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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!

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  • $\begingroup$ The result depends on the Mandelstam variables, as you wrote, but they are $s,t,u$ and at least two of them, $s,t$, are independent, right? So why are you only talking about $dt$ and not $ds$? I think that at the end, you are calculating the process at the parton level so PDF doesn't enter at all, as you correctly or optimistically anticipated, and $s,t$ as well as $p,E$ (?) describe properties of the elementary particles (and partons) and there is a simple transformation from some variables to others, right? $\endgroup$
    – Luboš Motl
    Commented Apr 28, 2016 at 11:52
  • $\begingroup$ I only talk about $dt$ because I derived the differential cross section this way. The general formula for a 2->2 body scattering problem is: $\frac{d\sigma}{dt} =\frac{1}{64 \pi s p_i^{2,*}} |M|^2$. But how can I transform $\frac{d\sigma}{dt}$ to $\frac{d\sigma}{dp_t}$. But as I wrote these lines I have an idea. I will try it out. Thanks for your answer. $\endgroup$
    – Shawn Hellmann
    Commented Apr 28, 2016 at 17:39
  • $\begingroup$ Okay, I tried several methods, but I have no idea. $\endgroup$
    – Shawn Hellmann
    Commented Apr 29, 2016 at 10:06

1 Answer 1

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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.

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  • $\begingroup$ I struggle with the understanding of the second and third equation. Where do they come from? Btw. Thanks for your answer. $\endgroup$
    – Shawn Hellmann
    Commented Apr 29, 2016 at 11:32
  • $\begingroup$ @ShawnHellmann write $\vec p_i=\vec p_i^L+\vec p_i^T$ with $\vec p_i^L\cdot \vec p_j^T=0$. $\endgroup$
    – AccidentalFourierTransform
    Commented Apr 29, 2016 at 11:42
  • $\begingroup$ Okay, I tried this. I am referring to your last equation: $\frac{d\sigma}{d p_3^T} = \frac{d\sigma}{dt} \frac{dt}{dp_T^3}$. Now: $\frac{dt}{dp_T^3} = 2p_1^L \frac{1}{2 \sqrt{p_3^2-p_3^{T,2}}}\cdot (-2 p_3^T)$. But this is zero, isn't it? And why can I replace $p_3^L$ in your second equation. It's not a vector. How I do I make sure that $\vec p_1^L \vec p_3^L = p_1^L p_3^L $ holds? $\endgroup$
    – Shawn Hellmann
    Commented Apr 29, 2016 at 13:16
  • $\begingroup$ The last equality follows tautologically because if you define the longitudinal parts of the 3-vectors as vectors at all, they are 1-dimensional vectors, so the inner product of the vectors is the same as the product of the "only components". No, the derivative of $t$ with respect to $p^T$ surely isn't zero. $\endgroup$
    – Luboš Motl
    Commented Apr 29, 2016 at 16:37
  • $\begingroup$ okay, that is clear now. But where is my problem in thinking? Depends the energy on the momentum? Edit: oooohhh. Energy-momentum-relation of special relativity. Does it work then? $\endgroup$
    – Shawn Hellmann
    Commented Apr 29, 2016 at 18:16

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