on dislocation models of grain boundariesdislocation models of grain boundaries was suggested by...
TRANSCRIPT
On dislocation models of grain boundaries
Victor L. Berdichevsky
Mechanical Engineering, Wayne State University, Detroit MI 48202 USA
(Dated: January 23, 2010)
1
Abstract
Dislocation models of grain boundaries was suggested by J.M. Burgers (1940) and W.L. Bragg
(1940). The �rst quantitative study of these models was given by W.T. Read and W. Shockley
(1950). They obtained a formula for the dependence of the grain boundary energy on the misori-
entation of the neighboring grains, which became a cornerstone of the grain boundary theory. The
Read-Shockley formula was based on a proposition that the grain boundary energy is the sum of
energies of the two sets of dislocations that come from the two neighboring grains. This proposition
was proved under an assumption on a quite special geometry of the slip planes. This paper aims
to show that the assumption is not necessary and the proposition holds for arbitrary geometry of
slip planes. Another goal of this paper is to provide all basic formulas of the theory: though the
dislocation model of grain boundaries is considered in all treatises on dislocation theory, a complete
analysis, including the relations for lattice rotations and displacements, has not be given. This
analysis shows, in particular, that continuum theory does not yield the proper relations for the
lattice misorientations, and these relations must be introduced by an independent ansatz.
Keywords: grain boundary, grain boundary energy, Read-Shockley
I. INTRODUCTION
Dislocation models of grain boundaries was suggested by J.M. Burgers (1940) and W.L.
Bragg (1940). The �rst quantitative study of these models was given by W.T. Read and W.
Shockley (1950). The key result of their classical paper is a formula for the dependence of
the grain boundary energy per unit area, ; on the misorientation of the neighboring grains,
�;
= 0�(1 + ln�m�) (1)
where 0 and �m are constants. Note the contribution of the logarithmic term for small
�: Actually, the relation (1) is satisfactory for � . 15�; and for larger misorientations the
grain boundary energy is approximately constant (see Friedel, 1964, Hirth and Lothe, 1982,
Humphreys and Hatherly, 2004).
To obtain (1) Read and Shockley modeled the grain boundary by two in�nite sets of edge
2
dislocations (Fig. 1).
FIG. 1: Dislocation model of the grain boundary.
Double lines in this Figure denote the slip lines, the symbol ? marks the position of theedge dislocation and indicates the direction of the extra half-plane. The displacement �eld
from the initial state of a perfect crystal to the current state has a jump on the slip planes.
The dislocations on the right side of the grain boundary are equally spaced with the distance
h1 and inclined to the horizontal axis for an angle �: The dislocations on the left side of the
grain boundary are spaced with another distance, h2; and inclined for an angle �: The crystal
is assumed to be isotropic, homogeneous and unbounded. Computation of energy of such a
dislocation ensemble is a problem of two-dimensional elasticity theory. The derivation of (1)
was based on the following feature of the dislocation array: the interaction energy between
the two sets of dislocations is zero. Then the total energy density of the array is equal to the
sum,
= 1 + 2; (2)
where 1 and 2 are the functions of the characteristics of only the �rst and the second sets of
dislocations, respectively1. Formula (2) was justi�ed by Read and Shockley for � = �+ �=2:
It remained unclear, if it holds for � 6= �+ �=2: This paper aims to prove (2) for any � and
�:
1 It is tempting to call 1 and 2 the self-energies of the two sets, however, they are not: energy densities
of both dislocation sets are in�nite because each dislocation set creates non-zero stresses at in�nity. It is
possible to consider though 1 and 2 as some "regularized" energies (see further Section 6).
3
Computation of energy of an in�nite set of dislocations in an unbounded space is not a
well-posed problem: di¤erent limit transitions to the unbounded space can yield di¤erent
answers. This is caused by an unlimited growth of energy of one dislocation when the body
size increases inde�nitely. We de�ne the energy of the dislocation array of Fig. 1 in the
following way. We consider a strip of the thickness L; which is in�nite in x�direction. Thestrip is sketched in Fig. 2. The strip contains a �nite number of dislocations. The faces of
FIG. 2: Setting of the problem for the dislocation model of grain boundaries.
the strip are stress free. The stresses at x = �1 are zero. We de�ne energy density of the
grain boundary as the limit,
= limL!1
ELL;
where EL is the elastic energy of the strip with the thickness L (an appropriate cut o¤ at the
dislocation core to have a �nite value of elastic energy is assumed). We show that is �nite
if the vertical component of total Burgers vector is zero2:
sin�
h1=sin �
h2: (3)
2 If l is a large distance along the array, then l=h1 and l=h2 are the numbers of the dislocations of the two sets.
The vertical components of the total Burgers vectors of the sets are � b sin� l=h1 and b sin (� � �) l=h2;respectively. The sum of these two numbers vanishes when (3) holds.
4
We show also that there is a universal function of the dislocation density of the set, n;
and the inclination angle, �;
� (n; �) =�b2
4� (1� �)n
�cos2 �+ ln
2
�"n
�; (4)
such that
1 = � (n1; �) ; 2 = � (n2; �) ; (5)
n1 and n2 being the numbers of dislocations per unit length, n1 = 1=h1; n2 = 1=h2; " the
size of the dislocation core, b the magnitude of Burgers vector. It is assumed that the crystal
is modeled by an isotropic elastic body with shear modulus � and Poisson�s coe¢ cient �:
The reason why the interaction energy vanishes is quite peculiar. It turns out that the
interaction energy per unit length is proportional to the number,
cos (� � �) limN!1
1
2N
NXn=�N
ln
�2
����sin��h1h2 n+ �
������ : (6)
Here � is some shift. If � = �=2 + �; the interaction energy is, obviously, zero. If � 6=�=2 + �; then one has to �nd the limit in (6). In a generic case, the distances h1 and h2 are
incommensurable, i.e. there are no integers, p and q; such that
ph1 = qh2: (7)
Then the numbers, �h1n=h2+�; n = 0;�1;�2:::; form a dense set on the circle. The densityof this set is constant due to the invariance with respect to shift for any angle (adding any
number to �): Therefore, for any periodic function, f (x) ; with the period 2�; the sum of the
form (6) can be replaced by the average value over the circle,
limN!1
1
2N
NXn=�N
f
��h1h2
n+ �
�=1
2�
Z �
��f (x) dx:
For the function in (6), ln 2 jsin xj ; as is known3,Z �
��ln [2 jsin xj] dx = 0:
Therefore, the interaction energy is zero for � 6= �+ �=2 as well.
