On the $q$-Laplace transform in the non-extensive statistical physics

0.11 MB·English

by  Won Sang Chung

On the q-Laplace transform in the non-extensive statistical physics
Won Sang Chung∗
Department of Physics and Research Institute of Natural Science,
College of Natural Science, Gyeongsang National University, Jinju 660-701, Korea
(Dated: January 24, 2013)
Inthispaper,q-Laplacetransformsrelatedtothenon-extensivethermodynamicsareinvestigated
by using the algebraic operation of the non-extensive calculus. The deformed simple harmonic
problem is discussed byusing the q-Laplacetransform.
3
I. INTRODUCTION
1
0
2 Boltzman-Gibbsstatisticalmechanicsshowshowfastmicroscopicphysicswithshort-rangeinteractionhasas effect

n on much larger space-time scale. The Boltzman-Gibbs entropy is given by
a
J W W
1
S =−k p lnp =k ln (1)
3 BG i i
p
2 i=1 i=1 i
X X

] where k is a Boltzmanconstant, W is a totalnumber ofmicroscopicpossibilities of the systemandpi is a probability
h of a given microstate among W different ones satisfying W p =1. When p = 1 , we have S =klnW.
p i=1 i 1 W BG
Boltzman-Gibbstheoryisnotadequateforvariouscomplex,natural,artificialandsocialsystem. Forinstance,this
- P
h theory does not explain the case that a zero maximal Lyapunov exponent appears. Typically, such situations are
t governed by power-laws instead of exponential distributions. In order to deal with such systems, the non-extensive
a
m statistical mechanics is proposed by C.Tsallis [1,2]. The non-extensive entropy is defined by
[ S =k(ΣWpq−1)/(1−q) (2)
q i i

