Vector invariant

The vector invariant is a vector property that can be assigned to a second order tensor . The components of a tensor refer to dyads of vectors , which in turn can be represented component by component with respect to a vector space basis. When the basis is changed, the components of the vectors change in a characteristic way but not the amounts of the vectors. The amount of a vector is therefore invariant to a change in the basis. The vector invariant of a tensor is also invariant to a change in the basis. Because the cross product goes into the definition, the vector invariant is only defined in three dimensions.

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

Second level tensors are used as linear mapping from geometric vectors to geometric vectors, which are generally rotated and stretched in the process, see figure on the right. The vector invariant is used to describe rotations: it is the axis of rotation around which an orthogonal tensor rotates a vector, and the angular velocity is proportional to the vector invariant of the velocity gradient .

definition

The vector invariant of a dyad of vectors from the three-dimensional Euclidean vector space arises by replacing the dyadic product " " with the cross product "×": ${\displaystyle {\vec {i}}}$ ${\displaystyle {\vec {a}}\otimes {\vec {b}}}$ ${\displaystyle {\vec {a}},\,{\vec {b}}\in \mathbb {V} }$ ${\displaystyle \mathbb {V} }$ ${\displaystyle \otimes }$

${\displaystyle {\vec {i}}({\vec {a}}\otimes {\vec {b}})={\vec {a}}\times {\vec {b}}\,.}$

If the vector is parallel to the vector , then the dyad is symmetric and the vector invariant vanishes. The replacement of the dyadic product by the cross product in a dyad can be achieved with the "scalar cross product" with the unit tensor 1 : ${\displaystyle {\vec {b}}}$ ${\displaystyle {\vec {a}}}$

${\displaystyle \mathbf {1} \cdot \!\!\times ({\vec {a}}\otimes {\vec {b}})=\left(\sum _{i=1}^{3}{\hat {e}}_{i}\otimes {\hat {e}}_{i}\right)\cdot \!\!\times ({\vec {a}}\otimes {\vec {b}}):=\sum _{i=1}^{3}({\hat {e}}_{i}\cdot {\vec {a}}){\hat {e}}_{i}\times {\vec {b}}={\vec {a}}\times {\vec {b}}\,.}$

The vectors stand for an orthonormal basis and “·” for the scalar product defined in the Euclidean vector space . For a second level tensor T , which can always be represented as a sum of dyads, the vector invariant is determined accordingly ${\displaystyle {\hat {e}}_{1,2,3}}$

${\displaystyle \mathbf {T} _{\times }:={\vec {i}}(\mathbf {T} ):=\mathbf {1\cdot \!\!\times T} \,.}$

The spelling is from Altenbach (2012). With regard to the orthonormal basis , the following is written specifically: ${\displaystyle \mathbf {T} _{\times }}$ ${\displaystyle {\hat {e}}_{1,2,3}}$

${\displaystyle \mathbf {T} =\sum _{i,j=1}^{3}T_{ij}{\hat {e}}_{i}\otimes {\hat {e}}_{j}\quad \rightarrow \quad {\vec {i}}(\mathbf {T} )={\begin{pmatrix}T_{23}-T_{32}\\T_{31}-T_{13}\\T_{12}-T_{21}\end{pmatrix}}}$

Invariance

The invariance of the vector invariant can be demonstrated by transforming the shape

${\displaystyle \mathbf {U} :=\sum _{i=1}^{3}{\vec {u}}_{i}\otimes {\vec {u}}^{i}\,,}$

are the unit tensors and do not change the tensor for the product with a tensor. The vectors must form a vector space basis and are the dual basis for this . A second order tensor with components with respect to any two basic systems and is given ${\displaystyle {\vec {u}}_{1,2,3}}$ ${\displaystyle {\vec {u}}^{1,2,3}}$ ${\displaystyle T_{ij}}$ ${\displaystyle {\vec {a}}_{1,2,3}}$ ${\displaystyle {\vec {b}}_{1,2,3}\,:}$

${\displaystyle \mathbf {T} =\sum _{i,j=1}^{3}T_{ij}{\vec {a}}_{i}\otimes {\vec {b}}_{j}\,.}$

With transformations U and V the tensor with components with respect to the bases resp. ${\displaystyle {\vec {u}}_{1,2,3}}$ ${\displaystyle {\vec {v}}_{1,2,3}\,:}$

