The cross section of top quark pair production is dominated at the LHC by gluon-gluon fusion, whereas at Tevatron, quark-antiquark annihilation is more prevalent. Why is this?

I know the fundamental difference between these colliders is that the LHC collides $pp$ pairs and the Tevatron collides $p\bar{p}$ pairs, so is it related to the parton distribution functions in some way?

  • $\begingroup$ A small part of the answer is that antiprotons have antiquarks, while protons have quarks, so there are fewer q-qbar collisions possible, but IIRC the bigger factor is indeed the PDFs, which tilts much more towards the gluon sea at higher energies. $\endgroup$
    – jwimberley
    Apr 15, 2015 at 10:47

1 Answer 1


In order to create a top quark pair you need at least an energy of $2 m_t$.

At Tevatron, the energy available is about 1 TeV per beam, where beam1 is a proton beam and beam2 an anti-proton beam. For simplicity, let's pick-up a parton with an energy $m_t = 173$ GeV from a proton and another parton with $E=m_t$ from the anti-proton. The ratio of the energy of the partons with respect to the (anti-)proton will be of the order of $x=0.17$ ($x$ is the Bjorken variable). Looking at the parton distribution function (a.k.a. PDF), you'll see that such relatively large $x$ is dominated by valence quarks and thus $u,u,d$ in proton and $\bar{u},\bar{u},\bar{d}$ in anti-proton. (An example of PDF is given in figure 19.4 of this link: http://pdg.lbl.gov/2014/reviews/rpp2014-rev-structure-functions.pdf ). Hence, the production of $t\bar{t}$ from quark-anti-quark is likely to happen at Tevatron.

At LHC, the situation is different because the energy of the beams is about 6.5 times larger. Hence, $x$ value is decreased by this factor. Typically, for $x$ below 0.1, you'll see that the probability to pick-up a gluon is much larger than a quarks (or anti-quarks). Notice, that the PDF of the gluon is divided by a factor 10 in the plot! So LHC collisions are largely dominated by gluon fusion production.


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