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Conserved Charges of Stress Energy Tensor

  1. Apr 3, 2014 #1
    Hello, Hi There

    I am trying to obtain the relations of the conserved charges of the stress tensor, it has 4, one is the hamiltonian and the other three are the momentum components.

    [itex]\vec{P}=-\int d^3y \sum_i{(-\pi_i(y) \nabla \phi_i(y))}[/itex]

    And i have to prove the conmutators

    [itex][\phi_i(x),\vec{P}]=-i \nabla\phi(x)[/itex] and [itex][\pi_i(x),\vec{P}]=i \nabla \pi_i(x)[/itex]

    I got the first one just fine

    [itex][\phi_i(x),\vec{P}]=-\int d^3 y \sum_j{[\phi_i(x),\pi_j(y)]\nabla \phi_j(y)}=
    -\int d^3 y \sum_j{i \delta_{ij} \delta^{(3)}(\vec{x}-\vec{y}) \nabla \phi_j(y)}=-i\nabla\phi_i(x) [/itex]



    But the second one is driving me crazy

    [itex][\pi(x),\vec{P}]=-\int d^3 y \sum_j{[\pi_i(x),\pi_j(y)]\nabla \phi_j(y)} [/itex]

    That conmutator is zero, ¿what i am doing wrong? how can those don't conmute.

    Also, whats the meaning of this relations

    Thans for the time
     
  2. jcsd
  3. Apr 3, 2014 #2
    Why are you taking the ##\phi## part out in the second case,in first commutation it works because ##\phi## will commute with other ##\phi## but in the second case it will be ##\pi## which will be taken outside because ##\pi's## will commute.Use by part in second commutation to shift the derivative on ##\pi## and then it's easy.
     
  4. Apr 3, 2014 #3
    Yes you are right andrien, i can't believe i didnt notice it myself, many thanks!, i was just considering it as a number for not clear reasons, thanks again.
     
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