Orthogonal tensor

Linear transformation of a vector by a tensor T . ${\displaystyle {\vec {v}}}$

Rotation of a vector about the rotational axis with angle by an orthogonal tensor Q . ${\displaystyle {\vec {v}}}$ ${\displaystyle {\vec {n}}}$ ${\displaystyle \alpha }$

Orthogonal tensors are unit-free second- order tensors that perform a rotation or rotation mirroring in Euclidean vector space . In continuum mechanics only rotations are considered, because rotational reflections do not occur in the physical laws of the macroscopic world determined by gravity .

Second level tensors are used here as linear mappings from geometric vectors to geometric vectors, which are generally rotated and stretched in the process, see figure above right. In the case of an orthogonal tensor that represents a rotation or a rotation mirroring, the stretching is omitted, so that the amount of the vector is not changed during the transformation, see the figure below on the right. Orthogonal tensors are usually denoted by the symbol Q or R , where R mostly stands for the rotation tensor in the polar decomposition of the deformation gradient .

With regard to the standard basis , orthogonal tensors can be written like orthogonal matrices and also have analogous properties. In contrast to matrices, however, the coefficients of a tensor reference a basic system of the underlying vector space, so that the coefficients of the tensor change in a characteristic way when the basic system changes. Every tensor has invariants that remain unchanged when the basis system changes. In the case of an orthogonal tensor, these invariants provide information about the angle of rotation, the axis of rotation and whether the tensor represents a rotation or a mirrored rotation.

Orthogonal tensors appear in the Euclidean transformation , with which the relationship between arbitrarily moving reference systems and the physical quantities present in them is described. In material theory , orthogonal tensors help to set up reference system-invariant material equations. In addition, the directional dependence of a material ( transverse isotropy , orthotropy ) is described with orthogonal tensors.

definition

Orthogonal tensors are second order tensors Q , for which the following applies:

${\displaystyle \mathbf {Q} ^{-1}=\mathbf {Q} ^{\top }}$  or   ${\displaystyle \mathbf {Q^{\top }\cdot Q} =\mathbf {Q\cdot Q} ^{\top }=\mathbf {1} }$

The superscript −1 denotes the inverse , (·) ^⊤ the transposed tensor and 1 the unit tensor . Because of

${\displaystyle 1=\operatorname {det} (\mathbf {1} )=\operatorname {det} (\mathbf {Q^{\top }\cdot Q} )=\operatorname {det} (\mathbf {Q} ^{\top })\operatorname {det} (\mathbf {Q} )=\operatorname {det} {(\mathbf {Q} )}^{2}}$

is

${\displaystyle \operatorname {det} (\mathbf {Q} )=\pm 1}$

An orthogonal tensor that represents a pure rotation is actually called orthogonal and has the determinant +1. If det ( Q ) = -1 the tensor performs a rotational mirroring. Because reflections are not considered in mechanics, det ( Q ) = +1 there.

Rigid body movements

Velocity field (black) of a rigid body (gray) along its path (light blue) is made up of the velocity of the center of gravity (blue) and the speed of rotation (red)

Every rigid body movement can be broken down into a translation and a rotation. Any stationary or moving point and the center of gravity of the body are suitable as a center of rotation, see figure on the right. Let be the time-fixed difference vector between a particle of the rigid body and its center of gravity at a point in time . The translation of the body can then be represented with its movement of the center of gravity (with ) and its rotation with an orthogonal tensor that is dependent on time but not on location (with ). Translation and rotation taken together define the motion function of the particle : ${\displaystyle {\vec {r}}({\vec {X}})={\vec {X}}-{\vec {S}}}$ ${\displaystyle {\vec {X}}}$ ${\displaystyle {\vec {S}}}$ ${\displaystyle {t}_{0}}$ ${\displaystyle {\vec {s}}(t)}$ ${\displaystyle {\vec {s}}(t_{0})={\vec {S}}}$ ${\displaystyle \mathbf {Q} (t)}$ ${\displaystyle \mathbf {Q} ({t}_{0})=\mathbf {1} }$ ${\displaystyle {\vec {\chi }}({\vec {X}},t)}$ ${\displaystyle {\vec {X}}}$

