How can beta plus decay be possible?

$$\beta +$$ decay is where a proton gets turned into a neutron and a positron and a neutrino.

However, a neutron is heavier than a proton, so obviously this reaction is endothermic. So then, why does it happen? I've seen an explanation here in the question How can a proton be converted to a neutron via positron emission and yet gain mass?

It describes that the final binding energy of the nucleus increases, thus making it possible by becoming more stable. But what actually causes the reaction to go in the first place? It's like saying this ball will go down the hill because it will lose energy - what gives it the nudge required? Is it something like energy from external gamma rays or something?

• If the energy barrier in your picture was low enough I would expect the ball to tunnel through it to the true ground state from the false one its in. – jacob1729 Jun 25 at 9:49
• Beta plus is possible because it happens when the proton is inside a nucleus, not when the proton is free. And when you calculate the energy balance, you must take into account the fact that the proton is part of the nucleus. When you do that, you find that the beta decay is indeed possible. – AWanderingMind Jun 25 at 10:08
• Your tunneling question applies to all decays, no? Heard of Gamow? – Cosmas Zachos Jun 25 at 10:46
• "Is it something like energy from external gamma rays or something?" No, in an unstable nucleus, radioactive decay is spontaneous, it doesn't need an external trigger. Don't forget that the nucleons in a nucleus have kinetic energy and momentum, they aren't stuck together in a static ball, despite the numerous pictures you have undoubtedly seen. – PM 2Ring Jun 25 at 12:03
• @Ben Crowel . I would agree, but the OP's notional potential well in isospace (I gather) may well be trying to evoke QM isocharge "leakage". It is a bad picture, but not one on which to base one's doubts. – Cosmas Zachos Jun 25 at 14:03

You write that

However, a neutron is heavier than a proton, so obviously this reaction is endothermic.

That's true for free neutrons and free protons, which is why free neutrons are unstable against beta decay but free protons aren't. However, the nuclear environment is much more complicated than the vacuum, and when thinking about beta-decay (of either sign) in a heavy nucleus, the free-particle masses aren't the right parameter to consider. What matters is whether the mass of the entire system is increased or decreased by the beta decay.

One hand-waving way$$^\dagger$$ to think about the energetics in positive beta decay is to remember that protons have positive electrical charge and repel each other. So a nucleus with "too many" protons will have more energy stored in its electric field than a nucleus with the same number of nucleons (protons and neutrons inclusive) but less total positive charge. An observer outside of the nucleus can't distinguish between the energies due to the masses of the constituent particles, the positive (repulsive) energy stored in the electric field, and the negative (attractive) energy of the strong-interaction field which binds the nucleus together --- all of these contributions just add up to make the total mass-energy of the nucleus. If a charged-current weak interaction can decrease this total mass-energy by transforming a constituent neutron into a proton, then that process is exothermic.

$$^\dagger$$I often describe concepts in nuclear physics using hand-waving analogies and, months or years afterwards, get really interesting clarifications in the comments from other users who are more cautious than I am. I love those and I look forward to them.

But what actually causes the reaction to go in the first place? It's like saying this ball will go down the hill because it will lose energy - what gives it the nudge required?

There isn't a potential barrier in beta decay, whether it's beta minus or beta plus decay. In both cases the decay is slow simply because the transition probability is so slow.

In what follows I'm going to over simplify a bit so take care about interpreting this too literally. Suppose we have two states $$\psi_1$$ and $$\psi_2$$ then for a transition to occur between them there must be an operator that mixes the states. So for example in the decay of an excited atom the electric field operator associated with the electric field of light mixes up the $$1s$$ and $$2p$$ states, and this is why a $$2p$$ can decay to a $$1s$$ and emit a photon. Conversely, the electric field operator does not mix the $$1s$$ and $$2s$$ states and that's why the $$2s \to 1s$$ transition is forbidden. Anyhow the size of the mixing determines the probability of the transition. This probability is calculated using Fermi's golden rule.

Consider first beta decay, because that's what Fermi originally considered. Then we have an initial state of a neutron and a final state of the proton, electron and antineutrino. Fermi used an approximate calculation that glossed over the fine details and calculated the decay probability for an isolated neutron. The calculation is described in the Wikipedia article on the Fermi interaction, but we don't need to worry about the details. The result is that the decay probability is small, and that's why the free neutron takes so long (15 minutes - eternity by nuclear physics standards!) to decay.

If you do the same calculation for a free proton decaying by beta plus decay then you'll get the answer zero because as you say that would violate conservation of energy. In principle we could do the calculation for the proton in a nucleus, but in practice the system is far too complicated to do this calculation from first principles. However we can take the experimentally measured energy difference between the undecayed and decayed states and plug this into the calculation as a parameter, and doing this we would end up with the observed long lifetime. The overall energy change is negative for the reasons discussed in the question you linked. The transition probability is slow simply because the mixing between the initial and final states is exceedingly small. Ultimately this is due to the fact the weak force is, well, weak.

