# Nonlinear finite elements/Kinematics - spectral decomposition
## Spectral decompositions

Many numerical algorithms use spectral decompositions to compute material behavior.

### Spectral decompositions of stretch tensors

Infinitesimal line segments in the material and spatial configurations are related by
    $d\mathbf {x} ={\boldsymbol {F}}\cdot d{\boldsymbol {X}}={\boldsymbol {R}}\cdot ({\boldsymbol {U}}\cdot d{\boldsymbol {X}})={\boldsymbol {V}}\cdot ({\boldsymbol {R}}\cdot d{\boldsymbol {X}})~.$
So the sequence of operations may be either considered as a stretch of in the material configuration followed by a rotation or a rotation followed by a stretch.
Also note that
    ${\boldsymbol {V}}={\boldsymbol {R}}\cdot {\boldsymbol {U}}\cdot {\boldsymbol {R}}^{T}~.$
Let the spectral decomposition of ${\boldsymbol {U}}$ be
    ${\boldsymbol {U}}=\sum _{i=1}^{3}\lambda _{i}~{\boldsymbol {N}}_{i}\otimes {\boldsymbol {N}}_{i}$
and the spectral decomposition of ${\boldsymbol {V}}$ be
    ${\boldsymbol {V}}=\sum _{i=1}^{3}{\hat {\lambda }}_{i}~\mathbf {n} _{i}\otimes \mathbf {n} _{i}~.$
Then
    ${\boldsymbol {V}}={\boldsymbol {R}}\cdot {\boldsymbol {U}}\cdot {\boldsymbol {R}}^{T}=\sum _{i=1}^{3}\lambda _{i}~{\boldsymbol {R}}\cdot ({\boldsymbol {N}}_{i}\otimes {\boldsymbol {N}}_{i})\cdot {\boldsymbol {R}}^{T}=\sum _{i=1}^{3}\lambda _{i}~({\boldsymbol {R}}\cdot {\boldsymbol {N}}_{i})\otimes ({\boldsymbol {R}}\cdot {\boldsymbol {N}}_{i})$
Therefore the uniqueness of the spectral decomposition implies that
    $\lambda _{i}={\hat {\lambda }}_{i}\quad {\text{and}}\quad \mathbf {n} _{i}={\boldsymbol {R}}\cdot {\boldsymbol {N}}_{i}$
The left stretch (${\boldsymbol {V}}$) is also called the spatial stretch tensor while the right stretch (${\boldsymbol {U}}$) is called the material stretch tensor.

### Spectral decompositions of deformation gradient

The deformation gradient is given by
    ${\boldsymbol {F}}={\boldsymbol {R}}\cdot {\boldsymbol {U}}$
In terms of the spectral decomposition of ${\boldsymbol {U}}$ we have
    ${\boldsymbol {F}}=\sum _{i=1}^{3}\lambda _{i}~{\boldsymbol {R}}\cdot ({\boldsymbol {N}}_{i}\otimes {\boldsymbol {N}}_{i})=\sum _{i=1}^{3}\lambda _{i}~({\boldsymbol {R}}\cdot {\boldsymbol {N}}_{i})\otimes {\boldsymbol {N}}_{i}=\sum _{i=1}^{3}\lambda _{i}~\mathbf {n} _{i}\otimes {\boldsymbol {N}}_{i}$
Therefore the spectral decomposition of ${\boldsymbol {F}}$ can be written as
    ${\boldsymbol {F}}=\sum _{i=1}^{3}\lambda _{i}~\mathbf {n} _{i}\otimes {\boldsymbol {N}}_{i}$
Let us now see what effect the deformation gradient has when it is applied to the eigenvector ${\boldsymbol {N}}_{i}$.
We have
    ${\boldsymbol {F}}\cdot {\boldsymbol {N}}_{i}={\boldsymbol {R}}\cdot {\boldsymbol {U}}\cdot {\boldsymbol {N}}_{i}={\boldsymbol {R}}\cdot \left(\sum _{j=1}^{3}\lambda _{j}~{\boldsymbol {N}}_{j}\otimes {\boldsymbol {N}}_{j}\right)\cdot {\boldsymbol {N}}_{i}$
From the definition of the dyadic product
    $(\mathbf {u} \otimes \mathbf {v} )\cdot \mathbf {w} =(\mathbf {w} \cdot \mathbf {v} )~\mathbf {u}$
Since the eigenvectors are orthonormal, we have
    $({\boldsymbol {N}}_{j}\otimes {\boldsymbol {N}}_{j})\cdot {\boldsymbol {N}}_{i}={\begin{cases}0&{\mbox{if}}~i\neq j\\{\boldsymbol {N}}_{i}&{\mbox{if}}~i=j\end{cases}}$
Therefore,
    $\left(\sum _{j=1}^{3}\lambda _{j}~{\boldsymbol {N}}_{j}\otimes {\boldsymbol {N}}_{j}\right)\cdot {\boldsymbol {N}}_{i}=\lambda _{i}~{\boldsymbol {N}}_{i}{\text{no sum on}}~i$
That leads to
    ${\boldsymbol {F}}\cdot {\boldsymbol {N}}_{i}=\lambda _{i}~({\boldsymbol {R}}\cdot {\boldsymbol {N}}_{i})=\lambda _{i}~\mathbf {n} _{i}$
So the effect of ${\boldsymbol {F}}$ on ${\boldsymbol {N}}_{i}$ is to stretch the vector by $\lambda _{i}$ and to rotate it to the new orientation $\mathbf {n} _{i}$.
We can also show that
    ${\boldsymbol {F}}^{-T}\cdot {\boldsymbol {N}}_{i}={\cfrac {1}{\lambda _{i}}}~\mathbf {n} _{i}~;~~{\boldsymbol {F}}^{T}\cdot \mathbf {n} _{i}=\lambda _{i}~{\boldsymbol {N}}_{i}~;~~{\boldsymbol {F}}^{-1}\cdot \mathbf {n} _{i}={\cfrac {1}{\lambda _{i}}}~{\boldsymbol {N}}_{i}$

