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12The Yield-Strain in Shear Banding Amorphous Solids

Ratul Dasgupta, H. George E. Hentschel and Itamar ProcacciaDepartment of Chemical Physics, The Weizmann Institute of Science, Rehovot 76100, Israel.

Dept of Physics, Emory University, Atlanta GA. 30322

(Dated: November 7, 2012)

In recent research it was found that the fundamental shear-localizing instability of amorphoussolids under external strain, which eventually results in a shear band and failure, consists of ahighly correlated array of Eshelby quadrupoles all having the same orientation and some density. In this paper we calculate analytically the energy E(, ) associated with such highly correlatedstructures as a function of the density and the external strain . We show that for strains smallerthan a characteristic strain Y the total strain energy initially increases as the quadrupole densityincreases, but that for strains larger than Y the energy monotonically decreases with quadrupoledensity. We identify Y as the yield strain. Its value, derived from values of the qudrupole strengthbased on the atomistic model, agrees with that from the computed stress-strain curves and broadlywith experimental results.

I. INTRODUCTION

Amorphous solids are obtained when a glass-former iscooled below the glass transition [13] to a state whichon the one hand is amorphous, exhibiting liquid like or-ganization of the constituents (atoms, molecules or poly-mers), and on the other hand is a solid, reacting elasti-cally (reversibly) to small strains. There is a large vari-ety of experimental examples of such glassy systems, andtheoretically there are many well studied models [46]based on point particles with a variety of inter-particlepotentials that exhibit stable supercooled liquids phaseswhich then solidify to an amorphous solid when cooledbelow the glass transition. Typically all these materials,both in the lab and on the computer, exhibit a so-calledyield-stress above which the material fails to a plasticflow. In previous research [710] it was pointed out thatdepending on the protocol of cooling to the glass state,the plastic response of the system can be either via ho-mogeneous flow or via shear bands. The former obtainstypically when the quenching to the glass state is fast,whereas the latter when the quench is slow. In thelatter case when the stress exceeds some yield stress, thesample, rather than flowing homogeneously in a plasticflow, localizes all the shear in a plane that is at 45 degreesto the compressive stress axis, and then breaks along thisplane [1].

In recent work [11] it was argued that the fundamentalinstability that gives rise to shear bands is the appearanceof highly correlated lines of Eshelby quadrupoles (andsee below for precise definition) which organize the non-affine displacement field of an amorphous solid such thatthe shear is highly localized along a narrow band. Howthis fundamental instability results in macroscopic shearbands, why these appear in 45 degrees to the principalstress axis, and what determines the difference in plasticresponse between fastly and slowly quenched glasses areall subjects of this paper. We will also present an ab-initio calculation of the yield stress at which an amor-phous solid is expected to response plastically with shearlocalization.

In Sect. II we review briefly the type of numericalsimulations that we do and explain the basic facts aboutplasticity of amorphous solids. Section III exhibits thefundamental solution of an Eshelby quadrupolar plasticevent. This section is not particularly new but is im-portant for our purposes in setting the stage and thenotation for the next Sect. IV in which we computethe energy of N such Eshelby quadrupoles in the elasticmedium. We show explicitly that as a function of the ex-ternal strain (or the resulting stress) there is a thresholdvalue at which a bifurcation occurs. Below this value onlyisolated Eshelby quadrupoles can appear in the system,leading to localized plastic events. Above this thresh-old a density of such quadrupoles can appear, and whenthey do appear they are highly correlated, preferring toorganize on a line at 45 degrees to the principal sheardirection. In Sect. V we present the analytic estimateof the yield strain, and demonstrate a satisfactory agree-ment with the numerical simulations. Finally, in Sect.VI we provide a summary of the most important resultsof the paper and offer a discussion of the road ahead.

II. PLASTICITY IN AMORPHOUS SOLIDS

AND SIMULATIONS

As a background to the calculations in this paper weneed to briefly review recent progress in understandingplasticity in amorphous solids [5, 6, 1214]. Below wedeal with 2-dimensional systems composed of N pointparticles in an area A, characterized by a total energyU(r1, r2, rn) where ri is the position of the ith parti-cle. Generalization to 3-dimensional systems is straight-forward if somewhat technical. The fundamental plasticinstability is most cleanly described in athermal (T = 0)and quasi-static (AQS) conditions when an amorphoussolid is subjected to quasi-static strain, allowing the sys-tem to regain mechanical equilibrium after every differen-tial strain increase. Higher temperatures and finite strainrates introduce fluctuations and lack of mechanical equi-librium which cloud the fundamental physics of plastic

2instabilities with unnecessary details [12].In our AQS numerical simulations we use a 50 50

binary Lennard-Jones mixture to simulate the shear lo-calization discussed in this work. The potential energyfor a pair of particles labeled i and j has the form

Uij(rij) = 4ij

[(ijrij

)12(ijrij

)6+A0

+ A1

( rijij

)+A2

( rijij

)2], (1)

where the parametersA0, A1 andA2 are added to smooththe potential at a scaled cut-off of r/ = 2.5 (up to thesecond derivative). The parameters AA, BB and ABwere chosen to be 2 sin(/10), 2 sin(/5) and 1 respec-tively and AA = BB = 0.5, AB = 1(see [15]). Theparticle masses were taken to be equal. The sampleswere prepared using high-temperature equilibration fol-lowed by a quench to zero temperature (T = 0.001) (see[16]). For shearing, the usual athermal-quasistatic shearprotocol was followed where each step comprises of anaffine shift followed by an non-affine displacement usingconjugate gradient minimization. The simulations wereconducted in two dimensions (2d) and employed Lees-Edwards periodic boundary conditions. This implies thata square sample of size L2 remains so also after strain.Samples were generated with quench rates ranging from3.2106 to 3.2102 ( in LJ units ), and were strainedto greater than 100 percent. Simulations were performedon system-sizes ranging from 5000 to 20000 particles witha fixed density of = 0.976 (in LJ units). The simula-tions reported in the paper have 10000 particles and aquench-rate of 6.4 106 (in LJ units).We choose to develop the theory for the case of exter-

nal simple shear since then the strain tensor is traceless,simplifying some of the theoretical expressions. Apply-ing an external shear, one discovers that the response ofan amorphous solids to a small increase in the externalshear strain (we drop tensorial indices for simplicity)is composed of two contributions. The first is the affineresponse which simply follows the imposed shear, suchthat the particles positions ri = xi, yi change via

xi xi + yi xiyi yi yi. (2)

