STUDY OF DEFECTS IN TWO BEAM CONDITIONS II

Francisco Lovey

Centro Atómico Bariloche Instituto Balseiro

PASI on Transmission Electron Microscopy

Santiago de Chile, July 2006

OsornoPuntiagudo

Images of crystalline defects

If the jth atom is moved from its position in the perfect crystal by a translation R(r), the potential at any point in the deformed crystal can be identified with the value of the potential at the position [r- R(r)] in the non deformed crystal. The potential in the deformed crystal can be expanded as a Fourier series, as for the non deformed crystal

ab r

-R r( )

ab r

R r( )

P erfec t c ry s ta l D e fo rm e d c ry sta l

g

R(r)-rg.gr i2

def e V )(V

The Fourier coefficient would change as

g.R(r)gg

i-2e UU me2

hUV

2

gg

g

R(r)-rg.gr i2

def e V )(V

To include the anomalous absorption

g.R(r)gg

i-2e UU

In the same way the coefficients of the wave function will change as

j

jjj )(i2)(i2)()( eeC)(k g

rgkrg.Rkg

kr .

)(i2)()( eCCjj rg.Rk

gk

g

col1def

col CQC

g

Q i2e 00 1

grR ).(g

gg

gg i2-)j(

)j(0

)j(

)j(0

i2-eC

C

C

C

e 0

0 1

g

Q-1i2-e 0

0 1

The basic eigenvalue equation

C

C )

K

g1(K2

C

C

Ui2KS UiU

UiU U(j)

(j)0

n

nn(j)

(j)0

0

0 j)(

abs

ggggg

gg

col)j(

abscolabs CCA

0

0

n

nn

abs

Ui2KS UiU

UiU U

Kg

1K2

1

ggg

gg

A

No defects

The eigenvalue equation

defcol

)j()j(defcol

def iq( CCA )

QAQA s-1 abdef col1def

col CQC

0i2-

i20

n

n

def

Ui2KS eUiU

eUiU U

gK

1K2

1

ggg

gg

g

g

A

g

Qi2e 0

0 1

grR ).(g

In the presence of defects can be written as

col)j(

abscolabs CCA

The wave function for the perfect crystals (including absorption) was

u C C -1D

z)iq(i2 e z(j)

(nearly normal incidence)

In the presence of defects it becomes

uCQCQ -1-1 1

Dγdef )ziqi(2 z(j)

e

(2)

g(1)g

(2)0

(1)0

C CC C

C

01

e 00 1

C CC C

e 0

0 e C CC C

e 00 1

g)2()2(

)1()1(

g i220

20

1g

10

z)iq(i2

z)iq(i2

20

1g

20

10

i2-0 g

g

Qi2e 0

0 1

Using the column approximation we shall assume that in each column

)rR R(z)(

The wave functions can be expressed as

The matrix Cdef=Q-1C makes Adef diagonal

defcol

)j()j(defcol

def iq( CCA )From

iq)( D(j)z

j)defdef-1def (CAC

1defD

)j(z

(j)defdef )(iq CCA

1defD

)j(z

(j)defdef )(iq CCA

uCQCQ -1-1 1

Dγdef )ziqi(2 z(j)

e

u e zi2def defA

From the operator series expansion

uA e)z( zi2def def

The variation in the thickness between z and z+dz is

)z( )zi2I()z( e)dzz( defdefdefzi2def def

AA

def defdef

i2 dz

d

Az

z)z(

)zz(

u )0(

0

i2-

i2

0

def

2

1iS e

1i

1

2

1

e 1

i1

2

1

2

1

K2

1

ggg

gg

g

g

A

def defdef

i2 dz

d

A

Wave function in the presence of defects

gR .(z)g

Nearly normal incidence, centrosymmetric crystals

gg

KU

gg

KU

)(F

Kv cell

gg

The Howie-Whelan equations

gg.R

gg

e

1i

1i

dz

d (z)i20

0

0

ggg.R

gg

g

iS2 e 1

i1

idz

d

00

(z)i2-

When g.R = 0, 1, 2,… the image of the defect vanishes

The Howie-Whelan equations can be written in an equivalent form by making the transformation:

VQ 1def

dz

d

dz

d

dz

d 11def V

QVQ

g

Q i2e 00 1

1

Dg

1

i2dz

d

QQ

dz

)z(dg

g.R

VAV

Dgi2dz

d

VAV

Dgi2dz

d

g

g-g

V2

i

2

1V

2

ii2

dz

dV0

0

0

Equivalent form of the Howie-Whelan equations

gg

gg

g g.RV

dz

))z((dS(

2

iV

2

i

2

1i2

dz

dV

00

The term plays the same role as Sg, representing the

local curvature of the atomic planes.dz

))z((d g.R

Dislocations

There are, basically, two types of dislocations: screw and edge dislocations. Dislocations combining both characteristics are called mixed dislocations

