Micromechanics of composites/Proof 7
Curl of the transpose of the gradient of a vector edit
Let be a vector field. Show that
Proof:
The curl of a second order tensor field is defined as
where 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}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d691445dbdf4b3f66eacd88098e412c9c5f5e524)
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}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3a8eba99b2e076a199cc0380b3d2e70d76491477)
In the above a quantity represents the -th component of a vector, and the quantity represents the -th components of a second-order tensor.
![{\displaystyle [~]_{i}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/686ebbfab1501ed07622b29407ad9eaa2c9200ba)
![{\displaystyle [~]_{ip}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7d5f3a6167a7f95ae491958faab9092dc4cec041)
Therefore, in index notation, the curl of a second-order tensor can be expressed as
![{\displaystyle [{\boldsymbol {\nabla }}\times {\boldsymbol {S}}]_{ip}=e_{ijk}{\frac {\partial S_{pk}}{\partial x_{j}}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b786a08dc372c47adb4994a682efb79d89c8802e)
Using the above definition, we get
![{\displaystyle [{\boldsymbol {\nabla }}\times {\boldsymbol {S}}^{T}]_{ip}=e_{ijk}{\frac {\partial S_{kp}}{\partial x_{j}}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0127ac19bf1d019dc7b1187d5d4ed8e206357ea0)
If , 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}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bd50999dad5938bddeeaca49629213bee3905dd3)
Therefore,
