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(9.2.42)

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2.5 Monte Carlo Simulations of Sticky Disks

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(4dT) 1f J(rik - d) J(rjk - d) rij sinB rjk 2d = ( -d )2

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(9.2.43)

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where ark = rikdrikdBjk and r;k = r[k +r[j -2rikrij cos Bjk have been used to obtain the second equality in (9.2.43). rij is the distance between particles i and j, and Bjk is the angle between the line segments Tij and Tik connecting particles i with particle j and with particle k, respectively. Note that P2 = 0 when rij > 2d, since particle k cannot form a double-bond with praticles i and j simultaneously when their distance is greater than two diameters. P2 has an integrable singularity at rij = 2d for any pair of disks i and j. Note that the total "transition probability" defined by Seaton and Glandt has the dimension of "volume" which is similar to the "effective volume" used by Kranendonk and Frenkel [1988]. The definitions of unnormalized total transition probabilities do not depend on the previous binding state of the displaced particle k. However, during an actual Monte Carlo step the binding state of a displaced disk may change from its present ;3-bond configuration to a new ;3'-bond configuration. Similarly, in order to calculate the unnormalized total transition probabilities, a list of all pairs of particles with which the test particle k could form a double-bond, has to be established. Then, the new binding state ;3' for the test disk k is selected according to the normalized probability

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(9.2.44)

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It is noted that the magnitude of P(;3') depends on the binding structure of the system prior to the disk k being moved. The transitions involve all three binding states. In the following, we describe the Monte Carlo process.

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Step 1. The Monte Carlo simulations for sticky disks also begin with N unbounded disks regularly placed within a unit square without overlapping. To create the oth (initial) realization of random distribution, the periodic particle positions are randomized by using the shuffling method described in Section 2.1.

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Step 2. To generate sticky disks, we do not shuffle the test disk within a small square of [o,~] x [0, ~]. Instead, the disk k makes a transition according to the breaking of its current binding state and the formation of new binding state with other disks. We evaluate the transition probability of particle k. Before changing the test disk k's binding state, we have to build a list of all possible pairs with which the test disk k could form a double-bond. We compute the normalized probabilities P((3/) according to the respective unnormalized total transition probability P(31 for (3/ = 0, 1,2. We also store the array of probabilities for the pair of disks (i, j)

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p(2)(i,j) = P2 P2

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(9.2.45)

Step 3. A random number generator, between and 1, is used to generate a random number r and compared with the obtained P((3/) of the current system to decide the new binding state of the test disk. Three obvious possibilities are: (3/ = 0, 1,2 corresponding to unbounded, single-bond, and double-bond state, respectively. If r ::; P(O), then (3/ = is chosen. If P(O) < r ::; P(O) + P(l), then (3/ = 1 is chosen. If P(O) + P(l) < r ::; 1, then (3/ = 2 is chosen.

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Step 4. The possible new location of disk k, (x~, Y~) is then determined as follows, depending on which (3/ is chosen is Step 3. If (3/ =0, the coordinates (x~, of the test disk is determined using two random number generators, each between and 1, for its x and y coordinates. If (3/ = 1, an integer random number i with equal probabilities, i = 1, 2, ... ,N, i:f- k, is generated to select the potential single-bond candidate disk. A random angle generator, between and 27f, is employed to determine . Then