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

0.11 MB·English

by  Jeppe C. Dyre

Point defect in solids: Shear dominance of the far-field energy
Jeppe C. Dyre
DNRF centre of viscous liquid dynamics “Glass and time,
” IMFUFA, Building 27, Department of Sciences,
Roskilde University, Postbox 260, DK-4000 Roskilde, DENMARK
(Dated: February 2, 2008)
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It is shown that the elastic energy far from a point defect in an isotropic solid is mainly shear
0
0 elasticenergy. Thecalculation,whichisbasedonastandarddipoleexpansion,showsthatnomatter
2 howlargeorsmallthebulkmodulusiscomparedtotheshearmodulus,lessthan10%ofthedistant
point defect energy is associated with volumechanges.
n
a PACSnumbers: 61.72.Ji
J

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An isotropic solid has two independent elastic constants, the shear modulus G and the bulk modulus K. It has

] been suggested that when a physical property depends on both G and K, the dependence on the shear modulus is
i
c often the most important [1]. Examples of this “shear dominance” were given by Granato in his important paper
s
- from 1992 [2]; they include the fact that defect energies vary only slightly with the bulk modulus, but are linearly
rl dependentontheshearmodulus(forCuheestimatedthatonlyabout3%oftheenergyofaninterstitialisbulkelastic
mt energy). In Ref. [1], which dealt with the non-Arrhenius viscosity of viscous liquids and its possible explanation in
terms of temperature-dependent instantaneous elastic constants [3], the following result was proved: If temperature
.
t dependence is quantified in terms of log-logderivatives, atleast 92%of the temperature dependence of the molecular
a
m vibrational mean-square displacement over temperature comes from the instantaneous shear modulus, whereas at
most 8% is due to that of the instantaneous bulk modulus.
-
d Howgeneralis “sheardominance”? Is itcoincidental,oris ita fairlygeneraltheme ofthree-dimensionalelasticity?
n Ofcourse,apropertylikethePoissonratiodependsequallyonthebulkandshearmoduli,sosheardominancecannot
o
be a completely general phenomenon. Nevertheless, there are severalexamples of it so it deserves to be investigated.
c
Asonecontributiontosheddinglightonthisquestion,webelowcalculatethemaximumratioofbulkelasticenergyto
[
shear elastic energy far from an arbitrarypoint defect in an isotropic solid. Based on a standarddipole expansion[4]
5
itisshownthatlessthan10%ofthedistantelasticenergyisbulkelasticenergy,i.e.,associatedwithdensity changes.
v
Although our main objective is to identify the dominant contribution to the elastic energy far from a point defect,
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5 it should be noted that according to Zener’s “strain energy model” [5] most energy associated with a point defect
7 is elastic. Elastic energies of point defects fall of rapidly (as r−6), as evident from Eshelby’s famous solution of the
6 problem of an elipsoidal inclusion [6, 7, 8, 9, 10, 11, 12], a result which is general. This means, of course, that most
0
of the elastic energy is located close to the point defect and that a dipole expansion is not realistic for calculating
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the total defect energy. Nevertheless,our results may be taken as an indicationof what contributes mostto the total
0
/ defect energy, the shear or the bulk elastic energies in the defect surroundings.
t
a The nature ofthe far-fielddeformationis alsoimportantforunderstanding andmodeling the long-rangeproperties
m ofdefect-defect interactions. If, forinstance,the displacementfieldisdominatedbysheardisplacements,adefectwill
- only interact weakly with one far away that is dilational close to its center. Some time ago Andreev discussed the
d
unique topological characteristics of point defects in three dimensions [13]. A comprehensive review of point defect
n
properties focusing on the vacancy was given by Kraftmakher [14]. The question of the nature of the deformation
o
c far from a point defect is relevant also for applications in materials science, e.g., for understanding fracture [9]. In
: its macroscopic description the question of the far-field properties of a point defect has been studied because of its
v
importantforunderstandingthemechanicalpropertiesofpolycrystalsandcomposites[4,6,15]. InthiscontexOnaka
i
X recently calculatedthe elastic shearenergydue to a macroscopicpointdefect (an “inclusion”)[16], but did notstudy
r the generalcase, leavingundetermined what is our focus here, the ratio between shear and bulk energy (see also Ref.
a
[12]). Garikipati and co-workers[17] recently discussed the role of continuum elasticity in determining the formation
energy of a point defect utilizing Eshelby’s result for the work done by an external stress during the transformation
of an inclusion [6], an interesting paper that also elicidates the limitations of this approachas compared to atomistic
calculations. We mention these works also to emphasize the continuing interest in the far-field properties of point
defects and their macroscopic analogs.
