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Is the entropy Sq extensive or nonextensive

a r X i v :c o n d -m a t /0409631v 1 [c o n d -m a t .s t a t -m e c h ] 23 S e p 2004

Is the entropy S q extensive or nonextensive?

Constantino Tsallis

Santa Fe Institute,1399Hyde Park Road,

Santa Fe,New Mexico 87501,USA

and

Centro Brasileiro de Pesquisas Fisicas,

Rua Xavier Sigaud 150,

22290-180Rio de Janeiro-RJ,Brazil

(Dated:February 2,2008)

The cornerstones of Boltzmann-Gibbs and nonextensive statistical mechanics respectively are the entropies S BG ≡?k W i =1p i ln p i and S q ≡k (1? W i =1p q i )/(q ?1)(q ∈R ;S 1=S BG ).Through them we revisit the concept of additivity,and illustrate the (not always clearly perceived)fact that (thermodynamical)extensivity has a well de?ned sense only if we specify the compo-sition law that is being assumed for the subsystems (say A and B ).If the composition law is not explicitly indicated,it is tacitly assumed that A and B are statistically independent .In this case,it immediately follows that S BG (A +B )=S BG (A )+S BG (B ),hence extensive,whereas S q (A +B )/k =[S q (A )/k ]+[S q (B )/k ]+(1?q )[S q (A )/k ][S q (B )/k ],hence nonextensive for q =1.In the present paper we illustrate the remarkable changes that occur when A and B are specially corre-lated .Indeed,we show that,in such case,S q (A +B )=S q (A )+S q (B )for the appropriate value of q (hence extensive),whereas S BG (A +B )=S BG (A )+S BG (B )(hence nonextensive).We believe that these facts substantially improve the understanding of the mathematical need and physical origin of nonextensive statistical mechanics,and its interpretation in terms of e?ective occupation of the W a priori available microstates of the full phase space.In particular,we can appreciate the origin of the following important fact.In order to have entropic extensivity (i.e.,lim N →∞S (N )/N <∞,where N ≡number of elements of the system ),we must use (i)S BG ,if the number W e?of e?ectively occupied microstates increases with N like W e?～W ～μN (μ≥1);(ii)S q with q =1?1/ρ,if W e?～N ρ

INTRODUCTION

A quantity X (A )associated with a system A is said additive with regard to a (speci?c)composition of A and

B if it satis?es

X (A +B )=X (A )+X (B ),

(1)

where +inside the argument of X precisely indicates that composition.For example,suppose we partition the inte-rior of a single closed bottle in two parts.If no chemical or other reactions occur between the gas molecules that might be inside the bottle,nor between these molecules and the bottle itself (and its internal physical partition),the number of gas molecules is an additive quantity with regard to the elimination of the partition surface.The same happens with the total energy of an ideal gas,where all interactions have been neglected,including the grav-itational one.More trivially,the total height of various

<∞.(4)

Clearly,all quantities that are additive with regard to a given composition,also are extensive with regard to

2 that same composition(and lim N→∞X(N)/N=X(1)),

whereas the opposite is not necessarily true.For ex-

ample,the total energy,the total entropy and the total

magnetization of the standard Ising ferromagnetic model

with N spins on a square lattice are extensive but not

additive quantities.In other words,they are asymp-

totically additive,but not strictly additive.Of course,

there are quantities that are neither additive nor even

extensive.They are called nonextensive.All types of

behaviors can exist,such as X(N)∝Nγ(γ≥0).For

instance,thermodynamical quantities that,with regard

to some speci?c composition,exhibitγ=0are called

intensive.Such is the case of the temperature,pres-

sure,chemical potential and similar quantities in a great variety of(thermodynamically equilibrated)systems ob-served in nature.A less trivial example of nonexten-sive quantity emerges within a spatially homogeneous d?dimensional classical gas whose N particles(exclu-sively)interact through a two-body interaction poten-tial that is strongly repulsive at short distances whereas it is attractive at long distances,decaying like1/rα(r≡distance between two particles),and0≤α/d.The total potential energy of such a system corresponds[1] toγ=2

?α/d if0≤α/d<1(i.e.,nonextensive), and toγ=1forα/d>1(i.e.,extensive).The to-tal potential energy of this particular model has a loga-rithmic N-dependance(i.e.,nonextensive)at the limiting valueα/d=1.The Lennard-Jones model for gases corre-sponds to(α,d)=(6,3),and has therefore an extensive total energy.In contrast,if we assume a cluster of stars gravitationally interacting(together with some physical mechanism e?ectively generating repulsion at short dis-tances),we have(α,d)=(1,3),hence nonextensivity for the total potential energy.The physical nonextensivity which naturally emerges in such anomalous systems is, in some theoretical approaches,desguised by arti?cially dividing the two-body coupling constant(which has in fact no means of“knowing”the total number of particles of the entire system)by N1?α/d.For the particular case α=0this yields the widely(and wildly!)used division by N of the coupling constant,typical for a variety of mean?eld approaches.See[2]for more details. Boltzmann-Gibbs(BG)statistical mechanics is based on the entropy

S BG≡?k

i=1

p i ln p i,(5)

with

i=1

p i=1,(6)

where p i is the probability associated with the i th micro-scopic state of the system,and k is Boltzmann constant. In the particular case of equiprobability,i.e.,p i=1/W (?i),Eq.(5)yields the celebrated Boltzmann principle (as named by Einstein[3]):

S BG=k ln W.(7)From now on,and without loss of generality,we shall take k equal to unity.

