For two-body decay, in the center of mass frame, final particle distribution is,(adsbygoogle = window.adsbygoogle || []).push({});

$$

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!

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

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