Spin Tensors as Infinitesimal Changes of Rotations edit page

Spin tensors are skew symmetric tensors that can be used to describe small rotational changes. Let us consider an arbitrary reference rotation

rot_ref = rotation.byEuler(10*degree,20*degree,30*degree);

and perturb it by a rotation about the axis (123) and angle delta=0.01 degree. Since multiplication of rotations is not commutative we have to distinguish between left and right perturbations

delta = 0.01*degree;
rot_123 = rotation.byAxisAngle(vector3d(1,2,3),delta);
rot_right = rot_123 * rot_ref;
rot_left = rot_ref * rot_123;

We may now ask for the first order Taylor coefficients of the perturbation as delta goes to zero which we find by the formula

\[ T = \lim_{\delta \to 0} \frac{\tilde R - R}{\delta} \]

T_right = (rot_right.matrix - rot_ref.matrix) ./ delta
T_left = (rot_left.matrix - rot_ref.matrix) ./ delta
T_right =
   -0.4002   -0.4146    0.7724
    0.5727   -0.5873   -0.2035
   -0.2484    0.5297   -0.1218
T_left =
   -0.5399   -0.6025    0.5816
    0.7530   -0.5816    0.1368
   -0.2648    0.1140    0.0122

Both matrices T_right and T_left are elements of the tangential space attached to the reference rotation rot_ref. Those matrices are characterized by the fact that they becomes skew symmetric matrices when multiplied from the left or from the right with the inverse of the reference rotation

S_right_L =  matrix(inv(rot_ref)) * T_right
S_right_R = T_right * matrix(inv(rot_ref))

S_left_L =  matrix(inv(rot_ref)) * T_left
S_left_R = T_left * matrix(inv(rot_ref))
S_right_L =
   -0.0000   -0.5892    0.4501
    0.5893   -0.0001   -0.6709
   -0.4501    0.6710   -0.0001
S_right_R =
   -0.0001   -0.8018    0.5345
    0.8018   -0.0001   -0.2672
   -0.5345    0.2673   -0.0000
S_left_L =
   -0.0001   -0.8018    0.5345
    0.8018   -0.0001   -0.2672
   -0.5345    0.2673   -0.0000
S_left_R =
   -0.0001   -0.9575    0.2758
    0.9575   -0.0001    0.0850
   -0.2758   -0.0850   -0.0000

A skew symmetric 3x3 matrix S is essentially determined by its entries \(S_{21}\), \(S_{31}\) and \(S_{32}\). Writing these values as a vector \((S_32,-S_{31},S_{21})\) we obtain for the matrices S_right_R and S_left_L exactly the rotational axis of our perturbation

vector3d(spinTensor(S_right_R)) * sqrt(14)

vector3d(spinTensor(S_left_L))  * sqrt(14)
ans = vector3d (y↑→x)
  x y z
  1 2 3
 
ans = vector3d (y↑→x)
  x y z
  1 2 3

For the other two matrices those vectors are related to the rotational axis by the reference rotation rot_ref 

rot_ref * vector3d(spinTensor(S_right_L)) * sqrt(14)

inv(rot_ref) * vector3d(spinTensor(S_left_R)) * sqrt(14)
ans = vector3d (y↑→x)
  x y z
  1 2 3
 
ans = vector3d (y↑→x)
  x y z
  1 2 3

The Functions Exp and Log

The above definition of the spin tensor works well only if the perturbation has small rotational angle. For large perturbations the matrix logarithm log provides the correct way to translate rotational changes into skew symmetric matrices

% define a large pertubation with rotational angle 1 radiant
delta = 1;
rot_123 = rotation.byAxisAngle(vector3d(1,2,3),1);

S = log(rot_ref * rot_123,rot_ref, SO3TangentSpace.rightSpinTensor); S  * sqrt(14)


S = log(rot_123 * rot_ref,rot_ref, SO3TangentSpace.leftSpinTensor); S  * sqrt(14)
ans = spinTensor (y↑→x)
  rank: 2 (3 x 3)
 
  0 -3  2
  3  0 -1
 -2  1  0
 
ans = spinTensor (y↑→x)
  rank: 2 (3 x 3)
 
  0 -3  2
  3  0 -1
 -2  1  0

Again the entries \(S_{21}\), \(S_{31}\) and \(S_{32}\) exactly coincide with the rotational axis multiplied with the rotational angle.

