The effect of strong surface energy anisotropy on migrating grain-boundary grooves Donghong Min and Harris Wong Citation: J. Appl. Phys. 100, 053523 (2006); doi: 10.1063/1.2336980 View online: http://dx.doi.org/10.1063/1.2336980 View Table of Contents: http://jap.aip.org/resource/1/JAPIAU/v100/i5 Published by the American Institute of Physics. Additional information on J. Appl. Phys. Journal Homepage: http://jap.aip.org/ Journal Information: http://jap.aip.org/about/about_the_journal Top downloads: http://jap.aip.org/features/most_downloaded Information for Authors: http://jap.aip.org/authors Downloaded 15 May 2013 to 160.94.45.157. This article is copyrighted as indicated in the abstract. Reuse of AIP content is subject to the terms at: http://jap.aip.org/about/rights_and_permissions
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The effect of strong surface energy anisotropy on migrating grain-boundarygroovesDonghong Min and Harris Wong Citation: J. Appl. Phys. 100, 053523 (2006); doi: 10.1063/1.2336980 View online: http://dx.doi.org/10.1063/1.2336980 View Table of Contents: http://jap.aip.org/resource/1/JAPIAU/v100/i5 Published by the American Institute of Physics. Additional information on J. Appl. Phys.Journal Homepage: http://jap.aip.org/ Journal Information: http://jap.aip.org/about/about_the_journal Top downloads: http://jap.aip.org/features/most_downloaded Information for Authors: http://jap.aip.org/authors
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The annealing of a strain-free polycrystalline solid in-duces grain growth, which reduces the total grain-boundaryenergy of the solid. Larger grains grow at the expense ofsmaller ones that shrink, and as a result the grain boundariesmigrate. If a grain boundary ends at a free surface, a groovewill form at the tip to reduce the combined surface and grain-boundary energies. This groove may hinder the movement ofthe grain boundary because groove displacement requiresmass transport and energy dissipation. If the grain boundaryis perpendicular to the free surface and stationary, the groovegrows deeper and wider with time.1,2 If the grain boundary isinclined and migrates, the groove either moves smoothlywith the grain boundary or moves in a “jerky” fashion.3–5 Tostudy grain growth properly, the effect of grain-boundarygrooves on grain-boundary migration must be understood.This effect is particularly important in thin films owing to thelarge surface to volume ratio.
Thermal grooving was first modeled by Mullins.1 Heconsidered a stationary grain boundary that is perpendicularto the free surface. He assumed that the groove surface hassmall slopes and the surface energy is isotropic. He com-pared various mass-transport mechanisms, such as surfacediffusion, bulk diffusion, and evaporation and condensation,and found that surface diffusion dominates for small groovesat temperatures far below the melting temperature.1,2,6 In thatcase, grooving follows a similarity law and the groove widthand height increase with time t as t1/4. Mullins’s model hasbeen extended in several directions. Zhang and Wong7 ana-
a�Author to whom correspondence should be addressed; electronic mail:
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lyzed the linear stability of the self-similar groove profile andfound it stable. The small-slope assumption has beeneliminated.8–12 Strong surface energy anisotropy has beenadded to the bicrystal with symmetric13 or asymmetric14
crystallographic orientations. The t1/4 similarity law holds inall the extended studies.
Mullins also studied the effect of grain-boundarygrooves on grain-boundary migration.3 His model consists ofa horizontal free surface and an inclined grain boundary thatmakes an angle � with the vertical axis. He assumed that thegrain boundary migrates at constant speed and sought asteady-state solution of the groove profile in a referenceframe moving with the grain boundary. Surface diffusion istaken to be the dominant mass-transport mechanism. Hefound that a solution exists if �=�c=�b /6�, where �b and �are the isotropic surface energy �per unit area� of the grainboundary and the free surface, respectively. Mullins sug-gested that a grain boundary will become stuck at the freesurface if ���c. After the grain boundary is pinned at thetip, the rest of the grain boundary still migrates and the angle� at the groove root will increase. Once ���c, the grainboundary tip can detach from the groove and migrate until���c. At which point, the grain-boundary tip will be pinnedagain. This can explain the jerky motion observed in experi-ments. Mullins solved the migrating-groove profile assuming�c�1. His solution has been generalized to finite �c by Minand Wong15 for isotropic groove surfaces. Kris et al.16 tookthe horizontal free surface behind the migrating groove as afacet to bring anisotropy into the problem. The rest of thegroove surface is still isotropic and is connected to the facetat a kink with continuous mass flux and chemical potential.
They also assumed that the groove surface has small slopes.
053523-2 D. Min and H. Wong J. Appl. Phys. 100, 053523 �2006�
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This work considers both sides of the groove surface to beanisotropic and studies the complete range of crystallo-graphic orientations of the bicrystal. In addition, the groovesurface has finite slopes to allow for facets with various ori-entations.
Mullins’s analysis3 does not truly combine grain-boundary grooving and grain-boundary migration. Recently,Zhang and Wong7,17 investigated the coupled grooving andmigration of an inclined grain boundary and found that thegrain-boundary tip is never pinned. In their work, the freesurface is horizontal and the grain boundary is initiallystraight and deviates from the vertical axis by an angle ��1. They assumed that the surface energy is isotropic andthe groove evolves by surface diffusion. They found that thecoupled motion can be separated into two time regimes. Inregime I, the grain-boundary tip turns vertically. In regime II,the grain boundary relaxes following two different paths de-pending on the inclination angle �. If ���c �=�b /6��, thegrain-boundary tip at �x0 ,y0� moves by surface diffusion:�x0 ,y0���t1/4 , t1/4� as t→�, which is slow and probably dif-ficult to observe �x0 and y0 denote the horizontal and verticalpositions of the grain-boundary tip�. If ���c, the grain-boundary groove moves by migration: �x0 ,y0���t1/2 , t1/6� ast→�. Hence, the groove moves faster, but is shallower. Thisdetailed analysis of the coupled motion shows that the grain-boundary groove is never pinned, but may slow down con-siderably if ���c. Vilenkin et al.18 simulated the coupledmotion numerically for a periodic array of grains in a thinfilm undergoing annealing. Kanel et al.19 used the same nu-merical method to study the shrinkage of a single grain par-tially immersed in a solid. They also did not observe pinningof the grain boundary by the groove.
Migrating grain-boundary grooves play a critical role inthe “quarter-loop” method of measuring grain-boundary mo-bility. In this method, the grain boundary in a bicrystal runsparallel to a free surface over a major portion of the bicrystalbefore it turns to intercept the free surface and forms aquarter-loop �see Discussion�.20 Upon heating at a tempera-ture of interest, the grain boundary migrates to reduce thegrain-boundary surface energy. Since the driving force isconstant, the grain boundary soon reaches a constant speed.The reduced mobility can be inferred by measuring the speedand the grain-boundary profile, and by comparing with amodel. However, previous models have neglected the grain-boundary groove.20 As a result, the grain-boundary tip mustalways be perpendicular to the free surface. Min and Wong15
incorporated the effect of the groove into the quarter-loopmethod. They found that the groove turns the grain-boundarytip by an angle � away from being perpendicular to the freesurface. The angle � depends only on the dihedral angle.They solved the groove and grain-boundary profiles for thecomplete range of �. They also considered the case that theisotropic surface energy on one side of the groove differsslightly from the other and calculated the inclination angle �in terms of the energy difference. In their analysis, they as-sumed that the groove moves at constant speed. The exis-tence of such a traveling-wave solution has been confirmedlater by Kanel et al.21,22
In this work, we consider a migrating groove with strong
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surface energy anisotropy on both sides of the groove. Ourmodel consists of a horizontal free surface with an inclinedgrain boundary that is migrating at constant speed V towardthe left, as depicted in Fig. 1. A groove forms at the tip of thegrain boundary and migrates with it by surface diffusion. Thesurface energies on both sides of the groove are anisotropicand have fourfold symmetry. The crystallographic orienta-tions of the bicrystal can be asymmetric and are denoted by�+ and �− �Fig. 1�. The governing equations and boundaryconditions for the groove motion are listed in Sec. II and aremade dimensionless in Sec. III. They are solved by a shoot-ing method as described in Sec. IV. Previous isotropicmodels3,15 show that the grain-boundary inclination angle �depends only on the dihedral angle �Fig. 1�. Here, we findthat � also depends on �+ and �−, and is determined for thefull range of the parameters �Sec. V�.
