Polar Decomposition

Any matrix can be decomposed into a sum of symmetric and antisymmetric matrices, but \mathbf{F} can be decomposed into a product of two matrices (one symmetric and one orthogonal)

(1)   \begin{equation*} \mathbf{F=V \cdot R = R \cdot U} \end{equation*}

\mathbf{U} and \mathbf{V} are called the Right Stretch Tensor and Left Stretch Tensor due to their respective positions (relative to \mathbf{R}) in eq. 1.

\mathbf{V}=\mathbf{V}^{T} (symmetric)
\mathbf{U}=\mathbf{U}^{T} (symmetric)
\mathbf{R}^{T}=\mathbf{R}^{-1} (orthogonal)

A proof of the orthogonality of \mathbf{R} is given in Appendix A.3.

(2)   \begin{equation*} \mathbf{C=F}^{T} \cdot \mathbf{F=U \cdot} \underbrace{\mathbf{R}^{T} \cdot \mathbf{R}}_{\mathbf{I}} \cdot \mathbf{U=U}^{2} \end{equation*}

(3)   \begin{equation*} \mathbf{B=V}^{2} \end{equation*}

\mathbf{R} is a rigid body rotation, while \mathbf{U} or \mathbf{V} each stretch and rotate (they each contain both normal and shear deformations, typically)

\mathbf{U} and \mathbf{V} have the same eigenvalues. Eigenvalues do not have an “order,” per se, but since \mathbf{U} and \mathbf{V} typically have different eigenvectors, one should be cautious when assuming equivalency of eigenvalues. In the principal directions of \mathbf{U} or \mathbf{V}, \mathbf{U} or \mathbf{V} contains no shear deformation (see Fig). We’ll look at actual stresses, later.

To find the principal directions of \mathbf{U} and \mathbf{V}, we must solve the eigenvalue problem:

\mathbf{U \cdot n}=\lambda \mathbf{n}
\underbrace{\mathbf{R \cdot U}}_{\mathbf{F}} \cdot \mathbf{n}=\mathbf{R \cdot}\lambda \mathbf{n} (pictured)
\mathbf{F = V \cdot R} \longrightarrow \mathbf{V \cdot (}\underbrace{\mathbf{R \cdot n}}_{\mathbf{m}})=\lambda (\underbrace{\mathbf{R \cdot n}}_{\mathbf{m}})
\mathbf{m_i}=\mathbf{R \cdot n_i}



e.x. Polar Decomposition

The deformed equilibrium configuration of a body defined by the deformation mapping:

x_1=X_1+3X_2 , x_2=X_2 , x_3=X_3


a) \mathbf{F} and \mathbf{C}
b) Eigenvalues and eigenvectors of \mathbf{C}
c) \mathbf{U} and \mathbf{U^{-1}}
d) \mathbf{R}
e) \mathbf{V}

a) \mathbf{F} = \begin{bmatrix} \frac{\partial x_1}{\partial X_1} & \frac{\partial x_1}{\partial X_2} & \frac{\partial x_1}{\partial X_3}\\ \frac{\partial x_2}{\partial X_1} & \frac{\partial x_2}{\partial X_2} & \frac{\partial x_2}{\partial X_3}\\ \frac{\partial x_3}{\partial X_1} & \frac{\partial x_3}{\partial X_2} & \frac{\partial x_3}{\partial X_3} \end{bmatrix} = \begin{bmatrix} 1 & 3 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}

note: \mathbf{F} will typically be a function of X_1, X_2, X_3 and we’ll be interested in the value of \mathbf{F} at a particular point. Here, it doesn’t matter.

