Micromechanics of composites/Proof 9

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Relation between axial vector and strain
Let $$\mathbf{u}$$ be a displacement field. Let $$\boldsymbol{\varepsilon}$$ be the strain field (infinitesimal) corresponding to the displacement field and let $$\boldsymbol{\theta}$$ be the corresponding infinitesimal rotation vector. Show that

\boldsymbol{\nabla} \times \boldsymbol{\varepsilon} = \boldsymbol{\nabla} \boldsymbol{\theta} ~. $$
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Proof:

The infinitesimal strain tensor is given by

\boldsymbol{\varepsilon} = \frac{1}{2}(\boldsymbol{\nabla}\mathbf{u} + \boldsymbol{\nabla}\mathbf{u}^T) ~. $$ Therefore,

\boldsymbol{\nabla} \times \boldsymbol{\varepsilon} = \frac{1}{2}[\boldsymbol{\nabla} \times (\boldsymbol{\nabla}\mathbf{u}) + \boldsymbol{\nabla} \times (\boldsymbol{\nabla}\mathbf{u}^T)] ~. $$ Recall that

\boldsymbol{\nabla} \times (\boldsymbol{\nabla}\mathbf{u}) = \mathbf{0}\qquad\text{and}\qquad \boldsymbol{\nabla} \times (\boldsymbol{\nabla}\mathbf{u}^T) = \boldsymbol{\nabla} (\boldsymbol{\nabla} \times \mathbf{u}) ~. $$ Hence,

\boldsymbol{\nabla} \times \boldsymbol{\varepsilon} = \frac{1}{2}[\boldsymbol{\nabla} (\boldsymbol{\nabla} \times \mathbf{u})] ~. $$ Also recall that

\boldsymbol{\theta} = \frac{1}{2} \boldsymbol{\nabla} \times \mathbf{u}~. $$ Therefore,

{ \boldsymbol{\nabla} \times \boldsymbol{\varepsilon} = \boldsymbol{\nabla} \boldsymbol{\theta} \qquad \square } $$