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Amorphous packings of hard spheres in large space dimension Giorgio Parisi 1,2 and Francesco Zamponi 3 ∗ † 1Dipartimento di Fisica, Università di Roma “La Sapienza”, P.le A. Moro 2, 00185 Roma, Italy 2INFM – CRS SMC, INFN, Università di Roma “La Sapienza”, P.le A. Moro 2, 00185 Roma, Italy 6 3Laboratoire de Physique Théorique de l’Ećole Normale Supérieure, 0 24 Rue Lhomond, 75231 Paris Cedex 05, France 0 (Dated: February 6, 2008) 2 In a recent paper we derived an expression for the replicated free energy of a liquid of hard n spheres based on the HNC free energy functional. An approximate equation of state for the glass a andanestimateoftherandomclosepackingdensitywereobtainedind=3. Hereweshowthatthe J HNCapproximationisnotneeded: thesameexpressioncanbeobtainedfromthefulldiagrammatic 5 expansion of thereplicated free energy. Then,we consider theasymptotics of thisexpression when 2 the space dimension d is very large. In this limit, the entropy of the hard sphere liquid has been computed exactly. Using this solution, we derive asymptotic expressions for the glass transition ] density and for the random close packing density for hard spheres in large space dimension. h c e m I. INTRODUCTION - t a Thestudyofamorphouspackingsofhardspheresisrelevantforalargeclassofproblems,includingliquids,glasses, t colloidaldispersions,granularmatter,powders,porousmedia,etc. [1,2,3,4,5,6,7,8,9,10,11,12,13]. Nevertheless, s . the question whether a glass transition exists for a liquid of identical hard spheres in finite dimension is still open. t a Recently, a quantitative description of the glass transition in structural glasses has been obtained by means of m the replica trick [14, 15, 16, 17, 18, 19]. The latter is applied in the context of a variational approximation - the - hypernetted chain (HNC) approximation - that leads to a suitable free energy functional for simple liquids [20, 21]. d This method was successfully applied to Lennard-Jones systems [14, 15, 16, 17] and, more recently, to hard spheres n in space dimension d=3 [18, 19]. Quantitative estimates of the glass transition temperature (or density) and of the o equation of state of the glass have been obtained in this way. Moreover, for hard spheres an estimate of the random c [ close packing density, i.e. the maximum density of the amorphous configurationsof the system, has been obtained as the value of the density where the pressure of the glass diverges [19]. 1 In this approximation the glass transition turns out to be similar to the 1-step replica symmetry breaking (1RSB) v transitionthathappens inaclassofmean-fieldspinglassmodels andindeedthe replicastrategydescribedabovewas 3 7 inspiredbytheexactsolutionofthesemodels[22,23,24,25]. However,forfinitedimensionalmodelswithshortrange 5 interactions the picture emerging from the replica-HNC approach should be modified by non-perturbative activated 1 processes (for a detailed discussion see e.g. [26, 27, 28]). The results obtained in d=3 have often been considered as 0 a kind of “mean-field” approximation. 