1-dimensional polymer in gravitational field

[WannabeNewton]

Homework Statement

A rubber band at temperature $T$ is fastened at one end to a peg and supports from its other end a weight $W$. Assume as a simple microscopic model of the rubber band that it consists of a linked polymer chain of $N$ segments joined end to end; each segment has length $a$ and can be oriented either parallel or antiparallel to the vertical direction. Find an expression for the resultant mean length $\bar{l}$ of the rubber band as a function of $W$. (Neglect the kineti energies or weights of the segments themselves, or any interaction between the segments.)

The Attempt at a Solution

Let the positive vertical direction be pointing upwards and let the zero of gravitational potential energy be set at the plane of the peg from which the rubber band hangs. The problem is quite easy at face value but I keep getting the wrong answer. Presumably when the problem says that the polymer chains "can be oriented either parallel or antiparallel to the vertical direction" they mean with the linked ends of the polymer chains kept fixed, otherwise the problem would be trivial. For any given microstate of this system, let $n$ be the number of polymer chains oriented antiparallel to the vertical direction so that the length of the rubber band for this given microstate is $l_n = (2n - N)a$; the total energy of the system for this microstate is then $E_n = -Wl_n$. The canonical partition function will thus be $Z = \sum_{n = 0}^N e^{\beta W l_n}$ and the average length will be $\bar{l} = \sum_{n = 0}^N l_n e^{\beta W l_n}/Z = \sum_{n = 0}^N \frac{1}{\beta}\partial_{W}e^{\beta W l_n}/Z = \frac{1}{\beta}\partial_W \ln Z$.

All that's left then is a calculation of $Z = e^{-\beta WNa}\sum_{n = 0}^{N}e^{2\beta Wna}$ which is a geometric series
hence $Z = e^{-\beta WNa}(\frac{e^{2\beta W Na}e^{2\beta Wa} - 1}{e^{2\beta W a}-1} )= \frac{e^{\beta W a(1+N)} - e^{-\beta Wa(1 + N)}}{e^{\beta Wa} - e^{-\beta Wa}} = \frac{\sinh \beta Wa(1 + N)}{\sinh \beta W a}$
therefore $\bar{l} =\frac{1}{\beta}\partial_W \ln Z= (1 + N)a\coth \beta Wa(1 + N) - a\coth \beta Wa$.

Incidentally this is also the partition function for a single atom in an external uniform magnetic field $B$ if we replace $N$ with spin $J$ and $Wa$ with $\mu B$ where $\mu$ includes the gyromagnetic factor.

However, according to the book, the answer is supposed to be $\bar{l} = Na \tanh \beta Wa$ which makes much more sense than the answer I got because my answer doesn't have the correct behavior in the appropriate limiting cases e.g. it is divergent when $W\rightarrow 0$ even though it should give $\bar{l}\rightarrow 0$ as $W\rightarrow 0$, which the book's answer does, since without any weight present the most probable microstate is the one for which there are equal numbers of polymer chains parallel and antiparallel to the vertical direction and as we know the average will agree with the most probable microstate.

However I cannot for the life of me figure out where the mistake in my solution is. Could anyone help me in finding my error? Thank you in advance.

 

[Oxvillian]

Hi WannabeNewton!

If I understand you correctly, you have a problem counting the microstates here. For each $n$, there are zillions of ways of arranging all the bits in the chain to add up to the corresponding length. Looks to me like you're only counting one in your partition function.

Incidentally this is also the partition function for a single atom in an external uniform magnetic field $B$ if we replace $N$ with spin $J$ and $Wa$ with $\mu B$ where $\mu$ includes the gyromagnetic factor.

The way you're counting the states this is true - but I think the correct analogy is that of a paramagnet in an external field, which we can do by finding the single-particle partition function $Z_1$ first and then saying that $Z = Z_1^N$.

 

[WannabeNewton]

Ah yes that was definitely a silly mistake on my part. I was calculating things as if I were in the microcanonical ensemble wherein one does not count degeneracies and sums over energy levels instead of possibly degenerate energy states but used a canonical partition function wherein degeneracies are counted and one sums over possibly degenerate energy states instead of energy levels. I've fixed it now and got the right answer, thank you very much Oxvillian!

And yes you're right it was akin to the problem of calculating the partition function of a simple two-level system of spins in an external magnetic field.

 

[WannabeNewton]

Hi Oxvillian! I was wondering if you could answer a quick question I had on a problem which is quite strongly related to the one above. I've attached the problem.

Now the problem itself is very straightforward once one makes sense of the initial setup but it's the initial setup that I have issues with. Specifically, if we forget that the word "tension" is ever mentioned in the problem and imagine $X_L$ to be an externally applied force pulling the chain molecule at both ends along the horizontal axis for a given configuration (denoted $L$) then one can just use the Gibbs ensemble for the external force $X_L$ to get the average $\langle X \rangle = \frac{1}{\beta}\partial_L \ln \mathcal{Z}$, where $\mathcal{Z}$ is the Gibbs partition function with, in this case, zero internal (potential) energy since the chains are free to rotate about the joints. This gives the right answer.

But the problem states that $\langle X \rangle$ is the average tension between the strings. This doesn't make sense to me. First of all tension isn't an externally applied force (whereas for the Gibbs ensemble we need an externally applied force) at least not in the usual sense of the term. That being said, we know the tension in a massless string will match the externally applied force that causes it so one could argue along those lines but the tension will be along the string and not along the horizontal axis depicted in the diagram so why would the above procedure give the tension in the chain molecule? It's basically the projection of the tension along the horizontal axis, is it not? And this projection is not anything physically meaningful as far as the usual interpretation of tension goes.

Thanks in advance.

 

[Oxvillian]

Hi WannabeNewton! Sorry for the slowness in noticing this post.

Not sure if I can be much help here. Are you worried about the tension not being an intensive property? I would think that it is in much the same way that the pressure in a gas is - we can talk about the pressure at a point in the gas as well as the pressure on the walls of the container. Perhaps dividing up the chain into lots of little subchains and thinking about the tension in each would help.

On the other hand I don't think you want to make your subchains too small because then the concept of tension kind of disappears. Tension in this sense is an emergent property like temperature - you need a decent number of links/molecules/whatever before the concept even makes sense. Talking about the temperature of a single atom, for example, is a bit silly.