3 Gradsteyn, Ryzhik, 1994, #4.384.15.
5
The Read-Shockley formula is obtained from (2)-(5) in the following way. It is shown in
Section 3 that the �rst set of dislocations causes the lattice rotation,
�1 =b
h1cos� = bn1 cos�: (8)
The lattice rotation produced by the second set is4 (Section 5)
�2 =�bh2cos � = bn2 jcos �j : (9)
The dislocation densities are linked by the condition (3),
n1 sin� = n2 sin �: (10)
Therefore, both dislocation densities can be expressed in terms of the lattice misorientation,
� = �1 + �2:
We have from (8)-(10):
bn1 =� sin �
sin � cos�+ sin� jcos �j
bn2 =� sin�
sin� jcos �j+ sin � cos�: (11)
Substitution of (11) in (4), (5) yields the Read-Shockley formula and the expression for the
constants 0; �m in terms of the angles �; �:
If h1 and h2 are commensurable, then the interaction energy is not zero. For example, for
are h1 = h2 the number (6) is equal to
cos (� � �) ln 2 jsin�j :
In such cases the grain boundary energy depends on one more parameter of dislocation
geometry, the shift between dislocation sets y0; introduced in Section 5.
Studying of the validity of Read-Shockley formula (1) for arbitrary � and � requires a
complete analysis of the problem, and we give here all necessary basic formulas, some of
which, like formulas for displacements and lattice rotations, are lacking in the treatises on
the theory of dislocations. We argue that the proper relation for the lattice misorientation
does not follow from the continuum theory and must be introduced by an additional ansatz.
4 Since � > �=2; cos� < 0:
6
II. PRELIMINARIES FROM COMPLEX ANALYSIS. SINGLE DISLOCATIONS
Complex presentation of solutions. To obtain the stress and displacement �elds of dislo-
cation sets we will use Kolosov�s presentation of the solutions of two-dimensional elasticity
equations in terms of functions of complex variable z = x + iy (Muskhelishvili, 1977, sect.
32):
u+ iv =1
2�
�{'� z'0 � �0
�xx + �yy = 4Re'
0 (12)
�yy � �xx + 2i�xy = 2 f�z'00 + �00g :
Here u and v are the x; y�components of the displacement vector, �xx; �xy and �yy the
components of the stress tensor, { = 3 � 4�; ' and � analytical functions of z, prime
denotes the derivative with respect to z; bar means the complex conjugation, Re the real
part. Addition of a function c0iz + c1 to ' and a function c2z + c3 to �; where c0 is a real
constant and c1; c2; c3 are some complex constants, does not change the stress state. Such
addition yields the change of displacements for a rigid motion. Formulas (12) correspond to
the plane strain state.
The stress function, ; which is introduced by the formulas
�xx =@2
@y2; �xy = �
@2
@x@y; �yy =
@2
@x2; (13)
is expressed in terms of ' and � as
= Re f�z'+ �g : (14)
The derivatives of the stress function can be computed from the relation
@
@x+ i
@
@y= '+ z'0 + �0: (15)
Change of coordinates. Equations (2.1) are not invariant with respect to changes of the
coordinate system. If z1 and z are the old and the new coordinates, respectively, and z is
obtained from z1 by shift for a constant, z0; and rotation for an a angle, �;
z = z0 + z1ei�; z1 = (z � z0) e
�i�; (16)
7
then '(z) and �(z) of the new coordinate system are linked to '1(z1) and �1(z1) of the old
coordinate system by the formulas (Muskhelishvili, 1977, sect. 38)
'(z) = ei�'1�(z � z0) e
�i�� (17)
�(z) = �1�(z � z0) e
�i��� z0ei�'1
�(z � z0) e
�i�� :The left hand sides of (12) are the components of displacements and stresses in the
z�coordinate system. If (12) is written in z1�coordinate system, then the left hand sidesare the displacement and stress components in z1�coordinates. They are linked to the com-ponents of the z�coordinate system by the tensorial law of transformation.
One horizontal edge dislocation. Consider �rst the displacement and stress �elds caused by
one edge dislocation with the slip line coinciding with the positive x�axis. The correspondingfunctions ' and � are:
' = �im ln z; � = im z ln z: (18)
Here m is the number,
m =�b
4� (1� �)=
�b
� (1 + {); (19)
and the branch of logarithm is selected by the condition,
0 6 Im ln z = arg z 6 2�;
while the angle arg z increases counterclockwise. The corresponding displacement �eld is:
u+ iv =b
2�
�arg z +
1
2 (1� �)
xy
x2 + y2� i
�1� 2�2 (1� �)
ln jzj+ 1
4 (1� �)
x2 � y2
x2 + y2
��: (20)
The displacement �eld has a jump on the positive x�axis. The displacement beneaththe axis is directed away from the origin. This corresponds to an extra half-plane of atoms
on the positive y�axis. We see from (20) that the y�component of displacements growslogarithmically at in�nity. This makes the description of in�nite sets of dislocations not
quite elementary.
Note that the stress function of the dislocation,
= Re f�z'+ �g = Re f�z (�im ln z) + imz ln zg = 2my ln jzj ;
is continuous on the slip line.
8
Let us �nd the stress component �xx :
�xx =@2
@y2= 2m
y
x2 + y2
�1 +
2x2
x2 + y2
�:
On the plane x = 0
�xx =2m
y:
We see that the surface force develops which pushes the material away from the plane on the
extra half-plane (y > 0) and contract material on the missing half-plane (y < 0) :
Inclined dislocation at a point z0: Let now the dislocation is placed at a point z0, and its
slip plane is rotated for an angle � (Fig. 3).
FIG. 3: Notation for inclined dislocation.
In the coordinate system fx1; y1g attached to the dislocation, we have according to (18)
'1 = �im ln z1; � = im z1 ln z1; z1 = x1 + iy1: (21)
To �nd the displacements and stresses in the (x; y)�coordinate system we set the link
between z�coordinate and z1�coordinate (16) and use the transformation rule (17). We getdropping the terms corresponding to the rigid motion
' (z) = �imei� ln (z � z0) (22)
9
� = im�e�i� (z � z0) + ei�z0
ln (z � z0) :
Function � enters only in the formula for the stress function (14). Displacements, stress and
derivatives of the stress function depend on �0; which for a single inclined dislocations is
�0(z) = ime�i� ln (z � z0) + imei�z0
z � z0+ ime�i�: (23)
The last term in (23), an additive constant, can be dropped: it corresponds to a rigid motion.
We obtain
u+ iv =mi
2�
�ei��ln (z � z0)� { ln (z � z0)
�� e�i�
z � z0z � z0
�: (24)
We see that both components of displacements are discontinuous, as it must be, because
u and v are the displacement components in z�coordinates. For the projections of the
displacements on the tangent and normal directions of the slip line, (u+ iv) e�i�; we get the
previous relations (20).
The stress function,
= �2m Im�e�i� (z � z0)
�ln jz � z0j ;
is continuous on the slip line.
III. DISLOCATION WALLS
In this Section we �nd the stress and displacement �elds of a periodic set of dislocations
placed on the y�axis with the period h: All dislocations have the same Burgers vector inclinedto the x�axis for an angle � (Fig. 14). First of all we have to remark on the reason for �xingthe position of slip lines of this set.
Slip line locations. It is well known that the stress states of dislocation sets are determined
uniquely by the dislocation positions. It is known also that the smooth strain �elds determine
the smooth displacements up to a rigid motion. For dislocations, the dislocation positions do
not determine the displacement �elds because the strains and displacements are not smooth.
