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Pauli-Lubanski pseudovector

  1. Nov 30, 2011 #1
    Hi can anyone help me prove the result of W2 of the Pauli-Lubanski pseudovector :



    This is very new to me and I've read I must use terms such as J13 and P3

    Where totally antisymmetric symbol is defined by:
    [itex]\epsilon[/itex]1234=1 and [itex]\epsilon[/itex]1243=-1
     
  2. jcsd
  3. Nov 30, 2011 #2

    dextercioby

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    Well, what have you tried ?? As per the guidelines, you must post your attempt.
     
  4. Nov 30, 2011 #3
  5. Nov 30, 2011 #4

    dextercioby

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    Ok, where did you fail ? Post or attach a scan of your work.
     
  6. Nov 30, 2011 #5
    Well here is my answer:

    W2= 0.5 M[itex]\mu\nu[/itex] M[itex]\mu\nu[/itex] P2 + M[itex]\mu\rho[/itex] M[itex]\nu\rho[/itex] P[itex]\mu[/itex] P[itex]\nu[/itex]

    Where the Pauli-Lubanski pseudovector given was:

    Wμ= - 0.5 [itex]\epsilon[/itex][itex]\mu\nu\rho\sigma[/itex] J[itex]\nu\rho[/itex] P[itex]\sigma[/itex]
     
  7. Nov 30, 2011 #6

    dextercioby

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    I get 6 terms when I expand

    [tex] W_2 = -\frac{1}{2} \epsilon_{2\nu\sigma\rho} M^{\nu\sigma}P^{\rho} = -\frac{1}{2} \left( \epsilon_{2013} M^{01}P^{3} + \epsilon_{2031} M^{03}P^{1} + \mbox{4 other terms}\right) [/tex]

    I don't think you can regroup them the way you did.
     
  8. Nov 30, 2011 #7
    Ok Im a little confused, could you go into more detail regarding your expansion? Im struggling to see how you got those terms! :confused:
     
  9. Nov 30, 2011 #8

    dextercioby

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    Well, the epsilon for a fixed mu (=2) can only take 3 values in 3! combinations. 0,1,3 and the other 5 combinations. That's why from the possible terms you have only 6 remaining.
     
  10. Nov 30, 2011 #9
    OK so the other four terms would be:

    ε2301 M30P1
    ε2310 M31P0
    ε2130 M13P0
    ε2103 M10P3
     
  11. Nov 30, 2011 #10

    dextercioby

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    Yes. Now the epsilons are +/-1 and you can regroup alike terms based on antisymmetry of M.
     
  12. Nov 30, 2011 #11
    OK so the antisymmetry rule again is:

    Mab=M-ab
     
  13. Nov 30, 2011 #12

    dextercioby

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    M_ab = - M_ba you mean...

    EDIT: Yes, exactly. ;)
     
    Last edited: Nov 30, 2011
  14. Nov 30, 2011 #13
    :rofl: Thats what I meant...

    So M31= - M13 for example...
     
  15. Nov 30, 2011 #14
    So using that information and that:

    ε1234=1 and ε1243=-1

    W2= -1/2 (-J10P3+ J30P1+J10P3 +J31P0-J30P1-J31P0)

    =-1/2 (0)
     
  16. Nov 30, 2011 #15

    dextercioby

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    No, no, no. It's 2 times minus for 3 terms in the bracket, one minus from epsilon, one minus from M, so the terms don't have opposite signs, but equal (choose all 3 with plus). So you have 3 times double contribution. The 2 can be then factored and cancelled with the 2 in the denominator.
     
  17. Nov 30, 2011 #16
    I understand so you end up with:

    W2= -(J10P3+J30P1+J31P0)

    Correct?
     
  18. Nov 30, 2011 #17

    dextercioby

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    Looks ok. Half the permutations of 0,1,3.

    EDIT for post below: Don't mention it.
     
  19. Nov 30, 2011 #18
    So I have the correct answer, thats great. Thanks for all your help :smile:
     
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