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Two body decay particle distribution and its Lorentz transformation

  1. Jan 20, 2014 #1
    For two-body decay ##A\rightarrow B+C##, if A is polarized, it is clear that we have:
    ##\frac{dN}{d\Omega}\propto 1+\alpha \cos\theta^*##, for final particle distribution.
    where, ##\theta^*## is the angle between the final particle's momentum ##p^*## and the polarization vector of ##A## in the rest frame of ##A##.

    And using ##d^3p^* = p^{*2}d\Omega dp^*##, we can rewrite the distribution formula in terms of ##\frac{dN}{d^3p^*}##.

    The question is, when we go to the laboratory frame that ##A## is moving with an arbitrary momentum ##\vec{p}_A##, what does ##\frac{dN}{d^3p}## looks like?
    I know that this is just an Lorentz transformation of arbitrary direction, but I failed to get the final expression, I feel it is too complicated.
     
  2. jcsd
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