## Polar decomposition

The w:Polar decomposition theorem states that any second order tensor whose determinant is positive can be decomposed uniquely into a symmetric part and an orthogonal part.

In continuum mechanics, the deformation gradient ${\displaystyle {\boldsymbol {F}}}$  is such a tensor because ${\displaystyle \det(\mathbf {F} )>0}$ . Therefore we can write

${\displaystyle {\boldsymbol {F}}={\boldsymbol {R}}\cdot {\boldsymbol {U}}={\boldsymbol {V}}\cdot {\boldsymbol {R}}}$

where ${\displaystyle {\boldsymbol {R}}}$  is an orthogonal tensor (${\displaystyle {\boldsymbol {R}}\cdot {\boldsymbol {R}}^{T}={\boldsymbol {\mathit {1}}}}$ ) and ${\displaystyle {\boldsymbol {U}},{\boldsymbol {V}}}$  are symmetric tensors (${\displaystyle {\boldsymbol {U}}={\boldsymbol {U}}^{T}}$  and ${\displaystyle {\boldsymbol {V}}={\boldsymbol {V}}^{T}}$ ) called the right stretch tensor and the left stretch tensor, respectively. This decomposition is called the polar decomposition of ${\displaystyle {\boldsymbol {F}}}$ .

Recall that the right Cauchy-Green deformation tensor is defined as

${\displaystyle {\boldsymbol {C}}={\boldsymbol {F}}^{T}\cdot {\boldsymbol {F}}}$

Clearly this is a symmetric tensor. From the polar decomposition of ${\displaystyle {\boldsymbol {F}}}$  we have

${\displaystyle {\boldsymbol {C}}={\boldsymbol {U}}^{T}\cdot {\boldsymbol {R}}^{T}\cdot {\boldsymbol {R}}\cdot {\boldsymbol {U}}={\boldsymbol {U}}\cdot {\boldsymbol {U}}={\boldsymbol {U}}^{2}}$

If you know ${\displaystyle {\boldsymbol {C}}}$  then you can calculate ${\displaystyle {\boldsymbol {U}}}$  and hence ${\displaystyle {\boldsymbol {R}}}$  using ${\displaystyle {\boldsymbol {R}}={\boldsymbol {F}}\cdot {\boldsymbol {U}}^{-1}}$ .

### How do you find the square root of a tensor?

If you want to find ${\displaystyle {\boldsymbol {U}}}$  given ${\displaystyle {\boldsymbol {C}}}$  you will need to take the square root of ${\displaystyle {\boldsymbol {C}}}$ . How does one do that?

We use what is called the spectral decomposition or eigenprojection of ${\displaystyle {\boldsymbol {C}}}$ . The spectral decomposition involves expressing ${\displaystyle {\boldsymbol {C}}}$  in terms of its eigenvalues and eigenvectors. The tensor product of the eigenvectors acts as a basis while the eigenvalues give the magnitude of the projection.

Thus,

${\displaystyle {\boldsymbol {C}}=\sum _{i=1}^{3}\lambda _{i}^{2}~{\boldsymbol {N}}_{i}\otimes {\boldsymbol {N}}_{i}}$

where ${\displaystyle \lambda _{i}^{2}}$  are the principal values (eigenvalues) of ${\displaystyle {\boldsymbol {C}}}$  and ${\displaystyle {\boldsymbol {N}}_{i}}$  are the principal directions (eigenvectors) of ${\displaystyle {\boldsymbol {C}}}$ .

Therefore,

${\displaystyle {\boldsymbol {U}}^{2}=\sum _{i=1}^{3}\lambda _{i}^{2}~{\boldsymbol {N}}_{i}\otimes {\boldsymbol {N}}_{i}}$

Since the basis does not change, we then have

${\displaystyle {\boldsymbol {U}}=\sum _{i=1}^{3}\lambda _{i}~{\boldsymbol {N}}_{i}\otimes {\boldsymbol {N}}_{i}}$

Therefore the ${\displaystyle \lambda _{i}}$  can be interpreted as principal stretches and the vectors ${\displaystyle {\boldsymbol {N}}_{i}}$  are the directions of the principal stretches.

#### Exercise:

If

${\displaystyle {\boldsymbol {U}}=\sum _{i=1}^{3}\lambda _{i}~{\boldsymbol {N}}_{i}\otimes {\boldsymbol {N}}_{i}}$

show that

${\displaystyle {\boldsymbol {U}}^{2}={\boldsymbol {U}}\cdot {\boldsymbol {U}}=\sum _{i=1}^{3}\lambda _{i}^{2}~{\boldsymbol {N}}_{i}\otimes {\boldsymbol {N}}_{i}~.}$

### Example of polar decomposition

Let us assume that the motion is given by

{\displaystyle {\begin{aligned}x_{1}&={\cfrac {1}{4}}\left[4~X_{1}+(9-3~X_{1}-5~X_{2}-X_{1}~X_{2})~t\right]\\x_{2}&=X_{2}+(4+2~X_{1})~t\end{aligned}}}

The adjacent figure shows how a unit square subjected to this motion evolves over time.