More directly this disorientation vector may be computed from two rotations using the options SO3TangentSpace.rightVector and SO3TangentSpace.leftVector 

v = log(rot_ref * rot_123,rot_ref,SO3TangentSpace.rightVector); v * sqrt(14)

v = log(rot_123 * rot_ref,rot_ref,SO3TangentSpace.leftVector); v * sqrt(14)
ans = vector3d (y↑→x)
  x y z
  1 2 3
 
ans = vector3d (y↑→x)
  x y z
  1 2 3

The other way round

Given a skew symmetric matrix S or a disorientation vector v we may use the command exp to apply this rotational perturbation to a reference rotation rot_ref 

% the truth
rot_ref * rot_123

% using a disorientation vector
exp(v,rot_ref,SO3TangentSpace.rightVector)

% using a spin tensor
exp(S,rot_ref,SO3TangentSpace.rightSpinTensor)
ans = rotation
 
  Bunge Euler angles in degree
     phi1     Phi    phi2
  94.0497 29.4341 358.508
 
 
ans = rotation
 
  Bunge Euler angles in degree
     phi1     Phi    phi2
  94.0497 29.4341 358.508
 
 
ans = rotation
 
  Bunge Euler angles in degree
     phi1     Phi    phi2
  94.0497 29.4341 358.508
% the other truth
rot_123 * rot_ref

% using a disorientation vector
exp(v,rot_ref,SO3TangentSpace.leftVector)

% using a spin tensor
exp(S,rot_ref,SO3TangentSpace.leftSpinTensor)
ans = rotation
 
  Bunge Euler angles in degree
     phi1     Phi    phi2
  74.5535 51.5051 9.61062
 
 
ans = rotation
 
  Bunge Euler angles in degree
     phi1     Phi    phi2
  74.5535 51.5051 9.61062
 
 
ans = rotation
 
  Bunge Euler angles in degree
     phi1     Phi    phi2
  74.5535 51.5051 9.61062

Disorientations under the presence of crystal symmetry

Under the presence of crystal symmetry the order whether a rotational perturbation is applied from the left or from the right. Lets perform the above calculations step by step in the presence of trigonal crystal symmetry

cs = crystalSymmetry('321');

% consider an arbitrary rotation
ori_ref = orientation.byEuler(10*degree,20*degree,30*degree,cs);

% next we disturb rot_ref by a rotation about the axis (123)
mori_123 = orientation.byAxisAngle(Miller(1,2,-3,3,cs),1);

% first we multiply from the right
ori = ori_ref * mori_123
ori = orientation (321 → y↑→x)
 
  Bunge Euler angles in degree
     phi1     Phi    phi2
  79.3156 43.3785  9.9013

Computing the right tangential vector gives us the disorientation vector in crystal coordinates

v = log(ori,ori_ref,SO3TangentSpace.rightVector); round(v)

exp(v,ori_ref,SO3TangentSpace.rightVector)
ans = Miller (321)
  h  k  i  l
  1  2 -3  3
 
ans = orientation (321 → y↑→x)
 
  Bunge Euler angles in degree
     phi1     Phi    phi2
  79.3156 43.3785  9.9013

computing the left tangential vector gives us the disorientation vector in specimen coordinates

v = log(ori,ori_ref,SO3TangentSpace.leftVector)
S = log(ori,ori_ref,SO3TangentSpace.leftSpinTensor)
exp(v,ori_ref,SO3TangentSpace.leftVector)
v = vector3d (y↑→x)
         x        y        z
  0.161601 0.428476 0.888985
 
S = vector3d (y↑→x)
         x        y        z
  0.161601 0.428476 0.888985
 
ans = orientation (321 → y↑→x)
 
  Bunge Euler angles in degree
     phi1     Phi    phi2
  79.3156 43.3785  9.9013