Thermal grooving experiments have been conducted us-ing polycrystalline Ni �Ref. 23� and alumina.24 After anneal-ing, grain-boundary grooves appeared on polished surfacesof the polycrystalline materials. The groove profiles weremeasured by atomic force microscopy. A region was identi-fied on the surface using artificial or naturally occurringmarkers, and the grain-boundary groove profiles and loca-tions were reexamined after an additional heat treatment.These experiments show that migrating grain-boundarygrooves are much more asymmetric than stationary grooves.In addition, a migrating grain boundary always moves to-ward the grain with the larger root angle. �The root angle isthe angle between the groove surface and the horizontal axisat the groove root, as represented by �+ and �− in Fig. 1.� Weobtain a good fit of the alumina groove profile using ouranisotropic model �Sec. VI�. We discuss the implications ofthe results in Sec. VII and conclude this work in Sec. VIII.
II. MATHEMATICAL FORMULATION
When an atom is transferred from a planar surface to acurved surface of curvature , the increase in chemical po-tential is given by the anisotropic Gibbs-Thompson equation
6
FIG. 1. A sketch of a grain boundary migrating to the left with constantinclination angle � and speed V. The crystallographic orientations of thebicrystal are defined by facet angles �+ and �−. Arc-length s along thegroove surface starts at the groove root where x=0. The surface unit normaln makes an angle with the y axis. At the groove root, =�+ for the rightgroove surface and =−�− for the left, and �+ and �− are called the rootangles.
�Appendix A�,
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053523-3 D. Min and H. Wong J. Appl. Phys. 100, 053523 �2006�
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� = ��� +d2�
d2� , �1�
where � is the atomic volume and � is the anisotropic sur-face energy per unit area. The orientation of the surface isspecified by the angle between the surface normal and thevertical axis �Fig. 1�. The average drift velocity of surfaceatoms is found from �1� using the Nernst-Einstein relationas1
U = −Ds
kT
��
�s, �2�
where Ds is the surface diffusion coefficient, kT is the ther-mal energy per atom, and s is the arc length along the sur-face. This drift velocity generates a surface current
J = U , �3�
where is the number of diffusing surface atoms per unitarea. When surface currents converge at a point, a local massbalance demands that the surface rises at that point with anormal velocity6
un = C�2
�s2��� +d2�
d2� , �4�
where C=Ds �2 /kT. This equation governs the evolution oftwo-dimensional solid surfaces when the mass is conveyedby surface diffusion and when the surface energy isanisotropic.13,25
A. Surface energy anisotropy
To describe the surface stiffness �+d2� /d2, we use arecently developed facet model, in which the surface stiff-ness is prescribed as a function of ,26,27
� +d2�
d2 = 2�0F�� , �5�
where �0=�eH /2� is the isotropic surface energy per unitarea ��e is the chemical potential of an equilibrium crystalwith a vertical symmetric plane and H is the height of theequilibrium crystal along the symmetric plane�. The functionF�� can be viewed as the radius of curvature of the equi-librium crystal surface made dimensionless by H �AppendixA�. In this way, a facet is described by the Dirac delta func-tion in F��, with the weight of the delta function equal tothe length of the facet plane. Previous models of surfaceenergy prescribe �=���. However, the surface stiffness �+d2� /d2 can become negative when the anisotropy isstrong. This can induce ill-posedness in surface evolutionproblems.27 By modeling the surface stiffness directly and bylimiting F�0, the ill-posedness is avoided. The limit F�0holds for temperatures at or above the roughening tempera-ture of the bicrystal.
To facilitate numerical calculations, the delta function is27
replaced by a spike function,
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F�� = R + L i=−N
N �cosh� − i
��−1
, �6�
where R, L, and � are constants. This periodic function hasspikes centered at =i �i are the orientations of facetplanes on the equilibrium crystal�. To model a groove sur-face, F�� is only needed in −����. Thus, N is chosensuch that the spike at −N−1 or N+1 has negligible contribu-tion to F�� in the domain of interest. The width and heightof the spike are controlled by ���1� and L, respectively.Since F�� can be viewed as the radius of curvature of theequilibrium crystal surface, L is therefore the radius of cur-vature of the facet planes, and R that of the corners of theequilibrium crystal, all made dimensionless by H.27 As thesharpness of the corners on an equilibrium crystal dependson temperature,28–30 R can be used to study the effect oftemperature. As R→0, the temperature decreases and ap-proaches the roughening temperature of the bicrystal. Theparameters R, L, and � are not independent because the func-tion F�� obeys two integral constraints as follows:27
�0
�
F��sin d = 1, �7a�
�0
�
F��cos d = 0. �7b�
They give L��1/2-R��−1 as �→0.27 Therefore, � can beinterpreted physically as a measure of the curvature of facetplanes on the equilibrium crystal. The spike-function modelcan be used to fit an equilibrium crystal shape to extract itsanisotropic surface energy.27
The surface energy per unit area � can be found by solv-ing �5�,27
�
�0= 2�Ye cos − Xe sin � �8�
where
Xe = − �0
F cos d , �9a�
Ye =1
2− �
0
F sin d �9b�
are the surface coordinates of the equilibrium crystal madedimensionless by H.
In this work we consider surface energies with fourfoldsymmetry. Thus, the equilibrium crystal is a square, i.e., i
= i� /2. We take �=0.005. With this narrow spike width, weonly need N=3 to get an accurate value of F�� in the do-main −����. Different values of R are considered. Foreach R, L is calculated numerically by substituting �6� into�7�. The equilibrium crystal shape is plotted in Fig. 2�a� forR=0.5, 0.05, and 0.0005. �The corresponding values for L=0, 56.97, and 63.23.� The surface positions are found to theaccuracy of 12 significant digits by evaluating the integralsin �9� using the Romberg method. When R=0.5, F=R and
the equilibrium crystal is a circular rod. As R decreases, the
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053523-4 D. Min and H. Wong J. Appl. Phys. 100, 053523 �2006�
D
corners become sharper and the crystal approaches a square.In Fig. 2�b�, the normalized surface energy � /�0 is graphedas a function of in a polar plot. Only a quadrant is shownowing to symmetry. The complete surface energy polar plotconsists of four arcs. A cusp develops at the meeting pointbetween two arcs when R is small. As R increases, the cuspsmoothens and the surface energy polar plot approaches acircle.
B. Quasisteady governing equations
The crystallographic orientation of the bicrystal is con-trolled by rotating the equilibrium crystal counterclockwiseby angle �, as illustrated in Fig. 1. This rotation is incorpo-rated into F�� by changing the argument from to −�.Thus, �4� becomes
un = B�2
�s2 �2F� − ��� , �10�
where B=C�0. The orientation �or facet� angle � is different
FIG. 2. �a� A quadrant of square crystals computed with �=0.005 and vari-ous R. Cartesian coordinates �Xe, Ye� are defined with the origin at thesquare center and are nondimensionlized by the height H of the crystalsalong the Ye axis. The parameter R is the nondimensionalized radius ofcurvature of the corners. The �-function solution yields a perfect square andcan be viewed as the limiting case of �→0 with R=0. �b� A quadrant ofsurface energy polar plots for the square crystals in �a�. The isotropic surfaceenergy per unit area �0 is the value of � when R=0.5.
for two sides of the bicrystal. As shown in Fig. 1, �=�+ for
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the right side and �=−�− for the left. In general, the twofacet angles are not equal.
