Answer Happy

1.1.- Hizme (e-) TI (2011 m.). (2)*() (3) Again, we won't make a big deal about the factors of (25). Note that the factors are each manifestly Lorentz invariant. In this problem we will use this expression to determine the phase space for a process with two final state particles, | dll= |(27)+'" (Q – 11 – va) dºpy (27)6(p– m?) d*py (27)863 – mi), Here p and p2 are the four-momenta of the two final state particles of masses m and ma. 1.1 Integration over on-shell conditions Perform the integration over the two on-shell conditions with respect to the energies of the final state momenta, dry and do. You should end up with rap, dp2 8"-) | all = SE CE (27)*5ºYQ – P – pa). At this point, the final state energies E, and E, are functions of the three-momenta. For example, E (P) = m + pi Hint: it may be helpful to recall the relation 8y- /u)) 8(x - f-'())) \df/d.x