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.

This post imported from StackExchange Physics at 2014-04-13 11:21 (UCT), posted by SE-user Love Learning