Scale invariant forces in 1d shuffled lattices

0.24 MB·English

by  Andrea Gabrielli

APS/123-QED
Scale invariant forces in 1d shuffled lattices
Andrea Gabrielli1,2
1 Istituto dei Sistemi Complessi - CNR, Via dei Taurini 19, 00185-Rome (Italy) and
2
SMC-INFM, Department of Physics, University “La Sapienza” of Rome, P.le Aldo Moro 2, 00185-Rome, Italy
(Dated: February 2, 2008)
6
0 In this paper we present a detailed and exact study of the probability density function P(F) of
0 thetotalforceF actingonapointparticlebelongingtoaperturbedlatticeofidenticalpointsources
2 of a power law pair interaction. The main results concern the large F tail of P(F) for which two
casesaremainlydistinguished: (i)Gaussian-likefastdecreasingP(F)forlatticewithperturbations
n
forbidding any pair of particles to be found arbitrarily close to oneeach other; (ii) L´evy-likepower
a
law decreasing P(F) when this possibility is instead permitted. It is important to note that in the
J
secondcasetheexponentofthepowerlawtailofP(F)isthesameforallperturbation(apartfrom
4
very singular cases), and is in an one to one correspondence with the exponent characterizing the

behaviorof thepair interaction with the distance between the two particles.
]
h
c PACSnumbers: 02.50.-r,05.40.-a,61.43.-j
e
m
I. INTRODUCTION ever the exact results we present in 1d are suggestive of
-
t thebehaviorofthesamequantitiesinhigherdimensions.
a
Infactonecansee [11]thatthe changeofspatialdimen-
t
s Theknowledgeofthestatisticalpropertiesoftheforce sion only renders calculation not explicitly performable,
. acting on a particle belonging to a gas and exerted by
t keeping qualitatively the behavior we present below.
a all the other particles provides important information in
m
many physical contexts and applications. Typical exam-
- ples are: (i) distribution of the gravitational force in a
d II. DEFINITIONS AND FORMALISM
gas of masses in cosmological and stellar astrophysical
n
o applications [1, 2, 3], (ii) distribution of molecular and Inorderto approachinthe properwaythe problemof
c dipolar interactions [4] in gas of particles, (iii) theory of
the global force probability distribution in a perturbed
[ defects in condensed matter physics [5], and (iv) granu-
lattice of particles interacting via a power law spatially
lar materials [6, 7]. The first seminal work in this field
2 decreasingpairinteraction,letusconsiderfirstlyagasof
v was due to Chandrasekhar [1] and deals, among many
such identical particles with microscopic density
5 subjects, with the gravitational force probability distri-
6 bution in a homogeneous Poisson spatial distribution of
n(x)= δ(x x ),
3 identical particles. By studying the characteristic func- − i
6 i
tion of the sum of the stochastic forces due to the single X
0
5 particles, the probability distribution of this total force wherexi isthepositionoftheith particle. Letusassume
is exactly found to be given by the so-called Holtzmark
0 thatthe averagenumber density n = n(x) >0(where
0
/ distribution, which is a three-dimensional analog of the theaverage ... istobeintendedasanehnsemibleaverage)
at one-dimensional fat tailed stable L´evy distributions. In is well definhedi(i.e., the particle distribution is uniform
m [2,4,8,9]approximatedextensions,todifferentbranches
on sufficiently large scales [9]). We then suppose that
of physics, of this approach can be found for more com-
- particles interact via a pair force f(x) depending on the
d plex particle distribution (i.e., point processes) obtained
mutual pair distance x as
n by perturbing a homogeneous Poisson point process. In
o this paper wepresentastudy ofthe totalforce probabil- x
c f(x)= C .
: ity distribution for a very different class of spatial par- − xα+1
v ticle distributions (i.e., point processes), the perturbed | |
Xi lattices of point particles, in the case in which the pair This means that f(x) gives the force exerted by a par-
interaction decays spatially as a general power law. We ticle in the origin on another particle in x (the force is
r
a thinkthatthisstudycanbeveryusefulforapplicationin attractive if C > 0, and repulsive if C < 0). Therefore
both solid state physics (e.g., in the case of Coulomb or the force-field in the point x of the space will be
dipolarpairinteraction)[4],andcosmologywheren-body
x x y x
gravitationalsimulations (introduced to study the prob- (x)=C i− C dyn(y) − ,
lemof“structureformation”duetogravitationalcollapse F i |xi−x|α+1 ≡ Z |y−x|α+1
X
from primordial cosmological mass density fluctuations) (1)
are performed usually starting from suitable perturbed where the last integral is over all the space. However,
lattice initial conditions [10]. We limit the study to the fromthelastexpressionofEq.(1),beingn >0,wehave
0
one-dimensionalcaseinordertoavoiddifficulties related that for a givenrealizationofthe stochastic density field
to the anisotropies of higher dimensional lattices. How- n(x), the infinite volume limit of (x) is not univocally
F
2
defined for α 1 (i.e. the integral in absolutely diverg- value obtained by Eq. (2) (i.e., subtracting the effect of
≤
ing,anditsvaluedependonhowthislimitistaken). The the background). This preliminary discussionof the sta-
same feature is presentinhigher spatialdimensions. For tistical definiteness of the force is useful to justify the
F
example in d =3 the same problem is present for α 3 symmetrical way in which we take the infinite volume
≤
and in general α d in d dimensions. For instance this limit in the one-dimensional shuffled lattice case we an-
≤
is the case of the gravitationalforce in a self-gravitating alyze in the rest of the paper. The statistical properties
homogeneousgasofidenticalmasses[1],andofCoulomb of we will find in this peculiar way (no background
F
interaction in the one component plasma (OCP) [12] of andsymmetricallimit) coincide withthoseofthe casein
identical electrical charges both in the disordered and which the effect of a negative background is considered
the ordered (i.e., the Coulomb lattice [13]) phases. This independently of the way in which the infinite volume
problemiswellknownincondensedmatterphysicsabout limit is taken,whichtherefore canbe consideredthe real
the OCP.However,inthis case,the problemis automat- physical case. Moreover, from the above considerations
ically solved by the presence in the physical system of we deduce that, as in the shuffled lattice S(k) k2 at
∼
a uniform background charge density n (x) = n with smallk(see[16]),theresultsarevalidforallvaluesα>0
b 0
−
opposite signwithrespectto the identicalchargedparti- in d=1.
clesandsuchthattoconserveglobalchargeneutralityin Let us take, therefore, a 1d regular chain of 2N +1
the system. Once the attractive force of the background unitary mass particles with a lattice spacing a > 0 (we
is considered on a charged particle together with the re- willtakeeventuallythelimitN + ),i.e.,theposition
pulsiveforcesexertedbytheotherparticles,theproblem ofthenth particleisX =na. T→here∞forethemicroscopic
n
of the infinite volume limit of is solved and its value density can be written as
F
unique. For what concerns the self-gravitating systems,
in Newtoniangravitationananalogofthe uniform back- N
groundoftheOCP(i.e. anegativeuniformmassdensity nin(x)= δ(x na).
