Thursday, 27 August 2015

homework and exercises - Poisson brackets and angular momentum


I'm trying to find [Mi,Mj] Poisson brackets.


{Mi,Mj}=l(MiqlMjplMiplMjql)


I know that:


Mi=ϵijkqjpk


Mj=ϵjnmqnpm


and so:


[Mi,Mj]=l(ϵijkqjpkqlϵjnmqnpmplϵijkqjpkplϵjnmqnpmql)



=lϵijkpkδjlϵjnmqnδmllϵijkqjδklϵjnmpmδnl


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 {Mi,Mj}=piqj instead of {Mi,Mj}=qipjpiqj


Where am I wrong? Could you give me some hints to continue?



Answer



You are confusing in the index, such calculations must be carried out very carefully. I would start with your difention. Mi=ϵijkqjpk


Mp=ϵpnmqnpm

{Mi,Mp}=l(MiqlMpplMiplMpql)


First term


=ϵijkpkδjlϵpnmqnδml=ϵilkpkϵpnlqn=(1)ϵlikpk(1)2ϵlpnqn=ϵlikpkϵlpnqn=(δipδknδinδkp)pkqn


Here I used the antisymmetry of ϵlik and equation ϵijkϵimn=δjmδknδjnδkm


Second term



Absolutely the same calculations. =ϵijkqjδklϵpnmpmδnl=ϵijlqjϵplmpm=ϵplmpmϵijlqj=ϵlpmpmϵlijqj=(δpiδmjδpjδmi)pmqj=


Make the change m=k,j=n. Then


=(δpiδknδpnδki)pkqn


All together


{Mi,Mp}=(δipδknδinδkp)pkqn+(δpiδknδpnδki)pkqn=δinδkppkqnδpnδkipkqn=ppqipiqp=qipppiqp


No comments:

Post a Comment

Understanding Stagnation point in pitot fluid

What is stagnation point in fluid mechanics. At the open end of the pitot tube the velocity of the fluid becomes zero.But that should result...