Unit tensor

A unit tensor in continuum mechanics is the linear mapping of each vector onto itself. The unit tensor is a dimensionless one-field tensor because it maps the vectors from a Euclidean vector space into the same vector space. Furthermore, the unit tensor is symmetric , orthogonal and unimodular . The coefficients of the second order unit tensor are called metric coefficients.

Unit tensors occur frequently in continuum mechanics . The unit tensor of the second order occurs in the strain tensors and the unit tensor of the fourth order in many material models (e.g. in Hooke's law ). Because of its importance, this article therefore deals with the three-dimensional Euclidean vector space and the second order unit tensor. The fourth order unit tensor is only mentioned in the chapter of the same name. A generalization to spaces of any finite dimension is possible in a simple manner.

definition

A Euclidean vector space and the set of linear mappings from to are given . Then the unit tensor is defined as ${\displaystyle \mathbb {V} ^{3}}$ ${\displaystyle {\mathcal {L}}(\mathbb {V} ^{3},\mathbb {V} ^{3})}$ ${\displaystyle \mathbb {V} ^{3}}$ ${\displaystyle \mathbb {V} ^{3}}$ ${\displaystyle \mathbf {1} }$

${\displaystyle {\begin{array}{ll}\mathbf {1} \in {\mathcal {L}}(\mathbb {V} ^{3},\mathbb {V} ^{3}):&\mathbb {V} ^{3}\rightarrow \mathbb {V} ^{3}\\&{\vec {v}}\mapsto {\vec {v}}\end{array}}}$.

Spellings

The characters “1”, “I” or “E” are used for the unit tensor . As typestyle is letter with double prime ( ), bold ( ), lower ( ) or scan ( ) is used. In index notation, this unit tensor agrees with the Kronecker delta . ${\displaystyle \mathbb {I} }$ ${\displaystyle \mathbf {1} }$ ${\displaystyle {\underline {\underline {1}}}}$ ${\displaystyle {\overline {\overline {1}}}}$ ${\displaystyle \delta _{ij}}$

Tensors fourth stage can be presented with the attached four, for example: . ${\displaystyle {\stackrel {4}{\mathbf {1} }}}$

In this article, the second order unit tensor and the fourth order unit tensor are used. ${\displaystyle \mathbf {1} }$ ${\displaystyle {\stackrel {4}{\mathbf {1} }}}$

properties

Because the identity of tensors can be proven via the bilinear form , every tensor is true for the ${\displaystyle \mathbf {T} }$

${\displaystyle {\vec {u}},{\vec {v}}\in \mathbb {V} ^{3}\quad \rightarrow \quad {\vec {u}}\cdot \mathbf {T} \cdot {\vec {v}}={\vec {u}}\cdot {\vec {v}}}$

identical to the unit tensor. Because of

${\displaystyle \mathbf {1} \cdot {\vec {v}}={\vec {v}}\quad \rightarrow \quad {\vec {v}}=\mathbf {1} ^{-1}\cdot {\vec {v}}}$

the unit tensor is equal to its inverse and because of

${\displaystyle {\vec {u}}\cdot \mathbf {1} \cdot {\vec {v}}=(\mathbf {1} ^{\top }\cdot {\vec {u}})\cdot {\vec {v}}={\vec {u}}\cdot {\vec {v}}\quad {\text{für alle}}\quad {\vec {u}},{\vec {v}}\in \mathbb {V} ^{3}}$

the unit tensor is also symmetric. From the last two properties it follows that the unit tensor is also orthogonal . Because the unit tensor does not reflect a vector (converted into the negative vector), the unit tensor is actually orthogonal, which is why it represents the "rotation" by 0 °. So its determinant is equal to one

${\displaystyle \mathrm {det} (\mathbf {1} )=1}$

therefore the unit tensor is unimodular. The unit tensor is the neutral element in the tensor product "·":

${\displaystyle \mathbf {A} \in {\mathcal {L}}(\mathbb {V} ^{3},\mathbb {V} ^{3})\quad \rightarrow \quad \mathbf {1\cdot A} =\mathbf {A\cdot 1} =\mathbf {A} }$.

The Frobenius scalar product of two tensors A and B is formed using the trace A  : B  : = Sp ( A ^T · B ). The scalar product of the unit tensor with another tensor of the second order gives its trace:

${\displaystyle \mathbf {A} \in {\mathcal {L}}(\mathbb {V} ^{3},\mathbb {V} ^{3})\quad \rightarrow \quad \mathbf {1:A} =\operatorname {Sp} (\mathbf {A} )}$.

