It is shown that the elastic energy far from a point defect in an isotropic
solid is mainly shear elastic energy. The calculation, which is based on a
standard dipole expansion, shows that no matter how large or small the bulk
modulus is compared to the shear modulus, less than 10% of the distant point
defect energy is associated with volume changes.Comment: Brief not

A. F. Andreev

C. Zener

J. D. Eshelby

Jeppe C. Dyre

L. D. Landau

P. A. Varotsos

T. Mura

Y. Kraftmakher

English

arXiv

Crossref

Dominance of shear elastic energy far from a point defect in a solid

Udgivelsesdato: martsIt is shown that the elastic energy far from a point defect in an isotropic solid is mainly shear elastic energy. The calculation, which is based on a standard dipole expansion, shows that less than 10% of the distant point defect energy is associated with volume changes, no matter how large or small the bulk modulus is compared to the shear modulus

Dyre, Jeppe

Roskilde Universitet

RoskildeUniversityDominance of shear elastic energy far from a point defect in a solidDyre, JeppePublished in:Physical Review B. Condensed Matter and Materials PhysicsDOI:10.1103/PhysRevB.75.092102Publication date:2007Document VersionPublisher's PDF, also known as Version of recordCitation for published version (APA):Dyre, J. (2007). Dominance of shear elastic energy far from a point defect in a solid. Physical Review B.Condensed Matter and Materials Physics, 75(9), 092102-1 - 092102-4.https://doi.org/10.1103/PhysRevB.75.092102General rightsCopyright and moral rights for the publications made accessible in the public portal are retained by the authors and/or other copyright ownersand it is a condition of accessing publications that users recognise and abide by the legal requirements associated with these rights.            • Users may download and print one copy of any publication from the public portal for the purpose of private study or research.            • You may not further distribute the material or use it for any profit-making activity or commercial gain.            • You may freely distribute the URL identifying the publication in the public portal.Take down policyIf you believe that this document breaches copyright please contact rucforsk@ruc.dk providing details, and we will remove access to thework immediately and investigate your claim.Download date: 27. Mar. 2021Dominance of shear elastic energy far from a point defect in a solidJeppe C. DyreDNRF Centre of Viscous Liquid Dynamics “Glass and Time,” IMFUFA, Department of Sciences, Roskilde University,Postbox 260, DK-4000 Roskilde, DenmarkReceived 16 May 2006; revised manuscript received 22 January 2007; published 12 March 2007It is shown that the elastic energy far from a point defect in an isotropic solid is mainly shear elastic energy.The calculation, which is based on a standard dipole expansion, shows that less than 10% of the distant pointdefect energy is associated with volume changes, no matter how large or small the bulk modulus is comparedto the shear modulus.DOI: 10.1103/PhysRevB.75.092102 PACS numbers: 61.72.JiAn isotropic solid just has two independent elastic con-stants, the shear modulus G and the bulk modulus K. It hasbeen suggested that when a physical property depends onboth G and K, the dependence on the shear modulus is oftenthe most important.1 Examples of this “shear dominance”were given by Granato2 in his profound paper from 1992;they include the fact that defect energies vary only slightlywith the bulk modulus, but are linearly dependent on theshear modulus for Cu, he estimated that only about 3% ofthe energy of an interstitial is bulk elastic energy. In Ref. 1which dealt with the non-Arrhenius viscosity of viscous liq-uids and its possible explanation in terms of temperature-dependent instantaneous elastic constants,3 the following re-sult was proved: If temperature dependence is quantified interms of log-log derivatives, at least 92% of the temperaturedependence of the molecular vibrational mean-square dis-placement over temperature comes from the instantaneousshear modulus; at most 8% is due to that of the instantaneousbulk modulus.The question is how general shear dominance is: Is itcoincidental or is it a fairly general theme of three-dimensional elasticity? Of course, a property such as thePoisson ratio depends equally on the bulk and shear moduli,so it cannot be a completely general phenomenon. Neverthe-less, there are so many examples that shear dominance de-serves to be investigated. As one contribution