How Does the Equipartition Theorem Apply to Diatomic Molecules in a Box?

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SUMMARY

The discussion centers on the application of the Equipartition Theorem to a system of N noninteracting diatomic molecules in a box, as outlined in Kerson Huang's text. The Hamiltonian for a single molecule is expressed as H=\frac{1}{2m}(\vec{p_1}^2+\vec{p_2}^2) +\frac{K}{2}(|\vec{r_1}-\vec{r_2}|^2). The internal energy derived from the partition function is E=\frac{9}{2}Nk_BT, suggesting 9 degrees of freedom, which conflicts with the expected 6 degrees of freedom for diatomic molecules. The discrepancy arises from the treatment of degrees of freedom in the Hamiltonian formulation.

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  • Understanding of the Equipartition Theorem
  • Familiarity with Hamiltonian mechanics
  • Knowledge of partition functions in statistical mechanics
  • Concept of degrees of freedom in molecular systems
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  • Study the derivation of the Equipartition Theorem in classical systems
  • Learn about Hamiltonian dynamics and its applications in statistical mechanics
  • Explore the concept of degrees of freedom in diatomic molecules
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Students and researchers in physics, particularly those focusing on statistical mechanics, thermodynamics, and molecular dynamics. This discussion is beneficial for anyone seeking to understand the nuances of the Equipartition Theorem as applied to diatomic molecules.

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[SOLVED] Equipartition Theorem

1. Kerson Huang, P 7.2:
Consider a classical system of N noninteracting diatomic molecules in a box of volume V at temperature T. The Hamiltonian for a single molecule is taken to be

H=\frac{1}{2m}(\vec{p_1}^2+\vec{p_2}^2) +\frac{K}{2}(|\vec{r_1}-\vec{r_2}|^2).

Obtain the internal energy and show that it is consistent with equipartition theorem.

2. Homework Equations and attempt
I used the formulae for partition function

Q_N = \frac{1}{N!}Q_1

where

Q_1 = \int d \omega e^{-\beta H}

Further, I moved to the center of mass frame, and wrote the Hamiltonan in this form:

H= \frac{P_{cm}^2}{4m} + \frac{p^2}{4m} +\frac{Kr^2}{2}

where now \vec{r}=\vec{r_1}-\vec{r_2} and the mass of two atoms the same, while we all know what center of mass frame is (it is not too imprtant for my question to show all the way of derivations, I guess).

After all, using E =-\frac{\partial}{\partial \beta} \ln Q_N

I found that E=\frac{9}{2}Nk_BT.

3. Question
According to equipartition theorem and using my last modified Hamiltonian looks I have 9 degrees of freedom and everything looks like being consistent. However, as far as I know, the diatomic molecule has 6 degrees of freedom (in case it is not rigid molecule). If I think, like that, then I am missing the factor (instead of 6 I have 9). Can you help me, where I am wrong?
 
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Gravitino said:
1. Kerson Huang, P 7.2:
According to equipartition theorem and using my last modified Hamiltonian looks I have 9 degrees of freedom and everything looks like being consistent. However, as far as I know, the diatomic molecule has 6 degrees of freedom (in case it is not rigid molecule). If I think, like that, then I am missing the factor (instead of 6 I have 9). Can you help me, where I am wrong?
I am guessing that the quantum consideration gave wrong result, the fact that the relative distance actually does not vary from 0 to infinity in phase space in classical universe gave probably a weird result...
 

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