Nonextensive statistical mechanics,?rst introduced in 1988[4,5,6](see[7,8,9,10,11,12,13,14,15]for re-views),is based on the so-called“nonextensive”entropy S q de?ned as follows:

S q≡

1? W i=1p q i

1?q

(z∈R;z>0;ln1z=ln z).(10) The inverse function,the q-exponential,is given by

e z q≡[1+(1?q)z]1/(1?q)(e z1=e z)(11) i

f the argument1+(1?q)z is positive,and equals zero otherwise.

The present paper is entirely dedicated to the analysis of the additivity or nonadditivity of S BG and of its gen-eralization S q.However,following a common(and some-times dangerous)practice,we shall from now on cease distinguishing between additive and extensive,and use exclusively the word extensive in the sense of strictly ad-ditive.

II.THE CASE OF TWO SUBSYSTEMS

Consider two systems A and B having respectively W A and W B possible microstates.The total number of possi-ble microstates for the system A+B is then in principle W≡W A+B=W A W B.We emphasized the expression “in principle”because,as we shall see,a more or less se-vere reduction of the full phase space might occur in the presence of strong correlations between A and B.

We shall use the notation p A+B

(i=1,2,...,W A;j= 1,2,...,W B)for the joint probabilities,hence

W A

i=1

W B

j=1

p A+B

=1.(12)

The marginal probabilities are de?ned as follows:

p A i≡

W B

j=1

p A+B

,(13)

hence

W A

i=1

p A i=1,(14)

3 and

p B i≡W A

i=1

p A+B

,(15)

hence

W B

j=1

p B j=1.(16) These quantities are indicated in the following Table.

\B1...

p A+B 12p A+B

1W B

p A1

p A+B 22p A+B

2W B

p A2

.........

p A+B W A2p A+B

W A W B

p A W

p B1...

12p?1

21?p

1?p1

It can be trivially veri?ed that Eq.(21)is not satis?ed. Therefore,for this special correlation,S BG is nonexten-sive.It can also be veri?ed that,for q=0and only for q=0,the following additivity is satis?ed:

S0(A+B)=S0(A)+S0(B),(24) therefore S0is extensive.Indeed S0(A+B)=2S0(A)= 2.We immediately see that,depending on the type of correlation(or lack of it)between A and B,the entropy which is extensive(reminder:as previously announced, we are using here and in the rest of the paper“extensive”to strictly mean“additive”)can be S BG or a di?erent one.

Before going on,let us introduce right away the dis-tinction between a priori possible states(in number W) and allowed or e?ective states(in number W e?).Let us consider the above case of two equal binary subsys-tems A and B and consequently W=4.If they are independent(i.e.,the q=1case),their generic case corresponds to0

Let us further construct on the above observations.Is it possible to unify,at the level of the joint probabilities, the case of independence(which corresponds to q=1) with the specially correlated case that we just analyzed (which corresponds to q=0)?Yes,it is possible.Con-sider the following Table:

4 A

\B1

p?f q(p)p

1?2p+f q(p)1?p

q?1

,(26)

where we have used the fact that A=B.In other words, we are facing a whole family of entropies that are exten-sive for the respective special correlations indicated in the Table just above.

Let us proceed and generalize the previous examples to two-state systems A and B that are not necessarily equal. The case of independence is trivial,and is indicated in the following Table:

\B1

p A1p B2p A1

p A2p B2p A2

p B1

1p A1+p B1?1

21?p A1

1?p B11

We verify that Eq.(24)is satis?ed.Is it possible to unify the above anisotropic q=1and q=0cases?Yes,it is. The special correlations for these cases are indicated in the following Table:

0.25

0.5

0.75

FIG.1:The function f q(p),corresponding to the two-system A=B case(with W A=W B=2),for typical values of q∈[0,1].A few typical nontrivial(q,f q(1/2))points are (0.4,0.043295),(0.5,0.064765),(0.6,0.087262),(0.7,0.111289), (0.8,0.138255),(0.9,0.171838),(0.99,0.225630).It can be easily veri?ed that these values satisfy the relation 21?q?[f q(1/2)]q?[(1/2)?f q(1/2)]q=1/2,which is the simple form that takes Eq.(25)for the p=1/2particular case.We also remind the trivial values f0(1/2)=0and f1(1/2)=1/4.

