Recall that $\mathrm{d}LIPS_2$ (one particle decaying into two particles of the same mass) is given by

$$\mathrm{d}LIPS_2 = \frac{\vert{\bf k_1'}\vert}{16\pi^2\sqrt{s}}\mathrm{d}\Omega_{cm}.$$

In a given decay, is all the angle dependence included in dLIPS? If I recall correctly, this does not need be the case, or else the integration over the angles would always be trivial.

What about the general case of one particle going to $n$ particles. $$\mathrm{d}LIPS_n$$

What about scattering cross sections?


In the case of a particle decaying to two identical particles, this is true. There is no angle-dependence in the scattering amplitude, so the integration is indeed trivial. When calculating scattering cross sections, this is in general not true: the scattering amplitude can depend on the phase space angle. It is also not true for the general $n$-particle decay. The angle dependence has to be evaluated separately for each case, it is not guaranteed that it is only contained in the phase space measure.

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  • $\begingroup$ 1->3 particles then? $\endgroup$ – Your Majesty Apr 13 '14 at 11:21
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    $\begingroup$ Excuse me, what is your question? $\endgroup$ – Frederic Brünner Apr 13 '14 at 11:32
  • $\begingroup$ I updated the question. I mean what about the angle dependence for 1 going to several (>2) particles. $\endgroup$ – Your Majesty Apr 13 '14 at 11:36
  • $\begingroup$ @LoveLearning: I have added lines addressing the general case. $\endgroup$ – Frederic Brünner Apr 22 '14 at 19:12

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