The displacement �elds depend on the location of the discontinuities. This is illustrated in
Fig. 4. In this Figure an edge dislocation is placed in a rectangle. Two possible choices of
the slip lines are shown in Fig. 4a and b. It is obvious that the corresponding displacement
10
FIG. 4: Dependence of the displacements �eld on the location of slip lines.
�elds are not linked by an orthogonal transformation. Only if the dislocation is in the center
of the rectangle, the displacement of Fig. 4b can be obtained from that of Fig. 4a by an
orthogonal transformation. It is, however, an improper transformation, the mirror image
with respect to the vertical axis. Since we are going to obtain not only the stress �elds but
the displacement �elds as well, it is essential to specify the positions of slip lines.
Regularization of ' (z). Consider an in�nite set of dislocations inclined for an angle � and
placed at the points z0+ ihn; n = 0;�1;�2; ::: The problem is linear, and the displacementand stress �elds of the set are equal to the sum of the corresponding �elds of individual
dislocations. As follows from (22) and (23), we have to �nd the sums,
S1 (z) �+1X�1
ln (z � z0 + ihn) and S2(z) �+1X�1
z0 � ihn
z � z0 + ihn: (25)
Obviously, the sums are diverging. We are going to regularize these sums by adding or
substructing in�nite terms to obtain some meaningful ad hoc functions ' and �0: Then we
will check that the regularized sums indeed solve the problem formulated in Introduction.
Consider the �rst sum (25). Let us di¤erentiate it over z. We get a diverging series,
dS1 (z)
dz�
+1X�1
1
z � z0 + ihn:
It is known that the addition of an in�nite constant,
�+1X�1
(ihn)�1 ;
and a proper rearrangement of the terms,+1X
n=�1
�1
z � z0 + ihn� 1
ihn
�;
11
makes this series converging. It is known also that
+1Xn=�1
�1
z + i�n� 1
i�n
�= coth z:
Therefore,+1X�1
�1
z � z0 + ihn� 1
ihn
�=
�
h
+1X�1
1
(z�z0)�h
+ i�n� 1
i�n
!=�
hcoth
�
h(z � z0) : (26)
So, we determine S1 (z) from the di¤erential equation
dS1 (z)
dz=�
hcoth
�
h(z � z0) + const: (27)
It is convenient to choose the constant in (27) in such a way that the right hand side of (27)
tends to zero along the lines y = const as x! +1: Then
dS1 (z)
dz=�
h
�coth
�
h(z � z0)� 1
�: (28)
Let us consider �rst a slightly simpli�ed equation
dA (z)
dz= coth z � 1: (29)
The right hand side of (29) has the poles, 1= (z � i�k) ; k = 0;�1;�2; :::; at the pointsz = i�k: Therefore, A (z) has at these points the logarithmic singularities. To select a unique
single-valued solution we have to specify the cuts in z�plane that prevent the multiplicityof the values. Choosing di¤erent cuts we get di¤erent solutions of the di¤erential equation.
Let us take, for example, a straight forward solution of (29)
A (z) = ln sinh z � z: (30)
For a branch of logarithm that has been chosen (0 6 arg z 6 2�) ; the corresponding cutsare shown in Fig. 5a.
This is seen from the expansion of sinh z in vicinity of points 2i�k and i (� + 2�k) :
sinh (2�ik +�z) = sinh�z = �z; sinh (i� + 2�ik +�z) = sinh (i� +�z) = ��z;
12
FIG. 5: Domains of function (30) (a), and function (31) (b).
ln�z and ln (��z) have the jumps along the positive and negative x�axis, respectively.Since the jumps will describe the jumps of the displacements on the slip lines, function (30)
corresponds to dislocations with the geometry of the slip lines depicted in Fig. 5a. We are
going to construct the solution with the slip lines located in the right half-plane, as in Fig.
5b. Therefore, the solution sought di¤ers from (30).
So, we seek the solution of (29) with the jumps of its imaginary part on the lines shown
in Fig. 5b. Consider an in�nite strip jyj 6 �=2: We will call it the basic strip. In the basic
strip we de�ne a function,
A0 (z) = ln sinh z � z � ln 12:
We deducted from function (30) a constant to have a function which decays exponentially
along each line y = const; 0 6 jyj 6 �=2; as x ! +1: Function A0 (z) solves our problem
in the strip. Function A0 (z) is not periodic over y : at the points x+ i�2 and x� i�2it takes
di¤erent values,
A0
�x+
i�
2
�= ln sinh
�x+
i�
2
�� x� i�
2� ln 1
2
= ln (i coshx)� x� i�
2� ln 1
2= �x� ln 1
2
13
A0
�x� i�
2
�= ln sinh
�x� i�
2
�� x+
i�
2� ln 1
2
= ln (�i coshx)� x+i�
2� ln 1
2= �x+ 2i� � ln 1
2:
Consider function A1 (z) = A0 (z � i�) : It is de�ned in the strip �=2 6 y 6 3�=2: The
derivative of A1 (z) isdA1dz
= coth (z � i�)� 1 = coth z � 1:
Therefore, A1 (z) satis�es equation (29) in the strip.
On the line y = �=2 A1 (z) and A0 (z) have di¤erent values:
A1
�x+ i
�
2
�= A0
�x� i
�
2
�which di¤ers from A0
�x+ i�
2
�:
The di¤erence is a constant, 2i�: First derivatives of A0 (z) and A1 (z) coincide at the line
y = �=2: All higher derivatives are derivatives of coth z; and, thus, coincide as well. Therefore,
to construct an analytical continuation of A0 (z) onto the strip �=2 6 Im z 6 3�=2; we haveto rede�ne A1 (z) by deducting a constant:
A1 (z) = A0 (z � i�)� 2i�:
Continuing this process we arrive at the following analytical function:
A(z) = ln sinh (z � i�n)� (z � i�n)� 2i�n (31)
for �n� �=2 6 y 6 �=2 + �n; n = 0;�1;�2; :::
For numerical simulations, it is convenient to write down this function in another form. For
y > 0; we introduce function
n+(y) =
8<: IntegerPart(y) if y � IntegerPart(y) < 12
IntegerPart(y) + 1 if y � IntegerPart(y) > 12
and for any y we set
n(y) = sign y n+ (jyj) :
Alternatively, we can de�ne n(y) by the equation
dn(y)
dy=
+1Xk=�1
�
�y � 1
2+ k
�;
14
where �(y) is the ��function. Then
A(z) = ln sinh
�z � i�n
�Im[z]
�
����z � i�n
�Im[z]
�
�� ln 1
2� 2i�n
�Im[z]
�
��: (32)
Now everything is prepared to construct S1(z) which is a solution of (28). We put
S1(z) = A��h(z � z0)
�:
Accordingly,
'(z) = �imei�A��h(z � z0)
�: (33)
Regularization of �0(z): As follows from (23), to obtain �0(z) for dislocation wall, we have
to regularize, in addition to the sum S1(z); the diverging sum
S2(z) �+1X�1
z0 � ihn
z � z0 + ihn:
Since
z0 � ihn = 2x0 � z + (z � z0 + ihn) ;
where x0 � Re z0; dropping an in�nite constant, we get
S2(z) � (2x0 � z)+1X�1
1
z � z0 + ihn� (2x0 � z)
dS1(z)
dz:
So, we set
S2(z) = (2x0 � z)�
hA0��h(z � z0)
�:
Finally,
�0 (z) = ime�i�A��h(z � z0)
�+ imei� (2x0 � z)
�
hA0��h(z � z0)
�: (34)
Function �(z) can be written in terms of function B(z) which is the solution of the equation,
dB(z)
dz= A(z);
but we do not need �(z) in what follows.
