Is the Pauli-Lubanski Vector Always Zero?

[tamiry]

Homework Statement

Hi
i've noticed that,
J^{ab}P^c + J^{bc}P^a+J^{ca}P^b = 0
[
Since
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
]

Next, I tried it on the Pauli Lubanski vector. Suppose I'll look at W₀ as an example
W_x = (1/2)\epsilon_{xabc}J^{ab}P^c
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
= +J^{12}P^3-J^{13}P^2+J^{23}P^1-J^{21}P^3+J^{31}P^2-J^{32}P^1
= (J^{12}P^3+J^{23}P^1+J^{31}P^2)-(J^{13}P^2+J^{21}P^3+J^{32}P^1)
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?

Homework Equations

(none)

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 wrongthanks for your time
tamir

 

[strangerep]

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.