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This is a massive oversimplification and anthropomorphizes elementary particles, but hopefully satisfies your curiosity. The reason for the worsening stability is because a free neutron has more potential energy than one bound to a proton. In the same way that a muon will "preferably" decay into an electron by interacting with a W-, if you were able to have a down quark floating in space by itself (you can't, for unrelated reasons,) that down quark would "want" to decay into a much lighter up quark by exchanging a W-. In both of these cases, the elementary particle loses mass through the interaction. Inside the free neutron, that's what happens, too. A down quark interacts with a W- to transform itself into an up quark; the neutron is now a proton, with the W- carrying away some of the mass and electric charge. However, neutrons and protons are more complex than just quarks. They're also big bundles of gluons, exchanging colour charge amongst each other. These gluons have mass and you need energy to create them. They actually account for most of the mass of a neutron/proton. A bound neutron and proton are exchanging these gluons amongst each other, too. Suppose you had a bound neutron and proton. If the neutron were to decay into a proton, the electric charge of the particles (now both +1) would cause them to repel; you can imagine either of two things might happen: the protons will now move apart _or_ they'll stick together. In the former, you need kinetic energy and in the latter, you're going to need more gluons to hold the whole thing together. Both of those cases requires more energy (==mass) than the neutron would "lose" by converting to a proton. So, if you can imagine the potential energy landscape, the bound neutron is at a local minimum (lower than a free neutron) |
Why can't the W-'s kinetic energy be smaller than whatever the neutron would lose by converting to a proton? There's no quantization preventing that here, right? My guess here is that it would not have (for the lack of a better term) sufficient "escape velocity", meaning that it would fall back into the neutron again, but wouldn't that imply (a) this is more like a dynamic equilibrium than a static one, where neutrons turn into protons and back into neutrons repeatedly, and (b) quantum uncertainty (Heisenberg, tunneling, etc.) should still mean that bound neutrons should still decay once in a while, and perhaps (c) a neutron next to a proton might randomly "swap" places once in a while if a W- from one gets pulled into the other one? (And if any of these is the case, then where do they draw the line to declare that a bound neutron is "stable"?)