${\displaystyle {\begin{aligned}{\vec {\chi }}({\vec {X}},t)=&{\vec {s}}(t)+\mathbf {Q} (t)\cdot ({\vec {X}}-{\vec {S}})\\\rightarrow \;{\vec {X}}-{\vec {S}}=&\mathbf {Q} ^{\top }(t)\cdot ({\vec {\chi }}({\vec {X}},t)-{\vec {s}}(t))\end{aligned}}}$

The speed of the particle is then

${\displaystyle {\begin{aligned}{\dot {\vec {\chi }}}({\vec {X}},t)=&{\dot {\vec {s}}}(t)+{\dot {\mathbf {Q} }}(t)\cdot ({\vec {X}}-{\vec {S}})\\=&{\dot {\vec {s}}}(t)+{\dot {\mathbf {Q} }}(t)\cdot \mathbf {Q} ^{\top }(t)\cdot ({\vec {\chi }}({\vec {X}},t)-{\vec {s}}(t))\\\rightarrow {\vec {v}}({\vec {x}},t)=&{\dot {\vec {s}}}(t)+{\boldsymbol {\Omega }}(t)\cdot ({\vec {x}}-{\vec {s}}(t))\end{aligned}}}$

The vector is the position of the particle at time t and is its velocity at time t . The transition from the upper to the lower equation changes from the Lagrangian to the Eulerian representation of motion. The tensor is skew symmetric : ${\displaystyle {\vec {x}}={\vec {\chi }}({\vec {X}},t)}$ ${\displaystyle {\vec {v}}({\vec {x}},t)={\dot {\vec {\chi }}}({\vec {X}},t)}$ ${\displaystyle {\boldsymbol {\Omega }}:={\dot {\mathbf {Q} }}\cdot \mathbf {Q} ^{\top }}$

${\displaystyle {\boldsymbol {\Omega }}+{\boldsymbol {\Omega }}^{\top }={\dot {\mathbf {Q} }}\cdot \mathbf {Q} ^{\top }+\mathbf {Q} \cdot {\dot {\mathbf {Q} }}^{\top }=(\mathbf {Q\cdot Q} ^{\top }{\dot {)\,}}={\dot {\mathbf {1} }}=\mathbf {0} \quad \rightarrow \quad {\boldsymbol {\Omega }}^{\top }=-{\boldsymbol {\Omega }}}$

and therefore has a dual vector with the property: ${\displaystyle {\vec {\omega }}}$

${\displaystyle {\boldsymbol {\Omega }}\cdot {\vec {v}}={\vec {\omega }}\times {\vec {v}}}$  for all   ${\displaystyle {\vec {v}}}$

Inserting the dual vector into the velocity field leads to Euler's velocity equation

${\displaystyle {\vec {v}}({\vec {x}},t)={\dot {\vec {s}}}(t)+{\vec {\omega }}(t)\times ({\vec {x}}-{\vec {s}}(t))}$

which does not contain a visible tensor. Only in the cross product, which corresponds to a tensor transformation, there is still an indication of a tensor.

Transformation properties

Vector transformation

An orthogonal tensor rotates vectors, because the scalar product of any two vectors remains under the linear mapping with Q :

${\displaystyle (\mathbf {Q} \cdot {\vec {u}})\cdot (\mathbf {Q} \cdot {\vec {v}})={\vec {u}}\cdot \mathbf {Q^{\top }\cdot Q} \cdot {\vec {v}}={\vec {u}}\cdot {\vec {v}}}$

In particular, with  : ${\displaystyle {\vec {v}}={\vec {u}}}$

${\displaystyle {|\mathbf {Q} \cdot {\vec {u}}|}^{2}=(\mathbf {Q} \cdot {\vec {u}})\cdot (\mathbf {Q} \cdot {\vec {u}})={\vec {u}}\cdot {\vec {u}}={|{\vec {u}}|}^{2}}$

therefore an orthogonal tensor Q does not change the Frobenius norm of a vector. Because the axis of rotation is mapped to itself in a pure rotation, the axis of rotation of the rotation is an eigenvector of an actually orthogonal tensor Q with eigenvalue one: ${\displaystyle {\vec {n}}}$

