Can anyone please tell me the expression for the curl of curl of a rank 2 covariant tensor? I've been going through a lot of books and sources and have not found an exact expression.
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$\begingroup$ Do you mean $\nabla\times(\nabla\times D)$ or the (colloquially written) $\nabla\times D\times\nabla$ which occurs in the compatibility condition of linearized elasticity theory? $\endgroup$– ccornMar 11, 2018 at 13:21
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$\begingroup$ In the latter case, use $$\epsilon^{hij}\epsilon_{kmn} = \begin{vmatrix} {\delta^h}_k & {\delta^h}_m & {\delta^h}_n \\ {\delta^i}_k & {\delta^i}_m & {\delta^i}_n \\ {\delta^j}_k & {\delta^j}_m & {\delta^j}_n \end{vmatrix}$$ $\endgroup$– ccornMar 11, 2018 at 21:24
1 Answer
Remark
I looked for exactly the same reference after your question on curl of the metric tensor. I too have had little luck but at the moment I am going with this analogy? observe that for some $\vec{c}$:
$$ (\vec{a}\times\vec{b})\cdot\vec{c} = (\vec{a}\otimes\vec{b} - \vec{b}\otimes\vec{a})\vec{c}$$
where $\otimes$ denotes the tensor product. If we are allowed to take this analogy written in component form (Im an engineer not a mathematician so dont take me seriouly! note, $\nabla_{(.)}$ denotes covariant differentation):
$$[\nabla\times S \cdot \vec{c}] = [\nabla{S} - (\nabla{S})^T]\vec{c} $$
A little more manipulation with a similar analogy can give the form ($\epsilon$ is the levi-civita symbol):
$$ [\nabla \times S]^j_{i} = \nabla_lS_{mi}\epsilon^{lmj}$$
Going back to your previous question, this logic would mean the curl of the metric would be zero.