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I am attempting to follow the solution provided by (Mark Viola) here (Kronecker delta and Levi-Civita symbol). I do not understand the following: Step 1 & 2

where the initial Levi-Civita symbols are expanded from only 1 term into 3.

I am assuming this is a lack of dot product and/or cross product knowledge. Any help would be appreciatetd.

user825535
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    (+1) for posting your question. I've posted a solution. Please let me know how I can improve my answer. I really want to give you the best answer I can. – Mark Viola Sep 16 '20 at 02:50

1 Answers1

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We begin with the identity

$$\begin{align} \varepsilon_{abc}\varepsilon_{k\ell m}&=(\hat x_a\cdot(\hat x_b\times \hat x_c))(\hat x_k\cdot(\hat x_\ell\times \hat x_m))\\\\ \tag1 \end{align}$$

In $(1)$, all indices are fixed.

Now, there are two cases. In Case $(i)$, none of the indices $(k,\ell,m)$ is equal to either of the other two and hence the number $a$ is equal to one of $k$, $\ell$, or $m$. In Case $(ii)$, two (or all three) of the indices $(k,\ell,m)$ are equal, in which case $\varepsilon_{k\ell m}=0$.

In Case $(i)$, we can write $\hat x_a$ as the sum

$$\hat x_a=\delta_{ak}\hat x_k+\delta_{a\ell}\hat x_\ell+\delta_{am}\hat x_m\tag 2$$

where $\delta_{ij}$ is the Kronecker Delta and we are not summing over the indices.

Note that we can substitute $(2)$ in $(1)$ in both Cases $(i)$ and $(ii)$ since for Case $(ii)$, $\varepsilon_{k\ell m}=0$. Therefore, enforcing this substitution reveals

$$\begin{align} \varepsilon_{abc}\varepsilon_{k\ell m}&=\left(\left(\delta_{ak}\hat x_k+\delta_{a\ell}\hat x_\ell+\delta_{am}\hat x_m\right)\cdot(\hat x_b\times \hat x_c)\right)(\hat x_k\cdot(\hat x_\ell\times \hat x_m))\\\\ \tag3 \end{align}$$

Finally, exploiting the property of the scalar triple product

$$\hat x_k\cdot (\hat x_\ell\times \hat x_m)=\hat x_\ell\cdot (\hat x_m\times \hat x_k)=\hat x_m\cdot (\hat x_k\times \hat x_\ell)$$

leads to the coveted eqality

$$\begin{align} \left(\hat x_a\cdot(\hat x_b \times \hat x_c)\right)\left(\hat x_k\cdot(\hat x_{\ell} \times \hat x_m)\right)&=\delta_{ak}\left((\hat x_b \times \hat x_c)\cdot \hat x_k\hat x_k\cdot(\hat x_{\ell} \times \hat x_m)\right)\\\\ &+\delta_{a\ell}\left((\hat x_b \times \hat x_c)\cdot \hat x_{\ell}\hat x_{\ell}\cdot(\hat x_m\times \hat x_k)\right)\\\\ &+\delta_{am}\left((\hat x_b \times \hat x_c)\cdot \hat x_{m}\hat x_{m}\cdot(\hat x_k\times \hat x_{\ell})\right) \end{align}$$

as was to be shown!

Mark Viola
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  • Hi @Mark, really appreciate you taking the time to provide such an in-depth answer. I understand the two different cases (i) and (ii). I also understand the use of the scalar triple product and substitution.

    However, I am not quite sure how we are arriving at EQN (2) and how x_a can be written in that form.

    EDIT: After mulling it over for a while longer, I seem to realize that since you are using the Kronecker Delta property (0 or 1), then only the term in which a = one of (k, l, m) will remain. That feels very genius and I understand it now. Thank you so much.

     – user825535 Sep 16 '20 at 15:55
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    You're welcome. My pleasure. As for the decomposition of $x_a$ note that the index $a$ must be one of $k$, $\ell$, or $m$. And the Kronecker $\delta_{ai}=1$ only when $a=i$. Otherwise it is $0$. Another way to proceed is to write $x_a=A\hat x_k+Bx_\ell+C\hat x_m$. To find $A$, $B$, and $C$, take the inner product of both sides with $\hat x_k$, $\hat x_\ell$, and $\hat x_k$, respectively. The left hand side will be a Kronecker Delta $\delta_{ak}$, $\delta_{a\ell}$, $\delta_{am}$ while the right hand side will be $A$, $B$, $C$, respectively. – Mark Viola Sep 16 '20 at 16:25