# Rubber Band Boltzmann Statistics

## Homework 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).

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

## 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?

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

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

TSny
Homework Helper
Gold Member
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.
Yes. Looks good.
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.
Yes, heating the stretched rubber band tends to make the band contract.
And if we define the average energy as $\left< E \right> = -LF$, then the average energy will also decrease as T increases.
$\left< E \right> = -LF$ is a deduction rather than a definition. Are you sure the average energy decreases 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?
Yes.

$\left< E \right> = -LF$ is a deduction rather than a definition. Are you sure the average energy decreases as T increases?
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?

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