Main invariant

The main invariants of a tensor are the coefficients of its characteristic polynomial .

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. In the same way, the main invariants of the tensor are invariant to a change in base, hence the name.

The main invariants of symmetric tensors play a central role in materials theory. An important requirement for material models is derived from the fact that a moving observer always measures the same material behavior as a stationary one. This property is called material objectivity . The movement of an observer is described mathematically as a change in the reference system and thus as a change in the vector space basis. The main invariants are therefore quantities that all observers perceive in the same way and that are therefore suitable for material modeling. Examples of material models that use the main invariants are Hooke's law , hyperelasticity, and plasticity theory .

The representation is made in three dimensions for second-order tensors, but can be generalized to more dimensions in a simple manner.

definition

A second order tensor is given . Then its characteristic polynomial is: ${\displaystyle \mathbf {T} }$

${\displaystyle p(\lambda ):=\mathrm {det} (\mathbf {T} -\lambda \mathbf {1} )=-\lambda ^{3}+\mathrm {I} _{1}\lambda ^{2}-\mathrm {I} _{2}\lambda +\mathrm {I} _{3}}$.

Therein the determinant , 1 is the unit tensor , a real or complex number and the coefficients are the three main invariants ${\displaystyle \mathrm {det} }$ ${\displaystyle \lambda }$ ${\displaystyle \mathrm {I} _{1,2,3}}$

${\displaystyle {\begin{aligned}\mathrm {I} _{1}:=&\mathrm {Sp} (\mathbf {T} )\\\mathrm {I} _{2}:=&{\frac {1}{2}}[\mathrm {Sp} {(\mathbf {T} )}^{2}-\mathrm {Sp} (\mathbf {T} \cdot \mathbf {T} )]=\mathrm {Sp(adj} (\mathbf {T} ))=\mathrm {Sp(cof} (\mathbf {T} ))\\\mathrm {I} _{3}:=&\mathrm {det} (\mathbf {T} )\end{aligned}}}$

The operator provides the trace of its argument, is the adjunct and the cofactor ${\displaystyle \mathrm {Sp} }$ ${\displaystyle \mathrm {adj} (\mathbf {T} )}$ ${\displaystyle \mathrm {cof} (\mathbf {T} )}$

${\displaystyle \mathrm {adj} (\mathbf {T} ):=\mathrm {cof} (\mathbf {T} )^{\top }:=\mathbf {T\cdot T} -\mathrm {I} _{1}\mathbf {T} +\mathrm {I} _{2}\mathbf {1} =\mathrm {det} (\mathbf {T} )\mathbf {T} ^{-1}}$

where the latter identity is only valid if the tensor is invertible and therefore is. ${\displaystyle \mathrm {det} (\mathbf {T} )\neq 0}$

Calculation of the main invariants

For tensors of the second level, the addition and multiplication is defined with a scalar, which is why the set of all tensors of the second level forms a vector space that has vector space bases that consist of dyads, which in turn are calculated with the dyadic product of two vectors. Let be the vector space of the geometric vectors. Then the vector space is the second order tensors that map vectors from into the . With regard to a vector space basis des , each tensor can be represented component-wise and from these components the main invariants can be calculated, which are of course independent of the choice of basis. ${\displaystyle \otimes }$ ${\displaystyle \mathbb {V} ^{3}}$ ${\displaystyle {\text{Lin}}(\mathbb {V} ^{3},\mathbb {V} ^{3})}$ ${\displaystyle \mathbb {V} ^{3}}$ ${\displaystyle \mathbb {V} ^{3}}$ ${\displaystyle {\text{Lin}}(\mathbb {V} ^{3},\mathbb {V} ^{3})}$

Main invariants in components with respect to the standard basis

Be the standard basis of and ${\displaystyle {\hat {e}}_{1,2,3}}$ ${\displaystyle \mathbb {V} ^{3}}$

${\displaystyle \mathbf {T} =\sum _{i,j=1}^{3}T_{ij}{\hat {e}}_{i}\otimes {\hat {e}}_{j}={\begin{pmatrix}T_{11}&T_{12}&T_{13}\\T_{21}&T_{22}&T_{23}\\T_{31}&T_{32}&T_{33}\end{pmatrix}}}$

a tensor with the components with respect to this standard basis. Then it is calculated ${\displaystyle T_{ij}}$

