Polymer Chain in Statistical Mechanics

[Silviu]

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!

 

[TSny]

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 $\langle r_z \rangle$?