First some preliminaries: We use the standard Einstein summation convention. The point defect is located at the
origin, is modelled as follows: Imagine a small sphere of radius R surrounding the point defect with all atoms within
thesphereremoved. Theeffectofthedefectonthesurroundingsarerepresentedbyasuitabledistributionofexternal
forces F applied to the surface of the sphere. We define a tensor λ as the following integral over the surface of the
i ij
2
sphere (where dA is the area element):
λ = F (y) y dA. (1)
ij I i j
|y|=R
Because the force distribution models the effect of the atoms within the sphere on the surroundings, the angular
momentum of the force distribution must be zero. This implies that λ is symmetric: λ =λ .
ij ji
Sharp brackets hi , referred to as “averages,” denote integrations over the surface of a sphere with radius r ≫ R
r
centred at the origin. The following identities become useful later on:
r2
hx x i = δ (2)
i j r ij
3
and
r4
hx x x x i = δ δ +δ δ +δ δ . (3)
i j k l r ij kl ik jl il jk
15 (cid:16) (cid:17)
If (λxx) is an abbreviation of λ x x , we find from these identities that
ij i j
r2
h(λxx)i = trλ (4)
r
3
and
r4
h(λxx)2i = (trλ)2+2tr(λ2) . (5)
r
15(cid:16) (cid:17)
Moreover,if λx is the vector whose i’th component is λ x , we have
ij j
r2
h(λ2xx)i = h(λx)2i = tr(λ2). (6)
r r
3
As mentioned, far from the defect both the bulk and the shear elastic energies vary with the distance from the
originas r−6. When one averagesovera sphere with radius r ≫R, the result for both the bulk and the shear elastic
energymustbeascalarvaryingwithdistanceasr−6 thatisofsecondorderintheforcesF . Itfollowsfromthebelow
i
calculation that these two scalars are both uniquely determined by G, K, and the λ-matrix. Consequently, because
these two scalar functions are of second order in the forces, the ratio of bulk to shear elastic energy must have the
following general structure:
Bulk elastic energy A(trλ)2+Btr(λ2)
= . (7)
Shear elastic energy C(trλ)2+Dtr(λ2)
Definingα=tr(λ2)/(trλ)2 thisratiois(A+Bα)/(C+Dα). Thequantityαvariesbetween1/3and∞: Bynormalizing
we may assume that trλ = 1; if the eigenvalues are denoted by µ we thus have µ +µ +µ = 1 and consequently
i 1 2 3
α = µ2 +µ2 +µ2. From this it follows that α varies between 1/3 and ∞. Since the energy ratio of Eq. (7) is
1 2 3
a monotonous function of α and since perfect isotropy (α = 1/3) implies zero bulk elastic energy (in this case the
displacementfieldisradiallysymmetricvaryingasr−2 whichimpliesnovolumechanges),themaximumbulktoshear
elastic energy ratio arises in the limit α → ∞. Thus in the below calculation we may ignore all terms with trλ and
keep only terms with a tr(λ2) factor.