\B1

p A1?f q(p A1,p B1)p A1

1?p A1?p B1+f q(p A1,p B1)1?p A1 p B1

say p A+B

11,which is determined once for ever.More

explicitly,we have that p A+B

12=p A1?p A+B

,p A+B

p B1?p A+B

11,p A+B

=1?p A1?p B1?p A+B

Eq.(27)recovers Eq.(25)as the particular instance p A1=p B1.And we can easily verify that,for0≤q≤1,

S q(A+B)=S q(A)+S q(B).(28) So,we still have extensivity for the appropriate value of q,i.e.,the value of q which has been chosen in Eq.(27) to de?ne the function f q(x,y)re?ecting the special type of correlations assumed to exist between A and B.In other words,when the marginal probabilities have all the information,then the appropriate entropy is S BG.But this happens only when A and B are independent.In all the other cases addressed within the above Table,the important information is by no means contained in the marginal probabilities,and we have to rely on the full set of joint probabilities.In such cases,S BG is nonextensive, whereas S q is extensive.

Before closing this section dedicated to the case of two systems,let us indicate the Table associated to the q=0 entropy for arbitrary systems A and B:

\B1...

p B2p B W

p A1

00p A2

.........

00p A W

p B1...

6 A

\B1

p A+B+C

121

(p A+B+C

112

)

2p A+B+C

211

(p A+B+C

222

)

The corresponding AB?marginal probabilities are in-

dicated in the Table below:

\B1

p A+B

which of course reproduces the situation we had for the

two-system(A+B)problem.This is to say p A+B

11=

p A+B+C 111+p A+B+C

112

,and so on.

A.Three independent subsystems

Consider?rst the case where all three subsystems A and B are binary and statistically independent,i.e.,such that the joint probabilities are given by

p A+B+C

ijk

=p A i p B j p C k(?(i,j,k)).(32) The corresponding Table is of course as follows

\B1

p A1p B2p C1

(p A1p B1p C2)

2p A2p B1p C1

(p A2p B2p C2)

We immediately verify that

S BG(A+B+C)=S BG(A)+S BG(B)+S BG(C)(33)

Therefore,S BG is extensive.Consistently,S q is,unless q=1,nonextensive.

B.Three specially correlated subsystems

Consider now that the three binary subsystems are cor-related as indicated in the next Table(with p A1+p B1+ p C1>2):

\B1

1?p B1

(1?p C1)

21?p A1

(0)

We easily verify that

S0(A+B+C)=S0(A)+S0(B)+S0(C).(34) For example,if A=B=C and2/3

Let us next unify the q=1and the q=0cases.We heuristically found the solution.It is indicated in the following Table:

\B1

?f q(p A1,p B1)

?p C1(p A1+p B1)

?p A1f q(p B1,p C1)

[p A1(1?p B1?p C1?f q(p A1,p C1)?f q(p B1,p C1)

p C1(1?p A1?p B1

?p B1f q(p A1,p C1)

[p B1(1?p A1?p C1

+f q(p A1,p B1))]

where the function f q(x,y)is de?ned in Eq.(27).In-terestingly enough,it has been possible to?nd a three-subsystem solution in terms of the two–subsystem and one-system ones.More explicitly,we have,for example,

that p A+B+C

111

=f q(p A1,p C1)+f q(p B1,p C1)?p C1(p A1+p B1)+

p C1f q(p A1,p B1)=p A+C

+p A+B

?p C1(p A1+p B1)+p C1p A+B

, and similarly for the other seven three-subsystem joint probabilities.Of course,all eight joint probabilities asso-ciated with the above Table are nonnegative;whenever the values of(p A1,p B1,p C1)replaced within one or the other of these analytic expressions yield negative numbers,the corresponding probabilities are to be taken equal to zero. The AB?marginal probabilities precisely recover the joint probabilities of the previously discussed two-system (A+B)Table.For example,[f q(p A1,p C1)+f q(p B1,p C1)?p C1(p A1+p B1)+p C1f q(p A1,p B1)]+[f q(p A1,p B1)+p C1(p A1+p B1)?f q(p A1,p C1)?f q(p B1,p C1)?p C1f q(p A1,p B1)]=f q(p A i,p B1), [?f q(p A1,p B1)+p A1(p B1+p C1)?p A1f q(p B1,p C1)]+[p A1(1?p B1?p C1+f q(p B1,p C1))]=p A1?f q(p A1,p B1),and so on. Finally,we verify that

S q(A+B+C)=

7 =S q(A)+S q(B)+S q(C)(35)

For the particular case A=B=C,the above Table

becomes

\B1

2p2?f q(p)?pf q(p)

[2p2?f q(p)?pf q(p)]

22p2?f q(p)?pf q(p)

[(1?p)(1?2p+f q(p))]

where we have used f q(p,p)=f q(p).

For the generic case of three subsystems with W A,

W B and W C states respectively,we have that W=

W A W B W C,whereas in the appropriate asymptotic sense

we expect W e?=[W1?q

A +W1?q

+W1?q

?2]1/(1?q)≤W

for0≤q≤1(the equality generically holds only for q=1).In the particular instance A=B=C,this expression becomes W e?=[3W1?q

?2]1/(1?q).

