• Support PF! Buy your school textbooks, materials and every day products Here!

Rubber Band Boltzmann Statistics

  • Thread starter samjohnny
  • Start date
  • #1
84
1

Homework Statement


Boltzmann Stats.JPG


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?
 

Answers and Replies

  • #2
TSny
Homework Helper
Gold Member
12,240
2,757
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
84
1
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?
 
  • #4
TSny
Homework Helper
Gold Member
12,240
2,757
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.
 
  • #5
84
1
##\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?
 
  • #6
TSny
Homework Helper
Gold Member
12,240
2,757
Yes, that's right. (Edit: Perhaps better to say that <E> becomes less negative.)
 

Related Threads for: Rubber Band Boltzmann Statistics

  • Last Post
Replies
4
Views
4K
  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
1
Views
11K
  • Last Post
Replies
6
Views
4K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
3
Views
3K
Replies
0
Views
2K
  • Last Post
Replies
3
Views
3K
Top