1
v The non-extensive entropy has attracted much interest among the physicist, chemist and mathematicians who study
0 the thermodynamicsofcomplex system[3]. When the deformationparameterq goes to 1,Tsallisentropy(2) reduces
8 to the ordinary one (1). The non-extensive statistical mechanics has been treated along three lines:
4
5
1. Mathematical development [4, 5, 6]
.
1 2. Observation of experimental behavior [7]
0
3. Theoretical physics ( or chemistry) development [8]
3
1
: The basis of the non-extensive statistical mechanics is q -deformed exponential and logarithmic function which
v
is different from those of Jackson’s [9]. The q-deformed exponential and q-logarithm of non-extensive statistical
i
X mechanics is defined by [10]
r
a t1−q−1
ln t= , (t>0) (3)
q
1−q
1
eq(t)=(1+(1−q)t)1−q, (x,q ∈R) (4)
where 1+(1−q)t>0.
Fromthe definition of q-exponentialand q-logarithm,q-sum, q-difference,q-productand q-ratioare definedby [ 5,
6]
x⊕y = x+y+(1−q)xy
x−y
x⊖y =
1+(1−q)y
∗Electronicaddress: [email protected]
2
x⊗y = [x1−q+y1−q−1]1−1q
x⊘y = [x1−q−y1−q+1]1−1q (5)
It can be easily checked that the operation ⊕ and ⊗ satisfy commutativity and associativity. For the operator ⊕,
the identity additive is 0, while for the operator ⊗ the identity multiplicative is 1. Indeed, there exist an analogy
between this algebraic system and the role of hyperbolic space in metric topology [11]. Two distinct mathematical
tools appears in the study of physical phenomena in the complex media which is characterized by singularities in a
compact space [12].
For the new algebraic operation, q-exponential and q-logarithm have the following properties:
ln (xy) = ln x⊕ln y e (x)e (y)=e (x⊕y)
q q q q q q
ln (x⊗y) = ln x+ln y e (x)⊗e (y)=e (x+y)
q q q q q q
ln (x/y) = ln x⊖ln y e (x)/e (y)=e (x⊖y)
q q q q q q
ln (x⊘y) = ln x−ln y e (x)⊘e (y)=e (x−y) (6)
q q q q q q
From the associativity of ⊕ and ⊗, we have the following formula :
1
t⊕t⊕t⊕···⊕t= {[1+(1−q)t]n−1} (7)
1−q
n times
| {z }
t⊗n =t⊗t⊗t⊗···⊗t=[nt1−q−(n−1)]1−1q (8)
n times
| {z }
II. Q-LAPLACE TRANSFORM
In this section, we find the q-Laplace transform related to the non-extensive thermodynamics. From the relation
e (x)e (y)=e (x⊕y), (9)
q q q
the q-analogue of enx, (n∈Z) is given by
1
eq(n⊙t)=[eq(t)]n =eq [(1+(1−q)t)n−1] =(1+(1−q)t)1−nq (10)
1−q
(cid:18) (cid:19)
Then we have
e (0⊙t)=1 (11)
q
and the inverse of e (n⊙t) is e ((−n)⊙t). For this adoption, q- Laplace kernel is defined by
q q
1
eq((−s)⊙t)=[eq(t)]−s =eq [(1+(1−q)t)−s−1] =(1+(1−q)t)−1−sq (12)
1−q
(cid:18) (cid:19)
Therefore q-Laplace transform is defined by
∞
L (F(t))= [e (t)]−sF(t)dt, (s>0) (13)
s q
Z0
Limiting q →1, the eq.(13) reduces to an ordinary Laplace transform. Form now on we assume that s is sufficinetly
large.
Since, for two functions F(t) and G(t), for which the integral exist
L (aF(t)+bG(t))=aL (F(t))+bL (G(t)), (14)
s s s
the q-Laplace transform is linear.
For F(t)=tN, (N =0,1,2,···) , we have the following result.
3
Theorem 1 For sufficiently large s , when q <1, the following holds:
N!
L (tN)= (15)
s
(s;1−q)
N+1
where
1 (n=0 )
(a;Q)n = n (a−kQ) (n≥1) (16)
(cid:26) k=1
Q
Proof. Let us assume that the eq.(15) holds for tN. Then,
∞
L (tN+1) = [e (t)]−stN+1dt
s q
Z0∞
= (1+(1−q)t)−1−sqtN+1dt
Z0 ∞
= (1+(1−q)t)−1−sq+1tN+1 + N +1 ∞(1+(1−q)t)−s−1(−1−qq)tNdt
1−q−s s−(1−q)
" #0 Z0
N +1
= L (tN+1)
s−(1−q) s−(1−q)
(N +1)N!
=
(s−(1−q))(s−(1−q);1−q)
N+1
(N +1)!
= (17)
(s;1−q)
N+2
In a similar way, we can obtain the q-Laplace transform for e (a⊙t) which is given by
q
e (a⊙t)=[e (t)]a
q q
Theorem 2 For sufficiently large s , when q <1, the following holds:
1
L (e (a⊙t))= (18)
s q
s−a−(1−q)
Proof. It is trivial.
In order to obtain q-Laplace transform for the trigonometric function, we need q-analogue of Euler identity. The
q-Euler formula is given by
e (ia⊙t)=C (a⊙t)+iS (a⊙t), (19)
q q q
where q-cosine and q-sine functions are defined by
a
C (a⊙t)=cos( ln(1+(1−q)t))
q
1−q
a
S (a⊙t)=sin( ln(1+(1−q)t)) (20)
q
1−q
and we used the following identity.
pi =eilnp =coslnp+isinlnp (21)
4
Indeed, q-cosine and q-sine functions can be expressed in terms of q-exponential as follows:
1
C (a⊙t)= [e (ia⊙t)+e ((−ia)⊙t)
q q q
2
1
S (a⊙t)= [e (ia⊙t)−e ((−ia)⊙t) (22)
q q q
2i
Then we have the q-Laplace transform for q-sine and q-cosine functions :
Theorem 3 For sufficiently large s , when q <1, the following holds:
s−(1−q)
L (C (a⊙t))=
s q (s−(1−q))2+a2
a
L (S (a⊙t))= (23)
s q (s−(1−q))2+a2
Proof. It is trivial.
The eq.(23) can be written in terms of the ordinary sine and cosine functions :
a s−(1−q)
L (cos[ ln(1+(1−q)t)])=
s 1−q (s−(1−q))2+a2
a a
L (sin[ ln(1+(1−q)t)])= (24)
s 1−q (s−(1−q))2+a2
The q-sine function and q-cosine function have the following zeros:
S (1⊙t )=S (t )=0, C (1⊙u )=C (u )=0, (25)
q n q n q n q n
where
tn =lnqenπ, un =lnqe(n+12)π, n∈Z (26)
From the zeros of the q-sine function and q-cosine function, we have the following Theorem:
Theorem 4 For sufficiently large s , when q <1, the following holds:
∞
t t
S (t)=t 1− 1− (27)
q ln ejπ ln e−jπ
j=1(cid:18) q (cid:19)(cid:18) q (cid:19)
Y
∞
t t t
C (t)= 1− 1− 1− (28)
q ln e−π/2 ln e(j+1/2)π ln e−(j+1/2)π
(cid:18) q (cid:19)j=1(cid:18) q (cid:19)(cid:18) q (cid:19)
Y
Proof. From the zeros of the q-sine function , we can set
∞
S (t) t t
q
=A 1− 1−
t ln ejπ ln e−jπ
j=1(cid:18) q (cid:19)(cid:18) q (cid:19)
Y
5
Because limt→0 Sqt(t) =1, we have A=1, which proves the eq.(27). Similarly we can easily prove the eq.(28).
Theorem 4 can be also written as follows :
Theorem 5 For sufficiently large s , when q <1, the following holds:
∞ 2
(1−q)t
S (t)=t 1+(1−q)t− (29)
q  2sinh(j(1−q)π)! 
jY=1 2
 
