How can I prove $\nabla \cdot \color{green}{(\mathbf{F} {\times} \mathbf{G})} $ with Einstein Summation Notation?
Source: Stewart. Calculus: Early Transcendentals (6 edn 2007). p. 1068. §16.5, Exercise #27.
$\nabla \cdot \color{green}{(\mathbf{F} {\times} \mathbf{G})} = \partial_h\color{green}{\epsilon_{hij}F_iG_j}$
$ = \epsilon_{hij}\partial_h[F_iG_j]$
$ = \color{purple}{\epsilon_{hij}G_j\partial_hF_i} \color{red}{+} \epsilon_{hij}F_i\partial_hG_j $
$\color{purple}{1. \text{How do I determine whether } \epsilon_{hij}G_j\partial_hF_i = \mathbf{G} \times \nabla \mathbf{F} \text{ or } \mathbf{G} \cdot (\nabla \times \mathbf{F}) ?}$
$\color{darkred}{\text{3. The answer shows a negative sign $-$, not +. What did I miss?}}$ I read this.
My guess One answer to my question 1 is that the divergence of a vector field, ie $\nabla \mathbf{F}$, is always a scalar. So taking its cross product is nonsensical. This is why $ \color{purple}{ \mathbf{G} \times \nabla \mathbf{F} } $ is wrong.
First Attempt Recanted due to celtschk's Answer:
I tried to compute $\operatorname{div}(\mathbf{F} \times \mathbf{G})$ by considering the $j$th component term in the sum of the divergence operator:
$\color{red}{[}\nabla \cdot \color{green}{(\mathbf{F} {\times} \mathbf{G})}\color{red}{]}_\color{red}{\LARGE{j}} = \partial_\color{red}{\LARGE{j}}{\epsilon_{hi\color{red}{\LARGE{j}}}}F_{\huge{\color{green}{i}}}G_{\huge{\color{green}{i}}} = \color{green}{\epsilon_{hi\color{red}{\LARGE{j}}}}\partial_\color{red}{\LARGE{j}}[F_{\huge{\color{green}{i}}}G_{\huge{\color{green}{i}}}] = \color{purple}{\epsilon_{hi\color{red}{\LARGE{j}}}G_i\partial_\color{red}{\LARGE{j}}F_i} \color{brown}{+} \color{gray}{\epsilon_{hi\color{red}{\LARGE{j}}}F_i\partial_\color{red}{\LARGE{j}}G_i} $
Supplement due to Muphrid and celtschk:
$1.1.$ Without geometric calculus or wedge products or any more advanced topics, how can I divine to move $G_h$ to the front, as Muphrid did?
$3.1.$ In $\color{purple}{F_i\partial_hG_j}$, the order of the subscripts is $\color{purple}{(i, h, j)}$ But in $\epsilon_{hij}$, the order is $(h, i, j)$. Thus, I must rotate $(h, i, j)$ to obtain $\color{purple}{(i, h, j)},$ by swapping $\color{purple}{h}$ and $\color{purple}{i}$ once $\Longrightarrow \epsilon_{hij} = -\epsilon_{\color{purple}{ihj}} $. This yields the required answer.
Must [the order of the subscripts of the components] = [the order of the subscripts in the Levi-Civita symbol]? Why does this work?
Supplement due to Muphrid's Comment on July 17th:
$3.2.$ How can I divine to pull $G_j$ to the front, and NOT the back?
You did: $\color{purple}{\epsilon_{hij}G_j\partial_hF_i} = \color{purple}{(\epsilon_{hij}\partial_hF_i)G_j} = \color{purple}{G_j\epsilon_{\color{magenta}{jhi}}\partial_hF_i = \mathbf{G} \cdot (\nabla \times \mathbf{F})}. $
I would've computed: $\color{purple}{\epsilon_{hij}G_j\partial_hF_i} = \color{purple}{(\epsilon_{hij}\partial_hF_i)G_j = (\nabla \times \mathbf{F}) \cdot \mathbf{G}. \text{ But this is wrong! }}$
Supplement due to celtschk & Muphrid's Comments on July 22nd
From celtschk: It is also not strictly required that the order of indices maps the order in the vector expression. However it helps to do that because it's less error prone...
From Muphrid: Yes, the order of indices in the Levi-Civita matters...
$3.3. $ Don't these two comments contradict? Must [the order of the subscripts of the components] = [the order of the subscripts in the Levi-Civita symbol] ?
