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.

REFERENCES

Berdichevsky, V.L., 2006. Continuum theory of dislocations revisited. Continuum Me-

chanics and Thermodynamics, 18, 195-222.

Berdichevsky, V.L., 2009. Variational principles of continuum mechanics. Springer Verlag,

Berlin, Heidelberg.

Bragg, W.L., 1940. Discussion. Proceedings of Physical Society, 52, 54-55.

Burgers, J.M., 1940. Geometrical considerations concerning the structural irregularities

to be assumed in a crystal. Proceedings of Physical Society, 52, 23-33.

Friedel, J., 1964. Dislocations. Pergamon Press.

Gradsteyn, I.S., Ryzhik, I.M., 1994. Tables of Integrals, Series, and Products. Academic

Press, New York.

37

Hill, R., 1950. The mathematical theory of plasticity. Oxford University Press.

Hirth, J.P., Lothe, J., 1982. Theory of dislocations. John Wiley and Sons.

Humphreys, F.J., Hatherly, M., 2004. Recrystallization and related annealing phenomena.

Elsevier.

Muskhelishvili, N.I., 1977. Some basic problems of the mathematical theory of elasticity:

fundamental equations, plane theory of elasticity, torsion and bending. Noordho¤ Interna-

tional, Leyden.

Nye, J.F., 1953. Some geometrical relations in dislocated crystals. Acta Metallurgica, 1,

153-162.

Read, W.T., Shockley, W., 1950. Dislocation models of crystal grain boundaries. Physical

Review, 78, 3, 275-289.

Weertman, J., Weertman, J.R., 1966. Elementary dislocation theory. The MacMillan Co.,

N.Y.

38