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I have two questions on the computation of $\nabla \times (\nabla \times \mathbf{A}) $ with Einstein summation notation based on http://www.physics.ohio-state.edu/~ntg/263/handouts/tensor_intro.pdf. It considers the $i$th component. All colours have been added by me. $$ (\color{green}{\nabla} \color{red}{\times} \color{purple}{(\nabla \times \mathbf{A})})_i = \color{red}{\epsilon_{ijk}}\color{green}{\partial_j} \color{purple}{(\nabla \times \mathbf{A})}_k \tag{*}$$

$$ = \color{red}{\epsilon_{ijk}}\color{green}{\partial_j} \color{purple}{\epsilon_{klm}\partial_lA_m}$$ $$ = \epsilon_{ijk}\epsilon_{klm}\partial_j \partial_lA_m $$ $$ = \epsilon_{ijk}\epsilon_{lmk}\partial_j \partial_lA_m $$ $$ = (\delta_{il}\delta_{jm} - \delta_{im}\delta_{jl}) \partial_j \partial_lA_m $$ Now, choose $j \rightarrow m$ and $ l \rightarrow i $ in the first term and let $j \rightarrow l$ and $ m \rightarrow i $ in the second:

$$ = \partial_m \partial_iA_m - \color{blue}{\partial_l \partial_lA_i}$$ $$ = \partial_i(\partial_m A_m) - \color{blue}{(\partial_l \partial_l\mathbf{A})_i}\tag{**}$$

$\Large{\text{Question 1}}$. In (*), where does the subscript $k$ on the RHS come from? Aren't we looking at the $i$th component? I know that $\mathbf{a} \times \mathbf{b} = a_gb_h\epsilon_{ghi}\mathbf{\hat{e_i}} = (..., \underbrace{a_gb_h\epsilon_{ghi}}_{i\text{th component}}, ...)\Longrightarrow (\mathbf{a} \times \mathbf{b})_i = a_gb_h\epsilon_{ghi} $.

$\Large{\text{Question 2}}$. In (**), how does $ \color{blue}{\partial_l \partial_lA_i = (\partial_l \partial_l\mathbf{A})_i} $ ?

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2 Answers 2

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I like your format.

Fisrt question: (*) is just applying the definition of $$ (\mathbf{a} \times \mathbf{b})_i = a_j b_k \epsilon_{jki} = a_j b_k \epsilon_{ijk} $$ Let $\mathbf{a} = \nabla = (\partial_1, \partial_2,\partial_3)$, and $a_j = \partial_j$. $\mathbf{b} = \nabla \times \mathbf{A}$ here, hence $$ (\nabla\times(\nabla\times \mathbf{A}))_i = \partial_j(\nabla\times \mathbf{A})_k \epsilon_{ijk} $$

Notice repeated subscripts get canceled, so here $i$ ceases to appear as a subscript on the RHS of (*), but it's in the Levi-Civita symbol. Also, $k$ is just a dummy summation subscript. What you have is still the $i$-th component of $\nabla\times(\nabla\times \mathbf{A})$, for subscript $i$ is in the permutation symbol.

Second question: notice $$\partial_l \partial_l A_i = \sum_{l=1}^3 \partial_l \partial_l A_i $$ while $$ \partial_l \partial_l \mathbf{A} = \sum_{l=1}^3 \partial_l \partial_l (\sum_{i=1}^3 A_i \mathbf{e}_i) = \sum_{i=1}^3\left(\sum_{l=1}^3 \partial_l \partial_l A_i\right)\mathbf{e}_i $$ therefore $$(\partial_l \partial_l \mathbf{A})_i = \partial_l \partial_l A_i$$

Finally the identity: $$ \nabla\times(\nabla\times \mathbf{A}) = -\Delta \mathbf{A} + \nabla \nabla\cdot\mathbf{A}. $$ I don't know if there is a direct proof without expanding the whole thing as sum.

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  • $\begingroup$ Thank you very much for your detailed answer. I've just 1 follow-up question. You wrote that "Notice repeated subscripts get canceled...," but what got cancelled in the equations preceding your sentence? I don't see any terms that became cancelled? $\endgroup$
    – user53259
    May 6, 2013 at 15:47
  • $\begingroup$ @LaPrevoyance Pardon the ambiguity in my answer. "repeated subscripts get canceled" means whenever we have repeated subscript $a_i b_i c_j = \sum^n_{i=1} a_i b_i c_j = d_j$, whose subscript follows $j$, has no subscript $i$ any more. The proper way to phrase this should be "repeat indices are summed out and disappeared in the result". $\endgroup$
    – Shuhao Cao
    May 6, 2013 at 15:54
  • $\begingroup$ Thank you for your response. Last question: So had you been referring to the subscript $i$? It doesn't appear on the RHS of (*) as a subscript of the whole expression anymore, but only in the Levi-Civita symbol? Also, thank you for liking my formatting! $\endgroup$
    – user53259
    May 6, 2013 at 16:09
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Based on Shuhao Cao's detailed and helpful answer, I realised that a simpler way to understand (**) is by first grasping that $[\mathbf{F}]_i = F_i $.

This is true because $\mathbf{F} = \sum_{i = 1}F_i\mathbf{\hat{e_i}} \Rightarrow [\mathbf{F}]_i = F_i$.

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