This affine response results in nonzero forces between theparticles (in an amorphous solid) and these are relaxedby the non-affine response ui which returns the systemto mechanical equilibrium. Thus in total ri ri + ui.The nonaffine response ui solves an exact (and modelindependent) differential equation of the form [5, 17]

duid

= H1ij j (3)

where Hij 2U(r1,r2,rn)

rirjis the so-called Hessian ma-

trix and i 2U(r1,r2,rn)

riis known as the non-affine

FIG. 1: (Color Online). Left panel: the localization of thenon-affine displacement onto a quadrupolar structure whichis modeled by an Eshelby inclusion, see right panel. Rightpanel: the displacement field associated with a single Eshelbycircular inclusion of radius a, see text. The best fit parametersare a 2.5 and 0.1. To remove the effect of boundaryconditions, the best fit is generated on a smaller box of size(x, y) [25.30, 75.92]

force. The inverse of the Hessian matrix is evaluated af-ter the removal of any Goldstone modes (if they exist). Aplastic event occurs when a nonzero eigenvalue P of Htends to zero at some strain value P . It was proven thatthis occurs universally via a saddle node bifurcation suchthat P tends to zero like P P [14]. For valuesof the stress which are below the yield stress the plasticinstability is seen [5] as a localization of the eigenfunctionof H denoted as P which is associated with the eigen-value P , (see Fig. 1 left panel). While at = 0 all theeigenfunctions associated with low-lying eigenvalues aredelocalized, P localizes as P (when P 0) on aquadrupolar structure as seen in Fig. 1 left panel for thenon-affine displacement field when the plastic instabilityis approached. These simple plastic instabilities involvethe motion of a relatively small number of particles (say20 to 30 particles) but the stress field that is released hasa long tail.When the strain increases beyond some yield strain,

the nature of the plastic instabilities can change in a fun-damental way [10]. The main analytic calculation thatis reported in Sect. IV shows that when the stress builtin the system is sufficiently large, instead of the eigen-function localizing on a single quadrupolar structure, itcan now localize on a series of N such structures,which are organized on a line that is at 45 degreesto the principal stress axis, with the quadrupolar

structures having a fixed orientation relative to

the applied shear. Fig. 2 shows the non-affine fieldthat is identical to the eigenfunction which is associatedwith this instability, clearly demonstrating the series ofquadrupolar structures that are now organizing the flowsuch as to localize the shear in a narrow strip aroundthem. This is the fundamental shear banding instability.Note that this instability is reminiscent of some chainlikestructure seen in liquid crystals, arising from the orien-tational elastic energy of the anisotropic host fluid [18],and ferromagnetic chains of particles in strong magneticfields [19]. The reader should note that the event shown

3FIG. 2: (Color Online). Left panel: The nonaffine displace-ment field associated with a plastic instability that results ina shear band. Right panel: the displacement field associatedwith 7 Eshelby inclusions on a line with equal orientation.Note that in the left panel the quadrupoles are not preciselyon a line as a result of the finite boundary conditions and therandomness. In the right panel the series of N = 7 Eshelbyinclusions, each given by Eq. (18) and separated by a distanceof 13.158, using the best fit parameters of Fig. 1, have beensuperimposed to generate the displacement field shown.

in Fig. 2 will move the particles only a tiny amount, andit is the repeated instability where many such event hitat the same region which results in the catastrophic eventthat is seen as a shear band in experiments. Neverthelessthis is the fundamental shear localization instability andin the sequel we will have to understand why repeatedinstabilities hit again and again in the same region. Weshould stress that this is not inevitable, for samples thatare prepared by a fast quench these instabilities appearat random places adding up to what seems to be a ho-mogeneous flow. For further discussion of this point seeSect. VI.

III. DISPLACEMENT IN 2D FOR A CIRCULAR

INCLUSION

As said above, the shear localizing instability appearsonly when the stress exceeds a threshold. To explainwhy, we turn now to analysis. As a first step we modelthe quadrupolar stress field which is associated with thesimple plastic instability as a circular Eshelby inclusion[5].

A. Circular Inclusion

We consider a 2d circular inclusion that has beenstrained into an ellipse using an eigenstrain or a stress-free strain which we take to be traceless ie

= 0

[20]. Here and below repeated indices imply a summationin 2-dimensions. A general expression for such a tracelesstensor can be written in terms of a unit vector n and ascalar as

= (2nn ) . (4)

We also assume that a homogenous strain acts glob-

ally (which in our case also triggers the local transforma-tion of the inclusion). This strained ellipsoidal inclusionboth feels a traction exerted by the surrounding elas-tic medium resulting in a constrained strain c in theinclusion, and itself exerts a traction at the inclusion-elastic medium interface resulting in the originally un-strained surroundings developing a constrained strain

field c(~X). Here and below ~X stands for an arbitrary

cartesian point in the material which for this purpose isapproximated as a continuum.A fourth-order Eshelby tensor S can be defined

which relates the constrained strain in the inclusion cto the eigenstrain viz.

c = S. (5)

