# How the two-body decay momentum distribution transform in lab frame?

1. Jun 5, 2014

### Chenkb

For two-body decay, in the center of mass frame, final particle distribution is,
$$W^*(\cos\theta^*,\phi^*) = \frac{1}{4\pi}(1+\alpha\cos\theta^*)$$
We have the normalization relation , $\int W^*(\cos\theta^*,\phi^*)d\cos\theta^* d\phi^*=1$.

And we also know that in CM frame $p^*$ is a constant, say, $p^*=C^*$.

So, the final particle momentum distribution can be write as(I'm not sure),
$$W^*(\cos\theta^*,\phi^*,p^*) = \frac{1}{4\pi}(1+\alpha\cos\theta^*)\delta(p^*-C^*)$$
If the above momentum distribution in CM frame is right,
then what does it look like in the lab frame,
$$W(\cos\theta,\phi,p)=???$$
assume that the mother particle moves with velocity $\beta$ along $z$ axis.
I know it is just a Lorentz transformation, but how to handle this, especially the $\delta$-function.

Best Regards!

2. Jun 7, 2014

### Einj

You only need to transform the angle according to the usual formula (see for example http://www.maths.tcd.ie/~cblair/notes/specrel.pdf). p* is usually a combination of the masses of the decaying particle and of the decay product so it remains the same under Lorentz transformation.