How to prove this equality using Einstein summation (index notation)

Question

I have to prove this equality using index notation (Einstein summation), but I don't know how to proceed from here:

$\nabla(A \cdot B)=A \times (\nabla \times B)+(A \cdot \nabla)B+B \times (\nabla \times A)+(B \cdot \nabla)A$

$\partial_i(a_j b_j)$ -> Product rule

$(\partial_i a_j)b_j+a_j(\partial_i b_j)$ -> What should I do here?

Thank you! All the best.

I would try to go from the curls to the rest of the operators (that is, I would pass all the curls to one side of the equation and proceed from that. A vector product or a curl can be written in terms of the permutation tensor, which has 3 properties to remember:

it changes sign with odd permutations of its indexes

it doesn't change with even permutations of its indexes

and https://en.wikipedia.org/wiki/Levi-Civita_symbol#Index_and_symbol_values — Manuel Pena, Jan 18 '23 at 22:40
Does this answer your question? Proving $A\times(\nabla\times B)+B\times(\nabla\times A)=\nabla(A\cdot B)-(A\cdot\nabla)B -(B\cdot\nabla)A$ with Einstein summation — Kurt G., Jun 25 '23 at 16:30

score 0 · Answer 1 · answered Jun 25 '23 at 16:12

Observe that \begin{align*} &\mathbf A \times (\boldsymbol \nabla \times \mathbf B) + (\mathbf A \boldsymbol \cdot \boldsymbol \nabla)\mathbf B + \mathbf B \times(\boldsymbol \nabla \times \mathbf A) + (\mathbf B \boldsymbol \cdot \boldsymbol \nabla)\mathbf A \\ ={}&\varepsilon_{ijk}A_i\left[\boldsymbol \nabla \times \mathbf B\right]_j \mathbf e_k + A_i \partial_i B_k \mathbf e_k + \varepsilon_{ijk} B_i \left[\boldsymbol \nabla \times \mathbf A \right]_j \mathbf e_k + B_i \partial_i A_k \mathbf e_k \tag{1}\\ ={}& \varepsilon_{jki} \varepsilon_{jpq} A_i \partial_p B_q \mathbf e_k + A_i \partial_i B_k \mathbf e_k + \varepsilon_{jki} \varepsilon_{jpq} B_i \partial_p A_q \mathbf e_k + B_i \partial_i A_k \mathbf e_k \tag{2}\\ ={}& \left(\delta_{kp} \delta_{iq} - \delta_{kq} \delta_{ip} \right) \left(A_i \partial_p B_q \mathbf e_k + B_i \partial_p A_q \mathbf e_k\right) + A_i \partial_i B_k \mathbf e_k + B_i \partial_i A_k \mathbf e_k \tag{3}\\ ={}& A_i \partial_k B_i \mathbf e_k - A_i \partial_i B_k \mathbf e_k + B_i \partial_k A_i \mathbf e_k - B_i \partial_i A_k \mathbf e_k + A_i \partial_i B_k \mathbf e_k + B_i \partial_i A_k \mathbf e_k \tag{4}\\ ={}& A_i \frac{\partial B_i}{\partial x_k} \mathbf e_k + B_i \frac{\partial A_i}{\partial x_k} \mathbf e_k \tag{5} \\ ={}& \partial_k (A_i B_i) \mathbf e_k \tag{6} \\ ={}& \boldsymbol \nabla(\mathbf A \boldsymbol \cdot \mathbf B), \tag{7} \end{align*} where

$(1)$ follows since $\mathbf F \times \mathbf G = \epsilon_{ijk} F_i G_j \mathbf e_k$ and $(\mathbf F \boldsymbol \cdot \boldsymbol \nabla) = F_i \partial_i$;
$(2)$ follows since $\boldsymbol \nabla \times \mathbf F = \varepsilon_{ijk}\partial_i F_j \mathbf e_k$;
$(3)$ follows using the $\varepsilon$-contraction identity $\varepsilon_{ijk} \varepsilon_{ipq} = \delta_{jp} \delta_{kq} - \delta_{jq} \delta_{kp}$;
$(4)$ follows using the $\delta$-contraction identity $\delta_{ij}a_j = a_i$;
$(5)$ follows by cancelling equal terms;
$(6)$ follows by the product rule;
$(7)$ follows since $\boldsymbol \nabla \phi = \partial_i\phi \mathbf e_i$.

How to prove this equality using Einstein summation (index notation)

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