IV.ENLARGING THE SCENARIO

A.The case of N subsystems

The three-system case discussed above is a generic one under the assumption that W A=W B=W C=2.We have not attempted to generalize its corresponding spe-cial correlation Table to the generic(W A,W B,W C)case, and even less to the even more generic case of N such sys-tems(A1,A2,...,A N).It is clear however that,assuming that this(not necessarily trivial)task was satisfactorily accomplished,the result would lead to

S q(N

r=1A r)=

r=1

S q(A r),(36)

where q=1if all N systems are mutually independent, i.e.,

p A1+A2+...+A N i1i2...i N =

r=1

p A r i

(?(i1,i2,...,i N)),(37)

and q=1otherwise.This is to say,if we have indepen-dence,the only entropy which is extensive is S BG.If we do not have independence but the special type of(collec-tive)correlations focused on in this paper instead,then only S q for a special value of q is extensive.

For the case of independence,the generic composition law for S q is given by

ln[1+(1?q)S q(N

r=1A r)]=

r=1

ln[1+(1?q)S q(A r)],

(38)

or,equivalently,

1+(1?q)S q(

r=1

A r)=

r=1

[1+(1?q)S q(A r)].(39)

Eq.(38)exhibits in fact the well known(monotonic)

connection between S q and the Renyi entropy S R q≡

ln W i=1p q i /(1?q)= ln[1+(1?q)S q] /(1?q)(we

remind that,for independent systems,S R q is extensive,

?q).

We have generically W= N r=1W A r,which corre-

sponds of course to the total number of a priori pos-

sibly occupied states(i.e.,whose joint probabilities are

generically nonzero)for the generic q=1case.In

contrast,the generic q=0case has only W e?=

( N r=1W A r)?(N?1)nonzero joint probabilities.These

are p A1+A2+...+A N

11 (1)

=( N r=1p A r1)?(N?1)≥0,

p i

11 (1)

=p A1i

(i1=2,3,...,W A

),p1i

11 (1)

=p A2i

(i2=2,3,...,W A

),p111 (i)

=p A N

i N

(i N=2,3,...,W A

The generic q=1and q=0cases can,analogously to

what has been done before,be uni?ed through W e?=

( N r=1W1?q A r)?(N?1) 1/(1?q)≤W(0≤q≤1),where

the equality holds only for q=1.In the particular in-

stance A1=A2=...=A N≡A,this expression becomes

W e?=[NW1?q

?(N?1)]1/(1?q).

Furthermore,for N equal subsystems(a quite frequent

case,as already mentioned),Eq.(36)becomes

S q(N)=NS q(1),(40)

where the change of notation is transparent.This is an

extremely interesting relation since it already has the

shape that accomodates well within standard thermo-

dynamics,even if the entropic index q is not necessar-

ily the usual one,i.e.,q=1.It is allowed to think

that Clausius would perhaps have been as satis?ed with

this relation as he surely was with the same relation but

with S BG!One might also quite safely speculate that

if the system is such that its Table of joint probabili-

ties is not exactly of the type we have discussed here,

but close to it,then we might have,not exactly rela-

tion(40)but rather only asymptotically S q(N)∝N.In

other words,as long as the system belongs to what we

may refer to as the q?universality class,we should ex-

pect lim N→∞S q(N)/N<∞,in total analogy with the

usual BG case.

To geometrically interpret Eq.(40),we may consider

the case of equal probabilities in the allowed phase space,

i.e.,in that part of phase space which is expected to have,

not necessarily W microstates,but generically W e?mi-

crostates(with W e?≤W).The e?ective number W e?

is expected(at least in the N>>1limit)to be precisely

the number of all those states that the special collective

correlations allow to visit.So,if we assume equal prob-

abilities in Eq.(40)(i.e.,p A1+A2+...+A N

i1i2...i N

=1/W e?),we

8 obtain

ln q W e?≡

(W e?)1?q?1

2W A

1p10

200

...0...

W A00

p2p W

This Table corresponds to p A+B

ij =p iδij.Its generaliza-

tion to N equal systems is trivial:p i

1i2...i N

=p i

if all N

indices coincide,and zero otherwise.The corresponding entropy therefore asymptotically approaches the relation

S?∞(N)=S?∞(1)(?N),(48) thus corresponding toρ=0as anticipated.It appears then that all cases equivalent(through permutations)to the above Table,should yield the same limit q→?∞.

C.Connection with the Borges-Nivanen-Le

Mehaute-Wang q?product

Let us mention at this point an interesting connection that can be established between the present problem and the q?product introduced by L.

Nivanen,A.Le Mehaute and Q.A.Wang and by E.P.Borges[17].It is de?ned as follows:

x×q y≡(x1?q+y1?q?1)1/(1?q)(x×1y=xy).(49) It has the elegant,extensive-like,property

ln q(x×q y)=ln q x+ln q y,(50) to be compared with the by now quite usual, nonextensive-like,property

ln q(xy)=ln q x+ln q y+(1?q)(ln q x)(ln q y).(51) This type of structure was since long(at least since 1999)being informally discussed by A.K.Rajagopal, E.K.Lenzi,S.Abe,myself,and probably others.But only very recently it was beautifully formalized[17]. It has immediately been followed and considerably ex-tended by Suyari in a relevant set of papers[18].