It's tempting to ask what actually happens during the decay, and as PM 2Ring says in his answer it is possible to draw Feynmann diagrams showing the reactions of the valence quarks. But you need to be very cautious about taking this too literally. The fundamental particles like electrons and neutrinos emerge from quantum field theory in the limit where interactions between particles are weak. In the interior of a hadron the interactions are strong and they mix up the particles so the quantum field state cannot simply be described as a sum of distinct particles. You may have heard it said that the hadron contains many virtual particles, but these aren't really particles - it's just a way of describing the field state as a sum of particle states. The Fermi calculation glosses over the (hideously complicated) details and just uses overall energy changes.

Here's a Feynman diagram of $$\beta+$$ decay from the Wikipedia page on Beta decay:

The leading-order Feynman diagram for β+ decay of a proton into a neutron, positron, and electron neutrino via an intermediate W+boson.

As that article says, beta decay is a consequence of the weak force. Nucleons are composed of up quarks and down quarks, and the weak force allows a quark to change type by the exchange of a W boson and the creation of an electron/antineutrino or positron/neutrino pair.

We can model this process in terms of virtual particles. Because it is subject to the weak force, a quark constantly emits virtual W bosons. These bosons have a very short lifetime, and the quark usually re-absorbs the boson almost immediately.

However, if the energy conditions of the nucleus are suitable, there's a small probability that the W boson isn't re-absorbed but instead decays.

Whether the energy conditions are suitable depends on the configuration of the nucleus. In particular, a nucleus with an excessive number of protons is prone to $$\beta+$$ decay. The nuclear shell model goes into the details of how nucleons behave in the nucleus.

In $$\beta+$$ decay, an up quark in a proton emits a $$W+$$ boson, becoming a down quark. The $$W+$$ boson then decays to a positron and a neutrino before the down quark manages to re-absorb it.

Please bear in mind that this is just a model. The W boson is an internal line of a Feynman diagram, so it's a virtual particle, a mathematical calculation tool, it doesn't have to obey all of the rules of a real particle. In particular, it doesn't have to satisfy the usual energy-momentum relation. Please see On shell and off shell for further details.

• This doesn't help, because it introduces a W boson. The W boson is said to have a mass of 80.379 GeV/c². Saying it's only a virtual particle that's "off shell" doesn't get us anywhere. – John Duffield Jun 25 at 19:10
• @JohnDuffield Virtual particles give us a convenient way to describe interactions, and a practical way to perform calculations that would be otherwise formidable using the full quantum field theory. – PM 2Ring Jun 25 at 19:29
• PM 2Ring: noted. I've tried to give my own answer to this question. – John Duffield Jun 25 at 20:18

It is very important to understand, that there are two cases:

1. free proton, this is called an inverse beta decay, and it needs an incident anti neutrino of sufficient energy

2. proton inside a nucleus, in this case the energy is a little bit more (neutrino) then what you would expect based on the mass of the proton and the neutron and the positron

Now the reason the beta plus decay (inside a nucleus) happens, is because of what you are saying, stability. The nucleus becomes more stable by having relatively more neutrons (then before the conversion). The nucleus transforms towards more stability. The reason why it is becoming more stable has to do with the ratio of protons to neutrons. The nucleus is more stable when protons and neutrons number is closer to equal. Why? This has to do with the strong force and the isospin.

The daughter nucleus has a greater binding energy (and therefore a lower total energy) then the mother nucleus. Lower total energy is what every composite particle in the universe (disregarding external effects) is moving towards.

So there are two main reasons the nucleus needs to go to more stability:

1. isospin, this is very important to understand. In history, the neutron and the proton were considered the same particle, just different isospin. Quarks were discovered, and we know they do have a different internal structure, but the isospin quantum number remains in the SM. Now what is isospin and why do we call it that? Originally, the Pauli exclusion principle was thought (and still does, just is now not so relevant here) to say that two fermions (and neutrons and protons happen to be fermions) cannot occupy the same quantum state (they can't have all quantum numbers the same). So neutrons and protons had to be distinguished. There you have isospin. Now what does it mean after the discovery of quarks and the strong force and the residual strong force? A proton and another proton (or a neutron and another neutron) cannot come very close because:

1. the Pauli exclusion principle remains

2. the strong (and the residual strong) force was discovered, and was observed to become repulsive (the residual strong force) at very short distances

Now, why does it matter? Because if they are (neutrons and protons) at optimal distance, the strong force will keep them together with an extra power. What is this extra power? This is the residual strong force's phenomenon, and it causes one neutron and one proton to be able to become more stable (have a stronger bond) then two neutrons or two protons.

This means, that if you have one proton and one neutron, they will align (complement) their isospin, and therefore will become more stable.

If you have two neutrons (or just two protons), these will not be as stable, as if you have one neutron and one proton.

So if you have a nucleus, that has more protons then neutrons (proton-rich), then it will be stable, but it will move towards more stability by equalizing the number of protons and neutrons.