−
n (x) = n such to generate a repulsive force on the n=−N
b 0 X
−
particles) does not exist, and it has to be introduced in
Clearly the average density of particles in the system
the system artificially to regularize the problem (an ap-
is n = 1/a. We now apply an uncorrelated displace-
proach usually called Jeans’ swindle [14]). However this 0
ment field (i.e., a random shuffling) to this system, i.e.,
negative background comes out naturally, as an effect of
a random displacement U is applied to the generic nth
spaceexpansion,when the gravitationalmotionofparti- n
particle independently of the other particles. This dis-
cles is described, starting from the equations of general
placement field is completely characterized by the one-
relativity,incomovingcoordinatesinaquasi-uniformex-
displacement probability density function (PDF) p(u)
panding Einstein - De Sitter universe [15] which is the
(i.e., Prob(u U <u+du)=p(u)du). After the appli-
main model of universe used in cosmology. In practice ≤ n
cation of the displacements the new microscopic density
considering the presence of such balancing background
will be:
will give for the following expression:
F
N
y x
(x)=C dyδn(y) − , (2) n(x)= δ(x na un), (3)
F y xα+1 − −
Z | − | n=X−N
where δn(x) = n(x)−n0. This makes F to be defined the un’s being the realizations of the random variables
also for smaller α depending on the small k behavior of U all extracted from p(u) independently one of each
n
the power spectrum S(k) δ˜n(k)2 of the density other. We consider the case p(u)=p( u) for simplicity.
∼ | | Equation(3)saysthattheparticleorig−inallyinX =na,
field where δ˜n(k) is the FourieDr transfoErm of δn(x). By n
after the displacement will be in X =na+U . For the
n n
studying the large distance scaling behavior of the inte-
analysis of spatial density correlations in such a system
grated fluctuations of n(x) [9], it is simple to show that,
see [16]. Let us call q (x) the PDF of the position of the
assumingS(k) kβ atsmallk,inddimension isawell n
∼ F nth particle (i.e., Prob(x xn < x+dx) = qn(x)dx).
definedstatisticalquantity(i.e.,itsvaluedoesnotdepend ≤
Clearly it is given by
on the way in which the infinite volume limit is taken)
for α > (d β)/2 if β < 1, and α > (d 1)/2 if β 1
q (x) p(x na).
− − ≥ n
(see also the discussion in Appendix II on the definite- ≡ −
nessoftheforce respectivelyintheshuffledlatticeand Let us now assume, as above, that the nth particle
F
thehomogeneousPoissonparticledistributions). Notefi- creates a force-field in the point x of the type:
nally that takingthe infinite volume limit symmetrically
with respect to the point x on which the force is calcu- X x
n
f (x)=C −
lated the background gives a zero net force on the point n X xα+1
n
x. Therefore the value of obtained calculating Eq. (1) | − |
F
taking the infinite volume limit symmetrically with re- with α > 0 and C a constant. Therefore the total
spect to the point x gives automatically the well defined stochastic field generated at a generic point x of the
F
3
space by all the system particles is: However we show below how to reduce the problem to
that of a sum of independent stochastic terms, by in-
N X x troducing the concept of conditional probability density
n
(x)=C − . (4)
F X xα+1 function. The solution of this conditional problem will
n
n=X−N | − | give also the way to face the study of the first uncondi-
tional case given by Eq. (4).
Note that it is a sum of random variables. Let us call
About in Eq. (6), we want again to find the PDF
W (F) its PDF; it will be given by
0 F
W(F) of the value F of this force. As before, since the
∞ N displacements applied to the particles are independent
W (F)= dx p(x na) one of each other, we have the exact relation:
0 n n
Z Z−∞"n=−N − #
Y +∞ N
×δF −C−N,N xnxn−xxα+1 . W(F)=Z Z−∞ "n=Y−Ndxnp(xn−na)#
n6=0 | − |
X −N,N
  xn x0
Itisimmediatetoseethat (x)isthesumofindependent δ F C − ,
×  − x x α+1
random variables C(X Fx)/X xα+1. However, as n6=0 | n− 0|
n n X
− | − |
the PDF’s q (x) change with n, these variable are not  
n
that, througha simple changeof variables∆ =x x
identically distributed. As shown below, this, together n n− 0
or n=0, can be rewritten as
with the fact that needs not a normalization in N 6
F
to be well defined in the large N limit, are the reasons
+∞
why we do not obtain in general an exact Gaussian or W(F)= dx p(x )
0 0
L´evy limit [17, 18] for the W0(F). In order to study the Z−∞
asymptoticbehaviorinF andN,itis usualto introduce +∞ −N,N
theso-calledcharacteristicfunctionof ,i.e.,theFourier d∆ p(∆ +x na)
F  n n 0− 
transform (FT) of W0(F): Z Z−∞ n6=0
Y
 
∞ −N,N
Wˆ0(k) dF W0(F)eikF δ F C ∆n .