Let us now go back to the main topic of the present pa-per.Consider the following joint probabilities associated with N generic subsystems:

p A1+A2+...+A N

i1i2...i N

= 1?N+φ(q)i1i2...i N+N r=1(p A r i r)q?1 1/(q?1),

(52) whereφ(q)i

i2...i N

is a nontrivial function which ensures that

i1i2...i N

p A1+A2+...+A N

i1i2...i N

=1(53)

In the limit q→1,Eq.(52)must recover the independent-systems one,namely

p A1+A2+...+A N

i1i2...i N

r=1

p A r i

,(54)

which impliesφ(1)i

i2...i N

=0.

Notice that,excepting for the functionφ(q)i

i2...i N

,Eq.

(52)associates1/p A1+A2+...+A N

i1i2...i N

with q N r=1(1/p A r i r)with q N r=1x r≡[x11?q+x21?q+...+x N1?q?N+1]1/(1?q).

It follows from Eq.(52)that

(p A1+A2+...+A N

i1i2...i N

)q=(1?N+φ(q)i

i2...i N

)p A1+A2+...+A N

i1i2...i N

+p A1+A2+...+A N

i1i2...i N

r=1

(p A r i

)q?1(55)

hence

i1i2...i N

(p A1+A2+...+A N

i1i2...i N

)q=(1?N)

+ i1i2...i N p A1+A2+...+A N

i1i2...i N

r=1

(p A r i

)q?1,(56)

9 where we have imposed one more nontrival condition on

φ(q)i

1i2...i N

,namely that

i1i2...i N

p A1+A2+...+A N

i1i2...i N

φ(q)i

i2...i N

=0.(57)

One might naturally have the impression that no function

φ(q)i

1i2...i N

might exist satisfying simultaneously Eqs.(53)

and(57).This is not so however,at least for particular cases,since we have explicitly shown in the present paper solutions of this nontrivial problem.

Using the de?nition of S q in the left-hand member of the equality we obtain

(1?q)S q(N

r=1

A r)=

i1i2...i N p A1+A2+...+A N

i1i2...i N

r=1

(p A r i

)q?1?N.(58)

Let us now introduce in Eq.(58)the de?nition of marginal probabilities,namely

p A r i

i1i2...i r?1i r+1...i N p A1+A2+...+A N

i1i2...i N

.(59)

We obtain

(1?q)S q(N

r=1A r)=

r=1

(p A r i

)q?N.(60)

Using once again the de?nition of S q on the right-hand member,we?nally obtain

S q(N

r=1A r)=

r=1

S q(A r)(61)

as desired.

It should,however,be clear that this remarkable math-ematical fact by no means exhausts the problem of the search of explicit Tables of joint probabilities that would lead to extensivity of S q for nontrivial values of q.The constraints imposed by the de?nition itself of the concept of marginal probabilities are of such complexity that the search of solutions is by no means trivial,at least at our present degree of knowledge.Indeed,one easily appreci-ates this fact by looking at the explicit solutions indicated in Sections II.B and III.B.

V.CONCLUSIONS

Let us summarize the obvious conclusion of the present paper:Unless the composition law is speci?ed,the ques-tion whether an entropy(or some similar quantity)is or is not extensive has no sense.Allow us a quick di-gression.The situation is in fact quite analogous to the quick or slow motion of a body.Ancient Greeks consid-ered the motion to be an absolute property.It was not until Galileo that it was clearly perceived that motion has no sense unless the referential is speci?ed.In Galileo’s time,and even now,when no referential is indicated,one tacitly assumes that the referential is the Earth.In total analogy,when no composition law is indicated for ana-lyzing the extensivity of an entropy,one tacitly assumes that the subsystems that we are composing are indepen-dent.It is only—a big only!—in this sense that we can say that S BG is extensive,and that S q(for q=1)is nonextensive.

Once we have established the point above,the next natural question is:Are there classes of collective corre-lations for which we know which is the speci?c entropy to be extensive?(knowing,of course,that absence of all correlations leads to S BG).For this operationally impor-tant question,nontrivial illustrations on how the entropic form is dictated by the type of special collective correla-tions that might(or might not)exist in the system have explicitly presented in Section II.B and III.B.From this discussion,two vast categories of systems are identi?ed (at the most microscopic possible level,i.e.,that of the joint probabilities),namely those whose allowed phase space increases(in size)with N like an exponential or like a power-law,corresponding respectively to q=1and to q<1.

