## Pauli-Lubanski pseudovector

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:
$\epsilon$1234=1 and $\epsilon$1243=-1
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 Blog Entries: 9 Recognitions: Homework Help Science Advisor Well, what have you tried ?? As per the guidelines, you must post your attempt.
 Ive tried the method mentioned in the post below: http://www.physicsforums.com/showthread.php?t=245130

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## Pauli-Lubanski pseudovector

Ok, where did you fail ? Post or attach a scan of your work.
 Well here is my answer: W2= 0.5 M$\mu\nu$ M$\mu\nu$ P2 + M$\mu\rho$ M$\nu\rho$ P$\mu$ P$\nu$ Where the Pauli-Lubanski pseudovector given was: Wμ= - 0.5 $\epsilon$$\mu\nu\rho\sigma$ J$\nu\rho$ P$\sigma$
 Blog Entries: 9 Recognitions: Homework Help Science Advisor I get 6 terms when I expand $$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)$$ I don't think you can regroup them the way you did.
 Ok Im a little confused, could you go into more detail regarding your expansion? Im struggling to see how you got those terms!
 Blog Entries: 9 Recognitions: Homework Help Science Advisor 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.
 OK so the other four terms would be: ε2301 M30P1 ε2310 M31P0 ε2130 M13P0 ε2103 M10P3
 Blog Entries: 9 Recognitions: Homework Help Science Advisor Yes. Now the epsilons are +/-1 and you can regroup alike terms based on antisymmetry of M.
 OK so the antisymmetry rule again is: Mab=M-ab
 Blog Entries: 9 Recognitions: Homework Help Science Advisor M_ab = - M_ba you mean... EDIT: Yes, exactly. ;)
 Thats what I meant... So M31= - M13 for example...
 So using that information and that: ε1234=1 and ε1243=-1 W2= -1/2 (-J10P3+ J30P1+J10P3 +J31P0-J30P1-J31P0) =-1/2 (0)
 Blog Entries: 9 Recognitions: Homework Help Science Advisor 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.
 I understand so you end up with: W2= -(J10P3+J30P1+J31P0) Correct?
 Blog Entries: 9 Recognitions: Homework Help Science Advisor Looks ok. Half the permutations of 0,1,3. EDIT for post below: Don't mention it.

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