≡Z−∞ ×  − n6=0 |∆n|α+1
N ∞ y x  X 
= dyp(y na)exp iCk − (.5)
− y xα+1 Let us analyze the behavior of the conditional PDF
n=Y−NZ−∞ (cid:18) | − | (cid:19) P(F;x0), conditioned to the fact that the particle on
which the force is evaluated is at X =x :
By studying the small k behavior of the single integrals 0 0
in Eq. (5) and taking appropriately the limit N
→ ∞ +∞ −N,N
we can deduce the moments and the large F behavior of
P(F;x )= d∆ p(∆ +x na)
W (F). 0  n n 0− 
0 Z Z−∞ n6=0
However, we will study a slightly different and more Y
 
difficultproblem,whichisofparticularinterestifwewant −N,N
∆
n
to study the dynamics of the system particles under the δ F C . (7)
×  − ∆ α+1
effect of only this mutual force. We study directly the n6=0 | n|
X
statisticalpropertiesofthestochasticforceactingonone  
generic system particle. In particular we calculate the Inthis way,oncex0 is fixed, the totalforce is the sum
F
total force acting on the particle initially located at the of independent contributions
origin of the space and displaced to X =U :
0 0 ∆
n
f =C . (8)
−N,N −N,N Xn X0 n |∆n|α+1
= f (X )=C − . (6)
F n 0 Xn X0 α+1 It is interesting to see in which cases satisfies the cen-
nX6=0 nX6=0 | − | trallimittheorem. We willseethatitFneversatisfiesthis
In this way, taking the limit N , we get the sym- theoremevenwhenitsPDFisrapidlydecreasingatlarge
→ ∞
metricinfinitevolumelimitwithnonegativebackground values. More precisely, we will see that even in this last
of Eq.(1) with x=X , which is, as explained in the first case its PDF is dependent on the details of p(u).
0
paragraphof this section, equal to Eq.(2) with arbitrary For the sake of simplicity of notation let us assume
wayoftakingtheinfinitevolumelimit,andthereforegiv- in the rest of the paper that C = 1 (the repulsive case
ingtheforceonanyparticlebelongingtothesystemwith C = 1 will be trivially deduced from this). By per-
−
the uniform balancing background. Note that now, be- forming the simple change of variables given by Eq. (8),
cause of the presence of the variable X in each term of it is possible to find the conditional (i.e., conditioned to
0
the sum (6), is no more a sum of independent terms. X = x ) PDF g (f;x ) of the single stochastic force
0 0 n 0
F
4
generated by the particle in X on the particle fixed in 1. Non-Overlapping condition (NOC): No parti-
n
X =x : cle can be found arbitrarilyclose to any other par-
0 0
ticle; i.e., the supports[20] respectively of p(u) and
f −1−1/α f
of p(u na), for all integer n = 0, have an empty
g (f;x )= | | p +x na . (9)
n 0 α · f 1+1/α 0− overlap−. The main case of phy6sical interest in this
(cid:18)| | (cid:19)
class of displacement fields is when 0 < u <a/2
Clearly also the forces fn so distributed are independent such that p(u)=0 for u >u ; ∃ 0
ofoneeachother. Thesupportofgn(f;x0)canbesimply | | 0
deduced from the one of p(u).
2. Overlapping condition(OC):Particlescancross
oneeachotherandatleastonepairofparticlescan
befoundarbitrarilyclosetooneeachother;i.e.,the
III. LARGE F ANALYSIS AND LIMIT
supports respectively of p(u) and of p(u na), for
THEOREMS −
at least an integer n = 0, have a non-zero overlap.
6
The main case of physical interest in which this
Let us take the FT of Eq. (7) in order to evaluate
happens is when ǫ > 0 such that p(u) > 0 for all
the conditional characteristic function of the stochastic ∃
u <a/2+ǫ.
variable : | |
F
+∞ We will see that in the first case we obtain a rapidly de-
Pˆ(k;x0) dF P(F;x0)eikF (10) creasingW(F)eventhoughthereisnoconstrainttoward
≡
Z−∞ GaussianityinthelargeN limit,whileinthesecondcase
−N,N +∞ x wehaveapowerlawtailedW(F)similarlytothatofthe
= dxp(x+x na)exp ik .
0− xα+1 three-dimensional Holtzmark distribution [1].
n6=0 Z−∞ (cid:18) | | (cid:19)
Y
Note that the quantity
IV. DETAILED ANALYSIS OF EQ. (10)
+∞ x
dxp(x+x na)exp ik (11)
Z−∞ 0− (cid:18) |x|α+1(cid:19) Let us analyze the single factor of Eq. (10) which,
+∞ as aforementioned, is the FT of the conditional PDF
= dfgn(f;x0)eikf ≡{eikf}n;x0, gn(f;x0) of the force felt by the particle in X0 = x0
Z−∞ due to only the particle in X :
n
whereg (f;x ),givenbyEq.(9),istheconditionalPDF
n 0
ofthe fieldf feltbythe particleinx dueto the particle +∞ x
0 dxp(x+x na)exp ik
in xn, is the conditional charact+er∞istic function of this Z−∞ 0− (cid:18) |x|α+1(cid:19)
field f. Moreover a(f) = dfg (f;x )a(f). It
{ }n;x0 −∞ n 0 = exp ik x , (12)
is important to note that, if x = 0 (i.e., the particle
on which we calculate the force0Ris stuck at the origin), (cid:28) (cid:18) |x|α+1(cid:19)(cid:29)n,x0
the condition p(u)= p( u) on the displacements of any
particle would imply − where hs(x)in,x0 denotes the average of the function
s(x) over the shifted PDF h (x) = p(x + x
n,x0 0 −
+∞ x na). In practice, if we indicate with simply s(u) =
dxp(x+na)exp ik +∞ h i
xα+1 dup(u)s(u), then we can say that
Z−∞ (cid:18) | | (cid:19) −∞
+∞ x R
= dxp(x−na)exp −ik xα+1 . hs(x)in,x0 =hs(u+na−x0)i .
