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How do protons decay?

  1. Mar 18, 2013 #1
    Hi, so we are studying quantum physics at college and I just want to understand this concept of decay. Please could someone help me?
     
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  3. Mar 18, 2013 #2

    DrChinese

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    Welcome to PhysicsForums, waterliyl!

    As far as anyone knows, protons do not decay. Neutrons can decay into a proton plus additional particles.

    What is prompting your question?
     
  4. Mar 18, 2013 #3
    but i thought, sometimes protons decay into neutrons, electrons and anti neutrinos?
     
  5. Mar 18, 2013 #4

    jtbell

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    Actually, in order to conserve charge, it would have to be proton --> neutron + positron + neutrino.

    We've never seen a free proton decay, and not for lack of searching! But if it does happen, it will be via some exotic process involving new particles and/or interactions. The process above can't happen for a free proton because a neutron is heavier than a proton. You'd have to pump energy into a proton somehow in order to make it happen.

    However, in certain atomic nuclei, a proton can decay this way. It's often called beta+ decay or positron emission. This happens even though a neutron is heavier than a proton, because in nuclear decay, it's the mass of the entire nucleus before and after that counts, and in these cases the final nucleus is lighter than the initial nucleus because of different binding energies.
     
    Last edited: Mar 18, 2013
  6. Mar 18, 2013 #5
    yes, i should have clarified. I'm sorry, I am talking about nuclear decay... I don't know why it happens though? I mean... something about exchange particles comes to mind and weak interactive forces but I don't think I understand exactly what happens?
     
  7. Mar 18, 2013 #6

    jtbell

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    It goes via the weak interaction. First ##p \rightarrow W^+ + n## where the W is "virtual", then ##W^+ \rightarrow e^+ + \nu_e##.

    Actually it's quarks inside the proton/neutron that are involved: first ##u \rightarrow W^+ + d##, then the virtual W+ "decays" like above.
     
  8. Mar 18, 2013 #7
    Thanks! That's really helpful :)

    But I guess it can happen the other way too right? I don't understand how one particle can change into another particle unless all particles are made of sort of a base of particles, which are quarks I guess but I thought electrons and neutrinos are leptons, and therefore don't have quarks so why do particles change into them?
     
  9. Mar 18, 2013 #8

    mathman

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    Never ask "why". They just do. Current physics allows leptons to be created, in pairs consisting of a particle and an antiparticle. For example: the neutron decay ends up (after intermediate steps) as a proton + electron (particle)+ antineutrino (antiparticle).
     
  10. Mar 18, 2013 #9

    jtbell

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    Sometimes we can answer "why" in terms of an over-arching principle of some kind. But then that raises another "why" question, of course, namely, why is that principle true?

    The rules by which quarks and leptons can be created/destroyed/converted can be "explained" by (derived from) various types of gauge symmetry: SU(2) x U(1) for the unified weak and electromagnetic interactions, and SU(3) for the strong interaction. But why those particular symmetries? Nobody knows for sure... yet. That's what research into Grand Unified Theories is about.
     
    Last edited: Mar 18, 2013
  11. Mar 18, 2013 #10
    But why do we follow these principals ourselves? For instance we think know that an electron neutrino exists but actually, we just know that there is something out there which matches up with these equations and because logically if it has the characteristics of what we think we have found, then we have found it. but aren't we then just finding things because we are looking for it - therefore, creating these "explanations" up for ourselves?
     
  12. Mar 19, 2013 #11

    mathman

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    Without trying to address the specific question, these theories are accepted because they very accurately agree with measured information.
     
  13. Mar 20, 2013 #12
    In the case of nuclear isotopes, the reason why they decay is because there is an alternative state of lower energy that can be "reached" by one of the available types of interaction - in this case, a weak decay.
     
  14. Mar 22, 2013 #13
    Most Grand Unified Theories predict that free protons will decay. They also predict that neutrons will decay by essentially the same process, but for free neutrons, that decay channel is teeny teeny teeny tiny compared to the weak-interaction decay channel. These processes will also make protons and neutrons in nuclei decay, so neutron decay might be observed in otherwise-stable nuclei.

    There have been lots of searches for proton decay since the 1980's, and the latest experimental lower limit is about 10^(32) years. That's getting close to what some GUT's predict.
     
  15. Mar 31, 2013 #14
    If we're going to get into free proton decay, it should probably be noted that the standard model actually has a channel for proton decay. It's just incredibly suppressed (even more than the GUT modes would be), as it involves the non-perturbative electroweak instanton configurations known as sphalerons.
     
  16. Mar 31, 2013 #15
    How do sphalerons violate baryon number? I've seen them implicated in matter-antimatter asymmetry.
     
  17. Mar 31, 2013 #16
    Strictly, they violate B+L while conserving B-L. Think of them as creating an effectively coupling of three quarks and one (anti)lepton.
     
  18. Mar 31, 2013 #17
    I don't see how they violate baryon number and lepton number, because at first sight, the Standard Model conserves both of them.
     
  19. Mar 31, 2013 #18
    It does conserve them perturbatively. But, in a similar way to how a chiral anomaly can break what looks like a good symmetry of a model once you consider quantum corrections, non-perturbative gauge configurations can violate certain global symmetries. The upshot here is that what looks like a good symmetry at the level of the Lagrangian is not always realized in the full quantum mechanical, non-perturbative theory.
     
  20. Apr 26, 2015 #19
    But protons are supposed to be stable due to baryon number conservation, right? But, if the B number is actually broken by the chiral anomaly... are protons still considered stable in Standard Model? why?
     
  21. Apr 26, 2015 #20
    Whoa, didn't think I'd see this thread resurrected...

    Anyway, I didn't say that the chiral anomaly violates B, I said that the V+L violation from sphalerons is like what happens with the chiral anomaly in that it's an effect that isn't explicit in the model but, rather, comes about due to some sort of field configuration that the dynamics of the model admits non-trivially.

    All that said, I may as well go ahead and clarify what I was originally saying here. While my statement that, in effect, sphalerons couple three quarks and a(n) (anti)lepton is basically right - at least inasmuch as I was explaining how they replace baryons with antileptons - what I didn't include is that the full sphaleron configuration actually involves nine quarks and three (anti)leptons, meaning that B and L each change by [itex]\pm 3[/itex]. So, strictly, the correct conclusion is that a proton is stable but protons are not. Basically, if you have two protons, a sphaleron can decay them to an anti-proton and three positrons. And, since the total mass of the final state is smaller than the initial, this can happen spontaneously.

    The upshot of this is that, in principle, sphalerons could reduce the net baryon number of the universe to 0, 1, or -1.

    All that said, for all intents and purposes, it's safe to consider protons to be effectively stable in the Standard Model, since the time scales for sphalerons to be relevant are enormous.
     
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