I wanted to use the Levi-Civita permutation symbol, $\epsilon$ and the Kronecker delta, $\delta$ to prove that $$\vec a\times (\vec b\times\vec c) = (\vec a\cdot\vec c)\vec b - (\vec a\cdot\vec b)\vec c$$However I became confused as to what to do at the end and I had the minus signs the wrong way round. Here's what I've done so far: $$\vec a\times (\vec b\times\vec c) = a_i\hat e_i\times(\epsilon_{nlm}b_lc_m\hat e_n)$$ $$=\epsilon_{nlm}a_ib_lc_m\hat e_i\times\hat e_n$$ $$=\epsilon_{inx}\epsilon_{nlm}a_ib_lc_m\hat e_x$$ where $x$ is just an arbitrary index I put in.$$\epsilon_{inx}\epsilon_{nlm} = (\delta_{il}\delta_{xm} - \delta_{im}\delta_{xl})$$ $$\vec a\times (\vec b\times\vec c) = a_ib_lc_m\hat e_x(\delta_{il}\delta_{xm} - \delta_{im}\delta_{xl})$$ $$=a_ib_lc_m\hat e_x\delta_{il}\delta_{xm} - a_ib_lc_m\hat e_x\delta_{im}\delta_{xl}$$ $$=a_lb_lc_m\hat e_m - a_mb_xc_m\hat e_x$$This is the part where I got stuck on. I know that $a_lb_l = (\vec a\cdot \vec b)$ and that $c_m\hat e_m = \vec c$ but when I put this in I obtain$$(\vec a\cdot\vec b)\vec c-(\vec a\cdot\vec c)\vec b$$ which is the correct answer times $-1$. Does anyone know where I could've gone wrong? Perhaps in working out $\epsilon_{inx}\epsilon_{nlm}$?

  • $\begingroup$ You want to check if $\epsilon_{inx} \epsilon_{n\ell m} \stackrel{?}{=} \delta_{i\ell} \delta_{xm} - \delta_{im} \delta_{x\ell}$. Just check it with indices. Set $i=\ell=1$ and $x=m=2$ so that we check $\epsilon_{1n2} \epsilon_{n12} = \epsilon_{132} \epsilon_{312} = - 1 \stackrel{?}{=} \delta_{11} \delta_{22} = 1$. Clearly it is off by a sign. $\endgroup$ – Prahar Mitra Aug 20 '17 at 17:21
  • $\begingroup$ @Prahar So to clarify $\epsilon_{inx}\epsilon_{nlm} = (\delta_{im}\delta_{xl} - \delta_{il}\delta_{xm})$? Because that would make this work nicely :p $\endgroup$ – Jordan Abbott Aug 20 '17 at 17:23
  • $\begingroup$ This is something you should very very easily be able to check even by brute force (i.e. by specifically checking it for each index value) so that's what you should do if you need convincing. $\endgroup$ – Prahar Mitra Aug 20 '17 at 17:25
  • $\begingroup$ Okay thanks for your comment I'll keep that in mind next time I try something like this :) $\endgroup$ – Jordan Abbott Aug 20 '17 at 17:26

I find it easiest to permute the indices to get the unit vector index at the end.

$$\epsilon_{inx} = \epsilon_{xin}$$ $$ \epsilon_{nlm} = \epsilon_{lmn}$$ Now it's easy to see that when x & l are the same, you get +1, and when x/m and i/l match, it's -1

$$\epsilon_{xin}\epsilon_{lmn} = (\delta_{xl}\delta_{im} - \delta_{xm}\delta_{il})$$

Which explains your missing $-$ sign.

  • $\begingroup$ Thanks that'll help me a lot when it comes to tackling similar problems in the future :p $\endgroup$ – Jordan Abbott Aug 20 '17 at 17:28

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