# Continuum mechanics/Curl of a gradient of a vector

## Curl of the gradient of a vector - 1

Let ${\displaystyle \mathbf {v} }$  be a vector field. Show that

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

Proof:

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

${\displaystyle ({\boldsymbol {\nabla }}\times {\boldsymbol {S}})\cdot \mathbf {a} ={\boldsymbol {\nabla }}\times ({\boldsymbol {S}}^{T}\cdot \mathbf {a} )}$

where ${\displaystyle \mathbf {a} }$  is an arbitrary constant vector. Substituting ${\displaystyle {\boldsymbol {\nabla }}\mathbf {v} }$  into the definition, we have

${\displaystyle [{\boldsymbol {\nabla }}\times ({\boldsymbol {\nabla }}\mathbf {v} )]\cdot \mathbf {a} ={\boldsymbol {\nabla }}\times ({\boldsymbol {\nabla }}\mathbf {v} ^{T}\cdot \mathbf {a} )~.}$

Since ${\displaystyle \mathbf {a} }$  is constant, we may write

${\displaystyle {\boldsymbol {\nabla }}\mathbf {v} ^{T}\cdot \mathbf {a} ={\boldsymbol {\nabla }}(\mathbf {v} \cdot \mathbf {a} )={\boldsymbol {\nabla }}\varphi }$

where ${\displaystyle \varphi =\mathbf {v} \cdot \mathbf {a} }$  is a scalar. Hence,

${\displaystyle [{\boldsymbol {\nabla }}\times ({\boldsymbol {\nabla }}\mathbf {v} )]\cdot \mathbf {a} ={\boldsymbol {\nabla }}\times ({\boldsymbol {\nabla \varphi )}}~.}$

Since the curl of the gradient of a scalar field is zero (recall potential theory), we have

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

Hence,

${\displaystyle [{\boldsymbol {\nabla }}\times ({\boldsymbol {\nabla }}\mathbf {v} )]\cdot \mathbf {a} =\mathbf {0} \qquad \qquad \forall ~~\mathbf {a} ~.}$

The arbitrary nature of ${\displaystyle \mathbf {a} }$  gives us

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

## Curl of the transpose of the gradient of a vector

Let ${\displaystyle \mathbf {v} }$  be a vector field. Show that

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

Proof:

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

${\displaystyle ({\boldsymbol {\nabla }}\times {\boldsymbol {S}})\cdot \mathbf {a} ={\boldsymbol {\nabla }}\times ({\boldsymbol {S}}^{T}\cdot \mathbf {a} )}$

where ${\displaystyle \mathbf {a} }$  is an arbitrary constant vector. If we write the right hand side in index notation with respect to a Cartesian basis, we have

${\displaystyle [{\boldsymbol {S}}^{T}\cdot \mathbf {a} ]_{k}=[\mathbf {b} ]_{k}=b_{k}=S_{pk}~a_{p}}$

and

${\displaystyle [{\boldsymbol {\nabla }}\times \mathbf {b} ]_{i}=e_{ijk}{\frac {\partial b_{k}}{\partial x_{j}}}=e_{ijk}{\frac {\partial (S_{pk}~a_{p})}{\partial x_{j}}}=e_{ijk}{\frac {\partial S_{pk}}{\partial x_{j}}}~a_{p}=[({\boldsymbol {\nabla }}\times {\boldsymbol {S}})]_{ip}~a_{p}~.}$

In the above a quantity ${\displaystyle [~]_{i}}$  represents the ${\displaystyle i}$ -th component of a vector, and the quantity ${\displaystyle [~]_{ip}}$  represents the ${\displaystyle ip}$ -th components of a second-order tensor.

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

${\displaystyle [{\boldsymbol {\nabla }}\times {\boldsymbol {S}}]_{ip}=e_{ijk}{\frac {\partial S_{pk}}{\partial x_{j}}}~.}$

Using the above definition, we get

${\displaystyle [{\boldsymbol {\nabla }}\times {\boldsymbol {S}}^{T}]_{ip}=e_{ijk}{\frac {\partial S_{kp}}{\partial x_{j}}}~.}$

If ${\displaystyle {\boldsymbol {S}}={\boldsymbol {\nabla }}\mathbf {v} }$ , we have

${\displaystyle [{\boldsymbol {\nabla }}\times {\boldsymbol {\nabla }}\mathbf {v} ^{T}]_{ip}=e_{ijk}{\frac {\partial }{\partial x_{j}}}\left({\frac {\partial v_{k}}{\partial x_{p}}}\right)={\frac {\partial }{\partial x_{p}}}\left(e_{ijk}{\frac {\partial v_{k}}{\partial x_{j}}}\right)={\frac {\partial }{\partial x_{p}}}\left([{\boldsymbol {\nabla }}\times \mathbf {v} ]_{i}\right)=[{\boldsymbol {\nabla }}({\boldsymbol {\nabla }}\times \mathbf {v} )]_{ip}~.}$

Therefore,

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