External tensor product

In mathematics, the outer tensor product is a special product of two dyads, which consist of two vectors linked with the dyadic product . For the outer tensor product, cross products of the vectors are formed, so that this tensor product is restricted to three-dimensional spaces. Because the cross product "×" occurs twice in the outer tensor product, it is written here with the symbol "#". With the outer tensor product, the main invariants, the cofactor and the adjuncts of a tensor can be elegantly expressed and the cross product of vectors transformed with a tensor can be given. The designation “outer tensor product” is derived from the second name “outer product” of the cross product of vectors. Sometimes the dyadic product of tensors is also referred to as the “outer tensor product”. The naming here follows W. Ehlers.

definition

Let four vectors be given from the three-dimensional Euclidean vector space . Then the outer tensor product "#" with the dyadic product " " is defined via: ${\displaystyle {\vec {a}},\,{\vec {b}},\,{\vec {g}},\,{\vec {h}}\in \mathbb {V} ^{3}}$ ${\displaystyle \mathbb {V} ^{3}}$ ${\displaystyle \otimes }$

${\displaystyle ({\vec {a}}\otimes {\vec {g}})\#({\vec {b}}\otimes {\vec {h}}):=({\vec {a}}\times {\vec {b}})\otimes ({\vec {g}}\times {\vec {h}})}$

Second order tensors are sums of dyads. Be and vector space bases . Then every second order tensor can be A as a sum ${\displaystyle {\vec {a}}_{1,2,3},\,{\vec {b}}_{1,2,3},\,{\vec {g}}_{1,2,3}}$ ${\displaystyle {\vec {h}}_{1,2,3}}$

${\displaystyle \mathbf {A} =A^{ij}{\vec {a}}_{i}\otimes {\vec {g}}_{j}=A^{*ij}{\vec {b}}_{i}\otimes {\vec {h}}_{j}}$

with components A ^ij and A ^{* ij} to be determined . In this equation as well as in the following, Einstein's summation convention is to be used, according to which all indices occurring twice in a product, here i and j, are to be summed from one to three. The outer tensor product of two second order tensors is then:

${\displaystyle \mathbf {A} \#\mathbf {B} =\left(A^{ij}{\vec {a}}_{i}\otimes {\vec {g}}_{j}\right)\#\left(B^{kl}{\vec {b}}_{k}\otimes {\vec {h}}_{l}\right)=A^{ij}B^{kl}\left({\vec {a}}_{i}\times {\vec {b}}_{k}\right)\otimes \left({\vec {g}}_{j}\times {\vec {h}}_{l}\right)}$

Coordinate-free representation

Without reference to dyads, the outer tensor product of two tensors A and B can be written symbolically with the unit tensor 1 as

${\displaystyle {\begin{aligned}\mathbf {A} \#\mathbf {B} =&[\operatorname {Sp} (\mathbf {A} )\operatorname {Sp} (\mathbf {B} )-\operatorname {Sp} (\mathbf {A\cdot B} )]\mathbf {1} \\&+[\mathbf {A\cdot B} +\mathbf {B\cdot A} -\operatorname {Sp} (\mathbf {A} )\mathbf {B} -\operatorname {Sp} (\mathbf {B} )\mathbf {A} ]^{\top }\end{aligned}}}$

Because if these tensors are noted in relation to the standard basis ê _1,2,3 , for example, then with the Levi-Civita symbol : ${\displaystyle \mathbf {A} =A_{ij}{\hat {e}}_{i}\otimes {\hat {e}}_{j}}$ ${\displaystyle \varepsilon _{ijk}:=({\hat {e}}_{i}\times {\hat {e}}_{j})\cdot {\hat {e}}_{k}}$

${\displaystyle {\begin{aligned}\mathbf {A} \#\mathbf {B} =&(A_{ij}{\hat {e}}_{i}\otimes {\hat {e}}_{j})\#(B_{kl}{\hat {e}}_{k}\otimes {\hat {e}}_{l})\\=&A_{ij}B_{kl}({\hat {e}}_{i}\times {\hat {e}}_{k})\otimes ({\hat {e}}_{j}\times {\hat {e}}_{l})\\=&\varepsilon _{ikm}\varepsilon _{jln}A_{ij}B_{kl}{\hat {e}}_{m}\otimes {\hat {e}}_{n}\end{aligned}}}$