Now for an inclusion of arbitrary shape the constrainedstrain c, stress

c , and displacement field u inside

the inclusion are in general functions of space. For ellip-soidal inclusions, however, it was shown by Eshelby [2123] that the Eshelby tensor and the constrained stressand strain fields inside the inclusion become indepen-dent of space. We work here with a circular inclusionwhich is a special case of an ellipse and hence for such aninclusion, the Eshelby tensor is [2123]

S =4 18 (1 ) +

3 48 (1 ) ( + ) ,(6)

where is the Poissons ratio. Note that this is the Es-helby tensor for an inclusion in 2-dimensions. It is thesame as a cylindrical inclusion in 3-dimensions under aplane strain [23]. From Eqs. (5) and (6), we obtain

c =

[4 18 (1 ) +

3 48 (1 ) ( + )

]

=3 44 (1 )

For a traceless eigenstrain. (7)

The total stress, strain and displacement field inside thecircular inclusion is then given by

I = c +

=

3 44 (1 )

+

I = c + C

(c +

)uI = u

c + u

=

[3 44 (1 )

+

]X .

(8)

Here the super-script I indicates the inclusion and the denotes the eigenstress which is linearly related tothe eigenstrain by C, and which for anisotropic elastic medium simplifies further using

C + ( + ) , (9)to:

= 2 +

(10)

=E

1 + +

E(1 + ) (1 2)

,

4where and are the Lames parameters. One can eitherchoose the two Lames coefficients or E and as the twoindependent material parameters. The relations betweenthem are given by

= E2(1+) , =E

(1+)(12) (11)

E = (3+2)+ , =

2(+) (12)

These relations are correct in 3-dimensions with planestrain conditions and therefore also in 2-dimensions [23].The stress in the inclusion can now be written down

in terms of independent variables using Eq. (10) by

I = C(c +

)=

E1 +

c +E

(1 + ) (1 2)c

E1 +

+E

1 + (13)

as and are traceless. Note that Eq. (7) implies

that for a traceless eigenstrain, the constrained straininside the inclusion viz. c is also traceless and thus we

obtain from Eq. (13),

I =E

1 + c

E1 +

+E

1 +

=E

1 +

3 44 (1 )

E1 +

+E

1 +

=E

4 (1 + ) (1 ) +

E1 +

(14)

B. Constrained Fields in the Elastic Medium

In the surrounding elastic medium the stress, strain anddisplacement fields are all explicit function of space andcan be written

m(~X) = c(

~X) +

m( ~X) = c( ~X) +

um ( ~X) = uc( ~X) + u

( ~X) .

(15)

In order to compute the displacement field uc(~X) in

the isotropic elastic medium we need to solve the Lame-Navier equation

E2 (1 + ) (1 2)

2ucXX

+E

2 (1 + )

2ucXX

= 0,(16)

as there are no body forces present in our calculation.The constrained fields in the inclusion will supply theboundary conditions for the fields in the elastic mediumat the inclusion boundary. Also as r the con-strained displacement field will vanish.All solutions of equation Eq. (16) also obey the higher

order bi-harmonic equation

4ucXXXX

= 22uc = 0. (17)

Thus our objective is to construct from the radial so-lutions of the bi-laplacian equation Eq. (17) derivativeswhich also satisfy Eq. (16). Note that Eq. (17) is only anecessary (but not a sufficient) condition for the solutionsand Eq. (16) still needs to be satisfied. The calculationis presented in Appendix A, with the final result

uc( ~X) = (18)

4(1 )(ar

)2 [2(1 2) +

(ar

)2 ][2nn ~X X

]

+

2(1 )(ar

)2 [1

(ar

)2 ][2(n ~X)2r2

1]X .

C. Fit to the data

Armed with this analytic expression we return now toour numerics, cf. Fig. 1, and fit the two parametersin Eq. (18) to the data of the displacement exhibitedby a single localized plastic event. The result of thisprocedure is a 2.5 and 0.1. The quality of thisfit can be judged from the right panel of Fig. 1 wherewe exhibit the form of Eq. (18) with the parametersfitted to the displacement field in the left panel. Alsothe value of a appears reasonable since it means thatabout a2 20 particles are involved in the core of therelaxation event. On physical grounds this is about theright order of magnitude.We will keep these parameters fixed in all our calcula-

tions below. The reader should note that this is an ap-proximation when there are multiple quadrupoles in thesystem, since they influence each other and the solutionleading to Eq. (18) should be repeated in the presence ofmany quadrupoles. We expect however that the changesin the parameters should not be large when the densityof the quadrupoles is small. We will always work in thesmall density limit a2 1 where is the area densityof quadrupoles N/L2.

IV. THE ENERGY OF N ESHELBY

INCLUSIONS EMBEDDED IN A MATRIX

A. Notation

Having the form of a single quadrupole, Eq. (18),we turn now to the calculation of the energy associ-ated with N quadrupoles embedded in an elastic ma-trix. Once computed, we will show later that for largestrains the minimum of this energy is obtained for a lineof quadrupoles all having the same orientation. Fromnow on we use the notation that u, u . The en-ergy of N Eshelby inclusions embedded in a linear elasticmedium (or matrix) N , is given by the expression

E=1

2

Ni=1

V

(i)0

(i)

(i)dV +

1

2

V

Ni=1 V

(i)0

(m)

(m) dV(19)

5FIG. 3: Schematic of four Eshelby inclusions embedded in a matrix m

where the superscript i indicates the index of the inclu-sion and m indicates the matrix. We evaluate Eq. (19)in Appendix B. The result can be expressed in termsof four contributions: Emat which is the contribution ofthe strained matrix, E which is the energy of the N

qudrupoles in the external strain, Eesh which representsthe self energy of the N quadrupoles (their cost of cre-ation) and lastly Einc represents the energy of interactionbetween the inclusions. Explicitly

