How can Ishow that $(\vec a \times \vec b) \cdot (\vec a \times \vec b)=|\vec a|^2|\vec b|^2-(\vec a \cdot \vec b)^2$ using index notation?

Question

I'm trying to use index notation (i.e. Einstein summation notation) in order to show that $(\vec a \times \vec b) \cdot (\vec a \times \vec b)=|\vec a|^2|\vec b|^2-(\vec a \cdot \vec b)^2$.

Here's what I've done so far, but I'm stuck at a dead end: $$\begin{align}(\vec a \times \vec b) \cdot (\vec a \times \vec b)&=(\varepsilon_{ijk}a_jb_k) (\varepsilon_{ijk}a_jb_k) \\ \ \\&=\varepsilon_{ijk}\varepsilon_{ijk}a_ja_jb_kb_k \\ \ \\ &=[\delta_{jj}\delta_{kk}-\delta_{jk}\delta_{kj}] a_ja_jb_kb_k \\ \ \\ &=[3\cdot 3-\underbrace{\delta_{kk}}_{3} \ ]a_ja_jb_kb_k \\ \ \\&=6a_ja_jb_kb_k\end{align}$$ but where do I go from here? Thanks

Answer 1

To clarify my remark in comments: Using any summation index more than twice renders the summation convention is insensible. This becomes obvious if you write in terms of $\Sigma's$: $$(\vec{a}\times \vec{b})^2=\left(\sum_{j,k=1}^3\epsilon_{ijk}a_j b_k\right)^2\neq \sum_{j,k=1}^3(\epsilon_{ijk}a_j b_k)^2.$$ What is correct is $$(\vec{a}\times \vec{b})^2=\left(\sum_{j,k=1}^3\epsilon_{ijk}a_j b_k\right)\left(\sum_{l,m=1}^3\epsilon_{ilm}a_l b_m\right)= \sum_{jklm}^3 \epsilon_{ijk}a_j b_k\epsilon_{ilm}a_l b_m.$$

So a second pair of indices is absolutely crucial in carrying out the algebra correctly.

Answer 2

Per my comment (which had a typo and should have said $b_jb_m$) \begin{align} \varepsilon_{ijk}a_ib_j\hat{e}_k\cdot\varepsilon_{lmn}a_lb_m\hat{e}_n &= \varepsilon_{ijk}\varepsilon_{lmn}a_ib_ja_lb_m(\hat{e}_k\cdot\hat{e}_n)\\ &= \varepsilon_{ijk}\varepsilon_{lmn}a_ib_ja_lb_m\delta_{kn}\\ &= \varepsilon_{ijk}\varepsilon_{lmk}a_ib_ja_lb_m\\ &= (\delta_{il}\delta_{jm} - \delta_{im}\delta_{jl})a_ib_ja_lb_m \end{align} When $i=l$ and $j=m$, we have $a_i^2b_j^2$. What happens when $i=m$ and $j=l$?

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One of the problems you faced by indexing the Levi-Civita symbol both as $\varepsilon_{ijk}$ is when you take the dot product. Upon taking the dot product, you have $$ \hat{e}_k\cdot\hat{e}_k = 1 $$ which leads $\varepsilon_{ijk}\varepsilon_{ijk} = 6$. If you would have had $\varepsilon_{ijk}\varepsilon_{ijn}$, it would have let to same problem since $\hat{e}_k\cdot\hat{e}_n = \delta_{kn}$