Arc-length coordinates are used here to accommodatethe surface morphologies. The grain boundary makes anangle � with the vertical axis and moves with constant ve-locity V to the left, as shown in Fig. 1. The groove surface islocated by Cartesian coordinates �z ,y�, with z along the freesurface and y pointing upward. The surface normal makes anangle with the y axis. The surface coordinates �z ,y� andangle are functions of time t and arc-length s and arerelated by geometry,
�z
�s= cos , �11a�
�y
�s= sin . �11b�
If r=zi+yj is the position vector of a surface point, then�r /�t is the surface velocity at constant arc length. Given thenormal vector n=−i �sin �+ j�cos �, we can calculate un
=n ·�r /�t and =�s ·n=−� /�s. Equation �10� then be-comes
− �sin ��z
�t+ �cos �
�y
�t= − B
�2
�s2�2F� − ���
�s . �12�
This equation governs the evolution of two-dimensional freesurfaces with surface energy anisotropy.13,14
Define a coordinate x that moves with the groove �Fig.1�,
x = z + Vt . �13�
In this moving frame, �12� becomes
− �sin �� �x
�t− V� + �cos �
�y
�t
= − B�2
�s2�2F� − ���
�s . �14�
When the motion is steady in the moving frame, the aboveequation reduces to
− Bd2
ds2�2F� − ��d
ds = V sin . �15a�
The geometric relations in �11� are transformed to
�x
�s= cos , �15b�
�y
�s= sin . �15c�
There are three dependent variables: , x, and y, and theydepend only on s. Since the groove profile on the right sideof the groove is different from that on the left, each depen-dent variable is separated into two branches: right ��� andleft ���.
Boundary conditions are that at the groove root s=0,
x+ = x− = 0, �15d�
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053523-5 D. Min and H. Wong J. Appl. Phys. 100, 053523 �2006�
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y+ = y− = y0, �15e�
+ = �+, �15f�
− = − �−, �15g�
where y0 is the depth of the groove root and �+ and �− arethe root angles of the right and left surfaces made with the xaxis �Fig. 1�. Equations �15d� and �15e� specify continuity ofthe surface profile. The chemical potential in �1� and thesurface current in �3� also need to be continuous at thegroove root: �+=�−=�0 and J+=J−=J0. Thus,
− 2F�+ − �+��+
�s= − 2F�− + �−�
�−
�s=
�0
��0, �15h�
�
�s�2F�+ − �+�
�+
�s =
�
�s�2F�− + �−�
�−
�s
=J0kT
Ds ��0, �15i�
where the material properties Ds and are assumed to be thesame on both sides of the groove. As s→ ±�,
± → 0, �15j�
y± → 0, �15k�
i.e., the free surface is not disturbed far from the groove.A local force balance at the groove root gives Herring’s
boundary condition �Appendix A�,31
�b cos � = ����+−�+sin �+ + ����−−�−
sin �−
+ d�
d
�+−�+
cos �+ − d�
d
�−−�−
cos �−,
�16a�
�b sin � = ����+−�+cos �+ − ����−−�−
cos �−
− d�
d
�+−�+
sin �+ − d�
d
�−−�−
sin �−,
�16b�
where d� /d is the bending tension acting normal to thesurface tension � in the direction of increasing �AppendixA�. Notice that � on the right side of the groove root can bedifferent from � on the left. Given a bicrystal, �b, �, and �are specified and �+ and �− are solved from �16�. We find,however, that it is more convenient to use � and �+ as inputparameters. Thus, in this work, �+ and � are specified, �− isdetermined from �15�, then �b and � are calculated from�16�.
III. NONDIMENSIONALIZATION
The variables are nondimensionalized by �B /V�1/3,
which is the only length scale in the problem,
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S =s
�B/V�1/3 , X =x
�B/V�1/3 , Y =y
�B/V�1/3 . �17�
The governing equations become
d2
dS2�2F� − ��d
dS = − sin , �18a�
dX
dS= cos , �18b�
dY
dS= sin , �18c�
F�� = R + L i=−N
N �cosh� − i
��−1
. �18d�
The normalized boundary conditions are as follows. At S=0,
X+ = X− = 0, �18e�
Y+ = Y− = Y0, �18f�
+ = �+, �18g�
− = − �−, �18h�
− 2F�+ − �+�d+
dS= − 2F�− + �−�
d−
dS= C0, �18i�
d
dS�2F�+ − �+�
d+
dS =
d
dS�2F�− + �−�
d−
dS = M0,
�18j�
where Y0=y0�B /V�−1/3, C0=�0�B /V�1/3 /��0, and M0
=J0kT�B /V�2/3 /Ds ��0 are the dimensionless groove-rootdepth, chemical potential, and surface current to be deter-mined. As S→ ±�,
± → 0, �18k�
Y± → 0. �18l�
The grain-boundary and free surface energies per unit area�b and � in �16� are made dimensionless by the isotropic freesurface energy per unit area �0.
The groove height is found to be related to the surfacecurrent. If sin in �18c� is substituted into �18a�, the result-ing equation can be integrated once to yield
Y = −d
dS�2F� − ��
d
dS , �19�
where the boundary conditions Y →0 and →0 as S→ ±�have been imposed to eliminate the integration constant.Thus, the dimensionless groove height is the same as thedimensionless surface current except by a sign. Conse-quently, Y0=−M0 at the groove root. Hence, we only need tosolve for and X. Once =�S� is determined, Y is found
by �19�.
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053523-6 D. Min and H. Wong J. Appl. Phys. 100, 053523 �2006�
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IV. NUMERICAL METHOD
To solve �18�, we first specify the radius function F byassigning �=0.005, N=3, and i= i� /2. We consider R=0.5, 0.05, and 0.0005. For each R, L is calculated numeri-cally by substituting �6� into �7�. The function F in �18a� alsorequires the facet angles �+ and �− for the right and left sidesof the bicrystal. We choose different �+ and �− to study theireffects. Thus, F in �18a� is completely specified.
Equation �18� is solved by a fourth-order Runge-Kuttamethod.32 The dependent variables and X are functions ofarc-length S and are determined separately for the right side�S�0� and the left side �S�0�. The right-side solution isfound by shooting form infinity. As S→� �Appendix B�,
→ e−S/�2R�1/3−
1
156e−3S/�2R�1/3
, �20a�
X → S . �20b�
To start the integration, S is assigned a value �say, 2�. Theintegration proceeds in the direction of decreasing S until reaches a predetermined value of �+ �e.g., �+=10°�. Then theintegration is stopped and S, −2Fd /dS, d�2Fd /dS� /dS,and X are calculated by interpolation. By varying the startingS position and the step size, these variables are found to theaccuracy of 16 significant digits using a quadruple precisionFORTRAN program. The interpolated values of S and X areused to shift the whole curve so that the end point is at S=X=0. The interpolated values of −2Fd /dS andd�2Fd /dS� /dS yield C0 and M0 �=−Y0�.
The left-side integration starts from S=X=0 with calcu-lated values of C0 and M0 from the right side. However, theroot angle �− at S=0 is unknown and is found by shooting; itis varied until tends to zero as S→−� �say, at S=−10�. Byvarying the ending S position and the step size, �− is foundto be accurate to six significant digits. Thus, the input param-eters in the numerical integration are F and �+. The otherparameters such as �−, C0, and M0 are determined by theintegration. Once �+ and �− are known, �b /�0 and � arefound from �16�. This completes the solution.
V. RESULTS
A. Grooves with �+™�+ and �−™�−
With strong surface energy anisotropy, the groove profilecan still be smooth if ��+−�+��� and ��−−�−���. This isbecause the surface normal angle increases from zero atpositive infinity S to =�+ at the groove root �Fig. 1�, andthen from =−�− at the groove root to zero at negativeinfinity S, so that never reaches the facet angles at =�+
and =−�−. Since the groove surface misses the facet ori-entations, the groove profile is smooth and is similar to theisotropic solution. To illustrate this point, we consider a casein which �=0.005, �+=45°, �−=60°, and �+=7°. Since��+−�+���, the surface normal angle is sufficiently farfrom the facet angles that a smooth groove is obtained, asshown in Fig. 3 for R=0.5, 0.05, and 0.0005. All the profilesare smooth, even for the highly anisotropic case of R
=0.0005. Furthermore, the profiles are self-similar; the nu-
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merical integrations yield the same �−=14.038°. This is be-cause when ��−��±�� and ����±, the surface stiffnessF�−���R in �18d� and it can be removed from �18� bydefining
S =S
�2R�1/3 , X =X
�2R�1/3 , Y =Y
�2R�1/3 . �21�
Thus, �18� is transformed to
d3
dS3= − sin , �22a�
dX
dS= cos , �22b�
dY
dS= sin . �22c�
At S=0,
X+ = X− = 0, �22d�
Y+ = Y− = Y0 =Y0
�2R�1/3 , �22e�
+ = �+, �22f�
− = − �−, �22g�
−d+
dS= −
d−
dS= C0 =
C0
�2R�2/3 , �22h�
d2+
dS2=
d2−
dS2= M0=
M0
�2R�1/3 . �22i�
¯
FIG. 3. Nondimensionlized groove profiles when the root angles are farfrom the facet angles. The surface energy is computed with �=0.005 andvarious R. The calculated root angle on the left side is �−=14.038° fordifferent R. The anisotropic grooves have the same shape as the isotropicgroove �R=0.5� except that the size is reduced by a factor of �2R�1/3. Mul-lins’s small-slope isotropic solution in �24� is also plotted for comparison.