${\displaystyle {\begin{aligned}\mathrm {I} _{1}(\mathbf {T} )=&T_{11}+T_{22}+T_{33}\\\mathrm {I} _{2}(\mathbf {T} )=&T_{11}T_{22}+T_{11}T_{33}+T_{22}T_{33}-T_{12}T_{21}-T_{13}T_{31}-T_{23}T_{32}\\\mathrm {I} _{3}(\mathbf {T} )=&T_{11}(T_{22}T_{33}-T_{23}T_{32})+T_{12}(T_{23}T_{31}-T_{21}T_{33})\\&+T_{13}(T_{21}T_{32}-T_{22}T_{31})\end{aligned}}}$

Major invariants in components with respect to a general basis

Let and be any two basic systems of and ${\displaystyle {\vec {a}}_{1,2,3}}$ ${\displaystyle {\vec {b}}_{1,2,3}}$ ${\displaystyle \mathbb {V} ^{3}}$

${\displaystyle \mathbf {T} =\sum _{i,j=1}^{3}T^{ij}{\vec {a}}_{i}\otimes {\vec {b}}_{j}={\begin{pmatrix}T^{11}&T^{12}&T^{13}\\T^{21}&T^{22}&T^{23}\\T^{31}&T^{32}&T^{33}\end{pmatrix}}_{{\vec {a}}_{i}\otimes {\vec {b}}_{j}}}$

a tensor with the components related to these bases. Then it is calculated ${\displaystyle T^{ij}}$

${\displaystyle \mathrm {I} _{1}(\mathbf {T} )=\sum _{i,j=1}^{3}T^{ij}{\vec {a}}_{i}\cdot {\vec {b}}_{j}}$

${\displaystyle \mathrm {I} _{2}(\mathbf {T} )={\frac {1}{2}}\sum _{i,j,k,l=1}^{3}T^{ij}T^{kl}[({\vec {a}}_{i}\cdot {\vec {b}}_{j})({\vec {a}}_{k}\cdot {\vec {b}}_{l})-({\vec {a}}_{i}\cdot {\vec {b}}_{l})({\vec {a}}_{k}\cdot {\vec {b}}_{j})]}$

${\displaystyle \mathrm {I} _{3}(\mathbf {T} )={\begin{vmatrix}T^{11}&T^{12}&T^{13}\\T^{21}&T^{22}&T^{23}\\T^{31}&T^{32}&T^{33}\end{vmatrix}}{\begin{vmatrix}{\vec {a}}_{1}&{\vec {a}}_{2}&{\vec {a}}_{3}\end{vmatrix}}{\begin{vmatrix}{\vec {b}}_{1}&{\vec {b}}_{2}&{\vec {b}}_{3}\end{vmatrix}}}$

where the last two determinants correspond to the late products of the basis vectors.

Relation to the outer tensor product

The outer tensor product # is defined by means of dyads via

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

With this and the Frobenius scalar product " " of tensors the three main invariants get the representations ${\displaystyle :}$

${\displaystyle {\begin{aligned}\mathrm {I} _{1}(\mathbf {T} )&={\frac {1}{2}}(\mathbf {T} \#\mathbf {1} ):\mathbf {1} \;,\\\mathrm {I} _{2}(\mathbf {T} )&={\frac {1}{2}}(\mathbf {T} \#\mathbf {T} ):\mathbf {1} \;,\\\mathrm {I} _{3}(\mathbf {T} )&={\frac {1}{6}}(\mathbf {T} \#\mathbf {T} ):\mathbf {T} \;.\end{aligned}}}$

Connection with other invariants

Eigenvalues

The eigenvalues ​​of a second order tensor are the solutions of its characteristic polynomial and also invariants. According to Vieta's theorem: ${\displaystyle \lambda _{1,2,3}}$ ${\displaystyle p(\lambda )=0}$

${\displaystyle {\begin{aligned}\mathrm {I} _{1}=&\lambda _{1}+\lambda _{2}+\lambda _{3}\\\mathrm {I} _{2}=&\lambda _{1}\lambda _{2}+\lambda _{2}\lambda _{3}+\lambda _{3}\lambda _{1}\\\mathrm {I} _{3}=&\lambda _{1}\lambda _{2}\lambda _{3}\end{aligned}}}$.

