# Do ALL hadrons eventually decay into protons?

I have two sources, one (school textbook) is telling me that ''all hadrons eventually decay into a proton'', whereas a few other people have told me that ''only bayrons decay into protons''. I want to know because I originally wanted to find out whether mesons decay into protons, and mesons obviously are part of the hadron family, but they're not bayrons.

So I'm confused, which is true?

Dick
Homework Helper
I have two sources, one (school textbook) is telling me that ''all hadrons eventually decay into a proton'', whereas a few other people have told me that ''only bayrons decay into protons''. I want to know because I originally wanted to find out whether mesons decay into protons, and mesons obviously are part of the hadron family, but they're not bayrons.

So I'm confused, which is true?

Only baryons decay into protons. pi mesons are lighter than protons. That would be an impossible decay even if baryon number weren't conserved.

Only baryons decay into protons. pi mesons are lighter than protons. That would be an impossible decay even if baryon number weren't conserved.

I guess my textbook is wrong then, thats 3 people against the textbook now :p

anyway, what you said about pi mesons being lighter than proton, is new to me. So this a rule then, that a particle can only decay into things which have a smaller mass than itself?

by the way, you know 'rest energy' (in MeV units)? if I convert it to joules and then sub it into the e = mc^2 equation, that would give me the mass of the particles, right?

jtbell
Mentor
So this a rule then, that a particle can only decay into things which have a smaller mass than itself?

Yes. If a particle could decay into a heavier one, where would the extra rest-energy come from?

you know 'rest energy' (in MeV units)? if I convert it to joules and then sub it into the e = mc^2 equation, that would give me the mass of the particles, right?

Yes, you get the mass in kg. However, if you ask most physicists (at least most particle physicists) "what's the mass of [some particle]", they'll give it to you in MeV or keV units anyway, for example 938.3 MeV/c2 for a proton. Those units are more convenient for many of the calculations that they do.

Dick
Homework Helper
I guess my textbook is wrong then, thats 3 people against the textbook now :p

anyway, what you said about pi mesons being lighter than proton, is new to me. So this a rule then, that a particle can only decay into things which have a smaller mass than itself?

by the way, you know 'rest energy' (in MeV units)? if I convert it to joules and then sub it into the e = mc^2 equation, that would give me the mass of the particles, right?

Sure it's a rule. You can't decay into something heavier and still conserve energy. And yes, converting to joules and substituting will give you the mass in kg. It's good practice, but if you are going to do this a lot it's helpful to just remember some conversions. Like .5 MeV comes out pretty close to the electron mass and 940 MeV is close to the proton mass.

all makes sense now, thanks guys!

mfb
Mentor
To be precise: All free baryons decay to protons (or antiprotons).
Neutrons in nuclei can be stable.

All mesons eventually decay to photons, leptons and (rare) baryons, with only stable particles (proton, bound neutron, photon, electron, all 3 neutrino types and their antiparticles) in the final state.

To be precise: All free baryons decay to protons (or antiprotons).
Neutrons in nuclei can be stable..

and how do neutrons become 'free'?

I've read that if a nucleus gets too big (containing too many nucleons, or is protons more specifically? idk), then it'll become unstable, is this when neutrons become 'free'?

also while we're onto this topic, what is the minimum number of neutrons that a nucleus must have? (I know that for hydrogen, it can contain no neutrons, but is this the same for other elements?)

All mesons eventually decay to photons, leptons and (rare) baryons, with only stable particles (proton, bound neutron, photon, electron, all 3 neutrino types and their antiparticles) in the final state.

K-mesons decay into Pi mesons

Pi mesons then decay into muons and their respective neutrino/anti-neutrino (leptons). Is this right?

At least thats how I understand it, so where does photon come into all this?

bearing in mind I'm in my last year of school (A-levels), so you don't have to go into too much detail if its too advance for what me

mfb
Mentor
and how do neutrons become 'free'?
Some nuclei can release neutrons on their own, usually together with fission or directly after fusion.
Some nuclei are stable with neutrons - if you do not shoot other particles on them, they will stay like that forever (assuming no proton decay*).

*if protons decay, all baryons and all nuclei will decay to other particles

also while we're onto this topic, what is the minimum number of neutrons that a nucleus must have? (I know that for hydrogen, it can contain no neutrons, but is this the same for other elements?)
This is given by the proton drip line, which lists the first unbound isotopes per element.

