One-Dimensional Percolation Theory

In the tried and true methods physicists always employ when trying to solve a problem, we will first simplify the problem. The most simple percolation problem is a one-dimensional problem. Now we just have a chain of sites. Each site has two nearest neighbors: one to the left and one two the right. Each site is still occupied with a probability p.

We can solve for the critical value of p for which infinite clusters appear. If p=1, then every site is occupied, so clearly there is an infinite cluster. If p is less than 1, and the chain is quite large, there will be at least one unoccupied site. In one-dimension, if there is just one unoccupied site, it breaks the cluster, and there is no long a single cluster that spans the entire chain. Therefore, the critical probability is equal to one.

This is a somewhat trivial result. Nevertheless, let uss continue with the analysis as we can still extract properties that appear in higher dimensions.

Let us start by asking a question: If the probability of one site being occupied is p, what is probability of two sites next to each other being occupied? Since the probability of each site being occupied is independent, it is just p2. This generalizes for the probability that s specific sites are occupied to ps. The probability that a sites neighbor is empty is (1-p). Combining the above results, the probability a site is the left edge of a cluster of length s is ps(1-p)2. Figure 3 shows a cluster of five sites

Figure 3

If we have a long chain of length L, how many clusters of length 5 will there be? Well, if we assume the chain is very long, we can ignore the effects of the end of chain. Then the probability of each site beng the left edge of a cluster of length s is p5(1-p)2 and can be at any one of the L sites. Thus the total number of clusters of size 5 is Lp5(1-p)2. It is normally more practical to talk about the number of clusters per lattice site, which is the above expression divided by L, or just p5(1-p)2. For clusters of s sites, we define ns, normalized cluster, number as the number of clusters with size s per lattice site. That is

ns= p5(1-p)2 (Eqn. 1)

The probability that a site is a part of a cluster of size s is actually higher than the normalized cluster number, because it doesn’t have to be at the left end. It can be at any of the s sites. There fore the probability a singe site is in a cluster of site n is nss. Since each occupied site has to belong to a cluster of some size, if we add up the probabilities of all the different sizes, it must be equal to the probability that the site is occupied. That is

Σns = p (Eqn. 2)

Where Σ represents summing over all possible values of s. It is a worthwhile excersie to perform the above calculation to verify I am telling the truth!