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Why does LHC use a $pp$ collision and not $p\overline{p}$ collision, when you actually only need one ring in $p\overline{p}$ as compared to two in $pp$. Please answer in easy language (I am not much familiar with Bjorken variables as such).

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It's much easier to supply a source of protons than antiprotons which annihilate with everything around it unless kept in high vacuum. The only difference is the slight charge asymmetry that you need to account for, but proton-proton colliders are really mostly gluon colliders anyway, which are the same between protons and anti-protons.

As far as the effect of the charge asymmetry, this comes down to the quark content of the protons. For instance, in order to make a Z boson via the Drell-Yan interaction, one needs a quark-antiquark pair (e.g. an up/upbar). But since protons don't have any valence antiquarks, the charge asymmetry requires the anti-quark to be pulled from the "sea quarks." If your collider has low energy, then the sea quarks will generally not have enough momentum to make heavy particles. This is the hit you take when working with proton-proton collisions.

In general, the absence of anti-protons means that you need to pull anti-quarks from the vacuum which, in turn, implies they will tend to have lower energy.

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  • $\begingroup$ Then why does Tevatron uses ppbar collision? $\endgroup$ – kbg Mar 9 '17 at 17:41
  • $\begingroup$ Tthere is not the charge asymmetry I mentioned before. So it's somewhat preferable, but you also need to consider the engineering complexity and cost you add to your experiment. $\endgroup$ – Bobak Hashemi Mar 9 '17 at 17:44
  • $\begingroup$ @kbg the Tevatron aimed at much lower energies than the LHC, everything less difficult and less expensive, from magnets to vacuum. $\endgroup$ – anna v Mar 9 '17 at 19:49
  • $\begingroup$ @BobakHashemi, what do you mean by the "charge asymmetry"? To which of the two cases do you refer? $\endgroup$ – Helen Apr 17 '17 at 4:21
  • $\begingroup$ Hi Helen, I've updated my answer with a small elaboration. Hopefully that helps. Protons have charge $e$ whereas anti-protons have charge $-e$, so a ppbar collider will have net charge 0 in each interaction whereas a pp collider will have net charge $+2e$ (which is not symmetric between positive and negative as is 0). $\endgroup$ – Bobak Hashemi Apr 17 '17 at 17:37
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Using the the LHC's record energy per proton, 6.5 TeV and the rest energy of a proton, 938.257 MeV, the fractional difference between using protons and antiprotons is about

$\frac{2*938.257 MeV}{6.5 TeV}=0.000289$

So it looks like the energy they would gain by using antiprotons is insignificantly small.

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