K-mesons decay into Pi mesons

Pi mesons then decay into muons and their respective neutrino/anti-neutrino (leptons). Is this right?
That is the most common decay mode. And muons decay into electron + anti-electronneutrino + muonneutrino.

At least thats how I understand it, so where does photon come into all this?
Neutral pions usually decay into two photons, and other decays can emit a photon as well. This does not mean that every decay chain involves photons, but they are possible, so I listed them.

The http://pdglive.lbl.gov/listings1.brl?quickin=Y [Broken] has lists of decay modes and their relative probability (branching fraction).

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Some nuclei can release neutrons on their own, usually together with fission or directly after fusion.
Some nuclei are stable with neutrons - if you do not shoot other particles on them, they will stay like that forever (assuming no proton decay*).

*if protons decay, all baryons and all nuclei will decay to other particles

This is given by the proton drip line, which lists the first unbound isotopes per element.

That is the most common decay mode. And muons decay into electron + anti-electronneutrino + muonneutrino.

Neutral pions usually decay into two photons, and other decays can emit a photon as well. This does not mean that every decay chain involves photons, but they are possible, so I listed them.

The http://pdglive.lbl.gov/listings1.brl?quickin=Y [Broken] has lists of decay modes and their relative probability (branching fraction).

once again, thanks for explaining

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Hadrons have a property called the Baryon number which is conserved by all everyday reactions*.

Protons, neutrons and other baryons have Baryon numbers of 1 (with -1 for the corresponding antiparticles).

All mesons have Baryon number 0. Therefore, mesons can never decay into baryons unless a matching number of anti-baryons are also produced.

So it's possible that the final decay state of a heavy meson could contain a proton, but it would then also have to include an anti-proton.

Most commonly, though, the ultimate decay products of mesons are photons and/or leptons (electrons and neutrinos).

* The above all stems from quark number conservation. The only exceptions would be in physics beyond the Standard Model, eg GUTs that contain interactions that can change individual quarks into leptons. None of the currently known interactions can do this, as the electric charge of the particle would also have to change by a fractional amount and there are no currently known force-carrier bosons that can 'carry away' fractional electric charges.

Hadrons have a property called the Baryon number which is conserved by all everyday reactions*.

* The above all stems from quark number conservation. The only exceptions would be in physics beyond the Standard Model, eg GUTs that contain interactions that can change individual quarks into leptons. None of the currently known interactions can do this, as the electric charge of the particle would also have to change by a fractional amount and there are no currently known force-carrier bosons that can 'carry away' fractional electric charges.

This is actually not correct. The Standard Model, by itself, has non-perturbative gauge configurations called "sphalerons" which violate both the conservation of baryon and lepton numbers, but leave the difference between the two conserved. In other words, these processes can convert three quarks into an anti-lepton (or two quarks into an anti-quark and an anti-lepton, etc.)

OK. But these sphaleron transitions don't seem to be observed in any experiments I'm aware of. Is this because they would require even higher energy levels than (say) LHC can muster, or is there a bit more to it than that. Unfortunately, this area is some way beyond my own current understanding.

OK. But these sphaleron transitions don't seem to be observed in any experiments I'm aware of. Is this because they would require even higher energy levels than (say) LHC can muster, or is there a bit more to it than that. Unfortunately, this area is some way beyond my own current understanding.

Yes. They're pretty much inaccessible at ordinary scales. But, they're pretty strongly implicated to have some involvement in the matter/antimatter asymmetry. To wit, the availability of sphaleron transitions allows the entire asymmetry to be generated in either the baryon or lepton sector, since the B+L violating transitions will carry it over into the other.

mfb
Mentor
Yes. They're pretty much inaccessible at ordinary scales. But, they're pretty strongly implicated to have some involvement in the matter/antimatter asymmetry. To wit, the availability of sphaleron transitions allows the entire asymmetry to be generated in either the baryon or lepton sector, since the B+L violating transitions will carry it over into the other.
Even better: Sphalerons conserve B-L, but violate B+L in a natural way.
Our universe has large positive B and L values, but the difference can be small, or maybe even zero?
You need an asymmetry in the whole system, but no additional asymmetry of baryons or leptons.

Dick