Micromechanics of composites/Proof 8

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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} ~. $$
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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 } $$