${\displaystyle {\begin{aligned}\mathbf {T} =&\mathbf {U\cdot T\cdot V} ^{\top }\\=&\sum _{i,j,k,l=1}^{3}({\vec {u}}_{k}\otimes {\vec {u}}^{k})\cdot (T_{ij}{\vec {a}}_{i}\otimes {\vec {b}}_{j})\cdot ({\vec {v}}^{l}\otimes {\vec {v}}_{l})\\=&\sum _{i,j,k,l=1}^{3}\underbrace {[T_{ij}({\vec {u}}^{k}\cdot {\vec {a}}_{i})({\vec {v}}^{l}\cdot {\vec {b}}_{j})]} _{:=T_{kl}^{*}}{\vec {u}}_{k}\otimes {\vec {v}}_{l}\\=&\sum _{k,l=1}^{3}T_{kl}^{*}{\vec {u}}_{k}\otimes {\vec {v}}_{l}\end{aligned}}}$

The superscript "T" stands for transposition . This is calculated

${\displaystyle {\begin{aligned}{\vec {i}}(\mathbf {U\cdot T\cdot V} ^{\top })=&\sum _{k,l=1}^{3}T_{kl}^{*}{\vec {u}}_{k}\times {\vec {v}}_{l}\\=&\sum _{i,j,k,l=1}^{3}[T_{ij}({\vec {u}}^{k}\cdot {\vec {a}}_{i})({\vec {v}}^{l}\cdot {\vec {b}}_{j})]{\vec {u}}_{k}\times {\vec {v}}_{l}\\=&\sum _{i,j=1}^{3}T_{ij}{\vec {a}}_{i}\times {\vec {b}}_{j}\\=&{\vec {i}}(\mathbf {T} )\,.\end{aligned}}}$

The vector invariant therefore deserves its name.

Axiality and objectivity

If the above transformations are identical and orthogonal, i.e. if they have the property , then for a tensor T given as above we get : ${\displaystyle \mathbf {Q\cdot Q} ^{\top }=\mathbf {1} }$

${\displaystyle {\begin{aligned}{\vec {i}}(\mathbf {Q\cdot T\cdot Q} ^{\top })=&{\vec {i}}\left(\mathbf {Q} \cdot \left(\sum _{i,j=1}^{3}T_{ij}{\vec {a}}_{i}\otimes {\vec {b}}_{j}\right)\cdot \mathbf {Q} ^{\top }\right)\\=&{\vec {i}}\left(\sum _{i,j=1}^{3}T_{ij}(\mathbf {Q} \cdot {\vec {a}}_{i})\otimes (\mathbf {Q} \cdot {\vec {b}}_{j})\right)\\=&\sum _{i,j=1}^{3}T_{ij}(\mathbf {Q} \cdot {\vec {a}}_{i})\times (\mathbf {Q} \cdot {\vec {b}}_{j})\\=&\operatorname {det} (\mathbf {Q} )\mathbf {Q} \cdot \sum _{i,j=1}^{3}T_{ij}{\vec {a}}_{i}\times {\vec {b}}_{j}\\=&\operatorname {det} (\mathbf {Q} )\mathbf {Q} \cdot {\vec {i}}(\mathbf {T} )\,.\end{aligned}}}$

If Q is actually orthogonal , then its determinant det is equal to one and the vector invariant is objective, because it transforms in a Euclidean transformation like an objective vector.

In a rotary mirror to apply and, therefore, ${\displaystyle {\vec {i}}}$ ${\displaystyle \mathbf {Q} \cdot {\vec {i}}=-{\vec {i}}\,,\;\operatorname {det} \mathbf {Q} =-1}$

${\displaystyle {\vec {i}}(\mathbf {Q\cdot T\cdot Q} ^{\top })={\vec {i}}(\mathbf {T} )\,.}$

Vectors with this property in a rotational mirroring are axial vectors .

