Micromechanics of composites/Proof 7

Curl of the transpose of the gradient of a vector
Let $$\mathbf{v}$$ be a vector field. Show that

\boldsymbol{\nabla} \times (\boldsymbol{\nabla}\mathbf{v}^T) = \boldsymbol{\nabla} (\boldsymbol{\nabla} \times \mathbf{v}) ~. $$

 Proof:

The curl of a second order tensor field $$\boldsymbol{S}$$ is defined as

(\boldsymbol{\nabla} \times \boldsymbol{S})\cdot\mathbf{a} = \boldsymbol{\nabla} \times (\boldsymbol{S}^T\cdot\mathbf{a}) $$ where $$\mathbf{a}$$ is an arbitrary constant vector. If we write the right hand side in index notation with respect to a Cartesian basis, we have

[\boldsymbol{S}^T\cdot\mathbf{a}]_k = [\mathbf{b}]_k = b_k = S_{pk}~a_p $$ and

[\boldsymbol{\nabla} \times \mathbf{b}]_i = e_{ijk}\frac{\partial b_k}{\partial x_j} = e_{ijk}\frac{\partial (S_{pk}~a_p)}{\partial x_j} = e_{ijk}\frac{\partial S_{pk}}{\partial x_j}~a_p = [(\boldsymbol{\nabla} \times \boldsymbol{S})]_{ip}~a_p ~. $$ In the above a quantity $$[~]_i$$ represents the $$i$$-th component of a vector, and the quantity $$[~]_{ip}$$ represents the $$ip$$-th components of a second-order tensor.

Therefore, in index notation, the curl of a second-order tensor $$\boldsymbol{S}$$ can be expressed as

[\boldsymbol{\nabla} \times \boldsymbol{S}]_{ip} = e_{ijk}\frac{\partial S_{pk}}{\partial x_j} ~. $$ Using the above definition, we get

[\boldsymbol{\nabla} \times \boldsymbol{S}^T]_{ip} = e_{ijk}\frac{\partial S_{kp}}{\partial x_j} ~. $$ If $$\boldsymbol{S} = \boldsymbol{\nabla}\mathbf{v}$$, we have

[\boldsymbol{\nabla} \times \boldsymbol{\nabla}\mathbf{v}^T]_{ip} = e_{ijk}\frac{\partial }{\partial x_j} \left(\frac{\partial v_k}{\partial x_p}\right) = \frac{\partial }{\partial x_p}\left(e_{ijk}\frac{\partial v_k}{\partial x_j}\right) = \frac{\partial }{\partial x_p}\left([\boldsymbol{\nabla} \times \mathbf{v}]_i\right) = [\boldsymbol{\nabla} (\boldsymbol{\nabla} \times \mathbf{v})]_{ip} ~. $$ Therefore,

{ \boldsymbol{\nabla} \times (\boldsymbol{\nabla}\mathbf{v}^T) = \boldsymbol{\nabla} (\boldsymbol{\nabla} \times \mathbf{v}) \qquad \square } $$