Amount of a tensor

The magnitude of a tensor

${\displaystyle \left\|\mathbf {T} \right\|:={\sqrt {\mathbf {T} :\mathbf {T} }}:={\sqrt {\mathrm {Sp} (\mathbf {T} ^{\top }\cdot \mathbf {T} )}}}$,

defined with the Frobenius norm and the Frobenius scalar product “:”, can generally not be represented with the three main invariants. But it works with symmetrical or skew-symmetrical tensors. ${\displaystyle \left\|(\cdot )\right\|}$

For symmetric tensors , i. H. the tensor is identical to its transposed one , and therefore ${\displaystyle \mathbf {T} =\mathbf {T} ^{\top }}$

${\displaystyle {\begin{aligned}\mathrm {I} _{2}&={\frac {1}{2}}\left[\mathrm {Sp} {(\mathbf {T} )}^{2}-\mathrm {Sp} (\mathbf {T} \cdot \mathbf {T} )\right]={\frac {1}{2}}\left[\mathrm {I} _{1}^{2}-\mathrm {Sp} (\mathbf {T} ^{\top }\cdot \mathbf {T} )\right]={\frac {1}{2}}\left[\mathrm {I} _{1}^{2}-{\left\|\mathbf {T} \right\|}^{2}\right]\\\Rightarrow \quad \left\|\mathbf {T} \right\|&={\sqrt {\mathrm {I} _{1}^{2}-2\,\mathrm {I} _{2}}}\;.\end{aligned}}}$

For skew-symmetric tensors, and therefore and ${\displaystyle \mathbf {T} =-\mathbf {T} ^{\top }}$ ${\displaystyle \mathrm {Sp} (\mathbf {T} )=0}$

${\displaystyle {\begin{aligned}\mathrm {I} _{2}&={\frac {1}{2}}\left[\mathrm {Sp} {(\mathbf {T} )}^{2}-\mathrm {Sp} (\mathbf {T} \cdot \mathbf {T} )\right]={\frac {1}{2}}\mathrm {Sp} (\mathbf {T} ^{\top }\cdot \mathbf {T} )={\frac {1}{2}}{\left\|\mathbf {T} \right\|}^{2}\\\Rightarrow \quad \left\|\mathbf {T} \right\|&={\sqrt {2\,\mathrm {I} _{2}}}\;.\end{aligned}}}$

Traces of the powers of a tensor

The three main invariants can also be represented with the traces of the powers of a tensor, which are also invariants. Be

${\displaystyle \mathrm {J} _{n}:=\mathrm {Sp} (\mathbf {T} ^{n}):=\mathrm {Sp} (\underbrace {\mathbf {T} \cdot \mathbf {T} \cdot \ldots \cdot \mathbf {T} } _{n{\text{-mal}}}),\quad n=1,2,3,\ldots \;,}$

then applies

${\displaystyle {\begin{aligned}\mathrm {I} _{1}&={\mathrm {J} }_{1}\;,\\\mathrm {I} _{2}&={\frac {1}{2}}({\mathrm {J} }_{1}^{2}-{\mathrm {J} }_{2})\;,\\\mathrm {I} _{3}&={\frac {1}{6}}(2{\mathrm {J} }_{3}+{\mathrm {J} }_{1}^{3}-3{\mathrm {J} }_{1}{\mathrm {J} }_{2})={\frac {1}{3}}(\mathrm {J} _{3}+3\mathrm {I} _{1}\mathrm {I} _{2}-\mathrm {I} _{1}^{3})\;.\end{aligned}}}$

Derivatives of the main invariants

In hyperelasticity , the deformation energy that has to be applied to deform a body is sometimes modeled as a function of the main invariants of the strain tensor . The stresses then result from the derivation of the deformation energy according to the strain tensor, for which the derivatives of the main invariants according to the strain tensor are required. It is therefore worthwhile to provide these derivations.

The derivative of a scalar-valued function according to the tensor is the tensor for which applies ${\displaystyle f(\mathbf {T} )}$ ${\displaystyle \mathbf {T} }$ ${\displaystyle \mathbf {A} }$

${\displaystyle \mathbf {A} :\mathbf {H} =\left.{\frac {\mathrm {d} f(\mathbf {T} +s\mathbf {H} )}{\mathrm {d} s}}\right|_{s=0}\quad \forall \;\mathbf {H} \in {\text{Lin}}(\mathbb {V} ^{3},\mathbb {V} ^{3})}$

You then also write

${\displaystyle {\frac {\mathrm {d} f}{\mathrm {d} \mathbf {T} }}=\mathbf {A} }$.

