Vibrational free energy in the high temperature limit


by mSSM
Tags: energy, free, limit, temperature, vibrational
mSSM
mSSM is offline
#1
May8-12, 12:11 PM
P: 24
This doesn't really qualify as a proper physics question, but I don't know where to put this... I have been thinking about this for about 2 hours now, and just don't get the correct approximation.

For the free energy of the vibrational degrees of freedom of a diatomic ideal gas I have:
[tex]
F_{\text{vib}} = NT \log{\left( 1 - \mathrm{e}^{-\hbar\omega/T} \right) }
[/tex]

Now, for the high temperature regime, [itex]\hbar\omega \ll T[/itex], I know the result looks like:
[tex]
F_{\text{vib}} = -NT\log{T} + NT \log{(\hbar\omega)} - N \frac{1}{2} \hbar\omega
[/tex]

But I have no idea how to arrive there... I thought, one could rewrite it, to see where the terms should arrive from:
[tex]
F_{\text{vib}} = NT \log{\left( \frac{\hbar\omega}{T} \right)} - N T \left( \frac{1}{2} \frac{\hbar\omega}{T} \right)
[/tex]

But also that didn't get me far... since we have [itex]\hbar\omega/T \ll 1[/itex], I thought I could give it a try and expand the argument of the logarithm in terms of [itex]\hbar\omega/T[/itex] around [itex]0[/itex], but that only gave me the first part (upon expansion to first order; the zeroth order, obviously, drops):
[tex]
F_{\text{vib}} = NT \log{\left( \frac{\hbar\omega}{T} \right)}
[/tex]

Expanding to second order gives the one-half term, but everything else then doesn't quite make sense, since I can't just separate that...:
[tex]
F_{\text{vib}} = NT \log{\left( \frac{\hbar\omega}{T} - \frac{1}{2} \left( \frac{\hbar\omega}{T} \right)^2 \right)}
[/tex]

I also thought about working with limits, but I figure that does not really make sense (first, unit wise, and second it didn't lead me anywhere sensible).


Any ideas?


EDIT: Just to add: I am aware that in the high-temperature limit I can get the above expression by replacing the sum in the partition function for the vibrational states by an integral over the momenta and coordinates. (I didn't write the parition function down explicitly, but it is precursor of the first free energy I wrote down - the argument of the logarithm in that expression can be written as a geometric series, which then gives the partition function.)
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