study of defects in two beam conditions ii
DESCRIPTION
STUDY OF DEFECTS IN TWO BEAM CONDITIONS II. Puntiagudo. Osorno. Francisco Lovey Centro Atómico Bariloche Instituto Balseiro. PASI on Transmission Electron Microscopy Santiago de Chile, July 2006. Images of crystalline defects. - PowerPoint PPT PresentationTRANSCRIPT
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
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.