This is how it is calculated:

${\displaystyle {\begin{aligned}\left.{\frac {\mathrm {d} \mathrm {I} _{1}(\mathbf {T} +s\mathbf {H} )}{\mathrm {d} s}}\right|_{s=0}=&\mathrm {I} _{1}(\mathbf {H} )=\mathbf {1} :\mathbf {H} \rightarrow {\frac {\mathrm {d} \mathrm {I} _{1}(\mathbf {T} )}{\mathrm {d} \mathbf {T} }}=\mathbf {1} \\\left.{\frac {\mathrm {d} \mathrm {I} _{2}(\mathbf {T} +s\mathbf {H} )}{\mathrm {d} s}}\right|_{s=0}=&\left.{\frac {\mathrm {d} }{\mathrm {d} s}}{\frac {1}{2}}[\mathrm {I} _{1}(\mathbf {T} +s\mathbf {H} )^{2}-\mathrm {I} _{1}((\mathbf {T} +s\mathbf {H} )^{2})]\right|_{s=0}\\=&\left.{\frac {1}{2}}[2\mathrm {I} _{1}(\mathbf {T} +s\mathbf {H} )\mathrm {I} _{1}(\mathbf {H} )-2\mathrm {I} _{1}((\mathbf {T} +s\mathbf {H} )\cdot \mathbf {H} )]\right|_{s=0}\\=&\mathrm {I} _{1}(\mathbf {T} )\mathrm {I} _{1}(\mathbf {H} )-\mathrm {I} _{1}(\mathbf {T} \cdot \mathbf {H} )\\\rightarrow {\frac {\mathrm {d} \mathrm {I} _{2}(\mathbf {T} )}{\mathrm {d} \mathbf {T} }}=&\mathrm {I} _{1}(\mathbf {T} )\mathbf {1} -\mathbf {T} ^{\top }\end{aligned}}}$

With the characteristic polynomial and the determinant product theorem shows

${\displaystyle {\begin{aligned}\mathrm {det} (\mathbf {T} +s\mathbf {H} )=&\mathrm {det} \left(s\mathbf {T} \cdot \left({\frac {1}{s}}\mathbf {1} +\mathbf {T} ^{-1}\cdot \mathbf {H} \right)\right)=\mathrm {det} (\mathbf {T} )s^{3}\mathrm {det} \left(\mathbf {T} ^{-1}\cdot \mathbf {H} +{\frac {1}{s}}\mathbf {1} \right)\\=&\mathrm {det} (\mathbf {T} )s^{3}\left({\frac {1}{s^{3}}}+{\frac {1}{s^{2}}}\mathrm {I} _{1}(\mathbf {T} ^{-1}\cdot \mathbf {H} )+{\frac {1}{s}}\mathrm {I} _{2}(\mathbf {T} ^{-1}\cdot \mathbf {H} )+\mathrm {I} _{3}(\mathbf {T} ^{-1}\cdot \mathbf {H} )\right)\\=&\mathrm {det} (\mathbf {T} )[1+s\mathrm {I} _{1}(\mathbf {T} ^{-1}\cdot \mathbf {H} )+s^{2}\mathrm {I} _{2}(\mathbf {T} ^{-1}\cdot \mathbf {H} )+s^{3}\mathrm {I} _{3}(\mathbf {T} ^{-1}\cdot \mathbf {H} )]\end{aligned}}}$

The derivative is calculated from this

${\displaystyle {\begin{aligned}\left.{\frac {\mathrm {d} }{\mathrm {d} s}}\mathrm {det} (\mathbf {T} +s\mathbf {H} )\right|_{s=0}=&\mathrm {det} (\mathbf {T} )\mathrm {I} _{1}(\mathbf {T} ^{-1}\cdot \mathbf {H} )\\\rightarrow {\frac {\mathrm {d} \mathrm {I} _{3}(\mathbf {T} )}{\mathrm {d} \mathbf {T} }}=&\mathrm {det} (\mathbf {T} )\mathbf {T} ^{\top -1}\end{aligned}}}$.

This derivation only exists if T is invertible, i.e. det ( T ) ≠ 0.

Applications

The following examples show the use of the main invariants in material theories and often used material models:

1. Hooke's law : The stress tensor is calculated from the strain tensor according to . This contains the shear modulus and the Poisson's ratio . ${\displaystyle {\boldsymbol {\sigma }}}$ ${\displaystyle {\boldsymbol {\varepsilon }}}$ ${\displaystyle {\boldsymbol {\sigma }}=2G\left({\boldsymbol {\varepsilon }}+{\frac {\nu }{1-2\nu }}{\color {red}\mathrm {I} _{1}({\boldsymbol {\varepsilon }})}\mathbf {1} \right)}$ ${\displaystyle G}$ ${\displaystyle \nu }$
2. Hyperelasticity : The deformation energy density in the Neo-Hooke model is . There is a material parameter and the left Cauchy-Green tensor . ${\displaystyle \psi }$ ${\displaystyle \psi =\mu ({\color {red}\mathrm {I} _{1}(\mathbf {b} )}-3)}$ ${\displaystyle \mu }$ ${\displaystyle \mathbf {b} }$
3. Plasticity theory , strength theory : The v. Mises equivalent stress is a function of the second major invariant of the stress deviator . ${\displaystyle {\boldsymbol {\sigma }}_{v}={\sqrt {-3{\color {red}\mathrm {I} _{2}({\boldsymbol {\sigma }}^{\mathrm {D} })}}}}$ ${\displaystyle {\boldsymbol {\sigma }}^{\mathrm {D} }:={\boldsymbol {\sigma }}-{\frac {\color {red}\mathrm {I} _{1}({\boldsymbol {\sigma }})}{3}}\mathbf {1} }$
4. Incompressibility : Here the third Hauptinvariante of is the deformation gradient at each material point constant: . ${\displaystyle \mathbf {F} }$ ${\displaystyle {\color {red}\mathrm {I} _{3}(\mathbf {F} )}\equiv 1}$

example

The proof of the invariance of the trace of a tensor is provided. Let and be any two basic systems of and ${\displaystyle {\vec {a}}_{1,2,3}}$ ${\displaystyle {\vec {b}}_{1,2,3}}$ ${\displaystyle \mathbb {V} ^{3}}$

${\displaystyle \mathbf {T} =\sum _{i,j=1}^{3}T_{ij}{\vec {a}}_{i}\otimes {\vec {b}}_{j}\rightarrow \mathrm {Sp} (\mathbf {T} )=\sum _{i,j=1}^{3}T_{ij}{\vec {a}}_{i}\cdot {\vec {b}}_{j}}$.

When changing to other bases and with dual bases and the new components are calculated accordingly ${\displaystyle {\vec {c}}_{1,2,3}}$ ${\displaystyle {\vec {d}}_{1,2,3}}$ ${\displaystyle {\vec {c}}^{1,2,3}}$ ${\displaystyle {\vec {d}}^{1,2,3}}$ ${\displaystyle T_{ij}^{\mathrm {*} }}$

${\displaystyle {\begin{aligned}\mathbf {T} =&\sum _{i,j=1}^{3}T_{ij}{\vec {a}}_{i}\otimes {\vec {b}}_{j}=\sum _{i,j,k,l=1}^{3}T_{ij}({\vec {a}}_{i}\cdot {\vec {c}}^{k}){\vec {c}}_{k}\otimes ({\vec {b}}_{j}\cdot {\vec {d}}^{l}){\vec {d}}_{l}\\=:&\sum _{k,l}T_{kl}^{\ast }{\vec {c}}_{k}\otimes {\vec {d}}_{l}=:\mathbf {T} ^{\ast }\\\rightarrow T_{kl}^{\ast }=&\sum _{i,j=1}^{3}T_{ij}({\vec {a}}_{i}\cdot {\vec {c}}^{k})({\vec {b}}_{j}\cdot {\vec {d}}^{l})\end{aligned}}}$

The track with the new components therefore results in too ${\displaystyle T_{ij}^{\ast }}$

${\displaystyle {\begin{aligned}\mathrm {Sp} (\mathbf {T} ^{\ast })=&\mathrm {Sp} \left(\sum _{k,l=1}^{3}T_{kl}^{\ast }{\vec {c}}_{k}\otimes {\vec {d}}_{l}\right)=\sum _{k,l=1}^{3}T_{kl}^{\ast }{\vec {c}}_{k}\cdot {\vec {d}}_{l}\\=&\sum _{i,j,k,l=1}^{3}T_{ij}({\vec {a}}_{i}\cdot {\vec {c}}^{k})({\vec {b}}_{j}\cdot {\vec {d}}^{l}){\vec {c}}_{k}\cdot {\vec {d}}_{l}\\=&\sum _{i,j,k,l=1}^{3}T_{ij}({\vec {a}}_{i}\cdot {\vec {c}}^{k}){\vec {c}}_{k}\cdot ({\vec {b}}_{j}\cdot {\vec {d}}^{l}){\vec {d}}_{l}=\sum _{i,j=1}^{3}T_{ij}{\vec {a}}_{i}\cdot {\vec {b}}_{j}=\mathrm {Sp} (\mathbf {T} )\end{aligned}}}$

what was to be shown.

See also

• Continuum mechanics
• Euclidean transformation
• Formula collection tensor algebra
• Tensoranalysis formula collection

literature

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