As S→ ±�,
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053523-7 D. Min and H. Wong J. Appl. Phys. 100, 053523 �2006�
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± → 0, �22j�
Y± → 0. �22k�
The root angles �+ and �− are unchanged. The rescaled prob-lem �22� is exactly the same as the isotropic case studied byMin and Wong.15 An asymptotic solution of the isotropicproblem in the limit of small slope yields �Appendix C�
�− = 2�+ +5
14�+
3 +270 065
1 258 712�+
5 + ¯ � �0. �23�
When �+=7°, this expansion gives �−=14.038°, in agree-ment to the last digit with the numerical results of the aniso-tropic but smooth grooves in Fig. 3. The isotropic grooveprofile was also expanded to the same order of accuracy byMin and Wong.15 The leading-order solution is the same asMullins’s,3
Y+ = −1
3��+ + �−�e−X, �24a�
Y− = −2
3��+ + �−�eX cos��
3−
�3
2X� . �24b�
This is plotted in Fig. 3 and agrees well with the case of R=0.5.
To verify the scaling in �22�, the groove profile for R=0.0005 is multiplied by �2R�−1/3 and plotted in Fig. 3. Itagrees to at least six significant digits with the isotropic case�R=0.5�. Since R reflects the effect of temperature, thisseems to suggest that a migrating smooth anisotropic groovebecomes shallower in depth and narrower in width as thetemperature decreases, but the groove shape remains thesame. Unfortunately, the effect of temperature is more com-plex because as R decreases, � and d� /d also vary. Thus,the normalized grain-boundary energy �b /�0 and grain-boundary angle � must also change according to �16�. For thecase in Fig 3, as we decrease R=0.5, 0.05, and 0.0005,�b /�0=0.36, 0.32, and 0.32, and �=3.5°, 6.2°, and 6.5°, re-spectively. The root angles �+�=7° � and �−�=14.038° � arefixed in these calculations. In a migrating-groove experi-ment, we must first select a bicrystal and this fixes �+, �−,and �b /�0. As the temperature drops, R decreases, but �b /�0
does not necessarily follow the above calculated values. In-stead, �b /�0 will vary based on the material properties’ de-pendence on temperature. If this experimentally measured�b /�0 value is substituted into �16�, �+ and �− usually cannotbe kept fixed and have to adjust accordingly. Consequently,as the temperature decreases, the groove shape as well as thegroove size will change. Depending on the bicrystal materialand orientations, the change in shape could be insignificant�such as the case in Fig. 3� or dramatic �from smooth tofaceted�.
B. Strong anisotropy: Outline
For the rest of the paper, we study grooves with stronglyanisotropic surface energies and set �=0.005 and R=0.0005. We choose �+=15°, 45°, and 75°, and for each �+
value, we take �−=15°, 30°, 45°, 60°, and 75°. Once �+ and
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�− are fixed, we increase �+ from zero to �+ and beyond andrecord results of groove profile and �−. With �+ and �−
known, � and �b /�0 can be determined from �16�. We firstpresent groove depth Y0 and chemical potential C0 as a func-tion of �+ in Fig. 4 for various �+. We then plot the left rootangle �− vs �+ in Fig. 5 for all the cases of �+ and �−. Thebehavior of �− is explained by examining a couple of specialcases in Fig. 6. We then proceed to show, for all �+ and �−
cases, solutions of � in Fig. 7 and �b /�0 in Fig. 8.There are two sets of �+ and �− that need to be men-
tioned. When �+=�−, the bicrystal has symmetric crystallo-graphic orientations and some results show special features.When �++�−=90°, the bicrystal becomes a single crystalbecause the grain boundary vanishes owing to the fourfoldsymmetry of the surface energies. This singular case exhibitspeculiar behavior for �.
C. Groove depth Y0 and chemical potential C0
Figure 4 presents Y0 and C0 as a function of �+ for �+
=15°, 45°, and 75°. Since Y0 and C0 are determined from theintegration of the groove profile on the right side, they de-pend only on �+ and �+ and are independent of �−. If the
FIG. 4. �a� Groove depth Y0 and �b� chemical potential C0 at the groove rootas a function of the right root angle �+ for �+=15°, 45°, and 75°. Thesurface energy is computed with �=0.005 and R=0.0005. When �+��+, Y0
and C0 vary almost linearly with �+. When �+��+, Y0 and C0 begin to
change rapidly. The isotropic solutions Y0 and C0 listed in Eqs. �25� and �26�are plotted in dashed lines. The rescaled values Y0�2R�1/3 and C0�2R�2/3 areplotted in solid circles. Note that the data are independent of �− and that thesurface current at the groove root M0=−Y0.
surface energy is isotropic, we find �Appendix C�
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053523-8 D. Min and H. Wong J. Appl. Phys. 100, 053523 �2006�
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Y0 = − �+ +2
39�+
3 −98
78 585�+
5 + ¯ � Y0, �25�
C0 = �+ −1
78�+
3 +79
628 680�+
5 + ¯ � C0. �26�
These asymptotic series are derived in the limit �+→0, butare accurate for the complete range of �+. The isotropic so-
lutions Y0 and C0 are plotted as dashed lines in Fig. 4 and donot agree with the anisotropic results even for �+��+. How-
ever, if we rescale the isotropic solutions and plot Y0�2R�1/3
¯ 2/3
FIG. 5. The left root angle �− vs the right root angle �+ for different facetangles: �+=15° �a�, 45° �b�, and 75° �c�. For each �+, the left facet angle �−
varies from 15° to 75°. The surface energy is computed with �=0.005 andR=0.0005. The isotropic case �−=�0 in �23� is also plotted �dotted lines� asa reference.
and C0�2R� in Fig. 4 �solid circles�, then the anisotropic
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results follow the rescaled isotropic curves when �+��+.This again confirms the scaling in Eqs. �22e� and �22h�.When �+��+, facets begin to form on the groove surfaceand the groove becomes larger and deeper. The mass flux atthe groove root M0=−Y0. Thus, to move this large groove,the mass flux at the groove root must also increase propor-tionally, and so must the chemical potential C0.
D. Left root angle �−
The left root angle �− is calculated from �18� and plottedas a function of the right root angle �+ for �+=15° �Fig.5�a��, 45° �Fig. 5�b��, and 75° �Fig. 5�c��. For each �+, weconsider �−=15°, 30°, 45°, 60°, and 75°. As shown in Fig. 3,the groove with strongly anisotropic surface energy issmooth if �+��+ and �−��−. The groove shape is the sameas the isotropic case, i.e., they have the same �+ and �−.Thus, the isotropic solution is a useful reference to comparewith. The left root angle �− for the isotropic problem is givenin �23�. Although �23� is an asymptotic solution in the limitof small slope, it agrees with the finite-slope numerical solu-tion for the complete range of �+ and is plotted in Fig. 5 asa reference.15 Figure 5�a� shows results of �− for �+=15°and various �−. When �+ is small, �− increases with �+
following �0 in �23�. There is a competition between �+ and�− as to which one will reach the corresponding facet anglefirst. If �+��+ and �−��−, such as the cases for �−=45°,60°, and 75°, then �− will deviate from �0 and turn almostvertically to reach �−. On the other hand, if �−��− and�+��+, such as the case for �−=15°, then �− will stay rela-tively uniform until �+ reaches �+. Thus, once a facet ap-pears on one side of the groove, it will also appear on theother side. As a rule of thumb, since �0�2�+, we can con-clude that if �−�2�+, then �− will reach �− first and willdeviate from �0 by remaining constant. If �−�2�+, then �+
will reach �+ first and �− will deviate from �0 by turningvertically. This can also explain the behavior of �− in Figs.