Dual vector and cross product matrix

Every skew-symmetric tensor T can have

${\displaystyle \mathbf {T} \cdot {\vec {v}}={\vec {t}}\times {\vec {v}}\quad {\text{für alle}}\quad {\vec {v}}\in \mathbb {V} }$

a dual vector can be assigned. The dual vector is proportional to the vector invariant: ${\displaystyle {\vec {t}}}$

${\displaystyle {\vec {t}}=-{\frac {1}{2}}{\vec {i}}(\mathbf {T} )=-{\frac {1}{2}}\mathbf {1\cdot \!\!\times T} \,.}$

With regard to the orthonormal basis , it is written specifically: ${\displaystyle {\hat {e}}_{1,2,3}}$

${\displaystyle \mathbf {T} =\sum _{i,j=1}^{3}T_{ij}{\hat {e}}_{i}\otimes {\hat {e}}_{j}\quad \rightarrow \quad {\vec {t}}=-{\frac {1}{2}}\sum _{i,j=1}^{3}T_{ij}{\hat {e}}_{i}\times {\hat {e}}_{j}=\left({\begin{array}{c}T_{32}\\T_{13}\\T_{21}\end{array}}\right)\,.}$

The tensor T can be reconstructed from its dual vector using. In linear algebra , the matrix that is analogously assigned to the vector is called the cross product matrix . ${\displaystyle \mathbf {T} ={\vec {t}}\times \mathbf {1} }$ ${\displaystyle {\vec {t}}}$

properties

The following properties of the vector invariant can be demonstrated with elementary tensor algebra. Let x be any number, any vector, and any second order tensors. Then: ${\displaystyle {\vec {v}}}$ ${\displaystyle \mathbf {S} ,\,\mathbf {T} }$

${\displaystyle {\begin{aligned}{\vec {i}}(\mathbf {T} ^{\top })=&-{\vec {i}}(\mathbf {T} )\\{\vec {i}}(\mathbf {S+T} )=&{\vec {i}}(\mathbf {S} )+{\vec {i}}(\mathbf {T} )\\{\vec {i}}(x\mathbf {T} )=&x\,{\vec {i}}(\mathbf {T} )\\{\vec {i}}(\mathbf {S} \#\mathbf {T} )=&\mathbf {S} \cdot {\vec {i}}(\mathbf {T} )+\mathbf {T} \cdot {\vec {i}}(\mathbf {S} )\\\mathbf {T} \cdot {\vec {i}}(\mathbf {T} )=&\mathbf {T} ^{\top }\cdot {\vec {i}}(\mathbf {T} )={\vec {i}}(\mathbf {T} )\cdot \mathbf {T} \\{\vec {i}}(\operatorname {cof} (\mathbf {T} ))=&\mathbf {T} \cdot {\vec {i}}(\mathbf {T} )\\{\vec {i}}(\mathbf {T} ^{-1})=&-{\frac {1}{\operatorname {det} (\mathbf {T} )}}\mathbf {T} \cdot {\vec {i}}(\mathbf {T} )\quad {\text{falls}}\quad \operatorname {det} (\mathbf {T} )\neq 0\\{\vec {i}}(\mathbf {T\cdot S\cdot T} ^{\top })=&\operatorname {cof} (\mathbf {T} )\cdot {\vec {i}}(\mathbf {S} )\\{\vec {i}}(\mathbf {T} )\times {\vec {v}}=&(\mathbf {T^{\top }-T} )\cdot {\vec {v}}\\{\vec {i}}({\vec {v}}\times \mathbf {1} )=&-2{\vec {v}}\end{aligned}}}$

Here "#" is the outer tensor product ,

${\displaystyle \operatorname {cof} (\mathbf {T} ):=\mathbf {T} ^{\top }\cdot \mathbf {T} ^{\top }-\operatorname {I} _{1}(\mathbf {T} )\mathbf {T} ^{\top }+\operatorname {I} _{2}(\mathbf {T} )\mathbf {1} ={\frac {1}{2}}\mathbf {T} \#\mathbf {T} }$

the cofactor and I _1,2 principal invariants of the tensor T .

From the first two properties it follows that only the skew-symmetrical part of a tensor contributes something to the vector invariant and that symmetrical tensors have the zero vector as a vector invariant.

See also

Formula collection tensor algebra

Footnotes

1. ↑ The scalar cross product is defined with vectors via ${\displaystyle {\vec {a}},\,{\vec {b}},\,{\vec {c}},\,{\vec {d}}\in \mathbb {V} }$
   ${\displaystyle ({\vec {a}}\otimes {\vec {b}})\cdot \!\!\times ({\vec {c}}\otimes {\vec {d}}):=({\vec {b}}\cdot {\vec {c}}){\vec {a}}\times {\vec {d}}\,.}$

literature

• H. Altenbach: Continuum Mechanics . Springer, 2012, ISBN 978-3-642-24118-5 , pp. 34 .