Poisson brackets of angular momentum components


by bznm
Tags: poisson
bznm
bznm is offline
#1
Dec27-12, 05:27 AM
P: 62
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 Levi-Civita symbol in the first summation is zero... And if I go on, I obtain $$[M_i, M_j]=-p_iq_j$$ instead of $$[M_i, M_j]=q_ip_j-p_iq_j$$

Where am I wrong? :| Could you say to me how to go on? Thanks a lot!
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madness
madness is offline
#2
Dec27-12, 07:15 AM
P: 606
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.
bznm
bznm is offline
#3
Dec30-12, 02:44 AM
P: 62
Quote Quote by madness View Post
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.
Thank you!


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