The product of two Levi-Civita symbols depends on the determinant

${\displaystyle {\begin{aligned}\varepsilon _{ikm}\varepsilon _{jln}=&{\begin{vmatrix}\delta _{ij}&\delta _{il}&\delta _{in}\\\delta _{kj}&\delta _{kl}&\delta _{kn}\\\delta _{mj}&\delta _{ml}&\delta _{mn}\end{vmatrix}}\\=&\delta _{ij}\delta _{kl}\delta _{mn}+\delta _{il}\delta _{kn}\delta _{mj}+\delta _{in}\delta _{kj}\delta _{ml}\\&-\delta _{ij}\delta _{kn}\delta _{ml}-\delta _{il}\delta _{kj}\delta _{mn}-\delta _{in}\delta _{kl}\delta _{mj}\end{aligned}}}$

with the Kronecker delta δ _ij . This results in:

${\displaystyle {\begin{array}{rcl}\mathbf {A} \#\mathbf {B} &=&(\delta _{ij}\delta _{kl}\delta _{mn}+\delta _{il}\delta _{kn}\delta _{mj}+\delta _{in}\delta _{kj}\delta _{ml}\\&&-\delta _{ij}\delta _{kn}\delta _{ml}-\delta _{il}\delta _{kj}\delta _{mn}-\delta _{in}\delta _{kl}\delta _{mj})A_{ij}B_{kl}{\hat {e}}_{m}\otimes {\hat {e}}_{n}\\&=&A_{ii}B_{kk}{\hat {e}}_{m}\otimes {\hat {e}}_{m}+A_{ij}B_{ki}{\hat {e}}_{j}\otimes {\hat {e}}_{k}+A_{ij}B_{jl}{\hat {e}}_{l}\otimes {\hat {e}}_{i}\\&&-A_{ii}B_{kl}{\hat {e}}_{l}\otimes {\hat {e}}_{k}-A_{ij}B_{ji}{\hat {e}}_{m}\otimes {\hat {e}}_{m}-A_{ij}B_{kk}{\hat {e}}_{j}\otimes {\hat {e}}_{i}\\&=&\operatorname {Sp} (\mathbf {A} )\operatorname {Sp} (\mathbf {B} )\mathbf {1} +\mathbf {A^{\top }\cdot B^{\top }} +\mathbf {B^{\top }\cdot A^{\top }} \\&&-\operatorname {Sp} (\mathbf {A} )\mathbf {B} ^{\top }-\operatorname {Sp} (\mathbf {A\cdot B} )\mathbf {1} -\operatorname {Sp} (\mathbf {B} )\mathbf {A} ^{\top }\end{array}}}$

what corresponds to the identity given at the beginning.

properties

The coordinate-free display shows:

${\displaystyle {\begin{array}{rcl}\mathbf {1} \#\mathbf {1} &=&2\,\mathbf {1} \\\mathbf {A} \#\mathbf {1} &=&\operatorname {Sp} (\mathbf {A} )\mathbf {1} -\mathbf {A} ^{\top }\\(\mathbf {A} \#\mathbf {B} )^{\top }&=&(\mathbf {A} ^{\top })\#(\mathbf {B} ^{\top })\end{array}}}$

Associativity

The outer tensor product is not associative :

${\displaystyle \mathbf {A} \#(\mathbf {B} \#\mathbf {C} )\neq (\mathbf {A} \#\mathbf {B} )\#\mathbf {C} }$

as the example B = C = 1 shows:

${\displaystyle {\begin{aligned}\mathbf {A} \#(\mathbf {1} \#\mathbf {1} )=&\mathbf {A} \#2\mathbf {1} =2\operatorname {Sp} (\mathbf {A} )\mathbf {1} -2\mathbf {A} ^{\top }\\(\mathbf {A} \#\mathbf {1} )\#\mathbf {1} =&[\operatorname {Sp} (\mathbf {A} )\mathbf {1} -\mathbf {A} ^{\top }]\#\mathbf {1} \\=&2\operatorname {Sp} (\mathbf {A} )\mathbf {1} -[\operatorname {Sp} (\mathbf {A} ^{\top })\operatorname {Sp} (\mathbf {1} )-\operatorname {Sp} (\mathbf {A^{\top }\cdot 1} )]\mathbf {1} \\&-[\mathbf {A^{\top }\cdot 1} +\mathbf {1\cdot A^{\top }} -\operatorname {Sp} (\mathbf {A} ^{\top })\mathbf {1} -\operatorname {Sp} (\mathbf {1} )\mathbf {A} ^{\top }]^{\top }\\=&\operatorname {Sp} (\mathbf {A} )\mathbf {1} +\mathbf {A} \end{aligned}}}$

Commutativity

The outer tensor product is commutative :

${\displaystyle \mathbf {A} \#\mathbf {B} =\mathbf {B} \#\mathbf {A} }$

as can be seen from the coordinate-free representation.