However,it should be clear that the present paper is only exploratory in what concerns this hard task.Indeed, we have not found the generic answer for N(not neces-sarily equal)systems,and we have basically concentrated only on the interval0≤q≤1.We do not even know without doubt if the answer is unique(excepting of course for trivial permutations),or if it admits a variety of forms all belonging to the same universality class of nonexten-sivity(i.e.,sharing the same value of the entropic index q).Even worse,we still do not know what speci?cally happens in the structure of the allowed phase space in the(thermodynamically)most important limit N→∞, or in the frequent limit W A→∞(which would provide a precise geometrical interpretation to a formula such as W e?=[NW1?q

?(N?1)]1/(1?q)for say0≤q≤1).It is precisely this structure which is crucial for fully under-standing nonextensive statistical mechanics and its re-lated applications in terms on nonlinear dynamical sys-tems.For example,an interesting situation might occur if we compare the distribution which optimizes S q(N) and then consider N>>1,with the distribution corre-sponding to having?rst considered N>>1in S q(N)and only then optimizing.We certainly expect the thermody-namic limit and the optimization operation to commute for a system composed by N independent(or nearly in-dependent)subsystems.But the situation seems to be more subtle if our system was composed by N subsys-tems correlated in that special,collective manner which demands q=1in order to have entropy extensivity.Such a situation would be consistent with a property which emerges again and again[7,8,9,10,11,12,13,14,15]

for nonextensive systems,namely that the N→∞and the t→∞limits do not necessarily commute.One more relevant issue concerns what speci?c dynamical nature is required for a physical system to“live”,in phase space, within a structure close to one of those that we have presently analyzed.It is our conjecture that this would occur for nonlinear dynamical systems whose Lyapunov spectrum is either zero or close to it,i.e.,under circum-stances similar to the edge of chaos,where many of the so called complex systems are expected to occur.We leave all these questions as open points needing further progress.

Let us?nally mention the following point.It is by no means trivial to?nd sets of joint probabilities(asso-ciated to relevant statistical correlations)that produce very simple marginal probabilities(such as p and1?p for binary variables)and which simultaneously admit the imposition(as we have done here)of strict additivity of the corresponding entropy.This has been possible for S q.This might be in principle possible as well for other entropic forms.The fact however that,like S BG,S q si-multaneously(i)admits such solutions,(ii)is concave (?q>0),(iii)is Lesche-stable,and(iv)leads to?nite entropy production per unit time[12],constitutes—we believe—a strong mathematical basis for being physi-cally meaningful in the thermostatistical sense.

Acknowledgments

It is with pleasure that I acknowledge very fruitful dis-cussions with S.Abe,C.Anteneodo,F.Baldovin,E.P. Borges,J.P.Crutch?eld,J.D.Farmer,M.Gell-Mann, H.J.Haubold,L.Moyano,A.K.Rajagopal,Y.Sato and D.R.White.I have also bene?ted from a question put long ago by M.E.Vares related to the possible di?erence between W and W e?.

[1]C.Tsallis,Chaos,Solitons and Fractals13,371(2002).

[2]C.Anteneodo and C.Tsallis,Phys.Rev.Lett.80,5313

(1998);C.Anteneodo,Physica A342,112(2004). [3]A.Einstein,Annalen der Physik33,1275(1910)[“Usu-

ally W is put equal to the number of complexions...In order to calculate W,one needs a complete(molecular-mechanical)theory of the system under consideration.

Therefore it is dubious whether the Boltzmann prin-ciple has any meaning without a complete molecular-mechanical theory or some other theory which describes the elementary processes.S=R

英语阅读系列·有趣的字母

A biscuit can be a bear. 一块饼干可以是小熊。 A biscuit can be a butterfly. 一块饼干可以是蝴蝶。 But…If birds are nearby…但是…..如果小鸟恰好在旁边的话…… The biscuits can only be birds.饼干就只是小鸟啦。 Lesson 3 Cool Cat 酷猫卡里 Cary is a cool cat. 卡里是只酷酷的小猫。 Cary can cut carrots like this. 卡里能像这样切胡萝卜。 Cary can climb a coconut tree like this. 卡里能像这样爬树。 Cary can clean a crocodile like this. 卡里能像这样清理鳄鱼。 Cary can catch flies like this. 卡里能像这样抓苍蝇。 Can Cary color like this? 卡里能像这样涂色吗？ Yes! Cary is a cool cat. 是的！卡里是只酷酷的小猫。 Lesson 4 Dancing Dad 爸爸爱跳舞

一年级攀登英语一级A教学计划

一年级攀登英语一级A教学计划一年级攀登英语一级A教学计划 1、激发学生学习英语的兴趣,培养他们学习英语的积极态度,使他们初步建立学习英语的自信心。 2、培养学生具有一定的语感和良好的语音、语调及书写基础,以及良好的学习习惯。 3、培养学生的观察、记忆、思维、想象和创造能力。一年级攀登英语一级A教学计划 (20**学年度第一学期) 本学期我担任一年级4个班的攀登英语教学工作,现我对本学期教学工作做如下计划: 一、学生情况分析一年级的孩子的年龄多为6周岁,具有好奇、好活动、爱表现、善模仿等特点。他们的学习兴趣很浓,接受能力、模仿能力很强,学习习惯初步养成,因此在本学期应注重培养学生良好的学习习惯,训练学生的听说能力,调动他们的自主能动性、积极性,营造互帮互助,共同学习英语的语境。二、教学目标 1、激发学生学习英语的兴趣,培养他们学习英语的积极态度,使他们初步建立学习英语的自信心。 2、培养学生具有一定的语感和良好的语音、语调及书写基础,以及良好的学习习惯。

