• An occupation probability p = 0.523 is used in a site percolation simulation on an infinite square lattice, is it possible for an infinite cluster to occur in the simulation? What about a bond simulation  on the same lattice?
For site percolation simulation on an infinite square lattice, pc=0.592746.
p < pc, therefore not possible.

For bond simulation, pc=0.5
p > pc, so it is possible.

• Calculate the average cluster size S for the square lattice given in Figure 4-7. Please provide detailed steps for the calculation.

Average cluster size

S = ( sum s = 1 to infinity  ns s2 ) / ( sum s = 1 to infinity  ns 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.

• Write down the percolation probability P and average cluster size S as a function of the occupation probability for a simple cubic lattice near the critical point.
•

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

• An occupation probability p = 0.623 is used in a site percolation simulation on an infinite square lattice, determine its percolation probability, average cluster size and correlation length near the critical point.
•

The critical exponents for this lattice are:
beta=5/36, gamma=43/18, nu=4/3

P = 0                                         for p < pc
P = (p-pc)beta = (p-pc)5/36   for p > pc

S =  |p-pc|-gamma =  |p-pc|-43/18
xi  =  | p - pc | -nu =  | p - pc | -4/3

• The exponents tau and sigma for a lattice are 0.48 and 2.31 respectively, derive the other exponents beta and gamma.
•

Exponent relations:

beta =  tau - 2/sigma
gamma =  3 - tau/sigma

beta = 0.48 - 2/2.31 =
gamma = 3 - 0.48/2.31 =

• Prove the following relationship:
(1-p)2 sum s = 1 to infinity  ( s ps ) =
(1-p)2 (p d/dp)2  sum s = 1 to infinity  ( ps )

d/dp (ps ) = s ps-1
d/dp (pd/dp (ps )) =d/dp( s ps )=s2 ps-1
pd/dp (pd/dp (ps )) =d/dp( s ps )=s2 ps
sum s = 1 to infinity pd/dp (pd/dp (ps )) = sum s = 1 to infinity s2 ps

(1-p)2 (p d/dp)2  sum s = 1 to infinity  ( ps ) =
(1-p)2 sum s = 1 to infinity  ( s ps )

•  Given that:
S= (1-p)2 sum s = 1 to infinity  ( s2  ps )
S = (1 + p) /  (1 - p)        (p < pc)

sum s = 1 to infinity  ns s2  =
(1-p)2 sum s = 1 to infinity  ( s ps ) =
(1-p)2 (p d/dp)2  sum s = 1 to infinity  ( ps )  =
(1-p)2 (p d/dp)2 (p/(1-p)) =

(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  ns s  = p

S = ( sum s = 1 to infinity  ns s2 ) / ( sum s = 1 to infinity  ns s )

= p(1+p)/(1-p)/p = (1+p)/(1-p)

•   From   g(r) = pr   and g(r) = exp(-r/xi), derive: xi  ~ 1/(pc - p)

Given that:

p=exp(lnp)

and

-lnp ~ 1-p for p -> 1

and pc -> 1 (hence: -lnp ~ pc-p)

we have:

g(r) = pr = (exp(lnp))r = exp(rlnp) = exp[-r(pc-p)] =

= exp[-r/(1/(pc-p))]

Therefore:

xi  ~ 1/(pc - p)

• What is the procedure for percolation simulation?

• We first generate a random percolation configuration by assigning 1 for occupation, and 0 for empty at each site.
• Compute physical quantities from this configuration and store them in some places.
• Generate the next configuration and repeat the  computation and storage.
• After N configurations, compute the average and statistical errors etc.