Screw Edge

The strain field of the dislocations is calculated on the base of the linear elasticity theory

Elasticity

The fundamental equation of the linear elasticity yields

klijklij c Where are the stresses (force per unit area) acting on the ith face in the jth direction

ij

The coefficients are the elastic constants

ijklc

The strains are defined as kl

k

l

l

kkl x

u

x

u

2

1

The displacement field around dislocations

k

l

l

kkl x

u

x

u

2

1

The uk’s functions are the component of the displacement fields at the coordinates x1, x2 and x3. Thus the uk’s are the components of the vector R(r) at the point r(x1, x2, x3).

lkkl

At rest there can not be torque on the element, so that

provided there are not internal torques.

jiij

jilkijlkjiklijkl cccc

The elastic constants

klijklij c jiij lkkl

Contracted matrix notation as , where m and n are each indices corresponding to a pair of indices ij and kl, according to the following equivalency:

mnc

ij or kl 11 22 33 23 31 12 32 13 21

m or n 1 2 3 4 5 6 7 8 9

.................. ..............cc cc

cc cc

231246232344

112212111111

21

13

32

12

31

23

33

22

11

9991

81

71

61

51

41

333231

232221

1918 1716 14 1312 11

21

13

32

12

31

23

33

22

11

c ........................................ c........................................... c............................................ c........................................... c

........................................... c........................................... c

............................. c c c

.............................. c c cc cc cc c c c

klijklij c

There are 81 elastic constants, however because of the symmetry only 21 elastic constants are independent.

Reduced 6x6 representation

12

31

23

33

22

11

6656 46 362616

5655 45 352515

4645 44 342414

363534332313

262524232212

1615 14 1312 11

12

31

23

33

22

11

c cc c c c c cc c c c c cc c c c c c c c c c c c c c c c c cc c c c

ijij 2

Only three different elastic constants

12

31

23

33

22

11

44

44

44

111212

121112

1212 11

12

31

23

33

22

11

c 0 0 0 0 0

0 c 0 0 0 0

0 0 c 0 0 0

0 0 0 c c c

0 0 0 c c c

0 0 0 c c cFor cubic crystals

Crystal Number of elastic constants

Cubic 3

Hexagonal 6

Orthorhombic 9

Monoclinic 13

In isotropic crystals only two elastic constants are needed:

)cc(2

1c 121144

12c

Shear modulus

(Lamé constant)

)(2

11

33

11

22

The Poisson modulus

Displacement field of screw dislocation in an isotropic media

Since only a shear deformation along the Burgers vector direction (z direction) on the (x,z) plane has been introduced, we have:ux = uy = 0.

In an isotropic media the displacement uz increases uniformly with the angle , from uz = 0 for = 0 to uz = b for  = 2

2

b),r(u z

y

xtan

2)y,x( 1b

R

The extinction condition requires g.R = 0, 1, .. g.b = 0

r

x

y

Displacement field of edge dislocation in an isotropic media

Planar strain: uz = 0 and 0z

u i

The solution for the displacement field gives

)1(4

2cosrln

)1(2

21

)1(4

2sinzebbR

The extinction condition requires

g.R = 0, 1, .. g.b = 0

g.(b^ez) = 0 z // eg

Extinction in isotropic crystals

Possible Burgers vectors

[011] and [101]

g b g.b g.b^e

111

[110]

[011]

[101]

0

0

0

≠ 0

≠ 0

≠ 0

111

[110]

[011]

[101]

1

1

1

≠ 0

≠ 0

≠ 0

2

1

2

1

2

1

2

1

2

1

2

1

Straight dislocations in anisotropic elastic media

In mechanical equilibrium no net force can act on the volume element, giving the expression

0xx

uc

lj

k2

ijkl

The x3 axis is parallel to the dislocation line direction. The elastic constants are referred to this system. In an infinite media the displacements, the strain, and stresses are all independent on x3

The partial differential equation has a solution of the type

)(fAu kk 21 x px

Computer electron micrographs and defects identification. Head, Humble, Clarebrough, Morton, Forwood, North Holland, 1973.

