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nrqed

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## Main Question or Discussion Point

Does anyone know how to prove that [itex] W^2 = 0 [/itex] for massless particles, where W is the Pauli-Lubanski vector

[tex] W^\mu = \frac{1}{2} \epsilon^{\mu \nu \rho \sigma} M_{\rho \sigma} P_\nu [/tex]

??

The books I have either give no proof, or they rely on the expression obtained in the massive case,

[tex] W^2 = - m^2 \vec{J}^2 [/tex]

However, the derivation of this formula relies entirely on working in the rest frame of the particle, so it makes no sense to infer anything about the massless limit from this result.

I have found that, in general,

[tex] W^2 = - \frac{1}{2} P^2 M_{\mu \nu} M^{\mu \nu} + M_{\mu \nu} M^{\sigma \nu} P^\mu P_{\sigma} [/tex]

In the massless limit, the first term is zero but I can't see why the second term is also zero. I have considered working in the frame

[tex] p_\mu = (E,0,0,E) [/tex]

but I still don't get zero. The antisymmetry of the Lorentz generators M does not help.

Thanks in advance

[tex] W^\mu = \frac{1}{2} \epsilon^{\mu \nu \rho \sigma} M_{\rho \sigma} P_\nu [/tex]

??

The books I have either give no proof, or they rely on the expression obtained in the massive case,

[tex] W^2 = - m^2 \vec{J}^2 [/tex]

However, the derivation of this formula relies entirely on working in the rest frame of the particle, so it makes no sense to infer anything about the massless limit from this result.

I have found that, in general,

[tex] W^2 = - \frac{1}{2} P^2 M_{\mu \nu} M^{\mu \nu} + M_{\mu \nu} M^{\sigma \nu} P^\mu P_{\sigma} [/tex]

In the massless limit, the first term is zero but I can't see why the second term is also zero. I have considered working in the frame

[tex] p_\mu = (E,0,0,E) [/tex]

but I still don't get zero. The antisymmetry of the Lorentz generators M does not help.

Thanks in advance