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my question concerns the kinematics of 2 to 2 particle scattering. I refer to Peskin and Schroeder eq.17.59 going from this expression


to this


He uses $2\pi\,p_T\,dp_T=d^2p_T$ but I fail to see how that is true?

Thank you for your insight!

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Well you are in two dimensions so: $d^2p_T = dp^x_Tdp^y_T = p_T dp_Td\theta$, where $\theta$ is the angle in the plane $(x,y)$ and $p_T$ the radial coordinate. Assuming isotropy you can integrate over $\theta \in [0,2\pi]$ giving what you are looking for. – Learning is a mess Nov 13 '12 at 13:29
Thank you! My confusion was coming from exactly this part $d^2p_T=dp_T^xdp_T^y$, now I understand. – user15943 Nov 13 '12 at 13:58
@user15943: If you have solved your problem, it would be good if you could write an answer, so that the post doesn't end up as an orphan. – Qmechanic Nov 16 '12 at 18:02

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