∞ 2
t (1−q)t
C (t)= 1− 1+(1−q)t− (30)
q (cid:18) lnqe−π/2(cid:19)j=1" (cid:18)2sinh(12(1−q)(j+1/2)π)(cid:19) #
Y
Proof. It is trivial from the formula coshx−1=2sinh2 x.
2
III. Q-LAPLACE TRANSFORM AND DIFFERENTIAL EQUATION
Nowwediscusstheq-differentialequation. Themainpurposeofq-Laplacetransformisinconvertingq-differential
equation into simpler forms which may be solvedmore easily. Like the ordinary Laplace transform, we can compute
the q-Laplace transformation of derivative by using the definition of the q-Laplace trnsform, which is given by
Ls(F′(t))=sLs+1−q(F(t))−F(0) (31)
An extension gives
L (F′′(t))=s(s+1−q)L (F(t))−sF(0)−F′(0) (32)
s s+2(1−q)
Generally, we have following theorem :
Theorem 6 For sufficiently large s , when q <1, the following holds:
n−1
Ls(F(n)(t))=[s;1−q]nLs+n(1−q)(F(t))− [s;1−q]n−1−iF(i)(0), (33)
i=0
X
where F(0)(0)=F(0) and
1 (n=0 )
[a;Q]n = n (a+kQ) (n≥1) (34)
(cid:26) k=1
Q
Proof. Let us assume that the eq.(34) holds for n. Then,
∞
Ls(F(n+1)(t)) = (1+(1−q)t)−1−sqF(n+1)(t)dt
Z0
= sLs+1−q(F(n)(t))−F(n)(0)
n−1
= s{[s+1−q;1−q]nLs+(n+1)(1−q)(F(t))− [s+1−q;1−q]n−1−iF(i)(0)}−F(n)(0)
i=0
n X
= [s;1−q]n+1Ls+(n+1)(1−q)(F(t))− [s;1−q]n−iF(i)(0) (35)
i=0
X
6
We have another formula for the q-Laplace transform of derivative as follows:
F(t)
L (F′(t))=sL −F(0) (36)
s s
1+(1−q)t
(cid:18) (cid:19)
An extension gives
F(t)
L (F′′(t))=s(s+1−q)L −sF(0)−F′(0) (37)
s s (1+(1−q)t)2
(cid:18) (cid:19)
Generally, we have the following theorem:
Theorem 7 For sufficiently large s , when q <1, the following holds:
n−1
F(t)
Ls(F(n)(t))=[s;1−q]nLs (1+(1−q)t)n − [s;1−q]n−1−iF(i)(0) (38)
(cid:18) (cid:19) i=0
X
Proof. It is not hard to prove Theorem 7.
Comparing Theorem 6 with Theorem 7, we have the following Lemma:
Lemma 8 For sufficiently large s , when q <1, the following holds:
F(t)
L =L (F(t)) (39)
s (1+(1−q)t)n s+n(1−q)
(cid:18) (cid:19)
Proof. It is trivial.
With the help of q-Laplace transform of the derivative, we can solve some differential equation. It is worth noting
that e (t) is not invariant under the ordinary derivative, instead it obeys
q
d 1
(e (t))= e (t) (40)
q q
dt 1+(1−q)t
Now consider the following differential equation:
F(t)
F′(t)= , F(0)=1 (41)
1+(1−q)t
It is evident that e (t) is a solution of the eq.(40).
q
Let us consider the vertical motion of a body in a resisting medium in which there again exists a retarding force
proportional to the velocity. Let us consider that the body is projected downward with zero initial velocity v(0)=0
in a uniform gravitationalfield. The equation of motion is then given by
d
m v =mg−kv(t) (42)
dt
This equation is not solved by using the q-Laplace transform , instead we solve the following equation :
d v(t)
m v =mg−k (43)
dt 1+(1−q)t
The solution of the eq.(43) is then given by
g
v(t)= 1−q+ k (1+(1−q)t−[eq(t)]−mk) (44)
m
7
Similarly, we can modify the harmonic problem whose equation of motion is given by
2
d x(t)
m x(t)=−mw2 , (45)
dt (1+(1−q)t)2
(cid:18) (cid:19)
where x(0)=A, dx (0)=0. The solution of the eq.(45) is then given by
