# Two body decay particle distribution and its Lorentz transformation

1. Jan 20, 2014

### Chenkb

For two-body decay $A\rightarrow B+C$, if A is polarized, it is clear that we have:
$\frac{dN}{d\Omega}\propto 1+\alpha \cos\theta^*$, for final particle distribution.
where, $\theta^*$ is the angle between the final particle's momentum $p^*$ and the polarization vector of $A$ in the rest frame of $A$.

And using $d^3p^* = p^{*2}d\Omega dp^*$, we can rewrite the distribution formula in terms of $\frac{dN}{d^3p^*}$.

The question is, when we go to the laboratory frame that $A$ is moving with an arbitrary momentum $\vec{p}_A$, what does $\frac{dN}{d^3p}$ looks like?
I know that this is just an Lorentz transformation of arbitrary direction, but I failed to get the final expression, I feel it is too complicated.