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Temperature and pressure in nonextensive thermostatistics

Temperature and pressure in nonextensive thermostatistics
Temperature and pressure in nonextensive thermostatistics

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Temperature and pressure in nonextensive thermostatistics Q.A.Wang,L.Nivanen,A.Le M′e haut′e ,Institut Sup′e rieur des Mat′e riaux et M′e caniques Avanc′e s ,44,Avenue F.A.Bartholdi,72000Le Mans,France and M.Pezeril Laboratoire de Physique de l’′e tat Condens′e ,Universit′e du Maine,72000Le Mans,France Abstract The de?nitions of the temperature in the nonextensive statistical thermodynamics based on Tsallis entropy are analyzed.A de?nition of pressure is proposed for nonadditive systems by using a nonadditive e?ective volume.The thermodynamics of nonadditive photon gas is discussed on this basis.We show that the Stefan-Boltzmann law can be preserved within nonextensive thermodynamics.PACS :05.20.-y,05.70.-a,02.50.-r 1Introduction The nonextensive statistical mechanics (NSM)[1]based on Tsallis entropy is believed by many to be a candidate replacing Boltzmann-Gibbs statistics (BGS)for nonextensive or nonadditive systems which may show probability distributions di?erent from that of BGS.So according the common belief,

NSM,just as BGS,should be able to address thermodynamic functions and intensive variables like temperature T ,pressure P ,chemical potential μetc.Although the Legendre transformation between the thermodynamic functions is preserved in some versions of NSM with sometimes certain deformation,the de?nition of intensive variables is not obvious if the thermodynamic functions

1

such as entropy S,energy U or free energy F are nonadditive.There are sometimes misleading calculations usingβ=1/T= ?S

?V T or P=1V(for photon gas)without specifying the nonadditivity(or additivity)of each functions or noticing that additive internal energy U and volume V associated with nonadditive S and F will lead to non-intensive temperature or pressure which would make the thermodynamic equilibrium or stationarity impossible in the conventional sense.

On the other hand,within NSM,due to the fact that di?erent formalisms are proposed from di?erent statistics or information considerations,thermo-dynamic functions do not in general have the same nonadditive nature in di?erent versions of NSM.This has led to di?erent de?nitions of,among others,a physical or measurable temperatureβp which is sometimes equal toβ[2],sometimes equal toβmultiplied by a function of the partition func-tion Z q?1[3,4,5,6,7]or Z1?q[8,9]which keepsβp intensive,where q is the nonadditive entropy index1,or sometimes de?ned by deformed entropy and energy[9,10,11].This situation often results in confusion and misleading discussions of these temperatures[12]or other intensive variables[13],with-out knowing or mentioning the validity conditions relevant to them and the risk to have non intensive temperature or pressure.

The present paper tries to make a state of the art on this subject with brief discussions of the speci?cities of each formalism of NSM and the relevant consequences.It is hoped that this paper may o?er to the reader a global view of the situation and of some important questions which are still matters of intense investigation.

2The?rst de?nition of physical temperature of NSM

We look at a composite system containing two subsystems A and B,all having the same q as nonadditive entropy index.The entropy nonadditivity of the total system is given by

S(A+B)=S(A)+S(B)+(1?q)S(A)S(B).(1)

,(q∈R)[1]

1?q

2

This relationship is intrinsically connected with the product joint probability

p ij(A+B)=p i(A)p j(B),(2) or inversely,where i or j is the index of physical states for A or B.Eq.(2)has been intuitively taken as an argument for the independence of A and B and for the energy additivity of A+B.This additivity o?ers the?rst possibility to establish zeroth law and to de?ne temperature within NSM[3,4,5,6,7]. The intensive physical temperature is de?ned as

βp=1

w i p q i?S w i p q iβ.(3)

This de?nition is an universal model of NSM and not connected to any spe-ci?c statistical formalism.

If thisβp is applied to NSM having typically the power law distribution

p i=1

a with[·]≥0(4)

where E i is the energy of a system at state i and a is1?q or q?1according to the maximum entropy constraints of the formalism[8,14],there may be in general a con?ict between the product joint probability and the energy additivity condition due to the nonadditive energy E i(A+B)=E i(A)+ E j(B)?aβp E i(A)E j(B).So the validity of this thermostatistics strongly lies on neglecting E i(A)E j(B).

