Nonextensive Statistical Mechanics

3.3 Correlations, Occupancy of Phase-Space, and Extensivity of S q 75We shall refer to this choice as the stretched exponential model. If α = 1, itrecovers the previous case, i.e., the independent model. If α = 0 and 0 < p < 1, wehave that all probabilities vanish for fixed N, excepting r N0 = p and r NN = 1 − p .All these models, with the unique exception of the independent model, involvecorrelations. These correlations might however be not strong enough in order torequire an entropy different from S BG if we seek for extensivity. Let us be moreprecise. The entropy of the N-system is given byS q (N) = 1 − ∑ Nn=0N!(N−n)! n! [r Nn] q(S 1 (N) =−N∑N!)(N − n)! n! r Nn ln r Nn .q − 1n=0(3.131)The question we want to answer is the following: Is there a value of q such thatS q (N) is extensive, i.e., such that lim N→∞ S q (N)/N is finite?The answer is trivial for the independent model. The special value of q is simplyunity. Indeed, in that case, we straightforwardly obtainS BG (N) = NS BG (1) =−N[p ln p + (1 − p)ln(1− p)] . (3.132)In this simple case, the BG entropy is not only extensive but even additive.Numerical calculation has shown that the answer still is q = 1 for the Leibnitztriangle, and for the stretched model with p > 0 and α>0. All these examples areillustrated in Fig. 3.14.Table 3.9 Anomalous probability sets: d = 1(top) andd = 2(bottom). The left number withinparentheses indicates the multiplicity (i.e., Pascal triangle). The right number indicates the correspondingprobability. The probabilities, noted r N,n , asymptotically satisfy the Leibnitz rule, i.e.,lim N→∞r N,n+r N,n+1r N−1,n= 1(∀n). In other words, the system is, in this sense, asymptotically scaleinvariant.Notice that the number of triangle elements with nonzero probabilities grows like N,whereas that of zero probability grows like N 2(N = 0) (1, 1)(N = 1) (1, 1/2) (1, 1/2)(N = 2) (1, 1/2) (2, 1/4) (1, 0)(N = 3) (1, 1/2) (3, 1/6) (3, 0) (1, 0)(N = 4) (1, 1/2) (4, 1/8) (6, 0) (4, 0) (1, 0)(N = 0) (1, 1)(N = 1) (1, 1/2) (1, 1/2)(N = 2) (1, 1/3) (2, 1/6) (1, 1/3)(N = 3) (1, 3/8) (3, 5/48) (3, 5/48) (1, 0)(N = 4) (1, 2/5) (4, 3/40) (6, 3/60) (4, 0) (1, 0)Is it possible to have special correlations that make S q to be extensive only forq ≠ 1? The answer is yes. Let us illustrate this on two examples [199]. The first ofthem is neither strictly nor asymptotically scale-invariant. The second one is asymptoticallyinvariant. To construct both of them we start from the Leibnitz triangle,and then impose that most of the possible states have zero probability. Their initial