MTEX Toolbox

Misorientations

Misorientation describes the relative orientation of two grains with respect to each other. Important concepts are twinnings and CSL (coincidence site lattice),

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The misorientation angle
Misorientations
Coincident lattice planes
Twinning misorientations
Highlight twinning boundaries
Phase transitions

The misorientation angle

First we import some EBSD data set, compute grains and plot them according to their mean orientation. Next we highlight grain 70 and grain 80 

mtexdata twins
% use only proper symmetry operations
ebsd('M').CS = ebsd('M').CS.properGroup;
grains = calcGrains(ebsd('indexed'),'threshold',5*degree)
CS = grains.CS; % extract crystal symmetry

plot(grains,grains.meanOrientation,'micronbar','off')
hold on
plot(grains([74,85]).boundary,'edgecolor','w','linewidth',2)

text(grains([74,85]),{'1','2'})
 
grains = grain2d  
 
 Phase  Grains  Pixels    Mineral  Symmetry  Crystal reference frame
     1     126   22833  Magnesium       622       X||a*, Y||b, Z||c*
 
 boundary segments: 3892
 triple points: 122
 
 Properties: GOS, meanRotation
 
  I'm going to colorize the orientation data with the 
  standard MTEX colorkey. To view the colorkey do:
 
  colorKey = ipfColorKey(ori_variable_name)
  plot(colorKey)

After extracting the mean orientation of grain 70 and 80

ori1 = grains(74).meanOrientation
ori2 = grains(85).meanOrientation
 
ori1 = orientation  
  size: 1 x 1
  crystal symmetry : Magnesium (622, X||a*, Y||b, Z||c*)
  specimen symmetry: 1
 
  Bunge Euler angles in degree
     phi1     Phi    phi2    Inv.
  19.4774 78.0484 220.043       0
 
 
ori2 = orientation  
  size: 1 x 1
  crystal symmetry : Magnesium (622, X||a*, Y||b, Z||c*)
  specimen symmetry: 1
 
  Bunge Euler angles in degree
     phi1     Phi    phi2    Inv.
  71.4509 167.781 200.741       0

we may compute the misorientation angle between both orientations by

angle(ori1, ori2) ./ degree
ans =
   85.7093

Note that the misorientation angle is computed by default modulo crystal symmetry, i.e., the angle is always the smallest angles between all possible pairs of symmetrically equivalent orientations. In our example this means that symmetrisation of one orientation has no impact on the angle 

angle(ori1, ori2.symmetrise) ./ degree
ans =
   85.7093
   85.7093
   85.7093
   85.7093
   85.7093
   85.7093
   85.7093
   85.7093
   85.7093
   85.7093
   85.7093
   85.7093

The misorientation angle neglecting crystal symmetry can be computed by

angle(ori1, ori2.symmetrise,'noSymmetry')./ degree
ans =
  107.9464
  100.9175
   94.2925
  136.6180
  107.7066
  179.5952
  140.0598
  137.3717
  179.8061
  101.4414
  140.4168
   85.7093

We see that the smallest angle indeed coincides with the angle computed before.

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Misorientations

Remember that both orientations ori1 and ori2 map crystal coordinates onto specimen coordinates. Hence, the product of an inverse orientation with another orientation transfers crystal coordinates from one crystal reference frame into crystal coordinates with respect to another crystal reference frame. This transformation is called misorientation 

mori = inv(ori1) * ori2
 
mori = misorientation  
  size: 1 x 1
  crystal symmetry : Magnesium (622, X||a*, Y||b, Z||c*)
  crystal symmetry : Magnesium (622, X||a*, Y||b, Z||c*)
 
  Bunge Euler angles in degree
     phi1     Phi    phi2    Inv.
  149.581 94.2922 150.134       0

