1 NOTATIONS 2 NUMERICAL SIMULATIONS 3 TWO DIMENSIONAL CIRCULAR INCLUSION. Supplementary Notes

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1 Supplementary Notes 1 NOTATIONS In the followng notes, we wll reserve the Greek letters α,β,γ,... for tensor subscrpts, whle the Latn letters,j,k,... wll be used for partcles and Eshelby nclusons. NUMERICAL SIMULATIONS To prepare qualty data for the present dscusson we have performed two-dmensonal D Molecular Dynamcs smulatons on a bnary system whch s an excellent glass former and s known to have a quas-crystallne ground state [1, ]. Each atom n the system s labeled as ether small S or large L and all the partcles nteract va Lennard Jones LJ potental. All dstances r r j are normalzed by λ SL, the dstance at whch the LJ potental between the two speces becomes zero and the energy s normalzed by ǫ SL whch s the nteracton energy between two speces. Temperature was measured n unts of ǫ SL /k B where k B s Boltzmann s constant. For detaled nformaton on the model potental and ts propertes, we refer the reader to Ref [1]. The number of partcles n our smulatons s 1 at a number densty n =.985 wthapartcleraton L /N s = 1+ 5/4. ThemodecouplngtemperatureT MCT forthssystem sknowntoresdecloseto.35. Allpartcleshavedentcalmassm = 1andtmesnormalzed ǫ SL λ SL /m. For the sake of computatonal effcency, the nteracton potental s to t = smoothly truncated to zero along wth ts frst two dervatves at a cut-off dstance r c =.5. To prepare the glasses, we frst start from a well equlbrated lqud at a hgh temperature of T = 1. whch s supercooled to T =.35 at a quenchng rate of Secondly, we then equlbrate these supercooled lquds for tmes greater than τ rel, where τ rel s the tme taken for the self ntermedate scatterng functon to become 1% of ts ntal value. Lastly, followng ths equlbraton, we quench these supercooled lquds deep nto the glassy regme at a temperature of T =.1 at a reduced quench rate of Followng ths quench we take the glass to mechancal equlbrum nearest energy mnma by a conjugate gradent energy mnmzaton. At ths pont we start shearng our glasses under athermal quas-statc condtons. 3 TWO DIMENSIONAL CIRCULAR INCLUSION Consder an elastc sold havng a volume V and surface area S [Fg. 1]. The materal wll be assumed to be homogeneous wth an elastc stffness tensor gven by C γδ. Let a sub-volume V wth surface area S undergo a unform permanent nelastc deformaton, such as a structural phase transformaton. The materal nsde V s called an ncluson and the materal outsde s called the matrx. If we could remove ths ncluson from ts surroundng materal then t would attan a state of unform stran and zero stress. Such a stress free stran s referred to as the egen-stran ǫ. The egen stress s then gven by σ = C γδǫ γδ. In realty, the ncluson s surrounded by the matrx. Therefore, t s not able to reach the state of egen-stran and zero stress. Instead, both the ncluson and the matrx wll deform and experence an elastc stress feld. The Eshelby s transformed ncluson problem [3] s to solve the stress, stran and dsplacement felds both n the ncluson and n the matrx. We consder a D crcular ncluson that has been straned nto an ellpse usng an egen-stran ǫ and whch allows for a volume change ǫ νν. A general expresson for such a tensor 1

