# Micromechanics of composites/Proof 9

## Relation between axial vector and strain

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

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

Proof:

The infinitesimal strain tensor is given by

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

Therefore,

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

Recall that

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

Hence,

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

Also recall that

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

Therefore,

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