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I considered the $k$th component of $\text{curl $f\mathbf{F}$}$. $f$ is a scalar field and $\mathbf{F}$ a vector field.

$\color{green}{[}\nabla \times (fF)\color{green}{]} _{\LARGE{\color{green}{k}}} = \epsilon_{ij\LARGE{\color{green}{k}}}\partial_i(f\mathbf{F})_j $
$= \epsilon_{ij\LARGE{\color{green}{k}}}\partial_i(fF_j) \qquad \qquad \qquad \qquad (\text{since $\mathbf{(F)}_j :=$ the $j$th component of $\mathbf{F} = F_j$})$
$= \epsilon_{ij\LARGE{\color{green}{k}}}(F_j\partial_if + f\partial_iF_j) \qquad \qquad (\text{since $f$ scalar})$
$= \underbrace{\epsilon_{ij\LARGE{\color{green}{k}}}F_j\partial_if}_{\Large{\bigstar}} + f\underbrace{{\epsilon_{ij\LARGE{\color{green}{k}}}\partial_iF_j}}_{\LARGE{\color{green}{[}\nabla \times \mathbf{F}\color{green}{]}_{\LARGE{\color{green}{k}}}}} $

Hereafter, I refer only to the term with the star underneath. Since [$\color{#007FFF}{F_j}$ corresponding to $\color{#007FFF}{\mathbf{F}}$] appears before [$\color{#FF00FF}{\partial_if}$ corresponding to $\color{#FF00FF}{\nabla f}$], thus ${\epsilon_{ij\LARGE{\color{green}{k}}}\color{#007FFF}{F_j}\color{#FF00FF}{\partial_if}} = {\color{green}{[}\color{#007FFF}{\mathbf{F}} \times \color{#FF00FF}{\nabla f}\color{green}{]} _{\LARGE{\color{green}{k}}}}$.
But the answer states $\color{green}{[}\nabla f \times \mathbf{F}\color{green}{]} _{\LARGE{\color{green}{k}}}$. What went wrong?


$\large{\text{Supplement to Andrew D's response :}}$

Here's my understanding of your answer : In ${\epsilon_{ij\LARGE{\color{green}{k}}}F_j\partial_if}, \; {(i, j, \LARGE{\color{green}{k}})}$ (in the subscript of the Levi-Civita symbol) denotes the order of the components. So the $i$th component must appear first, and the $j$th component second.

However, since $(i, j, k) = \color{brown}{(j, k, i)}$, therefore $\epsilon_{ijk} = \epsilon_{\color{brown}{\LARGE{jki}}}$. $\color{brown}{\text{Now, $j$ precedes $i$, so wouldn't this result in the wrong order of the components?}}$


$\large{\text{2nd Supplement to Andrew D's Comment beneath his Answer :}}$

$\color{#3EB489}{\text{The variable in the permutation succeeding the variable that's not summed}}$ corresponds to the first component to appear. Here, $k$ denotes the component being analysed so is not summed. Since I am looking at $\color{brown}{(j, k, i)}$, $\color{#3EB489}{i}$ succeeds $k$ so the $\color{#3EB489}{i}$th component is the first. Therefore, ${\epsilon_{ijk}F_j\color{#3EB489}{\partial_{\LARGE{i}}}f}$ = ${\epsilon_{\color{brown}{\LARGE{jki}}}\color{#3EB489}{\partial_{\LARGE{i}}}F_jf} = \color{green}{[}\color{#3EB489}{\nabla f} \times \mathbf{F}\color{green}{]} _{\LARGE{\color{green}{k}}} $

However, this appears to discord with Steven Stadnicki's 2nd comment, according to which: $ {\epsilon_{\color{brown}{\LARGE{jki}}}F_j\partial_if} = {\color{green}{[}\mathbf{F} \times \nabla f\color{green}{]} _{\LARGE{\color{green}{k}}}}$?