${\displaystyle \mathbf {Q} \cdot {\vec {n}}={\vec {n}}}$

If Q is an improperly orthogonal tensor, then is

${\displaystyle \mathbf {Q} \cdot {\vec {n}}=-{\vec {n}}}$

Spat product and cross product

Spat, which is spanned by three vectors

The space product of three vectors is the volume of the space spanned by the vectors, see picture. If the three vectors are labeled as in the picture and transformed with an orthogonal tensor, the spatial product is calculated as: ${\displaystyle {\vec {a}},\,{\vec {b}},\,{\vec {c}}}$

${\displaystyle {\begin{aligned}(\mathbf {Q} \cdot {\vec {a}})\cdot [(\mathbf {Q} \cdot {\vec {b}})\times (\mathbf {Q} \cdot {\vec {c}})]=&\operatorname {det} {\begin{pmatrix}\mathbf {Q} \cdot {\vec {a}}&\mathbf {Q} \cdot {\vec {b}}&\mathbf {Q} \cdot {\vec {c}}\end{pmatrix}}\\=&\operatorname {det} \left[\mathbf {Q} \cdot {\begin{pmatrix}{\vec {a}}&{\vec {b}}&{\vec {c}}\end{pmatrix}}\right]\\=&\operatorname {det} (\mathbf {Q} )\operatorname {det} {\begin{pmatrix}{\vec {a}}&{\vec {b}}&{\vec {c}}\end{pmatrix}}\\=&\operatorname {det} (\mathbf {Q} )\,{\vec {a}}\cdot ({\vec {b}}\times {\vec {c}})\end{aligned}}}$

If the tensor is actually orthogonal, then the late product is not changed by it, otherwise the late product reverses its sign. Next follows:

${\displaystyle {\begin{array}{l}(\mathbf {Q} \cdot {\vec {a}})\cdot [(\mathbf {Q} \cdot {\vec {b}})\times (\mathbf {Q} \cdot {\vec {c}})]={\vec {a}}\cdot \mathbf {Q} ^{\top }\cdot [(\mathbf {Q} \cdot {\vec {b}})\times (\mathbf {Q} \cdot {\vec {c}})]=\operatorname {det} (\mathbf {Q} )\,{\vec {a}}\cdot ({\vec {b}}\times {\vec {c}})\\\rightarrow {\vec {a}}\cdot \left\{\mathbf {Q} ^{\top }\cdot [(\mathbf {Q} \cdot {\vec {b}})\times (\mathbf {Q} \cdot {\vec {c}})]-\operatorname {det} (\mathbf {Q} )({\vec {b}}\times {\vec {c}})\right\}=0\end{array}}}$

This applies to every vector , which is why the vector in the curly brackets disappears and opens ${\displaystyle {\vec {a}}}$

${\displaystyle (\mathbf {Q} \cdot {\vec {b}})\times (\mathbf {Q} \cdot {\vec {c}})=\operatorname {det} (\mathbf {Q} )\mathbf {Q} \cdot ({\vec {b}}\times {\vec {c}})}$

can be closed. Therefore, an actually orthogonal tensor can be extracted from the cross product, while a sign change still takes place with an improperly orthogonal tensor.

The volume element is calculated with the late product and the surface element is calculated with the cross product . In a rotary reflection both elements change sign, which is why they only transformation with a really orthogonal tensor Q invariant to Euclidean transformation are.

Tensor transformation

Let T be an arbitrary second order tensor that has an eigenvalue and an associated eigenvector , i.e. ${\displaystyle \lambda }$ ${\displaystyle {\vec {v}}}$

${\displaystyle \mathbf {T} \cdot {\vec {v}}=\lambda {\vec {v}}}$

holds and Q is an orthogonal tensor. Then

${\displaystyle \mathbf {Q\cdot T} \cdot {\vec {v}}=(\mathbf {Q\cdot T\cdot Q} ^{\top })\cdot (\mathbf {Q} \cdot {\vec {v}})=\lambda (\mathbf {Q} \cdot {\vec {v}})}$

So the tensor S  : =  Q · T · Q ^{⊤ has the} same eigenvalues ​​as T but the eigenvectors rotated with Q. It follows immediately that the main invariants and magnitudes of S and T agree.

