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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

$\frac{d^3\sigma}{dy_3dy_4dp_T}=x_1f_1(x_1)x_2f_2(x_2)\;2p_T\frac{d\sigma}{d\hat{t}}(1+2\to3+4)$

to this

$\frac{d^4\sigma}{dy_3dy_4d^2p_T}=x_1f_1(x_1)x_2f_2(x_2)\;\frac{1}{\pi}\frac{d\sigma}{d\hat{t}}(1+2\to3+4)$

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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3  
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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