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Or in proton-electron collision.

To destroy is to turn into other particles, not baryons. In context of the baryon asymmetry.

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+1 I don't see why this question is necessarily a bad one. –  Noldorin Nov 12 '10 at 23:05

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The way you used baryon made me wonder if you are unaware of the baryon nearly conservation law.

From the link:

The baryon number is nearly conserved in all the interactions of the Standard Model. 'Conserved' means that the sum of the baryon number of all incoming particles is the same as the sum of the baryon numbers of all particles resulting from the reaction. An exception is the chiral anomaly. However, sphalerons are not all that common. Electroweak sphalerons can only change the baryon number by 3.

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Thanks for the link, but.. "many of the open questions beyond the Standard Model will be addressed by the LHC" –  voix Nov 13 '10 at 0:10
    
It would be good if LHC had triggers for baryon number changing events, because I am not sure that at EW scales, you might not have some collision that generates an EW instanton locally. There have been arguments that it is hard to do in collisions, because you need a coherent Higgs lump, but it would be good to have experimental data. –  Ron Maimon Apr 11 '12 at 6:08

Most proton-proton collisions will be elastic: throw in two protons and two protons will come out, deflected at some angle. But the more interesting collisions are those where individual constituents of the proton (quarks, antiquarks, or gluons) interact. For instance, all the interesting high-energy proton-proton collisions at the LHC are really collisions of two quarks, or two gluons, or a gluon and a quark (or similar combinations involving antiquarks) coming from the two protons. The results of this "hard" collision often come out at a large angle away from the proton beam, while the "remnants" of the proton that weren't directly involved in the collision sail off down the beamline in roughly the same direction the proton was originally going. It's probably reasonable to say that the original proton was "destroyed" in this process, although some large fraction of its energy and its constituents keep moving in the same direction.

Of course, the word "destroyed" is a little fuzzy. The protons don't just disappear. There are a few constraints on the final result of the collision process: it must conserve momentum, energy, and baryon number. A proton has baryon number +1, as does a neutron and various heavier "hyperons," whereas an antiproton has baryon number -1. The momentum, energy, and baryon number can all be divided up in complicated ways. So you might argue that there's a sense in which you didn't "destroy" the protons, since at the end you still have to have a total of 2 baryons. But in general, the proton remnants moving down the beamline could join up with antiquarks and make mesons with no net baryon number, while baryons could form from the hard collision and move away at a large angle. In that case I wouldn't say there's any sense in which the original protons maintained their "identity," and I think it would be reasonable to say they were destroyed.

All things considered, though, it's probably best to not think in terms of the word "destroy" that you chose at all. Maybe the way to visualize it is like this: most proton-proton collisions are like two billiard balls that bounce off each other in an elastic way. But the collisions that are of the most interest at a collider are more like cases where two billiard balls hit each other and small pieces of each shear off and fly out at a big angle, but the bulk of each ball keeps moving in roughly the direction it started from.

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Thank you for your answer –  voix Nov 13 '10 at 0:17
    
The problem is that the proton can be destroyed by electroweak instantons, under weird circumstances, like high temperatures (kT of order 1-10TeV) or maybe even just high energy collisions (maybe CM energy as low as 10 TeV perhaps). The calculations are difficult for coherent things like instantons in particle collisions, and I don't know if collisions producing baryon number violation are ruled out. –  Ron Maimon Apr 11 '12 at 6:10

If changing the protons into something else counts as "destroying" it, then yes, this is what keeps stars burning.

In particular, two protons can interact and form a deuteron, a positron and an antineutrino and some energy.

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In higher symmetries than the Standard Model the proton can decay. At the moment the experimental limit of the decay half life is very stringent, about 10^34 years. In these theories it is allowed to decay to a neutral pion and a positron, as an example.

If such a decay is allowed by higher symmetries then even in collisions one could construct a diagram with very low probability which would allow the decay in a collision, thus having a baryon number change.

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You can't destroy a particle. Without involving more complex concepts such as colour, momentum and energy must be conserved, and this implies that you cannot destroy particles. You can produce new particles or radiation by colliding protons (or neutrons...), but, in the sense that they explode and disappear, it is impossible.

Nevertheless, you can get muons, gamma-rays, or, more interesting, you could be able to "see" the quarks inside the protons for a moment.

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I suspect you're alluding to the conservation of baryon and lepton numbers here, but this statement is imprecise and confusing to the point that it verges on being simply wrong. –  dmckee Mar 6 '11 at 1:08
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You can destroy a photon. That's what happens in the photoelectric effect. –  Ben Crowell Jul 24 '11 at 3:21

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