1. greater binding energy and lower total energy. A nucleus that has equal number (or closer to equal) of protons and neutrons will have a stronger residual force binding energy (nuclear force) because of isospin and because neutrons and protons are able to come closer then protons and protons (or neutrons and neutrons), and the strong force is very distance dependent, so the smaller nucleus (even if this size difference is very little) will cause a stronger binding energy

Now you are asking what triggers the conversion. This is because you have the misconception that the proton inside the nucleus is stable. It is not. You are looking at this the wrong way. You would think of the proton as a separate entity. It is not. It is part of the nucleus. The nucleus is not stable, because it is proton rich, and it has to transform into a less proton rich nucleus to become more stable. It is the nucleus itself that decays (we do not use the word decay here usually, but transformation). The way the nucleus becomes more stable is by converting the proton into a neutron.

You think of the nucleus as boolean. Stable or not stable. In QM, it is all about probabilities. The proton rich nucleus has its own quantum characteristics, wavefunction, and those give the proton rich nucleus a probability to transform into a more stable energy level, thus become less proton rich. Equilibrium is what the universe wants, and the nucleus is playing along the rules too, wanting to move as close to equilibrium as it can in terms of the number of neutrons and protons.

• Stability is a boolean variable, in QM as well as anywhere. If the system is in an unstable state, then it's in an unstable state, period. – Emilio Pisanty Jun 27 at 13:54

How can beta plus decay be possible?

Because it isn't really a decay. At least, not like beta minus decay is a decay. That's where you start with a neutron, which is perhaps a free neutron. Without interacting with anything else, after circa 15 minutes the free neutron decays into a proton, an electron, and an antineutrino. If you've got a free proton instead of a free neutron, you'll never see it decay into a neutron, a positron, and a neutrino.

$$\beta +$$ decay is where a proton gets turned into a neutron and a positron and a neutrino. However, a neutron is heavier than a proton, so obviously this reaction is endothermic.

I don't like to see the word endothermic here. This is a reaction where you have to add energy to the proton. You have to add energy equivalent to the following:

One positron

One neutrino

One electron

One antineutrino

The first two are emitted in beta plus decay, the latter two are emitted if your resultant neutron was a free neutron that decayed back to a proton. If you say the neutrino energy and the antineutrino energy is similar to the electron energy of 511keV, which is the same as the positron energy, you're talking about circa 2MeV. You also need some extra energy because the electron and the positron typically have considerable kinetic energy, so let's round that up to 3MeV. This might sound a lot, but it isn't.

The deuteron has a binding energy of 2.22 MeV. It consists of one proton and one neutron. The triton has a binding energy of 8.48 MeV. It consists of one proton and two neutrons. Adding one neutron to a deuteron more than doubles the binding energy per nucleon. One neutron makes a difference of 6.26 MeV. The triton can undergo beta decay to become a stable helion or helium-3 nucleus comprised of two protons and one neutron. It has a binding energy of 7.71 MeV. The helium-4 nucleus consists of two neutrons and two protons. The binding energy is 28.29 MeV. Adding one neutron to a helion more than doubles the binding energy per nucleon. One neutron makes a difference of 20.58 MeV. The important point to note is that as the nuclei get bigger, the binding energies get huge compared to the energy of the electron, positron, neutrino, and antineutrino. Now see this in the Wikipedia beta decay article:

"However, β+ decay cannot occur in an isolated proton because it requires energy, due to the mass of the neutron being greater than the mass of the proton. β+ decay can only happen inside nuclei when the daughter nucleus has a greater binding energy (and therefore a lower total energy) than the mother nucleus. The difference between these energies goes into the reaction of converting a proton into a neutron, a positron and a neutrino and into the kinetic energy of these particles".

If you somehow convert a proton into a neutron you increase the binding energy of the overall nucleus and you've got oodles of energy to spare.

So then, why does it happen? I've seen an explanation here in the question How can a proton be converted to a neutron via positron emission and yet gain mass? It describes that the final binding energy of the nucleus increases, thus making it possible by becoming more stable. But what actually causes the reaction to go in the first place? It's like saying this ball will go down the hill because it will lose energy - what gives it the nudge required? Is it something like energy from external gamma rays or something?

No. You aren't pushing a proton up a hill. The nucleus falls downhill. I think the best way to think of it is that is suffers a partial collapse. This converts a proton into a neutron that binds the other nucleons closer together and increases the binding energy. Conservation of charge and spin means the positron and the neutrino get ejected. This might sound unfamiliar, but note that Linus Pauling wrote about crystal packing in 1929 and came up with a close-packed spheron model for atomic nuclei in 1965. The Wikipedia nuclear binding energy article says physicists used to refer to a packing fraction calculation. In the Wikipedia atomic nucleus article you can read how packing protons and neutrons is like packing hard spheres. Now, you shouldn't think of protons and neutrons as billiard ball particles. It's the wave nature of matter, not the billiard ball nature of matter. However you should associate increased binding energy with a smaller nucleus. The nucleus really does suffer a partial collapse.