Can we destroy a proton?


by DiracPool
Tags: destroy, proton
DiracPool
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#1
Jan8-13, 03:00 AM
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Or, more generally, can we destroy a nucleon? I know we can convert between a proton and a neutron, but what about obliteration? If we try to separate a quark from a nucleon we cannot do so because we bud off an anti-quark and the original quark goes back in (or something like that). So we can't measure any free quarks. Does this mean that we can't bust open a nucleon? If we can, then what happens, the nucleon just become a bunch of neutrinos, lepton's, and photons? Are we removing protons from the universe's pool at the LHC?
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Drakkith
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#2
Jan8-13, 05:34 AM
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Sure, you can hit a nucleon so hard that you "rip" it apart. But the energy required to do so is so high that you actually end up CREATING new particles from the energy, resulting in a shower of new particles emerging from the collision. The higher the energy of the collision the more particles and the higher mass they generally are. There isn't a "pool" of particles we take from, they are actually created from energy. The energy doesn't disappear, it is converted into mass per E=MC2.
DiracPool
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Jan8-13, 05:40 AM
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There isn't a "pool" of particles we take from, they are actually created from energy.
OK, I get that, but when we add energy through acceleration to these protons in the LHC, collide them, and get the shower of new particles, is there ever an instance where the resultant shower does NOT include a proton or neutron in the product? That is, the product of the collision is a zoo of particles that are perhaps just leptons and photons, and/or maybe heavy quarks?

mfb
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#4
Jan8-13, 07:56 AM
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Can we destroy a proton?


is there ever an instance where the resultant shower does NOT include a proton or neutron in the product?
It is hard to measure this rate (due to experimental issues), but it is certainly possible to have no proton and neutron after the collision.
At the same time, you have to conserve baryon number. Those collisions have to have other baryons in the final state, and those decay to protons and/or neutrons after a while.
DiracPool
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#5
Jan8-13, 02:59 PM
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Those collisions have to have other baryons in the final state, and those decay to protons and/or neutrons after a while.
Interesting...so qualitatively you do not, in fact, lose a nucleon in the end result.
Drakkith
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#6
Jan8-13, 03:05 PM
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Quote Quote by DiracPool View Post
Interesting...so qualitatively you do not, in fact, lose a nucleon in the end result.
If it's not there after the collision, I would say you do. Even if you eventually get a few back after decays.
jtbell
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#7
Jan8-13, 03:17 PM
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As mfb said, you have to conserve baryon number. So you need at least one baryon at all times. Nucleons (protons and neutrons) aren't the only baryons.
Parlyne
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#8
Jan8-13, 03:30 PM
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Quote Quote by jtbell View Post
As mfb said, you have to conserve baryon number. So you need at least one baryon at all times. Nucleons (protons and neutrons) aren't the only baryons.
But, every free baryon eventually decays down to a proton. The only way to get rid of a proton is with an anti-proton (or other anti-baryon). Well, that is, unless you're at sufficiently high energy (and, probably, density) that non-perturbative weak processes are accessible, in which case it's possible to have a proton (or, perhaps, more correctly, its constituents) decay in a way that produces an anti-lepton. But, that's not something that happens at any noticeable rate at ordinary scales.
Aromal
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#9
Jan24-13, 08:15 PM
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sure.


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