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Pauli Lubanski vector always zero?

  1. Oct 17, 2014 #1
    1. The problem statement, all variables and given/known data
    Hi
    i've noticed that,
    [tex]J^{ab}P^c + J^{bc}P^a+J^{ca}P^b = 0[/tex]
    [
    Since
    [tex]J^{ab}P^c + J^{bc}P^a+J^{ca}P^b = x^a\partial^b\partial^c -x^b\partial^a\partial^c + x^b\partial^c\partial^a-x^c\partial^b\partial^a+x^c\partial^a\partial^b-x^a\partial^c\partial^b = 0[/tex]
    ]

    Next, I tried it on the Pauli Lubanski vector. Suppose I'll look at W0 as an example
    [tex]W_x = (1/2)\epsilon_{xabc}J^{ab}P^c [/tex]
    [tex]2W_0 = \epsilon_{0abc}J^{ab}P^c = \epsilon_{0123}J^{12}P^3+\epsilon_{0132}J^{13}P^2+\epsilon_{0231}J^{23}P^1+\epsilon_{0213}J^{21}P^3+\epsilon_{0312}J^{31}P^2+\epsilon_{0321}J^{32}P^1[/tex]
    [tex]= +J^{12}P^3-J^{13}P^2+J^{23}P^1-J^{21}P^3+J^{31}P^2-J^{32}P^1[/tex]
    [tex]= (J^{12}P^3+J^{23}P^1+J^{31}P^2)-(J^{13}P^2+J^{21}P^3+J^{32}P^1)[/tex]
    and in each parentheses we have an expression identical to the identity I started with.
    Therefore, W0 = 0.

    likewise I get for all the other W components. So, is W = 0 by definition?

    2. Relevant equations
    (none)

    3. The attempt at a solution
    the only thing I could have missed is the identity from above, but i looks quite solid. I don't see were did I go wrong


    thanks for your time
    tamir
     
  2. jcsd
  3. Oct 17, 2014 #2

    strangerep

    User Avatar
    Science Advisor

    You could get to that conclusion faster by noticing that
    $$
    \epsilon^{\mu\nu\lambda\rho} (x_\nu \partial_\lambda - x_\lambda \partial_\nu) \partial_\rho ~=~ 0 ~,
    $$because the partial derivatives commute.

    But you're only considering orbital angular momentum, hence it only applies to a spin-0 particle. If there's non-zero intrinsic spin, things are different.

    Check Wikipedia's entry on the Pauli--Lubanski vector.
     
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