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I read that all baryons apart from the proton itself decay into protons (why though?) and that mesons do not decay into protons due to having less mass than protons.

Thus it makes sense for the $\beta^-$ decay to occur as the neutron decays to the proton, but how does the $\beta^+$ decay take place if the proton does not decay at all?

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    $\begingroup$ Note that the restriction of mesons decaying to protons has nothing to do with mass ... there are mesons heavier than the proton, but they can't decay to protons because their valence quark content isn't compatible with that of a baryon. In principle a very heavy meson could have a proton anti-proton pair in its decay products, but that must be massively suppressed. $\endgroup$
    – dmckee --- ex-moderator kitten
    Commented Nov 18, 2014 at 2:21
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    $\begingroup$ Yes. The decay $J/\psi \to p \overline p$, for example, has been measured with a branching ratio of 0.2 % pdg.lbl.gov/2019/listings/rpp2019-list-J-psi-1S.pdf $\endgroup$
    – RogerJBarlow
    Commented Sep 8, 2020 at 11:04

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Under certain conditions, such as having a high charge/mass ratio (not enough neutron constituents) the nucleus will cast off some charge, but doesn't have enough energy to eject a whole proton. Effectively, in the field of a large mass, a positron and neutrino are ejected, the atomic number (Z) of the daughter nucleus is reduced by 1 compared to the parent, and the mass number (A) is unchanged. The daughter nucleus becomes more tightly bound (it has less mass energy than before). This reaction does not appear to take place without the presence of other mass.

The proton by itself is mass restricted from decaying into a neutron plus positron ($Q = -1.804$ MeV) or even electron capture ($Q = -782$ keV). But the proton-proton $\to e^+ + \nu$ has a $Q= +420$ keV, so there is enough mass-energy present in the center of mass for the deuteron and positron to form.

The neutrino appears because in addition to conserving energy, momentum, charge, and baryon number, something called lepton number is conserved in the reaction.

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  • $\begingroup$ The first paragraph is pretty garbled, and it's not clear to me what you have in mind there. It seems as though you think that mass-energy equivalence, along with the fact that nuclear masses aren't equal to $Zm_p+Nm_n$, implies that neutron number and proton number aren't well defined. That's not true. If you have in mind something about the fact that protons and neutrons are composites of quarks, then that isn't coming through clearly here, and in any case isn't relevant. In the second paragraph, I think you're confused about what it means to put a many-body wavefunction together [...] $\endgroup$
    – user4552
    Commented Nov 17, 2014 at 23:37
  • $\begingroup$ [...] out of a Slater determinant of single-particle wavefunctions. It doesn't mean that "there aren't any individual particles." In the final paragraph, the discussion of pp fusion doesn't address the question, which is about beta decay, not fusion reactions that involve a weak-interaction process. $\endgroup$
    – user4552
    Commented Nov 17, 2014 at 23:39
  • $\begingroup$ Not only are the neutron and proton numbers well defined, but their momentum states are sufficiently well correlated that when uses models (like Glauber models) that assume the nucleons have well defined positions it is a small but definite error to treat those positions as uncorrelated. $\endgroup$
    – dmckee --- ex-moderator kitten
    Commented Nov 18, 2014 at 2:20
  • $\begingroup$ The point of the original last paragraph (now next to last) is that while a proton by itself won't decay into a neutron and positron, simply bringing in another proton will allow the positron process to occur and produce a proton+neutron bound nucleus, even though it's not purely positron decay of $^2$H. $\endgroup$
    – Bill N
    Commented Nov 18, 2014 at 20:08

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