0
$\begingroup$

I am stuck in an intermediate step.

In order to evaluate the product of the Levi-Civita symbols, we use the identity $$ \epsilon_{m n j} \epsilon_{i j k}=\delta_{m k} \delta_{n i}-\delta_{m i} \delta_{n k} $$ and the properties of the Kronecker delta functions. Thus, we obtain $$ \begin{aligned} \mathbf{a} \times(\mathbf{b} \times \mathbf{c}) &=\epsilon_{m n j} \epsilon_{i j k} a_i b_m c_n \mathbf{e}_k \\ &=a_i b_m c_n\left(\delta_{m k} \delta_{n i}-\delta_{m i} \delta_{n k}\right) \mathbf{e}_k \\ &=a_n b_m c_n \mathbf{e}_m-a_m b_m c_n \mathbf{e}_n \\ &=\left(b_m \mathbf{e}_m\right)\left(c_n a_n\right)-\left(c_n \mathbf{e}_n\right)\left(a_m b_m\right) \\ &=\mathbf{b}(\mathbf{a} \cdot \mathbf{c})-\mathbf{c}(\mathbf{a} \cdot \mathbf{b}) \end{aligned} $$

http://people.uncw.edu/hermanr/qm/Levi_Civita.pdf

$\endgroup$
1
  • $\begingroup$ just replace $k,i$ with $m,n$ in the first and then $k,i$ with $n,m$ in the second i.e. use the definition of $\delta_{xy}$ $\endgroup$ Commented Oct 17, 2022 at 13:52

1 Answer 1

0
$\begingroup$

Here you have,

$$ \begin{aligned} \mathbf{a} \times(\mathbf{b} \times \mathbf{c}) &=\epsilon_{m n j} \epsilon_{i j k} a_i b_m c_n \mathbf{e}_k \\ &=a_i b_m c_n\left(\delta_{m k} \delta_{n i}-\delta_{m i} \delta_{n k}\right) \mathbf{e}_k \\ &=a_i b_m c_n\delta_{m k} \delta_{n i}\mathbf{e}_k-a_i b_m c_n\delta_{m i} \delta_{n k} \mathbf{e}_k \end{aligned} $$ Now recall that this expression includes bunch of summations. In the the first term, $\delta_{m k} \delta_{n i}$ makes all other terms zero unless $m=k$ and $n=i$. So all that remains is $a_n b_m c_n \mathbf{e}_m$. Similarly the second term $\delta_{m i} \delta_{n k}$ makes all other terms zero unless $m=i$ and $n=k$. So the secomd term is $a_m b_m c_n \mathbf{e}_n.$ That gives you

$$ \begin{aligned} a_i b_m c_n\delta_{m k} \delta_{n i}\mathbf{e}_k-a_i b_m c_n\delta_{m i} \delta_{n k} \mathbf{e}_k&=a_n b_m c_n \mathbf{e}_m-a_m b_m c_n \mathbf{e}_n \\ &=\left(b_m \mathbf{e}_m\right)\left(c_n a_n\right)-\left(c_n \mathbf{e}_n\right)\left(a_m b_m\right) \\ &=\mathbf{b}(\mathbf{a} \cdot \mathbf{c})-\mathbf{c}(\mathbf{a} \cdot \mathbf{b}) \end{aligned} $$

$\endgroup$

You must log in to answer this question.