Z−∞ (cid:18) | | (cid:19)
We want to study Eq. (11) in the limit of small k. Sim-
Consequently, if one fixes x =0 Eq. (10) becomes:
0
ilarly to what pointed out in the previous section, the
N +∞ x 2 small k behavior of exp ik x is different in
Wˆ(k)=n=1(cid:12)Z−∞ dxp(x−na)exp(cid:18)ik|x|α+1(cid:19)(cid:12) . the two cases in whDich, a(cid:16)s a|x|cαo+n1s(cid:17)eEqun,exn0ce of displace-
Y(cid:12) (cid:12) ments, the pair of particles initially in x=0 and x=na
(cid:12) (cid:12)
However the sh(cid:12)ift x0 = 0 of the particle initially a(cid:12)t the cannotorcanbefoundarbitrarilyclosetooneeachother,
6
origin, and on which we calculate the force, breaks this
i.e.,respectivelyifthesupportsofp(u)andp(u na)have
symmetry, which is anyway recovered when the aver- −
an empty or a non-zero overlap.
age over p(x ) is performed to proceed from P(F;x ) to
0 0 Letusstartwiththecase(i). If 0<u∗ < na/2such
W(F) (however we will see that this is a further source ∃ | |
thatp(u)=0for u u∗,theexponentofexp ik x
of noise when we calculate explicitly the variance of the | |≥ |x|α+1
force ). cantakeonly limited values in the integral(12(cid:16)) (i.e., th(cid:17)e
F
In order to proceed into the analysisof the PDF of , support of g (f;x ) is restricted to only a finite interval
n 0
F
we have to distinguish two basically different cases: of values of f). Note that in the given hypothesis (i), if
5
n > 0 the quantity x can take only strictly positive val- rewrite it as in Eq. (11):
ues, while if n < 0 it takes only strictly negative values.
In this case, if n>0, we can write: x +∞
exp ik dfg (f;x )eikf
xα+1 ≡ n 0
x (cid:28) (cid:18) | | (cid:19)(cid:29)n;x0 Z−∞
exp ik +∞ x
(cid:28) (cid:18) |x|α+1(cid:19)(cid:29)n,x0 = dxp(x+x0−na)exp ik xα+1
+∞ +∞ (ikx−α)m Z−∞ (cid:18) | | (cid:19)
= dxp(x+x na)
0− m! Note that in this case, differently from the previous one,
Z−∞ mX=0 the support of gn(f;x0) includes arbitrarily large values
+∞ (ik)m of f for which, using Eq. (9), we have
= x−αm | |
m! n,x0
mX+=∞0(ik)m (cid:10) (cid:11) gn(f;x0)= p(x0α−na)|f|−(α+1)/α+o(|f|−(α+1)/α).
= (u+na x )−αm , (13)
0
m! −
m=0 LetuscallM =[α−1]theintegerpartofα−1. Byusing
X (cid:10) (cid:11)
theresultspresentedinAppendixIwecanconcludethat
where
u∗ exp ik x (15)
(u+na−x0)−αm =Z−u∗dup(u)(u−x0+na)−αm. (cid:28) (cid:18) |x|α+1(cid:19)(cid:29)n;x0
(cid:10) (cid:11) M (ik)m
If instead n<0, Eq. (13) becomes = (u+na x0)−αm +Sn(k;x0),
m! −
m=0
X (cid:10) (cid:11)
x +∞ ( ik)m
exp ik = − ( x)−αm where Sn(k;x0) contains all the terms of order higher
(cid:28) (cid:18) |x|α+1(cid:19)(cid:29)n,x0 mX=0 m! (cid:10) − (cid:11)n,x0 than M, including the singular part of the small k ex-
= +∞ (−mik!)m (−u−na+x0)−αm , (14) kp.anBsyionusoinfgDeExqp.(cid:16)(3ik2)|x,|xαw+e1(cid:17)caEnn;fix0nawllhyicwhriisteo:f order 1/α in
m=0
X (cid:10) (cid:11)
S (k;x )= (16)
Note that, as p(u) = p( u), we have n 0
−
( u na+x0)−αm = (u na+x0)−αm . In (−1)(M+1)/2πp(x0−na) k1/α+o(k1/α) for odd M
bho−th −cases i h − i αΓ[(α+1)/α]cos α−1−Mπ
2

(cid:16) (cid:17)
(cid:12)(cid:12)(cid:12)(cid:12)(cid:28)(cid:18)|x|xα+1(cid:19)m(cid:29)n,x0(cid:12)(cid:12)(cid:12)(cid:12)<(|n|a+2u∗)−αm Here,αfoΓr[((αs−+im11))Mp/lα/i2]cπsiitpny((cid:16)x,0αw−−en12−ah)Maπv(cid:17)eke1x/cαlu+deod(kt1h/eα)cafsoerinevwenhiMch
for any m(cid:12) 0, and therefor(cid:12)e the series in Eq. (13) ab-
≥ exactly M = 1/α for which we have logarithmic correc-
solutely converges. It is very important to note that, if
tions in k to the above equations.
u∗ <a/2 (i.e., no pair ofparticles canbe found arbitrar-
ilyclosetooneeachother)allthe factorsinEq.(10)can
be represented as a Taylor series (13) to all orders m.
V. FINDING W(F)
As shown below in more detail, this implies that when
u∗ < a/2, W(F) has all finite moments and therefore is
rapidly decreasing at large F. Atthispointwecangofurtherandclassifythepossible
behaviors of P(F;x ) and W(F).
In the secondcase (ii) instead 0<ǫ<a/2 suchthat, 0
at least u satisfying na/2 ǫ∃< u < na/2+ǫ, we Basicallywe againdistinguish the following two cases:
∀ | | − | | | |
have p(u) > 0. In this case the particle initially at na
± 1. ǫ>0 such that u >a/2 ǫ one has p(u)=0;
and the particle initially at the origin can be found ar- ∃ ∀| | −
bitrarily close. This implies that the quantity x (i.e.,
2. 0<ǫ<a/2suchthat,atleast usatisfyinga/2
u x +na) in (12) is permitted to take arbitrary small ∃ ∀ −
0 ǫ< u a/2+ǫ, p(u)>0.