\mathbf{C=F}^{T} \cdot \mathbf{F}=\begin{bmatrix} 1 & 3 & 0\\ 3 & 10 & 0\\ 0 & 0 & 1\end{bmatrix}

b) \mathbf{C \cdot n}=\lambda \mathbf{n} ; det|-\lambda \mathbf{I+C}|=0
\longrightarrow \lambda_1=.092 ; \lambda_2=1 ; \lambda_3=10.91

[-\lambda_i \mathbf{I+C}]\mathbf{n_i}=0

i.e. for \lambda_1,

Choose arbitrary n_1 ; solve for n_2 from either equation ; normalize:

\begin{bmatrix}n_1 \\ n_2 \\ n_3\end{bmatrix}_{\lambda_1} = \begin{bmatrix}.957 \\ -.290 \\ 0 \end{bmatrix} ; \begin{bmatrix}n_1 \\ n_2 \\ n_3 \end{bmatrix}_{\lambda_2} = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} ; \begin{bmatrix}n_1 \\ n_2 \\ n_3 \end{bmatrix}_{\lambda_3} = \begin{bmatrix}.290 \\ .957\\ 0 \end{bmatrix}

c) [\mathbf{U}]_{\mathbf{n}}=\sqrt{[\mathbf{C}]_{\mathbf{n}}}=\begin{bmatrix}\sqrt{\lambda_1} & & \\& \sqrt{\lambda_2} & \\ & & \sqrt{\lambda_3}\end{bmatrix}=\begin{bmatrix}.303 & & \\& 1 & \\ & & 3.303\end{bmatrix}

[\mathbf{U}]_{\mathbf{e}}=[\boldsymbol{\Phi}][\mathbf{U}]_{\mathbf{n}}[\boldsymbol{\Phi}]^{T}=\begin{bmatrix} .56 & .83 & 0\\.83 & 3.05 & 0\\0 & 0 & 1\end{bmatrix} (symmetric)
where [\boldsymbol{\Phi}]=\begin{bmatrix} (n_1)_{\lambda_1}&(n_1)_{\lambda_2}&(n_1)_{\lambda_3} \\(n_2)_{\lambda_1}&(n_2)_{\lambda_2}&(n_2)_{\lambda_3} \\ (n_3)_{\lambda_1}&(n_3)_{\lambda_2}&(n_3)_{\lambda_3} \end{bmatrix}

It’s important to keep in mind that \mathbf{U}, \mathbf{E}, and \mathbf{C} have the same eigenvectors and that these eigenvectors (and any other strain direction or magnitude) are generally dependent on the particular point in space of interest (e.g. \mathbf{X_1}, \mathbf{X_2}, \mathbf{X_3}). For this simple example, \mathbf{F} is independent of \mathbf{X_1}, \mathbf{X_2}, and \mathbf{X_3} (i.e. the deformation is the same everywhere in the body – a “homogenous” deformation), but this would generally not be the case.

The eigenvalues occur in the direction of the eigenvectors, thus, \mathbf{n_1 \cdot E \cdot n_1}=\lambda_1 ; \mathbf{n_2 \cdot E \cdot n_2}=\lambda_2 ; \mathbf{n_3 \cdot E \cdot n_3}=\lambda_3. It makes sense that if we only know the eigenvectors and want to undo this transformation to bring us back to \mathbf{E}, then we need a transformation matrix that involves \mathbf{n_1},\mathbf{n_2},\mathbf{n_3}. We can see that \boldsymbol{\Phi} \cdot \mathbf{U} \cdot \boldsymbol{\Phi}^{T} is the opposite of the usual \boldsymbol{\Phi}^{T} \cdot \mathbf{U} \cdot \boldsymbol{\Phi}, and without going through the rigorous derivation of why \Phi_{i1}=(n_1)_{i}, \Phi_{i2}=(n_2)_{i}, \Phi_{i3}=(n_3)_{i}, it at least makes some sense intuitively.

d) [\mathbf{R}]_{\mathbf{e}}=[\mathbf{F}]_{\mathbf{e}}[\mathbf{U}]_{\mathbf{e}}^{T}=\begin{bmatrix} .55 & .83 & 0\\ -.83 & .55 & 0\\0 & 0 & 1\end{bmatrix} (orthogonal ; \mathbf{R}^{T}=\mathbf{R}^{-1})

e) [\mathbf{V}]_{\mathbf{e}}=[\mathbf{F}]_{\mathbf{e}}[\mathbf{R}]_{\mathbf{e}}^{T}=\begin{bmatrix}3.05 & .83 & 0\\.83 & .55 & 0\\0 & 0 & 1\end{bmatrix}