Distributive law

The outer tensor product is distributive over addition and subtraction:

${\displaystyle {\begin{aligned}\mathbf {A} \#(\mathbf {B+C} )=&\mathbf {A} \#\mathbf {B} +\mathbf {A} \#\mathbf {C} \\\mathbf {(A+B)} \#\mathbf {C} =&\mathbf {A} \#\mathbf {C} +\mathbf {B} \#\mathbf {C} \end{aligned}}}$

what can be proven in the coordinate-free representation.

Connection with the double cross product of tensors

H. Altenbach defines the double cross product of dyads as

${\displaystyle ({\vec {a}}\otimes {\vec {g}})\times \times ({\vec {b}}\otimes {\vec {h}}):=({\vec {g}}\times {\vec {b}})\otimes ({\vec {a}}\times {\vec {h}})=({\vec {g}}\otimes {\vec {a}})\#({\vec {b}}\otimes {\vec {h}})\,,}$

which differs from the outer tensor product only by the transposition of the first factor.

Isotropy

The outer tensor product of two tensors can be interpreted as a function of these tensors:

${\displaystyle \mathbf {f} (\mathbf {A,B} ):=\mathbf {A} \#\mathbf {B} }$

Let an arbitrary orthogonal tensor Q be given , for which the identity applies. Then applies ${\displaystyle \mathbf {Q^{\top }\cdot Q} =\mathbf {1} }$

${\displaystyle {\begin{array}{rcl}&&\mathbf {f} (\mathbf {Q\cdot A\cdot Q^{\top },Q\cdot B\cdot Q^{\top }} )\\&=&(\mathbf {Q\cdot A\cdot Q^{\top }} )\#(\mathbf {Q\cdot B\cdot Q^{\top }} )\\&=&[\operatorname {Sp} (\mathbf {Q\cdot A\cdot Q^{\top }} )\operatorname {Sp} (\mathbf {Q\cdot B\cdot Q^{\top }} )-\operatorname {Sp} ((\mathbf {Q\cdot A\cdot Q^{\top }} )\cdot (\mathbf {Q\cdot B\cdot Q^{\top }} ))]\mathbf {1} \\&&+(\mathbf {Q\cdot A\cdot Q^{\top }} )^{\top }\cdot (\mathbf {Q\cdot B\cdot Q^{\top }} )^{\top }+(\mathbf {Q\cdot B\cdot Q^{\top }} )^{\top }\cdot (\mathbf {Q\cdot A\cdot Q^{\top }} )^{\top }\\&&-\operatorname {Sp} (\mathbf {Q\cdot A\cdot Q^{\top }} )(\mathbf {Q\cdot B\cdot Q^{\top }} )^{\top }-\operatorname {Sp} (\mathbf {Q\cdot B\cdot Q^{\top }} )(\mathbf {Q\cdot A\cdot Q^{\top }} )^{\top }\\&=&\mathbf {Q} \cdot {\bigl \{}[\operatorname {Sp} (\mathbf {A} )\operatorname {Sp} (\mathbf {B} )-\operatorname {Sp} (\mathbf {A\cdot B} )]\mathbf {1} +\mathbf {A^{\top }\cdot B^{\top }} +\mathbf {B^{\top }\cdot A^{\top }} \\&&\quad \quad \quad -\operatorname {Sp} (\mathbf {A} )\mathbf {B} ^{\top }-\operatorname {Sp} (\mathbf {B} )\mathbf {A} ^{\top }{\bigr \}}\cdot \mathbf {Q} ^{\top }\\&=&\mathbf {Q} \cdot \mathbf {f} (\mathbf {A,B} )\cdot \mathbf {Q} ^{\top }\end{array}}}$

The outer tensor product is therefore an isotropic tensor function .