to shed lighton this question, we calculate below the maximum ratio ofbulk elastic energy to shear elastic energy far from an arbi-trary point defect in an isotropic solid. Based on a standarddipole expansion,4 it is shown that less than 10% of the dis-tant elastic energy is bulk elastic energy, i.e., associated withdensity changes.Although our main objective is to identify the dominantcontribution to the elastic energy far from a point defect, itshould be noted that according to Zener’s “strain energymodel”5 most energy associated with a point defect is elastic.Elastic energies of point defects fall off rapidly as r−6 asevident from Eshelby’s famous solution of the problem of anellipsoidal inclusion,6–12 a result which is general. Thismeans, of course, that most of the elastic energy is locatedclose to the point defect and that a dipole expansion is notrealistic for calculating the total defect energy. Nevertheless,our results may be taken as an indication of what contributesmost to the total defect energy, the shear or the bulk elasticenergies in the defect surroundings. Moreover, the nature ofthe far-field deformation is important for understanding andmodeling the long-range properties of defect-defect interac-tions. If, for instance, the displacement field is dominated byshear displacements, a defect will only interact weakly withanother defect far away if the latter is dilatational close to itscenter.Some time ago, Andreev discussed the unique topologicalcharacteristics of point defects in three dimensions.13 A com-prehensive review of point defect properties emphasizing thevacancy was given by Kraftmakher.14 The question of thenature of the deformation far from a point defect is relevantalso for applications in materials science, e.g., for under-standing fracture.9 In its macroscopic description, the ques-tion of the far-field properties of a point defect was studiedbecause of its importance for understanding the mechanicalproperties of polycrystals and composites.4,6,15 In this con-text, Onaka recently calculated the elastic shear energy dueto a macroscopic point defect an “inclusion”,16 but did notstudy the general case to find the ratio between shear andbulk energy see also Ref. 12. Garikipati et al.17 recentlydiscussed the role of continuum elasticity in determining theformation energy of a point defect utilizing Eshelby’s resultfor the work done by an external stress during the transfor-mation of an inclusion;6 this interesting paper also elucidatedthe limitations of this approach as compared to atomisticcalculations. We mention these works also to emphasize thecontinuing interest in the far-field properties of point defectsand their macroscopic analogies.First some preliminaries: We use the standard Einsteinsummation convention. The point defect is located at theorigin and is modeled as follows: Imagine a small sphere ofradius R surrounding the point defect with all atoms withinthe sphere removed. The effect of the defect on the surround-ings is represented by a suitable distribution of externalforces Fi applied to the surface of the sphere. We define thetensor ij as the following integral over the surface of thesphere where dA is the area element:ij = y=RFiyyjdA . 1Because the force distribution models the effect of the atomswithin the sphere on the surroundings, the angular momen-tum of the force distribution must be zero. This implies that is symmetric: ij = ji.PHYSICAL REVIEW B 75, 092102 20071098-0121/2007/759/0921024 ©2007 The American Physical Society092102-1Sharp brackets  r, referred to as “averages,” denote inte-grations over the surface of a sphere with radius rR cen-tered at the origin. The following identities become usefullater:xixjr =r23ij 2andxixjxkxlr =r415ijkl + ik jl + il jk . 3If xx is an abbreviation of ijxixj, we find from theseidentities thatxxr =r23tr  4andxx2r =r415tr 2 + 2 tr2 . 5Moreover, if x is the vector whose ith component is ijxj,we have2xxr = x2r =r23tr2 . 6As mentioned, far from the defect both the bulk and theshear elastic energies vary with the distance from the originas r−6. When one averages over a sphere with radius rR,the result for both the bulk and the shear elastic energy mustbe a scalar varying with distance as r−6 that is of secondorder in the forces Fi. It follows from the calculation belowthat these two scalars are both uniquely determined by G, K,and the  matrix. Consequently, because these two scalarfunctions are of second order in the forces, the ratio of bulkto shear elastic energy must have the following general struc-ture:Bulk elastic energyShear elastic energy=Atr 2 + B tr2Ctr 2 + D tr2. 