COEFIC DE ABSORCION ANOMALA (ANO) 9.000000E-02DESVIACION CONDICION DE BRAGG (W) 0.000000E+00ESPESOR DE LA MUESTRA (THICK) 4.000000COMIENZO DE LA INTEGRACION (START) -5.000000E-01FINAL DE LA INTEGRACION (FINISH) 4.500000SEPARACION DE DISLOCACIONES (SEP) 0.000000E+00PARAMETROS DE RED (SIDE) 3.000000 3.000000 3.000000NUM DEL 1er. VECTOR DE BURGERS (LB) 0 0 1DEN DEL 1er. VECTOR DE BURGERS (LD) 1LINEA DE LA 1er. DISLOCACION (LU) 1 1 1VECTOR DE DIFRACCION (LG) 1 -1 0DIRECCION DEL HAZ (LBM) -56 -56 9900

C11 1.291000C22 1.291000C33 1.291000C44 8.240000E-01C55 8.240000E-01C66 8.240000E-01C12 1.097000C13 1.097000C15 0.000000E+00C23 1.097000C25 0.000000E+00C35 0.000000E+00C46 0.000000E+00

When g is perpendicular to u and b the image is symmetric with respect to u.

b = 100 g = 1-1 0 u = 1 1 1

011

b = 010 g = 1-1 0 u = 1 1 1

b = 001 g = 1-1 0 u = 1 1 1

Simulations

011b = 001 g = -1-1 0 u = 1 1 1

b = 010 g = -1-1 0 u = 1 1 1

b = 100 g = -1-1 0 u = 1 1 1

020b = 001 g = 020 u = 1 1 1

b = 010 g = 020 u = 1 1 1

b = 100 g = 020 u = 1 1 1

When the line e is perpendicular to an elastic symmetry plane or parallel to un even-fold rotation axisExtinction appears: -for edge component when g is parallel to e and perpendicular to b- for the screw component when g is perpendicular to e

Extinctions in anisotropic crystals

Courtesy of E. Zelaya and J. Pelegrina

Weak beam image of superlattice dislocation

000000 g 2g 3g

Dark field image

dEEEE 1221

VCH = 1 mm / min

N = 1

N = 125

Wire = 0.5 mm

Pseudoelastic cycling of NiTi shape memory

Courtesy H. Soul

Dislocation density

ab

R1

R2

Stacking faults

Definition of the translation vectors

Images from stacking faults

Start from equation uCC 1-

D

z)iq(i2 z)j(

e

Include the transformation CQCC )z()z( -1

u)(e )z 1-1-

D

z)q(i21- z)j(

CQCQ

g

Qi2e 0

0 1

0

1

e 0

0 1

C C

C C

e 0

0 e

C C

C C

e 0

0 1

)(

)(g

z(2)

z(1)

g i2(2)(2)0

(1)(1)0

z)iqi(2

z)iqi(2

(2)(1)

(2)0

(1)0

i2-

0

g

g

ggg r

r

)0(0

)t( 10)t( 1g

)t(0 )t(g

t1t

t2

t´R

)(t

)t(

e 0

0 1

C C

C C

e 0

0 e

C C

C C

e 0

0 1

(t)

)t(

1

10

i2(2)(2)0

(1)(1)0

t)iqi(2

t)iqi(2

(2)(1)

(2)0

(1)0

i2-

0

g2z

(2)

2z(1)

ggg

g

ggg

0

1

C C

C C

e 0

0 e

C C

C C

)t(

)t((2)(2)

0

(1)(1)0

t)iqi(2

t)iqi(2

(2)(1)

(2)0

(1)0

1

10

2z(2)

1z(1)

g

g

ggg

)i(i22 )2()1(g e tcos-tcos1e sin

2

1tsin cositcos(t)

0

)i(

i2)2()1(

g

e tsin i-tcos cos

tsin i-tcos cos1e sin2

1tsin sin i

(t)

g

t2

1tt 1 (distance from the fault to the specimen center)

1w

i1w

12

g

2

g

ggSw gR .(z)g

Transmitted and diffracted amplitude

wcot

(t)0

(t)g

is symmetric with respect to the foil center

is not symmetric with respect to the foil center

From D.B.Williams and C.B.Carter. Transmission Electron Microscopy. Plenun Press, 1996.

T

B

-The fringes at the top are the same in BF and DF images -The fringes at the bottom are complementary

Symmetry of the stacking faults images

Stacking faults in close-packed structures

Image of the stacking faults

A. Condó, F.Lovey

Phil Mag. 79 (1999) 511

Definition of the stacking fault displacement vectors

Interaction of a basal plane fault with a (128) plane fault

Experimental and simulated images in two-beam conditions

A. Condó, F.Lovey

Phil Mag. 79 (1999) 511

Stacking fault images in two-beam condition

Another example

Summary

The imaging of defects in two-beam condition remains as a powerful technique to study defects.

Combination with HRTEM, LACBED and STEM, whenever possible, would be convenient.