6 Activatedprocessesshouldbecomelessrelevantonincreasingthespacedimension. Itisthennaturaltoaskwhether 0 this approximation describes better and better the true properties of the system (eventually becoming exact) when / t the space dimension is large as it is well known that the mean field approximation becomes exact in the d a → ∞ m limit. Moreover, the study of sphere packings in large space dimension is relevant for information theory and lot of effort has been devoted to finding the densest packing for d [29, 30]. Despite this effort, only some not very - → ∞ d restrictive bounds have been obtained, and it is still unclear whether the densest packings for d are amorphous →∞ n or crystalline. o Inthispaperwewill: i)showthatind thereplicaapproachpredictstheexistenceofaglasstransitiondensity c →∞ ρ and compute the value of this density; ii) compute the maximum density ρ of the amorphous packings (or glass : K c v states) in d . Unfortunately the value of ρ we find is within the current bounds for crystalline packings so we c i cannot addr→ess∞the problem whether the densest packings are amorphous or not. X The paper is organized as follows: first we briefly outline the method, without entering in details as they are r a discussed in [19]; then we present an improvement of the theory of [19], i.e. we show how to construct the small cage expansion without using the HNC approximation; finally we discuss the limit d of the theory. →∞ ∗ [email protected] † [email protected] 2 II. THE METHOD In the replica method one considers a liquid made ofm copies of the originalsystem, with the constraintthat each atom of a given replica must be close to an atom of the other m 1 replicas, i.e. that the replicated liquid must − be made of molecules of m atoms, each one belonging to a different replica. It was shown that this trick allows to compute all the properties of the glass, including the size of the cages, the vibrational entropy, etc., provided one is able to perform the analytical continuation to real m 1, see [14, 15, 16, 25] for a detailed discussion. Thus the ≤ problem is to compute the free energy of the replicated system for integer m and to continue the expression for real m. This strategyhasbeen firsttestedonmeanfieldspinglassmodels [15,25], andthen appliedto systemsofparticles interacting througha Lennard-Joneslike potential [14, 16, 17]. To compute the replicatedfree energy for a systemof particles,theideawastostartfromthestandardHNCfreeenergyforamolecularliquid[21]andexpanditinapower seriesofthe “cageradius”,that representsthe amplitude ofthe vibrations ofthe particles inthe glassstate [15]. The method was successful but could not be extended straightforwardly to hard spheres, because at some stage it was assumed that vibrations were harmonic, an approximation that clearly breaks down for hard core potentials. ThesmallcageexpansionofthereplicatedHNCfreeenergywasworkedoutin[19]. Itwasshownthatthetheoryis in reasonable quantitative agreement with numerical data even if the HNC approximation is a very poor description of the hardspheres liquid. In next section we will show that the result of [19] for the replicated free energy,obtained startingfromtheHNC freeenergyfunctional,canbederivedfromthe fulldiagrammaticexpansion,withouttheneed of neglecting any class of diagrams. III. SMALL CAGE EXPANSION Our approachwill be basedon the standarddiagrammatic approach(virialexpansion). It seems to us that similar results could also be obtained by a direct computation, however we think that it is more instructive to use the more familiar diagrammatic approach. We startfromthe diagrammaticexpansionof the canonicalfree energy as a function of the single-moleculedensity ρ(x)andoftheinteractionpotentialbetweenmolecules. Itcanbeobtainedfromthe grancanonicalpartitionfunction of the system via a Legendre transform [21]. Calling x (x ,...,x ) the coordinate of a molecule, the expression 1 m ≡ for the replicated free energy functional is [20, 21] (setting from now on β =1): Ψ[ρ(x),f(x,y)]= dxρ(x)[logρ(x) 1] + + + + + (1) − − (cid:1) (cid:2) (cid:3) (cid:4) (cid:5) ··· Z (cid:20) (cid:21) where the black circles represent a function