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Poisson brackets of angular momentum componentsby bznm
Tags: poisson 
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#1
Dec2712, 05:27 AM

P: 70

I want to find [M_i, M_j] Poisson brackets.
$$[M_i, M_j]=\sum_l (\frac{\partial M_i}{\partial q_l}\frac{\partial M_j}{\partial p_l}\frac{\partial M_i}{\partial p_l}\frac{\partial M_j}{\partial q_l})$$ I know that: $$M_i=\epsilon _{ijk} q_j p_k$$ $$M_j=\epsilon _{jnm} q_n p_m$$ and so: $$[M_i, M_j]=\sum_l (\frac{\partial \epsilon _{ijk} q_j p_k}{\partial q_l}\frac{\partial \epsilon _{jnm} q_n p_m}{\partial p_l}\frac{\partial \epsilon _{ijk} q_j p_k}{\partial p_l}\frac{\partial \epsilon _{jnm} q_n p_m}{\partial q_l})$$ $$= \sum_l \epsilon _{ijk} p_k \delta_{jl} \cdot \epsilon_{jnm} q_n \delta_{ml} \sum_l \epsilon_{ijk}q_j \delta_{kl} \cdot \epsilon_{jnm} p_m \delta_{nl}$$ Then I have thought that values that nullify deltas don't add any informations in the summations. And so, $$m=l, j=l$$ but so I obtain $$m=j$$. But if $$m=l$$, the second LeviCivita symbol in the first summation is zero... And if I go on, I obtain $$[M_i, M_j]=p_iq_j$$ instead of $$[M_i, M_j]=q_ip_jp_iq_j$$ Where am I wrong? : Could you say to me how to go on? Thanks a lot! 


#2
Dec2712, 07:15 AM

P: 611

You have 3 j's in the same term. Make sure your dummy indices (i.e. the ones that are summed over) are different from the variable indices. Use a different letter for each dummy index to avoid confusion.



#3
Dec3012, 02:44 AM

P: 70




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