# Micromechanics of composites/Proof 8

## Relation between axial vector and displacement

Let ${\displaystyle \mathbf {u} }$ be a displacement field. The displacement gradient tensor is given by ${\displaystyle {\boldsymbol {\nabla }}\mathbf {u} }$. Let the skew symmetric part of the displacement gradient tensor (infinitesimal rotation tensor) be

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

Let ${\displaystyle {\boldsymbol {\theta }}}$ be the axial vector associated with the skew symmetric tensor ${\displaystyle {\boldsymbol {\omega }}}$. Show that

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

Proof:

The axial vector ${\displaystyle \mathbf {w} }$ of a skew-symmetric tensor ${\displaystyle {\boldsymbol {W}}}$ satisfies the condition

${\displaystyle {\boldsymbol {W}}\cdot \mathbf {a} =\mathbf {w} \times \mathbf {a} }$

for all vectors ${\displaystyle \mathbf {a} }$. In index notation (with respect to a Cartesian basis), we have

${\displaystyle W_{ip}~a_{p}=e_{ijk}~w_{j}~a_{k}}$

Since ${\displaystyle e_{ijk}=-e_{ikj}}$, we can write

${\displaystyle W_{ip}~a_{p}=-e_{ikj}~w_{j}~a_{k}\equiv -e_{ipq}~w_{q}~a_{p}}$

or,

${\displaystyle W_{ip}=-e_{ipq}~w_{q}~.}$

Therefore, the relation between the components of ${\displaystyle {\boldsymbol {\omega }}}$ and ${\displaystyle {\boldsymbol {\theta }}}$ is

${\displaystyle \omega _{ij}=-e_{ijk}~\theta _{k}~.}$

Multiplying both sides by ${\displaystyle e_{pij}}$, we get

${\displaystyle e_{pij}~\omega _{ij}=-e_{pij}~e_{ijk}~\theta _{k}=-e_{pij}~e_{kij}~\theta _{k}~.}$

Recall the identity

${\displaystyle e_{ijk}~e_{pqk}=\delta _{ip}~\delta _{jq}-\delta _{iq}~\delta _{jp}~.}$

Therefore,

${\displaystyle e_{ijk}~e_{pjk}=\delta _{ip}~\delta _{jj}-\delta _{ij}~\delta _{jp}=3\delta _{ip}-\delta _{ip}=2\delta _{ip}}$

Using the above identity, we get

${\displaystyle e_{pij}~\omega _{ij}=-2\delta _{pk}~\theta _{k}=-2\theta _{p}~.}$

Rearranging,

${\displaystyle \theta _{p}=-{\frac {1}{2}}~e_{pij}~\omega _{ij}}$

Now, the components of the tensor ${\displaystyle {\boldsymbol {\omega }}}$ with respect to a Cartesian basis are given by

${\displaystyle \omega _{ij}={\frac {1}{2}}\left({\frac {\partial u_{i}}{\partial x_{j}}}-{\frac {\partial u_{j}}{\partial x_{i}}}\right)}$

Therefore, we may write

${\displaystyle \theta _{p}=-{\cfrac {1}{4}}~e_{pij}\left({\frac {\partial u_{i}}{\partial x_{j}}}-{\frac {\partial u_{j}}{\partial x_{i}}}\right)}$

Since the curl of a vector ${\displaystyle \mathbf {v} }$ can be written in index notation as

${\displaystyle {\boldsymbol {\nabla }}\times \mathbf {v} =e_{ijk}~{\frac {\partial u_{k}}{\partial x_{j}}}~\mathbf {e} _{i}}$

we have

${\displaystyle e_{pij}~{\frac {\partial u_{j}}{\partial x_{i}}}~=[{\boldsymbol {\nabla }}\times \mathbf {u} ]_{p}\qquad {\text{and}}\qquad e_{pij}~{\frac {\partial u_{i}}{\partial x_{j}}}~=-e_{pji}{\frac {\partial u_{i}}{\partial x_{j}}}=-[{\boldsymbol {\nabla }}\times \mathbf {u} ]_{p}}$

where ${\displaystyle [~]_{p}}$ indicates the ${\displaystyle p}$-th component of the vector inside the square brackets.

Hence,

${\displaystyle \theta _{p}=-{\cfrac {1}{4}}~\left(-[{\boldsymbol {\nabla }}\times \mathbf {u} ]_{p}-[{\boldsymbol {\nabla }}\times \mathbf {u} ]_{p}\right)={\frac {1}{2}}~[{\boldsymbol {\nabla }}\times \mathbf {u} ]_{p}~.}$

Therefore,

${\displaystyle {{\boldsymbol {\theta }}={\frac {1}{2}}{\boldsymbol {\nabla }}\times \mathbf {u} \qquad \square }}$