FIG. 6. Nondimensionlized groove profiles for �+=15°, �−=75°, and �+
=12° and 14.5°. The surface energy is computed with �=0.005 and R=0.0005. For �+=12°, the left root angle is found to be �−=42.165°; thegrain-boundary angle and energy are calculated from �16� as �=14.97° and�b /�0=0.018. For �+=14.5°, �−=74.629°, �=45.63°, and �b /�0=0.43.
5�b� and 5�c�.
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053523-9 D. Min and H. Wong J. Appl. Phys. 100, 053523 �2006�
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For the case of �+=15° and �−=75° in Fig. 5�a�, as �+
→�+, �− increases rapidly to reach �−. This rapid change of�− is investigated. For �+�11°, �− follows the isotropicsolution and the groove is smooth, but small, as illustrated inFig. 3. When �+=12°, �−=42.145° and is in the middle ofrapid increase. The groove profile is plotted in Fig. 6. Itshows that facets begin to appear but the corners are stillsmooth. When �+=14.5°, �−=74.629° and has turned to stayrelatively uniform. The groove is much larger with long fac-ets and sharp corners �Fig. 6�. To explain the rapid variation
FIG. 7. Grain-boundary inclination angle � vs right root angle �+ for dif-ferent facet angles: �+=15° �a�, 45° �b�, and 75° �c�. For each �+, the leftfacet angle �− varies from 15° to 75°. The surface energy is computed with�=0.005 and R=0.0005. The isotropic case �=�0 in �27� is also plotted�dotted lines� as a reference. The solid and open circles are the asymptoticsolutions in �28� for �+��+ and �−��−.
of �−, we examine the values of chemical potential C0 and
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surface current M0 �=−Y0� in Fig. 4. When �+��+, C0 andM0 increase linearly with �+. However, when �+��+, themagnitudes of C0 and M0 rise sharply. The large mass fluxand surface curvature increase the left root angle �− until itreaches the facet angle �−, which causes the left groove sur-face to be faceted.
For the case of �+=15° and �−=15° in Fig. 5�a�, �−
follows the isotropic curve until �+�6° and then stays rela-tively constant. This cannot be explained by the rapid varia-
FIG. 8. Grain-boundary energy per unit area �b normalized by the isotropicsurface energy per unit area �0 vs the right root angle �+ for different facetangles: �+=15° �a�, 45° �b�, and 75° �c�. For each �+, the left facet angle �−
varies from 15° to 75°. The surface energy is computed with �=0.005 andR=0.0005. The isotropic solution in �29� is also plotted �dotted lines� as areference. The solid and open circles are the asymptotic expansions in �30�for �+��+ and �−��−.
tions of C0 and M0 �=−Y0� in Fig. 4 because the variations
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053523-10 D. Min and H. Wong J. Appl. Phys. 100, 053523 �2006�
D
happen at �+��+=15°. The behavior of �− can be explainedby the continuity of chemical potential at the groove root in�18i�: C0=−2F��+−�+��+ /�s=−2F��−−�−��− /�s. Thenumerical integration is performed first for the right side ofthe groove and the computed C0 is a slow varying functionof �+ when �+�6°, as shown in Fig. 4. After C0 is foundfrom the right side, it is used to start the integration for theleft side. However, when �+�6°, �−→�− and F��−−�−�reaches a spike. The rapid increase of F��−−�−� forces�− /�s to decrease as their product stays relatively constant.Thus, the left groove surface becomes more planar at thegroove root. Since F��−−�−� is no longer constant, the res-caling in �21� does not work and the left groove surface mustdeviate from the isotropic solution. The left root angle �− canstay relatively uniform and still can generate the slow varia-tion in C0 because of the rapid increase of F as �−→�−.
When the right root angle �+=�+, the left root angle�−��− in general. Table I lists values of �− for different �−
when �+=�+=15°, 45°, and 75°. The data in Table I do notseem to follow a definite pattern. Two cases in Table I show�−��−+90°. This is explained in detail in the Discussionsection. These solutions have been excluded in Fig. 5.
E. Grain-boundary tip angle �
The grain-boundary tip inclination angle � is calculatedfrom �16� and plotted as a function of �+ for �+=15°, 45°,and 75° in Figs. 7�a�–7�c�, respectively. For each �+, �−
=15°, 30°, 45°, 60°, and 75°. If the surface energy is isotro-pic, �= ��−−�+� /2 and can be well approximated by �Ap-pendix C�
� =1
2�+ +
5
28�+
3 +270 065
2 517 424�+
5 + ¯ � �0. �27�
This asymptotic series is derived in the limit �+→0, but isaccurate for the complete range of �+. This isotropic solutionis also plotted as dotted lines in Fig. 7 as a reference. When�+��+ and �−��−, there is no facet on the groove surfaceand the groove shape follows the isotropic solution, as de-picted in Figs. 3 and 5. However, the grain-boundary incli-nation angle � is very different from the isotropic solutioneven when �+��+ and �−��−, as illustrated in Fig. 7 ex-cept for the unphysical case of �++�−=90°. This is becausewhen �+��+ and �−��−, the surface stiffness �+��� re-duces to a constant, and since the surface stiffness is whatappears in the governing equation �18�, the groove profile issmooth and similar to the isotropic groove. However, � iscalculated from �16�, which contains � and �� /�. These
terms vary nonlinearly with �+ and �−. We show in Appen-
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dix D that if �+��+ and �−��−, and if �++�−�90°, then
� →�− − �+
2, �28a�
but if �++�−=90°, then
� →�− − �+
2, �28b�
which is the same as the isotropic solution �Appendix C�.These asymptotic predictions are plotted in Fig. 7 as open�28a� and solid �28b� circles and agree with the numericalsolutions when �+��+ and �−��−. Therefore, the grain-boundary angle � deviates from the isotropic curve even for�+��+ and �−��−, except when �++�−=90°.
When �+��+, we also see that �=0 if the bicrystal issymmetric with �−=�+, ��0 if �−��+, and ��0 if �−
��+, except for the singular case of �++�−=90°. This logi-cal order is lost when �+��+ as � begins to oscillate wildly.
The grain-boundary angle � is insensitive to �− exceptwhen �++�−=90°. As shown in Fig. 7, if �++�−�90°, thecurves stay relatively constant until �+��+. The influence of�− is weak. Consider the case of �+=75° and �−=75° in Fig.5�c�. The curve of �− turns away from the isotropic solutionat �+�32° to stay relatively constant. However, the corre-sponding solution of � in Fig. 7�c� has no observable changewhen �+�32°. If �++�−=90°, then � deviates from the iso-tropic curve when �−��−, as demonstrated most clearly byFig. 7�c�. However, the case of �++�−=90° is unphysicalbecause the grain boundary vanishes. Thus, � in general isinsensitive to �−.
F. Grain-boundary energy
The grain-boundary energy per unit area �b is calculatedby �16� and presented in Fig. 8 as a function of �+. Theisotropic surface energy per unit area �0 is used to make �b
dimensionless. For all the cases of �+ and �−, Fig. 8 showsthat �b /�0 remains constant for �+��+ and increasessharply when �+��+. Thus, �b /�0 is insensitive to �−, sameas �. However, unlike �, �b /�0 does not follow the isotropicsolution even when �++�−=90°. The isotropic curve obeys15
�b
�0= 2 sin��+ + �−
2� , �29�
which is plotted as dotted lines in Fig. 8. It shows that noneof the cases follows the isotropic solution. As described in
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053523-11 D. Min and H. Wong J. Appl. Phys. 100, 053523 �2006�
D
�b
�0= 4R sin��+ + �−
2� , �30a�
if �++�−=90°. Notice that this expression differs from theisotropic solution only by a factor 2R. Since R�1, �b /�0
�1. This expression is plotted in Fig. 8 as solid circles. If�++�−�90°, then Appendix D gives
�b
�0= �2 sin��+ + �−
2−
�
4� . �30b�
This is plotted in Fig. 8 as open circles. These asymptoticpredictions agree with the numerical solutions when �+
��+ and �−��−.
