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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.

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    $\begingroup$ You can do that because of the product rule. $\partial_i (A_n A_n) = A_n \partial_i A_n + A_n \partial_i A_n = 2 A_n \partial_i A_n$. $\endgroup$ – Kaster Sep 15 '15 at 21:57
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    $\begingroup$ There's an error in your first equation. It should say either $(\partial_iA_l)A_l$ or $A_l\partial_iA_l$ instead of $\partial_iA_lA_l$. $\endgroup$ – joriki Sep 15 '15 at 21:58
  • $\begingroup$ It is indeed, it should be $A_l\partial_iA_l$ and it is in my notes. I did a typo here. But do the order change anything? $\endgroup$ – user269620 Sep 15 '15 at 22:09
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    $\begingroup$ @Danny actually it does when you have an operator, not just a multiplicand ($\partial$), that's why whenever ambiguity arises, you need to use parens. See my edits of the question. $\endgroup$ – Kaster Sep 15 '15 at 22:13
  • $\begingroup$ I see, so if I'm doing something with like vector elements i can just threat them like scalars but whenever there is an operator involved i need to be more careful? $\endgroup$ – user269620 Sep 15 '15 at 22:17
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\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}

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  • $\begingroup$ Thanks! What about the other question i had? $\endgroup$ – user269620 Sep 15 '15 at 22:10
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    $\begingroup$ @Danny, of course, $[\mathbf A]_i = A_i$. $\endgroup$ – Kaster Sep 15 '15 at 22:11
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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.

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  • $\begingroup$ I see your point now, but then the solution manual must be wrong. What does it mean when you have the operator to the right in the scalar product, is it commutative so i can switch them around? $\endgroup$ – user269620 Sep 16 '15 at 8:19
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    $\begingroup$ There are two scalar products in that last expression. Which one you mean? In any case, notice that you cannot just "move around" $\nabla$ as you see fit, if that's what you mean. Ex.: $\nabla\boldsymbol{A}^2$ is a normal vector, while $\boldsymbol{A}^2\nabla$ is an operator (the gradient times a constant) that needs to be applied on a function. You cannot add both terms just like that. $\endgroup$ – MASL Sep 16 '15 at 12:28
  • $\begingroup$ Yeah I understand now I think, thanks! $\endgroup$ – user269620 Sep 16 '15 at 15:27

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