What’s the spin of spin-1/2 particles?

The title of this post is nonsensical, isn’t it? The spin of spin-1/2 particles is 1/2. Yes, but let me ask you this: one half of what? What’s the unit here? And why would we take half of it?

If you are reading blogs like this, you will probably be able to answer that question: it should be a half unit of Planck’s (reduced) quantum of action. Spin comes in units of ħ/2, doesn’t it?

Maybe. Maybe not. The Planck-Einstein relation tells us that physical action comes in full units of h—not in half units. In fact, there is a rather vague quantum-mechanical rule that says angular momentum should be quantized in units of ħ, and it is based on the Planck-Einstein relation:Planck-EinsteinWe also get the E = ħω relation when substituting the frequency (f) in the E = hf relation for the rotational frequency of some spinning object with a tangential velocity = a·ω:Planck-Einstein 2The a is then some radius, and ω is just the angular frequency ω = 2πf. The point is this: why would angular momentum – that’s spin, isn’t it? – of elementary particles come in half-units of ħ? You won’t find an E = hf/2 or E = ħω/2 relation in a physics textbook. If you do, let me know.


It is just one of those quantum-mechanical rules one cannot really understand, isn’t it? And so we should just accept it and go along with the rest of the story, isn’t it?

Well… No. We don’t agree. It’s not just one of those rules. We should try to understand what this is all about.

The good news is this: we can. Moreover, it is actually not all that difficult. However, we should not get ahead of ourselves here. All will be clear in a minute − or so we hope when you are going through the trouble of reading us! − and so let us start nice and easily with some introductory remarks.

Very short introductory remarks

First, I should apologize to the early readers of the Matter page of this site. I had to re-write it a few times before it took its current shape. However, I am happy with it now (although I’ll probably keep tweaking it here and there). It explains all of the stable matter-particles – the electron and the proton, basically, but also the fairly stable muon-electron (a lifetime of 2.2 microseconds is decent at the sub-atomic scale): their mass (the equivalent mass of the energy in the oscillation), their magnetic moment, and their size (radius)—and it does so in terms of a ring current: if there’s a magnetic moment, there must be some current, right?

It’s actually very nice I can finally confirm the ring current model works for the proton as well. There was a factor 4 in the formula for its Compton radius (it is 4 times ħ/mc) which we couldn’t explain right away, but we now feel very comfortable in stating that factor is there because of the large angular momentum of a proton. So, yes, I thought it would be good to dedicate my very first post on this new site to that. [I’ve got an older site—but that was getting too large.] Indeed, I don’t have the intention to blog a lot here, but this is a result that is really worth highlighting, I feel.

OK. We’re done with the introduction now. Let us look at the equations because − when everything is said and done − they are all that matters, literally. Ideally, you should let them speak for themselves so you wouldn’t need to read us! However, it might save you some time if I talk you through them. 🙂

The ring current model and the radius of a proton

The explanation of the proton radius (about 0.83-0.84 fm—with or without the anomaly) is the same as the one we used to solve for the electron radius—except for an extra factor in the Planck-Einstein relation: that factor 4 we mentioned. So we use an E = 4ħω equation, which we need to justify, of course. The other two formulas (mass-energy equivalence and the tangential velocity formula) are the same. So we get this:

Proton radius derivation

How do we justify the E = 4ħω equation? We interpret it as the angular momentum of the proton being equal to four units of ħ, rather than one only. Is this legit? I just said there is no such thing as an E = ħω/2 relation, right? Right—even if, in the end, I actuall will show that the ħ/2 value has some meaning. Just note this: there is no issue with angular momentum coming in multiples of ħ. That’s how we explain electron orbitals, in fact. You should also remember the E = ω relation as the solution to the black-body radiation problem. So, yes, we think we’re fine. We can interpret this E = ω relation in terms of angular momentum: the (reduced) unit of physical action is also the (natural) unit for measuring angular momentum. And, yes, we think of the speed of light as the tangential velocity of the Zitterbewegung charge in the ring current, so the following derivation makes sense mathematically as well as physically:

proton L formula

We still need to explain it a bit, though. Intuitively, this rather large value for the angular momentum of a proton makes sense because it helps to explain the extraordinarily high mass density of a proton, which we associate with the equally enormous centripetal force inside of the proton, as evidenced by our calculations of the magnitude of the force inside an electron, muon and proton respectively, which we briefly present hereunder so as to remind the reader of the physicality of the situation:

ultimate correction

These are, effectively, enormous forces at the sub-atomic scale but – again – we need to explain equally monstrous mass densities, so we do think these values make sense, although we may want to think about incorporating the effects of mass densities on spacetime curvature for the muon and, surely, for the proton. These effects may or may not be relevant, indeed. [We don’t think they affect the results significantly because mass densities and radii are still far from the values you get out of Schwarzschild’s = 2Gm/c2 formula. Having said that, we do believe Dr. Burinskii has good reasons to doubt one can use the Schwarzschild radius formula for objects with angular momentum, a magnetic moment and other properties that do not apply when calculating, say, the Schwarzschild radius of the mass of a baseball.]

