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One-dimensional polymer (Statistical Physics)
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[QUOTE="Christopher Grayce, post: 6150429, member: 658133"] Gosh I'm sorry, I didn't read the problem carefully. I thought you were being asked to derive the tension. Now that I've read it more carefully, here's what I think is going on: The tension is the one-dimensional analogue of the pressure for an ordinary system. So you need to *replace* the pV term with an equivalent involving the tension (analogous to the pressure) and the length of the chain (analogous to the volume). Remember all these terms in the exponent are just the various things that can contribute to the energy. Normally you have the internal energy, which comes from the system itself, its internal degrees of freedom, and then you have "external" terms that come from the interaction of the system with the environment, and these usually have the form of a work term, i.e. some kind of "force" time some kind of "displacement." The canonical terms for the usual systems studied (e.g. volumes of gases) are pV (p = force, V = displacement) and uN (chemical potential = "force", mole number equals "displacement"), but if you have more complicated systems you can have more work-like terms, e.g. in an electric field with some dipole moment mu you have something like mu*E, and so forth. There's a discussion in H. B. Callen "Thermodynamics and an Introduction to Thermostatics", a classic graduate text, on this point that is clear and illuminating. In this case I think the work-like term you need is the tension times the elongation of the string, the elastic energy, and that should be like 1/2 t*(L-L0) if memory serves, with t = tension and L = length of polymer and L0 = unstretched (t=0) length of polymer. So for your case that works out to 1/2*t*n*(l1-l2). (Notice I'm assuming E1 > E2, because otherwise the tension comes out negative, we have to assume it takes work to stretch the polymer, from which it follows E1 > E2 -- that really should have been stated in the problem, but oh well.) So now your Hamiltonian looks like: E = nE1 + (N-n)E2 + 1/2 (l1-l2)*t*n which you can write entirely as a function of n plus the parameters E1,E2,l1,l2 plus the new intensive thermodynamic variable t, which is taking the place of p in the usual exponent. Your partition function Z then becomes a function of (n,t and T) plus the parameters. I think your deduction of the degeneracy factor (the number of combinations) is spot on. So then it just remains to do the sum (or turn it into an integral) to get Z, and then do two more to get <E> and <L>. You can write the E and L that goes in both sums in terms of n plus the parameters, so if you can do the Z sum at all, hopefully you can do those sums, too. Sometimes you can do tricks like write the sum/integral that gives you <E> as some kind of derivative with respect to a parameter of Z, and that *might* give you a simple first-order differential equation which is easier to solve than the sum/integral itself. I'd have to fiddle with the math to figure out how to approach it. A good check of any solution method would be to write some quick code on a computer and calculate Z, <E>, and <L> numerically. Sorry I didn't pay closer attention the first time, hope that's useful. [/QUOTE]
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One-dimensional polymer (Statistical Physics)
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