Micromechanics of composites/Proof 6

Curl of the gradient of a vector - 1
Let $$\mathbf{v}$$ be a vector field. Show that

\boldsymbol{\nabla} \times (\boldsymbol{\nabla}\mathbf{v}) = 0 ~. $$

Proof:

For a second order tensor field $$\boldsymbol{S}$$, we can define the curl as

(\boldsymbol{\nabla} \times \boldsymbol{S})\cdot\mathbf{a} = \boldsymbol{\nabla} \times (\boldsymbol{S}^T\cdot\mathbf{a}) $$ where $$\mathbf{a}$$ is an arbitrary constant vector. Substituting $$\boldsymbol{\nabla}\mathbf{v}$$ into the definition, we have

[\boldsymbol{\nabla} \times (\boldsymbol{\nabla}\mathbf{v})]\cdot\mathbf{a} = \boldsymbol{\nabla} \times (\boldsymbol{\nabla}\mathbf{v}^T\cdot\mathbf{a}) ~. $$ Since $$\mathbf{a}$$ is constant, we may write

\boldsymbol{\nabla}\mathbf{v}^T\cdot\mathbf{a} = \boldsymbol{\nabla} (\mathbf{v}\cdot\mathbf{a}) = \boldsymbol{\nabla} \varphi $$ where $$\varphi = \mathbf{v} \cdot \mathbf{a}$$ is a scalar. Hence,

[\boldsymbol{\nabla} \times (\boldsymbol{\nabla}\mathbf{v})]\cdot\mathbf{a} = \boldsymbol{\nabla} \times (\boldsymbol{\nabla \varphi)} ~. $$ Since the curl of the gradient of a scalar field is zero (recall potential theory), we have

\boldsymbol{\nabla} \times (\boldsymbol{\nabla \varphi)} = \mathbf{0} ~. $$ Hence,

[\boldsymbol{\nabla} \times (\boldsymbol{\nabla}\mathbf{v})]\cdot\mathbf{a} = \mathbf{0} \qquad\qquad \forall\mathbf{a} ~. $$ The arbitrary nature of $$\mathbf{a}$$ gives us

{ \boldsymbol{\nabla} \times (\boldsymbol{\nabla}\mathbf{v}) = \mathbf{0} \qquad \square } $$