Why do orthogonal matrices represent rotations

Orthogonal tensor

Linear mapping of a vector \$ \ vec {v} \$ by a tensor \$ \ mathbf {T} \$.
Rotation of a vector \$ \ vec {v} \$ around the axis of rotation \$ \ vec {n} \$ with angle \$ \ alpha \$ by an orthogonal tensor \$ \ mathbf {Q} \$.

Orthogonal tensors are unit-free tensors of the second order, which perform a rotation or rotation mirroring in Euclidean vector space. In continuum mechanics, which describes physical processes, only rotations are considered, because rotational reflections do not occur in nature.

Second-level tensors are used here as linear mapping of geometric vectors onto geometric vectors, which are generally rotated and stretched in the process, see figure above on the right. In the case of an orthogonal tensor that represents a rotation or rotation mirroring, there is no stretching, 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 symbols \$ \ mathbf {Q} \$ or \$ \ mathbf {R} \$, where \$ \ mathbf {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. Unlike matrices, however, the coefficients of a tensor refer to 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 Invariantsthat remain unchanged when the basic system is changed. 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 \$ \ mathbf {Q} \$, for which the following applies:

\$ \ mathbf {Q} ^ {- 1} = \ mathbf {Q} ^ \ top \ quad \ textsf {or} \ quad \ mathbf {Q ^ \ top \ cdot Q} = \ mathbf {Q \ cdot Q} ^ \ top = \ mathbf {I} \ ,. \$

The superscript \$ -1 \$ denotes the inverse, \$ \ left (\ cdot \ right) ^ \ top \$ the transposed tensor and \$ \ mathbf {I} \$ the unit tensor. Because of

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

is

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

An orthogonal tensor representing pure rotation becomes actually orthogonal called and has the determinant +1. With \$ \ operatorname {det} (\ mathbf {Q}) = - 1 \$ the tensor performs a rotational mirroring. Because reflections do not occur in nature, in a physical context \$ \ operatorname {det} (\ mathbf {Q}) = + 1 \$.

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 also the center of gravity of the body are suitable as a center of rotation, see figure on the right. Let \$ \ vec {r} (\ vec {X}) = \ vec {X} - \ vec {S} \$ the time-fixed difference vector between a particle \$ \ vec {X} \$ of the rigid body and its center of gravity \$ \ vec {S} \$ at a point in time \$ {t} _ {0} \$. The translation of the body can then with its center of gravity movement \$ \ vec {s} (t) \$ (with \$ \ vec {s} (t_0) = \ vec {S} \$) and its rotation with one of the time but not of the place dependent orthogonal tensor \$ \ mathbf {Q} (t) \$ (with \$ \ mathbf {Q} ({t} _ {0}) = \ mathbf {I} \$). Translation and rotation taken together define the motion function \$ \ vec {\ chi} (\ vec {X}, t) \$ of the particle \$ \ vec {X} \$:

\$ \ vec {\ chi} (\ vec {X}, t) = \ vec {s} (t) + \ mathbf {Q} (t) \ cdot (\ vec {X} - \ vec {S}) \ quad \ rightarrow \ quad \ vec {X} - \ vec {S} = \ mathbf {Q} ^ \ top (t) \ cdot (\ vec {\ chi} (\ vec {X}, t) - \ vec { s} (t)) \ ,. \$

The speed of the particle is then

\$ \ begin {array} {rcl} \ 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) + \ varvec {\ Omega} (t) \ cdot (\ vec {x} - \ vec {s} (t)) \ end {array} \$

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

\$ \ varvec {\ Omega} + \ varvec {\ Omega} ^ \ top = \ dot {\ mathbf {Q}} \ cdot \ mathbf {Q} ^ \ top + \ mathbf {Q} \ cdot \ dot {\ mathbf {Q}} ^ \ top = \ frac {\ text {d}} {\ text {d} t} (\ mathbf {Q \ cdot Q} ^ \ top) = \ dot {\ mathrm {I}} = \ mathbf {0} \ rightarrow \ boldsymbol {\ Omega} ^ \ top = - \ boldsymbol {\ Omega} \$

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

\$ \ boldsymbol {\ Omega} \ cdot \ vec {v} = \ vec {\ omega} \ times \ vec {v} \ quad \ text {for all} \ quad \ vec {v} \ ,. \$

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

\$ \ 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.

