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

by Chenkb
Tags: lorentz boost, two body decay
 P: 27 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!