Dot product with a third tensor

Forming the Frobenius scalar product ":" of the outer tensor product A # B with a third tensor C yields:

${\displaystyle {\begin{array}{rcl}(\mathbf {A} \#\mathbf {B} ):\mathbf {C} &=&[\operatorname {Sp} (\mathbf {A} )\operatorname {Sp} (\mathbf {B} )\mathbf {1} +\mathbf {A^{\top }\cdot B^{\top }} +\mathbf {B^{\top }\cdot A^{\top }} \\&&-\operatorname {Sp} (\mathbf {A} )\mathbf {B} ^{\top }-\operatorname {Sp} (\mathbf {A\cdot B} )\mathbf {1} -\operatorname {Sp} (\mathbf {B} )\mathbf {A} ^{\top }]:\mathbf {C} \\&=&\operatorname {Sp} (\mathbf {A} )\operatorname {Sp} (\mathbf {B} )\operatorname {Sp} (\mathbf {C} )+\operatorname {Sp} (\mathbf {B\cdot A\cdot C} )+\operatorname {Sp} (\mathbf {A\cdot B\cdot C} )\\&&-\operatorname {Sp} (\mathbf {A} )\operatorname {Sp} (\mathbf {B\cdot C} )-\operatorname {Sp} (\mathbf {C} )\operatorname {Sp} (\mathbf {A\cdot B} )-\operatorname {Sp} (\mathbf {B} )\operatorname {Sp} (\mathbf {C\cdot A} )\end{array}}}$

From this is the cyclical interchangeability

${\displaystyle (\mathbf {A} \#\mathbf {B} ):\mathbf {C} =(\mathbf {B} \#\mathbf {C} ):\mathbf {A} =(\mathbf {C} \#\mathbf {A} ):\mathbf {B} }$

readable.

Relationship with the main invariants

→ Main article : Main invariant

From and the cyclical interchangeability of the factors in the product follows: ${\displaystyle \mathbf {T} \#\mathbf {1} =\operatorname {Sp} (\mathbf {T} )\mathbf {1} -\mathbf {T} ^{\top }}$ ${\displaystyle (\mathbf {A} \#\mathbf {B} ):\mathbf {C} }$

${\displaystyle {\begin{array}{rclcl}(\mathbf {T} \#\mathbf {1} ):\mathbf {1} &=&3\operatorname {Sp} (\mathbf {T} )-\operatorname {Sp} (\mathbf {T} )&=&2\operatorname {I} _{1}(\mathbf {T} )\\(\mathbf {T} \#\mathbf {T} ):\mathbf {1} &=&(\mathbf {T} \#\mathbf {1} ):\mathbf {T} =\operatorname {Sp} (\mathbf {T} )^{2}-\operatorname {Sp} (\mathbf {T\cdot T} )&=&2\operatorname {I} _{2}(\mathbf {T} )\\(\mathbf {T} \#\mathbf {T} ):\mathbf {T} &=&\operatorname {Sp} (\mathbf {T} )^{3}+2\operatorname {Sp} (\mathbf {T} ^{3})-3\operatorname {Sp} (\mathbf {T} )\operatorname {Sp} (\mathbf {T} ^{2})&=&6\operatorname {I} _{3}(\mathbf {T} )\end{array}}}$

The functions I _1,2,3 are the three principal invariants of the tensor T .

Calculation of the cofactor and the adjuncts

The cofactor of an invertible tensor is the tensor , which according to Cayley-Hamilton's theorem ${\displaystyle \operatorname {cof} (\mathbf {T} )=\operatorname {det} (\mathbf {T} )\mathbf {T} ^{\top -1}}$

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

reads. The latter identity also applies to non-invertible tensors. The outer tensor product of a tensor with itself provides the double cofactor:

${\displaystyle {\begin{aligned}\mathbf {T} \#\mathbf {T} =&[\operatorname {Sp} (\mathbf {T} )^{2}-\operatorname {Sp} (\mathbf {T\cdot T} )]\mathbf {1} +2\mathbf {T^{\top }\cdot T^{\top }} -\operatorname {Sp} (\mathbf {T} )\mathbf {T} ^{\top }-\operatorname {Sp} (\mathbf {T} )\mathbf {T} ^{\top }\\=&2\operatorname {cof} (\mathbf {T} )\end{aligned}}}$

The adjunct is the transposed cofactor:

${\displaystyle \operatorname {adj} (\mathbf {T} )=\operatorname {cof} (\mathbf {T} )^{\top }={\frac {1}{2}}(\mathbf {T} \#\mathbf {T} )^{\top }={\frac {1}{2}}\left(\mathbf {T} ^{\top }\right)\#\left(\mathbf {T} ^{\top }\right)}$

Tensor product of two outer products

With ${\displaystyle \varepsilon _{pqu}\varepsilon _{stu}=\varepsilon _{upq}\varepsilon _{ust}=\delta _{ps}\delta _{qt}-\delta _{pt}\delta _{qs}}$

can

${\displaystyle {\begin{array}{rcl}(\mathbf {A} \#\mathbf {B} )\cdot (\mathbf {C} \#\mathbf {D} )&=&\varepsilon _{ikm}\varepsilon _{pqu}A_{ip}B_{kq}{\hat {e}}_{m}\otimes {\hat {e}}_{u}\cdot \varepsilon _{stv}\varepsilon _{jln}C_{sj}D_{tl}{\hat {e}}_{v}\otimes {\hat {e}}_{n}\\&=&\varepsilon _{ikm}\varepsilon _{pqu}\varepsilon _{stu}\varepsilon _{jln}A_{ip}B_{kq}C_{sj}D_{tl}{\hat {e}}_{m}\otimes {\hat {e}}_{n}\\&=&\varepsilon _{ikm}(\delta _{ps}\delta _{qt}-\delta _{qs}\delta _{pt})\varepsilon _{jln}A_{ip}B_{kq}C_{sj}D_{tl}{\hat {e}}_{m}\otimes {\hat {e}}_{n}\\&=&\varepsilon _{ikm}\varepsilon _{jln}A_{ip}C_{pj}B_{kq}D_{ql}{\hat {e}}_{m}\otimes {\hat {e}}_{n}+\varepsilon _{ikm}\varepsilon _{ljn}A_{ip}D_{pl}B_{kq}C_{qj}{\hat {e}}_{m}\otimes {\hat {e}}_{n}\\\rightarrow (\mathbf {A} \#\mathbf {B} )\cdot (\mathbf {C} \#\mathbf {D} )&=&(\mathbf {A\cdot C} )\#(\mathbf {B\cdot D} )+(\mathbf {A\cdot D} )\#(\mathbf {B\cdot C} )\end{array}}}$

can be calculated.

Transformation properties

Cross product

With the help of the outer tensor product, tensors can be "excluded" from the cross product of two vectors:

${\displaystyle (\mathbf {A} \cdot {\vec {u}})\times (\mathbf {A} \cdot {\vec {v}})={\frac {1}{2}}\mathbf {A} \#\mathbf {A} \cdot ({\vec {u}}\times {\vec {v}})=\operatorname {cof} (\mathbf {A} )\cdot ({\vec {u}}\times {\vec {v}})}$

This result is used when calculating the contents of deformed surfaces .

For proof, the cross product of two vectors with components with respect to the standard basis is represented by the Levi-Civita symbol:

${\displaystyle {\vec {u}}\times {\vec {v}}=u_{p}{\hat {e}}_{p}\times v_{q}{\hat {e}}_{q}=\varepsilon _{pqr}u_{p}v_{q}{\hat {e}}_{r}}$

Applying the outer tensor product of two tensors to this product yields:

${\displaystyle (\mathbf {A} \#\mathbf {B} )\cdot ({\vec {u}}\times {\vec {v}})=\varepsilon _{ikm}\varepsilon _{jln}A_{ij}B_{kl}({\hat {e}}_{m}\otimes {\hat {e}}_{n})\cdot \varepsilon _{pqr}u_{p}v_{q}{\hat {e}}_{r}=\varepsilon _{ikm}\varepsilon _{pqn}\varepsilon _{jln}A_{ij}B_{kl}u_{p}v_{q}{\hat {e}}_{m}}$

