I know that it is somehow related to the parton distribution functions, allowing specific reactions with gluons instead of quarks and anti-quarks, but I would really appreciate more detailed answers !
Thanks
3 Answers
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The difference in scattering cross sections is more evident the lower the energy of collisions. Fig 41.11. At the energies of TeV the probability of new physics observations is the same for both choices of collisions.
The reason is that at low energies the fact that the proton has three quarks and the anti proton three anti quarks predominates. Quark antiquark scattering at low energies has much higher cross section than quark quark due to the extra possibility of annihilation of the quarks. At low energy the gluon "sea" plays a small part. The higher the energy of interactions the higher the number of energetic gluons that scatter and finally at TEV energies that is what predominates and the two cross sections converge. Thus for physics it makes no difference whether one uses as targets protons or antiproitons, as far as discovery potential goes.
There may be some technical advantage in the construction, in that in principle the antiproton-proton beams can circulate in the same magnetic configuration as mirror images and make the magnet construction circuits simpler. I guess that the need for high luminosity made LHC a proton proton collider, since it is more difficult to store antiprotons. I would have to research this guess.
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$\begingroup$ But using proton proton isn't a way to reduce $q \bar q$ reaction and favour g g ? $\endgroup$– gdzApr 22, 2011 at 14:35
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$\begingroup$ My remark is related to higgs production by gluon fusion $\endgroup$– gdzApr 22, 2011 at 14:47
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1$\begingroup$ Within each, nucleon and antinucleon, the distributions of the gluon "sea" are the same, that is why at high energies it makes no difference which one uses, since then the percentage of energy carried by the original quarks of the incoming quarks/antiquarks is small in the interactions of interest ( high transverse momentum, i.e. deep inelastic) and the sea gluons predominate. $\endgroup$– anna vApr 23, 2011 at 4:30
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I would add to @anna's answer that $p\bar{p}$ collider such as the Tevatron is CP symmetric. This was one of the arguments for continuing the Tevatron Experiments. To quote from the proposal
Measurements that get a special advantage from the p-pbar environment. The primary example in this category is CP-violation, which strongly limits the range of allowed models of new physics up to scales of several TeV. There are good a priori reasons to expect the existence of some non-SM CP-violating processes, and finding them is of comparable importance to addressing electroweak symmetry breaking. Precision measurements at the 1% level or better are accessible at the Tevatron due to the CP-symmetric initial state (p-pbar), and symmetry of the detectors that allow cancellation of systematics. Some of these measurements already show tantalizing effects, like the recently published di-muon asymmetry result from the DZero experiment, showing the first indication of a deviation from the Standard Model picture of CP-violation. Other measurements are exploring a completely new field, as the recent CPVmeasurement with the D 0 mesons at CDF, yielding a substantial improvement in precision with respect to previous B-factories data. This has provided a proof of feasibility of an exciting program of precision measurement with a unique possibility to find anomalous interactions in up-type quarks. A non CP-related example in this category is the forward-backward asymmetry in top quark production. Current measurements by both CDF and DZero indicate an asymmetry above the Standard Model prediction. If this persists with more data, it can be interpreted as new dynamics. This is not an easy measurement to replicate in a proton-proton environment.
http://www.fnal.gov/directorate/Tevatron/Tevatron_whitepaper.pdf
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1$\begingroup$ This is an important addition, since this difference between proton-proton and antiproton-proton collisions exists at all energies. $\endgroup$– anna vApr 21, 2011 at 18:49
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From the machine side, a symmetric $p\bar{p}$ collider can have only one beampipe, so it is much simpler. On the other hand if you fill it with many bunches they will start to collide all around the machine. You may manage to separate their orbits, but they will still feel the fields reciprocally generated (long-range beam-beam interaction) that will limit the beam intensity. So, even if you could produce an arbitrary amount of $\bar{p}$ (which would still be a major limitation), you won't be able to fill the machine with a very high current.
With two separate beam pipes this problem is limited to small sections close to the interaction regions. Two rings allows you also to better optimise each beam. In the end you will achieve an higher maximum current and so luminosity. It is also possible to store the same species but also totally different species, like protons and lead ions. The price to pay is a much more expensive and complicated machine in which many system (and so the failure probability/downtime) are replicated twice.
