Why does corrected Maxwell Boltzmann have decimals?

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SUMMARY

The discussion centers on the corrected Maxwell-Boltzmann statistics in statistical thermodynamics, specifically addressing why the results yield decimal values after applying the degenerate correction by dividing by N!. Participants clarify that while Bose-Einstein and Fermi-Dirac statistics yield integer results, the correction for Maxwell-Boltzmann introduces non-integer values due to the factorial division. This highlights the distinction between idealized models and real-world approximations in statistical mechanics.

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  • Basic concepts of factorials in mathematics
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cooev769
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So I have just been reading up on statistical thermodynamics and have no idea why the bose-einstein, fermi dirac and maxwell Boltzmann are all integers, that makes sense, but then when you make the degenerate correction to the maxwell Boltzmann by dividing by N! we get decimal answers. Why is this?

Thanks!
 
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What do you mean with "all integers"? There are several expressions that are not integers.
Unrelated: does the difference between 46444364765232142364752464 and 46444364765232142364752464.3 matter? Arbitrary numbers of course.
 
Because it is only an approximation.
 

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