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 }