3、培养学生的观察、记忆、思维、想象和创造能力能在图片、手势、情境等非语言提示的帮助下,听懂清晰的话语和录音。三、教材分析本书共由五部分构成:本册教材学习内容包括:日常英语(Everyday English)、歌曲童谣(Songs and Chants).动画英语(Cartoon English)等 1、日常英语:与学生日常生活密切相关的情景对话,共12个话题,每个话题包括2-4句情景对话。 2、动画英语:以《迪士尼神奇英语》(上)VCD为基本学习材料。共6张VCD,12课,每课4个小节,由学生熟悉和喜欢的迪士尼动画片片段组合而成。 3、歌曲童谣:包括歌曲和童谣两大块,其中有精心制作的传统英文经典歌曲,也有攀登英语独创的童谣日常英语替换内容:以日常英语重点句型为主线,结合歌曲童谣和动画英语中高频出现的词汇进行的开放性交替。 5、节奏英语:将日常英语替换内容编成12段朗朗上口的韵文,并配以节奏明快的背景音乐形成的独特学习内容。四、教学措施 1、考虑到小学生好动爱玩的特点,以活动为课堂教学的主要形式,设计丰富多彩的教学活动,让学生在乐中学、学中用,从而保证学生英语学习的可持续性发展。 2、通过听、说、读、写、唱、游、演、画、做等形式,进行大量的语言操练和练习。

英语阅读系列·有趣的字母

攀登英语阅读系列·有趣的字母北京师范大学“认知神经科学与学习”国家重点实验室攀登英语项目组编著 Lesson1Frank the Rat 大老鼠弗兰克 Frank the rat is in a bag. 大老鼠弗兰克在袋子里呢。 Frank is in a hat. 他在圆顶帽子里呢。 Frank is in a pan. 他在平底锅里呢。 Frank is on an apple. 他在苹果上呢。 Frank is on a bat. 他在球棒上呢。 “Oh, no!” Frank is on a cat. “噢，糟了！”弗兰克落在了猫身上！ Lesson 2 The Biscuits 我们来做饼干 “I’m hungry!”“我饿了！” A biscuit…一块饼干可以是...... A biscuit can be a bus.一块饼干可以是公共汽车。 A biscuit canbe a bike. 和一块饼干可以是自行车。 A biscuit can be a boat. 一块饼干可以是小船。 A biscuit can be a banana. 一块饼干可以是香蕉。 A biscuit can be a bear. 一块饼干可以是小熊。 A biscuit can be a butterfly. 一块饼干可以是蝴蝶。 But…If birds are nearby…但是…..如果小鸟恰好在旁边的话…… The biscuits can only be birds.饼干就只是小鸟啦。 Lesson 3 Cool Cat 酷猫卡里 Cary is a cool cat. 卡里是只酷酷的小猫。 Cary can cut carrots like this. 卡里能像这样切胡萝卜。 Cary can climb a coconut tree like this. 卡里能像这样爬树。 Cary can clean a crocodile like this. 卡里能像这样清理鳄鱼。 Cary can catch flies like this.卡里能像这样抓苍蝇。 Can Cary color like this? 卡里能像这样涂色吗？ Yes! Cary is a cool cat. 是的！卡里是只酷酷的小猫。 Lesson 4 Dancing Dad 爸爸爱跳舞 Dad loves dancing! 爸爸爱跳舞！ Dad is dancing with the desk. 他和桌子一起跳。 Dad is dancing with the duck. 他和鸭子一起跳。 Dad is dancing with the door. 他和门一起跳。 Dad is dancing with the dog. 他和小狗一起跳。 Dad is dancing with the deer. 他和小鹿一起跳。 Where is Dad? 爸爸在哪儿呀？ Wow！哇！ What a dancing Dad! 真是个爱跳舞的爸爸！ Lesson 5 Red Ben 小本的红色世界 Ben likes red. 小本喜欢红色。 Ben paints the eggs red. 他把鸡蛋涂成红色。 Ben paints the eggplants red. 他把茄子涂成红色。 Ben paints the lemons red. 他把柠檬涂成红色。

(完整版)攀登英语-有趣的字母(中英文_纯文字_26篇全)