A more important question is this, however, what is the nature of these forces? Why this more massive version of an electron, and why can’t we think of the proton as an anti-muon? Why the different radius and force? Let us quickly summarize our thoughts on that.

The nature of the centripetal force

Our answer is and must remain speculative, of course, but we think the force inside of a muon and a proton are basically the same. Indeed, we think of the centripetal force in an electron as, essentially, being the same as the classical electromagnetic force that we know so well. Why? Because it only has an electric charge to grab onto: the Zitterbewegung charge has no other properties but its charge, which is electromagnetic. Even its mass must be entirely relativistic because, to reach lightspeed, it must have zero rest mass.

So what’s the force inside of a muon and a proton then? We need to do a lot more thinking and fleshing out of the idea but, essentially, we think it must be some strong(er) version of the electromagnetic force. Think of the strength of the electromagnetic force as being determined by some fundamental frequency—which has one or more higher modes.

Do we have any evidence of that? Not really, but we think the rather peculiar phenomenon of neutrinos shows we may be spot-on: whenever one suspects a strong or stronger force to be present, we have processes involving the emission and/or absorption of neutrinos. Photons carry electromagnetic energy from here to there, and so we think the neutrinos do the same for the strong(er) force: they ferry the ‘strong(er)’ energy around.

We will not dwell on this because we wrote this post to address the question in its title: how can we relate an angular momentum of 4 units of ħ to the spin-1/2 property of matter-particles? Let us talk about that now.

Electrons, muons and protons as spin-1/2 particles

The reader who has read some of our articles already will know we think analyses in terms of g-factors are very useful. In fact, we have always consistently maintained that we cannot directly observe the angular momentum of elementary particles. The spin-1/2 property of matter-particles is, therefore, essentially, a hypothesis which no one can prove. Indeed, the calculation of a g-factor always involves an assumption regarding the angular mass of the particle that we are looking at. We, therefore, think that the concept of a g-factor is not very scientific: what is the use of calculating some g-factor if one cannot directly observe the shape of an electron, or a muon, or a proton—or any sub-atomic particle, really?

Having said that, we do feel the q/m or m/q ratio is an interesting factor to look at when analyzing radii, masses and magnetic moments of elementary particles. It answers this basic question: how much charge do we get per mass unit? In fact, we find the reverse question even more interesting: how much mass (or equivalent energy) do we get per unit charge? Indeed, when someone would ask me to summarize the difference between an electron, muon and a proton, in one single number, I would probably refer to this m/q ratio. Why not the mass? Because the mass number itself doesn’t explain anything. In contrast, we can, effectively, directly relate the m/q or q/m ratio to the ring current model. Let us show you how by re-writing the usual g-factor equations for the electron and proton. We write it like this:

g-factors electron and proton

The informed reader will probably cry wolf here: this is not the conventional definition of the g-factor. He or she is right, but we let him cry. We ask the reader to think of the electron as a spin spin-1/2 because its real g-factor – as shown above – is 1/2. 🙂

What about the proton? That must be a spin-1/2 particle too, right? Relax. It is. We will show you why. We should first note a small inconvenience: the CODATA value for the magnetic moment of a proton is not a nice 2 or 4 number. The CODATA value differs from the theoretical value of the magnetic moment by a √2 factor. We think this factor is there because of precession but we have to be honest and inform the reader we have no hard facts here. We did contact Prof. Dr. Randolf Pohl − eminent expert on the proton radius puzzle ((he and his team established the 0.841 fm measurement of the proton radius back in 2010) as well as an equally eminent member of the CODATA Task Group for Fundamental Constants – but he has not come back to us yet (not on this, at least). Any case, we won’t dwell on this because we talk about this ad nauseam on our Matter page and so here we just want to present the basic logic and results of such talk. [In case the reader thinks this is questionable conduct, we have one answer only: we may be unreasonable, but we don’t think Nature is: a missing √2 factor or an extra 1/2 factor usually shows one’s on the right track − Feynman, Dirac and so many other Great Spirits said so! − so we’re not bothered by it!]

The point is this: we get a g-ratio that is four times than of an electron! So what’s the matter here—again, quite literally? The answer is simple: we showed the E = 4ħω implies a much larger angular momentum for the proton: four units of ħ, instead of just one. That makes sense in light of the mass and radius difference between a muon and a photon: a muon is, effectively, about 2.22 times the size of a proton (about 1.87 fm versus 0.84 fm), but the proton mass is about 8.88 times that of a muon. It must, therefore, pack a lot more angular momentum, right?

