# $\beta^+$ decay question

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?

• In a nucleus (not just a bare proton), beta-decay of a proton can be allowed by conservation of mass-energy. – Ben Crowell Nov 17 '14 at 21:38
• 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. – dmckee Nov 18 '14 at 2:21

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 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 [...] – Ben Crowell Nov 17 '14 at 23:37
• 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. – Bill N Nov 18 '14 at 20:08