Understanding Proton Decay: College Physics Help

In summary: Different numbers of particles?In summary, waterliyl says that protons do not decay, neutrons can decay into a proton plus additional particles, and free protons will decay.
  • #1
waterliyl
11
0
Hi, so we are studying quantum physics at college and I just want to understand this concept of decay. Please could someone help me?
 
Physics news on Phys.org
  • #2
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?
 
  • #3
but i thought, sometimes protons decay into neutrons, electrons and anti neutrinos?
 
  • #4
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:
  • #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?
 
  • #6
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.
 
  • #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?
 
  • #8
waterliyl said:
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?

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).
 
  • #9
mathman said:
Never ask "why". They just do.

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:
  • #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?
 
  • #11
waterliyl said:
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?
Without trying to address the specific question, these theories are accepted because they very accurately agree with measured information.
 
  • #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.
 
  • #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.
 
  • #14
lpetrich said:
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.

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.
 
  • #15
How do sphalerons violate baryon number? I've seen them implicated in matter-antimatter asymmetry.
 
  • #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.
 
  • #17
I don't see how they violate baryon number and lepton number, because at first sight, the Standard Model conserves both of them.
 
  • #18
lpetrich said:
I don't see how they violate baryon number and lepton number, because at first sight, the Standard Model conserves both of them.

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.
 
  • #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?
 
  • #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.
 
  • #21
Now I am curious if sphalerons are the only reason to consider protons to be stable in the Standard Model. Are they? or there are more "possible answers" to my previous question.
 
  • #22
So far as I know, sphalerons are the only source of B-violation in the Standard Model.
 
  • #23
Are other baryons or nuclei allowed to decay?
 
  • #24
I assume you're asking about B-violating decays specifically, as there are many B-conserving nuclear decays that are well-known.

Generically, sphalerons should be applicable to any process where B changes by 3 and B-L is conserved, provided that the process correctly conserves electric charge and has available phase space - which, here, is mostly to say that you shouldn't expect final state taus when you're decaying baryon pairs. (Depending on the exact details, though, there could be additional suppression due to flavor-changing processes; but, even that doesn't make a process impossible.)
 
  • #25
could you give some examples, please?

Thank you very much for your responses.
 
  • #26
The basic process I describe for two protons can easily enough be modified to cover the other possibilities. Basically, you can replace any proton in that process with a neutron as long as one of the positrons changes into an anti-neutrino; and, you can also replace every particle with its antiparticle. And, nothing, in principle, should prevent this from happening to nucleons while they're in a nucleus; so, you should be able to derive potential nuclear decays from that.

But, I should caution that all of this is speaking purely theoretically. Sphalerons have been shown to be a mathematical consequence of the Standard Model; but, they have never been observed in the laboratory.
 
  • #27
I am reading a textbook and have just seen a problem which I do not understand what it means, can I ask you?
 
  • #28
Breo said:
I am reading a textbook and have just seen a problem which I do not understand what it means, can I ask you?

what?
 
  • #29
what means that a fermion bilinear cancel for a Majorana fermion such that psy = psyc ?
 
  • #30
Breo said:
what means that a fermion bilinear cancel for a Majorana fermion

Where did you get that from?
 
  • #31
It is an old textbook in russian: Particle physics and Symmetries - Svörtynsky

The problem says:

Of all the fermion bilinears (scalar, pseudoscalar, vector, axialvector and tensor), which of them cancel for a Majorana fermion such that ##\psi = \psi^c##? Does a kinetic term of the form ## \bar{\psi}\gamma_{\mu}\partial^{\mu}\psi^c ## cancel?
 
  • #33
I still do not understand one thing. Probably is a small misconception.

If protons are generally considered to be stable because the baryon number symmetry protects them, but it is broken in the Standard Model at the quantum level (chiral anomaly, but still SM), why we still say that protons are stable in SM? (forget about GUTs)
 
  • #34
In last week's seminar, a professor said that the chiral anomaly will lead to a lifetime of $10^{70}$ years.

From this I conclude the following: The SU(5) GUT have lifetimes of sometimes less than $10^{35}$ years. Therefore the difference is so large, that one can consider the protons stable even with the chiral anomaly in the standard model.
 
  • #35
zombie post...
 

1. What is proton decay and why is it important in college physics?

Proton decay is a hypothetical process in which a proton, one of the fundamental particles that make up an atom, can spontaneously break down into smaller particles. This concept is important in college physics because it is a key component of the Grand Unified Theory, which seeks to explain the fundamental forces of the universe and unify them into a single framework.

2. How does proton decay occur?

According to current theories, proton decay can occur through a process called baryon number violation, in which the baryon number (a quantum number that represents the number of quarks in a particle) is not conserved. This can happen through the exchange of particles called X and Y bosons, which are predicted by the Grand Unified Theory.

3. Can proton decay be observed or measured?

So far, there is no experimental evidence for proton decay. However, scientists have been searching for it using large underground detectors, such as the Super-Kamiokande in Japan and the Deep Underground Neutrino Experiment in the United States. These experiments are looking for specific decay modes that would confirm the existence of proton decay.

4. What are the implications of proton decay for our understanding of the universe?

If proton decay is observed, it would have significant implications for our understanding of the universe. It would provide evidence for the Grand Unified Theory and help us better understand the fundamental forces and particles that make up our universe. It could also help explain the matter-antimatter asymmetry and the excess of matter over antimatter in the universe.

5. Are there any potential applications of understanding proton decay?

While there are currently no practical applications of understanding proton decay, it could lead to advancements in our understanding of the universe and potentially inspire new technologies. Additionally, the research and experiments conducted to study proton decay could have spin-off benefits in other areas of science and technology.

Similar threads

  • High Energy, Nuclear, Particle Physics
Replies
8
Views
1K
  • High Energy, Nuclear, Particle Physics
Replies
6
Views
538
  • High Energy, Nuclear, Particle Physics
Replies
1
Views
776
  • High Energy, Nuclear, Particle Physics
Replies
5
Views
2K
  • High Energy, Nuclear, Particle Physics
Replies
4
Views
2K
  • High Energy, Nuclear, Particle Physics
Replies
2
Views
444
  • High Energy, Nuclear, Particle Physics
Replies
14
Views
1K
  • High Energy, Nuclear, Particle Physics
Replies
4
Views
1K
  • High Energy, Nuclear, Particle Physics
Replies
5
Views
1K
Replies
4
Views
338
Back
Top