I'm having trouble understanding what this triple curl is:
$$ \boldsymbol \nabla \times \boldsymbol \nabla \times \boldsymbol \nabla \times \mathrm{A} =\quad ?$$
where $\mathrm{A}$ is a vector field. In component form I wrote:
$$ \alpha_i =(\boldsymbol \nabla \times \vec{\mathrm{A}})_i = \epsilon_{ijk} \partial_j \mathrm{A}_k $$
Then:
$$ \beta_m= (\boldsymbol \nabla \times \boldsymbol \nabla \times \vec{\mathrm{A}})_m = \epsilon_{mni} \partial_n \alpha_i = (\delta_{jm} \delta_{kn} - \delta_{jn} \delta_{km}) \partial_n \partial_j \mathrm{A}_k $$
and finally:
$$ (\boldsymbol \nabla \times \vec{\beta} )_a = \epsilon_{abm} \partial_b \beta_m = \epsilon_{abm} \partial_b ( (\delta_{jm} \delta_{kn} - \delta_{jn} \delta_{km}) \partial_n \partial_j \mathrm{A}_k ) $$
$$ (\boldsymbol \nabla \times \vec{\beta} )_a = \epsilon_{abm} (\delta_{jm} \delta_{kn} - \delta_{jn} \delta_{km})\partial_b ( \partial_n \partial_j \mathrm{A}_k ) $$
$$ ( \boldsymbol \nabla \times \vec{\beta})_a = \epsilon_{abm} (\delta_{jm} \delta_{kn} )\partial_b (\partial_n \partial_j \textrm{A}_k ) - \epsilon_{abm} ( \delta_{jn} \delta_{km})\partial_b (\partial_n \partial_j \mathrm{A}_k ) $$
$$ ( \boldsymbol \nabla \times \vec{\beta})_a = \epsilon_{abj} \partial_b ( \partial_k \partial_j \textrm{A}_k ) - \epsilon_{abk} \partial_b ( \partial_j \partial_j \mathrm{A}_k ) $$
I see a Laplacian on the last term, but don't know what to do from here, as in, how is the initial expression represented in terms of divergence, curl, Laplacian and gradient. How do I finish this?
I am aware of the identity:
$$ \boldsymbol \nabla \times \boldsymbol \nabla \times \mathbf{A}= \boldsymbol \nabla(\boldsymbol \nabla \cdot \mathbf{A}) - \nabla^2 \mathbf{A}$$
and I could just apply the curl to both terms, but isnt there another way to represent this?