**Preliminary note**: Since writing the post, I developed a more comprehensive paper. You can find it on my academia.edu site (click here). It’s a bit longer – and also more technical – than the post below. Have fun ! 🙂

According to common wisdom, we need to introduce a new charge – and, therefore, a new *force *– to explain why protons will stick together. But we have neutrons too, right? Can’t they serve as *glue*? Now *that*’s an idea. About 99.999866 per cent of helium on this planet consists of two protons and two neutrons: we write this isotope as ^{4}He. The only other *stable* isotope is ^{3}He, which consists of two protons and *one* neutron. Let me *google* this… This is what Wikipedia writes: “Within the nucleus, protons and neutrons are bound together through the nuclear force. *Neutrons are required for the stability of nuclei*, with the exception of the single-proton hydrogen atom.”[1]

So now we need to examine this glue: what is it? What’s the difference between a neutron and a proton? A proton is stable. Neutrons are only stable inside of a nucleus: free neutrons *decay*. Their mean lifetime is almost 15 minutes, so that’s almost *eternity *in atomic physics. *Almost*, but not quite: free neutrons are *transient *oscillations. Why are neutrons stable in a nucleus but not in free space? We think it’s the Planck-Einstein relation: two protons, two neutrons and two electrons – a helium atom, in other words – are stable because all of the angular momenta in the *oscillation *add up to (some multiple of) Planck’s (reduced) quantum of action. The angular momentum of a neutron in free space does not, so it has to fall apart in a (stable) proton and a (stable) electron – and then a neutrino which carries the remainder of the energy. Let’s jot it down:Let’s think about energy first. The neutron’s energy is about 939,565,420 eV. The proton energy is about 938,272,088 eV. The difference is 1,293,332 eV. That’s almost 1.3 MeV.[2] The electron energy gives us close to 0.511 MeV of that difference – so that’s only 40% – but its *kinetic *energy can make up for a lot of the remainder! We then have the neutrino to provide the change—the *nickel-and-dime*, so to speak.[3]

Is this *decay *reversible? It is: a proton can *capture *an electron and, somehow, *become *a neutron. It usually happens with proton-rich nuclei absorbing an inner atomic electron, usually from the K or L electron shell, which is why the process is referred to as K- or L-electron capture:Once again, we have a neutrino providing the *nickel-and-dime* to ensure energy conservation. It is written as the anti-particle of the neutrino in the neutron decay equation. Neutrinos and anti-neutrinos are neutral, so what’s the difference? The specialists in the matter say they have no idea and that a neutrino and an anti-neutrino might well be one and the same thing.[4] Hence, for the time being, we’ll effectively assume they’re one and the same thing: we might write both as ν_{e}. No mystery here—not for me, at least. Or not here and not right now, I should say: the neutrino is just a vehicle to ensure conservation of energy and momentum (linear and/or angular).

It is tempting to think of the proton as some kind of *atomic system *itself, or a positive *ion *to which we may add an electron so as to get a neutron. You’ll say: that’s the hydrogen atom, right? No. The hydrogen atom is much larger than a neutron: the Bohr radius of a hydrogen atom is about 0.53 *pico*meter (1 pm = 1´10^{–}^{12} m). In contrast, the radius of a neutron is of the order of 0.8 *femto*meter (1 fm = 1´10^{–}^{15} m), so that’s about 660 times smaller. While a neutron is much smaller, its energy (and, therefore, its mass) is significantly higher: the energy difference between a hydrogen atom and a neutron is about 0.78 MeV. That’s about 1.5 times the energy of an electron. The table below shows these interesting numbers.A good model of what a proton and a neutron actually *are*, will also need to explain why electron-positron pair production only happens when the photon is fired into a nucleus. The mainstream interpretation of this phenomenon is that the surplus kinetic energy needs to be absorbed by some heavy particle – the nucleus itself. My guts instinct tells me something else must be going on. Electron-positron pair production does seem to involve the *creation *of an electric charge out of energy. It puzzled Dirac (and many other physicists, of course) greatly.Let us think about sizes once more. If we try the mass of a proton (or a neutron—almost the same) in the formula for the Compton radius, we get this:That’s about 1/4 of the actual radius as measured in scattering experiments. We have a good rationale for calculating the Compton radius of a proton (or a neutron). It is based on the *Zitterbewegung *model for elementary particles: a pointlike charge whizzing around at the speed of light. For the electron, the charge is electric. For the proton or the neutron, we think of some *strong *charge and we, therefore, get a very different energy and, hence, a very different Compton radius.[5] However, a factor of 1/4 is encouraging but not good enough. If anything, it may indicate that a good model of a proton (and a neutron) should, besides some strong force, also incorporate the classical electric charge. It is difficult to think about this, because we think the pointlike electric charge has a radius itself: the *Thomson *or *classical *electron radius, which is equal to:This is about 3.5 times *larger *than the proton or neutron radius. It is even larger than the measured radius of the deuteron nucleus, which consists of a proton and a neutron bound together. That radius is about 2.1 fm. As mentioned above, this ‘back-of-the-envelope’ calculation of a Compton radius is encouraging, but a good model for a proton (and for a neutron) will need to explain these 1/4 or 3.5 factors.