7Defining =tr2 / tr 2, this ratio is A+B / C+D.The quantity  varies between 1/3 and : By normalizingwe may assume that tr =1; if the eigenvalues are denotedby i, we thus have 1+2+3=1 and =12+22+32.From this it follows that  varies between 1/3 and . Sincethe energy ratio of Eq. 7 is a monotonous function of  andsince perfect isotropy =1/3 implies zero bulk elastic en-ergy in this case, the displacement field is radially symmet-ric varying as r−2, which implies no volume changes, themaximum bulk to shear elastic energy ratio arises in the limit→. Thus in the calculation below, we may ignore allterms with tr  and keep only terms with a tr2 factor.Poisson’s ratio  is defined18 by =123K − 2G3K + G. 8If a force F is applied at the origin of an isotropic elasticcontinuum, the displacement field at the point x ,y ,z isgiven18 byux,y,z 3 − 4F + F · nnr, 9where r2=x2+y2+z2 and n= x ,y ,z /r is the unitvector pointing from the origin to x ,y ,z. It is convenientto introduce the variable	 	 2 − 4 , 10in terms of which Eq. 9 becomesui  	 + 1Fir+Fjnjnir. 11We proceed to perform a standard dipole expansion byfirst noting that if y is the coordinate for a point on the smallsphere surrounding the defect y  =R and x is the coordi-nate for the point of interest far away, to lowest order iny  / x  =R / x we have, if r	x,1x − y= x2 + y2 − 2x · y−1/2= r−11 − 2x · yr2−1/2= r−1 +x · yr3. 12Similarly,1x − y3= x2 + y2 − 2x · y−3/2= r−31 − 2x · yr2−3/2= r−3 + 3x · yr5. 13To calculate the displacement field at point x, we first notethat when the force Fiy is integrated over the small sphereradius R, the result is zero. Thus when one integrates overthe small sphere, the first term of Eq. 11 to lowest order in1/r becomes	 + 1y=RFiyx − ydA = 	 + 1y=RFiyx · yr3dA= 	 + 1r−3ijxj . 14Similarly, to lowest order the second term of Eq. 11 givesthe following contribution to the displacement field:BRIEF REPORTS PHYSICAL REVIEW B 75, 092102 2007092102-2y=RFjyxj − yjxi − yix − y3dA= y=RFjyr−3 + 3x · yr5 xj − yjxi − yidA= − r−3 jixj − r−3 j jxi + 3r−5 jlxjxlxi. 15Thus, if proportionality is replaced by equality for simplicityof notation—which is ok because we only wish to calculatean energy ratio and have already dropped the overall propor-tionality constant of Eq. 9—we findui = 	r−3ijxj − r−3 tr xi + 3r−5 jlxjxlxi. 16Next we calculate the stain tensor. First, note that ifk is the partial derivative with respect to xk, we havekr−n= −nr−n+2xk. Thus,kui = − 3	r−5ijxjxk + 	r−3ij jk + 3r−5 tr xixk− r−3 tr ik − 15r−7 jlxjxlxixk + 3r−5 jl jkxlxi+ 3r−5 jlxjlkxi + 3r−5 jlxjxlik. 17All terms in this expression vary with r as r−3. Consequently,this term is common to the bulk and shear elastic energiesand may be dropped from our calculation of their ratio. Thecalculation is simplified notationally by putting r=1 or,equivalently, replacing xi by xi /r. When this convention isadopted, the strain tensor is given by ignoring the factor 2 inthe strain tensor definition, uik= iuk+kui /2,uik = 2− tr  + 3xxik + 2	ik + 6 tr xixk − 30xxxixk− 3	ijxjxk + kjxjxi + 6kjxjxi + 6ijxjxk= 2− tr  + 3xxik + 2	ik + 6 tr  − 30xxxixk+ 6 − 3	ijxjxk + kjxjxi . 18Throwing out terms with tr , we end with the followingexpression for the strain tensor:uik = 6xxik + 2	ik − 30xxxixk+ 6 − 3	ijxjxk + kjxjxi . 19To calculate the bulk elastic energy, we need the trace ofthis which, when again terms with tr  are dropped, is givenas xixi=r2 is set equal to unity,tr u = 18xx − 30xx + 6 − 3	2xx = − 6	xx .20The bulk elastic energy density18 averaged over the spherewith radius r subsequently set equal to unity is, when use ismade of Eq. 5 and terms with tr  are ignored, given byBulk elastic energy =K2tru2r=K236	2xx2r=125K	2 tr2 . 21To find the shear elastic energy, we need the transversepart of the strain tensor, uik in terms of which the shearenergy density is Guikuik defined as:18uik 	 uik −13truik = 23 + 	xxik + 2	ik− 30xxxixk + 32 − 	ijxjxk + kjxjxi . 22Squaring and summing all elements of the transverse straintensor, which is required to calculate the shear elastic energy,lead to the following xixi=r2 is again set equal to unity:uikuik = 43 + 	2xx23 + 4	2 tr2 + 900xx2+ 92 − 	22x2 + 2xx2 + 8	3 + 	xxtr − 1203 + 	xx2 + 123 + 	2 − 	2xx2− 120	xx2 + 12	2 − 	22xx − 1802 − 	xx2xx . 23Averaging this expression ignoring all tr  terms leads touikuikr = 123 + 	2 215 + 4	2 + 900 215+ 182 − 	213+215 − 1203 + 	 215+ 243 + 	2 − 	215− 120	215+ 24	2 − 	13− 3602 − 	215tr2=257	2 + 12	 + 108tr2 . 