Poisson Brackets / Levi-Civita Expansion

[Bismar]

Hi,

I am stumped by how to expand/prove the following identity,

$$\{L_i ,L_j\}=\epsilon_{ijk} L_k$$

I am feeling that my knowledge on how to manipulate the Levi-Civita is not up to scratch.

Am i correct in assuming,

$$L_i=\epsilon_{ijk} r_j p_k$$
$$L_j=\epsilon_{jki} r_k p_i$$

Which follows on to,

$$\{L_i ,L_j\}=\{\epsilon_{ijk} r_j p_k,\epsilon_{jki} r_k p_i\}$$

And then I'm stuck. I'm assuming the Kronecker Delta identities with the Levi-Civita work into it in some way, but i do not understand how/why.

I can work it out if i expanded the Levi-Civita to such,

$$L_1=r_2 p_3 - r_3 p_2$$
$$L_2=r_3 p_1 - r_1 p_3$$
$$L_3=r_1 p_2 - r_2 p_1$$

But then that's trivial... :(

 

[dextercioby]

Please, use different summation indices.

$$\left\{\epsilon_{ilk} r_{l}p_{k}, \epsilon_{jmn}r_{m}p_{n}\right\}$$

Then, of course,

$$\left\{r_i, p_j\right\} = \delta_{ij}$$

 

[Bismar]

Sorry, I'm afraid i do not understand, where does that get you?

$$\epsilon_{ikl} \epsilon_{jmn}(\frac{dr_k p_l}{dr_o} \frac{dr_m p_n}{dp_o} - \frac{dr_m p_n}{dr_o} \frac{dr_k p_l}{dp_o})$$

$$= \epsilon_{ikl} \epsilon_{jmn}(\delta_{ko}\delta_{no}p_l r_m -\delta_{mo}\delta_{lo} p_n r_k)$$

If that's even right, which I'm sure isn't, I'm stuck again.

 

[vanhees71]

$$\begin{split}<br /> \{L_a,L_b\} &=\epsilon_{acd} \epsilon_{bef} \{x_c p_d,x_e p_f\} \\<br /> &= \epsilon_{acd} \epsilon_{bef} (\{x_c,x_e p_f \} p_d+x_c \{p_d,x_e p_f \}) \\<br /> &= \epsilon_{acd} \epsilon_{bef}(x_e p_d\delta_{cf} -x_c p_f \delta_{de}) \\<br /> &= \epsilon_{acd} \epsilon_{bec} x_e p_d - \epsilon_{acd} \epsilon_{bdf} x_c p_f \\<br /> &=[(-\delta_{ab} \delta_{de}+\delta_{ae} \delta_{db}) x_e p_d+ (\delta_{ab} \delta_{cf} - \delta_{af} \delta_{cb}) x_c p_f] \\<br /> &= -\delta_{ab} \vec{x} \cdot \vec{p} + x_a p_b+\delta_{ab} \vec{x} \cdot \vec{p} -x_b p_a \\<br /> &=x_a p_b-x_b p_a=\epsilon_{abc} L_c<br /> \end{split}$$