A mathematical proof[3]shown that this was possible,for a N-body sys-tem,if and only if q<1and N→∞.This is not to be forgotten.For the systems with q>1or with?nite size without thermodynamic limits,this additive energy model is not justi?ed.

Especially,when this model is applied to the formalism of NSM de-duced from the normalized expectation given by the escort probability U= i p q i E i

Z [1?(1?q)βp(E i?U)]1

Z

[1?(1?q)

β

1?q,(5)

Eq.(3)becomes

βp=1

?U

=Z q?1β.(6)

In this case,βp is not to be confounded withβalthough we have hereβ=?S

3

The ?rst formalism of NSM The ?rst formalism[1]of NSM maximizes entropy under the constraint U = i p i E i with normalized p i .The distribution function is given by

p i =1q ?1.(7)

The product probability implies the following nonadditivity of energy :

E i (A +B )=E i (A )+E j (B )?(q ?1)βp E i (A )E j (B )(8)and U (A +B )=U (A )+U (B )?(q ?1)βp U (A )U (B ).The temperature of this formalism is still given by Eq.(6)as brie?y discussed in [8].

The thermodynamic relations can be deduced from the basic expression of entropy of this formalism

S =

Z 1?q ?1q ?1+βp U where S p is an “auxiliary entropy”introduced

to write the generalized heat as dQ =T p dS p .The ?rst law reads dU =T p dS p ?dW .The free energy F is de?ned as

F =U ?T p S p =?T p Z q ?1?1

?T p V and Eqs.(5)and (10)with

Z = i [1?(q ?1)βp E i ]1q ?1=1? i p 2?q i

4The second formalism of NSM with unnor-

malized expectation

This formalism is deduced from the entropy maximum under the constraint U = i p q i E i with normalized p i [15].The distribution function is given by

p i =11?q .(12)

and the nonadditivity of energy by E i (A +B )=E i (A )+E j (B )?(1?q )βp E i (A )E j (B )and

U (A +B )=U (A )Z 1?q (B )+U (B )Z 1?q (A )+(q ?1)βp U (A )U (B ).(13)As discussed in [2],this is the only formalism of NSM in which the math-ematical framework of the thermodynamic relationships is strictly identical to that of BGS with βp =β.The heat is given by dQ =T dS ,the ?rst law by dU =T dS ?dW and the free energy by

F =U ?T S =?T Z 1?q ?1

Z [1?(1?q )βp E i ]

11?q .

The nonadditivity of energy is given by

U (A +B )=U (A )+U (B )+(q ?1)βp U (A )U (B ).

5

The de?nition of the physical temperatureβp in this formalism is discussed in[8,9]and reads

βp=Z1?q

?S

q?1

+βp Z q?1U(17) or S p=Z1?q S=Z1?q?1

1?q

,(18) dF=?S p dT p?dW where dW is the work done by the system.S p is given by[12]

S p=? i p q i p q?1i?1q?1.(19) which is concave only for q>1/2so that not to be maximized to get dis-tribution functions although its maximum formally leads to p i∝[1?(q?1)βp E i]1

?T although we can write F=U?T p S p=U?T S.In

addition,Z is not derivable with respect toβsince it is a self-referential function when written as a function ofβ.This calculation can be done for S only in the second formalism with unnormalized expectation and normalized probability associated toβ=1/T=?S

p i

and e i=ln[1+(q?1)βp E i]

Z e?βp e i

which is identical to Eq.(15).Within this framework,the temperature is β=?s

(q?1)βp

.(20)

6

In this deformed formalism,everything is just as in BGS.This mathematical framework has been used for the equilibrium problem of the systems having

di?erent q’s[10,11].

6Systems having di?erent q’s

The reader should have noticed that all the above discussions are based on the entropy nonadditivity given by Eq.(1)which is valid only for systems having the same index q.For systems A,B and A+B each having its own

q,this relationship breaks down even if the product joint probability holds. So for establishing the zeroth law,we need more general nonadditivity for entropy.A possible one is proposed as follows[10]:

(1?q A+B)S(A+B)=(1?q A)S(A)+(1?q B)S(B)(21)

+(1?q A)(1?q B)S(A)S(B)

which recovers Eq.(1)whenever q A+B=q A=q B.

