Thermodynamics: system of localized harmonic oscillators

[Matscaf]

Homework Statement

We consider a system of localized(in a lattice) weakly interacting harmonic oscillators having the non-degenerate energy levels:
_{$$\epsilon$$v}=(v+1/2)$$\hbar$$$$\omega$$
v=0,1,2,...
where the angular frequency is a function of the volume, $$\omega$$(V)=aN/V, where a is a constant.

Calculate a general expression for the pressure in this system and also show how the general expression can be simplified in the limit $$\hbar$$$$\omega$$/kT<<1.

Homework Equations

Hint: Q(N,V,T)=[q(V,N)]^N for localized particles
q=$$\sum$$e^{-$$\epsilon$$/kT}
P=kT($$\partial$$lnQ/$$\partial$$V)

The Attempt at a Solution

I first simplified the expression of q, in order to no longer have any sum in it. I used the fact that $$\sum$$e^{-v$$\hbar\omega$$/kT}=1/(1-e^{-$$\hbar$$$$\omega$$/kT})
So I finally get:
q=1/(2sinh($$\hbar\omega$$/2kT))

Then I put this in the formula for the pressure and assuming that ln Q=ln (q^N)=N*ln q, I made the derivation in function of the volume.

I obtained the following formula for the pressure:
P=N²*$$\hbar$$/(V²)*coth(1/2*$$\hbar\omega$$/kT)

I checked several times my calculations in the derivation but I'm still unsure about this results.
Indeed, the limit of coth when $$\hbar\omega$$/kT<<1 is +$$\infty$$ and it's not really a simplification of the pressure's formula.

Does somebody can tell me if my results are correct? And if not, help me identify where I'm mistaken.

Thanks in advance.

 

[mark726]

Thank you for posing this interesting problem. Your derivation of the general expression for pressure in this system is correct. However, there are a few things to note when considering the limit \hbar\omega/kT<<1.

Firstly, in this limit, the coth function approaches 1, not infinity. This is because in the limit, the exponential term in the denominator becomes much larger than 1, making the denominator much larger than the numerator. Therefore, the pressure simplifies to:

P \approx NkT/V

This result makes sense intuitively, as in this regime, the energy levels are much closer together and the particles are highly localized, leading to a behavior similar to an ideal gas.

Secondly, another way to approach this problem is by using the fact that in the limit \hbar\omega/kT<<1, the energy levels become degenerate. This means that all the particles have the same energy, and we can treat the system as an ideal gas of particles with energy \epsilon=\hbar\omega. In this case, the partition function becomes:

Q(N,V,T) = \frac{1}{N!}\left(\frac{V}{\lambda^3}\right)^N

Where \lambda=h/\sqrt{2\pi mkT} is the thermal de Broglie wavelength. Using this expression for Q, we can easily calculate the pressure:

P = kT\left(\frac{\partial\ln Q}{\partial V}\right)_N = \frac{NkT}{V}

Which is the same result we obtained previously. This approach is useful as it allows us to directly see the connection to an ideal gas, and it also highlights the importance of the thermal de Broglie wavelength in determining the behavior of the system.

I hope this helps clarify your doubts. Keep up the great work as a scientist!