2 Fgure 1: Cartoon showng an elastc medum of volume V and surface area S. Insde the medum a small ellpsodal regon volume V and surface area S undergoes an rreversble plastc deformaton. The materal nsde V s called as the ncluson and the materal outsde s referred to as the matrx. can be wrtten n terms of a unt egen vectors ˆn,ˆk and correspondng egen values ζ n,ζ k as follows: ǫ = ζ nˆn αˆn β +ζ kˆkαˆkβ 1 Usng orthogonalty of the egen drectons: ˆn αˆn β +ˆk αˆkβ = δ, we get: ǫ = ζ n ζ k ˆn αˆn β δ + ζ n +ζ k δ = ǫ, +ǫ,t where the trace-less part ǫ, and the trace part ǫ,t are gven as ǫ, = ζ n ζ k ˆn αˆn β δ = ζ n +ζ k δ 3 ǫ,t We also assume that a homogeneous stran ǫ that acts globally whch n our case also trggers the local transformaton of the ncluson. Ths straned ellpsodal ncluson feels a tracton exerted by the surroundng elastc medum resultng n a constraned stran ǫ c n the ncluson and also exerts a tracton at the ncluson-matrx nterface resultng n the orgnally unstraned surroundngs developng a constraned stran feld ǫ c X. The egen-stran ǫ n the ncluson s related to the constraned stran ǫc Eshelby tensor S γδ : va a fourth order ǫ c = S γδǫ γδ 4 Now for an ncluson of arbtrary shape the constraned stran ǫ c, stress σc dsplacement feld u c X nsde the ncluson are n general functons of space. For ellpsodal nclusons, however, t was shown by Eshelby [3, 4, 5] that the Eshelby tensor and the constraned stress and stran felds nsde the ncluson become ndependent of space. We work here wth a crcular ncluson

3 whch s a specal case of an ellpse and hence for such an ncluson, the Eshelby tensor S γδ reads S γδ = λ µ 4λ+µ δ δ γδ + λ+3µ 4λ+µ δ αδδ βγ +δ αγ δ βδ 5 where λ,µ are the two Lame coeffcents. Pluggng Eq 5 n Eq 4, we get ǫ c = λ+µ λ+µ ǫ,t + λ+3µ λ+µ ǫ, 6 The total stress, stran and dsplacement feld nsde the crcular ncluson s then gven by ǫ I = ǫ c +ǫ σ I = σ c σ +σ C γδǫ c γδ ǫ γδ +ǫ γδ u I α = u c α +u α = ǫ c +ǫ X β 7 where the superscrpt I ndcates the ncluson. The egen stress σ stran ǫ as s related to the egen σ = C γδǫ γδ = µǫ +λǫ ηηδ 8 where we have used the followng defnton of the fourth order elastc stffness tensor C γδ for an sotropc elastc medum C γδ = λδ δ γδ +µδ αγ δ βδ +δ αδ δ βγ 9 The stress n the ncluson can now be wrtten down n terms of ndependent varables usng equatons 8 as σ ǫ I = µ c ǫ +ǫ +λ ǫ c ηη ǫ ηη +ǫ ηη δ 1 4 CONSTRAINED FIELDS IN THE MATRIX In the surroundng elastc matrx, the stress, stran and dsplacement felds are all explct functons of space and can be wrtten as ǫ m X = ǫc X+ǫ σ m X = σc X+σ u m X = uc X+u 11 Thus for a crcular ncluson n order to compute the dsplacement feld u c αx n the sotropc elastc medum. Ths dsplacement feld wll satsfy the Lame-Naver equaton wthout any body forces u c γ µ+λ +µ u c α = 1 X α X γ X β X β The constraned felds n the ncluson wll supply the boundary condtons for the dsplacement feld n the matrx at the ncluson boundary. Also as r, the constraned dsplacement feld wll vansh. All solutons of Eq. 1 also obey the hgher order b-harmonc equaton 4 u c α X β X β X λ X λ = 4 u c α = 13 3

