# From infinitesimal momentum volume to infinitesimal rapidity & tranverse momentum

I'm trying to derive a relationship given in a paper which is used to obtained a differential cross-section distribution in function of rapidity and transverse momentum of final state particles,

$$\frac{d^3 p}{E}=dyd^2p_t.$$

Since $p_t$ is the transverse momentum, once I choose the beam to be in the $z$ direction, the momentum volume can be written as $d^3p=dp_zd^2p_t$ (where I used $dp_xdp_y=d^2p_t$). Now, there is a direct relationship between rapidity and the longitudinal momentum given by the well known formula, $$y=\frac{1}{2}\log\left({\frac{E+p_z}{E-p_z}}\right).$$

Differentiation with respect to $p_z$ returns $$\frac{dy}{dp_z}=\frac{E}{E^2-p_z^2},$$ or $$\frac{dp_z}{E}=\frac{E^2-p_z^2}{E^2}dy,$$ which, in the end, gives me, $$\frac{d^3p}{E}=\frac{E^2-p_z^2}{E^2}dyd^2p_t.$$

I'm doing something wrong but I can't see it.

P.S. No assumptions seemed to have been made except for the fact that they treat the scattering in the CM from the beginning.

My bad. The mistake is in the third equation $\frac{dy}{dp_z}$ because I treated the energy $E$ as if it was independent of $p_z$. The correct answer is, $$dy=dp_z\left(\frac{\partial y}{\partial p_z}+\frac{\partial y}{\partial E}\frac{dE}{dp_z}\right),$$ Using the fact that $E^2=p_t^2+p_z^2+m^2$ and that $p_t$, $p_z$ and $m$ are independent of each others, the expression between the parentheses is so that $$dy=\frac{dp_z}{E}.$$