I am looking at the proof of the following identity:
a x (b x c) = (a.c)b – (a.b)c
I have only just been introduced to Levi-Civita notation and the Kronecker delta, so could you please break down your answer using summations where possible.
My issue is with a specific line in the proof, so I will write out the proof and then state which line I am struggling to understand.
a x (b x c) $=\epsilon_{ijk}a_j$(b x c)$_k=\epsilon_{ijk}\epsilon_{klm}a_jb_lc_m$
Then applying the '$\epsilon-\delta$ Identity',
$\epsilon_{ijk}\epsilon_{klm}a_jb_lc_m=(\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl})a_jb_lc_m=a_jb_ic_j-a_jb_jc_j.$
The last line of this proof is the part I struggle with. I have tried expanding the bracket and changing the middle to
$(\delta_{il}b_l)\delta_{jm}a_jc_m-(\delta_{im}c_m)\delta_{jl}a_jb_l$
and then doing the summation of each separate bit but it doesn't seem to be getting me any closer to the final part.
Best Answer
The result should be a vector; you should either look at the $k$th component of $a\times (b\times c)$ or else add in an extra $e_k$ term: \begin{align*} u\times v = \epsilon_{ijk} u_j v_k e_i, \end{align*} taking the usual basis $e_1, e_2, e_3$ of $\mathbb{R}^3$. For fixed $i$, the $i$th component of $a\times (b\times c)$ is \begin{align*} a\times (b\times c) = \epsilon_{ijk} a_j (b\times c)_k = \epsilon_{ijk}\epsilon_{klm}\, a_j b_l c_m \end{align*} as stated. Furthermore, \begin{align*} \epsilon_{ijk}\epsilon_{klm} = \epsilon_{ijk}\epsilon_{lmk} = \delta_{il}\delta_{jm} - \delta_{im}\delta_{jl}, \end{align*} so \begin{align*} \epsilon_{ijk}\epsilon_{klm}\, a_j b_l c_m &= \delta_{il}\delta_{jm} a_j b_l c_m - \delta_{im}\delta_{jl} a_j b_l c_m = a_j b_i c_j - a_j b_j c_i, \end{align*} using the fact that $\delta_{ij} u_i v_j = u_i v_i$. (This is more or less the definition of $\delta$.) But \begin{align*} a_j b_i c_j - a_j b_j c_i = (a.c)b_i - (a.b)c_i, \end{align*} from which the result follows.