Micromechanics of composites/Proof 7

Curl of the transpose of the gradient of a vector

Let $𝐯$ be a vector field. Show that

$\mathrm{\nabla }\times (\mathrm{\nabla }{𝐯}^T)=\mathrm{\nabla }(\mathrm{\nabla }\times 𝐯).$

Proof:

The curl of a second order tensor field $𝑺$ is defined as

$(\mathrm{\nabla }\times 𝑺)\cdot 𝐚=\mathrm{\nabla }\times ({𝑺}^T\cdot 𝐚)$

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

$[{𝑺}^T\cdot 𝐚{]}_k=[𝐛{]}_k=b_k=S_{pk}{a}_p$

and

$[\mathrm{\nabla }\times 𝐛{]}_i=e_{ijk}\frac{\partial b_k}{\partial x_j}=e_{ijk}\frac{\partial (S_{pk}{a}_p)}{\partial x_j}=e_{ijk}\frac{\partial S_{pk}}{\partial x_j}{a}_p=[(\mathrm{\nabla }\times 𝑺){]}_{ip}{a}_p.$

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 $𝑺$ can be expressed as

$[\mathrm{\nabla }\times 𝑺{]}_{ip}=e_{ijk}\frac{\partial S_{pk}}{\partial x_j}.$

Using the above definition, we get

$[\mathrm{\nabla }\times {𝑺}^T{]}_{ip}=e_{ijk}\frac{\partial S_{kp}}{\partial x_j}.$

If $𝑺=\mathrm{\nabla }𝐯$, we have

$[\mathrm{\nabla }\times \mathrm{\nabla }{𝐯}^T{]}_{ip}=e_{ijk}\frac{\partial }{\partial x_j}\left(\frac{\partial v_k}{\partial x_p}\right)=\frac{\partial }{\partial x_p}\left(e_{ijk}\frac{\partial v_k}{\partial x_j}\right)=\frac{\partial }{\partial x_p}([\mathrm{\nabla }\times 𝐯{]}_i)=[\mathrm{\nabla }(\mathrm{\nabla }\times 𝐯){]}_{ip}.$

Therefore,

$\mathrm{\nabla }\times (\mathrm{\nabla }{𝐯}^T)=\mathrm{\nabla }(\mathrm{\nabla }\times 𝐯)◻$

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