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How the twobody decay momentum distribution transform in lab frame? 
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#1
Jun514, 04:35 AM

P: 27

For twobody 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
Jun714, 07:18 PM

P: 306

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.



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