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L11E01.tex
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L11E01.tex
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\documentclass[solutions.tex]{subfiles}
\xtitle
\begin{document}
\maketitle
\begin{exercise}
Confirm Eq. $(3)$. Also prove that
\[
V_iA_j - V_jA_i = \sum_k \epsilon_{ijk}\Bigl(\vec{V} \times \vec{A}\Bigr)_i
\]
\end{exercise}
Let's recall that the Levi-Civita symbol is defined by:
\[
\epsilon_{ijk} = \begin{cases}
+1 & \text{if } (i,j,k) \text{ is } (1,2,3), (2,3,1), \text{ or } (3,1,2), \\
-1 & \text{if } (i,j,k) \text{ is } (3,2,1), (1,3,2), \text{ or } (2,1,3), \\
0 & \text{if } i = j, \text{ or } j = k, \text{ or } k = i
\end{cases}
\]
Eq. $(3)$ refers to:
\[
\Bigl(\vec{V} \times \vec{A}\Bigr)_i = \sum_j\sum_k\epsilon_{ijk}V_jA_k
\]
So the idea is to express the components of the cross-product of
two $3$D vectors with the Levi-Civita symbol. Let's have a look at
the components of the cross-product of two vectors:
\begin{equation*}\begin{aligned}
\Bigl(\vec{V}\times\vec{A}\Bigr)_x &&=&& V_yA_z - V_zA_y \\
\Bigl(\vec{V}\times\vec{A}\Bigr)_y &&=&& V_zA_x - V_xA_z \\
\Bigl(\vec{V}\times\vec{A}\Bigr)_z &&=&& V_xA_y - V_yA_x \\
\end{aligned}\end{equation*}
\begin{remark} There's a typo in the book the last term should
contain an $A_x$, but the book says its an $A_z$. There's another
typo in the exercise actually; we'll get to it in a moment.
\end{remark}
Observe that somehow, all those components are "equivalent", or
"symmetric": for instance, we can get the second line from the
first, by renaming in the first $x$ by $y$, $y$ by $z$ and $z$ by
$x$. \\
This implies that to verify Eq. $(3)$, we can satisfy ourselves with
doing it only for one component, as the procedure would be exactly
similar for the two others. So, let's get going, for instance by
trying to prove the first line:
\begin{equation*}\begin{aligned}
\Bigl(\vec{V} \times \vec{A}\Bigr)_x &&=&& \sum_j\sum_k\epsilon_{xjk}V_jA_k \\
~ &&=&& \sum_k\underbrace{\epsilon_{xxk}}_{=0}V_xA_k
+\sum_k\epsilon_{xyk}V_yA_k
+\sum_k\epsilon_{xzk}V_zA_k \\
~ &&=&& \sum_k\Bigl(\epsilon_{xyk}V_yA_k+\epsilon_{xzk}V_zA_k\Bigr) \\
~ &&=&& \underbrace{\epsilon_{xyx}}_{=0}V_yA_x
+\underbrace{\epsilon_{xzx}}_{=0}V_zA_x
+ \underbrace{\epsilon_{xyy}}_{=0}V_yA_y+\epsilon_{xzy}V_zA_y
+ \epsilon_{xyz}V_yA_z+\underbrace{\epsilon_{xzz}}_{=0}V_zA_z \\
~ &&=&& \underbrace{\epsilon_{xzy}}_{=-1}V_zA_y +
\underbrace{\epsilon_{xyz}}_{=1}V_yA_z \\
~ &&=&& V_yA_z - V_zA_y \qed
\end{aligned}\end{equation*}
For similar reasons (symmetry), we only have to consider the case e.g.
$i=x$ of the remaining equation to be done with it, as the two others
would be derived identically, but for some systematic renaming. \\
We then have three sub-cases, depending on the value of $j$.
If $j=i (=x)$, then one one side:
\[ V_iA_j - V_jA_i = V_iA_i - V_iA_i = 0 \]
And on the other:
\[ \sum_k \underbrace{\epsilon_{iik}}_{=0}\Bigl(\vec{V} \times \vec{A}\Bigr)_i=0 \]
And so the equation holds. Now let's consider the case where $j=y$.
One one side we have:
\[ V_iA_j - V_jA_i = V_xA_y - V_yA_x \]
And on the other:
\begin{equation*}\begin{aligned}
\sum_k \epsilon_{xjk}\Bigl(\vec{V} \times \vec{A}\Bigr)_x &&=&&
\sum_k \epsilon_{xjk}\Bigl(V_yA_z - V_zA_y\Bigr) \\
~ &&=&& \sum_k \epsilon_{xyk}\Bigl(V_yA_z - V_zA_y\Bigr) \\
~ &&=&& \Bigl(V_yA_z - V_zA_y\Bigr)
(\underbrace{\epsilon_{xyx}}_{=0} + \underbrace{\epsilon_{xyy}}_{=0}
+ \underbrace{\epsilon_{xyz}}_{=1}) \\
~ &&=&& V_yA_z - V_zA_y
\end{aligned}\end{equation*}
Well, the computations are right, but obviously the result isn't! There's
a typo in the book: we're expected to prove:
\[
V_iA_j - V_jA_i = \sum_k \epsilon_{ijk}\Bigl(\vec{V} \times \vec{A}\Bigr)_k
\]
So, let's start again the development of the right hand side:
\begin{equation*}\begin{aligned}
\sum_k \epsilon_{xjk}\Bigl(\vec{V} \times \vec{A}\Bigr)_k &&=&&
\sum_k \epsilon_{xyk}\Bigl(\vec{V} \times \vec{A}\Bigr) \\
~ &&=&&
\underbrace{\epsilon_{xyx}}_{=0}\Bigl(\vec{V} \times \vec{A}\Bigr)_x
+\underbrace{\epsilon_{xyy}}_{=0}\Bigl(\vec{V} \times \vec{A}\Bigr)_y
+\underbrace{\epsilon_{xyz}}_{=1}\Bigl(\vec{V} \times \vec{A}\Bigr)_z \\
~ &&=&& \Bigl(\vec{V} \times \vec{A}\Bigr)_z\\
~ &&=&& V_xA_y - V_yA_x \qed
\end{aligned}\end{equation*}
Which corresponds to the left-hand side. Let's do it once more with $j=z$.
On one side:
\[ V_iA_j - V_jA_i = V_xA_z - V_zA_x \]
On the other:
\begin{equation*}\begin{aligned}
\sum_k \epsilon_{xjk}\Bigl(\vec{V} \times \vec{A}\Bigr)_k &&=&&
\sum_k \epsilon_{xzk}\Bigl(\vec{V} \times \vec{A}\Bigr) \\
~ &&=&&
\underbrace{\epsilon_{xzx}}_{=0}\Bigl(\vec{V} \times \vec{A}\Bigr)_x
+\underbrace{\epsilon_{xzy}}_{=-1}\Bigl(\vec{V} \times \vec{A}\Bigr)_y
+\underbrace{\epsilon_{xzz}}_{=0}\Bigl(\vec{V} \times \vec{A}\Bigr)_z \\
~ &&=&& -\Bigl(\vec{V} \times \vec{A}\Bigr)_y\\
~ &&=&& -(V_zA_x - V_xA_z) \\
~ &&=&& V_xA_z - V_zA_x \qed
\end{aligned}\end{equation*}
\end{document}