Exploring Casimir Operators and Rest Reference Conditions in Particle Physics

[noamriemer]

Hi guys!
There is something I would like to get your help with...

I am looking at the equation:

$W^{\mu}=-\frac{1}{2} \varepsilon^{\mu\nu\lambda\sigma}M_{\nu\lambda}p_{\sigma}$

Which is, if I understand correctly,a Casimir Operator.
Now, I wish to look at a particle in its rest reference, meaning,
$p_\mu=(m,0,0,0)$

Why would these conditions yield :
$W^\mu =\frac {1} {2} m\varepsilon^{\mu\nu\lambda0}M_{\nu\lambda}$
?
I can seem to understand how the indices change...

The next thing I want to do, is understand what happens if I take $m^2<0$

Why does this condition mean that the momentum vector would be
$p_\mu=(0,0,0,m)$
?
Thank you

 

[Nugatory]

The expression you've quoted uses the Einstein summation convention, in which repeated indices are summed over: AⁱB_i is a convenient short way of writing $\sum$AⁱB_i

And because the only non-zero element of p is p₀, when you do the summation over σ, all the terms are zero except the one in which σ is zero.

 

[noamriemer]

Thank you!

But why does
$p_{\mu}=(0,0,0,m)$ relate to $m^2<0$?

And likewise,

$p_{\mu}=(p,0,0,p)$ relate to $m=0$?

I understand why
$p_{\mu}=(m,0,0,0)$ relate to massive particle,
My logic here is $p_0=E$ and $E\approx m$
and $\vec{p}=0$ (because we are looking at the reference frame)
But same logic does not work for me regarding the two eq. above...
Thank you!