VI. COMPARISON WITH EXPERIMENT
Grain-boundary grooving on polished polycrystallinealumina has been studied using a combination of visible-light microscopy �VLM� and atomic force microscopy�AFM� by Munoz et al.24 A sample first underwent a heattreatment for 30 min at 1650 °C in air. Visible-light micros-copy was used to map the central region of the sample.Groove profiles located in the VLM maps were measured byAFM. Using again VLM, the same region was reexaminedby AFM after the sample had gone through another heattreatment for 1 h at 1650 °C in air. A typical AFM linegroove profile for a migrating grain boundary is copied fromtheir Fig. 4�b� and plotted in Fig. 9. The profile positionshave been normalized by the groove depth.
Since the crystallographic orientations of the bicrystalwere not measured in the experiment, it is not possible tohave a direct comparison. Instead, we will fix �=0.005 andvary R to obtain a best fit to the experimental groove profile.The facet angles �+ and �− are first taken as the indicatedvalues in the AFM picture and the root angle �+ is chosen
FIG. 9. Comparison between theory and experiment. The experimentalgroove profile is taken from Fig. 4�b� of Munoz et al. �see Ref. 24� andnormalized by the groove depth. The theoretical curve is also normalized bythe groove depth following �31� and is computed with �=0.005, R=0.03,�+=6.5°, �−=23.5°, and �+=6°. The left root angle is solved from �18� as�−=22.66°. The grain-boundary angle and normalized energy are calculatedfrom �16� as �=10.45° and �b /�0=0.23. Mullins’s isotropic solution listedin �24� is also plotted for �+=6°.
slightly smaller than �+. The left root angle �− is computed
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from �18�, and the grain-boundary angle � and normalizedenergy �b /�0 are calculated from �16�. We adjust R, �+, �−,and �+ until � and �b /�0 fall within the physically acceptableranges and the groove profile agrees reasonably well on bothsides of the groove. Since the migrating velocity V was notrecorded in the experiment, we cannot compute the dimen-sional groove positions. Thus, only the groove shape is com-pared in Fig. 9 by rescaling the groove profile by the groovedepth,
Xn =X
− Y0, �31a�
Yn =Y
− Y0. �31b�
The profile presented in Fig. 9 is calculated with �=0.005,R=0.03, �+=6.5°, �−=23.5°, and �+=6°. The left root angleis solved from �18� as �−=22.66°. The grain-boundary angleand normalized energy are determined from �16� as �=10.45° and �b /�0=0.23. The experimental profile on theright side of the groove shows a small peak, which is absentin our predicted profile. This suggests that the migratinggroove might be decelerating, because a decelerating grooveexhibits a prominent peak on the left side and a minor peakon the right side.17 Another possible source of discrepancy isthat the surface energy of alumina in the cross-sectionalplane may have sixfold symmetry instead of four.33 Given allthe uncertainties, the agreement between the experimentaland modeled profiles shown in Fig. 9 is encouraging. Wealso plot Mullins’s isotropic groove profile for �+=6° in Fig.9. It shows that the isotropic solution does not agree with theobserved migrating groove profile.
VII. DISCUSSION
A quadruple precision FORTRAN program is used to per-form the numerical integration. This high precision is neededto resolve the rapid variation in the spike function. Given thewidth of the spike �=0.005, a step size �s=10−4 is requiredto yield the desired accuracy of �− �six significant digits�.With this small step size, the discretization error of thefourth-order Runge-Kutta scheme is of order 10−16, whichcannot be resolved by the usual double precision variables,and we have to upgrade to quadruple precision.
The normalized profiles are computed by a shootingmethod. Since the right side starts from large positive S usingan asymptotic expansion, a solution always exists. The inte-grated normal angle increases monotonically with decreas-ing S and the integration is stopped when =�+. This mono-tonic variation of is observed even in the presence ofstrong anisotropy. Thus, a solution always exists even when�+��+, and increases rapidly near the groove root, asillustrated in Fig. 10�a� for �=0.005, R=0.0005, �+=45°,and �+=90°. On the left side of the groove, some solutionsare easier to find than others. When �−��−, such as the casepresented in Fig. 3, not much effort is needed to get theprofile. When �−��−, great care must be taken to locate thevalue of the shooting parameter �−. Sometimes the profile
will miss the first facet plane and catch the second one, i.e.,
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053523-12 D. Min and H. Wong J. Appl. Phys. 100, 053523 �2006�
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�−��−+90°, as shown in Fig. 10�b� for �+=�+=45° and�−=15°. �This is a case presented in Table I.� The computedleft root angle �−=104.18°, which is about 90° larger than�−. The calculated grain-boundary energies are unphysicallylarge ��b /�0�1.9� for the grooves presented in Fig. 10�a�and 10�b�. These unphysical solutions have been excluded inFigs. 5, 7, and 8.
The grain-boundary tip may form a ridge instead of agroove, as illustrated in Fig. 10�c� for �=0.005, R=0.0005,�+=75°, �−=45°, and �+=−14.5°. The ridge is obtained bystarting the shooting on the right side with the negativebranch of the asymptotic expansion �Appendix B�. In thiscase starts negative and decreases monotonically with Suntil =�+=−14.5°. The integration of the left profile re-mains the same and yields �−=−44.09°. The grain-boundarytip angle �=−24.42° and the grain-boundary energy ratio�b /�0=0.4954, as caluclated by �16�. Since the value of�b /�0 is reasonable, the ridge solution in Fig. 10�c� is physi-cally admissible. The ridge cannot exist if the surface energyis isotropic because at the ridge tip there is no upward-pointing force to balance the downward-pointing surface ten-sions of the free surface and the grain boundary. However, ifthe surface energy is anisotropic, the bending tension canprovide a lifting force �Appendix A�. The ridge also appearsin modeling of stationary grain boundaries,14 and has beenobserved in twin grain-boundary experiments.34,35
In this work, �+, �−, and �+ are specified and �− iscomputed by solving �18�. With �+ and �− determined, � and�b /�0 can then be calculated from �16�. In a migrating grain-boundary experiment, �+, �−, and �b /�0 are fixed once aparticular bicrystal is selected. The other parameters �+, �−,and � are measured. For most materials, 0��b /�0�0.5.36
This requirement forces �+��+ for most cases shown inFig. 8, indicating that most migrating grooves should be fac-eted if the surface energy is strongly anisotropic. For somecases, the constraint on �b /�0 yields negative �+, leading toridges instead of grooves �Fig. 10�c��. For some values of �+,�−, and �b /�0, such as �+=45°, �−=60°, and �b /�0=0.36,there is a whole range of �+ that can solve �16� and �18�.Within this range of �+, �− varies a lot �Fig. 5�, but � staysconstant �Fig. 7�. Since �− changes from the isotropic curveto relatively uniform values near �−, the migrating groovealso changes from smooth and small to faceted and large.Thus, the migrating-groove problem admits multiple solu-tions when the surface energy is strongly anisotropic. Thisdiffers from the isotropic problem in which no multiple so-lutions are found.15,21,22
Solution of the migrating-groove problem specifies thegrain-boundary tip angle �, which may not be compatiblewith the migrating grain boundary. As shown in Fig. 7, wecan have ��0 or ��0 depending on the values of �+, �−,and �b /�0. In the “quarter-loop” method for measuringgrain-boundary mobility, the grain boundary migrates to re-duce the grain-boundary energy, as depicted in Fig. 11�a�.The grain boundary represented by the solid line migrates tothe left with constant speed and is compatible with themigrating-groove solution with ��0 �Fig. 1�. However, forsome cases, such as �+=75°, �−=30°, and �b /�0=0.36, we
FIG. 10. Nondimensionlized groove profiles for �a� �+��+, �b� �+=�+ and�c� �+�0. The profiles are computed with �=0.005 and R=0.0005. In �a�,�+=45°, �−=30°, �+=90°, and the calculated left root angle �−=30.05°, thegrain-boundary angle �=−22.05°, and the normalized grain-boundary en-ergy �b /�0=2.16. In �b�, �+=�+=45° and �−=15°. The computations give�−=104.18°, �=2.19°, �b /�0=1.91. In �c�, �+=75°, �−=45°, and �+
=−14.5°. The calculations yield �−=−44.09°, �=−24.42°, and �b /�0
=0.4954.
