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While I was studying statistical mechanics, I saw this in the book that I'm following: We can divide the partition function into a product,

$$ \zeta = \zeta_\text{trans}\zeta_\text{int} $$

where $\zeta_\text{trans}$ and $\zeta_\text{int}$ correspond to translational and internal energy (rotational, vibrational, etc). The book said that, for monoatomic gases, we can just set $\zeta_\text{int} = 1$. It didn't explain why. The explicit expression for $\zeta_\text{int}$ is : $$ \zeta_\text{int} = \frac{1}{h^{l}}\int e^{-\beta H} $$ , $$l = n-3$$

where $n$ is the degrees of freedom. Why we can do this for the monoatomic gas?

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  • $\begingroup$ Where would internal energy come from for monatomic gases? $\endgroup$ Apr 6, 2020 at 23:27
  • $\begingroup$ From nowhere, my point is : what happens to the $$\zeta_{int}$$ integration? $\endgroup$ Apr 6, 2020 at 23:30
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    $\begingroup$ What are you integrating over in your expression for $\zeta_\mathrm{int}$? $\endgroup$ Apr 11, 2020 at 13:12
  • $\begingroup$ Hi, came upon the same question myself. Is the book you were using Schroeder? If so, can you point out where it is said that Z_int =1 ? $\endgroup$
    – Lopey Tall
    Jul 13, 2020 at 19:20

1 Answer 1

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In gas interatomic interactions are negligible. Therefore $H$ is always zero, and $e^{-\beta H} = 1$ for any temperature and any point. As the gas is monoatomic, all degrees of freedom are translational. Apparently the book authors choose integration range of each degree of freedom to be equal to Plank's constant $h$. Hence the integral equals $h^l$ and the interaction partition function $\zeta_{int} = 1$.

Only the total partition function is important. It's decomposition on various contributions is the matter of convenience.

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    $\begingroup$ You mean $e^{-\beta H} = 1$, right? $\endgroup$ Apr 11, 2020 at 12:17
  • $\begingroup$ Yes, thank you! $\endgroup$
    – user36313
    Apr 11, 2020 at 12:48

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