24Summarizing, we find thatBulk elastic energyShear elastic energy12/5K	22/5G7	2 + 12	 + 108=KG6	27	2 + 12	 + 108. 25In terms of the dimensionless variablek 	KG, 26we have 	=6/ 3k+1 which, when substituted into Eq. 25,leads toBulk elastic energyShear elastic energy2k9k2 + 8k + 4. 27The derivative of the fraction on the right with respect to k iszero when 29k2+8k+4=2k18k+8, implying k=2/3.Thus the maximum bulk elastic energy is when k=2/3. Inconclusion,BRIEF REPORTS PHYSICAL REVIEW B 75, 092102 2007092102-3Bulk elastic energyShear elastic energyBulk elastic energyShear elastic energyk = 23  110.28More typically, k=5/2 leads to a maximum bulk shear ratioof 20/337, which is roughly 6%.An important unsolved problem of condensed-matterphysics is the origin of the non-Arrhenius average relaxationtime of glass-forming liquids, where one generally observesan activation energy that increases quite a lot upon cooling.One class of theories are the elastic models recently re-viewed in Ref. 3, according to which the activation energy issome linear combination of the instantaneous bulk and shearmoduli. As regards the temperature dependence of the acti-vation energy, the shear modulus completely dominates bycontributing at least 90%1. A so-called flow event—a jump inconfiguration space from one to another potential-energyminumum—is usually localized in real space. Since the sur-rounding ultraviscous liquid on the short time scale may beregarded as a solid, the molecular displacements may be re-garded as those of a point defect in a disordered solid, al-beit. The main result of this Brief Report sheds light on thenature of flow events, because it once again emphasizesshear dominance—that the shear properties are much moreimportant than the bulk.In an interesting book from 1986, Varotsos andAlexopoulos VA discussed point defect energies in solids,as well as activation energies for point defect diffusion19.They concluded that these energies can always be written asthe isothermal bulk modulus times a microscopic volume.Thus, according to VA, point defect energies and activationenergies always scale with the bulk modulus. This result is atvariance with the above calculation. In practical terms, ofcourse, Poisson’s ratio usually does not vary that much, sothe bulk and shear moduli are roughly proportional. Thismeans that, in practice, it is not easy to determine whether itis the bulk or the shear modulus that controls things. VAexplicitly argued in their book that it is the bulk, not theshear modulus that is important, but they did not base this onprecise calculations similar to the above. We hope the abovecalculation may encourage other researchers to look into thequestion; perhaps the short-range contribution that was nottaken into account here could be bulk dominated. The ideathat many important physical properties are controlled bysimple quantities like the elastic moduli deserves to be fullyinvestigated.This work was supported by the Danish National Re-search Foundation’s Centre for Viscous Liquid Dynamics“Glass and Time.”1 J. C. Dyre and N. B. Olsen, Phys. Rev. E 69, 042501 2004.2 A. V. Granato, Phys. Rev. Lett. 68, 974 1992.3 J. C. Dyre, Rev. Mod. Phys. 78, 953 2006.4 T. Mura, Micromechanics of Defects in Solids, 2nd ed. Kluwer,Dordrecht, 1982.5 C. Zener, Trans. Am. Inst. Min., Metall. Pet. Eng. 147, 3611942.6 J. D. Eshelby, Proc. R. Soc. London, Ser. A 241, 376 1957.7 R. Kienzler and G. Herrmann, Acta Mech. 125, 73 1997.8 S. Onaka, Philos. Mag. Lett. 80, 367 2000.9 R. Kienzler and G. Herrmann, Mech. Res. Commun. 29, 5212002.10 S. F. Li, Int. J. Eng. Sci. 42, 1215 2004.11 S. Matthies and G. W. Vinel, Solid State Phenom. 105, 1132005.12 F. D. Fischer, H. J. Böhm, E. R. Oberaigner, and T. Waitz, ActaMater. 54, 151 2006.13 A. F. Andreev, JETP Lett. 62, 136 1996.14 Y. Kraftmakher, Phys. Rep. 299, 80 1998.15 J. D. Eshelby, Prog. Solid Mech. 2, 89 1961.16 S. Onaka, Philos. Mag. Lett. 85, 115 2005.17 K. Garikipati, M. L. Falk, M. Bouville, B. Puchala, and H. Naray-anan, J. Mech. Phys. Solids 54, 1929 2006.18 L. D. Landau and E. M. Lifshitz, Theory of Elasticity, 2nd ed.Pergamon, London, 1970.19 P. A. Varotsos and K. D. Alexopoulos, Thermodynamics of PointDefects and Their Relation with Bulk Properties North-Holland, Amsterdam, 1986.BRIEF REPORTS PHYSICAL REVIEW B 75, 092102 2007092102-4

Point defect in solids: Shear dominance of the far-field energy

Point defect in solids: Shear dominance of the far-field energy

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