Properties of A(z): Function A0(z);
A0(z) = coth z � 1;
15
is single-valued in the entire z�plane and periodic with the period i�: Function A0(z) havethe following limit behavior: for each �xed y and x! +1; A0(z) exponentially tend to zero,
A(z) = e�2xO(1); A0(z) = e�2xO(1); (35)
for each �xed y and x! �1;
A0(z) = �2 + e2xO(1): (36)
In the basic strip, A0(z) exponentially decays in the upper half of the basic strip, 0 6 y 6 �=2;
as x! +1; and for x! �1 behaves as
A0(z) = �2z + i� + e2xO(1): (37)
In the bottom part of the basic strip, � �26 y < 0; the asymptotics for x! �1 remain the
same as the asymptotics in the upper half and is given by (37). If x! +1; the asymptotics
in the bottom part is
A0(z) = 2�i+ e�2x0(1):
In the strip �2+�n 6 y 6 3�
2+�n; n = 0;�1;�2; :::; A(z) di¤ers from the values in the basic
strip for an additive constant 2i�n:
Function ReA(z) exponentially decays as x ! +1: Function ImA(z) tends to
�2i�n(y=�) as x! +1:
Derivatives of stress function. In what follows we will need to know also the derivatives
of the stress function. They are obtained by plugging (33), (34) in (15):
@
@x+ i
@
@y= 2mi
�ei�ReA
�� (z � z0)
h
�+ e�i� (x� x0)
�
hA0�� (z � z0)
h
�!: (38)
Stress state of dislocation wall. For derivatives of ' and �0 we have
'0 = �miei��hA0��h(z � z0)
�; '00 = miei�
��h
�2 1
sinh2��h(z � z0)
� ;�00 = m
(2 sin�
�
hA0 � iei�
��h
�2 2x0 � z
sinh2 �h(z � z0)
):
Plugging these in (12) we get the following relations for displacements and stresses:
u+iv =mi
2�
(ei�
A
�� (z � z0)
h
�� {A
�� (z � z0)
h
�!+ e�i�
�
h2 (x0 � x)A0
�� (z � z0)
h
�)(39)
16
�xx + �yy =4m�
hIm
�ei�A0
�� (z � z0)
h
��(40)
�yy � �xx + 2i�xy =4m�
h
(sin�A0
�� (z � z0)
h
�+ iei�
�
h
x� x0
sinh2 �h(z � z0)
): (41)
This is a solution of elasticity equations. It involves regularizations, in which we did not
pay attention to the meaning of "in�nite additions". Therefore, the physical meaning of the
solution is still to be established. Let us analyze (39)-(40). First of all, we have to check
that the singularities at the points z0 + ihn corresponds to the inclined dislocations indeed.
If z ! z0 + ihn; then A and A0 have the singularities,
A��h(z � z0)
�= ln�z; A0
��h(z � z0)
�=h
�
1
�z; �z = z � (z0 + ihn) :
Therefore, the displacements (39) have the asymptotics:
u+ iv =mi
2�
�ei��ln�z � { ln�z
�+ e�i�
2 (x0 � x)
�z
�:
The di¤erence from (24) is a translational motion in the direction ie�i� because
2 (x0 � x)
�z=
x� x0 + i (y0 � y)
(x� x0)� i (y � y0)� 1 = ��z
�z� 1:
Since A and A0 do not have the singularities other than z0 + ihn; the solution (39)-(41)
describes the in�nite array of dislocations. Note that stresses are periodic functions of y: The
displacements are not periodic because A(z) is not a periodic function.
If x! +1; then, according to (35), '0 ! 0: Therefore, from (40), �xx+ �yy ! 0: In (41)
the right hand side also goes to zero. Hence, all components of the stress tensor go to zero.
If x! �1; then from (36) and (40), (41)
�xx + �yyjx=�1 = �8m�
hsin�:
�yy � �xx + 2i�xyjx=�1 = �8m�
hsin�:
Therefore,
�xxjx=�1 = 0; �yyjx=�1 = �8m�
hsin�; �xyjx=�1 = 0: (42)
Consider two extreme cases, vertical extra half-planes and horizontal extra half-planes.
Vertical extra half-planes. If � = 0; i.e. all extra half-planes of dislocations are vertical,
the stresses exponentially decay as jxj ! 1: To see the meaning of the obtained solution in
17
a �nite strip, jyj 6 L=2; we consider the surface forces, �yy and �xy; on the line y = const:
The resultant of the surface forces can be found from the relations
+1Z�1
�yydx =
+1Z�1
@2
@x2dx =
@
@x
����x=+1
� @
@x
����x=�1
+1Z�1
�xydx = �+1Z�1
@2
@x@ydx = � @
@y
����x=+1
+@
@y
����x=�1
:
The limit values of derivatives of at x = �1 are zero as follows from (38). Therefore,
the resultant is zero. Stresses are periodic in y with the period h. For y = y0 + h=2; the
functions �yy and �xy, divided by 2�m=h; are shown in Fig. 6. For y 6= h=2 the stress
FIG. 6: Dependence of normalized stresses, �yyh=2m� and �xyh/ 2m�, on dimensionless coordinate
X = x�=h:
distributions would be slightly di¤erent. So, if we consider a strip with a �nite thickness L
and choose L=2 to be nh+ h=2; then, for any n, the stress and displacement �elds obtained
correspond to the solution of the elastic problem with loading shown in Fig. 7. The surface
load decay exponentially away from the dislocation wall on the distances of the order h.
Since the resultant of the surface load is zero, the stress state of such load in an elastic
body without dislocations decays as a second power of universe distance from loading area
(Muskhelishvili, 1977, Sect. 90). To get a solution for dislocations in an unloaded strip, a
localized stress �eld that "kills" the surface load should be added to the obtained one. Note
18
FIG. 7: Loading of the strip corresponding to the solution obtained; only the top and the bottom
parts of the strip are shown; the number of dislocations can be any.
that the magnitude of surface load is very large: for b=h = 0:01; � = 0:3; �yy ' 0:002�:
However, due to power decay of stresses, (h=r)2 ; r being the distance from the loading area,
for r > 10h; the additional stress �eld can be neglected, and the solution obtained can be
trusted.
