# Thermondynamics: Ideal Rubber Band

1. Dec 14, 2005

### alfredbester

Set-up: Rubber band modelled as an Ideal chain (polymer) with Nr (links) monomers pointing right, and Nl monomers pointing left each of length l.
Find expressions for the length of the band, its statistical weight (iI think that means the multiplicity?) and its entropy, in terms of Nr and Nl.
The length L is just, L = (Nr + Nl)l
Statistical Weight Ω(S(R))
Entropy is given by S(R) = k.ln (Ω(S(R)))
where R is a vector representing the net displacement of the ends.
I'm not sure how to get the statistical weight. I know the general formula is
Ω(N, a) = (a + N -1) / a!(N-1)!, is a = l in my case?

2. Dec 14, 2005

### StatusX

You want to figure out how many different values of Nr and Nl give each R. For example, if Nr=N and Nl=0 (ie, all the links are pointing right), then the net displacement is Nl. But this is the only way to get a displacement of Nl, so it has a degeneracy (multiplicity) of 1, and so an entropy of 0. The formula you mentioned at the end is close to the one you'll need. Look up the binomial distribution.

3. Dec 14, 2005

### alfredbester

Had a look at it, and this is the best I could come up with.

Ω(Nr - Nl, N) = N! / (Nr!.Nl!),
using stirling's approximation I got this to be

Ω(Nr - Nl, N) = 2^N. exp(-(Nr - Nl)^2 / 2N)

4. Dec 14, 2005

### alfredbester

Got that wrong, is it?

Ω(N, Nr) + Ω(N, Nl) = 2N! / {(0.5N + Nr)!(0.5N + Nl)!}

5. Dec 14, 2005

### StatusX

Your first equation looks right. Stirling's approximation, as I learned it, is ln(n!) = n ln(n) - n, which would mean n! = (n/e)^n (those should both be approximately equals signs). This can be used to get an estimate for the number of states and the entropy. Your second equation may be correct, but I'm not familiar with whichever form of stirlings approximation you used (there are a few of them). The equation in your most recent post doesn't really make sense to me.

Last edited: Dec 14, 2005
6. May 8, 2007

Greetings! I know this is an old thread but...

I would like to know how to get from

Ω(Nr - Nl, N) = N! / (Nr!.Nl!)

to

Ω(Nr - Nl, N) = 2^N. exp(-(Nr - Nl)^2 / 2N)

using Sterling's approximation or otherwise. I have been running into this wall:

S=k ln Ω(Nr - Nl, N) => S = k (N ln (N/(N-Nr)) + R ln ((N-Nr)/R))
after Sterling's approximation and subbing Nl=N-Nr.

This yields a (partials) dS/dNr = k ln ((N-Nr)/Nr).

I understand that f = -T (dNr/dL) (dS/dNr). Where the length, L = 2Nr - N.

So I expect dS/dNr to yield L under proper conditions. I don't see that happening, though. Where'd I go wrong? (Aside from choosing Physics...)

7. May 8, 2007

(Emoticons?!? How 'bout greek letters instead?)

Nevermind. I got it. It requires a more complete version of Sterling's Approximation:

N!= (N^N)*(e^-N)*sqrt(2*pi*N) <--- where this last square root term contains the L dependence (eventually).

Later.