Now is

${\displaystyle {\begin{array}{rcl}(\mathbf {A} \cdot {\vec {u}})\times (\mathbf {B} \cdot {\vec {v}})-(\mathbf {A} \cdot {\vec {v}})\times (\mathbf {B} \cdot {\vec {u}})&=&A_{ij}u_{j}{\hat {e}}_{i}\times B_{kl}v_{l}{\hat {e}}_{k}-A_{ij}v_{j}{\hat {e}}_{i}\times B_{kl}u_{l}{\hat {e}}_{k}\\&=&\varepsilon _{ikm}A_{ij}B_{kl}u_{j}v_{l}{\hat {e}}_{m}-\varepsilon _{ikm}A_{ij}B_{kl}u_{l}v_{j}{\hat {e}}_{m}\\&=&\varepsilon _{ikm}(\delta _{jp}\delta _{lq}-\delta _{lp}\delta _{jq})A_{ij}B_{kl}u_{p}v_{q}{\hat {e}}_{m}\\&=&\varepsilon _{ikm}\varepsilon _{pqn}\varepsilon _{jln}A_{ij}B_{kl}u_{p}v_{q}{\hat {e}}_{m}\\\rightarrow (\mathbf {A} \cdot {\vec {u}})\times (\mathbf {B} \cdot {\vec {v}})-(\mathbf {A} \cdot {\vec {v}})\times (\mathbf {B} \cdot {\vec {u}})&=&(\mathbf {A} \#\mathbf {B} )\cdot ({\vec {u}}\times {\vec {v}})\end{array}}}$

In the equation chain has been used. Specifically, with B = A, the relationship listed above is calculated . ${\displaystyle \varepsilon _{pqn}\varepsilon _{jln}=\varepsilon _{npq}\varepsilon _{njl}=\delta _{jp}\delta _{lq}-\delta _{lp}\delta _{jq}}$

Late product

In components related to the standard basis is calculated

${\displaystyle {\begin{array}{rcl}(\mathbf {A} \#\mathbf {B} ):\mathbf {C} &=&(A_{ij}{\hat {e}}_{i}\otimes {\hat {e}}_{j})\#(B_{kl}{\hat {e}}_{k}\otimes {\hat {e}}_{l}):C_{pq}{\hat {e}}_{p}\otimes {\hat {e}}_{q}=\varepsilon _{ikm}\varepsilon _{jln}A_{ij}B_{kl}C_{mn}\\&=&\varepsilon _{jln}(A_{ij}{\hat {e}}_{i}\times B_{kl}{\hat {e}}_{k})\cdot C_{mn}{\hat {e}}_{m}=\varepsilon _{jln}[(\mathbf {A} \cdot {\hat {e}}_{j})\times (\mathbf {B} \cdot {\hat {e}}_{l})]\cdot (\mathbf {C} \cdot {\hat {e}}_{n})\\\rightarrow (\mathbf {A} \#\mathbf {B} ):\mathbf {C} &=&\;\;\;[(\mathbf {A} \cdot {\hat {e}}_{1})\times (\mathbf {B} \cdot {\hat {e}}_{2})]\cdot (\mathbf {C} \cdot {\hat {e}}_{3})+[(\mathbf {A} \cdot {\hat {e}}_{2})\times (\mathbf {B} \cdot {\hat {e}}_{3})]\cdot (\mathbf {C} \cdot {\hat {e}}_{1})+[(\mathbf {A} \cdot {\hat {e}}_{3})\times (\mathbf {B} \cdot {\hat {e}}_{1})]\cdot (\mathbf {C} \cdot {\hat {e}}_{2})\\&&-[(\mathbf {A} \cdot {\hat {e}}_{2})\times (\mathbf {B} \cdot {\hat {e}}_{1})]\cdot (\mathbf {C} \cdot {\hat {e}}_{3})-[(\mathbf {A} \cdot {\hat {e}}_{3})\times (\mathbf {B} \cdot {\hat {e}}_{2})]\cdot (\mathbf {C} \cdot {\hat {e}}_{1})-[(\mathbf {A} \cdot {\hat {e}}_{1})\times (\mathbf {B} \cdot {\hat {e}}_{3})]\cdot (\mathbf {C} \cdot {\hat {e}}_{2})\end{array}}}$

Instead of the standard basis , any other orthonormal basis can also be used here.

See also

Formula collection tensor algebra

Individual evidence

1. ^ W. Ehlers: Supplement to the lectures, Technical Mechanics and Higher Mechanics . 2014, p. 24 f . ( uni-stuttgart.de [PDF; accessed on February 28, 2015]).
2. ^ H. Altenbach: Continuum Mechanics . Springer, 2012, ISBN 978-3-642-24118-5 , pp. 32 .