−
values up to zero, and therefore in the last expression | |≤
of Eq. (13) there would be an infinite number of diverg-
ing terms of the last series. In other words the Taylor
A. Case 1: fast decreasing W(F)
series sum in the second expression of Eq. (13) cannot
be exchanged with the average operation ... , and
h in,x0 In this case the system satisfies the NOC and all the
we expect a singular part in the small k expansion of
factors in Eq. (10) can be expanded in the Taylor series
the average exp ik x . In order to find it, we
|x|α+1 (13) and (14) for all different n.
n,x0
D (cid:16) (cid:17)E
6
We can then write It is simple to see that
+∞ 2 N
Pˆ(k;x) dF P(F;x)eikF (17) F = (u x+na)−2α (19)
≡ 2 − u x
ZN−∞ nX=1n(cid:10)(cid:10) (cid:11) (cid:11)
= ei(u−x+kna)α e−i(u+x+kna)α − (u−x+na)−α u (u+x+na)−α u x
nY=1D ED E 1(cid:10),(cid:10)N (cid:11) (cid:10) (cid:11) (cid:11) o
N +∞ (ik)m + (u x+na)−α (u+x+na)−α
= (u x+na)−αm − u− u
" m! − Xn<l(cid:10)(cid:2)(cid:10) (cid:11) (cid:10) (cid:11) (cid:3)
nY=1 mX=0 (cid:10) (cid:11) (u x+la)−α (u+x+la)−α ,
+∞( ik)l × − u− u x
−l! (u+x+na)−αl , where f(cid:2)o(cid:10)r clarity we hav(cid:11)e red(cid:10)efined (cid:11) (cid:3)(cid:11)
#
l=0
X (cid:10) (cid:11)
u∗
a(u) = dua(u)p(u)
where in the average ... over the displacement u we h iu −uu∗∗
hthaavtefurosemdEthqe. (s1y7m),mweetrhhyaipveroPpˆe(krt;y−px()u=) =Pˆ†p((k−;xu)),.wNhoertee hhahb((xu),ixx)=iuRiRx−u=∗d−xuu∗a∗(x−)upu∗(∗xd)udxb(u,x)p(u)p(x).
A† indicates the complex conjugate of A.
By calling again u∗ <a/2 the maximal permitted dis- It is matter of simplRe algeRbra to show that, for p(u) =
p( u), Eq. (19) can be rewritten as
placement for each particle (i.e., the support of p(u) is
−
included in [ u∗,u∗]), we can find Wˆ(k)= [W(F)] by
− F 2 N
simply calculating the following average F = (u x+na)−2α (u x+na)−α 2
2 − u− − u x
u∗ nX=1D(cid:10) (cid:11) (cid:10) (cid:11) E
1,N
Wˆ(k)= dxp(x)Pˆ(k;x). (18)
+ (u x+na)−α [ (u x+la)−α
Z−u∗ − u − u
n,l
X(cid:10)(cid:10) (cid:11) (cid:10) (cid:11)
It is simple to verify that, as Pˆ(k; x) = Pˆ†(k;x) and (u+x+la)−α ] (20)
− u x
−
p(u) = p(−u), the function Wˆ(k) is real and Wˆ(k) = We se(cid:10)e that the force(cid:11)va(cid:11)riance is composed of two differ-
Wˆ( k). TheTaylorexpansioninkofWˆ(k)isobtainable
− ent contributions: the former, given by the first sum in
from Eqs.(17) and (18). Since it is a real function and
Eq.(20),ismainlyduetothefluctuationsinthedisplace-
is the characteristic function only even powers of k are
ments u of all the sources of the force (in this term the
present. In particular the coefficient of the k2 term is
averageover x is only a smoothing operation), while the
2/2 where
−F latter, given by the second sum, is determined basically
by the fluctuations created by the stochastic displace-
+∞ ment x of the particle initially in the originon which we
h( )= dF W(F)h(F).
F evaluate the force (in this term is the averagesover u to
Z−∞
play a role of simple smoothing).
It is interesting and useful in applications to calculate
Actually, rigorously speaking, we should show that all
the coefficients of the Taylorexpansionof Wˆ(k) arecon- all the above expressions by evaluating all the terms in
the sums in Eq. (20) to the second order in (u x)/na.
vergent to finite values in the limit N . It is simple ±
to show it by expanding the terms (→u ∞x +na)−αm In order to do this, we use the following second order
0
h ± i Taylor expansion for B A:
of Eq. (17) in Taylor series of (u x0)/na for n 1 ≪
± ≥
whichisjustifiedbythefactthatinthe givenhypothesis
−γ
B
u + x0 2u∗ <a, and considering that (A+B)−γ =A−γ 1+
| | | |≤ A
(cid:18) (cid:19)
+∞ (ik)m(na)−αm +∞(−ik)l(na)−αl =1, n 1. =A−γ 1 γB + γ(γ+1) B 2+o B 2 .
m! × l! ∀ ≥ " − A 2 (cid:18)A(cid:19) (cid:18)A(cid:19) #
m=0 l=0
X X
From this, substituting respectively A with na and B
Therefore we conclude that we can write Wˆ(k) in the with u x, we have that
±
following form:
(u x+na)−γ
±
+∞ 2n u x γ(γ+1) u x 2
Wˆ(k)= ( 1)n F k2n. (na)−γ 1 γ ± + ± .