get ��0. This means that the grain boundary will form a
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053523-13 D. Min and H. Wong J. Appl. Phys. 100, 053523 �2006�
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loop that is more than a quarter of a circle, i.e., the grainboundary will turn more than 90° from the tip to the planarsection, as represented by the dashed line in Fig. 11�a�. Thisgrain-boundary shape does not minimize the grain-boundaryenergy since the exact quarter loop will have less grain-boundary area or energy. This incompatibility arises becausethe migrating-groove solution is obtained to minimize thefree surface energy for given crystallographic orientations��+, �−, and �b /�0� under the assumption of steady motionby surface diffusion. On the other hand, the grain boundarymigrates to minimize the grain-boundary energy. �Thedashed line in Fig. 11�a� is a mathematically correct solution,but is likely to be unstable.� These two driving forces maycompete against each other, resulting in the jerky motion ofmigrating grooves.3–5
The migrating-groove solutions with ��0 can be com-patible with steady grain-boundary migration for a differentgrain-boundary shape. Figure 11�b� shows a polycrystallinethin film with symmetric grains spanning across the film.Grain 1 shapes like a thin pancake and is shrinking to reducethe grain-boundary energy. If ��0, the grain boundarybulges out from the pancake �solid line in Fig. 11�b�� andwill migrate inward driven by the curvature.15,20 The groovesalso migrate inward and therefore the two motions are com-patible. Conversely, if ��0, the grain boundary curves in-ward �dashed line� and has to migrate outward driven by thecurvature. This is not compatible with the inward-movinggrooves and will again lead to the jerky motion.
VIII. CONCLUSIONS
This work studies the effect of surface energy anisotropyon migrating grain-boundary grooves. A delta-functionmodel is used to describe the anisotropic surface energies attemperatures above the roughening temperature of the bic-
FIG. 11. �a� The “quarter-loop” method for measuring grain-boundary mo-bility. The grain boundary migrates towards the left to reduce the grain-boundary energy. The solid-line grain boundary has tip angle ��0, whereasthe dashed line has ��0. The groove size has been exaggerated. �b� Apolycrystalline thin film with symmetric grains spanning across the film.Grain 1 shapes like a pancake and is shrinking. The dashed-line grainboundary has tip angle ��0, whereas the solid line has ��0.
rystal. The migrating-groove profile is located by arc-length
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coordinates, which allows finite groove slopes and overhang-ing groove profiles �Fig. 10�. The resulting quasisteady non-linear differential equations are solved by a shooting method.Since the groove translates, the groove profile is asymmetricabout the groove root and each side has to be determined bya different method. We find that the groove tilts the grain-boundary tip by an angle � away from being perpendicular tothe free surface. The angle � depends on the crystallographicorientations of the bicrystal, represented by the facet angles�+ and �− �Fig. 1�. We vary �+ and �− systematically tocover the full range of orientations. For most orientations,the groove is faceted. However, if the exposed groove sur-face does not contain a facet orientation, then the groove issmooth and has the same shape as the corresponding isotro-pic groove except that the size is reduced by a factor of�2R�1/3, where R is a parameter that decreases as the tem-perature drops and R=0 at the roughening temperature. Evenwhen the anisotropic groove is smooth, � generally does notfollow the isotropic solution, indicating that the anisotropyhas dramatic effects on the behavior of migrating grooves.Given a particular bicrystal, the crystallographic orientations�+ and �−, and the normalized grain-boundary energy �b /�0
are specified, and we find that the migrating-groove problemmay have a unique solution or multiple solutions. The grain-boundary tip angle � can be positive or negative, and the signmay not be compatible with steady grain-boundary migra-tion, resulting in the jerky motion observed in experiments.A migrating-groove profile measured on a polycrystallinealumina surface is fitted by our model and the profiles com-pare well.
ACKNOWLEDGMENTS
This work was supported by NSF �to one of the authors�H.W.�� under the CAREER program �Grant No. DMR-9984950� and under the Solid Mechanics and Materials En-gineering program �Grant No. CMS-0407785�. Another au-thor �D.M.� thanks LSU Graduate School for a dissertationfellowship.
APPENDIX A: DERIVATION OF HERRING’SBOUNDARY CONDITION BY THE VARIATIONALMETHOD
Herring derived the anisotropic Gibbs-Thompson for-mula by the method of virtual displacement.31 He applied thesame method to arrive at a boundary condition at the inter-ception point of three interfaces. The boundary conditionrepresents a force balance and contains d� /d, which hecalled a torque term. Mullins presented a derivation of theanisotropic Gibbs-Thompson formula by the variationalmethod,6 but the boundary condition was derived again bythe method of virtual displacement. Here, we adopt Mullins’svariational approach with the addition of mass conservation.This yields Herring’s boundary condition together with theanisotropic Gibbs-Thompson formula.
Consider a two-dimensional film on a smooth solid sub-
strate and in thermodynamic equilibrium with its own vapor,
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053523-14 D. Min and H. Wong J. Appl. Phys. 100, 053523 �2006�
D
as illustrated in Fig. 12. Since this closed system is in equi-librium, the Helmholtz free energy of the system is at a mini-mum and its variation is zero,
��0
x0
���1 + hx2�1/2 + �fs − �sg + �h�dx = 0, �A1�
where � represents the variation of a functional,37 h is thefilm height, x is a horizontal coordinate starting at the centerof the film, hx=dh /dx, the film-vapor surface energy per unitarea � is a function of the film slope, i.e., �=��hx�, �fs and�sg are the film-substrate and substrate-vapor surface energyper unit area, respectively, and x0 is the half-width of thefilm. Owing to symmetry, only half of the film is considered.The first term in the integral represents the surface energy ofthe film-vapor interface. The second and third terms repre-sent the net surface energy at the solid substrate; if the film-substrate interface lengthens, then the system gains film-substrate surface energy but loses substrate-vapor surfaceenergy. Conservation of mass is imposed by a Lagrange mul-tiplier �. Equation �A1� can be derived rigorously by the useof thermodynamics.38 Expansion of �A1� gives
�0
x0
��hdx + �0
x0 ��1 + hx2�1/2 d�
dhx+
�hx
�1 + hx2�1/2�hxdx
+ ���1 + hx2�1/2 + �fs − �sg�x0
�x0 = 0, �A2�
where the film-edge position x0 is allowed to vary �by �x0�.Since �hx=d��h� /dx, the second integral is expanded via in-tegration by parts,
�0
x0 �� −�hxx
�1 + hx2�3/2 − �1 + hx
2�1/2hxxd2�
dhx2
−2hxhxx
�1 + hx2�1/2
d�
dhx��h�dx + � �
�1 + hx2�1/2 + �fs − �sg
− hx�1 + hx2�1/2 d�
dhx
x=x0
�x0 − ��1 + hx2�1/2 d�
dhx
+�hx
�1 + hx2�1/2
x=0
��h�x=0 = 0, �A3�
where ��h�x=x0= �−hx�x=x0
�x0 has been invoked. Since �h isarbitrary, the above equation yields three equilibrium condi-tions. The first coefficient leads to an equation that governs
FIG. 12. A solid film on a substrate in equilibrium with its vapor. Thetwo-dimensional film is symmetric and contacts the substrate at x= ±x0. Asthe film is perturbed, the mass is conserved, but the contact-line positionscan change.
the equilibrium film profile in two dimensions,
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� =�hxx
�1 + hx2�3/2 + ��1 + hx
2�1/2hxxd2�
dhx2 +
2hxhxx
�1 + hx2�1/2
d�
dhx .