ρ(x) and the lines represent a function f(x,y) = e ϕ(x,y) 1, and − − ϕ(x,y) = ϕ(x y ) is the interaction potential between two molecules, with ϕ(r) = for r < D and 0 a | a − a| ∞ otherwise,the usualhardsphere interaction. The function f(x,y) is equalto 1 if x y <D for atleastone pair a a P − | − | a of spheres, and vanishes otherwise. The information on the inter-replica interaction that is used to build molecules is encoded in the function ρ(x). The generic diagram of the series expansion of Ψ represents an integral of the form 1 = dx dx dx ρ(x )ρ(x )ρ(x )f(x ,x )f(x ,x )f(x ,x ) . (2) 1 2 3 1 2 3 1 2 2 3 3 1 (cid:6) S Z where S is the symmetry number of the diagram [20]. For the one-molecule density we make the ansatz ρ(x)=ρ dX e−(xa2−AX)2 =ρ dXρ(u) , ρ(u)= e−(u2aA)2 (3) (√2πA)d (√2πA)d Z a Z a Y Y b b where ρ =N/V is the number density of molecules, X is the center of mass of the molecule and u x X. This a a ≡ − allows to compute exactly the ideal gas term of the free energy: 1 d d dxρ(x)[logρ(x) 1]=logρ 1+ (1 m)log(2πA)+ (1 m logm) . (4) N − − 2 − 2 − − Z We want to perform a power series expansion in √A of the interaction term. 3 Atthezerothorder,thefunctionsρ(u)areproductsofdeltafunctions,ρ(u)= δ(u ),sothatallthe coordinates a a x are equal to X. This gives a contribution a Q b b ρ3 = dX dX dX F(X ,X )F(X ,X )F(X ,X ) . (5) 1 2 3 1 2 2 3 3 1 (cid:7) S Z where F(X,Y) = θ(D X Y ) is represented by a dashed line and a vertex i without a black dot represents − −| − | the constant ρ and an integration over the corresponding X . This is exactly the contribution of the diagram to the i usualnon-replicatedfreeenergy. Thus,atthezerothorderthemolecularfreeenergysimplyreducestothefreeenergy [ρ,F(r)] of a liquid made by the center of masses of the molecules, with density ρ and an hard core interaction. F To discuss the first order correction in √A let us first insert Eq. (3) into (2): ρ3 = dX dX dX du du du ρ(u )ρ(u )ρ(u )f(X +u ,X +u )f(X +u ,X +u )f(X +u ,X +u ), (6) 1 2 3 1 2 3 1 2 3 1 1 2 2 2 2 3 3 3 3 1 1 (cid:8) S Z using the notation X +u=(X +u ,..b.,Xb+u )band 1 m f(X1+u1,X2+u2)=e− aϕ(X1−X2+u1a−u2a) 1 . (7) − P For smallA the u are smalltoo, u √A. The function f abovehas the propertythat, if X X differs from D by 1 2 ∼ | − | a quantity √A, it is independent of u ,u : in fact it is a constant equal to 1 if X X <D and 0 otherwise. 1 2 1 2 ≫ − | − | That is, if X X is not close to D, one has f(X +u ,X +u )=F(X ,X ) and the diagram does not give any 1 2 1 1 2 2 1 2 | − | contribution apart from the zeroth order discussed above (recall that duρ(u)=1). Thus, the corrections in √A due to a given diagram come from the region of the integration space where at least R two of the coordinates Xi, Xj connected by a link have distance D, Xi Xbj D+O(√A). Let us call such a pair | − |∼ a “singular pair”, and the link connecting them as a “singular link”. The regions of the integration space of the X i wheren pairsX , X aresingularhaveatotalvolume (√A)n. Thus the onlycontributionatorder√Acomes from i j ∼ n=1, i.e. only one singular pair. In all the non-singular links we can replace f(X +u ,X +u ) F(X ,X ) and we get i i j j i j → 1 = dx dx dx ρ(x )ρ(x )ρ(x )f(x ,x )f(x ,x )f(x ,x ) 1 2 3 1 2 3 1 2 2 3 3 1 (cid:9) S Z 3ρ3 = + dX dX dX du du ρ(u )ρ(u )f(X +u ,X +u ) F(X ,X ) F(X ,X )F(X ,X ) 1 2 3 1 2 1 2 1 1 2 2 1 2 2 3 1 3 (cid:10) S − Z (cid:20)Z (cid:21) = + , b b (cid:11) (cid:12) (8) where the wiggly line represents the function Q(X ,X )= du du ρ(u )ρ(u )f(X +u ,X +u ) F(X ,X ) , (9) 1 2 1 2 1 2 1 1 2 2 1 2 − Z whichisdifferentfrom0onlyif X X D+bO(√Ab). Thusforeachdiagramthecorrectionisobtainedbyreplacing 