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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.

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1 Answer 1

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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 $$dy=\frac{dp_z}{E}.$$

Thanks

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