攀登英语——有趣的字母 A: Frank the rat 老鼠弗兰克 Frank the rat is in a bag. 大老鼠弗兰克在袋子里。 Frank is in a hat. 弗兰克在圆顶帽子里。 Frank is in a pan. 弗兰克在平底锅里。 Frank is on an apple. 弗兰克在苹果里。 Frank is on a bat. 弗兰克在球棒上。 “Oh/no!”F r ank is on a cat. “哦?不！”弗兰克落在了猫身上。 B：The Biscuits 饼干 “I’m h ungry!”“我饿了!” A biscuit, —块饼干 A biscuit can be a bus. —块饼干可以是公共汽车。 A biscuit can be a bike. 可以是自行车。 A biscuit can be a boat. 可以是船。 A biscuit can be a banana. 可以是香蕉。 A biscuit can be a bear.可以是小熊。 A biscuit can be a butterfly.可以是蝴蝶。 But，if birds are nearby,但是，如果鸟儿靠近 The biscuits can only be birds.饼干只能是小鸟啦。 C： Cool Cat酷猫卡里 Cary is a cool cat.卡里是只酷酷的小猫。 Cary can cut carrots like this.卡里能像这样切胡萝卜。 Cary can climb a coconut tree like this. 卡里能像这样爬树。Cary can clean a crocodile like this.卡里能像这样淸洁鳄鱼。Cary can catch flies like this.卡里能像这样抓苍蝇。 Can Cary color like this? 卡里能像这样涂色吗？ Yes!是的 Cary is a cool cat.卡里是只酷酷的小猫。 D: Dancing Dad爸爸爱跳舞 Dad loves dancing!爸爸爱跳舞！ Dad is dancing with the desk.他和桌子一起跳。

攀登英语-有趣的字母(中英文_纯文字_26篇全)教案资料

攀登英语-有趣的字母(中英文_纯文字_26 篇全)

英语阅读系列·有趣的字母

英语阅读系列·有趣的字母攀登英语阅读系列?有趣的字母攀登英语阅读系列·有趣的字母北京师范大学“认知神经科学与学习”国家重点实验室攀登英语项目组编著 Lesson1 Frank the Rat 大老鼠弗兰克 Frank the rat is in a bag. 大老鼠弗兰克在袋子里呢。

Frank is in a hat. 他在圆顶帽子里呢。Frank is in a pan. 他在平底锅里呢。Frank is on an apple. 他在苹果上呢。Frank is on a bat. 他在球棒上呢。“Oh, no!” Frank is on a cat. “噢，糟了！”弗兰克落在了猫身上！ Lesson 2 The Biscuits 我们来做饼干 “I'm hungry!”“我饿了！” A biscuit…一块饼干可以是...... A biscuit can be a bus. 一块饼干可以是公共汽车。 A biscuit can be a bike. 和一块饼干可以是自行车。 A biscuit can be a boat. 一块饼干可以是小船。 A biscuit can be a banana. 一块饼干可以是香蕉。．攀登英语阅读系列?有趣的字母 A biscuit can be a bear. 一块饼干可以是小熊。 A biscuit can be a butterfly. 一

块饼干可以是蝴蝶。 But…If birds are nearby…但是…..如果小鸟恰好在旁边的话…… The biscuits can only be birds.饼干就只是小鸟啦。 Lesson 3 Cool Cat 酷猫卡里 Cary is a cool cat. 卡里是只酷酷的小猫。 Cary can cut carrots like this. 卡里能像这样切胡萝卜。 Cary can climb a coconut tree like this. 卡里能像这样爬树。 Cary can clean a crocodile like this. 卡里能像这样清理鳄鱼。 Cary can catch flies like this. 卡里能像这样抓苍蝇。 Can Cary color like this? 卡里能像这样涂色吗？ Yes! Cary is a cool cat. 是的！卡里是只酷酷的小猫。爸爸爱跳舞Dancing Dad Lesson 4 攀登英语阅读系列?有趣的字母

攀登英语-有趣的字母(中英文_纯文字_26篇全)

攀登英语——有趣的字母 A:Franktherat老鼠弗兰克 Franktheratisinabag.大老鼠弗兰克在袋子里。Frankisinahat.弗兰克在圆顶帽子里。 Frankisinapan.弗兰克在平底锅里。 Frankisonanapple.弗兰克在苹果里。 Frankisonabat.弗兰克在球棒上。 “Oh/no!”F rankisonacat.“哦?不！”弗兰克落在了猫身上。 B：TheBiscuits饼干 “I’mhungry!”“我饿了!” Abiscuit,—块饼干 Abiscuitcanbeabus.—块饼干可以是公共汽车。Abiscuitcanbeabike.可以是自行车。 Abiscuitcanbeaboat.可以是船。 Abiscuitcanbeabanana.可以是香蕉。 Abiscuitcanbeabear.可以是小熊。Abiscuitcanbeabutterfly.可以是蝴蝶。 But，ifbirdsarenearby,但是，如果鸟儿靠近Thebiscuitscanonlybebirds.饼干只能是小鸟啦。 C：CoolCat酷猫卡里 Caryisacoolcat.卡里是只酷酷的小猫。Carycancutcarrotslikethis.卡里能像这样切胡萝卜。Carycanclimbacoconuttreelikethis.卡里能像这样爬树。Carycancleanacrocodilelikethis.卡里能像这样淸洁鳄鱼。Carycancatchflieslikethis.卡里能像这样抓苍蝇。CanCarycolorlikethis?卡里能像这样涂色吗？ Yes!是的 Caryisacoolcat.卡里是只酷酷的小猫。 D:DancingDad爸爸爱跳舞 Dadlovesdancing!爸爸爱跳舞！Dadisdancingwiththedesk.他和桌子一起跳。