How do I calculate the differential cross section with respect to the transversal momentum? First of all, sorry for my English, my first language is German.
My problem is: I calculated the matrix element of the quark-gluon-Compton process (q+g -> gamma + q). With the kinematics of scattering in the center-of-mass frame, i could calculate dsigma/dt. This differential cross section only depends on the Mandelstam variables.
Later on I should include the parton distribution functions to calculate the differential cross section (with a Monte Carlo Simulation). My end result should be the differential cross section with respect to the transverse momentum of the direct photon. 
So how do i transform my dsigma/dt to dsigma/dp without including the PDFs? I want to know how I accomplish this task in general before messing around with the PDFs.
Thank you all!
 A: The process is discussed at the parton level – both in the initial form and the desired form – so the conversion cannot depend on PDFs.
Now, the Mandelstam variable $t$ is equal to
$$ t = (p^\mu_1 - p^\mu_3)^2 = m_1^2+m_3^2 -2 E_1 E_3 +2 \vec p_1\cdot \vec p_3 $$
in the "mostly minus" metric convention. The masses of particles are fixed and the total energy is given by the initial state. We need to find a relationship between $t$ and the transverse momentum $p_t$ and between their differentials. The inner product of the 3-vectors is the key to the transverse/longitudinal split:
$$\vec p_1\cdot \vec p_3 = p_1^L p_3^L + \vec p_1^T\cdot \vec p_3^T $$
The initial state has $\vec p^T_{1,2}=0$ so the second term may be dropped. On the other hand, the only scattering-angle-dependent quantity is 
$$p_3^L = \sqrt{p_{3,\rm max}^2 - (p_3^T)^2} $$
So
$$ t = m_1^2+m_3^2-2E_1E_3+2 p_1^L \sqrt{p_{3,\rm max}^2 - (p_3^T)^2} $$
This is a relationship between $t$ and $p^T$ and you may differentiate it to find the relationship between $dt$ and $dp^T$. There is a lot of uninsightful calculation need to be done and I think it should be done by you at the end.