What happens might be something like this: we fire an enormous amount of electromagnetic energy into a nucleus (the equivalent mass of the photon has to match the mass of the electron and the positron that’s being produced) and, hence, we destabilize the stable nucleus. However, Nature is strong. The strong force is strong. Some intermediate energy state emerges but Nature throws out the spanner in the works. The end result is that all can be analyzed, once again, in terms of the Planck-Einstein relation: we have stable particles, once again. [Of course, the positron finds itself in the anti-Universe and will, therefore, quickly disappear in the reverse process: electron-positron annihilation.]

But so that’s just a story right now. We need to develop it into a proper theory.

**Post scriptum**: We’ve calculated a Compton radius for the proton. If – in analogy with the electron model – we would (also) have a current inside, then we should be able to calculate that current. Let us limit ourselves to the electric current – because we don’t have much of an idea about what a strong current would represent. The circular electric current creates a magnetic moment. We got the right value for an electron:What do we get if we do a similar calculation for a pointlike charge moving around at the speed of light but in a much smaller loop – a loop measured in *femto*meter rather than picometer? The calculation below shows we get a similar result in terms of *structure *but note the result is expressed in terms of the *nuclear *magneton (m_{N}) which uses the proton mass, as opposed to the Bohr magneton, which uses the electron (rest) mass.Unsurprisingly, the actually measured value is different, and the difference is much larger than Schwinger’s a/2p fraction. To be precise, μ_{p} » 2.8·μ_{N}, so the *measured *value of the proton’s magnetic moment is almost three times that of its theoretical value. It should be no surprise to us – because we use a radius that’s 1/4 of what might be the actual radius of the loop. In fact, the measured value of the proton’s magnetic moment suggests the *actual *radius of the loop should be 2.8 times the theoretical Compton radius:Again, these results are not exact, but they’re encouraging: they encourage us to try to describe the proton in terms of some kind of hybrid model – something that mixes the classical electric charge with some strong charge. No need for QFT or virtual particles. 🙂

[1] https://en.wikipedia.org/wiki/Neutron.

[2] CODATA data gives a standard error in the measurements that is equal to 0.46 eV. Hence, the measurements are pretty precise.

[3] When you talk money, you need big and small denominations: banknotes versus coins. However, the role of coins could be played by photons too. Gamma-ray photons – produced by radioactive decay – have energies in the MeV order of magnitude, so they should be able to play the role of whatever change we need in an energy equation, right? Yes. You’re right. So there must be more to it. We see neutrinos whenever there is radioactive decay. Hence, we should probably associate them with that, but how exactly is a bit of a mystery. Note that the decay equation conserves linear, angular (spin) momentum and (electric) charge. What about the color charge? We’re not worried about the color charge here. Should we be worried? I don’t think so, but if you’d be worried, note that this rather simple decay equation does respect color conservation – regardless of your definition of what quarks or gluons might actually *be*.

[4] See the various articles on neutrinos on Fermi National Accelerator Laboratory (FNAL), such as, for example, this one: https://neutrinos.fnal.gov/mysteries/majorana-or-dirac/. The common explanation is that neutrinos and anti-neutrinos have opposite spin but that’s nonsensical: we can very well imagine one and the same particle with two spin numbers.

[5] See: Jean Louis Van Belle, *Who Needs Yukawa’s Wave Equation?*, 24 June 2019 (http://vixra.org/abs/1906.0384).