Calculation of orthogonal tensors

When calculating orthogonal tensors, three tasks can arise:

• How is the corresponding orthogonal tensor constructed from the axis of rotation and the angle of rotation?
• Which orthogonal tensor transforms two given, mutually twisted vector space bases into one another?
• What is the axis of rotation and the angle of rotation of a given orthogonal tensor?

These questions are answered in the following sections.

Axis of rotation and angle given

Let be a unit vector (of length one) and an angle. Then is the tensor ${\displaystyle {\hat {n}}}$ ${\displaystyle \alpha }$

${\displaystyle {\begin{array}{lcl}\mathbf {Q} &=&+\mathbf {1} +\sin(\alpha ){\hat {n}}\times \mathbf {1} +(\cos(\alpha )-1)(\mathbf {1} -{\hat {n}}\otimes {\hat {n}})\\&=&+{\hat {n}}\otimes {\hat {n}}+\cos(\alpha )(\mathbf {1} -{\hat {n}}\otimes {\hat {n}})+\sin(\alpha ){\hat {n}}\times \mathbf {1} \end{array}}}$

actually orthogonal and rotates around the axis with an angle of rotation . The cross product of with the unit tensor gives the skew-symmetric axial tensor of  : ${\displaystyle {\hat {n}}}$ ${\displaystyle \alpha }$ ${\displaystyle {\hat {n}}}$ ${\displaystyle {\hat {n}}}$

${\displaystyle {\hat {n}}\times \mathbf {1} ={\hat {n}}\times \left(\sum _{i=1}^{3}{\hat {e}}_{i}\otimes {\hat {e}}_{i}\right):=\sum _{i=1}^{3}({\hat {n}}\times {\hat {e}}_{i})\otimes {\hat {e}}_{i}={\begin{pmatrix}0&-n_{3}&n_{2}\\n_{3}&0&-n_{1}\\-n_{2}&n_{1}&0\end{pmatrix}}}$

when the components of with respect to the standard base are ê _1,2,3 . ${\displaystyle n_{1,2,3}}$ ${\displaystyle {\hat {n}}}$

In the case of a rotating mirroring it would be

${\displaystyle {\begin{array}{lcl}\mathbf {Q} &=&{\color {red}-}\mathbf {1} +\sin(\alpha ){\hat {n}}\times \mathbf {1} +(\cos(\alpha ){\color {red}+}1)(\mathbf {1} -{\hat {n}}\otimes {\hat {n}})\\&=&{\color {red}-}{\hat {n}}\otimes {\hat {n}}+\cos(\alpha )(\mathbf {1} -{\hat {n}}\otimes {\hat {n}})+\sin(\alpha ){\hat {n}}\times \mathbf {1} \end{array}}}$

In any case, the tensor Q has the trace and the skew-symmetric part

${\displaystyle {\begin{array}{rcl}\operatorname {Sp} (\mathbf {Q} )&=&\operatorname {Sp} [\operatorname {det} (\mathbf {Q} ){\hat {n}}\otimes {\hat {n}}+\cos(\alpha )(\mathbf {1} -{\hat {n}}\otimes {\hat {n}})+\sin(\alpha ){\hat {n}}\times \mathbf {1} ]\\&=&\operatorname {det} (\mathbf {Q} )+2\cos(\alpha )\\{\frac {1}{2}}(\mathbf {Q} -\mathbf {Q} ^{\top })&=&\sin(\alpha ){\hat {n}}\times \mathbf {1} =\sin(\alpha ){\begin{pmatrix}0&-n_{3}&n_{2}\\n_{3}&0&-n_{1}\\-n_{2}&n_{1}&0\end{pmatrix}}\end{array}}}$

The formula for Q given at the beginning can also be written with a rotation vector : ${\displaystyle {\vec {\alpha }}:=\alpha {\hat {n}}}$

${\displaystyle \mathbf {Q} =\mathbf {1} +{\frac {\sin(\alpha )}{\alpha }}{\vec {\alpha }}\times \mathbf {1} +{\frac {1-\cos(\alpha )}{{\alpha }^{2}}}({\vec {\alpha }}\otimes {\vec {\alpha }}-({\vec {\alpha }}\cdot {\vec {\alpha }})\mathbf {1} )=\exp({\vec {\alpha }}\times \mathbf {1} )}$