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    $\begingroup$ Can you see that it makes no difference whether you write $\varepsilon_{ijk}F_j\partial_if$ or $\varepsilon_{ijk}(\partial_if)F_j$? $\endgroup$ Jul 2, 2013 at 15:56
  • $\begingroup$ Also, to your last point: the critical element is that your final index $k$ needs to correspond to the last element of the Levi-civita symbol, not the middle one. Otherwise you would have $(a\times b)_k = \varepsilon_{ijk}a_ib_j = \varepsilon_{jki}b_ja_i = (b\times a)_k$. $\endgroup$ Jul 2, 2013 at 15:59
  • $\begingroup$ @StevenStadnicki: Thank you for your comments. Wrt your first comment: Does it make no difference because we are just multiplying 0the components therein? Wrt your second comment: Could you please see the 2nd supplement that I've added to my original Question? $\endgroup$
    – user53259
    Jul 3, 2013 at 14:02
  • $\begingroup$ I had hoped that for the rest of my life I would not have to witness such a cumbersome proof of an identity about ${\rm curl}(f{\bf F})$. In my opinion the "summation convention" was Einstein's biggest mistake. It prevented people from thinking what's going on here. $\endgroup$ Oct 3, 2013 at 10:57
  • $\begingroup$ @ChristianBlatter Einstein is reputed to have quipped that it was his greatest contribution to mathematics; what would you consider a less cumbersome proof? $\endgroup$
    – Rax Adaam
    Mar 9, 2018 at 0:54

3 Answers 3

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It isn't the ordering of $F_j$ and $\partial_if$ in the product of terms which determines what order the terms are in the cross product, it is the ordering of the suffix's which does so - as we have $\{i,j,k\}$ as our right-handed set (taken from the ordering of the suffices from the Levi-Civita symbol given), it means that we take the $\partial_if$ as being the first component of the cross product and $F_j$ as our second, so we get $[ \nabla f \times \mathbf F]_k$ as required.

If this doesn't make much sense, say so and I'll try and clarify what I'm saying.

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  • $\begingroup$ @LePressentiment Sure thing - I feel my original answer may have been slightly confusing: the important thing is that we are considering the order of the suffices after the k (or whatever the suffix we are not summing over may be), so we still take the $i$ first and then the $j$. Does that make sense? $\endgroup$
    – Andrew D
    Jul 2, 2013 at 15:40
  • $\begingroup$ Thank you very much again. Yes, but another issue appears to have sprung. Could you please look at the 2nd supplement added to my original post? $\endgroup$
    – user53259
    Jul 3, 2013 at 14:04
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Recall that $\vec{a} \times \vec{b} = - \, \vec{b} \times \vec{a}$.

Now consider each, using your definitions (I have included the unit vector $\hat{e}_i$ explicitly, simply for additional clarity; see below):

$$ \vec{a} \times \vec{b} = \epsilon_{i{\color{red}j}{\color{blue}k}} \, \hat{e}_i \, a_{\color{red}j} \, b_{\color{blue}k} = - \, \vec{b} \times \vec{a} = - \epsilon_{i{\color{red}j}{\color{blue}k}} \, \hat{e}_i \, b_{\color{red}j} \, a_{\color{blue}k}.$$

Recall that $\epsilon_{ijk}$ is perfectly anti-symmetric, which means that interchanging any two indices changes its sign (there are only 6 nonzero terms, so it's not long to verify this by plugging in actual values), for example:

$$\epsilon_{1{\color{red}2}3} = - \epsilon_{13{\color{red}2}}.$$

Or, in general, $-\epsilon_{i{\color{red}j}k} = \epsilon_{ik{\color{red}j}}$. Put this together with our result for $-\vec{b} \times \vec{a}$, above, we have:

$$- \, \vec{b} \times \vec{a} = - \epsilon_{i{\color{red}j}{\color{blue}k}} \, \hat{e}_i \, b_{\color{red}j} \, a_{\color{blue}k} = \epsilon_{i{\color{blue}k}{\color{red}j}}\, \hat{e}_i \,a_{\color{blue}k} \, b_{\color{red}j} = \vec{a} \times \vec{b},$$

as expected.

N.B. the ordering of the $a_{j/k}$ and $b_{j/k}$ terms themselves doesn't matter, it's the relative position of their index in $\epsilon_{ijk}$ that matters. ie.

$$\begin{align*}\vec{a} \times \vec{b} & = \epsilon_{ijk} \, \hat{e}_i \, a_j \, b_k \\ & = \epsilon_{ijk} \hat{e}_i \, b_k \, a_j \\ & = \epsilon_{jki} \hat{e}_j \, a_k \, b_i \\ & = \epsilon_{kij} \hat{e}_k \, a_i \, b_j \end{align*} $$

In the beginning, it can be easy to conflate $i$, $j$ and $k$ with the unit vector in Cartesian space, but they aren't the same thing.

