Register to reply

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

by Chenkb
Tags: lorentz boost, two body decay
Share this thread:
Jun5-14, 04:35 AM
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,
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!
Phys.Org News Partner Physics news on
Detecting neutrinos, physicists look into the heart of the Sun
Measurement at Big Bang conditions confirms lithium problem
Researchers study gallium to design adjustable electronic components
Jun7-14, 07:18 PM
P: 323
You only need to transform the angle according to the usual formula (see for example 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