Right. So what’s the meaning then of this mysterious spin-1/2 property? It simply is this: the ratio of (1) the product of the mass and the magnetic moment and (2) the product of the charge and the angular momentum is equal to 1/2—for an electron, for a muon, and for a proton. You can easily verify this by calculating the simplest of simple gyromagnetic ratios, which is just the ratio of the magnetic moment and the angular momentum and expressing it in terms of these natural q/m unit. You get the following rather amazing set of equations:

spin - one over two property

Brilliant, isn’t it? But still mysterious. Why the 1/2 factor? We think it is there because of the concept of the effective mass of the Zitterbewegung charge. The basic idea is this: the energy equipartition theorem tells us that we can think of the energy (or equivalent mass) of an electron (and a muon and a proton) as being equally split over (1) the relativistic or effective mass of the oscillating charge, which we write as mγ = m/2, and (2) the energy of the oscillating force fields that keeps the charge in its orbit or oscillatory motion. Hence, that 1/2 factor is a bit of a technicality in our model which we may want to get rid of by bringing it to the left-hand side of our equations. That’s easy enough, indeed. You get this:tautologies

Wow! Amazing formulas, isn’t it! Much nicer having the unity factor on the right-hand side, isn’t it? Frankly, that was our first reaction too, but now we actually don’t think so. The equations above are just tautologies: you can add and remove factors as you please. They, therefore, don’t say anything. [Remember our objective is to try to present the equations in a way they can speak for themselves, so you don’t need our explanations in the future.]

Indeed, we prefer the expressions with the 1/2 factor so as to remind us of the physics of the situation. Let us re-explain the equations above by re-inserting the basic formulas we started off with. Remember this:

1. The ring current model tells us the magnetic moment is generated by a real-life circular current and so it’s equal to the product of that current and the surface area of the loop: μ = I·πa2.

2. The angular momentum is equally real, and so we have the L = ω·I = ω·m·a2 for that.

3. Inserting both formulas in the (m·μ)/(q·L) ratio and cancelling the common factors, we get what we think of the simplest and, therefore, most beautiful expression of the ring current model of elementary particles:

ring current super formula

We can now also think about what factor we should pair with the 1/2 factor if we would want to move it over. You might think − because that looks so damn easy − it must be the 1/2 factor in the = ω/2π identity. However, that doesn’t make any sense in terms of the physicality of the situation that we are trying to model here. So what’s left? Because some half- or double-charge factor does not make sense, it must be some half- or double mass, right?

You’re right. So it is the mass factor in the numerator or the denominator?

[…] Come on! Think! Let the equation talk to you. Please! That’s how we learn. That’s how we truly understand. I can’t do it for you: if you’re reading what I write here, then it means you want to understand things like they really are—or… Well… I should be modest: like things might be—according to some kind of realistic interpretation of the core equations in physics, at least. I like to call it physics we can believe in. 🙂


By now, I guess you’d agree we should probably put the 1/2 factor in the formula for the angular momentum, right? So we should write L as L = m·a2/2. Why? We’ll let you think about that, but it is, indeed, because of the effective mass concept. It looks a bit bewildering but you should embrace it—if only because it pops up quite naturally in quantum physics. Richard Feynman, for example, gets the equation out of a quantum-mechanical analysis of how an electron could move along a line of atoms in a crystal lattice.

It is, therefore, not sacrilegious to write L as L = m·a2/2. It is, effectively, the orbital angular momentum of the pointlike Zitterbewegung charge inside of an electron, or a muon—and a proton! The elementary charge, really—positive or negative! In fact, it answers the question in the title of this post: why are spin-1/2 particles spin-1/2 particles?

We said we don’t like tautologies in physics, but the following set of equations – which are tautologies as well – may be enlightening:tautology

Now that is beautiful, I’d think. But we admit beauty is in the eye of the beholder. So, yes, we welcome any thoughts, questions, or remarks you might have! Especially when you’d see interpretations or nuances that we might have missed.

The nature of the strong force

How can we be so sure that the centripetal force inside of the muon and the force inside the proton are the same? The honest answer is this: we cannot be sure of that. We think it is, but that’s because muon disintegration also involves neutrinos—and neutrinos are always present when physicists talk about the strong(er) force.

Let us try to think this through: we basically think of the strong(er) force as some strong(er) version of the electromagnetic force that keeps an electron together, right? Now, applying the basic equations – for ring currents and angular momentum and, yes, some more basic ones (like the mass-energy equivalence and Planck-Einstein relation) – we get this:

Super formula

What does this tell us? It tells us we can analyze the difference between the force inside of a muon and a proton − a difference which we express as a ratio here − in terms of two factors:

  1. The difference between their radii: the radius of a muon is about 2.22 times larger than that of a proton.
  2. That ratio then gets multiplied by four, which is the factor which distinguishes their presumed angular momentum: ħ and 4ħ.

