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,

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

← Proof 6

Micromechanics of composites

Proof 8 →