 An example of a motion.

The deformation gradient is given by

${\displaystyle {\boldsymbol {F}}={\frac {\partial \mathbf {x} }{\partial {\boldsymbol {X}}}}\quad \implies \quad F_{ij}={\frac {\partial x_{i}}{\partial X_{j}}}}$

Therefore

{\displaystyle {\begin{aligned}F_{11}&={\frac {\partial x_{1}}{\partial X_{1}}}={\cfrac {1}{4}}\left[4+(-3-X_{2})~t\right]\\F_{12}&={\frac {\partial x_{1}}{\partial X_{2}}}={\cfrac {1}{4}}\left[(-5-X_{1})~t\right]\\F_{21}&={\frac {\partial x_{2}}{\partial X_{1}}}=2~t\\F_{22}&={\frac {\partial x_{2}}{\partial X_{2}}}=1\end{aligned}}}

At ${\displaystyle t=1}$  at the position ${\displaystyle {\boldsymbol {X}}=(0,0)}$  we have

${\displaystyle \mathbf {F} ={\begin{bmatrix}{\frac {\partial x_{1}}{\partial X_{1}}}&{\frac {\partial x_{1}}{\partial X_{2}}}\\{\frac {\partial x_{2}}{\partial X_{1}}}&{\frac {\partial x_{2}}{\partial X_{2}}}\end{bmatrix}}={\cfrac {1}{4}}{\begin{bmatrix}1&-5\\8&4\end{bmatrix}}}$

You can calculate the deformation gradient at other points in a similar manner.

#### Right Cauchy-Green deformation tensor

We have

${\displaystyle {\boldsymbol {C}}={\boldsymbol {F}}^{T}\cdot {\boldsymbol {F}}}$

Therefore,

${\displaystyle \mathbf {C} =\mathbf {F} ^{T}~\mathbf {F} ={\cfrac {1}{16}}{\begin{bmatrix}65&27\\27&41\end{bmatrix}}}$

To compute ${\displaystyle {\boldsymbol {U}}}$  we have to find the eigenvalues and eigenvectors of ${\displaystyle {\boldsymbol {C}}}$ . The eigenvalue problem is

${\displaystyle (\mathbf {C} -\lambda ^{2}~\mathbf {I} )\mathbf {N} =\mathbf {0} }$

where

${\displaystyle \mathbf {I} ={\begin{bmatrix}1&0\\0&1\end{bmatrix}}}$

To find the eigenvalues we solve the characteristic equation

${\displaystyle \det(\mathbf {C} -\lambda ^{2}~\mathbf {I} )=0}$

Plugging in the numbers, we get

${\displaystyle \det {\begin{bmatrix}{\cfrac {65}{16}}-\lambda ^{2}&{\cfrac {27}{16}}\\{\cfrac {27}{16}}&{\cfrac {41}{16}}-\lambda ^{2}\end{bmatrix}}=0}$

or

${\displaystyle \lambda ^{4}-{\cfrac {53}{8}}~\lambda ^{2}+{\cfrac {121}{16}}=0}$

This equation has two solutions

{\displaystyle {\begin{aligned}\lambda _{1}^{2}&={\cfrac {53}{16}}+{\cfrac {3}{16}}~{\sqrt {97}}=5.159\\\lambda _{2}^{2}&={\cfrac {53}{16}}-{\cfrac {3}{16}}~{\sqrt {97}}=1.466\end{aligned}}}

Taking the square roots we get the values of the principal stretches

${\displaystyle \lambda _{1}=2.2714\qquad \lambda _{2}=1.2107}$

To compute the eigenvectors we plug into the eigenvalues into the eigenvalue problem to get

${\displaystyle \left\{{\begin{bmatrix}65&27\\27&41\end{bmatrix}}-\lambda _{1}^{2}~{\begin{bmatrix}1&0\\0&1\end{bmatrix}}\right\}~{\begin{bmatrix}N_{1}^{(1)}\\N_{2}^{(1)}\end{bmatrix}}={\begin{bmatrix}0\\0\end{bmatrix}}}$