Emat 12()

() V = V

()xy

()xy =

V E22 (1 + )

(20)

E 12()

(Ni=1

(,i) V

(i)0

)= a2()xy

Ni=1

(,i)yx = a2E(1 + )

Ni=1

n(i)x n(i)y (21)

Eesh 12

Ni=1

(,i)

(c,i) V

(i)0 +

1

2

Ni=1

(,i)

(,i) V

(i)0 =

a2

2

Ni=1

((,i) (c,i)

)(,i) (22)

Einc 12

Ni=1

(,i)V(i)0

j 6=i

(c,j) (Rij)

= a2

2

ij

[(,i)

(c,j) (Rij) +

(,j)

(c,i) (Rij)

](23)

The above expressions are specific to 2D, for a global strain corresponding to simple shear under the linear

approximation. Thus xy =2 here, and the traceless eigenstrain takes the form

(,i)yx = 2n

(i)x n

(i)y .

The form of Einc is not final, and we bring it to its final form in Appendix C. The final result is

Einc = E()2a2

8(1 2)ij

(a

Rij

)2

[ 8

{(1 2) +

(a

Rij

)2}(4(n(i) n(j)

)(n(i) rij

)(n(j) rij

)

2(n(i) rij

)2 2

(n(j) rij

)2+ 1

)+ 4

(2 (1 2) +

(a

Rij

)2)(2(n(i) n(j)

)2 1

)

8(1 2

(a

Rij

)2)(2(n(i) rij

)2 1

)(2(n(j) rij

)2 1

)

+ 32

(1

(a

Rij

)2)((n(i) rij

)(n(j) rij

)(n(i) n(j)

)(n(i) rij

)2 (n(j) rij

)2)](24)

6where rij ~XijRij .Our task is now to find the configuration of N Eshelby

quadrupoles that minimize the total energy. Obviously,if the external strain is sufficiently large, we need tominimize E separately, since it is proportional to .The minimum of (21) is obtained for

n(i)x = n(i)y =

12. (25)

Substituting this result in Eq. (24) simplifies it consid-erably. We find

Einc = a2()2E

8(1 2)

(a

Rij)2{ 8[(1 2) + ( a

Rij)2] + 4[2(1 2) + ( a

Rij)2] 8[1 2( a

Rij)2][2(n r)2 1]2

+32[1 ( aRij

)2][(n r)2 (n r)4]}. (26)

We can find the minimum energy very easily. Denotex (n r)2, and minimize the expression A[2x 1]2 B[x x2]. The minimum is obtained at x = 1/2, orcos =

1/2. We thus conclude that when the line of

correlated quadrupole forms under shear, this line is in45 degrees to the compressive axis, as is indeed seen inexperiments. Of course there are two solutions for this,perpendicular to each other. Note that for other externalstrains which are not consistent with a traceless straintensor (or in 3-dimensions) this conclusion may change.

The physical meaning of this analytic result is thatit is cheaper (in energy) for the material to organize Nquadrupolar structures on a line of 45 degrees with thecompressive stress, all having the same orientation, thanany other arrangement of these N quadrupoles, includ-ing any random distribution. This explains why sucha highly correlated distribution appears in the strainedamorphous solid, and why it can only appear when theexternal strain (or the built-up stress) are high enough.This fact, in addition to the observation that such anarrangement of Eshelby quadrupoles organizes the dis-placement field into a localized shear, explains the originof this fundamental instability.

V. ESTIMATE OF YIELD-STRESS AND

NUMBER OF ESHELBY QUADRUPOLES

In this section we turn to estimate the yield stress andthe associated density of Eshelby quadrupoles. To thisaim we need to compute one other energy term that wasnot needed until now, namely Eesh which was the samefor all the configurations of the quadrupoles.

A. Expression for Eesh

The energy term Eesh was given by Eq. (22) which canbe re-written as

=a2

2

Ni=1

(,i) C

((,i) (c,i)

)(27)

=a2

2

Ni=1

(,i) C

((,i)

3 44(1 )

(,i)

), Cf. Eq. 7

=a2

2

Ni=1

(,i) C

(,i)

4(1 ) .

For an isotropic matrix, using Eq. (9), we have

C(,i) =

(,i) + ( + )

(,i)

= 2(,i) =

E1 +

(,i) (28)

for a symmetric traceless eigenstrain. Using Eq. (28) forcircular inclusions each of radius a in 2D, we obtain

Eesh =a2

2

E4(1 2)

Ni=1

(,i)

(,i) (29)

Now,

(,i)

(,i) = (

(,i))2(2n(i) n

(i)

)(2n

(i) n

(i)

)= 2((,i))2 (30)

For all eigenstrains equal (ie. (,i) = ), we have

Eesh =Ea2N ()24(1 2) (31)

7B. Einc for a line of equidistant quadrupoles with

the same polarization

At this point we need to compute the energy term Eincfor the special configurations of N quadrupoles that areequi-distant and with the same polarization, organized

in a line. In this case we have

n(i) = n, rij = r, and Rij = |j i|R (32)

Starting from Eq. (26) we specialize to the present situ-ation

Einc = E()2a2

8(1 2)N1i=1

Nj,j>i

( aR

)2 1(j i)2

{ 8

[(1 2) +

( aR

)2 1(j i)2

]

+ 4

[2 (1 2) +

( aR

)2 1(j i)2

] 8

[1 2

( aR

)2 1(j i)2

] [2(n r)2 1

]2+ 32

[1

( aR

)2 1(j i)2

])[(n r)2 (n r)4

]}(33)

For (n r)2 = 1/2, the above expression reduces to

Einc = E()2a2

2(1 2)N1i=1

Nj,j>i

[2

(j i)2( aR

)2 3

( aR

)4 1(j i)4

](34)