The exponential of the skew-symmetric matrix is defined and used for rotary matrices . ${\displaystyle {\vec {\alpha }}\times \mathbf {1} }$

Rotation vectors of other length can also be used:

${\displaystyle {\begin{array}{lcl}{\vec {\alpha }}=\tan \left({\dfrac {\alpha }{2}}\right)\;{\hat {n}}&\rightarrow &\mathbf {Q} =\mathbf {1} +{\dfrac {2}{1+{\vec {\alpha }}\cdot {\vec {\alpha }}}}({\vec {\alpha }}\times \mathbf {1} +{\vec {\alpha }}\otimes {\vec {\alpha }}-({\vec {\alpha }}\cdot {\vec {\alpha }})\mathbf {1} )\\[2ex]{\vec {\alpha }}=\sin(\alpha )\;{\hat {n}}&\rightarrow &\mathbf {Q} =\mathbf {1} +{\vec {\alpha }}\times \mathbf {1} +{\dfrac {1}{1+\cos(\alpha )}}({\vec {\alpha }}\otimes {\vec {\alpha }}-({\vec {\alpha }}\cdot {\vec {\alpha }})\mathbf {1} )\\[2ex]{\vec {\alpha }}=\sin \left({\dfrac {\alpha }{2}}\right)\;{\hat {n}}&\rightarrow &\mathbf {Q} =\mathbf {1} +2\cos \left({\dfrac {\alpha }{2}}\right){\vec {\alpha }}\times \mathbf {1} +2({\vec {\alpha }}\otimes {\vec {\alpha }}-({\vec {\alpha }}\cdot {\vec {\alpha }})\mathbf {1} )\end{array}}}$

The latter variant is based on the quaternions. In Büchter (1992) there is a detailed discussion of the various parameterization options for rotations.

Pre-image and image vectors are given

Given are three linearly independent vectors , which accordingly form a vector space basis . The dual basis for this is , so that ${\displaystyle {\vec {u}}_{1,2,3}}$ ${\displaystyle {\vec {u}}^{1,2,3}}$

${\displaystyle {\vec {u}}_{i}\cdot {\vec {u}}^{j}=\delta _{ij}:={\begin{cases}1&{\textsf {falls}}\;i=j\\0&{\textsf {sonst}}\end{cases}}\;,\quad i,j={1,2,3}}$

applies. The symbol is the Kronecker Delta . If the vector group emerges from the base by rotation , then there is an orthogonal tensor Q for which: ${\displaystyle \delta _{ij}}$ ${\displaystyle {\vec {v}}_{1,2,3}}$ ${\displaystyle {\vec {u}}_{1,2,3}}$

${\displaystyle \mathbf {Q} \cdot {\vec {u}}_{i}={\vec {v}}_{i}\,,\quad i=1,2,3}$

With the dyadic product " " of vectors, this tensor has the form: ${\displaystyle \otimes }$

${\displaystyle \mathbf {Q} =\sum _{i=1}^{3}{\vec {v}}_{i}\otimes {\vec {u}}^{i}}$

With the dual basis is calculated ${\displaystyle {\vec {v}}_{1,2,3}}$ ${\displaystyle {\vec {v}}^{1,2,3}}$

${\displaystyle \sum _{k=1}^{3}({\vec {v}}^{k}\otimes {\vec {u}}_{k})\cdot \mathbf {Q} ^{\top }=\sum _{i,k=1}^{3}({\vec {v}}^{k}\otimes {\vec {u}}_{k})\cdot ({\vec {u}}^{i}\otimes {\vec {v}}_{i})=\sum _{i=1}^{3}{\vec {v}}^{i}\otimes {\vec {v}}_{i}=\mathbf {1} }$

hence the two representations

${\displaystyle \mathbf {Q} =\sum _{i=1}^{3}{\vec {v}}_{i}\otimes {\vec {u}}^{i}=\sum _{i=1}^{3}{\vec {v}}^{i}\otimes {\vec {u}}_{i}}$

exist. The same tensor Q converts the dual bases into one another:

${\displaystyle \mathbf {Q} \cdot {\vec {u}}^{i}={\vec {v}}^{i}\,,\quad i=1,2,3}$

The determinant of the tensor is calculated with the above representations as follows:

${\displaystyle {\begin{array}{lcl}\operatorname {det} (\mathbf {Q} )&=&\operatorname {det} {\begin{pmatrix}{\vec {v}}_{1}&{\vec {v}}_{2}&{\vec {v}}_{3}\end{pmatrix}}\cdot \operatorname {det} {\begin{pmatrix}{\vec {u}}^{1}&{\vec {u}}^{2}&{\vec {u}}^{3}\end{pmatrix}}={\dfrac {\operatorname {det} {\begin{pmatrix}{\vec {v}}_{1}&{\vec {v}}_{2}&{\vec {v}}_{3}\end{pmatrix}}}{\operatorname {det} {\begin{pmatrix}{\vec {u}}_{1}&{\vec {u}}_{2}&{\vec {u}}_{3}\end{pmatrix}}}}\\&=&\operatorname {det} {\begin{pmatrix}{\vec {v}}^{1}&{\vec {v}}^{2}&{\vec {v}}^{3}\end{pmatrix}}\cdot \operatorname {det} {\begin{pmatrix}{\vec {u}}_{1}&{\vec {u}}_{2}&{\vec {u}}_{3}\end{pmatrix}}={\dfrac {\operatorname {det} {\begin{pmatrix}{\vec {u}}_{1}&{\vec {u}}_{2}&{\vec {u}}_{3}\end{pmatrix}}}{\operatorname {det} {\begin{pmatrix}{\vec {v}}_{1}&{\vec {v}}_{2}&{\vec {v}}_{3}\end{pmatrix}}}}=+1\end{array}}}$

because above a rotation and thus the same handedness of the bases was assumed. With a rotational mirroring, det ( Q ) = -1 and the handedness of the two bases would be different.

Given tensor

The axis of rotation of an orthogonal tensor Q is its vector invariant . Let the bases and their dual bases for i = 1,2,3 and the orthogonal tensor Q be defined as in the previous section. Then we get for the axis of rotation of Q : ${\displaystyle {\vec {n}}:=\mathbf {1} \cdot \!\!\times \mathbf {Q} }$ ${\displaystyle {\vec {u}}_{i}\,,{\vec {v}}_{i}}$ ${\displaystyle {\vec {u}}^{i}\,,{\vec {v}}^{i}}$

${\displaystyle {\begin{array}{rcl}{\vec {n}}=\mathbf {1} \cdot \!\!\times \mathbf {Q} &=&\displaystyle \sum _{i=1}^{3}\mathbf {1} \cdot \!\!\times ({\vec {v}}_{i}\otimes {\vec {u}}^{i})=\sum _{i=1}^{3}{\vec {v}}_{i}\times {\vec {u}}^{i}\\&=&\displaystyle \sum _{i=1}^{3}\mathbf {1} \cdot \!\!\times ({\vec {v}}^{i}\otimes {\vec {u}}_{i})=\sum _{i=1}^{3}{\vec {v}}^{i}\times {\vec {u}}_{i}\end{array}}}$

because the scalar cross product "· ×" with the unit tensor exchanges the dyadic product for the cross product. Because of

${\displaystyle {\begin{array}{lcl}\mathbf {Q} \cdot {\vec {n}}&=&\displaystyle \sum _{i=1}^{3}\mathbf {Q} \cdot ({\vec {v}}_{i}\times {\vec {u}}^{i})=\operatorname {det} (\mathbf {Q} )\sum _{i=1}^{3}(\mathbf {Q} \cdot {\vec {v}}_{i})\times (\mathbf {Q} \cdot {\vec {u}}^{i})\\&=&\displaystyle \operatorname {det} (\mathbf {Q} )\sum _{i,k=1}^{3}({\vec {u}}^{k}\cdot {\vec {v}}_{i}){\vec {v}}_{k}\times {\vec {v}}^{i}=\operatorname {det} (\mathbf {Q} )\sum _{k=1}^{3}{\vec {v}}_{k}\times {\vec {u}}^{k}=\operatorname {det} (\mathbf {Q} ){\vec {n}}\end{array}}}$

the vector invariant is actually an eigenvector and therefore parallel to the axis of rotation. In the matrix representation with the rows and columns of Q with respect to the standard basis ê _1,2,3 we get: ${\displaystyle {\vec {z}}_{1,2,3}}$ ${\displaystyle {\vec {s}}_{1,2,3}}$