Vector transformation

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

\$ (\ 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 \$ \ vec {v} = \ vec {u} \$:

\$ {| \ 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 \ ,, \$

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

\$ \ mathbf {Q} \ cdot \ vec {n} = \ vec {n} \ ,. \$

If \$ \ mathbf {Q} \$ is an improperly orthogonal tensor, then is

\$ \ 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 denoted as \$ \ vec {a}, \, \ vec {b}, \, \ vec {c} \$ as in the picture and transformed with an orthogonal tensor, the spatial product is calculated as follows:

\$ \ begin {array} {rcl} (\ 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 {array} \$

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

\$ \ 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 \$ \ vec {a} \$, which is why the vector in the curly brackets disappears and opens

\$ (\ 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. With a rotational mirroring, both elements change their sign, which is why they only change with a transformation with one actually orthogonal Tensor \$ \ mathbf {Q} \$ are invariant to a Euclidean transformation.

Tensor transformation

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

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

holds, and \$ \ mathbf {Q} \$ is an orthogonal tensor. Then

\$ \ 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 \$ \ mathbf {S}: = \ mathbf {Q \ cdot T \ cdot Q} ^ \ top \$ has the same eigenvalues ​​as \$ \ mathbf {T} \$ but the eigenvectors rotated with \$ \ mathbf {Q} \$. It follows immediately that the main invariants and amounts of \$ \ mathbf {S} \$ and \$ \ mathbf {T} \$ coincide.

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 each other?
• 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 \$ \ has {n} \$ a unit vector (of length one) and \$ \ alpha \$ an angle. Then is the tensor

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

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

\$ \ hat {n} \ times \ mathbf {I} = \ 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} \ ,, \$

if \$ n_ {1,2,3} \$ are the components of \$ \ hat {n} \$ with respect to the standard base \$ \ hat {e} _ {1,2,3} \$.

In the case of a rotating mirroring,

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

In any case, the tensor \$ \ mathbf {Q} \$ has the trace and the skew-symmetric part

\$ \ begin {array} {rcl} \ operatorname {Sp} (\ mathbf {Q}) & = & \ operatorname {Sp} [\ operatorname {det} (\ mathbf {Q}) \ hat {n} \ otimes \ hat {n} + \ cos (\ alpha) (\ mathbf {I} - \ hat {n} \ otimes \ hat {n}) + \ sin (\ alpha) \ hat {n} \ times \ mathbf {I} ] \ & = & \ operatorname {det} (\ mathbf {Q}) + 2 \ cos (\ alpha) \ \ frac {1} {2} (\ mathbf {Q} - \ mathbf {Q} ^ \ top) & = & \ sin (\ alpha) \ hat {n} \ times \ mathbf {I} = \ sin (\ alpha) \ begin {pmatrix} 0 & - n_3 & n_2 \ n_3 & 0 & - n_1 \ - n_2 & n_1 & 0 \ end {pmatrix} \ ,. \ end {array} \$

The above formula for \$ \ mathbf {Q} \$ can also be written with a rotation vector \$ \ vec {\ alpha}: = \ alpha \ hat {n} \$:

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

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

Rotation vectors with a different length can also be used:

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

The latter variant is based on the quaternions. In Büchter (1992)[1] 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 \$ \ vec {u} _ {1,2,3} \$, which accordingly form a vector space basis. The dual basis for this is \$ \ vec {u} ^ {1,2,3} \$, so that

\$ \ vec {u} _i \ cdot \ vec {u} ^ j = \ delta_ {ij}: = \ begin {cases} 1 & \ textsf {if} \; i = j \ 0 & \ textsf {otherwise} \ end {cases} \;, \ quad i, j = {1,2,3} \$

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

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

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

\$ \ mathbf {Q} = \ sum_ {i = 1} ^ 3 \ vec {v} _i \ otimes \ vec {u} ^ i \ ,. \$

With the base \$ \ vec {v} ^ {1,2,3} \$ dual to \$ \ vec {v} _ {1,2,3} \$, the calculation is