The establishment of zeroth law for this case has been discussed by using

the unnormalized expectations just as in the second formalism of NSM,i.e., u= i p q i e i with i p i=1[10],or u= i p i e i with i p q i=1[11].The reason for this is that these unnormalized expectations allow one to split the ther-

modynamics of the composite systems into those of the subsystems through

the generalized product joint probability p q A+B

ij (A+B)=p q A i(A)p q B i(B)if

i p i=1[or p ij(A+B)=p i(A)p i(B)if i p q i=1].This thermodynamic splitting is just a necessary condition for the statistical interpretation of the zeroth law.

In this case,the deformed entropy s and energy u are not necessarily additive as in the case of an unique q.In fact,when u= i p q i e i with i p i=1is used,their nonadditivities are given as follows

q A+B s(A+B)

i p q A i(A)+q B s(B)

ij p q A+B ij(A+B)=q A u(A) j p q B j(B).(23)

7

The temperature is given byβp=β=?s

?U here U= i p q i E i.The

thermodynamic relations are the same as in the second formalism of NSM or in BGS.

This de?nition of temperature can be discussed in another way.From Eq.(21),for a stationary state of(A+B)extremizing R(A+B),we have

(q A?1)dS(A)

i p i(B)=0.(24)

Now using the above mentioned product joint probability and the relationship i p q i=Z1?q+(1?q)βU,we get(1?q A)β(A)dU(A) i p i(B)=0which suggests following energy nonadditivity

(1?q A)dU(A)

i p i(B)=0(25)

as the analogue of the additive energy dU(A)+dU(B)=0of Boltzmann-Gibbs thermodynamics.Eq.(27)and Eq.(28)lead toβ(A)=β(B).

Summarizing the de?nitions of temperature,we haveβp=Z q?1β= Z q?1?S

?U for the normalized expectations

U= i p q i E i with i p q i=1.On the other hand,βp=β=?S

?V T.If we want the pressure to be intensive,V will be nonadditive.This is a delicate choice to make since nonadditive volume is nontrivial and not so easy to be understood as nonadditive energy or entropy.For a standard system,we tend to suppose additive volume as well as additive particle number.However,in view of the fact that the work dW is in general nonadditive,additive volume implies non intensive pressure P,which is impossible if the equilibrium or stationary state is established in the conventional sense for,e.g.a gas of

8

photons or of other particles.So,?rst of all,for the following discussion, let us suppose an intensive pressure P,i.e.,P(A)=P(B)at equilibrium or stationarity.

Intensive P implies nonadditive V.If one wants to suppose additive volume(the real one)and particle number N,V must be regarded as an e?ective volume,as a function of the real volume V p supposed additive.

In this case,a question arises about the nature of the work dW which is no more proportional to the real volume dV p.Is it a real work?Our answer is Yes because dW is supposed to contribute to the energy variation dU or dF according to the?rst law.A possibility to account for this work is that,for a nonextensive or nonadditive system,e.g.,a small system or a heterogeneous system,the surface/interface e?ects on the total energy, compared with the volume e?ect,are not negligible.When the pressure makes a small volume variation dV p,the work may be dW=P dV p+dWσwhere dWσis the part of work related to the surface/interface variation dσ. In general,the relationship dWσ~dσshould depend on the nature and the geometry of the system of interest.If we suppose a simple case where dWσ=αP d(σθ)andσ=γVηp(α,γ,ηandθare certain constants),the work can be written as dW=P dV p+αγP d(Vηθp)=P d[V p+αγVηθp]which means V=V p+αγVηθp.This example shows that a nonadditive e?ective volume can be used for nonextensive systems to write the nonadditive work in the form dW=P dV,just as in the conventional additive thermodynamics. 7.1A de?nition of pressure for NSM

Now let us come back to NSM.To determine the nonadditivity of the e?ective volume V with additive real volume V p,one has to choose a given version of NSM with given nonadditivity of entropy and energy.Without lose of generality,the following discussion will be made within the second formalism of NSM.From the entropy de?nition and nonadditivity Eq.(1)and the energy nonadditivity Eq.(13),we can write,at equilibrium or stationarity, dS(A+B)=[1+(1?q)S(B)]dS(A)+[1+(1?q)S(A)]dS(B)(26)