Stability of dislocation wall. It is instructive to �nd the force acting on a dislocation of the
ensemble. The force is de�ned as derivative of energy with respect to dislocation position on
the slip plane. Here by energy one means a regularized elastic energy, when the self-energy of
dislocation is removed. One can show that the derivative of energy is (see, e.g., Berdichevsky
2009, Sect. 6.10)@H
@�= ��ijnjbi:
Here @H=@� is the derivative of energy in the direction outside of the slip plane, nj the normal
to the slip plane, bi Burgers vector, �ij the stress �eld generated by all other dislocations of
the ensemble, summation over repeated indices is implied. Emphasize that Burgers vector
of a dislocation is not determined uniquely and makes sense only in conjunction with vector
nj : if one marks by the signs + and � the two sides of the slip surface and vector nj is
19
directed from the side � to the side +, then bi is the di¤erence of displacements of the sides+ and �. For another choice of the sides + and �, vectors nj and bi change directions, whilethe product binj remains the same. The force, as it must be, depends only on this product.
For the dislocation located at the origin n1 = 0; n2 = 1; b1 = �b; b2 = 0: Thus the force is
@H
@�= �0xyb;
where �0xy is the value of xy�stress component for the dislocation wall with the removeddislocation at the origin. To �nd �0xy, we have to deduct from the stress �eld of the dislocation
wall the stress �eld of the dislocation at the origin. From (18), (12) and (41) we have on the
slip line y = 0
�0xy =2m�2
h2x
sinh2 �xh
� 2mx:
The graph of this function is shown in Fig. 8. As follows from this graphs, @H=@� < 0 for
FIG. 8: Dimensionless shear stress �0xyh=2�m on a slip line y = 0 as a function of dimensionless
coordinate X = �x=h:
x > 0; @H=@� > 0 for x < 0; and @H=@� = 0 for x = 0: Therefore, the position of dislocation
at x = 0 is an equilibrium position. The ��direction (the direction outside of the slip plane)looks to the left. If the dislocation is shifted from the equilibrium to the right, @H=@� < 0,
i.e. energy at the origin is smaller. If the dislocation is shifted to the left, @H=@� > 0; and
again energy at the origin is smaller. This means that the equilibrium position is stable.
20
Horizontal extra half-planes. Consider another extreme case � = �=2: In this case the
extra half-planes of dislocations are horizontal and lie on the left of the array (Fig. 9). As
FIG. 9: A dislocation wall with horizontal extra half-planes; note the non-zero stresses at x = �1.
follows from (42) the stress �yy develops as x! �1;
�yy (1) = �8m�
h= � 2�b
h (1� �):
This stress is huge: for b=h = 0:01 it corresponds to the elastic strain "yy = �0:01: Thedependence of strains "yy on x for y = y0 +
12h is shown in Fig. 10. To get a universal curve
we use the dimensionless parameters "yyh /b and X = (x� x0)/h: The stress �eld obtained
can be considered as the limit as L!1 in the elastic problem sketched in Fig. 9.
To obtain the picture of deformation in an unloaded strip one has to add to the solution
obtained a solution of the elastic problem for the load shown in Fig. 11a. The resulting
deformed state is sketched in Fig. 11b. An essential di¤erence from the case of vertical
extra half-plane is that the deformed state depends on the strip thickness: in Fig. 11b, the
characteristic size of the deformed region in longitudinal direction is L. This emphasizes
an important feature of the solution obtained: by including the load at x = �1 we get a
universal solution which is independent on the strip thickness L.
Consider now the mirror image of the dislocation wall in Fig. 9 with respect to y�axis,i.e. the dislocation wall with vertical slip lines and the extra half-planes located on the right
21
FIG. 10: The dependence of dimensionless yy-component of strains, "yyh=b; on dimensionless x-
coordinate, X = (x� x0)=h; for y = y0 + h=2 and � = �=2:
FIG. 11: Auxiliary elastic problem (a) and a sketch of deformed state for unloaded strip with
dislocations (b).
side of the dislocation wall. At �rst glance, it should have zero stresses at x = �1 and
non-zero �yy at x = +1: This is not, however, the stress state which we constructed: we
selected the solution by setting stresses to be zero at x = +1: In our solution, the mirror
image of the dislocation wall of Fig. 9 corresponds to � = 3�=2: From (42) we get vertical
traction at x = �1; �yy = 8m�=h as shown in Fig. 12a. The relation to the mirror image
of the stress state of Fig. 9 is explained by the sketch of Fig. 12: one has to add a constant
22
stress state, �yy = 8m�=h; �xx = �xy = 0; to link our solution with the mirror image of the
stress state of Fig. 9.
FIG. 12: The stress state of the mirror image of the dislocation wall in Fig. 9 is the sum of the
stress state similar to that of Fig. 9 and tension along y�axis at in�nity.
As follows from (39)-(41), for the inclined dislocation array the stress �eld is a linear
combination of the �elds corresponding to the two choices of �; � = 0 and � = �=2:
Body rotation and displacements. To �nd the body rotations we have to compute the
derivatives of displacements. Di¤erentiating (39) with respect to x we get
@u
@x+ i
@v
@x=mi�
2�h
(ei��A0 � {A0
�� 2e�i�A0 � e�i�
�
h
2 (x0 � x)
sinh2 �h(z � z0)
): (43)
We see that the displacement gradient tends to zero at x = +1 along the lines y = const for
all lines except the slip lines. At the slip lines, displacements have a jump, �bei�: Computingthe change of displacements along y�axis for large positive x we get
(u+ iv) jy=L=2 � (u+ iv) jy=�L=2� �bL
hei�:
Therefore, due to the displacement jumps, the average shear for large positive x is
1
L
Z L=2
�L=2
@u
@ydy = � b
hcos�: (44)
At x = �1 the displacement gradient is �nite. We have from (43),
@v
@x
����x=�1
=m�
2�hcos� (1 + {) =
b
hcos�: (45)
23
Since "xy = 0 at x = �1; @v=@x = �@u=@y and
@u
@y
����x=�1
= � bhcos�: (46)
We see that the average shear on both sides of the dislocation wall, (44) and (46), are the
same as it must be. Otherwise, an overlap of material particles would occur.
Di¤erentiating (39) with respect to y and tending x to �1 we obtain
@u
@y+ i
@v
@y
����x=�1
= � bhei�: (47)
The real part of (47) coincides with (46) while the imaginary part gives the value of "yy at
�1 :
"yyjx=�1 = �b
hsin�: (48)
The left half of the body rotates as rigid body, the right half has a shear. The schematic
picture of the displacement �eld for � = 0 is shown in Fig. 13. We see that there is an overall
FIG. 13: Scheme of the displacement �eld caused by an array of dislocations with � = 0:
counterclockwise rotation of the body.
Emphasize that the positions of dislocations in Fig. 13 are shown in undeformed state (in
Lagrangian coordinates x, y). In the deformed state the dislocation wall is inclined. Within
24
the linear theory we ignore the di¤erence between the positions of dislocations in deformed
and undeformed states. The error induced in calculation of the stress state is of the order of
the angle of rotation.