Eigensystem

From the properties of the unit tensor it can immediately be deduced that every vector is eigenvector of the unit tensor with the associated eigenvalue one. Because every basis vector of any orthonormal basis of the underlying vector space is also the eigenvector of the unit tensor, the representations ${\displaystyle {\hat {v}}_{1,2,3}}$

${\displaystyle \mathbf {1} =\sum _{i=1}^{3}{\hat {v}}_{i}\otimes {\hat {v}}_{i}=\sum _{i,j=1}^{3}\delta _{ij}{\hat {v}}_{i}\otimes {\hat {v}}_{j}}$

to be used. This is where the dyadic product forms . ${\displaystyle \otimes }$

Representation methods with basis vectors

With respect to the standard basis , the unit tensor is called ${\displaystyle {\hat {e}}_{1,2,3}}$

${\displaystyle \mathbf {1} =\sum _{i=1}^{3}{\hat {e}}_{i}\otimes {\hat {e}}_{i}=\sum _{i,j=1}^{3}\delta _{ij}{\hat {e}}_{i}\otimes {\hat {e}}_{j}={\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}}}$

written so that it matches its matrix notation here . For another orthonormal basis with basis vectors it can be used as ${\displaystyle {\hat {v}}_{1,2,3}}$

${\displaystyle \mathbf {1} =\sum _{i=1}^{3}{\hat {v}}_{i}\otimes {\hat {v}}_{i}=\sum _{i,j=1}^{3}\delta _{ij}{\hat {v}}_{i}\otimes {\hat {v}}_{j}={\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}}_{{\hat {v}}_{i}\otimes {\hat {v}}_{j}}}$

be noted. Is an arbitrary basis in vector space and its dual basis , then is ${\displaystyle {\vec {g}}_{1,2,3}}$ ${\displaystyle {\vec {g}}^{1,2,3}}$

${\displaystyle \mathbf {1} =\sum _{i=1}^{3}{\vec {g}}_{i}\otimes {\vec {g}}^{i}=\sum _{i=1}^{3}{\vec {g}}^{i}\otimes {\vec {g}}_{i}=\sum _{i,j=1}^{3}({\vec {g}}_{i}\cdot {\vec {g}}_{j}){\vec {g}}^{i}\otimes {\vec {g}}^{j}=\sum _{i,j=1}^{3}({\vec {g}}^{i}\cdot {\vec {g}}^{j}){\vec {g}}_{i}\otimes {\vec {g}}_{j}}$.

If there is another arbitrary basis of the vector space and the dual basis for it, then the general representation applies: ${\displaystyle {\vec {a}}_{1,2,3}}$ ${\displaystyle {\vec {a}}^{1,2,3}}$

${\displaystyle {\begin{array}{rcl}\mathbf {1} &=&\displaystyle \sum _{i,j=1}^{3}({\vec {a}}_{i}\cdot {\vec {g}}_{j}){\vec {a}}^{i}\otimes {\vec {g}}^{j}=\sum _{i,j=1}^{3}({\vec {a}}^{i}\cdot {\vec {g}}_{j}){\vec {a}}_{i}\otimes {\vec {g}}^{j}\\&=&\displaystyle \sum _{i,j=1}^{3}({\vec {a}}^{i}\cdot {\vec {g}}^{j}){\vec {a}}_{i}\otimes {\vec {g}}_{j}=\sum _{i,j=1}^{3}({\vec {a}}_{i}\cdot {\vec {g}}^{j}){\vec {a}}^{i}\otimes {\vec {g}}_{j}\end{array}}}$.

Invariants

The three main invariants of the unit tensor are

${\displaystyle {\begin{array}{lll}\mathrm {I} _{1}&:=\operatorname {Sp} (\mathbf {1} )&=3\;,\\\mathrm {I} _{2}&:={\frac {1}{2}}(\operatorname {Sp} {(\mathbf {1} )}^{2}-\operatorname {Sp} (\mathbf {1} ^{2}))&=3\;,\\\mathrm {I} _{3}&:=\mathrm {det} (\mathbf {1} )&=1\;.\end{array}}}$

Because of this, these are also the main invariants of the n-th powers of the unit tensor. The trace of the unit tensor is equal to the dimension of the underlying vector space. ${\displaystyle \mathbf {1} =\mathbf {1\cdot 1} }$

The absolute value of the unit tensor is the square root of the dimension of the vector space:

${\displaystyle \left\|\mathbf {1} \right\|:={\sqrt {\operatorname {Sp} (\mathbf {1^{\top }\cdot 1} )}}={\sqrt {\operatorname {Sp} (\mathbf {1} )}}={\sqrt {3}}}$.