= i p q i(B) ?S(A)?V(A) U dV(A)

+ i p q i(A) ?S(B)?V(B) U dV(B)

9

= i p q i(B) ?S(A)?U(B) V dU(B)

+ i p q i(B) ?S(A)?V(A) S dV(A)

+ i p q i(A) ?S(B)?V(B) S dV(B)

=β P(A) i p q i(B)dV(A)+P(B) i p q i(A)dV(B) =0.

Here we have used dS(A) i p q i(B)=0,dU(A) i p q i(B)=0[2],and ?S?V S ?S i p q i(A)+dV(B)

i p q i is additive,just as dS i p q i.

It can be checked that this kind of calculation is also possible within other versions of NSM as long as the energy nonadditivity is determined by the product joint probability which is in turn a consequence of the entropy nonadditivity Eq.(1)or Eq.(21)postulated for Tsallis entropy.

7.2About nonadditive photon gas

Now let us suppose a nonadditive photon gas,which is possible when emission body is small.For example,the emission of nanoparticles or of small optical cavity whose surface/interface e?ect may be important.We have seen in the above paragraph that dU,dS and dV should be proportional to each other.This can be satis?ed by U=f(T)V and S=g(T)V.In addition,

we admit the photon pressure given by P=U

3f(T).From the?rst law

dU=T dS?P dV,we obtain

V ?f

?T

dT+gdV)?

1

?T =T?g

3

f=T g leading to1

?T

=4f ?V

)T=(?P

3

?f

in the conventional thermodynamics.This is because the thermodynamic functions here,though nonadditive,are nevertheless“extensive”with respect to the e?ective volume.This result contradicts what has been claimed for blackbody radiation on the basis of non intensive pressure[13],and is valid as far as the pressure is intensive.

Is intensive pressure always true?The?nal answer of course depends on experimental proofs which are still missing as far as we know.If pressure may be non intensive for nonadditive or nonextensive systems,the whole theory of thermodynamics must be reviewed.

8Conclusion

In summery,we have analyzed all the temperature de?nitions of NSM we can actually?nd in the literature.A de?nition of intensive pressure is proposed for nonextensive thermodynamics by using a nonadditive e?ective volume. The thermodynamics of a nonadditive photon gas is discussed on that ba-sis.It is shown from purely thermodynamic point of view that the Stefan-Boltzmann law can be valid within NSM in this case.

References

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15,537(2003)

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S.Abe,S.Martinez, F.Pennini and A.Plastino,Phys.Lett.A, 281,126(2001)

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S.Martinez,F.Pennini,and A.Plastino,Physica A,295,246(2001) [6]S.Martinez, F.Nicolas, F.Pennini,and A.Plastino,Physica A,

286,489(2000)

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[7]Raul Toral,Physica A,317,209(2003)

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di?erent nonextensive systems,submitted(2003),cont-mat/0310056 [11]Q.A.Wang,L.Nivanen,A.Le M′e haut′e and M.Pezeril,About the

stationary state of nonextensive systems of di?erent q indices,submitted (2003),cont-mat/0305398

[12]J.A.S.Lima,J.R.Bezerra and R.Silva,Chaos,Solitons&Fractals,

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[13]M.Nauenberg,Phys.Rev.E,67,036114(2002);

M.Nauenberg,Reply to C.Tsallis’“Comment on critique of q-entropy for thermal statistics by M.Nauenberg”,cond-mat/0305365v1

[14]F.Pennini,A.R.Plastino and A.Plastino,Physica A,258,446(1998);

C.Tsallis,R.S.Mendes and A.R.Plastino,Physica A,261,534(1999)

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[16]A.R′e nyi,Calcul de probabilit′e,(Dunod,Paris,1966)

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Thank you very much! Is this your glass No it isn't It is my glass. Here you are. Thank you very much! 情景二 模板对话: What's this It's a cake. What's that It's an orange. Are you hungry Oh! nice. May I have some Excuse me! Is that your apple May I have it 对话一: What's this Edda It's a cake. What's that It's an orange. 对话二: Edda! are you hungry Yes, what's this It's a cake. Oh! nice. 对话三: Mom, What's this It's a melon. May I have some Of course. 对话四: Excuse me, Is that your apple May I have it Please!

