Micromechanics of composites/Proof 8

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

Proof:

The axial vector $𝐰$ of a skew-symmetric tensor $𝑾$ satisfies the condition

$𝑾\cdot 𝐚=𝐰\times 𝐚$

for all vectors $𝐚$. 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 $\mathit{\omega }$ and $\mathit{\theta }$ 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 $\mathit{\omega }$ with respect to a Cartesian basis are given by

${\omega }_{ij}=\frac{1}{2}(\frac{\partial u_i}{\partial x_j}-\frac{\partial u_j}{\partial x_i})$

Therefore, we may write

${\theta }_p=-\frac{1}{4}{e}_{pij}(\frac{\partial u_i}{\partial x_j}-\frac{\partial u_j}{\partial x_i})$

Since the curl of a vector $𝐯$ can be written in index notation as

$\mathrm{\nabla }\times 𝐯=e_{ijk}\frac{\partial u_k}{\partial x_j}{𝐞}_i$

we have

$e_{pij}\frac{\partial u_j}{\partial x_i}=[\mathrm{\nabla }\times 𝐮{]}_p\text{and}{e}_{pij}\frac{\partial u_i}{\partial x_j}=-e_{pji}\frac{\partial u_i}{\partial x_j}=-[\mathrm{\nabla }\times 𝐮{]}_p$

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

Hence,

${\theta }_p=-\frac{1}{4}(-[\mathrm{\nabla }\times 𝐮{]}_p-[\mathrm{\nabla }\times 𝐮{]}_p)=\frac{1}{2}[\mathrm{\nabla }\times 𝐮{]}_p.$

Therefore,

$\mathit{\theta }=\frac{1}{2}\mathrm{\nabla }\times 𝐮◻$

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