We have

N1i=1

Nj=i+1

1

(j i)s =N1i=1

Nin=1

1

ns N (s) for N >> 1 , (35)

where (s) is the Riemann zeta function. Thus we obtain

Einc = E()2a2N

2(1 2)[2( aR

)2(2) 3

( aR

)4(4)

]. (36)

At this point we realize that the distance R betweenthe quadrupoles is not determined. We will choose Rby demanding that the line density of the quadrupolesremains invariant in the thermodynamic limit, or R L/N . Thus, with N/L, the energy density in thestrip L a reads

EincLa

= E()2a

2(1 2)[2(a)2(2) 3(a)4(4)

](37)

Similarly from equation 31, we obtain

EeshLa

=Ea()24(1 2) (38)

From equation (25) we have

E

La= aE

2 (1 + ) (39)

Thus the plastic energy density is given by

E(, )

La E

+ Eesh + EincLa

(40)

=E()24(1 2)

[A

(1

Y

)aB(a)3 + C(a)5

],

where A = 1, B = 4(2) and C = 6(4). Yis defined as

Y

2(1 ) . (41)

Eq. (40) is plotted, using the numerical values of theparameters found in our numerical simulations in Fig.VB for various values of . For < Y the minimumof the expression is attained always at = 0, and theshear localization cannot occur. Only at Y a newsolution opens up to allow a finite density of the Es-helby quadrupoles. Therefore Y is by definition the yieldstrain.

8FIG. 4: (Color Online). The total plastic energy for the cre-ation of an array of quadrupoles with line density for threevalues of : = Y 0.1, = Y 0.05, and = Y

VI. SUMMARY AND CONCLUSIONS

We have presented a theory of the fundamental insta-bility that leads to shear localization and eventually toshear bands. One remarkable observation is that the nat-ural plastic instability that occurs spontaneously in oursimulations results in a displacement field that is sur-prisingly close to the one made by an Eshelby circularinclusion, see Fig. 1. The best fit for the parameter a, ofthe order of 2.5 is in agreement with the intuitive beliefthat shear transformation zones involve 20-30 particles,as a2 would predict. Basing our analysis on this simi-larity we could develop an analytic theory of the energyneeded to create N such inclusions, whether scattered inthe system randomly or aligned and organized in highlycorrelated shear localized structure. We discover thatthe latter becomes energetically favorable when exceedsY 2(1) . In our system 0.215, and with our best

fit 0.1 we predict Y 0.07 which is right on the

mark as one can seen from Ref. [11].While we believe that our calculation of the energy of

N quadrupoles is accurate for densities such that a2 1, the interaction between the quadrupoles become muchmore involved for higher densities, and we avoided thiscomplication. The consequence is that we cannot pre-dict a-priory the critical density of our quadrupoles, andwe leave this interesting issue for future research. An-other issue that warrants further study is whether theparameters a and are material parameters which aredetermined by the small scale structure of the glass, andif so, how to estimate them a-priori. Also, are these pa-rameters dependent on the way the system is stressed,i.e. via shear or via tensile compression etc..Next we need to discuss the difference in behavior of

glasses that were quenched relatively quickly and thosethat were quenched relatively slowly. In our simulationswe find that the latter exhibit the fundamental shearlocalization instability again and again along approxi-mately the same line, accumulating displacement that

FIG. 5: (Color Online). Repeated shear localization insta-bilities in the case of slow quench. Upper panel: the energyvs. strain and the instabilities that were chosen for displayin the lower panel. The average non affine displacement fieldafter the instabilities that are marked in the upper panel. Thepoint to notice is that the instability falls repeatedly on thesame band, accumulating to a shear band.

develops into a shear band. On the other hand theformer tends every time to have plastic instabilities atdifferent places, sometime localized and sometime morecorrelated, and on the average this then appears like ahomogenous flow. This difference is stressed in Fig. 5where the case of slow quench is exhibited. It appearsthat fastly quenched systems may have plastic instabili-ties almost everywhere, and there is no special preferencefor one line or another. Slowly quenched systems are ini-tially harder to shear localize, but once it happens in aparticular (random) line it is likely to repeat again andagain along the same line. Making these words quantita-tive is again an issue left to future research. In particularit is interesting to find what is the precise meaning of fastand slow quenches, fast and slow compared to what, andhow this changes the microscopic local structure.Finally the effects of temperature and finite strain rates

on the present mechanism constitute a separate piece ofwork which of course is of the utmost importance. Likethe other open subject mentioned above, it will be takenup in future research.

Acknowledgments

This work had been supported in part by the IsraelScience Foundation, the German-Israeli Foundation andby the European Research Council under an ideasgrant.

9Appendix A: The displacement field of an Eshelby

circular inclusion

1. Solutions of the Lame-Navier equation

We look for linear combinations of derivatives of the ra-dial solutions of Eq. 17 which are linear in the eigenstrain and go to zero at large radii. In addition the termsmust transform as components of a vector field. Such asolution can be written down as

uc = A

ln r

X+B

3 ln r

XXX

+ C3(r2 ln r

)XXX

, (A1)

where X is the component of the position vector withthe origin at the center of the Eshelby quadrupole, and

r | ~X|. It can be checked that any other terms areeither zero, do not go to zero as r , or do not trans-form as components of a vector. We re-write equationEq. (16) as

(1

1 2)

2ucXX

+2uc

XX= 0 (A2)

which is the equation for the constrained displacementfield in the elastic matrix subject to appropriate bound-ary conditions. From Eq. (A1), we obtain

2ucXX

=A

X

[2

XXln r

]+B

3

XXX

[2

XXln r

]+C

3

XXX

[2

XXr2 ln r

].(A3)

We need the following identities

2

XX(ln r) = 0

2

XX

(r2 ln r

)= 4 ln r + 4

(A4)

