The geometry of the de Broglie wavelength

I thought I would no longer post stuff here but I see this site still gets a lot more traffic than the new one, so I will make an exception and cross-post an announcement of a new video on my YouTube channel. Indeed, yesterday I was to talk for about 30 minutes to some students who are looking at classical electron models as part of an attempt to try to model what might be happening to an electron when moving through a magnetic field. Of course, I only had time to discuss the ring current model, and even then it inadvertently turned into a two-hour presentation. Fortunately, they were polite and no one dropped out—although it was an online Google Meet. In fact, they reacted quite enthusiastically, and so we all enjoyed it a lot. So much that I adjusted the presentation a bit the next morning (which added even more time to it unfortunately) and published it online. So this is the link to it, and I hope you enjoy it. If so, please like it—and share it! 🙂

Oh! Forgot to mention: in case you wonder why this video is different than others, see my Tweet on Sean Carroll’s latest series of videos hereunder. That should explain it. 🙂

Post scriptum: I got the usual question from one of the students, of course: if an electron is a ring current, then why doesn’t it radiate its energy away? The easy answer is: an electron is an electron and so it doesn’t—for the same reason that an electron in an atomic orbital or a Cooper pair in a superconducting loop of current does not radiate energy away. The more difficult answer is a bit mysterious: it has got to do with flux quantization and, most importantly, with the Planck-Einstein relation. I will not be too explicit here (it is just a footnote) but the following elements should be noted:

1. The Planck-Einstein law embodies a (stable) wavicle: a wavicle respects the Planck-Einstein relation (E = h·f) as well as Einstein’s mass-energy equivalence relation (E = mc2). A wavicle will, therefore, carry energy but it will also pack one or more units of Planck’s quantum of action. Both the energy as well as this finite amount of physical action (Wirkung in German) will be conserved—cycle after cycle.

2. Hence, equilibrium states should be thought of as electromagnetic oscillations without friction. Indeed, it is the frictional element that explains the radiation of, say, an electron going up and down in an antenna and radiating some electromagnetic signal out. To add to this rather intuitive explanation, I should also remind you that it is the accelerations and decelerations of the electric charge in an antenna that generate the radio wave—not the motion as such. So one should, perhaps, think of a charge going round and round as moving like in a straight line—along some geodesic in its own space. That’s the metaphor, at least.

3. Technically, one needs to think in terms of quantized fluxes and Poynting vectors and energy transfers from kinetic to potential (and back) and from ‘electric’ to ‘magnetic’ (and back). In short, the electron really is an electromagnetic perpetuum mobile ! I know that sounds mystical (too) but then I never said I would take all of the mystery away from quantum physics ! 🙂 If there would be no mystery left, I would not be interested in physics. :wink: On the quantization of flux for superconducting loops: see, for example, There is other stuff you may want to dig into too, like my alternative Principles of Physics, of course ! 🙂  

The proton radius puzzle solved

I thought I’d stop blogging, but I can’t help it: I think you’d find this topic interesting – and my comments are actually too short for a paper or article, so I thought it would be good to throw it out here.

If you follow the weird world of quantum mechanics with some interest, you will have heard the latest news: the ‘puzzle’ of the charge radius of the proton has been solved. To be precise, a more precise electron-proton scattering experiment by the PRad (proton radius) team using the Continuous Electron Beam Accelerator Facility (CEBAF) at Jefferson Lab has now measured the root mean square (rms) charge radius of the proton as[1]:

rp = 0.831 ± 0.007stat ± 0.012syst fm

If a proton would, somehow, have a pointlike elementary (electric) charge in it, and if it it is in some kind of circular motion (as we presume in Zitterbewegung models of elementary particles), then we can establish a simple relation between the magnetic moment (μ) and the radius (a) of the circular current.

Indeed, the magnetic moment is the current (I) times the surface area of the loop (πa2), and the current is just the product of the elementary charge (qe) and the frequency (f), which we can calculate as f = c/2πa, i.e. the velocity of the charge[2] divided by the circumference of the loop. We write:F1Using the Compton radius of an electron (ae = ħ/mec), this yields the correct magnetic moment for the electron[3]:F2What radius do we get when applying the a = μ/0.24…´10–10 relation to the (experimentally measured) magnetic moment of a proton? I invite the reader to verify the next calculation using CODATA values:F3When I first calculated this, I thought: that’s not good enough. I only have the order of magnitude right. However, when multiplying this with √2, we get a value which fits into the 0.831 ± 0.007 interval. To be precise, we get this:

Of course, you will wonder: how can we justify the √2 factor? I am not sure. It is a charge radius. Hence, the electrons will bounce off because of the electromagnetic fields. The magnetic field of the current ring will be some envelope to the current ring itself. We would, therefore, expect the measured charge radius to be larger than the radius of the current ring (a). There are also the intricacies related to the definition of a root mean square (rms) radius.