1 2 | − |∼ one F-link with one Q-link in all the possible non-equivalent ways. This is equivalent to taking the derivative of the zeroth order diagram with respect to F(X ,X ), multiplying by Q(X ,X ) and integrating over X , X . Summing 1 2 1 2 1 2 the contribution of all the diagrams we then obtain the first order correction to the replicated free energy: δ ∆Ψ[ρ(x),f(x,y)]= dX dX F Q(X ,X ) . (10) 1 2 1 2 δF(X ,X ) Z 1 2 where the derivative with respectto F(X ,X ) ofthe free energyis takenatconstantρ. Recallingthat the canonical 1 2 free energy is the Legendre transform of the grancanonicalfree energy, it easy to prove that [20, 21] δ δlogZ ρ2 GC F = = Y(X ,X ) , (11) 1 2 δF(X ,X ) −δF(X ,X ) − 2 1 2 1 2 where the function Y(r) eϕ(r)G(r) is continuous for hard spheres. ≡ 4 The calculationofthe functionQ(X ,X )=Q(X X )wasalreadydone in[19]andis reportedinAppendix A: 1 2 1 2 | − | Q(r)=2√AQ Σ (D)δ(r D) , (12) m d − whereQ isananalyticfunctionofm,withQ 0.638(1 m)+o((1 m)2),seeAppendixA,andΣ (D)= 2πd/2Dd−1 m m ∼ − − d Γ(d/2) is the surface of a d-dimensional hypersphere of radius D. Substituting Eq.s (11), (12) in Eq. (10), and adding the correction coming from the ideal gas term, Eq. (4), we obtain Ψ[ρ(x),f(x,y)]= (ρ)+N d(1 m)log(2πA)+ d(1 m logm) ρ2√AQ Σ (D)V ∞drδ(r D)Y(r) m d F 2 − 2 − − − − (cid:20) (cid:21) Z0 d d = (ρ)+N (1 m)log(2πA)+ (1 m logm) Nρ√AQ Σ (D)Y(D) , m d F 2 − 2 − − − (cid:20) (cid:21) (13) where (ρ) is the free energy of the non-replicated liquid and Y(D) = G(D) is the value of the pair correlation F function at contact. This is the same result obtained in [19] from the expansion of the HNC free energy, but here it was obtained without need of the HNC approximation. This is important because the HNC approximationgives very poor results for hard spheres systems. Recalling that (ρ) = NS(ρ), where S(ρ) is the equilibrium entropy per particle of the liquid, we obtain the F − replicated free energy by optimizing with respect to A in Eq. (13): 1 d d Φ(m,ρ)= minΨ[ρ(x),f(x,y)]= S(ρ)+ (1 m)log[2πA ]+ (m 1 logm) , ∗ N A − 2 − 2 − − (14) 1 m D A (m)= − ∗ Q ρV (D)Y(D) m d p whereV (D)= 2πd/2Dd isthevolumeofasphereofradiusD. ThefunctionA (m)hasthemeaningofa“cageradius” d dΓ(d/2) ∗ as discussed above. This result holds in any space dimension d. In d = 3 we used the Carnahan-Starling expression [20] for the entropy S(ρ), which reproduces very well the numerical data for the equation of state of the hard sphere liquid. This was done in [19] on a phenomenological ground, but is fully justified by the present derivation in which no reference to the HNC approximation is done. It was shown that Eq. (14) predicts a glass transition density ρ which is in good agreement with numerical results. K Moreover,in the glass phase the cage radius decreases with the density and reaches 0 at a value ρ which is then the c maximum allowed density for an amorphous state, i.e. the random close packing density. For ρ ρ the pressure of c → the glass diverges. It was also shown that the averagenumber of neighbors of a given sphere at ρ is equalto z =2d. c Adetaileddiscussionandacomparisonwithnumericaldatacanbefoundin[19]. Herewewilldiscussthe predictions of Eq. (14) in the limit d . →∞ IV. ENTROPY OF THE LIQUID IN THE LIMIT OF LARGE DIMENSION The problem of computing the entropy of the hard sphere liquid for d was addressed in [31, 32], where the → ∞ same result was obtained in two independent ways. In [31] it was shown that the ring diagrams dominate the virial series order by order in ρ for large d, and the entropy was computed by a resummation of these diagrams; in [32] simple equations for