Rubber Band Boltzmann Statistics

1. Feb 19, 2016

samjohnny

1. The problem statement, all variables and given/known data

2. Relevant equations

$$Z(1) = \sum_{i=1}^{} e^{\frac{E_i}{K_bT}}$$ where $E_i$ is each of the possible energy states available to a single link (in this case the right and the left states).

$$P=\frac{\sum_{i=1}^{} e^{\frac{E_i}{K_bT}}}{Z}$$

3. The attempt at a solution

Hi all,

For part a) I obtained $Z(1) = 2cosh(\frac{lF}{K_bT})$ = the partition function for a single link.

For b) the probability of a single link to point to the right is: $P=\frac{exp[\frac{lF}{K_bT}]}{2cosh(\frac{lF}{K_bT})}$.

And for part c), the total partition function would be $Z=[Z(1)]^N$, where $Z(1)$ is as given in the answer for part a).

Is this all correct thus far?

For part d) however I'm unsure on how to proceed. Any ideas?

2. Feb 19, 2016

TSny

Looks good so far. For (d), try to come up with an expression for $\left< N_R \right>$ based on your result for (b).

3. Feb 19, 2016

samjohnny

Ah right, since we have the probability of a link pointing to the right, and we can yield the same for a link pointing to the left, we can simply compute the expectation/average value as $\left< N_R \right> = NP_R$ where $P_R$ is the result of part b). Doing the same for the $\left< N_L \right>$, gives $L=Nltanh(\frac{lF}{K_bT})$ as required.

For part e), am I right in saying that as the temperature is increased (i.e. as the argument of tanh gets smaller), the total L decreases. And if we define the average energy as $\left< E \right> = -LF$, then the average energy will also decrease as T increases. And since L decreases, there are more possible microstates of right/left pointing links, and so the entropy increases. Is this correct?

4. Feb 19, 2016

TSny

Yes. Looks good.
Yes, heating the stretched rubber band tends to make the band contract.
$\left< E \right> = -LF$ is a deduction rather than a definition. Are you sure the average energy decreases as T increases?
Yes.

5. Feb 20, 2016

samjohnny

Ah right, where we use the fact that $\left< E \right> = \left< N_R \right>E_R +\left< N_L \right>E_L$. So, we've determined that L decreases as the stretched band is heated. So then $LF$ must also get smaller. However, the fact that there is a minus sign ($\left< E \right> = -LF$) indicate that the Energy is in fact increasing (i.e. becoming more positive). Is that it?

6. Feb 20, 2016

TSny

Yes, that's right. (Edit: Perhaps better to say that <E> becomes less negative.)