大学生心理健康 口语对话

A: Hey xiaohan, have you heard that a student of FuDan University was murdered by his roommate? B: Yes, I have. I was shocked when I first heard the news. I even can’t image that a college student is killed by his roommate, they should be the best friends. A:So, who do you think should take the responsibility for this? B: Uh, The various factors causing problem, but I think the most important reason is that we have pay to much attention to students’ academic achievement, and have ignored the students’ psychological health education. A:I couldn’t agree more with you,there is no denying that as more and more students go to college for further educationpsychological problems are becoming serious among the college students. Especially for the freshmen, who are unadapted to the new surroundings of college life and at a loss for the new environment, is easier to sufferfrom psychological problem. B: I agree totally. And as far as I am concerned, college students are in the social-changing stage,their ways of dealing with problems andpersonal value are in contradiction with the social requirement. What’s more, various of temp tation coming from the society makes them suffer from the role conflict.

(浅谈基本单位名录库管理维护工作)

浅谈基本单位名录库管理维护工作 基本单位名录库是包含辖区内所有法人单位和产业活动单位的基本标识、主要属性、基本状态和主要经济指标等信息资料的数据库。科学管理基本单位名录库是有效开展统计调查、确保统计数据质量的基础。随着国家统计局统计工作现代化“四大工程”建设的不断推进,基本单位名录库基础作用的重要性也日益凸显。 一、基本单位名录库的作用 基本单位名录库是进行联网直报的“字典库”。基本单位名录库具有鉴别统计调查单位基本属性,界定单位报表类别的功能。单位必须在名录库中存在,且通过“四上”审批程序对企业基本信息审核通过之后,才能进入企业一套表平台。这种“要有数、先进库,要进库、走程序”的原则,能有效保证进入企业一套表单位来源的可靠性、真实性,同时从源头上避免企业数据专业间、地区间重复。 基本单位名录库是抽选样本和推算总体的“样本框”。名录库是规模以下工业、限额以下贸易业等各种抽样调查的样本平台,为样本轮换提供总体样本框,利用样本资料和名录库资料才能科学推断总体情况。 基本单位名录库是核查各类普查单位的“清单”。对各项普查活动来讲,基本单位名录库是准确确定普查对象总体、规范各类普查表实施范围的基础,是保证普查工作有序推进的必要条件。 基本单位名录库是开展统计服务的“基石”。通过基本单位的总

体状况,分析基本单位的地区分布、行业布局和产业结构等情况,形成政府进行宏观决策的基础信息库。这些基础信息是政府制定经济政策、调整经济结构、规划城乡建设的重要依据。 二、我县基本单位名录库管理维护工作现状 在非普查年份,省局根据税务部门按季度提供的企业纳税信息分区县进行整理,与工商信息合并整理后下发到县区。县统计局普查中心充分利用相关部门提供和专业调查获得的季度新增、变更、注销资料,按乡镇进行分解,及时反馈给乡镇。并按照分级负责、共同维护的工作机制,对全县新增单位进行调查、录入,对变更单位的各项指标进行维护,对注销单位的资料进行核实剔除,对跨地区、跨专业的单位重复问题进行处理,确保名录库管理维护的及时性、准确性和完整性。同时,结合名录库更新情况,进一步规范流程、强化责任,加强部门协作,做好“一套表”调查单位月度审批入库的工作。县统计局及时协调申报过程中出现的问题,严格把关审批资料,确保上报单位资料的准确性,确保符合条件的企业及时入库。 三、基本单位名录库管理维护中存在的主要问题 (一)部门软件通用性低,部门资源共享共建难度大。目前,统计部门更新维护名录库,所用资料主要从编办、民政、国税、地税、工商五部门收集。各部门根据自己的职能范围,建立了不同的单位信息库。但由于各部门的独立性、特殊性,各自使用的信息系统不一致,各部门基本单位的定义标准不统一和基本单位名录库指

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