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!