$3.4.$ Ought the order for both match? Why or why not?
Solution 1:
You are contracting the indices of $\mathbf F$ and $\mathbf G$ with each other; In $\mathbf G\cdot(\nabla\times\mathbf F)$ you'd contract it with the remaining index of $\epsilon$ (the indices of $\mathbf F$ and $\nabla$ are already contracted). However $\mathbf G\times\nabla \mathbf F$ does not make sense to me. $\partial_jF_i$ are the components of a rank-2 tensor; it might make sense to write that tensor as $\nabla\mathbf F$. However what doesn't make sense to me is to take the cross product. Looking closer to the expression, I notice that it doesn't make sense in index notation either because there are three occurrences of the index $i$ in the product; only one (free index) or two (summed index) are possible.
I don't see a difference to your first question. Just the roles of $\mathbf F$ and $\mathbf G$ are exchanged.
-
Your very first step is already wrong. You can easily see that by the fact that on the left hand side, $j$ is a free index, while on the right hand side, it is summed over (it appears twice in a product). A second mistake is the index $h$ which doesn't even appear on the LHS. A third mistake is the three occurrences of the index $i$.
Actually, I now even notice that your "zeroth" step also is wrong: The expression you are taking the $j$-th component from is a scalar, and therefore doesn't have a $j$-th component. (And yes, the index $j$ does mean the j-th component; anyway, considering the $j$-th term in isolation wouldn't make too much sense anyway).
Edit for 3.1 and 3.2:
Unless there's a derivative, which would determine the order. This is explained in the paragraph before "edit for 3.3 and 3.4". But otherwise, No, it is not required that $\text{[the order of the indices in the Levi-Civita symbol]}$
$= \text{[the order of the components in the component expression].}$
After all, they are just scalar factors (unless there's a derivative).
It is also not strictly required that [the order of indices in the Levi-Civita Symbol] maps the order in the vector expression. However it helps to do that because it's less error prone. What definitely matters is whether such a permutation is even or odd; this corresponds to the cross product being antisymmetric.
Putting the indices in the "right" order before conversion definitely helps with making no errors.
The second term in your 3rd equation is $\epsilon_{hij}F_i\partial_hG_j$. Since you have a derivative here, you have the additional constraint that the derivative has to come in front of what it derives, which in an expression of the form $\mathbf a\cdot(\mathbf b\times\mathbf c)$ can only happen if $\mathbf b$ is the derivative, and $\mathbf c$ is what is derived.
Therefore you want the index of $\partial$ second, and the index of $G$ third, so that those two appear in the vector product in exactly that way. That is, you want $\epsilon_{ihj}$ which you indeed get by exchanging $h$ and $i$, and therefore you get the sign change.
Edit for 3.3 and 3.4:
The order of the indices in the epsilon tensor matters, because reordering may change the sign. However, for permutations without a sign change (ie even ones), this order of the indices can change without affecting the final answer. Moreover, since the cross product is NOT commutative but the dot product is, thus in the vector expression, only the order of the vectors in the cross product matters, not the order in the dot product.
The order of the components in component form doesn't matter (except for derivative operators, of course). This means you can exchange the $A_i$, $B_j$ and $C_k$ any way you like (making sure you keep the correct index on each, of course) without changing the value: $$\epsilon_{ijk}A_iB_jC_k = \epsilon_{ijk}B_jA_iC_k = \epsilon_{ijk}B_jC_kA_i = \dots$$
Now if the order fits everywhere, then there are two possibilities for interpreting $\epsilon_{ijk}A_iB_jC_k$:
$$\begin{align} \epsilon_{ijk}A_iB_jC_k & = \color{#C33602}{\epsilon_{ijk}A_i(B_jC_k)} \qquad \text{ or } \color{#766303}{\qquad \epsilon_{ijk}(A_iB_j)C_k} \\ & = \color{#C33602}{\mathbf A\cdot(\mathbf B\times\mathbf C)} \qquad \text{ or } \color{#766303}{\qquad (\mathbf A\times\mathbf B) \cdot \mathbf C = \mathbf C \cdot (\mathbf A\times\mathbf B) } \end{align}$$ These two expressions are actually equal, due to the vector identity: $$\color{#C33602}{\mathbf A\cdot(\mathbf B\times\mathbf C)}=\mathbf B\cdot(\mathbf C\times\mathbf A) = \color{#766303}{\mathbf C\cdot(\mathbf A\times\mathbf B)}$$
So on both sides you can do odd permutations which require a sign change while even permutations which do NOT. On the LHS in terms of the components, a sign change equates with an odd permutation of the indices. On the RHS in terms of the vector expression, a sign change equates with an exchange of the two factors of the cross product (preceded and/or followed by one of the equivalent changes mentioned above, of course).