the pair correlation function g(r) were introduced, and solved in the limit d . In both → ∞ cases it was found that S(ρ) is given by the ideal gas term plus the first virial correction (i.e. by the Van der Waals equation). The equations introduced in [32] are indeed the minimal requirements for a pair distribution function g(r). If D is the sphere diameter, ρ=N/V is the density, h(r)=g(r) 1 and h(q) is its Fourier transform, one has: − g(r) 0 , ≥ g(r)=0 for r D , (15)  ≥ S(q)=1+ρh(q) 0 , ≥ Thefirstconditioncomesfromthefactthatg(r)isaprobability(theprobabilityoffindingaparticleatdistancergiven that there is a particle in the origin), the second from the fact that two particles cannot be at distance smaller than 5 D (due tothe hardcorerepulsion). ThethirdconditioniseasilyobtainedbyprovingthatS(q)=1+ρh(q)= ρ 2 , q | | where ρ is a Fourier component of the density fluctuations [20]. q (cid:10) (cid:11) In [32] a solution of Eq.s (15) for d was found. We set the sphere diameter D = 1 (defining V V (1) and d d → ∞ ≡ Y Y(1)) and following [32] we define d=2N +3, the reduced density ρ ρV and d ≡ ≡ logρ 2logρ ρ . (16) 1 ≡ N ≡ d 3 − Note that ρ is related to the packing fraction φ ρV (1/2), i.e. to the fraction of volume covered by the spheres, by d ≡ φ=2 dρ. It isfound thatforρ 1 the solutionofEq.s (15)is simplyg(r)=θ(r 1), while forρ>1it hasthe form − ≤ − g(r) = θ(r 1) 1+exp[ Nh (r)] with h (r) > 0, i.e. it is given by the step function plus an exponentially small 1 1 − { − } correction which can be explicitly computed, see [32]. The pressure is then found to be: P 1ρ for ρ 1 , =1+ 2 ≤ (17) ρ (21ρ(1+e−2NK(pc)) for ρ>1 , where K(p)=log(1+ 1 p2) 1 p2 logp , (18) − − − − and p the solution of p p c φ (p)=log(1+ 1 p2) 1 p2+1 log2=ρ . (19) 0 1 − − − − ItturnsoutthatK(p )>0for0 ρ <φ (1)=p1 log2=p0.3068... Essentiallythepressureisgivenbytheidealgas c 1 0 ≤ − contribution plus the first virialcorrection,with anothercorrectionwhich is exponentially small up to ρ =1 log2. 1 − Above the latter value a solution of Eq.s (15) could not be found. These results are strongly consistent with the results of [31] where it was shown that the resummation of the ring diagrams gives exactly Eq. (17) up to ρ = 1 log2. At this value of the density a pole develops that seems to 1 − correspond to a liquid instability (the Kirkwood instability [31]). It not clear if this instability implies that there are no solutions of Eq.s (15) (or of the HNC equations) for high densities,oritsimplyimpliesthatonehastolookformorecomplexsolutions. Althoughthequestionisveryinteresting from the mathematical point of view (maybe also in relation to the problem of finding the most dense lattices [33]), this question is not physically relevant in this contest. We will show later that the glass transition indeed preempts this instability that is therefore in a non-physical region of the density: this system becomes unstable toward replica symmetry breaking before reaching the Kirkwood instability. Using the exact relation P 1 dS =1+ ρY = ρ , (20) ρ 2 − dρ it turns out that 1ρ for ρ 1 , S(ρ)=1 logρ 2 ≤ (21) − −(12ρ(1+L(pc)e−2NK(pc)) for ρ>1 , where, recalling that ρdpc = 1 dpc = 1 , L(p) is such that dρ N dρ1 Nφ′0(pc) L(p) 2NK (p)L(p) 1 1 ′ ′ L(p)+ − =1 L(p) = . (22) Nφ′0(p) ⇒ ∼ 1−2K′(p)/φ′0(p) 21+√1−p2 1 p2 − Again, up to exponentially small corrections, the entropy of the liquid is given by the ideal gas term plus the first virialcorrection. Similarly,bycomparingEq.s(20)and(17)wefindthatY =1uptoexponentiallysmallcorrections. V. GLASS TRANSITION IN LARGE SPACE DIMENSION To locate the glass transition we substitute Eq. (21) in Eq. (14) and