Free eneryg of a harmonc oscillator

[stunner5000pt]

tau = Boltzmann constant times absolute temperature
A one dimensional harmonic oscillator has an infinite series of equally spaced energy states, with $\epsilon_{s} = s \hbar \omega$ where s is a positive integer or zero and omega is the calssical frequency of the oscillator.

a) Find the free energy of the system

b) Find the entropy an Heat capacity

well ok the parititon function here is

$$Z = \sum_{s=0}^{\infty} \exp(\epsilon_{s}/ \tau) = 1 + e^{\frac{\hbar \omega}{\tau}} + e^{\frac{2\hbar \omega}{\tau}} + e^{3\frac{\hbar \omega}{\tau}} + ... = \frac{e^{\frac{\hbar \omega}{\tau}}}{e^{\frac{\hbar \omega}{\tau}} -1}$$

so to find the free energy i simply have to find the natural log of Z and multiply that by -tau, which becomes
$$-\tau \left(\frac{\hbar \omega}{\tau} - \log (e^{\frac{\hbar \omega}{\tau}} -1 )\right)$$

can this simplify furhter though?

b) to find the netropy we simply find the partial derivative of the free energy wrt tau while holding V contsnat and we get the negative entrpy

for the specific heat capacaity i am not usre...
isnt $C_{V} = \frac{3}{2}$ since N = 1 as there is only one one dimension??

Please help

thank you for the help!

 

[Jhero]

Hello,

I can help you with your questions. Let's start with finding the free energy of the system. You are correct in your calculation, and the expression for the free energy can be simplified further by using the identity ln(x-1) = -ln(1-x). This will give you:

F = -\tau \left(\frac{\hbar \omega}{\tau} + \log (1-e^{-\frac{\hbar \omega}{\tau}} )\right)

Moving on to finding the entropy, you are correct that it can be found by taking the partial derivative of the free energy with respect to temperature (tau). The resulting expression will be:

S = \frac{\hbar \omega}{\tau} + \frac{\hbar \omega}{e^{\frac{\hbar \omega}{\tau}} -1}

Lastly, for the heat capacity, you are correct that it can be found by taking the partial derivative of the energy (which is the same as the free energy in this case) with respect to temperature. However, the expression for the heat capacity will depend on the dimensionality of the system. In this case, for a one-dimensional system, the heat capacity will be:

C_V = \frac{1}{2} k_B \left(\frac{\hbar \omega}{k_B T}\right)^2 \frac{e^{\frac{\hbar \omega}{k_B T}}}{(e^{\frac{\hbar \omega}{k_B T}} - 1)^2}

Where k_B is the Boltzmann constant.

I hope this helps. Let me know if you have any further questions.