Polymer Chain in Statistical Mechanics

  • Thread starter Silviu
  • Start date
  • #1
624
11

Homework Statement


A polymer chain consist of a large number N>>1 segments of length d each. The temperature of the system is T. The segments can freely rotate relative to each other. A force f is applied at the ends of the chain. Find the mean distance ##\textbf{r}## between the ends.

Homework Equations




The Attempt at a Solution


So when there was no force the result was 0. Now I am a bit confused. I wrote the partition function as: $$Z=(2 \pi \int_0^\pi d \theta sin (\theta) e^{\beta f d cos (\theta)})^N = 4 \pi \frac{sinh(\beta f d)}{\beta f d}$$. Now the probability of a given state is $$p_{state}=\frac{e^{\beta f d \sum_1^N cos(\theta_i)}}{Z}$$ and I am thinking to write $$<\textbf{r}> = \frac{(2 \pi \int_0^\pi f(\theta) d \theta sin (\theta) e^{\beta f d cos (\theta)})^N}{Z}$$. But I am not sure what this ##f(\theta)## should be. I want it to represent the vectorial distance between 2 neighboring points as a function of the angle between the direction of the force and d, but I am not sure how to write it. Should I do it separately for x, y and z and add them up or is there a way to write it directly? Also the solution I have uses: $$F=-NTlog(Z)$$ and then: $$L = -\frac{\partial F}{\partial f}$$ where L is the answer required. I am not sure I understand why. First, it looks like a scalar not a vector and why would the distance be given by that derivative? Thank you!
 

Answers and Replies

  • #2
TSny
Homework Helper
Gold Member
12,963
3,320
Now I am a bit confused. I wrote the partition function as: $$Z=(2 \pi \int_0^\pi d \theta sin (\theta) e^{\beta f d cos (\theta)})^N = 4 \pi \frac{sinh(\beta f d)}{\beta f d}$$.
OK, this looks right.

It might be helpful to note that ##Z## can be expressed alternately as $$Z = \int_0^{2\pi} d \phi_1 \int_0^\pi d \theta_1 sin \theta_1 \int_0^{2\pi} d \phi_2 \int_0^\pi d \theta_2 sin \theta_2 ... \int_0^{2\pi} d \phi_N \int_0^\pi d \theta_N sin \theta_N \, e^{\beta f d \sum_1^N cos(\theta_i)}$$
##\phi_i## is the azimuthal orientation of the ith segment. This expresses ##Z## as an integration over the microstates of the entire chain. A microstate of the chain is the specification of all of the individual ##\theta_i## and ##\phi_i##. Thus, the probability distribution function for the states of the chain is $$P(\theta_1, \phi_1, ... \theta_N, \phi_N) = \sin \theta_1 ... \sin \theta_N e^{\beta f d \sum_1^N cos(\theta_i)}$$

Regarding ##\langle \textbf{r} \rangle##, I would follow your suggestion of doing the components separately. $$\langle \textbf{r} \rangle =\langle r_x \rangle \mathbf{e_x} +\langle r_y \rangle \mathbf{e_y} + \langle r_z \rangle \mathbf{e_z}$$ Note that ##r_x = \sum_1^N \Delta x_i## where ##\Delta x_i## is the x-displacement of the ith segment. You can express ##\Delta x_i## in terms of ##\theta_i## and ##\phi_i##. You can then evaluate ##\langle r_x \rangle## using the probability distribution function ##P(\theta_1, \phi_1, ... \theta_N, \phi_N)##. Similarly for ##\langle r_y \rangle## and ##\langle r_z \rangle##.

Also the solution I have uses: $$F=-NTlog(Z)$$ and then: $$L = -\frac{\partial F}{\partial f}$$ where L is the answer required. I am not sure I understand why.
If you use the alternate expression for ##Z## as I gave above, what expression do you get for ##\frac{\partial F}{\partial f}##? Do you get something that is related to ##\langle r_x \rangle##, ##\langle r_x \rangle##, or ##\langle r_z \rangle##?
 

Related Threads on Polymer Chain in Statistical Mechanics

Replies
0
Views
4K
Replies
1
Views
803
Replies
5
Views
1K
Replies
20
Views
2K
  • Last Post
Replies
0
Views
2K
  • Last Post
Replies
7
Views
8K
  • Last Post
Replies
0
Views
2K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
0
Views
1K
  • Last Post
Replies
2
Views
2K
Top