Knowing the fact that $\nabla\times(\nabla\times A)=\nabla(\nabla\cdot A)-(\nabla\cdot\nabla)A$, can anyone help me in calculating $\nabla\times(\nabla\times(\nabla\times A))$ using Levi-Civita tensor?

  • 2
    $\begingroup$ The curl of a gradient is $0$, so only the second term sticks around. But it can't be represented any simpler than $-\nabla \times (\nabla^2 A)$, in general. Interestingly it is also equal to $-\nabla^2(\nabla \times A)$, but that's not any simpler to me. So if you start from the formula for $\nabla \times\nabla \times A$ you immediately get the result. $\endgroup$ – user137731 Sep 27 '16 at 0:15
  • $\begingroup$ If you want to use the LC tensor note that the expression you have is $\nabla\times C$ where $C = \nabla(\nabla\cdot A) - \nabla^2 A$. Write $C$ in index notation and use $\nabla\times C = \epsilon_{ijk}{\bf e}_i \nabla_j C_k$. For this problem this does not simplify things. Doing it like the comment above is simpler and more natural here. $\endgroup$ – Winther Sep 27 '16 at 0:53

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