n=0 − (2n)! ≃ " − na 2 (cid:18) na (cid:19) #
X
7
Moreover we have that (u x)2n+1 = 0 for any B. Case 2: power law tailed W(F)
± u x
integer n due to the symmetry p(u) = p( u), while we
have that (u x)2 (cid:10)(cid:10)= 2σ2 wher(cid:11)e(cid:11)σ−2 = u2 = Asshownabove,thisisthecaseinwhichtheOCissat-
± u x u
−uu∗∗duu2p(cid:10)((cid:10)u)isthe(cid:11)var(cid:11)ianceofthe singledispla(cid:10)cem(cid:11)ent. iisnfiiteida,l il.aet.t,icpearptoicslietsioanrsebpeeyromnidttethdetolimjuitmpofotuhteorfeltahteeidr
Therefore we can write
R unitary cell in such a way to be found arbitrarily close
σ2
(u x+na)−γ (na)−γ 1+γ(γ+1) , to some other particle. Note that this is always the case
± u x ≃ (na)2 when the support of p(u) is unlimited, i.e., if p(u) > 0,
(cid:20) (cid:21)
a(cid:10)n(cid:10)d (cid:11) (cid:11) u IR. However the same kind of W(F) is also ob-
∀ ∈
tained if u∗ > a/2 such that p(u) > 0, u [ u∗,u∗]
(u x+na)−α 2 (na)−2α 1+α(3α+2) σ2 . and zero o∃utside. The difference between∀thes∈e t−wo sub-
± u x≃ (cid:20) (na)2(cid:21) cases is only in the amplitude of the power law tail of
D(cid:10) (cid:11) E
Henceforth W(F) but not in its exponent. In general,if the particle
(u x+na)−2α (u x+na)−α 2 initially at the lattice site x = na is permitted, through
± u− ± u x displacements, to be found arbitrarily close to the par-
D(cid:10) α2σ2 (cid:11) (cid:10) (cid:11) E ticle initially at x = 0, it will contribute to the product
= . (10) through a factor of the type (15). If instead this is
(na)2(α+1)
not permitted, it will contribute to (10) through a fac-
Moreover tor of the form (13) or (14) depending respectively on
(u x+na)−α (u x+la)−α whether n > 0 or n < 0. In any case if M = [1/α], in
− u ± u x order to find the main terms of the small k expansion of
(cid:10)(cid:10)(na)−α(la)−α (cid:11) (cid:10) (cid:11) (cid:11) Wˆ(k) (so to determine the large F tail of W(F)), it is
≃
α2σ2 1 1 sufficienttotruncateallthesmallk expansionofthe dif-
1 +α(α+1)σ2 + ;
× ∓ (la)(na) (na)2 (la)2 ferentfactorsinEq.(10)atmosttotheorderM+1. For
(cid:20) (cid:18) (cid:19)(cid:21) the sake of simplicity, let us limit the discussion to the
from which
case in which strictly αM < 1 in such a way to exclude
(u x+na)−α [ (u x+la)−α (21) logarithmic corrections in k. We can write
− u − u
(cid:10)(cid:10) (u+x+la)−α(cid:11) ](cid:10) 2α2σ2(cid:11) . Pˆ(k;x0) (24)
− u x ≃ (la)α+1(na)α+1 (OC) M (ik)m x m
(cid:10) (cid:11) (cid:11) +A(n,x ,α)k1/α
Uobstinaignall these results in all the terms of Eq. (20), we ≃ Yn "mX=0 m! (cid:28)(cid:18)|x|α+1(cid:19) (cid:29)n,x0 0 #
2 N 1 N 1 2 (NOC) M+1(ik)m x m ,
F2 ≃α2σ2nX=1(na)2(α+1) +2"nX=1(na)α+1#  × Yl "mX=0 m! (cid:28)(cid:18)|x|α+1(cid:19) (cid:29)l,x0#
(22) where A(n,x ,α) is the coefficient of the term k1/α in
0
It is simple toverify that both sums in Eq. (22) arecon- Eq.(16),andthe firstproductonn is onthe pa∼rticlesin
vergingforN + forallα>0,forwhichwecanthen X withn=0whichcanbefoundarbitrarilyneartothe
n
→ ∞ 6
rewrite particleinX (i.e., satisfyingthe OCwithrespecttothe
0
F2 α2σ2 particleinX0), while the productonl is onthe particles
ζ(2α+2)+2ζ2(α+1) , (23) in X with l =0 which have a positive minimal distance
2 ≃ a2(α+1) l 6
tothesameparticle(i.e.,satisfyingtheNOCwithrespect
where ζ(t) is the Ri(cid:2)emann zeta function (n(cid:3)ote that for to the particlein X ). If p(u)>0 u IR allthe system
0
t 1+ wee have ζ(t) 1/(t 1)). Again the first term particles with n=0 are included i∀n th∈e first product. If
is→due to the fluctuatio≃ns in t−he position of the sources, instead p(u) > 06 for u [ u∗,u∗] with u∗ > a/2 and
while the second one is due to the fluctuations in the zerooutside, the firstpro∈du−ctinclude only contributions
position of the particle on which we are calculating the from the particles with 2u∗/a<n<2u∗/a and n=0,
force. In particular In Eq. (22) the generic term of the whiletheothersareinclu−dedinthe secondproduct.6The
firstsumgivetherelativeweightofthenth nearestneigh- large F behavior of P(F;x ) and consequently of W(F)
0
bor particles in determining the force on the particle in is completely determined by the singular term of order
X0. At last we can say that, in the case of displace- k1/α in the smallk expansionof Pˆ(k;x0). It is simple to
ments limited within a box well contained in a unitary see that up to the order k1/α
cell around the initial lattice position, we can approxi-
M
mate W(F) with a Gaussian PDF with zero mean and
Pˆ(k;x )= c (x )km+c (x )k1/α,
variance given by Eq. (23). However, as already pointed 0 m 0 1/α 0
out, there is no constraint, in the limit N , toward mX=0
→∞
rigorousGaussianityandnon-Gaussiancorrectionsarein where the the coefficients c (x ) can be deduced
m 0
general present. by counting from Eq. (24) (in particular c (x ) =
m 0
8
imFm(x0)/m!, where Fm(x0) = −+∞∞dF P(F;x0)Fm is VI. CONCLUSIONS
the mth moment of P(F;x ) and m M) and
0 R ≤
g g We havepresenteda detailedstudy of the PDF W(F)
n6=0
of the stochastic force generated by a randomly per-
c1/α = A(n;x0), turbed lattice of sourceFs of a scale invariant attractive
−2u∗/aX<n<2u∗/a pair interaction field f(x) = Cx/xα+1 with α > 0 at
− | |
where the formula includes also the case u∗ . The distance x from the source.
→ ∞
small k expansion up to the order k1/α of W(F) will be In general we distinguish two cases:
consequently
1. The NOC is satisfied and no pair of particles can
M befoundatanarbitrarilysmallreciprocaldistance;
Wˆ(k) b km+b k1/α,
m 1/α
≃
m=0 2. TheOCissatisfiedanditexistsatleastoneofsuch
X
pairs of particles.
where
+∞ im In the first case we have a fast decreasing W(F) similar
b = dx p(x )c (x )= m with m M
m 0 0 m 0 m!F ≤ toaGaussianPDFatlargeF,eventhoughnoconstraint
Z−∞ towardanexactGaussiancentrallimittheoremisfound.
+∞
b = dx p(x )c (x ), In the second case a power law tailed W(F) is found.
1/α 0 0 1/α 0
Z−∞ Theuniqueexponentofsuchpowerlawisdirectlyrelated
to the pair interaction exponent α, while its amplitude
where m = +∞dF W(F)Fm =
F −∞ depends also on the lattice spacing a (with respect to
−+∞∞dx0p(x0)Fm(x0) ≡ FRm(x0) . It is possible the unit distance through which we measure x in f(x))
and in general on the shape of the perturbations PDF
tRo evaluate explicitly b1/α bDy using EEq. (16).