IV. LATTICE MISORIENTATION AT DISLOCATION WALL
The issue of the lattice misorientation introduced by a dislocation wall is not as elementary
as it is presented in publications on the subject. Consider the simplest case of a dislocation
wall with � = 0: Such a wall can be interpreted as a grain boundary of two misoriented crystals
(see Hirth, Lothe, 1982, Sect. 19-2 or Weertman and Weertman, 1966, p.187). Geometrical
reasoning yields a simple relation between the misorientation angle, �; and dislocation density,
h�1 :
tan�
2=1
2
b
h: (49)
The argumentation is meaningful only for small b=h, and we take here the linearized
version of (49),
� =b
h: (50)
We are going to argue that this relation is an inheritance of the crystal nature of the
substance modeled: it does not follow from the continuum model of dislocation walls. Indeed,
consider the displacements caused by a dislocation wall within the continuum theory. These
are the displacements from the initial stress-free state, which is identi�ed with the perfect
(defect-free) lattice. As is seen from Fig. 13 both halves of the elastic body on the two
sides of the dislocation wall rotates counterclockwise. The angles of rotation are di¤erent:
according to (45) and (46) for � = 0; on the left of the dislocation wall the angle of rotation
is
�� =1
2
�@v
@x� @u
@y
�����x=�1
=b
h: (51)
On the right of the dislocation wall
v =@v
@x
����x=+1
= 0
while, according to (44), u (+1; y) grows linearly at average providing the rotation for the
25
angle
�+ =1
2
�@v
@x� @u
@y
�����x=+1
=1
2
b
h:
The rotations on the two sides of the dislocation wall are di¤erent because @v=@x = 0 at
x = +1 and @v=@x 6= 0 at x = �1 (the former can be always assumed, the latter is the
result of solution of the elasticity problem). The relative rotation of the two halves of the
elastic body is
� = �� � �+ =1
2
b
h: (52)
In continuum mechanics, the symmetry axes are assumed to be moving with continuum. If
one would accept this for the case under consideration, then the lattice misorientation would
be given by (52) and, thus, be di¤erent from (50). We see that the lattice misorientation
must be introduced in the continuum theory by an additional ansatz. As such one can use the
following proposition (Berdichevsky, 2006): a homogeneous �ow of dislocation does not cause
the lattice rotation. The dislocation wall depicted in Fig. 13 can be viewed as a result of a
homogeneous dislocation �ow arrived from the right half of the specimen. Therefore, there is
no lattice rotation on the right, while on the left the lattice deformation coincides with that
of the elastic body. On the left, the elastic body rotates for the angle (51). Accordingly, the
lattice misorientation is b=h and, thus, coincides with (50).
An equivalent way to introduce the lattice misorientation goes back to Nye (1953), who
suggested that the lattice rotation is caused by elastic distortion only. In modern notation,
the elastic distortion is de�ned as
�(e)ij =
@ui@xj
� �(p)ij ;
where �(p)ij is the plastic distortion. It is determined uniquely by dislocation geometry:
�(p)ij =
Xdislocations
binj� () :
Here is the slip plane, � () the ��function, nj the normal vector to the slip plane. In thecase under consideration only one component of the plastic distortion is non-zero,
�(p)12 = �b
Xa
� (La) ;
26
La being the slip lines of Fig. 13. Averaged over y plastic distortion coincides with averaged
over y gradient of displacements (44). Therefore, elastic distortion is zero on the right of the
dislocation wall. On the left of dislocation wall plastic distortion is zero, and elastic distortion
coincides with the displacement gradient. Hence, the lattice misorientation is given by (50).
In summary, the lattice rotation is an additional "entry" of a continuum theory.
If � 6= 0; then from (45) and (46) the lattice misorientation is
� =b
hcos�:
It is maximum for � = 0; decreases with the increase of � and becomes zero for � = �=2:
To understand the origin of the decline of the misorientation caused by the increase of �;
consider the schematic picture of displacements on the dislocation wall in Fig. 14. In the
FIG. 14: Schematic picture of displacements at the dislocation wall in case of inclined dislocations.
right half-plane the smoothed displacement gradient is constant, and, ignoring the change of
displacements caused by dislocation cores, we show at the wall the displacement �eld which
27
develops at large positive x. At a point x = 0, y = R > 0 the displacement �eld has the
coordinates�!u R =
��bR
hcos�;�bR
hsin�
�:
Here R=h is the number of dislocations on the segment [0; R] : The horizontal component of�!u R causes the rotation of the left half-plane. The angle of rotation of the left half-plane
is � (b=h) cos�: The larger �, the smaller "the part of �!u R" producing the rotation. The
vertical component of �!u R linearly depends on y as �y(b=h) sin�: This corresponds to thevertical strain "yy in the left half-plane, � (b=h) sin�, in full agreement with (48).
V. DISLOCATION MODEL OF GRAIN BOUNDARY: DOUBLE-WALL
Stress �elds. We return now to the dislocation model of the grain boundary shown in Fig.
1. We put two sets of dislocations on the y�axis: Dislocations of the �rst set are placed at thepoints y = h1n; n = 0;�1;�2; :::; dislocations of the second set at the points y = y0 + h2n;
n = 0;�1;�2; :::: We choose the Burgers vectors as shown in Fig. 1 (they are determinedby the positions of extra half-planes). The displacements and stresses of the �rst set can be
determined by the formulas of the previous Section for z0 = 0, while for the second set one
has to change in these formulas b by �b; put h = h2 and make a shift for iy0 :
'1 = �miei�A��z
h1
�; �01 = mi
�e�i�A
��z
h1
�� ei�z
�
h1A0��z
h1
��; (53)
'2 = miei�A
�� (z � iy0)
h2
�(54)
�02 = mi
��e�i�A
�� (z � iy0)
h2
�+ ei�z
�
h2A0�� (z � iy0)
h2
��: (55)
The total �elds is given by the sums,
' (z) = '1 (z) + '2 (z) ; �0 (z) = �01 (z) + �02 (z) :
The displacement and stress �elds can be found by plugging (53), (54), (55) into (12) or by
summation of the �elds (39)-(41) corresponding to the two sets of dislocations:
u+ iv =mi
2�
(ei�
A
��z
h1
�� {A
��z
h1
�!� e�i�
�
h12xA0
��z
h1
�
28
�ei� A
�� (z � iy0)
h2
�� {A
�� (z � iy0)
h2
�!+ e�i�
�
h22xA0
�� (z � iy0)
h2
�)(56)
�xx + �yy = 4m Im
�ei�
�
h1A0��z
h1
�� ei�
�
h2A0�� (z � iy0)
h2
��(57)
�yy � �xx + 2i�xy = 4m
(sin�
�
h1A0��z
h1
�+ iei�
��
h1
�2x
sinh2 �zh1
� sin � �h2A0�� (z � iy0)
h2
�� iei�
��
h2
�2x
sinh2 �(z�iy0)h2
): (58)
We begin the analysis of (56)-(58) with an obvious note that the stresses tend to zero
exponentially as x! +1: If x! �1; the stresses are �nite. From (57) and (36)
�xx + �yyjx=�1 = �8m��sin�
h1� sin �
h2
�:
Therefore, in order for the grain boundary not to produce long range stresses, it is necessary
thatsin�
h1=sin �
h2: (59)
As was explained in Introduction, this condition means that the vertical component of the
total Burgers vector is zero. It is seen from (58) that (59) is also a su¢ cient condition for
vanishing of all components of stresses at x = �1: Symbolically, the superposition is shown
in Fig. 15. Due to (59), the stresses are exponentially decaying away from the dislocation
FIG. 15: A symbolic picture showing how the superposition of two stress �elds results in a stress
�eld localized near the dislocation wall.
wall. The characteristic size of the decay is max fh1; h2g : In what follows we consider thegrain boundaries for which (59) holds true.
