Assuming that the Einstein frequency $\omega$ for some crystal depends on the molar volume:

$\omega=\omega_o - A\log(\frac{v}{v_o})$

Is there a particular method to follow through in order to obtain the entropy $S$ of this system in the microcanonical formalism? I have tried everything I know but nothing seems to be working.

  • $\begingroup$ The formula you've been given for $\omega_0$ seems over-elaborate. Since it requires huge pressures to decrease the volume of a solid by even 1%, it seems pretty safe to assume that $\frac {V}{V_0}$ differs from 1 by less than about 1%, so $\text{ln} \frac{V}{V_0} =\frac{V-V_0}{V_0}$ to a very good approximation, Thus we might as well write$$\omega = \omega_0 - B(V-V_0)$$ in which B is a constant. $\endgroup$ – Philip Wood Mar 31 '18 at 18:41

One of the difficulties with answering Statistical Mechanics questions is knowing which approach the questioner is familiar with. For the micro canonical approach we usually derive the entropy for a crystal of N atoms with a total of q quanta ($\hbar \omega$) of energy over and above the zero point energy. The number of ways of distributing the $q$ quanta among 3$N$ oscillators is $$\Omega=\frac {(N-1+q)!}{(N-1)!\ q!}$$Dropping the "1"s, taking ln of both sides, using Stirling's formula (crude version) and tidying gives $$\text{ln}\ \Omega=[(3N+q)\ \text{ln}(3N+q)-3N\ \text{ln} 3N-q\ \text{ln} q].$$ So $$S=k\ \text{ln} \Omega=k[(3N+q)\ \text{ln}(3N+q)-3N\ \text{ln} 3N-q\ \text{ln} q].$$ But q is related to the internal energy of the crystal by$$U=\hbar \omega \left(\frac{3}{2} N+q\right),$$and you have been given an equation for $\omega$. If you want S in terms of temperature, don't forget that $$T=\left(\frac{\partial U}{\partial S}\right)_V.$$ This is the micro canonical approach. It's a bit of a pig to use in this case, and you'll note that I've left you to do the algebra needed to express S in terms of the variables you want.

The algebra is easier if you're content to restrict yourself to the high temperature case, $q>>3N$.

Note You may be concerned that we don't seem to have used an actual ensemble. The reason for apparently not doing so is that it simply takes us in a circle, as we'll now see…

For an ensemble of $N_{ens}$ identical solids each with the same energy, and therefore the same value of $N$ and $q$, the number of ways of achieving this is simply $Ω^{N_{ens}}$. So the entropy of the ensemble is $S_{ens}=k\ \text{ln} Ω^{N_{ens}}=k\ N_{ens}\ \text{ln} \Omega\ $. So the entropy per system is $\frac{1}{N_{ens}}\ S_{ens} = k\ \text{ln}\ \Omega$.

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  • $\begingroup$ @Jepsilon I've expanded my original answer. Does it answer your question? $\endgroup$ – Philip Wood Mar 28 '18 at 7:45

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