How to integrate the phase space volume for 2 -> 2 scattering processes?

Question

In the QFT book from Schwartz it is stated that $$ d\sigma = \frac{1}{4E_1E_2|\vec{v}_1-\vec{v}_2|}|\mathcal{M}|^2 d\Pi_{\text{LIPS}}\tag{5.22} $$ where $$ d\Pi_{\text{LIPS}} = (2\pi)^4\delta^4(\sum p)\frac{1}{4E_3E_4}\frac{d^3p_3}{(2\pi)^3}\frac{d^3p_4}{(2\pi)^3}\tag{5.26} $$ is the lorentz-invariant phase space volume in the case of 2 -> 2 scattering. In the following the phase space volume is calculated in terms of d$\Omega$ by initially integrating over $\vec{p}_4$. Similar approaches can be seen in pretty much every other textbook like in chapter 4.5 of Peskin&Schroeder. Now my question is why is this allowed in the case of the Matrix element squared carrying dependence of for example $\vec{p}_4$ via a Mandelstam-variable? In none of the books it is stated that the procedure does not hold for cases like these. Any help is highly appreciated!

Answer 1

The dependence on one of the momentum variables is easily cancelled by virtue of the dirac delta which ensures conservation of 4-momentum. In general, the modulus square of the amplitude will depend on the momentum of the particles involved in the process but what one does is the integrate first on one of the output momenta, as you say. If we consider the process with momenta $p_1,p_2 \rightarrow p_3,p_4$, then the integration on $p_3$, for example, goes something like this

$$\int |M(p_1,p_2,p_3,p_4)|\delta(p_1+p_2-p_3-p_4)\,\mathrm{d}^4p_3\,\mathrm{d}^4p_4 = \int |M(p_1,p_2,\color{red}{p_4-p_1-p_2},p_4)|)\,\mathrm{d}^4p_4\tag{1}$$

and only then, if possible, one proceeds with the second integration.

Of course, in realty, the calculation is more difficult than what is portrayed in equation $(1)$, but the concept is the same.