Register to reply 
How the twobody decay momentum distribution transform in lab frame? 
Share this thread: 
#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: 323

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.



Register to reply 
Related Discussions  
Two body decay particle distribution and its Lorentz transformation  High Energy, Nuclear, Particle Physics  0  
Rotational Kinetic Energy of Body in another body reference frame  Classical Physics  9  
Rest frame angular distribution of meson decay into two photons  Advanced Physics Homework  6  
Two body decay and momentum  High Energy, Nuclear, Particle Physics  3  
Lab frame versus center of momentum frame  High Energy, Nuclear, Particle Physics  6 