We are nowgin the situatiogn to connect the singular p(u). InparticularinthiscaseW(F)hasapowerlawtail
term b k1/α of the small k expansion of Wˆ(k) to the with the same exponent as the stable L´evy distribution
1/α
found in the Poisson case (see Appendix II) but with a
large F tail of W(F) by using directly the arguments in
reducedamplitude, eventhough,analogouslyto the case
Appendix I. This gives simply
(i), no constrainthas been found towardthe stable L´evy
W(F) BF−1−1/α distribution.
≃
Some further general considerations have can now be
with
done:
1 +∞ n6=0
B = dx p(x ) p(x na). (25) Inthecaseinwhichtheprobabilityoffindingarbi-
α 0 0 0− •
Z−∞ −2u∗/aX<n<2u∗/a trarilyclosetooneeachother,thelargeF behavior
ofW(F)isbasicallydeterminedbythe smallxbe-
Note that if the support of p(x na) is much larger
0 havior of f(x) and not at all by the the fact if it
−
thana andp(u)is smooth(i.e., approximatelyconstant)
is long range or not. Therefore if we considered a
on the scale a, we can approximate Eq. (25) with
fast decreasing f(x) but with the same divergence
1 p(0)a in x = 0 we would have deduced the same con-
B = − . clusions about the exponent of the large F tail of
αa
W(F) [21].
Note that this last approximated expression is not de-
pendent on the details of p(u) for u = 0. Finally, we For this reason, even if we consider a lattice per-
6 •
can observe that we have obtained a power law tailed turbed by correlated displacements, we expect to
W(F) characterizedby the same exponent of the case of obtain the partition into the two cases (i) and (ii)
ahomogeneousPoissonparticledistributionpresentedin above considered depending on the possibility or
Appendix II. The only differences are the two following: not to find pair of particles arbitrarily close.
The amplitude of this power law tail is reduced
• ActuallydifferentcasesforthelargeF tailofW(F)
in the shuffled lattice with respect to that of the •
between the Gaussian-like “fast decreasing” and
Poisson particle distribution, given by Eq. (37), of
L´evy-like “power law” tailed PDF with exponent
a factor
β =(α+1)/αarepossibleinveryparticularcases.
+∞ n6=0 These cases correspond to the choice of p(u) such
dx0p(x0) ap(x0 na) 1 p(0)a. that p(u) > 0 exactly for u [ a/2,a/2] and zero
− ≃ − ∈ −
Z−∞ −2u∗/aX<n<2u∗/a outside. By changing the limit behaviors of p(u)
when u a/2 we can obtain different large F
Intheshuffledlatticewehavethispowerlawtailfor behavior→sof±W(F). Inparticularifp(a/2)>0and
•
each α > 0, while in the Poisson case the problem finite(weconsiderp(u)=p( u))wehavethesame
is not well defined for α 1/2 (see Appendix II). case as (ii) described above−with β =(α+1)/α. If
≤
9
instead p(a/2) = 0, depending on the behavior of Using the general relation valid for 0<β <1
p(u)foru (a/2)− wewillhavedifferentvaluesof
β but in g→eneral larger than (α+1)/α. If, finally, ∞ qβ π
dq = , (29)
pw(aay/2in)t=egr+a∞ble()ininsugecnhearawlawyetohbattapi(nuβ)rsemmaallienrstahnayn- Z0 1+q2 2cos β2π
(α+1)/α (but always > 1 so that W(F) remains (cid:16) (cid:17)
we can conclude
integrable).
Bπ
hˆ(k)= kβ−1+o(kβ−1). (30)
Appendix I: Fourier transform of power law tailed Γ(β)cos βπ
2
PDFs
(cid:16) (cid:17)
If instead of using the integral representation (28) one
We are interested in the small k behavior of the char- used
acteristic function fˆ(k) of a given power law tailed PDF
1 ∞
f(x) which for large x behaves as Ax−α with α > 1. x−β = dzzβ/2−1e−x2z,
Let us call [α] = n ≥| |1 the integer p|a|rt of α. In this | | Γ β2 Z0
hypothesisfˆ(k)hasaregularTaylorexpansionuptothe
(cid:16) (cid:17)
order n 1 followed by a singular term proportional to one should obtain the alternative expression containing
kα−1: − only Gamma functions:
+∞ n−1 (ik)m
fˆ(k) dxf(x)eikx = xm+fˆs(k), (26) B√πΓ 1−β
≡Z−∞ m=0 m! hˆ(k)= 2 kβ−1+o(kβ−1).
X 2βΓ(cid:16)β (cid:17)
where xm = +∞dxxmp(x) and fˆ(k) contains the sin- 2
−∞ s (cid:16) (cid:17)
gular part ofRfˆ(k) and at small k is an infinitesimal of Note that the coefficient of the term kβ−1 is real and
order α 1 in k (if α is an integer it contains also loga-
positive. Another important case is when the function
−
rithmic corrections).
h(x) has an odd non-integrable power law tail, i.e.:
Now we show that effectively at sufficiently small k,
fˆs(k) ∼ Bkα−1 (where now a(k) ∼ b(k) means that h(x)=B|x|−β[2θ(x)−1]+h0(x), (31)
lim [a(k)/b(k)] = 1) giving an explicit expression for
k→0
B as a function of both the amplitude A and the expo- where θ(x) is the usual Heaviside step function, B > 0,
nent α. 0<β <1, andh (x) the same of Eq.(27). By using the
0
First of all, let us study the case of a function h(x) same integral transformation leading to Eq. (30), we in
that can be written as this case we obtain:
h(x)=B|x|−β +h0(x), (27) hˆ(k)=i Bπ kβ−1+o(kβ−1).
where B > 0, 0 < β < 1 and h (x) is a smooth func- Γ(β)sin βπ
0 2
tion, integrable in x = 0 and such that xβh (x) 0
0 (cid:16) (cid:17)
→
for x . This means that h(x) presents an even At this point we can go back to the problem of finding
| | → ∞
power law tail. In this case the small k behavior of the dominant small k contribution of the term fˆ(k) in
hˆ(k)= +∞dxh(x)eikx is completelydeterminedbythe Eq.(26)forthePDFf(x)decayingatlarge x asAsx−α.