However, if
♦ the order of indices matches in the Levi-Civita symbol and the corresponding vector components
♦ and the order of the vector names matches in the component expression and the vector expression,
then you are on the safe side without putting much thought into it.
Solution 2:
By definition
$\nabla \cdot (F\times G)=\epsilon_{jik}\partial_j(F_iG_k)= \epsilon_{jik}\partial_jF_i G_k+\epsilon_{jik}F_i\partial_j G_k=(\nabla \times F) \cdot G -F\cdot (\nabla \times G),$
as $(\nabla \times F)_k=\epsilon_{kji}\partial_j F_i=\epsilon_{jik}\partial_j F_i,$ while $(\nabla \times G)_i=\epsilon_{ijk}\partial_j G_k=-\epsilon_{jik}\partial_j G_k.$
I used
$$\epsilon_{jik}=\epsilon_{kji}, $$ $$\epsilon_{ijk}=-\epsilon_{jik}. $$
Solution 3:
1) It's $\mathbf G \cdot (\nabla \times \mathbf F)$. This is a case where vector calculus unfortunately makes life more difficult for you. One way to explain it would be as follows.
You have $\epsilon_{hij} (\partial_h F_i) G_j$. You can permute the indices of the Levi-Civita to make the expression into something more readily recognizable:
$$\epsilon_{hij} (\partial_h F_i) G_j = [\epsilon_{jhi} \partial_h F_i] G_j$$
The term in square brackets is a curl, and so it can be represented as some vector $A_j = \epsilon_{jhi} \partial_h F_i$.
$$G_j [\epsilon_{jhi} \partial_h F_i] = G_j A_j = \mathbf G \cdot \mathbf A = \mathbf G \cdot (\nabla \times \mathbf F)$$
3) The answer probably shows a minus sign because they reorder the indices of the Levi-Civita symbol. You didn't write this, but it seems clear to me they intended this because of how they write the order of the components.
$$\epsilon_{hij} F_i \partial_h G_j = -\epsilon_{hji} F_i \partial_h G_j = - - \mathbf F \cdot (\nabla \times \mathbf G)$$
The minus sign follows because $hji$ is an odd permutation of $hij$.
What follows below is purely for mathematical interest.
I said that "vector calculus makes life difficult" because it's not immediately obvious how to convert the index notation identity you're given into a meaningful, coordinate free result again. All the indices ought to be on the same footing, but for some reason, they're not. It's the $G_h$ you have to pull out, and it's not immediately obvious that this must be the case.
One way to avoid this problem is to not use index notation at all. This is often done using differential forms, but I'll use something else called geometric calculus instead.
The identity you're trying to prove takes the following form in geometric calculus:
$$\nabla \wedge (\mathbf F \wedge \mathbf G) = (\nabla \wedge \mathbf F) \wedge \mathbf G - \mathbf F \wedge (\nabla \wedge \mathbf G)$$
These "wedge products" I use here are associative and antisymmetric. They do not result in vectors, but they are much better behaved than cross products because they are associative. Indeed, the proof becomes ridiculously easy. First, you apply the product rule:
$$\nabla \wedge (\mathbf F \wedge \mathbf G) = (\nabla \wedge \mathbf F) \wedge \mathbf G + \dot \nabla \wedge \mathbf F \wedge \dot{\mathbf G}$$
The overdot tells me that, in the second term, $\mathbf G$ is being differentiated, even though $\nabla$ isn't next to it. The associativity of the wedge lets me swap $\nabla$ and $\mathbf F$ at the cost of a minus sign.
$$(\nabla \wedge \mathbf F) \wedge \mathbf G + \dot \nabla \wedge \mathbf F \wedge \dot{\mathbf G} = (\nabla \wedge \mathbf F) \wedge \mathbf G - \mathbf F \wedge \nabla \wedge \mathbf G$$
No index gymnastics required. Wedge products make a lot of the arbitrary identities in 3d vector calculus make more sense.