Thus we obtain from equation A3,

2ucXX

= 4C3 ln r

XXX(A5)

Similarly, the expression for

2ucXX

=2

XX

[A

ln r

X+B

3 ln r

XXX+ C

3(r2 ln r

)XXX

]

=

X

[A

2 ln r

XX+B

4 ln r

XXXX+ C

4(r2 ln r

)XXXX

]

(A6)

which can be re-written (noting that the second and thirdterms involve the laplacian for which we have identities

from Eqs. (A4) as

2ucXX

=

X

[A

2 ln r

XX+ C

2 (4 ln r + 4)

XX

]

= (A+ 4C) 3 ln r

XXX(A7)

10

Plugging expressions (A5) and (A7) in Eq. (A2), we thusobtain

(A+4C)12

3 ln rXXX

+ 4C3 ln r

XXX= 0

[A+4C12 + 4C

]

3 ln rXXX

= 0 C = A8(1) (A8)

We can thus re-write Eq. (A1) as

uc = A

ln r

X+B

3 ln r

XXX A

8 (1 )

3(r2 ln r

)XXX

(A9)

The following identities are now required:

ln r

X=

Xr2

3 ln r

XXX=

2r2 (X +X +X) + 8XXXr6

3(r2 ln r

)XXX

=2r2 (X +X +X) 4XXX

r4(A10)

Using these relations we can re-write Eq. A9 as

uc = A

Xr2

+B

[2r2 (X +X +X) + 8XXXr6

]

A8 (1 )

[2r2 (X +X +X) 4XXX

r4

]

= AXr2

[2B

r4+

A

4 (1 ) r2] (X +X +X) +

[8B

r6+

A

2 (1 ) r4]XXX(A11)

Remembering that the eigenstrain is traceless,we find that (X +X +X) = 2X using which we

can simplify Eq. (A11) to obtain

uc =[Ar2 4B

r4 A2(1)r2

]X +

[8Br6

+ A2(1)r4

]X

XX

=[Ar2

122(1) 4Br4

]X +

[8Br6

+ A2(1)r4

]X

XX (A12)

At r = a (the radius of the circular inclusion), the form of expression Eq. (A12) must match the form of the constraineddisplacement field of the inclusion which from Eq. (7) is 344(1)

X . Thus the co-efficient of the second term in

expression (A12) must go to zero at the inclusion boundary, which gives us

8Ba6

+ A2(1)a4 = 0

B = a2A16(1) (A13)

Thus we have

uc =[Ar2

122(1) 4r4 a

2A16(1)

]X +

[8r6

a2A16(1) +

A2(1)r4

]X

XX

= A4r2(1)

[2 (1 2) + a2

r2

]X +

A2r4(1)

[1 a2

r2

]X

XX

(A14)

And the value of uc at r = a should match the value obtained from Eq. (7), implying

3 44 (1 ) =

A

4 (1 ) a2 [2 (1 2) + 1]3 44 (1 ) =

A (3 4)4 (1 ) a2 A = a

2 (A15)

11

The expression for uc becomes:

uc =1

4 (1 )(a2

r2

)[2 (1 2) +

(a2

r2

)]X +

1

2 (1 )(a2

r2

)[1

(a2

r2

)]

XXXr2

(A16)

From Eq. 4, we have = (2nn ) and thus

X = [2n

(n ~X

)X

]X

X =

X (2nn )X = [2(n ~X

)2 r2

](A17)

allowing us to write the final vectorial expression for the displacement field:

~uc(~X)=

4 (1 )(a2

r2

)[2 (1 2) +

(a2

r2

)] [2n(n ~X

) ~X

]

+

2 (1 )(a2

r2

)[1

(a2

r2

)]2(n ~X

)2r2

1

~X

(A18)

We can also derive expressions for the constrained stress and strain fields. Noting that

f(r)

X= f (r)

r

X= f (r)

Xr

(n ~X

)X

= n (A19)

12

we obtain

ucX

=

X

[

4 (1 )(a2

r2

){2 (1 2) +

(a2

r2

)}{2n

(n ~X

)X

}

+

2 (1 )(a2

r2

){1

(a2

r2

)}2(n ~X

)2r2

1

X

]

=

4 (1 ){4 (1 2)

(a2

r4

) 4

(a4

r6

)}{2n

(n ~X

)X

}X

+

4 (1 ){2 (1 2)

(a2

r2

)+

(a4

r4

)}{2nn }

+

2 (1 ){(

2a2

r4

)+

(4a4

r6

)}2(n ~X

)2r2

1

XX

+

2 (1 ){(

a2

r2

)(a4

r4

)}4(n ~X

)n

r2

4(n ~X

)2X

r4

X

+

2 (1 ){(

a2

r2

)(a4

r4

)}2(n ~X

)2r2

1

uc

X=

4 (1 )[ 4

(a2

r2

){(1 2) +

(a2

r2

)}2n

(n ~X

)r

Xr

Xr

+

(a2

r2

){2 (1 2) +

(a2

r2

)}{2nn }

4(a2

r2

){1 2

(a2

r2

)}2(n ~X

)2r2

1

XXr2

+8

(a2

r2

){1

(a2

r2

)}(n ~X

)n

r(n ~X

)2r2

Xr

Xr

+2

(a2

r2

){1

(a2

r2

)}2(n ~X

)2r2

1

]

(A20)

13

Thus the expression for c 12(ucX

+ucX

)is

c( ~X) =

8 (1 )[ 4

(a2

r2

){(1 2) +

(a2

r2

)}{2

(n ~Xr

)(nXr

+nXr

) 2XX

r2

}

+2

(a2

r2

){2 (1 2) +

(a2

r2

)}{2nn }

8(a2

r2

){1 2

(a2

r2

)}2(n ~X

)2r2

1

XXr2

+8

(a2

r2

){1

(a2

r2

)}(n ~Xr

)(Xnr

+Xnr

) 2

(n ~X

)2r2

XXr2

+4

(a2

r2

){1

(a2

r2

)}2(n ~X

)2r2

1

]

