Finding the exact critical probabilities is possible for some lattices, but is often very challenging. Luckily there are other computational tools at our disposal. We may be able to find relationships between the critical probabilities of different lattices even without knowing the values.
A simple transformation is to remove edges from a lattice. Removing edges can only increase the sizes of the clusters, so if the original lattice percolates, the new lattice will as well. In particular, if \(p\) is the original critical probability, and \(q\) is the critical probability of the modified lattice, we have \(q\le p\). Consider the example below:
We have removed several edges from a triangular lattice, displaying the new clusters. Try setting the probability very low, so that almost all edges are blocked. What does the lattice with the deleted edges look like?
It has exactly the same structure as a (slanted) square lattice!
This is a definitive proof of something you might have observed before: the square lattice critical probability is smaller than the triangular lattice critical probability! While we could (and did) run computer simulation to find this result, in other cases it might not be so easy. Theoretical results, even imprecise ones like this one, are powerful because they reveal absolute limitations, and tell us about physical systems we might not even have imagined, let alone run experiments on.
Another edge-deletion transformation turns the square lattice into a hexagonal lattice.
These transformations suggest some relationships between critical probabilities. To make a square lattice, we delete a third of the edges from the triangular lattice, so that if a fraction p of the edges were open before, \(\frac{1}{3}+\frac{2}{3}p\) are open afterwards. Similarly, we delete a quarter of the edges from the square lattice, so if \(q\) were open before, \(\frac{1}{4}+\frac{3}{4} q\) are open afterwards. If the resulting lattices percolate, it suggests that the critical probability for the triangular lattice is roughly \(\frac{1}{3}+\frac{2}{3}p_S\), and similarly that the critical probability of the square lattice is roughly \(\frac{1}{4}+\frac{3}{4} p_H\). We will revisit this shortly, after we explore an even more creative transformation!
One interesting class of transformation comes from duality. Duality refers to a special pairing between different physical systems with related parameters, which can take on many different forms. For lattice systems, duality transformations may come from considering the vertices instead of the cells.1 Instead of grouping the cells into clusters connected by open edges, we group vertices into clusters connected by the blocked edges:
What happens to the right-most lattice system as you change the open edge probability? Does it seem to have a critical point? How do the sizes of large clusters in both graphs relate?
You should notice that when a vertex cluster becomes large, it encloses many small clusters of cells. Similarly, when there is a large cluster of cells, there are many small islands of blocked edges inside of it. Furthermore, cell clusters and vertex clusters cannot pass through each other. All of these observations, taken together, suggest that exactly one type of cluster percolates at any given probability.
In order to use this duality transformation to learn about percolation, we need to relate the vertex problem to the cell problem. The vertices are arranged in a lattice with some edges present, and are clustered together, with vertices in the same cluster if they are connected together by a sequence of edges. This is exactly the same as our description of the percolation problem, but with "cell" replaced by "vertex" and the flow through "open edges" replaced by flow along edges that were "blocked" in the cell problem. Thus if the original lattice has probability p, its dual is a percolation system on a new lattice with probability 1-p. Now we need to figure out the structure of the dual lattices.
Look back at the square lattice's dual transformation above. How many edges come out of each vertex? How do they connect together?
We can see that the square lattice is self-dual, meaning that it is its own dual! Similarly, if you play around with the simulation below, you can see that the dual of the triangular lattice is a hexagonal lattice.
Our reasoning about the cell- and vertex-clusters can lead us to more relationships between critical probabilities. If a lattice has critical probability p, the vertex clusters on that lattice percolate for probabilities below p, and so, accounting for closed edges becoming open edges after the transformation, the dual lattice will have critical probability 1-p. For the self-dual square lattice, these probabilities must be equal, which is only possible if the critical probability is 1/2. The duality between the triangular and hexagonal lattices similarly tells us that their critical probabilities sum to 1. All together we have: $$p_S = \frac{1}{2},\qquad p_T+p_H=1,\qquad p_T\approx \frac{1}{3}+\frac{2}{3}p_S,\qquad p_S \approx \frac{1}{4}+\frac{3}{4}p_H,$$ where \(\approx\) means approximately equal, from our bond deletion estimates. These are more than we need; any three of these equations will allow us to determine the values. Solving, we find
$$p_T \approx \frac{2}{3},\qquad p_S = \frac{1}{2},\qquad p_H\approx \frac{1}{3} $$ which agrees with our experimental results. However, we did not need any experiments obtain this result. All we needed to do was reason about the structure of the lattice percolation problem, leading us to some relationships which on their own didn't seem terribly useful. Yet, putting them together, we found excellent estimates of the real behavior!
One important class of transformations are renormalization groups. In statistical mechanics, one way we can make physical systems more manageable is to use averages instead of tracking every single detail. In a magnet, for example, we might consider the average direction that magnetic moments are pointing over a small region. If the moments are all aligned, our average will point in that direction. If the moments are oriented randomly, they will average out to 0. Thus an average still captures information about collective behavior of the system. To perform an average, however, we need to decide what scale it is over - do we include 10 atoms, or 100, or 1000? We expect that different choices will still describe the system more or less correctly, but they will lead to differences in the equations that describe the system: a group of 1000 magnetic moments pointing in the same direction has a larger effect on its neighbors than just 10 would, so the effective strength of the interaction will have to be scaled appropriately. This leads to another sort of non-obvious symmetry, which allows us to pick out phase transitions as fixed points which look essentially the same at all averaging scales.
We can adapt this idea to percolation.2 We will group cells of our lattice into small groups, and then determine rules for the edges of those groups being open or blocked. For the square lattice, we take two-by-two blocks of cells, and the right/bottom edges are open if the top/left cells can flow through that side. Click on edges in lattices below to toggle them between open and blocked.
Let's focus on the bottom edge (the other will be the same). The two edges on the right side of the original lattice square wont change anything, nor will the edge dividing the top two cells, as we only care about whether either top cell can connect through the bottom. If all five remaining edges are open, the top cells can clearly connect. This happens with probability \(p^5\). You can check that blocking any one edge wont stop this. Exactly one edge is blocked with probability \(5p^4(1-p)\), the 5 because there are five choices of edge to add, and the \(1-p\) to account for the one blocked edge. Through similar reasoning, we find there are 8 ways to block exactly 2 edges while still allowing the cells to connect, 2 ways to block exactly 3, and that any choice of 4 or 5 edges results in the bottom edge being blocked.
Putting it together, we find that the new probability is
$$ \begin{align}P(p)&=p^5+5p^4(1-p)+8p^3(1-p)^2+2p^2(1-p)^3 \\&= p^2(2p^3-5p^2+2p+2) \end{align}$$
For the triangular lattice, we can form new 4-cell triangles, choosing edges to be open when the center cell can connect through them. Counting possibilities, we find the probability in the new lattice is
$$ Q(p) = p^4+4p^3(1-p)+2p^2(1-p)^2 = p^2(2-p^2) $$