29
Displacements. Let us �nd the rotations. Di¤erentiating (56) over x; we obtain:
@u
@x+ i
@v
@x=mi
2�
(ei�
�
h1
A0��z
h1
�� {A0
��z
h1
�!
�2e�i� �h1A0��z
h1
�+ e�i�
��
h1
�22x
sinh2 �zh1
�ei� �h2
A0�� (z � iy0)
h2
�� {A0
�� (z � iy0)
h2
�!+ 2e�i�
�
h2A0�� (z � iy0)
h2
�
�ei���
h2
�22x
sinh2 �(z�iy0)h2
):
For x! �1;
@u
@x+ i
@v
@x=mi
2�
�2� (1 + {)
�cos�
h1� cos �
h2
�� 2�i (3� {)
�sin�
h1� sin �
h2
��:
Due to (59),@u
@x
����x=�1
= 0;
while@v
@x
����x=�1
= b
�cos�
h1� cos �
h2
�: (60)
Since 2"xy = @u=@y + @v=@x = 0 at x = �1 and the body at x = +1 is at rest @v=@x at
x = �1 determines the angle of rotation of the left half of the body with respect to the right
half. Two homogeneous �ows of dislocation did not disturb the crystal orientation, therefore
the angle of the lattice misorientation on the grain boundary in the deformed state is also
given by (60). Since � > �2; cos � < 0; and angle of misorientation � is equal to the sum of
misorientation angles (8) and (9).
Resultants. Let us establish one more feature of the stress state: �rst derivatives of
exponentially tend to zero at x = �1. Indeed, from (38)
@
@x+ i
@
@y= 2mi
(�ei�ReA
��z
h1
�+ e�i�x
�
h1A0��z
h1
�(61)
+ei� ReA
�� (z � iy0)
h2
�� e�i�x
�
h2A0�� (z � iy0)
h2
��:
30
The derivatives of are exponentially small as x ! +1. For x ! �1; according to (36),
A(z) w �2z + i�; A0(z) w �2; and (61), up to exponentially small terms, becomes
@
@x+ i
@
@y= 2mi
��ei�
��2�xh1
�� e�i�
2�x
h1+ ei�
��2�xh2
�+ e�i�
2�x
h2
�: (62)
The left hand side of (62) is zero due to (59).
An immediate consequence of this feature is vanishing of resultant of forces acting along
the cuts y = const :Z +1
�1�yydx =
Z +1
�1
@2
@x2dx = lim
a!1
@
@x
����x=a
� @
@x
����x0�a
!= 0:
Z +1
�1�xydx = �
Z +1
�1
@2
@x@ydx = � lim
a!1
�@
@y
����x=a
� @
@y
����x=�a
�= 0:
The total moment is, in general, non-zero.
Vanishing of the resultant allows us to claim that (56)-(58) give the solution of the problem
formulated in Introduction. Indeed, consider the strip shown in Fig. 2. We present the
solution of the problem as the sum of (56)-(58) and an addition. The addition must null
�yy and �xy at the faces of the strip, generated by (57)-(58). At the faces, �yy and �xy tend
exponentially to zero as x ! �1; and, thus, are localized near the y�axis. Besides, theyhave zero resultant. Therefore, the additional stresses decay as jrj�2 ; r is the distance fromthe localization area of �yy and �xy at the faces. Hence, at each �xed point inside the strip
the solution tends to (56)-(58) as L ! 1: The energy of the additional stresses is �nite
as L ! 1; and, thus, negligible in comparison with the energy of the plate, which is of
the order L: The interaction energy of stresses (57)-(58) and the additional stresses is also
�nite as L ! 1; because the region where the stresses (57)-(58) are not small is a strip
along the y�axis. Therefore, the two-dimensional integral over the strip of the product ofthe additional stresses and the stresses (57)-(58) is, actually, an integral over the strip, and,
thus, is �nite. Hence, the interaction energy between stresses (57)-(58) and the additional
stresses can be neglected as well, and (56)-(58) gives the solution sought.
Stability of dislocation double-walls. One can check that, in general, a dislocation double-
wall is not an equilibrium dislocation set. There are two exceptions: � = 0; � = 0; i.e. all
extra half-planes are in the plane of the wall, and � = �=4; � = 3�=4: Both such double-
walls are stable. This statement assumes that dislocations are treated in the framework
31
of isotropic elasticity and can move on the slip plane in any direction. Such a model is
not physically adequate. If a dislocation arrives on the grain boundary from one side, then
moving to another side could be hampered by other orientation of the slip planes. Apparently,
the study of stability of the dislocation arrays should include the crystal anisotropy and an
information on the change of dislocation energy when the direction of the slip plane changes.
To the writer�s knowledge these issues are still open. One can expect that incorporation of
anisotropy makes the dislocation double-walls stable for continuously changing � and �.
VI. ENERGY OF A TWO-DIMENSIONAL DISLOCATION ENSEMBLE
The energy of linear elastic body is the functionalZV
U ("ij) dV (63)
where U is the energy density per unit volume, V the region occupied by the elastic body,
"ij the components of the strain tensor, "ij = 12(@ui=@x
j + @uj=@xi) ; ui the components of
displacements (i; j = 1; 2; 3) :
Energy of a dislocation is de�ned as the minimum value of the functional (63) on the set
of all displacement �elds with a prescribed jump on some surface : [ui] = bi: By ['] we
denote the di¤erence of the values of ' on the two sides of : The vector bi is assumed to be
tangent to ; and, to be relevant to modeling of crystal deformations, the magnitude of bi
is constant and equal to interatomic spacing. To avoid the singularity at the dislocation line
�; the boundary of the slip surface ; one can smooth bi in a small vicinity of �: There are
also other ways to regularize energy.
A dual variational problem to �nd the energy of the dislocation is to minimize the integral5,ZV
U���ij�dV +
Z
�ijnjbidA; (64)
over all stress �elds obeying to the equilibrium equations
@�ij
@xj= 0 in V (65)
5 See details in Berdichevsky, 2009, Sect. 6.9.
32
continuity conditions on ; ��ij�nj = 0; (66)
and the boundary conditions,
�ijnj = 0 on @V: (67)
Here U� (�ij) is the complementary energy. Surface may consist of several pieces. This
corresponds to a system of dislocations. Then the minimum value of the functional (64) is
the negative energy of this system of dislocations.
FIG. 16: Dislocation con�guration in energy computation.