Hence, yes, we think it’s the same force: if the angular momentum wouldn’t be different, the proton force − sorry to use shorthand − would just be about 2.22 times stronger than the muon force.

Of course, it still begs the question: so why are their radii different? That amounts to asking this question: why can’t we analyze a proton as an anti-muon, or a muon as an anti-proton? I don’t have a theoretical answer to that question, but I do have a very practial one: if a proton was an anti-muon, or a muon was an anti-proton, then we wouldn’t have protons or muons: they’d be annihilating each other.

So, yes, we do think that the muon and the proton are products of the same force: they just have different angular momentum. That’s why a proton is not an anti-muon. It is as simple and as complicated as that, I am afraid.

So now we need a model to analyze two different modes of a two-dimensional oscillation of the elementary charge— in space and in time. We don’t have that model yet. :-/ But we’ve got a good ‘tool box’ here, don’t we? 🙂

Note: I’ll be honest and share all of my doubts here—if only because I know that’s the best way to arrive at some kind of ‘basic version of truth’. If the ratio between the muon’s spin − measured in full or half units of ħ − and the proton’s spin is 1/4, then where is the 1/2 or 1/3 ratio? Skipping niceties about 1, 2 or 1/2 factors here, the question is this: if we have ħ/2 or ħ and/or 4ħ or 4ħ/2 = 2ħ particles here, then where are the 2ħ or 3ħ particles?

My answer is this: the ħ/2 ratio makes sense because of the concept of the effective mass: the energy of our electron – or our muon—the reader will have noticed I am suffering from Stockholm syndrome because of my talk about ‘our’ muon or ‘our’ electron, right? 🙂 – is what it is, however: the energy of an elementary particle. So we can carve it up mathematically but not physically. There are no 2ħ or 3ħ particles, because they’d be ħ or 3ħ/2 particles which don’t fit into our two-dimensional oscillator model. Think of it like this: two modes of a two-dimensional oscillation—the relevant factor must be four rather than two, right?


OK. Let’s wrap this up. What’s the conclusion? The question was this: what’s the spin of spin-1/2 particles? Our answer is this:

  1. What is the meaning of spin in that question?
  2. See the question above: we cannot answer your question if you can’t define what you mean by spin.


That may sound a bit funny, but it is actually the problem of mainstream quantum physicists: they explain things they do not understand—the true nature of matter-particles or light—by introducing weird concepts they do not understand either! And spin is just one of them! 🙂 Don’t get us started on the other stuff! 🙂

Jean Louis Van Belle, 22 March 2020


4 thoughts on “What’s the spin of spin-1/2 particles?

  1. \documentstyle{article}
    In the original Dirac Equation we have
    \[ \left( \beta mc^2 + c \sum_{n=1}^{3} \alpha_n p_n \right) \psi(x,t)
    = i\hbar \frac{\partial \psi(x,t)}{\partial t}. \]
    We take it as granted that $c\neq0$ and divide both sides of the Dirac Equation by it. This gives us
    \[ \left( \beta mc + \sum_{n=1}^{3} \alpha_n p_n \right) \psi(x,t)
    = \left(\frac{i\hbar}{c}\right) \frac{\partial \psi(x,t)}{\partial t}. \]
    We can now reorganize the right hand side with regard to its scalars as
    \[ \left( \beta mc + \sum_{n=1}^{3} \alpha_n p_n \right) \psi(x,t)
    = \frac{i}{2\pi} \left(\frac{h}{c}\right) \frac{\partial \psi(x,t)}{\partial t} \]
    We can identify $i/2\pi$ based on the fundamentals of complex numbers easily. \\
    We can see that $h/c$, in its naked form, is related to the time-derivative of $\psi(x,t)$ by the chain rule. Perhaps the true natural constant is actually $h/c$ and that it is invariant under a change of frame-of-reference. \\

    We conclude that spin \mbox{\em does not} come in units of 1/2, however circles do come in two halves, an up half and a down half, if you will. If we think of natural spin as $\pm1$ we then have a \mbox{\em reduced} spin (in the spirit of the reduced Planck constant, $\hbar$) that comes in units of $\frac{1}{2}\pi$.

    SEE: https://stegman.us/mathstuf/unitspin.jpg


  2. Thank you for your polite consideration in posting my comment. I regret that I am fluent only in (Lamport-original) LaTeX.
    My efforts to insert LaTeX source code in HTML5 have been completely frustrated to date by arcane distractions.

    Oh, and the conclusion might be better stated as “units of 1 / pi”.


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