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Polymer Chain in Statistical Mechanics
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[QUOTE="TSny, post: 6016961, member: 229090"] 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##. 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##? [/QUOTE]
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