### Spectral decompositions of strains

Recall that the Lagrangian Green strain and its Eulerian counterpart are defined as
    ${\boldsymbol {E}}={\frac {1}{2}}~({\boldsymbol {F}}^{T}\cdot {\boldsymbol {F}}-{\boldsymbol {\mathit {1}}})~;~~{\boldsymbol {e}}={\frac {1}{2}}~({\boldsymbol {\mathit {1}}}-\left({\boldsymbol {F}}\cdot {\boldsymbol {F}}^{T}\right)^{-1})$
Now,
    ${\boldsymbol {F}}^{T}\cdot {\boldsymbol {F}}={\boldsymbol {U}}\cdot {\boldsymbol {R}}^{T}\cdot {\boldsymbol {R}}\cdot {\boldsymbol {U}}={\boldsymbol {U}}^{2}~;~~{\boldsymbol {F}}\cdot {\boldsymbol {F}}^{T}={\boldsymbol {V}}\cdot {\boldsymbol {R}}\cdot {\boldsymbol {R}}^{T}\cdot {\boldsymbol {V}}={\boldsymbol {V}}^{2}$
Therefore we can write
    ${\boldsymbol {E}}={\frac {1}{2}}~({\boldsymbol {U}}^{2}-{\boldsymbol {\mathit {1}}})~;~~{\boldsymbol {e}}={\frac {1}{2}}~({\boldsymbol {\mathit {1}}}-{\boldsymbol {V}}^{-2})$
Hence the spectral decompositions of these strain tensors are
    ${\boldsymbol {E}}=\sum _{i=1}^{3}{\frac {1}{2}}(\lambda _{i}^{2}-1)~{\boldsymbol {N}}_{i}\otimes {\boldsymbol {N}}_{i}~;~~\mathbf {e} =\sum _{i=1}^{3}{\frac {1}{2}}\left(1-{\cfrac {1}{\lambda _{i}^{2}}}\right)~\mathbf {n} _{i}\otimes \mathbf {n} _{i}$

#### Generalized strain measures

We can generalize these strain measures by defining strains as
    ${\boldsymbol {E}}^{(n)}={\cfrac {1}{n}}~({\boldsymbol {U}}^{n}-{\boldsymbol {\mathit {1}}})~;~~{\boldsymbol {e}}^{(n)}={\cfrac {1}{n}}~({\boldsymbol {\mathit {1}}}-{\boldsymbol {V}}^{-n})$
The spectral decomposition is
    ${\boldsymbol {E}}^{(n)}=\sum _{i=1}^{3}{\cfrac {1}{n}}(\lambda _{i}^{n}-1)~{\boldsymbol {N}}_{i}\otimes {\boldsymbol {N}}_{i}~;~~\mathbf {e} ^{(n)}=\sum _{i=1}^{3}{\cfrac {1}{n}}\left(1-{\cfrac {1}{\lambda _{i}^{n}}}\right)~\mathbf {n} _{i}\otimes \mathbf {n} _{i}$
Clearly, the usual Green strains are obtained when $n=2$.

#### Logarithmic strain measure

A strain measure that is commonly used is the logarithmic strain measure. This strain measure is obtained when we have $n\rightarrow 0$. Thus
    ${\boldsymbol {E}}^{(0)}=\ln({\boldsymbol {U}})~;~~{\boldsymbol {e}}^{(0)}=\ln({\boldsymbol {V}})$
The spectral decomposition is
    ${\boldsymbol {E}}^{(0)}=\sum _{i=1}^{3}\ln \lambda _{i}~{\boldsymbol {N}}_{i}\otimes {\boldsymbol {N}}_{i}~;~~\mathbf {e} ^{(0)}=\sum _{i=1}^{3}\ln \lambda _{i}~\mathbf {n} _{i}\otimes \mathbf {n} _{i}$