(A21)

allowing us to write the final expression for the constrained strain in the matrix

c(~X) =

4 (1 )[ 4

(a2

r2

){(1 2) +

(a2

r2

)}{(n ~Xr

)(nXr

+nXr

) XX

r2

}

+

(a2

r2

){2 (1 2) +

(a2

r2

)}{2nn }

4(a2

r2

){1 2

(a2

r2

)}2(n ~X

)2r2

1

XXr2

+4

(a2

r2

){1

(a2

r2

)}(n ~Xr

)(Xnr

+Xnr

) 2

(n ~X

)2r2

XXr2

+2

(a2

r2

){1

(a2

r2

)}2(n ~X

)2r2

1

]

(A22)

14

The trace of c is not zero in the elastic medium and is given by

c =

4 (1 )[ 4

(a2

r2

){(1 2) +

(a2

r2

)}2

(n ~Xr

)2 1

+ 0

4(a2

r2

){1 2

(a2

r2

)}2(n ~X

)2r2

1

+ 0

+ 4

(a2

r2

){1

(a2

r2

)}2(n ~X

)2r2

1

]

(A23)

c =

4 (1 )[ 4

(a2

r2

){(1 2) +

(a2

r2

)}2

(n ~Xr

)2 1

+ 4

(a2

r2

)(a2

r2

)2(n ~X

)2r2

1

]

c = (1 21

)(a2

r2

)2(n ~X

)2r2

1

(A24)

We are now in a position to calculate the constrained stress in the elastic medium due to the deformed inclusion.It is given by the expression

cij =E

1 + c +

E(1 + ) (1 2)

c =

E4 (1 2)

[.....

] E

(1 2)(a2

r2

)2(n ~X

)2r2

1

(A25)

where

[.....

]is the expression inside the square brackets in Eq. A22.

2. Constrained displacement field - Cartesian components

It proves useful to have the explicit cartesian components of the displacement field for computational and graphicalpurposes. If we consider the unit-vector n making an angle of with the positive direction of the x-axis, then the

15

cartesian components of equation A18 are :

ucx =

4 (1 )(a2

r2

)[2 (1 2) +

(a2

r2

)][2 cos (x cos+ y sin) x]

+

2 (1 )(a2

r2

)[1

(a2

r2

)][2 (x cos+ y sin)

2

r2 1

]x

ucx =

4 (1 )(a2

r2

)[2 (1 2) +

(a2

r2

)][x cos 2+ y sin 2]

+

2 (1 )(a2

r2

)[1

(a2

r2

)][(x2 y2) cos 2+ 2xy sin 2

r2

]x

ucy =

4 (1 )(a2

r2

)[2 (1 2) +

(a2

r2

)][2 sin (x cos+ y sin) y]

+

2 (1 )(a2

r2

)[1

(a2

r2

)][2 (x cos+ y sin)

2

r2 1

]y

ucy =

4 (1 )(a2

r2

)[2 (1 2) +

(a2

r2

)][x sin 2 y cos 2]

+

2 (1 )(a2

r2

)[1

(a2

r2

)][(x2 y2) cos 2+ 2xy sin 2

r2

]y

(A26)

Appendix B: Calculation of the energy of N quadrupoles

Eq. (19) can be re-written using 1/2 (u, + u,) as

E =1

4

Ni=1

V i0

(i)

(u(i), + u

(i),

)dV +

1

4

V

Ni=1 V

(i)0

(m)

(u(m), + u

(m),

)dV (B1)

Using the symmetry of the stress tensor, we obtain

E =1

2

Ni=1

V i0

(i)u

(i),dV +

1

2

V

Ni=1 V

(i)0

(m) u

(m),dV (B2)

We also have the identity

u, = (u), ,u = (u), (B3)

if there are no body forces. Thus we can write Eq. B2 as

E =1

2

Ni=1

V

(i)0

((i)u

(i)

),dV +

1

2

V

Ni=1 V

(i)0

((m) u

(m)

),dV (B4)

Using Gausss theorem to convert these volume integrals into area integrals, we obtain

E =1

2

Ni=1

Si0

(i)u

(i) n

(i) dS

Ni=1

1

2

Si0

(m) u

(m) n

(i) dS +

1

2

S

(m) u

(m) n

() dS (B5)

16

where n(i) and n() are unit normal vectors both pointing outwards respectively from the inclusion volume V i0 andthe matrix boundary. Eq. B5 can be rewritten as follows

E =1

2

S()

(m) u

(m) n

() dS +

1

2

Ni=1

S

(i)0

((i)u

(i) (m) u(m)

)n(i) dS

E = 12()

()

S()

Xn() dS +

1

2

Ni=1

S

(i)0

((i)u

(i) (m) u(m)

)n(i) dS

E = 12

V +

1

2

Ni=1

S

(i)0

((i)u

(i) (m) u(m)

)n(i) dS

(B6)

Thus we can write using the expressions earlier writtendown in Eq. (15)

(m)

(~X)=

() +

Ni=1

(c,i)

(~X)

(m)

(~X)=

() +

Ni=1

(c,i)

(~X)

u(m)

(~X)= u() (

~X) +

Ni=1

u(c,i)

(~X)

(B7)

where (c,i)

(~X)indicates the constrained strain at lo-

cation ~X in the matrix due to the eshelby labeled with

the index i etc. We also have for locations ~X inside theinclusions

(i)

(~X)=

() +

j 6=i

(c,j)(~X)+

(c,i) (,i)

(i)

(~X)=

() +

j 6=i

(c,j)(~X)+

(c,i) (,i)

u(i)

(~X)= u() +

j 6=i

u(c,j)

(~X)+ u(c,i) (,i) X

(B8)

where (,i) is the eigenstrain of the ith Eshelby and so on.Note that in the expression for the strain in the inclusiongiven by Eq. (B8) we have removed the eigenstrain from

the constrained strain (c,i) (,i) leaving only the elastic

contribution in order to calculate correctly the elasticcontribution to the energy. Using these expressions, theelastic energy of the system can be written from Eq. (B6)

E =1

2()

() V (B9)

+1

2

Ni=1

S

(i)0

((i)u

(i) (m) u(m)

)n(i) dS .