Notice this means you can cycle through the equivalent forms of epsilon until you find a form that "matches" the cross product. So, in your case:

$$\begin{align*} & \quad \epsilon_{ijk} F_j \partial_i f \\ & = \epsilon_{jki} F_j\partial_i f\\ & = \epsilon_{kij} F_j\partial_i f \end{align*}$$

Which we can now read as $\nabla f \times \vec{F}$ (or, equivalently: $-\, \vec{F} \times \nabla f$).


On including the unit vector in the definition of the cross product (because it seems there is some conflation between the indices $i$, $j$, $k$ and the basis vectors $\hat{i} = \hat{e}_1$, $\hat{j} = \hat{e}_2$ and $\hat{k}=\hat{e}_3$).

I chose to include the unit vector in my definition, because it removes this "components" expression. Instead, we can just give an expression for the (entire) cross product:

$$ \vec{a} \times \vec{b} = \epsilon_{ijk} \, \hat{e}_i \, a_j \, b_k.$$

This is the same as your definition; however, I've included the unit vector $\hat{e}_i$ explicitly. Notice, for example when $i = 1$, we get the first component of the cross-product:

$$ \epsilon_{123} \hat{e}_1 a_2b_3 + \epsilon_{132} \hat{e}_1 a_3b_2 = \hat{e}_1\, a_2b_3 - \hat{e}_1 \,a_3b_2 = (a_2b_3 - a_3b_2) \,\hat{e}_1,$$

and similarly for the other two components.

However, the naming of the indices is totally arbitrary, so we could equally write:

$$ \vec{a} \times \vec{b} = \epsilon_{lmn} \, \hat{e}_l \, a_m \, b_n, $$

or even,

$$ \vec{a} \times \vec{b} = \epsilon_{kji} \, \hat{e}_k \, a_j \, b_i, $$

because this represents every possible combination for all three indices equal to any one of 1, 2, or 3. (so $3^3 = 27$ different terms, where all but 6 are zero).

Now, once you've chosen your indices, order does matter, in particular all cyclic permutations [(1,2,3), (2,3,1), (3,1,2)] will have the same value, and all acyclic [(1,3,2), (3,2,1) and (2,1,3)] permutations will have the same value, i.e.

$$\vec{a} \times \vec{b} = \epsilon_{i{\color{blue}j}{\color{red}k}}\, \hat{e}_i \, a_{{\color{blue}j}} b_{\color{red}k} = \epsilon_{j{\color{blue}k}{\color{red}i}}\, \hat{e}_j \, a_{{\color{blue}k}}b_{\color{red}i} = \epsilon_{k{\color{blue}i}{\color{red}j}}\,\hat{e}_k \, a_{\color{blue}i} b_{\color{red}j}$$ whereas $$ \epsilon_{i{\color{red}k}{\color{blue}j}}\,\hat{e}_i \, a_{{\color{blue}j}} b_{\color{red}k} = \epsilon_{j{\color{red}i}{\color{blue}k}}\, \hat{e}_j \, a_{{\color{blue}k}}b_{\color{red}i} = \epsilon_{k{\color{red}j}{\color{blue}i}}\, \hat{e}_k \, a_{\color{blue}i} b_{\color{red}j} = \vec{b} \times \vec{a}$$

Hope this helps!

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Terms as $\epsilon_{ijk}a_ib_i$ do not carry any information on operations between $a_i$ and $b_j$, hence no information on e.g. commutativity. They really are meant as summing symbols (maybe with no meaning at al)l $F_j$ over indices which appear twice, no matter in what order these symbols are arranged: $$\epsilon_{ijk}a_ib_j=a_i\epsilon_{ijk}b_j=b_ja_i\epsilon_{ijk}=\dots$$ (Maybe, writing those kind of sums explicitly for e.g. $i,j,k=1,2,3$ helps.)

Concerning your 2nd supplement, you only need to keep in mind that $$(\nabla f)_i:=\partial_if:=\frac{\partial f}{\partial x^i}$$ denotes one single symbol with index. Therefore, $\epsilon_{ijk}\partial_iF_jf$ does not make sense in this context.

Finally, Steven Stadnicki's first comment is most important.

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