compute the equilibrium complexity [14, 15, 16, 19, 25]: ∂(Φ/m) d Σ(ρ)= m2 =S(ρ) log[2πA (1)]=S(ρ) S (ρ) , (23) ∗ vib ∂m − 2 − (cid:12)m=1 (cid:12) (cid:12) (cid:12) 6 600 400 200 Σ 0 -200 -400 0 0,02 0ρ,04 0,06 0,08 1 FIG. 1: The function Σ(ρ1), Eq. (25), for N =100. Our computation is based on Eq. (14) which comes from an expansion in powers of thecage radius and is valid only close to theKauzmann density where Σ=0. The dashed part of the curveis then unphysical. Unfortunately, the exact point where Eq. (14) breaks down could not be estimated. Thus in the figure the point where thecurvebecomes dashed has been chosen arbitrarily. whereS (ρ) dlog[2πA (1)]isthevibrationalcontributiontotheentropyoftheliquid; theglasstransitiondensity vib ≡ 2 ∗ (or Kauzmann density) ρ is defined by Σ(ρ) = 0. Note that the cage radius is, from Eq. (14), √A ρ 1; thus for K − ∝ ρ 0itis exponentiallylargeinN andthe smallcageexpansiondoes notmakesense. Forρ >0we get,neglecting 1 1 the≤terms related to K(p ) which are exponentially small in N, and using V eπ N 1 , c d ∼ N N√2πN 1 (cid:0) (cid:1) 1 S(ρ)=1 logρ ρ 1 Nρ +Nlog(eπ) NlogN log(√2πN3/2) eNρ1 , 1 − − 2 ∼ − − − − 2 (24) √2π √2π Svib(ρ)=dlog +dlog[ρ(1+e−2NK(pc))] 2Nlog (2N +3)Nρ1 , 0.638 ∼ 0.638 − and the equation for ρ is, neglecting for simplicity the terms growing slower than N, and defining α=2log √2π K 0.638 − log(eπ) 0.59, ∼ 1 0=Σ(ρ ) Nα NlogN +2Nρ (N +1) eNρ1 . (25) 1 1 ∼− − − 2 A plotofΣ(ρ )is reportedinFig.1. Let us firstneglectthe terms proportionalto N: thenthe equationΣ=0 gives 1 1 NlogN =2N2ρ eNρ1 , (26) 1 − 2 and a solution is simply Nρ = 1logN. However the derivative of Σ evaluated in this solution is 1 2 1 Σ′(ρ1)=2N2 NeNρ1 (27) − 2 and is positive, thus this value corresponds to the unphysical zero of Σ (see Fig. 1). The maximum of Σ is given by Σ =0, i.e. Nρ =log4N. Thus we look for a solution of Eq. (26) of the form ′ 1 eNρ1 =qN , (28) and we expect that q diverges with N. Substituting in Eq. (26) we get q =4log(q√N)=2logN +4logq , (29) which is solved iteratively by q =2logN +4log(2logN +4logq)=2logN +4log[2logN +4log(2logN +4logq)]=... . (30) Thus the value of the Kauzmann density is ρ =2NlogN +O(NloglogN) , K 1 (31) ρ = log(2NlogN +O(NloglogN)) . 1K N 7 A. Correction to the Kauzmann density Forfutureconvenienceitisusefultocomputethecorrectiontoρ duetothetermsO(N)wediscardedinEq.(25). K The full Eq. (25) is 1 NlogN +2N2ρ eNρ1 =Nα 2Nρ . (32) 1 1 − − 2 − We look again for a solution eNρ1 =Nq with q large. Then we obtain 4 4logq+2logN q =2α log(qN) . (33) − − N Therighthandsideiso(1)whilethelefthandsidedivergessotheleadingsolutionisq =q withq givenbyEq.(30). 0 0 We look for a solution q =q +q with q q : then the left hand side gives 0 1 1 0 ≪ q 4 4log(q√N) q =4log 1+ 1 q q 1 , (34) 1 1 − q − ∼ q − (cid:18) 0(cid:19) (cid:18) 0 (cid:19) then 4 q 1 =2α+O(N 1logN) , (35) 1 − q − (cid:18) 0 (cid:19) and finally as q logN 0 ∼ q = 2α+O((logN) 1) . (36) 1 − − The result for ρ is then K N ρ =Nq 2Nα+O =Nq 1.18N , (37) K 0− logN 0− (cid:18) (cid:19) with q given by Eq. (30). 0 B. Random close packing density Forρ>ρ the systemisin the glassphase andthe valuem thatoptimizes the free energy(per replica)is m <1 K ∗ ∗ [14, 15, 16, 19, 25]. m is the solution of Σ(ρ,m)=0, where ∗ ∂(Φ/m) Q Σ(ρ,m)=m2 =S(ρ) Nlog(2π)+(2N +3)Nρ Nm Nm(1 m) ′m 1 ∂m − − − − Qm (38) +Nlogm 2Nlog(1 m)+2NlogQ . m − − As Q π/4m for m 0 (see Appendix A), one can see from Eq. (14) that the cage radius A 0 for m 0. m ∗ ∼ → → → The random close packing density ρ is then the value of ρ at which m = 0 [19]. Then to compute ρ we have to p c ∗ c solve Σ(ρ,0)=0. Using Q π/4m, m ∼ p 1 Σ(ρ,0)=S(ρ)+(2N +3)Nρ +N 3log2 . (39) 1 2 − (cid:18) (cid:19) The condition Σ(ρ,0)=0, using Eq. (24), is similar to Eq. (32): 1 NlogN +2N2ρ1 eNρ1 =Nα′ 2Nρ1 , (40) − − 2 − with α =3log2 1 log(eπ) 0.56. The solution is given by Eq. (37) with α α: ′ − 2 − ∼− → ′ N ρ =Nq 2Nα +O =Nq +1.12N =ρ +2.3N , c 0− ′ logN 0 K (cid:18) (cid:19) (41) 1 ρ = log(Nq +1.12N) . 