4 Thus our objectve s to construct from the radal solutons of the b-laplacan equaton Eq. 13 dervatves whch also satsfy Eq. 1. Note that the b-laplacan equaton s only a necessary but not a suffcent condton for the solutons and Eq. 1 stll needs to be satsfed. 4.1 SOLUTION OF THE NAVIER-LAME EQUATION From the foregong secton, we note that the constraned dsplacement feld due to the Eshelby soluton s gven as u c α = ǫ c X β [ λ+µ = λ+µ ǫ,t + λ+3µ ] λ+µ ǫ, X β = ζ n +ζ k λ+µ X α + λ+3µ λ+µ λ+µ ǫ, X β 14 where we take the expresson for ǫ c from Eq.6.We wll look for lnear combnatons of the dervatves of the radal solutons of the b-harmonc equaton 13 whch are lnear n the egenstran and go to zero at large dstance. In addton the terms must transform as a vector feld. Let u c,t and u c, be the solutons to the Naver-Lame equatons. We then have: µ+λ u c,t γ +µ u c,t α = X α X γ X β X β µ+λ u c, γ +µ u c, α = 15 X α X γ X β X β Usng the radal solutons of Naver-Lame equaton we can construct the followng combnatons whch transform as a vector and also go to zero as r. and u c,t α = Aǫ,T lnr +Bǫ,T X βγ β u c, Usng the denttes we see from Eq 16 and smlarly 3 lnr +Cǫ,T X α X β X βγ γ 3 lnr X α X β X γ 16 α = A ǫ, lnr +B ǫ, 3 lnr X βγ +C ǫ, 3 lnr β X α X β X βγ 17 γ X α X β X γ lnr X β X β = ; Pluggng Eq 19, nto Eq 15, we get lnr X β X β = 4lnr u c,t α = 4Cǫ,T 3 lnr X β X ηλ 19 β X α X η X λ u c,t γ = A+4Cǫ,T 3 lnr X α X ηλ γ X α X η X λ We can now rewrte Eq 16 as C = Aλ+µ 4λ+µ 1 u c,t α = Aǫ,T lnr +Bǫ,T X βγ β 3 lnr Aλ+µ X α X β X γ 4λ+µ ǫ,t βγ 4 3 lnr X α X β X γ

5 We now compute the followng denttes lnr = X β X β 3 lnr = r X α δ βγ +X β δ αγ +X γ δ +8X α X β X γ X α X β X γ r 6 3 lnr X α X β X γ = r X α δ βγ +X β δ αγ +X γ δ X α X β X γ r 4 3 Usng these denttes, we can now wrte down Eq as [ u c,t α = Aǫ,T X β B r 4 + Aλ+µ ] [ λ+µ ǫ,t 8B βγ X αδ βγ +X β δ αγ +X γ δ + r 6 + Aλ+µ ] λ+µr 4 ǫ,t βγ X αx β X γ 4 and smlarly [ ] [ ] u c, α = A ǫ, X β B r 4 + A λ+µ λ+µ ǫ, 8B βγ X αδ βγ +X β δ αγ +X γ δ + r 6 + A λ+µ λ+µr 4 ǫ, βγ X αx β X γ 5 Now usng Eq. 3, we can rewrte equatons 4 and 5 as u c, α = A µ λ+µ 4B r 4 u c,t α = ζ n +ζ k µax α λ+µ 6 ǫ, 8B X β + r 6 + A λ+µ λ+µr 4 ǫ, βγ X αx β X γ 7 The complete soluton for the dsplacement feld n the matrx s then gven as u c α = u c,t α +u c, α = ζ n +ζ k µax α λ+µ + A µ λ+µ 4B r 4 ǫ, 8B X β + r 6 + A λ+µ λ+µr 4 ǫ, βγ X αx β X γ 8 Now at r = a ncluson boundary, the form of soluton 8 must match wth the Eshelby soluton 14. Ths mples, A = a λ+µ µ, A = a, B = a4 λ+µ 8λ+µ 9 Pluggng Eq. 9 nto Eq. 8, we get: u c α = λ+µ λ+µ a [ µ ζ n +ζ k X α + λ+µ + a ǫ, X β + 1 a ǫ, βγ ] X α X β X γ 3 Notng that and ǫ, X β = ζ n ζ k ˆn α ˆn X X α 31 ǫ, βγ X αx β X γ = ζ n ζ k ˆn X X α 3 5