Poisson’s ratio σ is defined [18] by
1 3K−2G
σ = . (8)
2 3K+G
3
If a force F is applied at the origin of an isotropic elastic continuum, the displacement field at the point (x′,y′,z′) is
given [18] by
(3−4σ)F+(F·n′)n′
u(x′,y′,z′) ∝ , (9)
r′
where r′2 = x′2 +y′2 +z′2 and n′ = (x′,y′,z′)/r′ is the unit vector pointing from the origin to (x′,y′,z′). It is
convenient to introduce the variable
Λ ≡ 2−4σ, (10)
in terms of which Eq. (9) becomes
Fi Fjn′jn′i
u ∝ (Λ+1) + . (11)
i r′ r′
We proceed to perform a standard dipole expansion by first noting that, if y is the coordinate for a point on the
small sphere surrounding the defect (|y| = R) and x is the coordinate for the point of interest far away, to lowest
order in |y|/|x|=R/|x| we have if r ≡|x|
1 = x2+y2−2x·y −1/2 =r−1 1−2x·y −1/2 =r−1+ x·y . (12)
|x−y| (cid:16) r2 (cid:17) r3
(cid:0) (cid:1)
Similarly
1 = x2+y2−2x·y −3/2 =r−3 1−2x·y −3/2 =r−3+3x·y . (13)
|x−y|3 (cid:0) (cid:1) (cid:16) r2 (cid:17) r5
To calculate the displacement field at point x we first note that when the force F (y) is integrated over the small
i
sphere radius R, the result is zero. Thus when one integrates over the small sphere, the first term of Eq. (11) to
lowest order in 1/r becomes
F (y) x·y
(Λ+1) i dA=(Λ+1) F (y) dA
I |x−y| I i r3
|y|=R |y|=R
= (Λ+1) r−3λ x . (14)
ij j
Similarly, to lowest order the second term of Eq. (11) gives the following contribution to the displacement field
F (y)(x −y )(x −y ) x·y
j j j i i dA = F (y) r−3+3 (x −y )(x −y )dA
I|y|=R |x−y|3 I|y|=R j (cid:16) r5 (cid:17) j j i i
= −r−3λ x −r−3λ x +3r−5λ x x x . (15)
ji j jj i jl j l i
Thus, if proportionality is replaced by equality for simplicity of notation – which is OK because we only wish to
calculate an energy ratio and have already dropped the overallproportionality constant of Eq. (9) – we find
u =Λr−3λ x −r−3trλ x +3r−5λ x x x . (16)
i ij j i jl j l i
Next we calculate the stain tensor. First, note that if ∂ is the partial derivative with respect to x , we have
k k
∂ r−n =(−n)r−(n+2)x . Thus
k k
∂ u = −3Λr−5λ x x +Λr−3λ δ +3r−5trλx x −r−3trλδ −15r−7λ x x x x
k i ij j k ij jk i k ik jl j l i k
+ 3r−5λ δ x x +3r−5λ x δ x +3r−5λ x x δ . (17)
jl jk l i jl j lk i jl j l ik
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All terms in this expression vary with r as r−3. Consequently, this term is common to the bulk and shear elastic
energiesandmaybe droppedfromourcalculationoftheir ratio. The calculationis simplified notationallyby putting
r =1(or,equivalently,replacingx byx /r). Whenthis conventionisadopted,thestraintensorisgivenby(ignoring
i i
the factor 2 in the strain tensor definition, u =(∂ u +∂ u )/2)
ik i k k i
u = 2(−trλ+3(λxx))δ +2Λλ +6trλx x −30(λxx)x x
ik ik ik i k i k
− 3Λ(λ x x +λ x x )+6λ x x +6λ x x
ij j k kj j i kj j i ij j k
= 2(−trλ+3(λxx))δ +2Λλ +(6trλ−30(λxx))x x
ik ik i k
+ (6−3Λ)(λ x x +λ x x ) (18)
ij j k kj j i
Throwing out terms with trλ we end with the following expression for the strain tensor
u =6(λxx)δ +2Λλ −30(λxx)x x +(6−3Λ)(λ x x +λ x x ) . (19)
ik ik ik i k ij j k kj j i
To calculate the bulk elastic energy we need the trace of this which, when again terms with trλ are dropped, is
given as (x x =r2 is put equal to unity)
i i
tru=18(λxx)−30(λxx)+(6−3Λ)2(λxx)=−6Λ(λxx). (20)
The bulkelasticenergydensity[18]averagedoverthe spherewithradiusr (subsequentlyput equaltounity)is,when
use is made of Eq. (5) and terms with trλ are ignored, given by
K K 12
Bulk elastic energy= h(tr(u))2i = 36Λ2h(λxx)2i = KΛ2tr(λ2). (21)
r r
2 2 5
To find the shear elastic energy we need the transverse part of the strain tensor, u⊥ (in terms of which the shear
ik
energy density is Gu⊥u⊥) defined as [18]: [18]
ik ik
1
u⊥ ≡ u − tr(u)δ
ik ik 3 ik
= 2(3+Λ)(λxx)δ +2Λλ −30(λxx)x x +3(2−Λ)(λ x x +λ x x ) . (22)
ik ik i k ij j k kj j i