\$ \ sum_ {k = 1} ^ 3 (\ vec {v} ^ k \ otimes \ vec {u} _k) \ cdot \ mathbf {Q} ^ \ mathrm {T} = \ 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 {I} \$

hence the two representations

\$ \ mathbf {Q} = \ sum_ {i = 1} ^ 3 \ vec {v} _i \ otimes \ vec {u} ^ i = \ sum_ {i = 1} ^ 3 \ vec {v} ^ i \ otimes \ vec {u} _i \$

are present. The same tensor \$ \ mathbf {Q} \$ converts the dual bases into one another:

\$ \ 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:

\$ \ 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 a rotation and thus the same handedness of the bases was assumed above. In the case of a rotational mirroring, \$ \ operatorname {det} (\ mathbf {Q}) = -1 \$ and the handedness of the two bases would be different.

Given tensor

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

\$ \ begin {array} {rcl} \ vec {n} = \ mathbf {I} \ cdot \! \! \ times \ mathbf {Q} & = & \ displaystyle \ sum_ {i = 1} ^ 3 \ mathbf { I} \ 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 {I} \ 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

\$ \ 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 lines \$ \ vec {z} _ {1,2,3} \$ and columns \$ \ vec {s} _ {1,2,3} \$ of \$ \ mathbf {Q} \$ with respect to the standard base \$ \ hat {e} _ {1,2,3} \$ results in:

\$ \ 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 {I} \ 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

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

Alternatively, axis of rotation \$ \ hat {n} = n_1 \ hat {e} _1 + n_2 \ hat {e} _2 + n_3 \ hat {e} _3 \$ and angle \$ \ alpha \$ from

\$ \ 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 ​​\$ {e} ^ {\ pm \ mathrm {i} \ alpha} \$ of \$ \ mathbf {Q} \$ are exponential functions of the angle.

Eigensystem

If three vectors \$ \ hat {q} _ {1,2,3} \$ are pairwise perpendicular to each other and the amounts have one, \$ \ has {q} _1 \$ the axis of rotation and \$ \ alpha \$ the angle of rotation of the tensor \$ \ mathbf {Q} \$, then it has the eigenvalues ​​and vectors

\$ \ begin {array} {lcllcl} \ lambda_1 & = & \ pm 1, & \ vec {v} _1 & = & \ hat {q} _1 \ \ lambda_2 & = & e ^ {\ mathrm {i} \ alpha} , & \ vec {v} _2 & = & \ frac {1} {\ sqrt2} (\ hat {q} _2- \ mathrm {i} \ hat {q} _3). \ \ lambda_3 & = & e ^ {- \ mathrm {i} \ alpha}, & \ vec {v} _3 & = & \ frac {1} {\ sqrt2} (\ hat {q} _2 + \ mathrm {i } \ hat {q} _3) \ end {array} \$

The number i is the imaginary unit and e the Euler's number. The vectors \$ \ hat {q} _ {2,3} \$ lie in the plane of rotation, are in this plane as long as \$ \ hat {q} _2 \ cdot \ hat {q} _3 = 0 \$ is guaranteed, but oriented as required. The representation results from this inherent system

\$ \ 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 \$ \ hat {q} _ {1,2,3} \$ 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.

Invariants

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

\$ \ 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

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

With the illustration above

\$ \ mathbf {Q} = \ sum_ {i = 1} ^ 3 \ vec {v} _i \ otimes \ vec {u} ^ i \$

the main invariants are calculated:

\$ \ 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, which is calculated with the unit tensor:

\$ \ vec {\ operatorname {i}} (\ mathbf {Q}) = \ mathbf {I \ 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:

\$ \ parallel \ mathbf {Q} \ parallel = \ sqrt {\ mathbf {Q}: \ mathbf {Q}} = \ sqrt {(\ mathbf {Q} ^ \ top \ cdot \ mathbf {Q}): \ mathbf {I}} = \ sqrt {\ mathbf {I}: \ mathbf {I}} = \ sqrt3 \ ,. \$