# One-dimensional polymer (Statistical Physics)

• CharlieCW

## Homework Statement

Consider a polymer formed by connecting N disc-shaped molecules into a onedimensional chain. Each molecule can align either its long axis (of length ##l_1## and energy ##E_1##) or short axis (of length ##l_2## and energy ##E_2##). Suppose that the chain is subject to tension ##\tau##.

a) Calculate the number of different ways of arranging the polymer such that there are n molecules aligned by its long axis.

b) Using the Gibbs canonical ensemble in 1-dimension find the average energy ##\langle E \rangle## and the average length of the chain ##\langle L \rangle##.

## Homework Equations

Gibbs canonical partition function

$$\Xi=\sum_{V_S}\sum_{i}exp[-\beta (E_i+pV_S)]$$

Hint: You can take the sum in the partition function as

$$\sum_{V_S}\sum_{i}=\sum_{\{ n \}}g(n)$$

where ##\{ n \}## denotes the states of the chain with ##n## molecules aligned by its long axis and ##g(n)## a degeneracy factor.

## The Attempt at a Solution

a) For the first problem, let's denote for simplicity a molecule in vertical position with ##0## and in horizontal position with ##1##. Since each molecule can only adopt two configurations ##(0,1)##, then ##N## molecules can adopt a total of ##2^N## possible configuration. Now if we want the number of ##n## molecules on state ##1##, it is the same to find the total number of ways of arranging n molecules in N spaces, that is:

$$g(n)={N \choose n}$$

b) I'm really not sure how to proceed on this one and how to even include the tension ##\tau##. I began by calculating the Gibbs canonical partition function as:

$$\Xi=\sum_{V_S}\sum_{i}exp[-\beta (E_i+pV_S)]$$

Using the hint, the sum can be simplified to:

$$\Xi=\sum_{\{ n \}}g(n)exp[-\beta (E_n+pV_S)]=\sum_{\{ n \}}{N \choose n}exp[-\beta (E_n+pV_S)]$$

The energy ##E_n## for the state with ##n## molecules aligned along it's axis is ##n(\epsilon_2-\epsilon_1)=n\Delta\epsilon##. On plus, it is subject to a tension which I think should be ##\tau (l_2-l_1)=\tau\Delta l##. I'm not sure if I should include this tension in the energy, or replace ##pV_S## by ##\tau \Delta l##.

After that, I have no idea on how to find the average energy and length.

Seems to me the only term in your exponent should be the energy due to the alignment. There's no role for pressure and volume here (if that's what that extra term is for). So all you need to do is write your partition function in terms of the length of the molecule, Z(l). Then the force (tension) will be dZ/dl. You might succeed in getting this in closed form either by looking up some sums of combinations, or perhaps approximating the sum by converting it to an integral (on the assumption that N >> 1, n >> 1, e << 1).

Seems to me the only term in your exponent should be the energy due to the alignment. There's no role for pressure and volume here (if that's what that extra term is for). So all you need to do is write your partition function in terms of the length of the molecule, Z(l). Then the force (tension) will be dZ/dl. You might succeed in getting this in closed form either by looking up some sums of combinations, or perhaps approximating the sum by converting it to an integral (on the assumption that N >> 1, n >> 1, e << 1).

Indeed, the term ##pV_S## is for the pressure and volume, but since the general formula was derived for a 3D recipient I was thinking about converting it to the one-dimensional case ##pV_S\rightarrow \tau L##. However, it also makes more sense that you mention to obtain the tension as ##dZ/dl##, as in that way I could relate the tension to the length of the chain.

I also checked how to solve the sums by approximating some series so I think I should be fine once I know how to write the partition function. However, I'm still unsure on how to express the partition function, that is, how to write the exponential in terms of the length and energy of the molecule. Here's my guess (correct me if I'm wrong).

From the statement, if I have ##n## molecules with length ##l_1##, then I have ##N-n## molecules with length ##l_2##, so that means that the total energy for a given state ##\{ n \}## should be ##E_n=nE_1+(N-n)E_2=n(E_1-E_2)+NE_2##, and thus,

$$exp[-\beta E_n]=exp[-\beta n(E_1-E_2)+NE_2]$$

Now, how could I include the length of the chain ##L##?

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.

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.

Don't worry, with your explanation I better understood the meaning of the terms in the exponential and I think I see more clearly how to deal with these kind of systems. So then my idea about considering the tension ##\tau## for the linear case was correct since, as you mentioned, it is part of the external work/energy exerted on the system, while the energies ##E_1## and ##E_2## are part of the internal energy of the system (and now I see how could I extend it if I consider other terms such as the chemical potential). I also appreciate the recommendation of the book, I've been mostly using Greiner but I wanted to check other approaches, so I'll check the one you suggested.

Now that I have the energy, indeed the rest is just a matter of algebra as I need to reduce the sums. I found in another website that, given by degeneracy factor, I will have sums of the form,

$$\sum_n {N \choose n} e^{-\beta n E_n}=\sum_n {N \choose n} x^n$$

with ##x=e^{-\beta E_n}## which, using the binomial theorem, can be reduced to,

$$\sum_n {N \choose n} x^n=(1+x)^n$$

As for the mean values, since the partition function is connected with the probability of states, from what you wrote and what I remember from statistics, I think I can calculate them using,

$$<E>=\frac{\sum_n Eexp[-\beta n E_n]}{\sum_n exp[-\beta n E_n]}$$

$$<L>=\frac{\sum_n Lexp[-\beta n E_n]}{\sum_n exp[-\beta n E_n]}$$

So as you said, it's just a matter of reducing the sums or taking the limit ##N>>1## so I can transform them to integrals and solve them.

Well I managed to solve it and I got that both the average energy and length follow a Fermi-Dirac like distribution. I think I'll post the solution during the weekend in case anyone finds it useful.