${\displaystyle \mathbf {Q} =\sum _{i=1}^{3}{\hat {e}}_{i}\otimes {\vec {z}}_{i}=\sum _{i=1}^{3}{\vec {s}}_{i}\otimes {\hat {e}}_{i}\quad \rightarrow \quad {\vec {n}}=\mathbf {1} \cdot \!\!\times \mathbf {Q} =\sum _{i=1}^{3}{\hat {e}}_{i}\times {\vec {z}}_{i}=\sum _{i=1}^{3}{\vec {s}}_{i}\times {\hat {e}}_{i}}$

The following relationships are known from the section # Rotation axis and angle given . The angle of rotation is calculated from the track

${\displaystyle \operatorname {Sp} (\mathbf {Q} )=\operatorname {det} (\mathbf {Q} )+2\cos(\alpha )}$

Alternatively, the axis and angle of rotation can be selected ${\displaystyle {\hat {n}}=n_{1}{\hat {e}}_{1}+n_{2}{\hat {e}}_{2}+n_{3}{\hat {e}}_{3}}$ ${\displaystyle \alpha }$

${\displaystyle {\frac {1}{2}}(\mathbf {Q} -\mathbf {Q} ^{\top })=\sin(\alpha ){\begin{pmatrix}0&-n_{3}&n_{2}\\n_{3}&0&-n_{1}\\-n_{2}&n_{1}&0\end{pmatrix}}\,,\quad {\sqrt {n_{1}^{2}+n_{2}^{2}+n_{3}^{2}}}=1}$

be determined.

The eigen system reveals that the two complex conjugate eigenvalues of Q are exponential functions of the angle. ${\displaystyle {e}^{\pm \mathrm {i} \alpha }}$

Eigensystem

If three vectors are perpendicular to each other in pairs and the magnitudes are one, the axis of rotation and the angle of rotation of the tensor Q , then this has the eigenvalues ​​and vectors ${\displaystyle {\hat {q}}_{1,2,3}}$ ${\displaystyle {\hat {q}}_{1}}$ ${\displaystyle \alpha }$

${\displaystyle {\begin{array}{lcllcl}\lambda _{1}&=&\pm 1,&{\vec {v}}_{1}&=&{\hat {q}}_{1}\\\lambda _{2}&=&e^{\mathrm {i} \alpha },&{\vec {v}}_{2}&=&{\frac {1}{\sqrt {2}}}({\hat {q}}_{2}-\mathrm {i} {\hat {q}}_{3}).\\\lambda _{3}&=&e^{-\mathrm {i} \alpha },&{\vec {v}}_{3}&=&{\frac {1}{\sqrt {2}}}({\hat {q}}_{2}+\mathrm {i} {\hat {q}}_{3})\end{array}}}$

The number i is the imaginary unit and e is Euler's number . The vectors lie in the plane of rotation, are in this, as long as this is guaranteed, but oriented as desired. The representation results from this inherent system ${\displaystyle {\hat {q}}_{2,3}}$ ${\displaystyle {\hat {q}}_{2}\cdot {\hat {q}}_{3}=0}$

${\displaystyle {\begin{array}{lcl}\mathbf {Q} &=&\pm {\hat {q}}_{1}\otimes {\hat {q}}_{1}+\cos(\alpha )({\hat {q}}_{2}\otimes {\hat {q}}_{2}+{\hat {q}}_{3}\otimes {\hat {q}}_{3})+\sin(\alpha )({\hat {q}}_{3}\otimes {\hat {q}}_{2}-{\hat {q}}_{2}\otimes {\hat {q}}_{3})\\&=&{\begin{pmatrix}\pm 1&0&0\\0&\cos(\alpha )&-\sin(\alpha )\\0&\sin(\alpha )&\cos(\alpha )\end{pmatrix}}_{{\hat {q}}_{i}\otimes {\hat {q}}_{j}}.\end{array}}}$

