Chapter 19: Entanglement Rates
There are so many ways to tangle. There have been efforts to classify tangling patterns in various ways — after all this is what we are doing when we name knots and make knot boards, or when we make knot tables by crossing number, or tables of torus knots and links, and so on.
Here we ask a basic question about the entanglement pattern: what is the rate of entanglement? That is, if you keep adding string to the entanglement pattern, how does the entanglement grow as the length of the string grows?
Any answer to this question will depend on how we measure entanglement. In previous chapters we have discussed several measures of entanglement complexity. There is ropelength, crossing number, several different kinds of stick number, and tile number. They are all different, but they all increase as the entanglement pattern becomes more complex.
All of these measures have a similar rubric. For any particular configuration of the string, the measure assigns a number. We then look for the minimum number assigned over all the configurations of a particular knot. For example, there are an infinite number of ways to tie a trefoil with a closed chain of sticks, but all of them require at least six sticks.
Now lets imagine we are making a longer and longer chain of sticks, how much will the entanglement increase? If we lay the sticks out in a straight line, we won’t tie any sort of knot, so the entangle will not increase no matter how long. we make the chain. On the other hand, if we pack a very long chain of sticks into a small ball it seems there could be lots of opportunity for entanglement. But we wouldn’t have to tie knots, maybe, we could pack the chain into the ball very carefully to avoid knotting.
We need to be little careful about what we mean by ‘packing’ here. If we start with a long closed unknotted chain, think say of gold chain necklace, where each little loop in the chain serves as a stick and the hasp is closed, and there are no knots in the chain, then when we pack the necklace into a small box we cannot introduce knots. As Jonathan Simon once observed, young children sometimes think you can tie a knot by simply squishing a length of line into a ball, but we more mature sorts know that doesn’t work. The necklace may seem tangled, but it is not knotted. You cannot knot the chain unless you undo the hasp to create open ends to manipulate.
On the other hand imagine building a chain of sticks in a small box by successively adding sticks to the end of the chain (this is how many biopolymers form). With this process you certainly could tie knots. We are are to focus for now on this second kind of process and return to the first later.
So we imagine that we are lengthening the string by adding on to the end(s). And we ask how the entanglement increases as we do this. Of course, as we just observed, it matters how we lengthen the string. One general principle is: a string restricted to a small box is more likely to entangle than a string that has lots of room to get away from itself.
Here is a simple example. Let’s say we have a long length of rope, and in it we tie an overhand (trefoil) knot at regular intervals. Then in this case the entanglement is growing linearly with the length. Each trefoil requires 3 crossings and let’s say about a foot of one inch diameter rope. And say we leave a foot between knots, then the rate is roughly 3 crossings every 2 feet. We might write E(L) = 3/2 L .
For a second example imagine we have a collection of circular rings of uniform diameter (say diameter 1) made of arbitrarily (infinitely) thin wire. If we have 2N rings, split them into two sets, with N in each pile. Now arrange them so that each ring in set A is linked with all the rings in set B. (See the illustration). How many necessary crossings do we have? If we take any pair of rings where one is in set A and one is in set B, they are linked and so require two crossings. There are N X N such pairs, so this is 2N^2 crossings. Each pair of rings requires 2 Pi length of wire. So the entanglement rate is
E(L) =(2N^2)/(2 Pi)
In the real world there is no such thing as an infinitely thin string. So we can ask, what sort of entanglement rates can we get with actual string or rope?
The example of linear growth above was constructed with a rope. so the linear case is realized. In our linked rings example, if we assume the wire has some thickness, the ring diameters have to grow in order to go around the wires that pass through their centers. If we assume the wire has thickness 1, then we can fit N^2 rings through a ring of diameter N. (See illustration). So in this configuration we have 2N^2 rings, each of length or order N. So the total length is of order N^3. Each ring in one set is linked with N^2 rings in the other set. There are N^2 rings in each set, so there are N^2 X N^2 crossings. N^4 crossings from N^3 length. So the entanglement is growing at rate L^4/3. This is in fact the fastest possible growth rate of entanglement for a string of some thickness, in other words, any real world string.
We can prove that the 4/3 exponent is the fastest possible rate of entanglement. The proof hinges on two aspects of the average crossing number. The first is that the average crossing number bounds the crossing number. This is easy to see. If the ACN were less than the crossing number, then there would be a conformation of the knot with a view that had fewer crossings than the crossing number, but the crossing number is defined as the fewest number of crossings in any conformation.
The second is in the beautiful geometry of the integral definition of the the ACN. The ACN was defined by Freedman and He, who modified Gauss’s inspired definition of the linking number of two space curves. Gauss realized that if you took two line elements (unit segments of line) in space, and asked whether they appeared to cross from an arbitrary viewpoint, this would depend on the inverse square of the distance between them. You can get a pretty good sense of this if you hold up two pencils and vary the distance between them, and also vary your viewpoint. Gauss was probably inspired by Newton, who used the notion of a ‘viewpoint’ approach and its connection to an inverse square law to great effect.
From the Gaussian/Newtonian school, the knot in space is broken up into a chain of segments, and we think of each of these segments as atoms, and these atoms affect each other with an inverse square force, and to find the ACN we sum the forces.
The thickness of the ‘rope’ means that these atoms cannot be too close together. We could think of them as marbles, and however the marbles are packed, their centers cannot be closer than 2 if the radii of the marbles are 1.
So imagine we have a bag of marbles, so the marbles are packed in space about as tightly as possible. And imagine there is an inverse square force between the centers of the marbles. And we wish to estimate the sum of the forces. Say we choose a marble near the center of the pack. How many marbles are within 1 in distance? Within 2? Imagine there are concentric spherical shells centered at our test marble, at radii 1, 2, 3,….M. How many marbles can fit between successive shells? Its bounded above by the volume of the thickened shell between them. That volume is bounded by the surface area of the outer shell 4 Pi m^2. So, just estimating for our test marble, because there is 4 pi m^2 volume at distance m, the contribution to the sum from that shell is 4 Pi m^2/m^2, just a constant. So now we just need to know how many shells to contain all of the marbles. Since the volume of a sphere is 4/3 Pi r^3 we set this equal to M, the total number of marbles, and so r, the number of shells, is M^(1/3). So this is the sum of the forces involving this test marble. There are M marbles, so the estimate is
M X M^(1/3) =M^(4/3)
of aThese are all topological measures, so it is possible for a particular 3D position of a loop to have many crossings when viewed from any viewpoint, but still be the unknot. and so have crossing number 0, because the crossing number is the minimum number of necessary crossings, and the crossings could be removed by moving the string without breaking open the loop.