In the present case the misorientation describes the coordinate transform from the reference frame of grain 80 into the reference frame of crystal 70. Take as an example the plane {11-20} with respect to the grain 80. Then the plane in grain 70 which alignes parallel to this plane can be computed by 

round(mori * Miller(1,1,-2,0,CS))
 
ans = Miller  
 size: 1 x 1
 mineral: Magnesium (622, X||a*, Y||b, Z||c*)
  h  2
  k -1
  i -1
  l  0

Conversely, the inverse of mori is the coordinate transform from crystal 70 to grain 80.

round(inv(mori) * Miller(2,-1,-1,0,CS))
 
ans = Miller  
 size: 1 x 1
 mineral: Magnesium (622, X||a*, Y||b, Z||c*)
  h  1
  k  1
  i -2
  l  0

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Coincident lattice planes

The coincidence between major lattice planes may suggest that the misorientation is a twinning misorientation. Lets analyse whether there are some more alignments between major lattice planes. 

%m = Miller({1,-1,0,0},{1,1,-2,0},{1,-1,0,1},{0,0,0,1},CS);
m = Miller({1,-1,0,0},{1,1,-2,0},{-1,0,1,1},{0,0,0,1},CS);

% cycle through all major lattice planes
close all
for im = 1:length(m)
  % plot the lattice planes of grains 80 with respect to the
  % reference frame of grain 70
  plot(mori * m(im).symmetrise,'MarkerSize',10,...
    'DisplayName',char(m(im)),'figSize','large','noLabel','upper')
  hold all
end
hold off

% mark the corresponding lattice planes in the twin
mm = round(unique(mori*m.symmetrise,'noSymmetry'),'maxHKL',6);
annotate(mm,'labeled','MarkerSize',5,'figSize','large','textAboveMarker')

% show legend
legend({},'location','SouthEast','FontSize',13);

we observe an almost perfect match for the lattice planes {11-20} to {-2110} and {1-101} to {-1101} and good coincidences for the lattice plane {1-100} to {0001} and {0001} to {0-661}. Lets compute the angles explicitly 

angle(mori * Miller(1,1,-2,0,CS),Miller(2,-1,-1,0,CS)) / degree
angle(mori * Miller(1,0,-1,-1,CS),Miller(1,-1,0,1,CS)) / degree
angle(mori * Miller(0,0,0,1,CS) ,Miller(1,0,-1,0,CS),'noSymmetry') / degree
angle(mori * Miller(1,1,-2,2,CS),Miller(1,0,-1,0,CS)) / degree
angle(mori * Miller(1,0,-1,0,CS),Miller(1,1,-2,2,CS)) / degree
ans =
    0.4489
ans =
    0.2193
ans =
   59.6757
ans =
    2.6092
ans =
    2.5387

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Twinning misorientations

Lets define a misorientation that makes a perfect fit between the {11-20} lattice planes and between the {10-11} lattice planes

mori = orientation.map(Miller(1,1,-2,0,CS),Miller(2,-1,-1,0,CS),...
  Miller(-1,0,1,1,CS),Miller(-1,1,0,1,CS))


% the rotational axis
round(mori.axis)

% the rotational angle
mori.angle / degree
 
mori = misorientation  
  size: 1 x 1
  crystal symmetry : Magnesium (622, X||a*, Y||b, Z||c*)
  crystal symmetry : Magnesium (622, X||a*, Y||b, Z||c*)
 
  Bunge Euler angles in degree
  phi1  Phi phi2 Inv.
   300    0    0    0
 
 
ans = Miller  
 size: 1 x 1
 mineral: Magnesium (622, X||a*, Y||b, Z||c*)
  h  1
  k  0
  i -1
  l  0
ans =
     0

and plot the same figure as before with the exact twinning misorientation.

