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

by Chenkb
Tags: lorentz boost, two body decay
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Chenkb
#1
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,
$$W(\cos\theta,\phi,p)=???$$
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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Einj
#2
Jun7-14, 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.


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