Simplifying entropy for a harmonic oscillator in the limit of large N

[Dixanadu]

Homework Statement

Hey guys,

So I have this equation for the entropy of a classical harmonic oscillator:

$\frac{S}{k}=N[\frac{Tf'(T)}{f(T)}-\log z]-\log (1-zf(T))$

where $z=e^{\frac{\mu}{kT}}$ is the fugacity, and $f(T)=\frac{kT}{\hbar \omega}$.

I have to show that, "in the limit of large N, this entropy becomes the following":

$\frac{S}{k}=N[1+\log(\frac{kT}{\hbar \omega})]=N[1+\log f(T)]$

Homework Equations

None that I know of

The Attempt at a Solution

So all I've done is plugged in the expression for f(T) and f'(T) into the entropy, to get this:

$\frac{S}{k}=N(1-\log z)-\log (1-z\frac{kT}{\hbar \omega})$

But i don't know what to do when N becomes large...

 

[Goddar]

Hi again...
So first, observe that from the summation form of Z (in your previous post):
N = z$\frac{\partial}{\partial z}$logZ
Use that with the explicit form of Z you found there, take the limit N>>1 and see what happens...
Also, in this limit what should you do with the terms that are not multiplied by N?

 

[Dixanadu]

Hey!

Yea you get $N=\frac{zf(T)}{1-zf(T)}$. If this is to become much greater than 1, then $zf(T) \rightarrow 1$. And anything that's not multiplied with N can take a hike! but i tried this, i don't get very far...but just to confirm, we only vary T, right? since we held $\mu , V$ constant when we found the entropy...?

 

[Goddar]

You have all you need to express the solution. You found that:
S/k = N(1 – logz) – log(1 –zf)
But then in the limit N >>1, you also found that the second term can be discarded and for the first:
zf = 1 (or in other terms z = 1/f)
Just put it together!

 

[Dixanadu]

Okay let me be a bit more specific with what I did.

Start with

$\frac{S}{k}= N[\frac{Tf'(T)}{f(T)}-\log z]-\log (1-zf(T))$

Using f(T) = kT/hbar ω, we get

$\frac{S}{k}=N[1-\log z]-\log (1-z\frac{kT}{\hbar \omega})$

Clearly $\log z = \frac{\mu}{kT}$, but the problem is the expression $-\log (1-zf(T))$...I tried to manipulate it as follows:

$-\log (1-zf(T))=\log (1-zf(T))^{-1}$, then using a first order taylor expansion:

$(1-zf(T))^{-1} \approx 1+zf(T)$

Then the only way you can get a log(zf(T)) term is if you ignore the 1...so finally i get this

$N[1-\log z]+\log(1+zf(T))$...but I am stuck!

 

[Dixanadu]

okay i'll try that...sorry i guess i was typing while you were typing!

 

[Dixanadu]

Yep i got it...thanks a bunch again!