Squaring and summing of all the elements of the transverse strain tensor, which is required to calculate the shear
elastic energy, leads to the following (x x =r2 is again put equal to unity):
i i
u⊥u⊥ = 4(3+Λ)2(λxx)23+4Λ2tr(λ2)+900(λxx)2+9(2−Λ)2[2(λx)2+2(λxx)2]
ik ik
+ 8Λ(3+Λ)(λxx)trλ−120(3+Λ)(λxx)2+12(3+Λ)(2−Λ)2(λxx)2−120Λ(λxx)2
+ 12Λ(2−Λ)2(λ2xx)−180(2−Λ)(λxx)2(λxx). (23)
Averaging this expression (ignoring all trλ-terms) leads to:
2 2 1 2
hu⊥u⊥i = 12(3+Λ)2 +4Λ2+900 +18(2−Λ)2( + )
ik ik r (cid:16) 15 15 3 15
2 2 2
− 120(3+Λ) +24(3+Λ)(2−Λ) −120Λ
15 15 15
1 2
+ 24Λ(2−Λ) −360(2−Λ) tr(λ2)
3 15(cid:17)
2
= 7Λ2+12Λ+108 tr(λ2). (24)
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(cid:0) (cid:1)
Summarizing, we find that
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Bulk elastic energy (12/5)KΛ2 K 6Λ2
≤ = . (25)
Shear elastic energy (2/5)G(7Λ2+12Λ+108) G 7Λ2+12Λ+108
In terms of the dimensionless variable
K
k ≡ , (26)
G
we have Λ=6/(3k+1) which, when substituted into Eq. (24), leads to
Bulk elastic energy 2k
≤ . (27)
Shear elastic energy 9k2+8k+4
The derivative of the fraction on the right with respect to k is zero when 2(9k2+8k+4) = 2k(18k+8), implying
k =2/3. Thus the maximum bulk elastic energy is when k =2/3. In conclusion,
Bulk elastic energy Bulk elastic energy 2 1
≤ k = ≤ . (28)
Shear elastic energy Shear elastic energy(cid:16) 3(cid:17) 10
More typically, k =5/2 leads to a maximum bulk shear ratio of 20/337 which is roughly 6%.
Oneimportantunsolvedproblemofcondensed-matterphysicsisthe originofthe non-Arrheniusaveragerelaxation
time of glass-forming liquids, where in most cases one observes an activation energy that increases quite a lot upon
cooling. Oneclassoftheoriesaretheelasticmodels(recentlyreviewedin[3])accordingtowhichtheactivationenergy
is some linear combination of the instantaneous bulk and shear moduli. As the regards the temperature dependence
of the activation energy the shear modulus completely dominates (contributing at least 90% [1]). A so-called flow
event – the jump in configuration space from one to another potential energy minumum – is usually localized in real
space. Since the surrounding ultraviscous liquid on the short time scale may be regarded as a solid, the molecular
displacements induced by a flow event may be regardedas those of a point defect (in a disorderedsolid, albeit). The
aboveresultshedslightonthenatureoffar-fieldflow-eventinduceddisplacementsbyemphasizing“sheardominance”
– that the shear modulus is much more important than the bulk.
In an interesting book from 1986 Varotsos and Alexopoulos (VA) discussed point defect energies in solids, as well
as activation energies for point defect diffusion [19]. VA concluded that these energies can always be written as the
(isothermal)bulk modulus times some microscopicvolume. Thus pointdefectenergiesandactivationenergiesalways
scale with the bulk modulus. This result is at variance with the above calculation. In practical terms, of course,
Poisson’s ratio usually does not vary very much, so the bulk and shear moduli are roughly proportional. This means
that in many cases it is difficult to determine whether the bulk or the shear modulus controls things. Admittedly,
VA explicitly argue in their book that it is the bulk, and not the shear modulus, which is important, but they do
not base this on calculations similar to ours (but, e.g., thermodynamic arguments). We do not claim that the above
calculationdisprovesVA,butitmayencourageotherresearcherstolookintothequestion. Surely,theideathatmany
important physical properties are controlled by as simple quantities as the elastic moduli deserves to be investigated
fully.
ThisworkwassupportedbytheDanishNationalResearchFoundation’sCentreforViscousLiquidDynamics“Glass
and Time.”
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