Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I have the cross-sections as a function of $\sqrt{s}$ for a process with a $u$-quark and $u$-antiquark in the initial state (eg.: $u \bar{u} \to e^- e^+$). I have a standard parton distribution function table (say, CTEQ). With these, how do I find the corresponding cross-sections for a process with protons in the initial state (e.g. $p p \to e^- e^+$ as at the LHC)?

Specifically, should I use some sort of convolution? How?

share|improve this question

1 Answer 1

You have a couple of things to think about here

  1. "How many" quarks (and anti-quarks) of initial momentum $\vec{p}$ are there?

  2. Given a pair with momenta $p_1$ and $p_2$ what is cross-section for scattering with invariant mass $\sqrt{s}$.

The first part is answered by the PDF. At LHC energies you can reasonably take the momentum of a parton to first order as $x p_p$ where $x$ is the Bjorken variable and $p_p$ is the momentum of the proton.

The second part you have already got modulo boosting from the parton CoM frame to the lab frame. But you can ditch the boost if you can express the cross-sections in Lorentz invariant terms (i.e. in terms of the Mandelstam variables.

In total you have an integration which is notionally of the form $$ \sigma(Q) = \int_{x_1} dx_1 F_q(x_1,Q) \int_{x_2} dx_2 F_\bar{q}(x_2,Q) \sigma_{q\bar{q}}\left(\sqrt{(\mathbf{p}_1 + \mathbf{p}_2)^2} \right) \quad , $$ where $\mathbf{p}_1$ and $\mathbf{p}_2$ are the 4-vectos of the two participating partons and $Q = -\sqrt{t}$. You can, of course, also integrate over $Q$ to get a total cross-section. For comparison against data you will probably need to do a little more work to get into detector coordinates.

For more precise results you need generalized PDFs so that you can take both transverse momenta and spins into account.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.