�A4�
The second and third terms in �A3� serve as boundary con-ditions for the equilibrium film profile. At x=0,
d�
dhx�1 + hx
2�1/2 +�hx
�1 + hx2�1/2 = 0. �A5�
At x=x0,
�
�1 + hx2�
+ �fs − �sg − hx�1 + hx2�1/2 d�
dhx= 0. �A6�
The surface normal angle is denoted by , which is theangle the film surface normal made with the vertical axis�Fig. 12�. From the definition,
tan = hx. �A7�
Thus,
d�
d= �1 + hx
2�d�
dhx, �A8a�
d2�
d2 = 2hx�1 + hx2�
d�
dhx+ �1 + hx
2�2d2�
dhx2 . �A8b�
Substitution of �A8� into �A4�–�A6� yields
� = �� +d2�
d2� , �A9�
where is the curvature of the film surface. This equation isrecognized as the anisotropic Gibbs-Thompson formula pre-sented in �1� and the Lagrange multiplier �=� /�, where �is the chemical potential and � is the atomic volume. Whenthe system is in equilibrium, �=�e and �A9� can be rear-ranged into �5�. The boundary condition �A5� at the symme-try plane x=0 becomes
cos d�
d+ � sin = 0. �A10�
Hence, at the symmetry plane d� /d=0, since =0. At thecontact line x=x0, �A6� becomes
� cos + �fs − �sg −d�
dsin = 0. �A11�
This is a force balance at the contact line and � is viewed assurface tension pointing away from the contact line. If thesurface tension is isotropic, ����� and �A11� reduces toYoung’s equation.39,40 If �=���, a bending tension d� /dacts perpendicular to � in the direction of increasing fromthe direction of �. �Note that sin �0 at x=x0.� When thebending-tension term is included in the horizontal and verti-cal force balances at the groove root, we get �16�.
The term d� /d has traditionally been described as atorque per unit area,6,31 because d� /d has the tendency tobend the surface towards the orientations with smaller �.There are two problems with this description. First, d� /d is
a vector with the direction normal to the film surface, but a
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053523-15 D. Min and H. Wong J. Appl. Phys. 100, 053523 �2006�
D
torque with the same bending effect is a vector pointing inthe direction of the contact line �Fig. 12�. Second, if d� /dis a torque about the contact line, then the moment arm willhave zero length and the torque will need to be unbounded.Thus, d� /d has the effect of a torque, but it is not a torque.Giving the bending effect, it seems that d� /d should bemore appropriately called the bending tension.
APPENDIX B: ASYMPTOTIC EXPANSIONOF THE GROOVE PROFILE AS S\�
As S→�, the free surface is almost flat and the surfacenormal angle �0 �Fig. 1�. If we consider cases in whichthe facet angle �+ is far from =0, i.e., �+��, then F�−���R and the constant R can be removed from �18a� bydefining
S =S
�2R�1/3 , X =X
�2R�1/3 . �B1�
The governing equation �18� then becomes
d3
dS3= − sin , �B2a�
dX
dS= cos . �B2b�
This is the same set of equations that governs the isotropicgroove profiles, and an asymptotic series has been obtainedin the limit →0,15
→ e−S −1
156e−3S, �B3a�
X → S . �B3b�
In terms of the original variables, this becomes
→ e−S/�2R�1/3−
1
156e−3S/�2R�1/3
, �B4a�
X → S , �B4b�
which is used to start the shooting from S→�. Note that thesolutions in �B4� multiplied by a negative sign also satisfy�B2�.
APPENDIX C: ASYMPTOTIC EXPANSIONS OF Y0, C0,�−, AND �
When the surface energy is isotropic, our previous workgives the right root angle �+ in the limit �→0 as15
�+ =1
3� −
5
1134�3 −
745
6 242 184�5 + ¯ , �C1�
where �=�++�− is the complementary dihedral angle. Herewe want to express �−=�−��+�. In the limit �+→0, the left
root angle �− is expanded as
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�− = k1�+ + k2�+3 + k3�+
5 + ¯ , �C2�
where k1, k2, and k3 are constants to be determined. Substi-tution of �C2� into �C1� and collection of terms yields
�+ =k1 + 1
3�+ + � k2
3−
5�k1 + 1�3
1134�+
3
+ � k3
3−
15�k1 + 1�2k2
1134−
745�k1 + 1�5
6 242 184�+
5 . �C3�
By comparing the coefficients, we get
�− = 2�+ +5
14�+
3 +270 065
1 258 712�+
5 + ¯ , �C4�
which is �0 listed in �23�.The groove depth Y0 and chemical potential C0 were
also expanded in the limit �→0 as15
Y0 = −1
3� +
31
4914�3 +
263 101
6 772 769 640�5 + ¯ , �C5�
C0 =1
3� −
4
819�3 −
338 603
3386384 820�5 + ¯ . �C6�
�Note that our C0 differs from that by Min and Wong15 by asign.� Since �=�++�−, we can substitute the expansion of�− in �C4� and get
Y0 = − �+ +2
39�+
3 −98
78 585�+
5 + ¯ , �C7�
C0 = �+ −1
78�+
3 +79
628 680�+
5 + ¯ . �C8�
These are listed as Y0 and C0 in �25� and �26�.From geometry, the grain-boundary tip inclination angle
� is related to the two root angles by15
� =1
2��− − �+� . �C9�
Substitution of �C4� into �C9� gives
� =1
2�+ +
5
28�+
3 +270 065
2 517 424�+
5 + ¯ , �C10�
which is �0 stated in �27�.
APPENDIX D: GRAIN-BOUNDARY ANGLE� AND NORMALIZED ENERGY �b /�0
The surface energy per unit area is given in �8�,
��� = 2�0�Ye��cos − Xe��sin � �D1a�
Differentiation of ��� yields
d�
d�� = − 2�0�Ye��sin + Xe��cos � . �D1b�
When these equations are substituted into �16�, we get, after
some simplification,
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053523-16 D. Min and H. Wong J. Appl. Phys. 100, 053523 �2006�
Since �Xe, Ye� in �9� depend on F in �6�, the above equationscan be further simplified,
�b cos � = 2�0��1
2− R��sin �+ + sin �−� + R�sin �+
+ sin �−� + L�0
�+−�+
cosh−1�
��cos�
+ �+�d + L�0
�−−�−
cosh−1�
��cos�
+ �−�d , �D3a�
�b sin � = 2�0��1
2− R��cos �+ − cos �−� + R
��cos �+ − cos �−� − L�0
�+−�+
cosh−1�
��
�sin� + �+�d + L�0
�−−�−
cosh−1�
��
�sin� + �−�d , �D3b�
where only the spike at =0 is kept in F�� because it is theonly one appearing in the domain of integration and becausethere is negligible overlapping between adjacent spikes.These equations are exact.
If the surface energy is isotropic, R=1/2, L=0, and �D3�becomes
�b cos � = �0�sin �+ + sin �−� , �D4a�
�b sin � = �0�cos �+ − cos �−� . �D4b�
Thus,
� =�− − �+
2, �D5a�
�b = 2�0 sin�+ + �−
2. �D5b�
These solutions are exact and serve as reference values inFigs. 7 and 8.
If �+��+ and �−��−, then the integrals in �D3� be-
come
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L�0
�±−�±
cosh−1�
��cos� + �±�d
� − �1
2− R�cos��±� , �D6a�
L�0
�±−�±
cosh−1�
��sin� + �±�d
� − �1
2− R�sin��±� , �D6b�
because the spike acts as the Dirac delta function.27 Equation�D3� simplies to
�b cos � = 2�0��1
2− R��sin �+ + sin �− − cos �+
− cos �−� + R�sin �+ + sin �−� , �D7a�
�b sin � = 2�0��1
2− R��cos �+ − cos �− + sin �+
− sin �−� + R�cos �+ − cos �−� . �D7b�
When �++�−=90°, �D7� is further reduced to
�b cos � = 2�0R�sin �+ + sin �−� , �D8a�
�b sin � = 2�0R�cos �+ − cos �−� . �D8b�
Thus,
� =�− − �+
2, �D9a�
�b = 4�0R sin��+ + �−
2� . �D9b�
The � expression is the same as the isotropic solution in �D5�and this explains the peculiar behavior of the case of �+
+�−=90° in Fig. 7. Even when � follows the isotropic solu-tion, �b does not because R�1. These asymptotic expan-sions are plotted in Figs. 7 and 8 and agree with the numeri-cal results.
If �++�−�90°, then �D7� becomes
�b cos � = �0�sin �+ + sin �− − cos �+ − cos �−� , �D10a�
�b sin � = �0�cos �+ − cos �− + sin �+ − sin �−� , �D10b�
because R�1. Thus,
� =�− − �+
2, �D11a�
�b = �2�0 sin��+ + �−
2−
�
4� . �D11b�
These solutions are also plotted in Figs. 7 and 8 and they
agree with the numerical results.
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053523-17 D. Min and H. Wong J. Appl. Phys. 100, 053523 �2006�
D
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