Because this system of equations is not linearly independent, we need another equation to solve this system of equations for ${\displaystyle N_{1}^{(1)}}$  and ${\displaystyle N_{2}^{(1)}}$ . This problem is eliminated by using the following equation (which implies that ${\displaystyle \mathbf {N} }$  is a unit vector)

${\displaystyle N_{2}^{(1)}={\sqrt {1-(N_{1}^{(1)})^{2}}}}$

Solving, we get

${\displaystyle \mathbf {N} _{1}={\begin{bmatrix}N_{1}^{(1)}\\N_{2}^{(1)}\end{bmatrix}}={\begin{bmatrix}0.8385\\0.5449\end{bmatrix}}}$

We can do the same thing for the other eigenvector ${\displaystyle \mathbf {N} _{2}}$  to get

${\displaystyle \mathbf {N} _{2}={\begin{bmatrix}N_{1}^{(2)}\\N_{2}^{(2)}\end{bmatrix}}={\begin{bmatrix}-0.5449\\0.8385\end{bmatrix}}}$

Therefore,

${\displaystyle {\boldsymbol {N}}_{1}\otimes {\boldsymbol {N}}_{1}=\mathbf {N} _{1}~\mathbf {N} _{1}^{T}={\begin{bmatrix}0.8385\\0.5449\end{bmatrix}}{\begin{bmatrix}0.8385&0.5449\end{bmatrix}}={\begin{bmatrix}0.7031&0.4569\\0.4569&0.2969\end{bmatrix}}}$

and

${\displaystyle {\boldsymbol {N}}_{2}\otimes {\boldsymbol {N}}_{2}=\mathbf {N} _{2}~\mathbf {N} _{2}^{T}={\begin{bmatrix}-0.5449\\0.8385\end{bmatrix}}{\begin{bmatrix}-0.5449&0.8385\end{bmatrix}}={\begin{bmatrix}0.2969&-0.4569\\-0.4569&0.7031\end{bmatrix}}}$

Therefore,

${\displaystyle {\boldsymbol {C}}=\lambda _{1}^{2}~{\boldsymbol {N}}_{1}\otimes {\boldsymbol {N}}_{1}+\lambda _{2}^{2}~{\boldsymbol {N}}_{2}\otimes {\boldsymbol {N}}_{2}\quad \implies \quad \mathbf {C} =5.159~{\begin{bmatrix}0.7031&0.4569\\0.4569&0.2969\end{bmatrix}}+1.466~{\begin{bmatrix}0.2969&-0.4569\\-0.4569&0.7031\end{bmatrix}}}$

We usually don't see any problem to calculate ${\displaystyle {\boldsymbol {C}}}$  at this point and go straight to the right stretch tensor.

#### Right stretch

The right stretch tensor ${\displaystyle {\boldsymbol {U}}}$  is given by

${\displaystyle {\boldsymbol {U}}=\lambda _{1}~{\boldsymbol {N}}_{1}\otimes {\boldsymbol {N}}_{1}+\lambda _{2}~{\boldsymbol {N}}_{2}\otimes {\boldsymbol {N}}_{2}\quad \implies \quad \mathbf {U} =2.2714~{\begin{bmatrix}0.7031&0.4569\\0.4569&0.2969\end{bmatrix}}+1.2107~{\begin{bmatrix}0.2969&-0.4569\\-0.4569&0.7031\end{bmatrix}}}$

or

${\displaystyle \mathbf {U} ={\begin{bmatrix}1.9565&0.4846\\0.4846&1.5256\end{bmatrix}}}$

We can invert this matrix to get

${\displaystyle \mathbf {U} ^{-1}={\begin{bmatrix}0.5548&-0.1762\\-0.1762&0.7114\end{bmatrix}}}$

#### Rotation

We can now find the rotation matrix by using th relation

${\displaystyle {\boldsymbol {R}}={\boldsymbol {F}}\cdot {\boldsymbol {U}}^{-1}}$

In matrix form,

${\displaystyle \mathbf {R} ={\cfrac {1}{4}}{\begin{bmatrix}1&-5\\8&4\end{bmatrix}}{\begin{bmatrix}0.5548&-0.1762\\-0.1762&0.7114\end{bmatrix}}={\begin{bmatrix}0.3590&-0.9334\\0.9334&0.3590\end{bmatrix}}}$

You can check whether this matrix is orthogonal by seeing whether ${\displaystyle \mathbf {R} ~\mathbf {R} ^{T}=\mathbf {R} ^{T}~\mathbf {R} =\mathbf {I} }$ .

You thus get the polar decomposition of ${\displaystyle {\boldsymbol {F}}}$ . In an actual calculation you have to be careful about floating point errors. Otherwise you might not get a matrix that is orthogonal.