dt
(cid:0) (cid:1)
q−1 q−1 q−1
x(t)=A(1+(1−q)t)2C3−2q rw2−( 2 )2⊙t)!+ w2−(q−1)2S3−2q rw2−( 2 )2⊙t! (46)
 2 
q
Theeqs.(43)and(45)seemtobetooartificialduetothefactor1+(1−q)t. Instead,wecanintroducetheq-derivative
[6] instead of the ordinary time derivative as follows:
F(t)−F(s) dF
D F(t)= lim =[1+(1−q)t]
t
s→t t⊖s dt
The Leibniz rule for q-derivative is as follows:
D [F(t)G(t)]=D [F(t)]G(t)+F(t)D [G(t)] (47)
t t t
Then the eq.(43) is replaced as follows :
mD v =mg−kv(t) (48)
t
The solution of the eq.(48) is then given by
mg
v(t)= (1−[eq(t)]−mk) (49)
k
Similarly, the eq.(45) is replaced as follows:
mD2x(t)=−mw2x(t) (50)
t
Using the q-Laplace transform, we get the solution of the eq.(50) :
x(t)=AC (w⊙t) (51)
q
Obtaining these solutions, we used the following theorem:
Theorem 9 For sufficiently large s , when q <1, the following holds:
n−1
Ls((1+(1−q)t)nF(n)(t))=[s−(n+1)(1−q);1−q]nLs(F(t))− [s−(n+1)(1−q);1−q]n−1−iF(i)(0) (52)
i=0
X
Proof. Let us assume that the eq.(52) holds for n. Then,
∞
Ls((1+(1−q)t)nF(n+1)(t)) = (1+(1−q)t)−1−sq+n+1F(n+1)(t)dt
Z0
= (s−(n+1)(1−q))L ((1+(1−q)t)nF(n)(t))−F(n)(0)
s
= (s−(n+1)(1−q)){[s−(n+1)(1−q);1−q] L (F(t))
n s
n−1
− [s−(n+1)(1−q);1−q]n−1−iF(i)(0)}−F(n)(0)
i=0
X
= [s−(n+2)(1−q);1−q] L (F(t))
n+1 s
n
− [s−(n+2)(1−q);1−q]n−iF(i)(0) (53)
i=0
X
8
The eq.(50) is also obtained by using the variational method whose Lagrangianis given by
1
L= dt (D x)2−U(x) (54)
t
2
Z (cid:18) (cid:19)
The equation of motion is then given by
∂L ∂L
D − =0, (55)
t
∂(D x) ∂x
(cid:18) t (cid:19)
where the momentum p is defined by
∂L
p= (56)
∂(D x)
t
For the harmonic potential U = 1mw2x2 , we have the eq.(50). This equation is rewritten by
2
(1+(1−q)t)2x¨+(1−q)(1+(1−q)t)x˙ =−w2x (57)
Replacing
1
η = ln(1+(1−q)t), (58)
1−q
the eq.(57) is then as follows :
∂2x
=−w2x(η) (59)
∂η2
Solving the eq.(59), we have
x(t)=Acoswη =AC (w⊙t) (60)
q
Here, let us investigate the times t when a body goes back to the initial position. This time is determined by
n
2πn
tn =lnqe w , (t=0,1,2,···) (61)
When q <1 and w>0, we have the following inequality:
tn+1−tn >tn−tn−1 (62)
Thus, the time when a body goes back to the initial position keeps increasing.
IV. CONCLUSION
Inthispaper,weusedthealgebraicoperationanddifferentialcalculusrelatedtothenon-extensivethermodynamics
to investigate the q-Laplace transform. We used the q-Laplace transform to solve some differential equation such as
harmonic oscillator problem.
We think that this work will be applied to some q-differential equation which might appear in the study of the
non-extensivestatisticalmechanics. We hope that these workandtheir relatedtopics willbe clearinthe nearfuture.
Refernces
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[6] E. Borges,Physica A 340 (2004) 95.
[7] S.Abe, A.Rajagopal, Science 300 (2003) 249.
[8] V.Latora, A,Rapisarda, A.Robledo, Science 300 (2003) 250.
[9] F. Jackson, Mess.Math. 38 , 57 (1909).
[10] C.Tsallis, Quimica Nova 17 (1988) 479.
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