# The entropic cost of tying knots in polymers

Imagine I take a polymer like polyethylene, of length $L$ with some number of Kuhn lengths $N$, and I tie into into a trefoil knot. What is the difference in entropy between this knotted polymer and a circular polyethylene unknot? Is there an approximate way of calculating this for knots of a given crossing number (or other topological invariant such as a Jones polynomial)?

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Why do you think it's a cost, and not a benefit? There is a way of calculating this--- figure out the statistics of prime decomposition of a random closed path. But its going to be peaked at something like a given number of prime knots per length. Will you accept an answer which is negative? –  Ron Maimon Aug 13 '11 at 2:53
@Ron Maimon, I would be very interested in any answer that helps with my intuition for when the unknot is no longer entropically favorable. For physical polymers of course, this isn't always going to be true. For example, consider any ideal knot, i.e. a knot tied using the shortest possible length of rope that has some thickness 'T'. –  TheSheepMan Aug 13 '11 at 18:41
is this a usefull link?:[Knots, bubbles, untying, and breathing: probing the topology of DNA and other biomolecules](cdsweb.cern.ch/record/832343/files/cer-002518715.pdf) –  Helder Velez Aug 14 '11 at 18:52
The unknot is only entropically favorable for short polymers, where you can calculate the entropy brute force, by simulating. For the long polymer limit, where you need a scaling analysis, the unknot would be unlikely. I was going to think about an answer in the long skinny polymer limit. –  Ron Maimon Aug 14 '11 at 19:41

The idea of knot inflation is that of A. Grosberg and it is tricky, but I'll try to explain some of it. Construction by way of scaling analysis is required to tackle this.

Suppose the volume of our polymer is given by $R$. Now fit a tube of fixed length, $L$, over the polymer; this tube has diameter, $D$, across its length that is just large enough to contain the polymer. Now the tube is inflated with a fixed $L$, then $D$ is uniformly allowed to increase until a maximum $D_{max}$ is reached. An aspect ratio, $p=L/D_{max}$, is then a weak topological invariant. Affine transformation of the inflated tube is done so that the polymer will fit in a sphere of size $R$, with $L_R$ and $D_R$ such that $L_R/D_R=p$ and $L_R D_R^2 \approx R^3$. He then uses this construction for finding the entropy.

The trick now is to assume that most of the available conformations of our polymer with a certain knot are also available to a trivial knot within our sphere of size $R$, giving $$S= S_{trivial} + \Delta S_{approx}$$

We can say that $S_{trivial}$ should be proportional to $N$ and so we can rewrite it as, $$S = sN + \Delta S_{approx}$$

with $s$ not a function of $p$ or $N$.

Now he confines an ideal knot in a cylinder (ties the polymer to both ends of the cylinder), and considers what will happen to the entropy if $p$, of the cylinder, is varied. He finds that $\Delta S_{approx}$ is proportional to $-p$. Finally, a polymer with the maximum amount of knots should give $S\approx 0$, so that means $p \approx N$ in that case, giving $$S = s(N-p)$$

Thus the entropy decreases with increasing knot complexity, $p$. This method is very crude, at best, but it does provide a qualitative approach to your query.

To see his full argument I'll have to refer you to the citation: A.Y. Grosberg “Entropy of a Knot: Simple Arguments About Difficult Problem,” In: Ideal Knots, Edited by A.Stasiak, V.Katrich, L.H.Kauffman, World Scientific, 1998, p. 129-142.

In that same article he extends his argument to calculate the probability of finding a knot as $$P(N) = N_0 e^{-N/N_0 - sp}$$