The eigenvalues ​​(here all equal to one) are also invariant.

Metric coefficients

The distance between two points with the position vectors

${\displaystyle {\vec {u}}=\sum _{i=1}^{3}u^{i}{\vec {g}}_{i}\quad {\text{und}}\quad {\vec {v}}=\sum _{i=1}^{3}v^{i}{\vec {g}}_{i}}$

with coordinates and with respect to any oblique basis system is calculated with the Skalarproduktnorm to ${\displaystyle u^{i}}$ ${\displaystyle v^{i}}$ ${\displaystyle {\vec {g}}_{1,2,3}}$

${\displaystyle |{\vec {u}}-{\vec {v}}|:={\sqrt {({\vec {u}}-{\vec {v}})\cdot ({\vec {u}}-{\vec {v}})}}={\sqrt {\sum _{i,j=1}^{3}(u^{i}-v^{i}){\vec {g}}_{i}\cdot (u^{j}-v^{j}){\vec {g}}_{j}}}={\sqrt {\sum _{i,j=1}^{3}({\vec {g}}_{i}\cdot {\vec {g}}_{j})(u^{i}-v^{i})(u^{j}-v^{j})}}}$.

This means that the products of the coefficients of the coordinate vector of the distance vector in the scalar product are weighted with the coefficients . In the representation ${\displaystyle u^{i}-v^{i}}$ ${\displaystyle {\vec {u}}-{\vec {v}}}$ ${\displaystyle {\vec {g}}_{i}\cdot {\vec {g}}_{j}}$

${\displaystyle \mathbf {1} =\sum _{i,j=1}^{3}({\vec {g}}_{i}\cdot {\vec {g}}_{j}){\vec {g}}^{i}\otimes {\vec {g}}^{j}}$

the coefficients are called metric coefficients because the metric of the vector space is specified with the scalar product norm . If the basis vectors are covariant (tangent vectors to the oblique coordinate system) then the scalar products are the covariant metric coefficients. Accordingly, the coefficients are then the contravariant metric coefficients. ${\displaystyle {\vec {g}}_{i}\cdot {\vec {g}}_{j}}$ ${\displaystyle {\vec {g}}_{1,2,3}}$ ${\displaystyle {\vec {g}}_{i}\cdot {\vec {g}}_{j}}$ ${\displaystyle {\vec {g}}^{i}\cdot {\vec {g}}^{j}}$

Fourth order unit tensor

The fourth-order unit tensor maps second-order tensors onto itself. If the second order tensors are the standard basis of the space of second order tensors, then is ${\displaystyle \lbrace \mathbf {E} _{m}\rbrace _{m=1,9}}$ ${\displaystyle {\mathcal {L}}(\mathbb {V} ^{3},\mathbb {V} ^{3})}$

${\displaystyle {\stackrel {4}{\mathbf {1} }}:=\sum _{m=1}^{9}\mathbf {E} _{m}\otimes \mathbf {E} _{m}}$

the unit tensor of the fourth order. Becomes

${\displaystyle \mathbf {E} _{m}:={\vec {e}}_{i}\otimes {\vec {e}}_{j},\quad i,j=1,2,3,\;m=3(i-1)+j}$

can be defined as usual

${\displaystyle {\stackrel {4}{\mathbf {1} }}:=\sum _{i,j=1}^{3}{\vec {e}}_{i}\otimes {\vec {e}}_{j}\otimes {\vec {e}}_{i}\otimes {\vec {e}}_{j}=\sum _{i,j,k,l=1}^{3}\delta _{ik}\delta _{jl}{\vec {e}}_{i}\otimes {\vec {e}}_{j}\otimes {\vec {e}}_{k}\otimes {\vec {e}}_{l}}$

to be written. If there is any basis of space and the dual basis for it, then applies ${\displaystyle \lbrace \mathbf {G} _{m}\rbrace _{m={1,9}}}$ ${\displaystyle {\mathcal {L}}(\mathbb {V} ^{3},\mathbb {V} ^{3})}$ ${\displaystyle \lbrace \mathbf {G} ^{n}\rbrace _{n={1,9}}}$

