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

  1. 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!
     
  2. jcsd
  3. 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.
     
Know someone interested in this topic? Share a link to this question via email, Google+, Twitter, or Facebook

0
Draft saved Draft deleted