% cycle through all major lattice planes
close all
for im = 1:length(m)
  % plot the lattice planes of grains 80 with respect to the
  % reference frame of grain 70
  plot(mori * m(im).symmetrise,'MarkerSize',10,...
    'DisplayName',char(m(im)),'figSize','large','noLabel','upper')
  hold all
end
hold off

% mark the corresponding lattice planes in the twin
mm = round(unique(mori*m.symmetrise,'noSymmetry'),'maxHKL',6);
annotate(mm,'labeled','MarkerSize',5,'figSize','large')

% show legend
legend({},'location','NorthWest','FontSize',13);

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Highlight twinning boundaries

It turns out that in the previous EBSD map many grain boundaries have a misorientation close to the twinning misorientation we just defined. Lets Lets highlight those twinning boundaries 

% consider only Magnesium to Magnesium grain boundaries
gB = grains.boundary('Mag','Mag');

% check for small deviation from the twinning misorientation
isTwinning = angle(gB.misorientation,mori) < 5*degree;

% plot the grains and highlight the twinning boundaries
plot(grains,grains.meanOrientation,'micronbar','off')
hold on
plot(gB(isTwinning),'edgecolor','w','linewidth',2)
hold off
  I'm going to colorize the orientation data with the 
  standard MTEX colorkey. To view the colorkey do:
 
  colorKey = ipfColorKey(ori_variable_name)
  plot(colorKey)

From this picture we see that large fraction of grain boudaries are twinning boundaries. To make this observation more evident we may plot the boundary misorientation angle distribution function. This is simply the angle distribution of all boundary misorientations and can be displayed with 

close all
plotAngleDistribution(gB.misorientation)

From this we observe that we have about 50 percent twinning boundaries. Analogously we may also plot the axis distribution

plotAxisDistribution(gB.misorientation,'contour')

which emphasises a strong portion of rotations about the (-12-10) axis.

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Phase transitions

Misorientations may not only be defined between crystal frames of the same phase. Lets consider the phases Magnetite and Hematite.

CS_Mag = loadCIF('Magnetite')
CS_Hem = loadCIF('Hematite')
 
CS_Mag = crystalSymmetry  
 
  mineral : Magnetite    
  symmetry: m-3m         
  a, b, c : 8.4, 8.4, 8.4
 
 
CS_Hem = crystalSymmetry  
 
  mineral        : Hematite          
  symmetry       : -3m1              
  a, b, c        : 5, 5, 14          
  reference frame: X||a*, Y||b, Z||c*

The phase transition from Magnetite to Hematite is described in literature by {111}_m parallel {0001}_h and {-101}_m parallel {10-10}_h The corresponding misorientation is defined in MTEX by 

Mag2Hem = orientation.map(...
  Miller(1,1,1,CS_Mag),Miller(0,0,0,1,CS_Hem),...
  Miller(-1,0,1,CS_Mag),Miller(1,0,-1,0,CS_Hem))
 
Mag2Hem = misorientation  
  size: 1 x 1
  crystal symmetry : Magnetite (m-3m)
  crystal symmetry : Hematite (-3m1, X||a*, Y||b, Z||c*)
 
  Bunge Euler angles in degree
  phi1     Phi    phi2    Inv.
   120 54.7356      45       0

Assume a Magnetite grain with orientation

ori_Mag = orientation.byEuler(0,0,0,CS_Mag)
 
ori_Mag = orientation  
  size: 1 x 1
  crystal symmetry : Magnetite (m-3m)
  specimen symmetry: 1
 
  Bunge Euler angles in degree
  phi1  Phi phi2 Inv.
     0    0    0    0

Then we can compute all variants of the phase transition by

symmetrise(ori_Mag) * inv(Mag2Hem)
 
ans = orientation  
  size: 48 x 1
  crystal symmetry : Hematite (-3m1, X||a*, Y||b, Z||c*)
  specimen symmetry: 1

and the corresponding pole figures by

plotPDF(symmetrise(ori_Mag) * inv(Mag2Hem),...
  Miller({1,0,-1,0},{1,1,-2,0},{0,0,0,1},CS_Hem))

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