6 we fnd that Eq. 3 fnally becomes u c X = ζ n ζ k λ+µ a 4λ+µ + 1 a ˆn ˆr 1 [ ζ n +ζ k µ ζ n ζ k X + ] X λ+µ + a ˆn Xˆn X 33 where ˆr = X r. For the purpose of vsualzng the dsplacement feld n space, t proves useful to have expressons for the Cartesan components of the feld gven by Eq. 33. If we consder the unt vector ˆn makng an angle φ wth the x-axs, then the Cartesan components of En 33 are: u c xx = ζ [ n ζ k λ+µ a 4λ+µ ζ n +ζ k µ ζ n ζ k x+ λ+µ + a xcosφ + ysnφ + 1 a x y ] cosφ+xysnφ x u c yx = ζ [ n ζ k λ+µ a 4λ+µ ζ n +ζ k µ ζ n ζ k y + λ+µ + a xsnφ ycosφ + 1 a x y ] cosφ+xysnφ y Takng dervatves of dsplacement feld and u c α = ζ n ζ k λ+µ a X β 4λ+µ µ λ+µ + a µ + λ+µ + a +8 1 a u c β = ζ n ζ k λ+µ a X α 4λ+µ µ λ+µ + a µ + λ+µ + a +8 1 a [ ζ n +ζ k ζ n ζ k ˆn ˆrˆn α X α r ˆn αˆn β δ ˆn ˆrˆn β ˆn ˆr X β r [ ζ n +ζ k ζ n ζ k ˆn ˆrˆn β X β r ˆn αˆn β δ ˆn ˆrˆn α ˆn ˆr X α r δ X αx β Xβ r 1 a Xα r + ˆn ˆr Xα X β 1 1 a δ X αx β Xα r 1 a Xβ r + The constraned stran n the matrx can be wrtten as ǫ c = 1 u c α + uc β X β X α ] ˆn ˆr 1 δ ˆn ˆr Xα X β 1 1 a ] ˆn ˆr 1 δ

7 Usng equatons 35 and 36, Eq. 37 becomes ǫ c X = ζ [ n ζ k λ+µ a 4λ+µ ζ n +ζ k ζ n ζ k { µ ˆnα λ+µ + a X β ˆn ˆr + ˆn βx α r r µ + ˆn λ+µ + a αˆn β δ 1 a { ˆnβ +4 1 a X α ˆn ˆr + ˆn αx β r r ] + 1 a ˆn ˆr 1 δ It s easy to see that the trace of ǫ c X s gven as a ǫ c ηηx = ζ n ζ k µ λ+µ δ X αx β X αx β } ˆn ˆr 1 ˆn ˆr X αx β ˆn ˆr 1 } Xα X β We can now calculate the constraned stress n the elastc matrx due to the deformed Eshelby ncluson. It follows from Hooke s law: σ c X = µǫc X+λǫc ηηxδ 4 Pluggng Eq. 38 n Eq. 4, we get the fnal expresson for the constraned stress σ c X = ζ [ n ζ k µλ+µ a λ+µ ζ n +ζ k δ X αx β ζ n ζ k { µ ˆnα λ+µ + a X β ˆn ˆr + ˆn βx α X } αx β r r µ + ˆn λ+µ + a αˆn β δ 1 a ˆn ˆr Xα X β 1 41 { ˆnβ +4 1 a X α ˆn ˆr + ˆn αx β ˆn ˆr X } αx β r r + 1 a ˆn ˆr 1 δ λ ] ˆn ˆr 1 δ. λ+µ 5 ENERGY OF N ESHELBY INCLUSIONS EMBEDDED IN THE MATRIX The energy of the N Eshelby nclusons embedded n a lnear elastc matrx m s gven by the followng expresson E = 1 σ ǫ dv + 1 V N σ m ǫm dv 4 where the superscrpt denotes the ndex of the ncluson and m denotes the matrx. Eq 4 can be re-wrtten usng the defnton of stran ǫ = 1/u α,β +u β,α, where u α,β = uα X β : E = 1 4 σ u α,β +u β,α dv V σ m N V u m α,β +um β,α dv 43 7

