New Approaches to Modern Physics/EXACT

118093 - New Approaches to Modern Physics

Advanced Statistical Mechanics

[Course Summary][Links to Lecture Notes and Homework][Announcements] [For Lecturers]
[JOAN ADLER's lectures]


  1. This lattice is unphysical on the microscopic scale, but actually relevant to models of internet connectivity and similar. There is research about percolation of fax machines and questions of internet connectivity that bear a resemblance to percolation issues. Anyone who finds this of interest can chose it for their targil question.

  2. We are viewing this solution because it is a useful test of percolation concepts. It originated with Flory in a polymer application.

  3. Let's start at the origin of the Bethe Lattice. In the figure below we have an origin, 3 branches and 6 subbranches.


    1. Our lattice has z=3, so there are z-1=2 bonds emanating from each new site in the outwards direction.

    2. Each bond leads to one neighbour, occupied with probability p, thus on average we have (z-1)p new occupied neighbours along which to continue the path.

    3. If (z-1)p is less than unity then the average number of new paths leading out to infinity decreases at each generation by this factor.

    4. Thus even if all z neighbours of an occupied origin are occupied (giving z chances to find a way out, and even if z is large, the probability of finding a continuous path of occupied neighbours goes to zero exponentially with path length if p is less than 1/(z-1) hence pc=1/(z-1)=1/2 for z=3.

    5. This argument is valid for both bond and site situations, and for a one dimensional lattice gives pc =1.

  4. We may now calculate P(p), the percolation probability.

    1. Consider Q, the probablity that an arbitrary site is not connected to infinity thru one fixed branch originating from the origin (or another site).

    2. The probability that two subbranches which start at a neighbour of the origin are not connected to infinity is Q2.

    3. pQ2 is the probability that the neighbour is occupied but not connected to infinity by any of its two subbranches. The neighbour is empty with probability (1-p) in which case even well-connected subbranches would not help it.

    4. Thus Q=1-p+pQ2 is the probability that this branch does not lead to infinity. This quadratic has two solutions, Q=1 and Q=(1-p)/p. The probability p-P(p) that the origin is occupied but not connected to infinity thru any of its 3 branches is pQ3. Thus P(p)=p(1-Q3) which gives zero for the first solution and gives P(p)/p=1-[(1-p)/p]3 for the other solution corresponding to pc=1/2.
  5. It is also possible to calculate the mean cluster size below pc by similar arguments, it is S(p)=(1+p)/(1-2p)
    see p. 30 of Stauffer and Aharony for details.

    what what