# Heuristic derivation of $W^\mu=\frac{1}{2}\epsilon^{\mu\nu\sigma\rho}P_\nu J_{\sigma\rho}$ using combination of physical and mathematical arguments

If a simple systematic way to derive or guess (either mathematically or by a combination of physical arguments and mathematics) that one of the Casimir operator of Poincare group is $W^2\equiv W_\mu W^\mu$ where $$W^\mu=\frac{1}{2}\epsilon^{\mu\nu\sigma\rho}P_\nu J_{\sigma\rho}\tag{1}.$$ In physics textbooks, (1) is given as a definition, and from which one can check that $W^2\equiv W_\mu W^\mu$ is indeed a Casimir. But I find this definition of $W^\mu$ to be quite non-trivial to guess. So I'm not looking for a rigorous derivation and if there are physical arguments to achieve this, it will do for me.

For a massive particle there is a reference frame at rest with it. In that reference frame, the four momentum $P_a$ has only its temporal component $a=0$ with value $m$. In that reference frame, looking at the formula you wrote, you see that $W^a$ has only three nonvanishing components $a=1,2,3$ with the evident meaning of angular momentum at rest with the particle (multiplied with the mass) if you take the meaning of $J_{0a}=-J_{a0}$ into account. The value of $W_aW^a$ is independent from the reference frame, so it can be computed at rest with the particle producing $m^2$ times the square of the angular momentum at rest with the particle: the squared spin times $m^2$.
Since the action of the Poincaré group is infinitesimally implemented by the generators of the group, the fact that $W_aW^a$ is an invariant just means that it commutes with all generators of the group, i.e., it is a Casimir operator.
The guess is based on the requirement of the translational invariance. Really, $J_{\mu\nu}$ doesn't commute with translations operator $P_{\mu}$. This means that the candidates $J_{\mu\nu}J^{\mu\nu}$, $\epsilon^{\mu\nu\alpha\beta}J_{\mu\nu}J_{\alpha\beta}$ on the role of the Casimir operator aren't translational-invariant.
In order to construct the translational-invariant Casimir operator, we can define $C = W_{\mu...}W^{\mu...}$, where $W_{\mu...}$ is translational-invariant operator. The only possible candidate (since $[P_{\mu},P_{\nu}] = 0$) is $$W^{\mu} \sim \epsilon^{\mu\nu\alpha\beta}P_{\nu}J_{\alpha\beta}$$