${\displaystyle {\stackrel {4}{\mathbf {1} }}=\sum _{m=1}^{9}\mathbf {G} _{m}\otimes \mathbf {G} ^{m}=\sum _{m=1}^{9}\mathbf {G} ^{m}\otimes \mathbf {G} _{m}=\sum _{m,n=1}^{9}(\mathbf {G} ^{m}\cdot \mathbf {G} ^{n})\mathbf {G} _{m}\otimes \mathbf {G} _{n}=\sum _{m,n=1}^{9}(\mathbf {G} _{m}\cdot \mathbf {G} _{n})\mathbf {G} ^{m}\otimes \mathbf {G} ^{n}}$

or with

${\displaystyle \mathbf {G} _{m}={\vec {a}}_{i}\otimes {\vec {g}}_{j},\quad i,j=1,2,3,\;m=3(i-1)+j}$

${\displaystyle \mathbf {G} ^{n}={\vec {a}}^{k}\otimes {\vec {g}}^{l},\quad k,l=1,2,3,\;n=3(k-1)+l}$

in the usual notation:

${\displaystyle {\begin{array}{rcl}{\stackrel {4}{\mathbf {1} }}&=&\displaystyle \sum _{i,j=1}^{3}{\vec {a}}_{i}\otimes {\vec {g}}_{j}\otimes {\vec {a}}^{i}\otimes {\vec {g}}^{j}=\sum _{i,j=1}^{3}{\vec {a}}^{i}\otimes {\vec {g}}^{j}\otimes {\vec {a}}_{i}\otimes {\vec {g}}_{j}\\&=&\displaystyle \sum _{i,j,k,l=1}^{3}({\vec {a}}^{i}\cdot {\vec {a}}^{k})({\vec {g}}^{j}\cdot {\vec {g}}^{l}){\vec {a}}_{i}\otimes {\vec {g}}_{j}\otimes {\vec {a}}_{k}\otimes {\vec {g}}_{l}\\&=&\displaystyle \sum _{i,j,k,l=1}^{3}({\vec {a}}_{i}\cdot {\vec {a}}_{k})({\vec {g}}_{j}\cdot {\vec {g}}_{l}){\vec {a}}^{i}\otimes {\vec {g}}^{j}\otimes {\vec {a}}^{k}\otimes {\vec {g}}^{l}\end{array}}}$.

example

The vectors

${\displaystyle {\vec {a}}_{1}={\begin{pmatrix}2\\3\\1\end{pmatrix}},\;{\vec {a}}_{2}={\begin{pmatrix}3\\-1\\2\end{pmatrix}},\;{\vec {a}}_{3}={\begin{pmatrix}-3\\0\\-2\end{pmatrix}}}$

form a base in and its dual base is ${\displaystyle \mathbb {V} ^{3}}$

${\displaystyle {\vec {a}}^{1}={\begin{pmatrix}2\\0\\-3\end{pmatrix}},\;{\vec {a}}^{2}={\begin{pmatrix}6\\-1\\-9\end{pmatrix}},\;{\vec {a}}^{3}={\begin{pmatrix}7\\-1\\-11\end{pmatrix}}}$.

So you get

${\displaystyle {\begin{array}{rcl}\mathbf {1} &=&\displaystyle \sum _{i=1}^{3}{\vec {a}}_{i}\otimes {\vec {a}}^{i}={\begin{pmatrix}2\\3\\1\end{pmatrix}}\otimes {\begin{pmatrix}2\\0\\-3\end{pmatrix}}+{\begin{pmatrix}3\\-1\\2\end{pmatrix}}\otimes {\begin{pmatrix}6\\-1\\-9\end{pmatrix}}+{\begin{pmatrix}-3\\0\\-2\end{pmatrix}}\otimes {\begin{pmatrix}7\\-1\\-11\end{pmatrix}}\\&=&{\begin{pmatrix}4&0&-6\\6&0&-9\\2&0&-3\end{pmatrix}}+{\begin{pmatrix}18&-3&-27\\-6&1&9\\12&-2&-18\end{pmatrix}}+{\begin{pmatrix}-21&3&33\\0&0&0\\-14&2&22\end{pmatrix}}={\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}}\end{array}}}$.

See also

• Identity matrix
• Metric tensor

literature

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