Micromechanics of composites/Proof 8

Relation between axial vector and displacement Let $\mathbf {u}$ be a displacement field. The displacement gradient tensor is given by ${\boldsymbol {\nabla }}\mathbf {u}$. Let the skew symmetric part of the displacement gradient tensor (infinitesimal rotation tensor) be ${\boldsymbol {\omega }}={\frac {1}{2}}({\boldsymbol {\nabla }}\mathbf {u} -{\boldsymbol {\nabla }}\mathbf {u} ^{T})~.$ Let ${\boldsymbol {\theta }}$ be the axial vector associated with the skew symmetric tensor ${\boldsymbol {\omega }}$. Show that ${\boldsymbol {\theta }}={\frac {1}{2}}~{\boldsymbol {\nabla }}\times \mathbf {u} ~.$

Proof:

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

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

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

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

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

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

or,

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

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

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

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

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

Recall the identity

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

Therefore,

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

Using the above identity, we get

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

Rearranging,

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

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

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

Therefore, we may write

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

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

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

we have

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

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

Hence,

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

Therefore,

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

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Micromechanics of composites

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