason why you are specifically interested in using the microcanonical ensemble? $\endgroup$ – taciteloquence Apr 4 '20 at 3:35
  • $\begingroup$ @taciteloquence My apologies for changing the question, I expected that it sufficiently contained the request to use that method for the derivation. I'm interested because I want to see whether the microcanonical derivation is consistent with the derivation through other means and also how it would be done that specific way. $\endgroup$ – Phy Apr 4 '20 at 16:36

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