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Some high-energy experiments (RHIC, LHC) use ion-ion collisions instead of proton-proton collisions. Although the total center-of-mass energy is indeed higher than p-p collisions, it might happen that the total energy per nucleon is actually lower. What are the advantages of using ion-ion collisions (e.g. gold-gold or lead-lead) instead of proton-proton collisions, considering the same accelerator?

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up vote 6 down vote accepted

As a clarification the energy per nucleon is always lower: for example, currently in the LHC the proton top energy is 3.5 TeV. Now the Pb energy is 3.5 TeV times Z so the energy per nucleon is 3.5*Z/A and A is greater than Z for every nucleus (except the proton where it is equal to one).

But the goal of ion-ion collision is not to increase the total energy or the energy per nucleon: it is to obtain a different type of collision.

It should be noted than in a proton-proton collision, the energy involved in the real collision process is variable: each quark and gluon carry a fraction of the energy of the proton, and hard collision involve a collision between a quark/gluon of one proton against a quark/gluon of the other.

In the case of ion-ion collision you have the same process: the energy is shared by the protons/neutrons and they can have different energies.

The goal of such collision is also to obtain a volume (bigger than in a p-p collision) with a very high energy density. In such a volume, a "state of matter" called quark-gluon-plasma is believed to be possibly created. The study of this QGP is one of the main goal of the ALICE experiment at the LHC.

A few references:

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+1 for mentioning the extended volume of colliding matter – David Z Nov 8 '10 at 23:53
"the energy per nucleon is always lower" Well, we can imagine a collision dominated by two $x > 1$ partons. Such events would, of course, be vanishingly rare. – dmckee May 19 '11 at 16:43
@dmckee: I think you interpreted that sentence to narrowly: He is trying to make a point about energy vs. energy per nucleon. Anyway, how would you produce $x>1$ for pp if you cannot get extra energy from the medium? – Benjamin Bannier May 19 '11 at 16:58

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