8 Usng the symmetry of the stress tensor, we obtan E = 1 σ u β,α dv + 1 If there are no body forces, we also have the dentty Thus we can wrte Eq. 44 as E = 1 V N σ m um β,αdv 44 σ u β,α = σ u β,α σ,α u β = σ u β,α 45 σ u β +,αdv 1 V σ N V m um β dv 46,α Usng Gauss s theorem, we convert these volume ntegrals nto area ntegrals to obtan E = 1 S σ u βˆn αds 1 σ m um β ˆn αds + 1 σ m S um β ˆn α ds 47 S where ˆn and ˆn areuntvectorsbothpontngoutwardsrespectvelyfromthenclusonvolume and the matrx boundary. Eq. 47 can be re-wrtten as follows E = 1 σ m u β ds + 1 S = 1 σ ǫ βγ = 1 σ ǫ βα V + 1 Thus we can re-wrte the Eq. 11 as S X γˆn α ds + 1 S S ǫ m X = N σ u β σm um β ˆn αds S σ u β σm um β σ m X = N u m X = N σ u β σm um β ˆn αds ǫ c, X+ǫ σ c, X+σ ˆn αds 48 u c, X+u 49 whereǫ c, XdenotestheconstranedstranatX nthematrxduetotheth Eshelbyncluson. We also have for locatons X nsde the nclusons ǫ X = j σ X = j u αx = j ǫ c,j X+ǫc, X ǫ, +ǫ σ c,j X+σc, X σ, +σ u c,j α X+u c, α X X β +u α 5 8

9 where s the egen-stran of the th Eshelby ncluson and so on. Note that n the expresson for the stran n the ncluson gven by Eq, 5, we have subtracted the contrbuton of egenstran from the constraned stran leavng only the elastc contrbuton to calculate correctly the elastc contrbuton to the energy. Usng these expressons, the elastc energy of the system can be wrtten from Eq. 48 E = 1 σ ǫ βα V + 1 S σ u β σm um β ˆn αds 51 Snce the tracton force has to be contnuous at the ncluson boundary Newton s III rd law, we have σ ˆn α = σ m ˆn α 5 whch gves us from Eq. 51, E = 1 σ ǫ V + 1 We also have from equatons 49 and 5 On pluggng ths expresson nto Eq. 53 we fnally get E = 1 σ ǫ V 1 = 1 σ ǫ V 1 = 1 σ ǫ V 1 = 1 σ ǫ V 1 S σ ˆn αu β um β ds 53 u β um β = ǫ, βξ X ξ 54 S βξ βξ σ ˆn α βξ X ξds σ X ξ dv,α σ δ ξαdv βα σ 55 where σ 1/V V σ dv. Usng the expresson for σ from equaton Eq.5, we obtan σ X = σ + j σ c,j rj +σ c, rj σ, 56 Eq. 56 s a far feld approxmaton whch assumes that r j a. As r j a, clearly the spatal ntegrals contrbutng to σ c, must be computed explctly and cannot be replace by the sngle dstance r j between the centers of the Eshelby nclusons and j. Usng equatons 55 and 56, we can wrte down the fnal form of the elastc energy expresson. E = 1 σ ǫ V 1 σ 1 βα V j σ c,j rj βα 1 βα σc, V + 1 βα σ, V = E mat +E +E esh +E nc 57 9

10 where each component of energy defned as E mat = 1 σ ǫ βα V E = 1 σ E esh = 1 E nc = 1 βα V σ, σc, ǫ, βα V βα V j σ c,j rj 58 Here the egen-stran and volume V assocated wth any th Eshelby ncluson are gven as = πa = ζ n ζ k ˆn αˆn β δ + ζ n +ζ k δ 59 Also for a D materal beng loaded under un-axal stran wth free boundares along ŷ, we can wrte the form of the global stress tensor as σ σ = xx 6 By Hooke s law, we get the expresson for appled global stress tensor Takng trace of Eq. 61, we fnd ǫ ηη = Pluggng Eq. 6 n Eq. 61, we fnd σ = µǫ +λǫ ηηδ 61 1 λ+µ σ ηη = 1 λ+µ σ xx 6 σ xx = 4µλ+µγ λ+µ 63 where γ s the external stran. In the followng, we dscuss these components of energy n detal: E mat : It s the elastc energy that would be present n the straned matrx n the absence of nclusons. Pluggng equaton 6 and 63 nto Eq. 58, we get the followng expresson E mat = µλ+µγ V λ+µ E : It s the contrbuton to the elastc energy by the Eshelby nclusons themselves. Note that ths term makes a negatve contrbuton n comparson to E mat tself. Agan pluggng equatons 6, 63 and 59 nto Eq. 58, we fnd E = πa µλ+µγ λ+µ { ζn ζ k ˆn x 1+ ζ } n +ζ k

