$\textbf{A} \times (\nabla \times \textbf{A})$ with index notation

Question

So i have this expression

$\textbf{A} \times (\nabla \times \textbf{A})$

and I'm starting to get pretty familiar to this index notation but its one step i can't get the hang of when i look in the solution manual.

Anyway i get this

$[\textbf{A} \times (\nabla \times \textbf{A})]_i = \ldots\ \text{some steps}\ \ldots = (\partial_iA_l)A_l-A_i \partial_mA_m = \frac{1}{2}\partial_i({A_lA_l) -A_i\partial_mA_m}$

What happens here with the first term in this last step?

And another question, can you always pull out the free index i and put it outside? You see the answer is this

$[\frac{1}{2}grad\,A^2-\textbf{A}\,div\textbf{A}]_i$

and they simply pull out that i in the last term and put it outside the whole expression, can you always do it like that and when can you not do it?

I have the book Vector Calculus of Matthews and its pretty thin when it comes to clarify things.

Answer 1

\begin{align} [\mathbf A \times (\nabla \times \mathbf A)]_i &= \epsilon_{ijk} A_j (\nabla \times \mathbf A)_k = \epsilon_{kij} \epsilon_{kmn} A_j \partial_m A_n = (\delta_{im} \delta_{jn} - \delta_{in} \delta_{jm})A_j\partial_m A_n = \\ &= A_n \partial_i A_n - A_m \partial_m A_i = \frac 12 \partial_i (A_n A_n) - (A_m \partial_m) A_i = \frac 12 [\nabla\ (\mathbf A \cdot \mathbf A)]_i - [(\mathbf A \cdot \nabla) \mathbf A]_i \end{align}

Answer 2

Question 1) You clearly copied something wrong. Let's see. First let's get the index notation of the cross product between two vectors $\boldsymbol{A},\boldsymbol{B}$: $$[\boldsymbol{A}\times\boldsymbol{B}]^i=\epsilon^i_{jk}\boldsymbol{A}^j\boldsymbol{B}^k$$, where $e^i_{jk}$ is the Levi-Civitta tensor which is zero if any two of its indices take on the same value; $+1$ if they correspond to an even permutation and $-1$ to an odd permutation. Ex.: $\epsilon^1_{12}=0\,;\,\epsilon^1_{23}=1\,;\,\epsilon^2_{13}=-1$. The expression on the left means the $i$-th component of the vector $\boldsymbol{A}\times\boldsymbol{B}$.

Thus we have $$[\boldsymbol{A}\times (\nabla\times\boldsymbol{A})]^i=\epsilon^i_{jk}\boldsymbol{A}^j\epsilon^k_{lm}\nabla^l\boldsymbol{A}^m\,=\,\epsilon^i_{jk}\epsilon^k_{lm}\boldsymbol{A}^j\nabla^l\boldsymbol{A}^m$$. Notice we have a contraction of the $\epsilon$'s. Let's do the sum over $k$ first. Here we consider the rest of the indices fixed. Thus the only posibility for this not to vanish is that $i=l\,;\,j=m$ or the other way around with a change of sign, i.e., $\epsilon^i_{jk}\epsilon^k_{lm}=\delta^i_l\delta_{jm}\,-\,\delta^i_m\delta_{jl}$. Insert this into the last expression we get

$$[\boldsymbol{A}\times (\nabla\times\boldsymbol{A})]^i=\,(\delta^i_l\delta_{jm}\,-\,\delta^i_m\delta_{jl})\boldsymbol{A}^j\nabla^l\boldsymbol{A}^m\,=\,\,(\boldsymbol{A}^j\nabla^i\boldsymbol{A}_j\,-\,\boldsymbol{A}^j\nabla_j\boldsymbol{A}^i)\,=\,(\frac{1}{2}\nabla^i\boldsymbol{A}^2\,-\,\boldsymbol{A}^j\nabla_j\boldsymbol{A}^i)\,=\,[\frac{1}{2}\nabla\boldsymbol{A}^2-(\boldsymbol{A}\cdot\nabla)\boldsymbol{A}]^i$$ which is not exactly what you wrote. Check the wikipedia page.

Question 2): Make sure you follow correctly and in detail which indices are bound and which are free at each step and the argument above.