# 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,

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