This is a sequel to the question: Proof and interpretation of the following percolation theory result for $n\times n$ square grid

In the paper: The Birth of the Infinite Cluster:Finite-Size Scaling in Percolation

Theorem 3.2 basically states that with probability going to $1$ as $n\rightarrow\infty$, the largest open cluster in an $n\times n$ box is of size $\theta(p)n^2$, $\theta(p)$ being the percolation probability (where $p>p_c$). Moreover, Theorem 3.1 part (iii) states that the expectation value of size of the second largest cluster is of sub linear order.

This result is for bond percolation. However, does there exist any analogous result for site percolation?


1 Answer 1


Concerning the relative size of the largest cluster $M_1$ and the second largest cluster $M_2$, for site percolation in an $n\times n$ lattice with site-occupation probability $p$, one has the scaling law

$$\langle M_1/M_2\rangle = C_0 + C_1 (p-p_c)n^{1/\nu}+{\cal O}(p-p_c)^2$$

with critical exponent $\nu=4/3$ and percolation threshold $p_c=0.59$. The coefficients $C_0$ and $C_1$ depend on the boundary condition. See, for example, arXiv:1504.07712

The averages of $M_1$ and $M_2$ separately scale as $$\langle M_1\rangle=n^{2-\beta/\nu}[a_0+a_1(p-p_c)n^{1/\nu}+{\cal O}(p-p_c)^2],$$ $$\langle M_2\rangle=n^{2-\beta/\nu}[b_0+b_1(p-p_c)n^{1/\nu}+{\cal O}(p-p_c)^2],$$ with $\beta=5/36$. Hence the ratio of the averages, $\langle M_1\rangle/\langle M_2\rangle$, follows the same scaling law as the average $\langle M_1/M_2\rangle$ of the ratio, upon substitution of $C_0\mapsto a_0/b_0$, $C_1\mapsto (a_1 b_0-a_0 b_1)/b_0^2$.

So I would conclude that this scaling for site percolation differs from the $n^2$ scaling for bond percolation.

  • $\begingroup$ Hey, thanks! :) This one looks interesting. Checking the paper! $\endgroup$
    – user123818
    May 3, 2018 at 22:21
  • $\begingroup$ Any idea if the second largest cluster has sublinear or logarithmic scaling? Maybe it's derivable from here $\endgroup$
    – user123818
    May 3, 2018 at 22:23
  • $\begingroup$ added the scaling of the averages of $M_1$ and $M_2$ separately; they follow the same $n$-dependence. $\endgroup$ May 3, 2018 at 22:31
  • $\begingroup$ Hmm. This seems to be a bit weird. According to the size distribution plots I had made (programmed), there seemed to be only one huge cluster and other smaller clusters for $p$'s beyond 0.5. Although my plots were only for small systems of size 100x100. I'll perhaps plot it for 1000x1000 and re-check. I was really sort of expecting some sublinear scaling for the smaller clusters. $\endgroup$
    – user123818
    May 3, 2018 at 22:36
  • 1
    $\begingroup$ These results give the size of the cluster in the scaling window at $p_c$ (that is, where $p$ approaches $p_c$ sufficiently quickly as $n\rightarrow\infty$); the $O(n^2)$ size scaling referred to in the question is for (fixed) $p>p_c$. $\endgroup$
    – j.c.
    May 3, 2018 at 22:57

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