1c 0 N 8 C. Lindemann ratio To check the consistency of the small cage expansion,it is interesting to estimate the Lindemann ratio in the glass phase, when ρ 2NlogN dlogd. The Lindemann ratio L for a given solid phase is the ratio between the typical ∼ ∼ amplitude of vibrations around the equilibrium positions and the mean interparticle distance. In our framework it can be defined as L ρ1/d√A (42) ≡ so that, using √A 1/ρ from Eq. (14), and ρ=ρV dlogd, one has d ∼ ∼ 1 L 1 , (43) ∼ √dlogd ≪ which is consistent with the assumption that vibrations are very small. D. Is the densest packing amorphous in large d? Wecancompareourpredictionforthemaximumdensityofamorphouspackingsρ dlogdwiththebestavailable c ∼ bounds on the density of crystalline packings. The corresponding packing fraction scales as φ 2 ddlogd. Unfor- c − ∼ tunately, the best lower bound for periodic packings is the Minkowski bound ρ 1, while the best upper bound is ∼ the Blichfeldt’s one, ρ 2d/2 [29, 30]. Our result for ρ lies between these bounds so we cannot give an answer to ∼ c the question whether the densest packings of hard spheres in large d are amorphous or crystalline. Hopefully better bounds onthe density ofcrystalline packingswill address this questionin the future. Howeverit is difficult to escape to the impression that the values of the densities of crystalline laminated lattices [29] up to d=50 suggest that there are lattices where ρ goes to a non-zero limit for infinite d. It is however quite possible that this is a preasymptotic 1 effect. It would be very interesting to find the density of laminated lattices in larger dimensions. It would be also interesting to find out the maximum value of the density for which the inequalities (15) have a solution,allowingalsodelta functions inh(p): those equationsarevalidalsoforregularlattices;howeverthe problem ishardanditisnotclearhowtoattacktheproblemmathematically(itisalinearprogrammingprobleminaninfinite dimensional space also for finite dimensions). VI. CONCLUSIONS Making use of the results of [19, 31, 32] we showed that the small cage expansion of the replicated free energy in large space dimension predicts a 1-step replica symmetry breaking transition. We obtained the asymptotic behavior of the glass transition density and of the random close packing density as d . It is worth to note that all the → ∞ phenomenarelatedtotheglasstransitionhappenisaregionofdensitieswhichisveryclosetoρ=1,inthesensethat ρ =(d/2) 1logρisboundedbyo(logd/d),whiletheKirkwoodinstability[31,32]happensforρ =1 log2=0.30..., 1 − 1 − i.e. well beyond the interesting range of values of ρ . In other words, the region of densities where the Kirkwood 1 instability happens is never reached due to the glass transition happening at lower density. It would be interesting to estimate the corrections at finite d to the asymptotic expressions. For the simplified model(15)thecorrectionsareexponentiallysmallindasfoundin[32]. Itisthenreasonablethattheexactexpression for the liquid entropy as well differs from the asymptotic solution of (15) by exponentially small terms, and the same might happen for the replicated free energy. If this is the case, the results we presentedin this paper should be exact for d . We hope that future work will address this point. →∞ Acknowledgments We are pleased to thank G. Biroli, J.