For bond simulation, pc=0.5
p > pc, so it is possible.
S = ( sum _{s = 1 to infinity} n_{s} s^{2} ) / ( sum _{s = 1 to infinity} n_{s} s )
clusters in Figure 4-7:
cluster class s ns
1
1 1
2
4 1
3
6 1
4
46 1
substitute above {s, ns} pairs into the formula.
For a simple cubic lattice, its critical exponents are:
beta=0.41, gamma=1.80
P = 0
for p < pc
P = (p-pc)^{beta} = (p-pc)^{0.41 }
for p > pc
S = |p-pc|^{-gamma} = |p-pc|^{-1.8}
^{}
^{}
^{}
The critical exponents for this lattice are:
beta=5/36, gamma=43/18, nu=4/3
S = |p-pc|^{-gamma} = |p-pc|^{-43/18}
xi = | p - pc | ^{-nu} = |
p - pc | ^{-4/3}
Exponent relations:
beta = tau - 2/sigma
gamma = 3 - tau/sigma
beta = 0.48 - 2/2.31 =
gamma = 3 - 0.48/2.31 =
(1-p)^{2} (p d/dp)^{2} sum _{s
= 1 to infinity} ( p^{s} ) =
(1-p)^{2} sum _{s = 1 to infinity }
( s^{2 } p^{s} )
(1-p)^{2} (p d/dp)( p/(1-p)2) =
(1-p)2 p [ 1/(1-p)2 + 2P/(1-p)3 ] =
p(1+p)/(1-p)
sum _{s = 1 to infinity} n_{s}
s = p
S = ( sum _{s = 1 to infinity} n_{s} s^{2} ) / ( sum _{s = 1 to infinity} n_{s} s )
= p(1+p)/(1-p)/p = (1+p)/(1-p)
p=exp(lnp)
and
-lnp ~ 1-p for p -> 1
and pc -> 1 (hence: -lnp ~ p_{c}-p)
we have:
g(r) = p^{r} = (exp(lnp))^{r} = exp(rlnp) = exp[-r(p_{c}-p)] =
= exp[-r/(1/(p_{c}-p))]
Therefore:
xi ~ 1/(p_{c} - p)