The handedness of the vector group determines the direction of rotation of the rotation around the axis of rotation . If the vector group is right-handed , the angle measures counterclockwise, otherwise clockwise around the axis of rotation. ${\displaystyle {\hat {q}}_{1,2,3}}$

Invariants

→ Main article : Main invariant

If the angle of rotation of the orthogonal tensor is Q , then: ${\displaystyle \alpha }$

${\displaystyle {\begin{array}{lcl}\operatorname {Sp} (\mathbf {Q} )&=&\operatorname {det} (\mathbf {Q} )+2\cos(\alpha )\\\operatorname {I} _{2}(\mathbf {Q} )&=&\operatorname {det} (\mathbf {Q} )\cdot \operatorname {Sp} (\mathbf {Q} )\\\operatorname {det} (\mathbf {Q} )&=&\pm 1\end{array}}}$

because the second main invariant is the trace of the cofactor

${\displaystyle \operatorname {cof} (\mathbf {Q} ):=\operatorname {det} (\mathbf {Q} )\mathbf {Q} ^{\top -1}:=\operatorname {det} (\mathbf {Q} )\mathbf {Q} }$

With the above illustration

${\displaystyle \mathbf {Q} =\sum _{i=1}^{3}{\vec {v}}_{i}\otimes {\vec {u}}^{i}}$

the main invariants are calculated:

${\displaystyle {\begin{array}{lcl}\operatorname {Sp} (\mathbf {Q} )&=&\displaystyle \sum _{i=1}^{3}{\vec {v}}_{i}\cdot {\vec {u}}^{i}\\\operatorname {I} _{2}(\mathbf {Q} )&=&\displaystyle \left(\sum _{i=1}^{3}{\vec {v}}_{i}\cdot {\vec {u}}^{i}\right){\frac {\operatorname {det} {\begin{pmatrix}{\vec {v}}_{1}&{\vec {v}}_{2}&{\vec {v}}_{3}\end{pmatrix}}}{\operatorname {det} {\begin{pmatrix}{\vec {u}}_{1}&{\vec {u}}_{2}&{\vec {u}}_{3}\end{pmatrix}}}}\\\operatorname {det} (\mathbf {Q} )&=&\displaystyle {\frac {\operatorname {det} {\begin{pmatrix}{\vec {v}}_{1}&{\vec {v}}_{2}&{\vec {v}}_{3}\end{pmatrix}}}{\operatorname {det} {\begin{pmatrix}{\vec {u}}_{1}&{\vec {u}}_{2}&{\vec {u}}_{3}\end{pmatrix}}}}=\pm 1\end{array}}}$

As given in the #Tensor section , the vector invariant is the axis of rotation that is calculated with the unit tensor:

${\displaystyle {\vec {\operatorname {i} }}(\mathbf {Q} )=\mathbf {1\cdot \!\!\times Q} =\sum _{i=1}^{3}{\vec {v}}_{i}\times {\vec {u}}^{i}=\sum _{i=1}^{3}{\vec {v}}^{i}\times {\vec {u}}_{i}}$

The Frobenius norm of an orthogonal tensor is always equal to the root of the space dimension:

${\displaystyle \parallel \mathbf {Q} \parallel ={\sqrt {\mathbf {Q} :\mathbf {Q} }}={\sqrt {(\mathbf {Q} ^{\top }\cdot \mathbf {Q} ):\mathbf {1} }}={\sqrt {\mathbf {1} :\mathbf {1} }}={\sqrt {3}}}$

See also

• Rotation matrix , orthogonal matrix , orthogonal mapping
• Quaternion
• Special orthogonal group , special linear group
• Formula collection tensor algebra

Footnotes

1. ↑ N. Büchter: Merging the degeneration concept and the shell theory for finite rotations . 1992 ( PDF version, archived on 2014-10-19 - Report No. 14 of the Institute for Structural Analysis at the University of Stuttgart).

literature

• H. Altenbach: Continuum Mechanics . Springer, 2012, ISBN 978-3-642-24118-5 .
• J. Hanson: Rotations in three, four and five dimensions . 2011, arxiv : 1103.5263 (English, original title: Rotations in three, four, and five dimensions .).