For edge dislocations in an elastic body unbounded in x3-direction, the problem becomes
two-dimensional: stresses and displacements are functions of two coordinates, x1 and x2: Dis-
placements have a jump on the slip line L, which is the cross-section of and (x1; x2)�plane.The general solution of equilibrium equations (65) is given by (13). Continuity of surface
forces on L means that� ;��= const (Greek indices run though values 1; 2; comma in indices
denotes partial derivative). Indeed,
�����n� = e��e��
� ;��
�n� = �e��� �
� ;��;�= �e��
d� ;��
ds= 0: (68)
Here e�� are Levi-Civita symbols (e11 = e22 = 0; e12 = �e21 = 1) ; n� and � � = �e��n� thenormal and tangent vectors to the slip plane, s are length along the slip line. In general,
33
� ;��6= 0: However, for the con�guration of the slip lines shown in Fig. 16,
� ;��= 0: This
con�guration is characterized by the condition that all dislocations came into the body from
the free surface. To show that� ;��= 0 on the slip lines, we note that similarly to (68),
;� = const on the free surface. Therefore,� ;��= 0 on each slip line that crosses the free
surface. Since� ;��= const along a slip line,
� ;��= 0 on each slip line. The slip lines
of the dislocation ensemble considered cross the free surface. Therefore, ;� are continuous
functions in V .
Note that the signs of vectors n� and �� in (68) are coordinated with the choice of positive
direction on @V and the slip lines as counterclockwise: n� is directed from the side - of the
slip line to the side + (side � is passed �rst).For an isotropic homogeneous body the energy functional to be minimized has the form
J ( ) =
ZV
1
4�
� ;��
;�� � � (� )2�d2x� l ( )
l ( ) =
ZL
e��b�d ;�ds
ds: (69)
Minimum is sought over all functions with continuous derivatives on L and constant deriv-
atives on the boundary @V of the body. Since the functional is invariant with respect to shifts
of for a linear function, ;� at @V can be set equal to zero. Then the set of admissible
functions is linear, and we can apply Clapeyron�s theorem for the minimum value of J( ) :
E = � min : ;�=0 on @V
J( ) =1
2l�� �; (70)
where � is the minimizer of J( ):
The problem is linear, and the minimizer is the sum of minimizers corresponding to each
dislocation located at the points r(a): Denote by b(a)� , L(a); and (a) Burgers vector, slip line
and stress function of ath dislocation. Then,
� =NXb=1
(b);
and the energy of the ensemble, according to (70), becomes
E =1
2
NXa;b=1
ZL(a)
e��b(a)�d (b);�ds
ds:
34
We extract from this sum the self-energy of dislocations,
Es:e: =1
2
NXa=1
ZLa
e��b(a)�d (a);�ds
ds (71)
and the interaction energy
Eint =1
2
NXa 6=b
ZLa
e��b(a)�d (b);�ds
ds: (72)
In (71) b(a)� (s) is constant along La except a small vicinity of the point r(a) where it changes
from a constant value to zero at r(a): This vicinity models the dislocation core. A customary
way to de�ne the self-energy of a dislocation is, in the terms considered, to put the self-energy
of ath dislocation equal to
1
2
ZL0a
e��b(a)�d (a);�ds
ds+ dislocation core energy (73)
where L0a is the line La with a removed small vicinity of the point r(a):
Since b(a)� is a constant at L0a and (a);� is zero at the boundary of the body, the integral in
(73) is equal to
�12e��b(a)� (a);�
�r(a) + "� (a)
�; (74)
where " is a distance on the order of the core size and � (a) the tangent vector to L(a): The
number (74) tends to in�nity as " ! 0; therefore " can be chosen in such a way that (74)
incorporates the dislocation core energy.
In the interaction energy (72), computing the integrals and taking into account that (b);�
are smooth in a vicinity of points r(a) for b 6= a; and b(a)� are not constant only in a small
vicinity of r(a) on the order of b, we get
Eint = �1
2
Xa 6=b
e��b(a)� (b);��r(a)�:
Finally,
E = �12
NXa=1
e��b(a)� (a);��r (a) + "� (a)
�� 12
Xa 6=b
e��b(a)� (b);��r(a)�: (75)
35
VII. ENERGY OF GRAIN BOUNDARY
Let us �nd the elastic energy of the dislocation array of Fig. 1. We have two sets of
dislocations, one has the Burgers vector b(1)� ; and another one b(2)� : Let EL be the energy of
the plate of the height L: Introducing energy per unit length of grain boundary,
= limL!1
ELL;
and the numbers of dislocations of two sets in the plate N1 = L=h1; N2 = L=h2; we have
from (75)
= 1 + 2 + int;
int = 12 + 21
1 = �1
2e��b(1)�
1
h1lim
N1!1
1
N1
Xn
(1);��ih1n+ "ei�
�(76)
2 = �1
2e��b(2)�
1
h2lim
N2!1
1
N2
Xn
(2);��iy0 + ih2n+ "ei�
� 12 = �
1
2e��b(1)�
1
h1lim
N1!1
1
N1
Xn
(2);��ih1n+ "ei�
� 21 = �
1
2e��b(2)�
1
h2lim
N2!1
1
N2
Xn
(1);��iy0 + ih2n+ "ei�
�:
Here (1) and (2) are the stress functions of the two sets of dislocations, 1 and 2 the
self-energies of the two sets of dislocations, int the interaction energy of the two sets.
To compute the self-energy of the �rst set we note the relation
e��b� ;� = Im (b1 � ib2)� ;x + i ;y
�: (77)
For the �rst set b(1)1 � ib(1)2 = �be�i�: Using (38) we see that (1);� (ih1n+ "ei') have the same
values for all n. Thus
1 =bm
h1Im
"e�i�i
�e�i�ReA
��"
h1ei��+ e�i�
�" cos�
h1A0��"
h1ei��!#
;
and we obtain
1 =bm
h1
�ln2h1�"
+ cos2 �
�: (78)
36
Similarly, taking into account that b(2)1 � ib(2)2 = �be�i�;
2 =bm
h2
�ln2h2�"
+ cos2 �
�:
These relations yield (4) and (5). For the interaction energy we have
12 = �12bm cos (� � �)
h1; 21 = �21
bm cos (� � �)
h2
where �12 and �21 are the limits,
�12 = limN!1
1
2N
NXn=�N
ln 2
����sin �
h2(h1n� y0)
�����21 = lim
N!1
1
2N
NXn=�N
ln 2
����sin �
h1(h2n+ y0)
���� :First of all, if ��� = �=2; then we arrive at the Read-Shockley result: the interaction energy
is zero. If � � � 6= 0; then, to �nd the interaction energy, we have to �nd the limits, �12
and �21: As was discussed in the Introduction, due to ergodicity of the sequence of numbers
y0 + �k; k = 0;�1; :: on the circle, they are equal to the integral,
1
2�
Z �
��ln 2 jsin xj dx:
This integral is zero. Thus, �12; �21; 12; 21; are zero and the interaction energy vanishes,
as claimed.
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38