−∞ | | | |
Fourier transform of B x−β, i.e.: Notethatnowwecannotapplydirectlytheargumentwe
R | |
have used for the above function h(x). In fact if, from
+∞
hˆ(k) B dxx−βeikx one side, also in this case we can write
∼ | |
Z−∞
f(x)=Ax−α+f (x)
In order to perform this Fourier transform we introduce | | 0
the integral representation:
with xαf (x) 0for x ,fromthe othersideα>
0
| | → | |→∞
x−β = 1 ∞dzzβ−1e−|x|z, (28) 1 (for definiteness of probability) and f0(x) contains a
nonintegrablesingularityatx=0suchthattocancelthe
| | Γ(β)
Z0 non integrable contribution of the Ax−α term at small
whereΓ(β)istheEulerGammafunction. UsingEq.(28) x. Inordertocircumventthis difficul|ty|weintroducethe
we can write: function
+∞ 1 ∞ +∞
dxx−βeikx = dzzβ−1 dxeikx−|x|z g(x)=xnf(x),
| | Γ(β)
Z−∞ Z0 Z−∞
2kβ−1 ∞ qβ where n is the integer part of α. In this way g(x) is
= dq .
Γ(β) 1+q2 similar to the function h(x) of Eq. (27) if n is even and
Z0
10
totheh(x)ofEq.(31)ifnisodd. Therefore,bydefining Let us now study the characteristic function
as usual gˆ(k)= +∞dxg(x)eikx, we can say that
−∞ +∞
Wˆ (k)= [W (F)]= dF W (F)eikF .
R Aπ P F P P
gˆ(k)= kα−n−1+o(kα−n−1) Z−∞
Γ(α n)cos α−nπ
− 2 By taking the FT of Eq. (34) we obtain
if n is even, and (cid:0) (cid:1)
N
L/2 dx x
gˆ(k)=i Aπ kα−n−1+o(kα−n−1) WˆP(k)= L exp ik xα+1
Γ(α n)sin α−nπ "Z−L/2 (cid:18) | | (cid:19)#
− 2
if n is odd. Now in orde(cid:0)r to fin(cid:1)d the singular part fˆ(k) By adding and subtracting 1 inside the square brackets,
s
and taking the limit L with N =ρ L we arrive at
offˆ(k) it is sufficientto integraten times gˆ(k) (the inte- →∞ 0
the final expression:
grationconstantsgivingriseto thefinite moments terms
of fˆ(k) in Eq. (26)). In this way we obtain +∞ t
Wˆ (k)=exp ρ k1/α dt 1 exp i .
fˆs(k) (32) P (cid:20)− 0 Z−∞ (cid:18) − (cid:18) |t|α+1(cid:19)(cid:19)(cid:21)
(−1)n/2Aπ kα−1+o(kα−1) for even n Note that
Γ(α)cos(α−nπ)
= 2 +∞ t ∞
dt 1 exp i =2 dt 1 cos(t−α)
 Γ((−α1))s(inn+(1α)/−22nAππ)kα−1+o(kα−1) for odd n, Z−∞2 ∞(cid:18) − (cid:18) |t|α+1(cid:19)(cid:19) Z0 (cid:0) − (cid:1)
where we have used the following property of Gamma = α duu−1−1/α(1−cosu),
function: Γ(x+1)=xΓ(x), hence[(α 1)...(α n)Γ(α Z0
n)]=Γ(α). Note that in both case th−e coeffici−ent of th−e where the last passage is due to the change of variable
term kα−1 is real. t−α = u. Let us now use, as in the previous appendix,
the integral representation:
1 ∞
Appendix II: Force PDF in a Poisson particle u−1−1/α = dzz1/αe−uz.
distribution Γ αα+1 Z0
Through this transform(cid:0)ation(cid:1) we arrive finally to the re-
Let us consider the case in which the particles are dis-
lation
tributed on the line interval ( L/2,L/2]of length L fol-
−
lowing a spatially stationary Poisson process with aver- +∞ t
dt 1 exp i
age density ρ0 > 0. We want to know the PDF WP(F) − tα+1
of the field Z−∞ (cid:18) (cid:18) | | (cid:19)(cid:19)
2 ∞ z−1+1/α
N = dz . (35)
F = xi (33) αΓ αα+1 Z0 1+z2
x α+1
i
Xi=1 | | Note that the last in(cid:0)tegra(cid:1)l is diverging for α 1/2, in-
≤
generated at the origin of the space by all the N system dicating that the problem is not well defined for these
particles(wecanconsiderN =ρ Lasthefluctuationsof valuesofαasF isnotawelldefinedstochasticquantity.
0
order√ρ L,duetothePoissonstatistics,arecompletely This means that the sum in Eq. (33) needs a L depen-
0
unimportant for this problem in the large L limit). We dent normalization to become a well defined stochastic
will follow the procedure to find W (F) in three dimen- variable. In fact, differently to the shuffled lattice case,
P
sions used by Chandrasekhar in [1] for the gravitational wherethetypicalmassfluctuationonregionsofsizeRis
force. Notethat,asthepositionsofdifferentparticlesare proportionalto R0, in the Poissoncase this is due to the
uncorrelated,thejointPDFp(x ,...,x )ofthepositions fact that such fluctuation is proportional to R1/2. The
1 N
of the N system particles is simply: field due to the mass fluctuation in a sphere of radius R
on the origin of the sphere is of order R−α for the shuf-
N fled lattice and R−α+1/2 in the Poisson point process.
p (x ,...,x )= p(x ),
N 1 N i This explains why for a shuffled lattice the problem is
iY=1 well defined for any α > 0 and not only for α > 1/2.
where p(x )=1/L. Therefore we can write This also says that for α < 1/2, in order to have a well
i
defined stochastic field also in the Poisson case, we have
L/2 N dx N x to divide the field in Eq. (33) by L−α+1/2 where L is
i i
WP(F)=Z Z−L/2"i=1 L #δF −j=1 |xj|α+1 . dthimeesnyssitoenms,sfiozre.whTichhetshaemseamaregmumasesntfluccatnuabteiounsseadreinred-
Y X
 (34) spectivelyproportionaltoR(d−1)/2intheshuffledlattice,