# Continuum mechanics/Curl of a gradient of a vector

## 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

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

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