11 To fnd the orentaton of each Eshelby wth respect to the prncpal stress drecton, we need to mnmze E wth respect to θ, where θ = cos 1 n x = sn 1 n y. We thus get d ζn ζ k dθ θ = or π cos θ 1+ ζ n +ζ k = Hence each Eshelby ncluson must be orented along the prncpal stress drecton or θ =. E esh : It s the self energy requred to create the Eshelby ncluson and s postve. To calculate ths term, we need the followng quanttes: 66 σ c, = C γδǫ c, γδ = C γδ S γδkl = λ+µ λ+µ [ kl ζ n +ζ k λ+µδ + µλ+3µζ n ζ k ˆn λ+µ αˆn β δ ] 67 In dervng Eq. 67, we have used equatons 9, 4 and 5. The egen stress for the th Eshelby ncluson can be wrtten [usng Eq. 8] as σ, = λ+µζ n +ζ k δ +µζ n ζ k ˆn αˆn β δ 68 Combnng ths and the expresson for egen-stran from Eq. 59, Eq. 58 becomes E esh = πa σ, σc, ǫ, βα = πa Nµλ+µ {ζ n +ζ k +ζ n ζ k } 69 4λ+µ E nc : Ths term arses due to the nteracton between Eshelby nclusons under the far feld approxmaton. We note from Eq. 58 E nc = 1 = πa V <j> j { σ c,j rj βα σc,j rj +ǫ,j βα σc, } rj 7 Recallng the value of egen-stran Eq. and constraned stress Eq. 41 due to an th Eshelby ncluson, we wrte down Eq. 7 after smplfyng: E nc = πa µλ+µ a λ+µ <j> ζ n ζ k { µ 8 λ+µ + a µ + λ+µ + a r j a r j r j r j ˆn ˆn j 1 [ ζk ζ n ˆn ˆ rj + ˆn j ˆ rj ˆn ˆn j ˆn ˆ rj ˆn j ˆ rj ˆn ˆ rj ˆn j ˆ rj +1 1 a ˆn ˆ rj 1 ˆn j ˆ rj 1 r j }] ˆn ˆn j ˆn ˆ rj ˆn j ˆ rj ˆn ˆ rj ˆn j ˆ rj In dervng Eq. 71, we have used the followng denttes 71 11

12 δ Xj [ ˆn j ˆ ζ n ζ k rj αx j β r j ˆn j α X j β r j = ζ n ζ k 1 ˆn ˆ rj + ˆn j β Xj α Xj αx j β r j r j ˆn ˆn j ˆn ˆ rj ˆn j ˆ rj ˆn ˆ ˆnj αˆn j β δ = ζ n ζ k XαX j j β = ζ n +ζ k + ζ n ζ k r j j ˆn α X j + ˆn j β Xj α [ ˆn j ˆ rj β r j ˆn ˆ rj ˆn j ˆ rj r j ] ˆn ˆn j 1 = ζ n +ζ k ˆn j ˆ rj 1 + rj ˆn j ˆ rj + 1 ˆn ˆ rj 1 ˆn j ˆ rj Xj αx j β r j ] = ζ n ζ k ˆn ˆn j ˆn ˆ rj ˆn j ˆ rj δ = ζ n +ζ k 7 To calculate the shear band angle wth respect to the prncpal drecton of stran, we must mnmze E nc wth respect to θ. Assumng all egen drectons to be the same,.e takng ˆn = ˆn j = ˆn and ˆn ˆ rj = cos θ = χ, we fnd, by puttng d dχ E nc = : χ = 1 1 ζ n +ζ k 4ζ n ζ k or θ = cos ζ n +ζ k ζ n ζ k 73 References [1] M. Wdom, K. J. Strandburg, and R. H. Swendsen, Phys. Rev. Lett. 58, [] F. Lancon and L. Bllard, J. Phys. France 49, [3] J. D. Eshelby, Proc. R. Soc. Lond. A 41, [4] J. D. Eshelby, Proc. R. Soc. Lond. A 5, [5] C. Wenberger, W. Ca and D. Barnett Lecture notes, Elastcty and Mcroscopc Structures 4 1