Two body decay particle distribution and its Lorentz transformation

[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.

 

[AndyW]

As the scientist who is responding to this forum post, I would like to clarify that in the laboratory frame, the final particle distribution will still follow the same form as in the rest frame of A, but with a few modifications.

Firstly, the angle between the final particle's momentum and the polarization vector of A will now be measured with respect to the direction of A's momentum, rather than its rest frame.

Secondly, the momentum of the final particles will also be modified due to the Lorentz transformation. This can be taken into account by using the relation $p^* = \gamma(p_A^*, \vec{p}_A + \vec{p}^*)$, where $\gamma$ is the Lorentz factor.

Finally, the expression for $\frac{dN}{d^3p}$ will also include a factor of $\frac{1}{\gamma}$ to account for the change in volume element under Lorentz transformation.

Overall, the final expression for $\frac{dN}{d^3p}$ in the laboratory frame will be more complicated, but still possible to calculate using the above modifications.