Since the traction force has to be continuous at the in-clusion boundary (Newtons third law), we have

(i) n

(i) =

(m) n

(i) at the inclusion boundary (B10)

which gives us from Eq. (B9),

E=1

2()

() V +

1

2

Ni=1

S

(i)0

(i)n

(i)

(u(i) u(m)

)dS(B11)

We also have from Eqs. (B7) and (B8),

u(i) u(m) = (,i) X . (B12)

On plugging this expression into Eq. (B11) gives us finally

E =1

2()

() V

1

2

Ni=1

S

(i)0

(i) n

(i)

(,i) XdS

= 12()

() V

1

2

Ni=1

(,i)

V

(i)0

((i)X

),dV

= 12()

() V

1

2

Ni=1

(,i)

V

(i)0

(i)dV

E = 12()

() V

1

2

Ni=1

V(i)0

(,i)

(i)

(B13)

where i (1/V i0 )V i0

(i)dV . Using the expression for

(i) from Eq. (B8), we obtain

(i)(

~X) () +j 6=i

(c,j) (Rij) + (c,i) (,i).(B14)

Eq. (B14) is a far field approximation that assumes thatRij a. As Rij a clearly the spatial integrals con-tibuting to c,i must be computed explicitly and cannotbe replaced by the single distance Rij between the centersof the Eshelby inclusions i and j.

Using expression (B13) we obtain

17

E =1

2()

() V

1

2()

(Ni=1

(,i) V

(i)0

) 1

2

Ni=1

(,i)

(c,i) V

(i)0 +

1

2

Ni=1

(,i)

(,i) V

(i)0

12

Ni=1

(,i)V(i)0

j 6=i

(c,j) (Rij)

E = Emat + E + Eesh + Einc. (B15)

where all these terms are defined in Eqs. (20)-(23).

Appendix C: The final form of Einc

We can also explicitly write Einc showing its linear dependence on the eigenstrain by using Eq. (23). This gives us

Einc = a2

2ij

[(2n(i) n

(i)

)(c,j) (Rij) +

(2n(j) n

(j)

)(c,i) (Rij)

]. (C1)

Plugging Eq. (A25) into the above equation, we find that the term inside the square braces in Eq. (C1) can be writtenas:

=E

(2n

(j) n

(j)

)4 (1 2)

[ 4

(a

Rij

)2{(1 2) +

(a

Rij

)2}

{(

ni ~X ijRij

)(n(i) X

(ij)

Rij+n(i) X

(ij)

Rij

) X

(ij) X

(ij)

(Rij)2

}

+

(a

Rij

)2{2 (1 2) +

(a

Rij

)2}{2n(i) n

(i)

}

4(

a

Rij

)2{1 2

(a

Rij

)2}2(n(i) ~X(ij)

)2(Rij)2

1

X(ij) X

(ij)

(Rij)2

+ 4

(a

Rij

)2{1

(a

Rij

)2}

(n(i) ~X(ij)

Rij

)(X

(ij) n

(i)

Rij+X

(ij) n

(i)

Rij

) 2

(n(i) ~X(ij)

)2(Rij)2

X(ij) X

(ij)

(Rij)2

+ 2

(a

Rij

)2{1

(a

Rij

)2}2(n(i) ~X(ij)

)2(Rij)2

1

]

E

1 2(a2

r2

)(2(n(i) ~X(ij))2

r2 1

)(2n(j) n

(j)

)

+ i j (C2)

where ~X(ij) indicates the vector joining the centers of the eshelby pair labeled as i and j, and i j in Eq. (C2)represents the term obtained by exchanging i and j.

18

In order to simplify Eq. (C2), we need the following identities

(2n(j) n

(j)

){( n(i) ~X(ij)Rij

)(n(i) X

(ij)

Rij+n(i) X

(ij)

Rij

) X

(ij) X

(ij)

(Rij)2

}+ i j

= 8n(i) n(j)(n(i) ~X(ij)

Rij

)(n(j) ~X(ij)

Rij

) 4

(n(i) ~X(ij)

Rij

)2 4

(n(j) ~X(ij)

Rij

)2+ 2 (C3)

(2n(j) n

(j)

)(2n(i) n

(i)

)+ i j = 4

[2(n(i) n(j)

)2 1

](C4)

(2n(j) n

(j)

) X(ij) X(ij)(Rij)2

+ i j = 22(n(j) ~X(ij)

Rij

)2 1

(C5)

(2n(j) n

(j)

)(n(i) ~X(ij)

Rij

)(X

(ij) n

(i)

Rij+X

(ij) n

(i)

Rij

) 2

(n(i) ~X(ij)

)2(Rij)2

X(ij) X

(ij)

(Rij)2

+ i j

= 8

(n(i) ~X(ij)

Rij

)(n(j) ~X(ij)

Rij

)(n(i) n(j)

) 8

(n(i) ~X(ij)

Rij

)2(n(j) ~X(ij)

Rij

)2(C6)

(2n(j) n

(j)

) + i j = 0 (C7)

Using these identities, we can write the final expression for the interaction energy in the form shown in Eq. (24).

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