Poisson brackets and angular momentum

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SUMMARY

The discussion focuses on the application of Poisson brackets in the context of angular momentum, specifically addressing the properties and calculations involving these mathematical constructs. The Poisson bracket is defined as [f, g] = ∂f/∂q_i ∂g/∂p_i - ∂f/∂p_i ∂g/∂q_i. The participant successfully demonstrated the identities [fg, h] = f[g, h] + [f, h]g and [Lj, Lk] = qjpk - qkpj, leading to the conclusion that [Li, |L|^2] = 0. The discussion also highlights the confusion between Poisson brackets and commutators, clarifying the notation used in different contexts.

PREREQUISITES
  • Understanding of Poisson brackets and their properties
  • Familiarity with angular momentum in classical mechanics
  • Knowledge of the permutation symbol (εijk) and its applications
  • Basic calculus, particularly partial derivatives
NEXT STEPS
  • Study the derivation and applications of the Poisson bracket in Hamiltonian mechanics
  • Explore the relationship between Poisson brackets and quantum mechanical commutators
  • Investigate the implications of angular momentum conservation in various physical systems
  • Learn about the role of the permutation symbol in tensor calculus and physics
USEFUL FOR

This discussion is beneficial for physics students, particularly those studying classical mechanics, as well as researchers and educators focusing on the mathematical foundations of angular momentum and phase space analysis.

Rhi
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Homework Statement



Let f(q, p), g(q, p) and h(q, p) be three functions in phase space. Let Lk =
εlmkqlpm be the kth component of the angular momentum.
(i) Define the Poisson bracket [f, g].
(ii) Show [fg, h] = f[g, h] + [f, h]g.
(iii) Find [qj , Lk], expressing your answer in terms of the permutation symbol.
(iv) Show [Lj , Lk] = qjpk−qkpj . Show also that the RHS satisfies qjpk−qkpj =
εijkLi. Deduce [Li, |L|2] = 0.
[Hint: the identity εijkεklm = δilδjm − δimδjl may be useful in (iv)]



Homework Equations

n/a

The Attempt at a Solution




i) [f,g]=\frac{\partial f}{\partial q_i}\frac{\partial g}{\partial p_i}\frac{\partial f}{\partial p_i}\frac{\partial g}{\partial q_i}

ii) easy to show from the definition in i)

iii) after a bit of working, I get εlmkql

iv) my working is quite long, but I get [Lj,Lk]=qjpk-qkpjijkLi as required.

The bit I'm having trouble with is the very last bit of the question, to deduce [Li, |L|2] = 0.

Since it's only a small part of the question, it seems as though this part should be fairly simple so maybe I'm overlooking something, but I don't get 0. This is my working:

[Li, |L|2]=[Li, LjLj]=Lj[Li, Lj]+[Li, Lj]Lj=2Lj[Li, Lj]

I'm not entirely sure where to go from here so any help (or pointing out of any glaring errors) would be great.
 
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I don't really see how in the last line you get to 2L_j[L_i,L_j] if this commutator contains a L_k and the L_subscripts don't commute.
 
I'm not really sure what you mean, there's no L_k involved in the last line? And I'm not really sure what you mean by a commutator, either..
 
Haha I was being infinitely stupid. I forgot you were talking about Poisson brackets. I have a lame excuse for it though namely that I usually use { , } for poisson and [ , ] for commutator. Now to redeem myself I will actually look at this last exercise. be back shortly!
 
So here it goes. Leave the last equality out and when you get

L_j[L_i,L_j]+[L_i,L_j]L_j = L_j\epsilon_{kij} L_k + \epsilon_{kij} L_kL_j = \epsilon_{kij}L_jL_k - \epsilon_{jik} L_kL_j = \epsilon_{kij}L_jL_k - \epsilon_{kij} L_jL_k = 0

Where in the equality before last I just relabel j to k and vice versa in the second summand.
 
Ah, I get it. That makes a lot of sense, cheers :)
 

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