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Pauli-lubanski pseudo-vecor

  1. Jul 15, 2008 #1
    Hi

    I’m wondering if anybody could help me to prove the result of W2 (square of the pauli-lubanski pseudo-vecor).
     
  2. jcsd
  3. Jul 15, 2008 #2
    Re: Pauli-Lubanski

    See first equation on page 117 of http://www.arxiv.org/abs/physics/0504062

    Eugene.
     
  4. Jul 15, 2008 #3
    Re: Pauli-Lubanski

    Thank you for your reply but in my case I need to start with

    Wa=(1/2)EabcdMbcPd

    And to end with

    W2=-(1/2)MabMabP2+MacMbcPaPb

    Could any body tell me how to derive this
     
  5. Jul 15, 2008 #4

    samalkhaiat

    User Avatar
    Science Advisor

    Re: Pauli-Lubanski

    use the identity

    [tex]
    \epsilon_{a}^{bcd}\epsilon^{a \bar{b} \bar{c} \bar{d}} = - \left| \begin{array}{ccc} \eta^{\bar{b}b} & \eta^{\bar{c}b} & \eta^{\bar{d}b} \\ \eta^{\bar{b}c} & \eta^{\bar{c}c} & \eta^{\bar{d}c} \\ \eta^{\bar{b}d} & \eta^{\bar{c}d} & \eta^{\bar{d}d} \end{array} \right|
    [/tex]

    and

    [tex]M_{ab} = - M_{ba}[/tex]


    regards


    sam
     
  6. Sep 8, 2008 #5
    Re: Pauli-Lubanski

    i tried to get an explicit form of this (Pauli-Lubanski pseudo vector) and i keep getting
    zero. i.e W=(0,0,0,0)...which would still make it an invariant; albeit a boring one.
    Looking at the form too, with M_ab=-M_ba it does seem that it should be zero.
    Can someone tell me what i might be doing wrong. Thanks
     
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