-P. Bouchaud, and S. Franz for useful discussions. One of us (F.Z.) benefited a lotfrom a discussionat the end ofhis seminarin LPTHE,Jussieuandwishes to thank allthe participantsfor their comments and suggestions. This work has been supported by the Research Training Network STIPCO (HPRN-CT-2002-00319). 9 APPENDIX A: THE FUNCTION Q(r) The function Q(r) has been defined in Eq. (9) as: Q(r)= du du ρ(u )ρ(u )f(r+u ,u ) F(r) , (A1) 1 2 1 2 1 2 − Z where F(r)=−θ(D−|r|) and f(r+u1,u2)=−1+b aeb−ϕ(r+u1a−u2a). Recalling that duρ(u)=1 we can rewrite Q R Q(r)= du1du2ρ(u1)ρ(u2) e−ϕ(r+u1a−u2a) θ(r D)=F0(r)m θ(rb D) , (A2) − | |− − | |− Z a Y where b b e−(u1)22+A(u2)2 e−4uA2 F (r)= du du θ(r+u u D)= du θ(r+u D) . (A3) 0 1 2 1 2 (√2πA)d | − |− (√4πA)d | |− Z Z From the expressions above we expect that Q(r) is non zero only if r D. | |∼ Let us first discuss the expansion of Q(r) in d=1. Defining 2 t 1 1 t erf(t) dxe x2 , Θ(t)= [1+erf(t)]= dxe x2 , (A4) − − ≡ √π 2 √π Z0 Z−∞ we have ∞ e−4uA2 r D du θ(r+u D)=Θ − . (A5) √4πA − √4A Z−∞ (cid:18) (cid:19) Note that for smallA, u is smalltoo,andas r D, the signofr+uis the sameas the signof r. Thus we canwrite | |∼ θ(r+u D) θ(r+u D)+θ( r u D) and | |− ∼ − − − − r D r+D r D F (r)=Θ − +Θ Θ | |− . (A6) 0 √4A − √4A ∼ √4A (cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19) where in the last step we neglected contributions of order exp( D2/A) for A 0, and − → m r D Q(r)= Θ | |− θ(r D) . (A7) √4A − | |− (cid:20) (cid:18) (cid:19)(cid:21) From the expression above it is easy to see that Q(r) is non zero only if r D √A, as expected. This | | − ∼ is because the function Θ(t) is exponentially close to θ(t) if t is large. Given a smooth function f(r), one has | | drQ(r)f(r) √Af(D) for small √A, i.e. Q(r) √Aδ(r D). We have now to compute the proportionality ∝ ∝ | |− factor. Defining the reduced variable t=(r D)/√4A, we have: R − ∞drQ(r)=2√A ∞ dt[Θ(t)m θ(t)] 2√AQ(A) 2√AQ +o(√A) , (A8) m − ≡ ∼ Z0 Z−√D4A where the function Q is the limit for A 0 of Q(A) and is given by m → Q = ∞ dt[Θ(t)m θ(t)] . (A9) m − Z−∞ It is easy to show that Q is a finite and smooth function of m for m=0, that m 6 Q =(1 m)Q +O[(m 1)2] , m 0 − − (A10) ∞ Q = dtΘ(t)logΘ(t) 0.638 , 0 − ∼ Z−∞ and that Q diverges as Q π/4m for m 0. Thus we get the result Q(r)=2√AQ δ(r D) in d=1. m m m ∼ → | |− p 10 In dimension d>1 we have, recalling that F (r) is rotationally invariant, for R= r: 0 | | drQ(r)= dr[F (r)m θ(r D)]=Ω ∞dRRd 1[F (R)m θ(R D)] , (A11) 0 d − 0 − | |− − − Z Z Z0 where Ω is the solid angle in d dimension, Ω =2πd/2/Γ(d/2). The function F (R) can be written as d d 0 e−4uA2 F (R)= du θ(Ri+u D) , (A12) 0 (√4πA)d | |− Z where i is the unit vector e.g. of the first direction in Rd. Forbsmall √A, the u are small too, and the function θ(ri+u D) is constantalongthe directionsorthogonalto i. Thus we canshowthat the integraloverthe variables | |− u ,µ=b1 is equal to 1 up to corrections of the order of exp( D/√A). We finally get: µ 6 − b b ∞ e−4uA21 R D F (R)= du θ(R+u D)=Θ − , (A13) 0 1 1 √4πA | |− √4A Z−∞ (cid:18) (cid:19) asinthe one dimensionalcase. Again,the function F (R)m θ(R D) islargeonlyforR D √Asoatthe lowest 0 order we can replace Rd 1 with Dd 1 in Eq. (A11). Finally−, using−that F (R) is given by−Eq.∼(A13) as in d= 1, we − − 0 get drQ(r)=Ω Dd 1 ∞dR[F (R)m θ(R D)]=Σ (D)2√AQ , (A14) d − 0 d m − − Z Z0 whereΣ (D)is the surfaceofad-dimensionalsphereofradiusD, Σ (D)=Ω Dd 1,i.e. inanydimensiondwe have d d d − Q(r)=2√AQ δ(r D) . (A15) m | |− [1] J. G. Berryman, Phys. Rev.A 27, 1053 (1983). [2] G. 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