I feel this cannot be a coincidence: the difference between our ‘theoretical’ value (0.83065 fm) and the last precision measurement (0.831 fm) is only 0.00035 fm, which is only 5% of the statistical standard deviation (0.007 fm). Proton radius solved?

Maybe. Maybe not. The concluding comments of Physics Today were this: “The PRad radius result, about 0.83 fm, agrees with the smaller value from muonic and now electronic hydrogen spectroscopy measurements. With that, it seems the puzzle is resolved, and the discrepancy was likely due to measurement errors. Unfortunately, the conclusion requires no new physics.” (my italics)

I wonder what kind of new physics they are talking about.

Jean Louis Van Belle, 24 January 2020

PS: I did make a paper out of this (see my or publications), and I shared it with the PRad team at JLAB. Prof. Dr. Ashot Gasparian was kind enough to acknowledge my email and thought “the approach and numbers are interesting.” Let us see what comes out of it. I need to get back to my day job. 🙂

[1] See: See also: and

[2] Zitterbewegung models assume an electron consists of a pointlike charge whizzing around some center. The rest mass of the pointlike charge is zero, which is why its velocity is equal to the speed of light. However, because of its motion, it acquires an effective mass – pretty much like a photon, which has mass because of its motion. One can show the effective mass of the pointlike charge – which is a relativistic mass concept – is half the rest mass of the electron: mγ = me/2.

[3] The calculations do away with the niceties of the + or – sign conventions as they focus on the values only. We also invite the reader to add the SI units so as to make sure all equations are consistent from a dimensional point of view. For the values themselves, see the CODATA values on the NIST website (

Wikipedia censorship

I started to edit and add to the rather useless Wikipedia article on the Zitterbewegung. No mention of Hestenes or more recent electron models (e.g. Burinskii’s Kerr-Newman geometries. No mention that the model only works for electrons or leptons in general – not for non-leptonic fermions. It’s plain useless. But all the edits/changes/additions were erased by some self-appointed ‘censor’. I protested but then I got reported to the administrator ! What can I say? Don’t trust Wikipedia. Don’t trust any ‘authority’. We live in weird times. The mindset of most professional physicists seems to be governed by ego and the Bohr-Heisenberg Diktatur.

For the record, these are the changes and edits I tried to make. You can compare and judge for yourself. Needless to say, I told them I wouldn’t bother to even try to contribute any more. I published my own article on the Vixrapedia e-encyclopedia. Also, as Vixrapedia did not have an entry on realist interpretations of quantum mechanics, I created one: have a look and let me know what you think. 🙂

Zitterbewegung (“trembling” or “shaking” motion in German) – usually abbreviated as zbw – is a hypothetical rapid oscillatory motion of elementary particles that obey relativistic wave equations. The existence of such motion was first proposed by Erwin Schrödinger in 1930 as a result of his analysis of the wave packet solutions of the Dirac equation for relativistic electrons in free space, in which an interference between positive and negative energy states produces what appears to be a fluctuation (up to the speed of light) of the position of an electron around the median, with an angular frequency of ω = 2mc2/ħ, or approximately 1.5527×1021 radians per second. Paul Dirac was initially intrigued by it, as evidenced by his rather prominent mention of it in his 1933 Nobel Prize Lecture (it may be usefully mentioned he shared this Nobel Prize with Schrödinger):

“The variables give rise to some rather unexpected phenomena concerning the motion of the electron. These have been fully worked out by Schrödinger. It is found that an electron which seems to us to be moving slowly, must actually have a very high frequency oscillatory motion of small amplitude superposed on the regular motion which appears to us. As a result of this oscillatory motion, the velocity of the electron at any time equals the velocity of light. This is a prediction which cannot be directly verified by experiment, since the frequency of the oscillatory motion is so high and its amplitude is so small. But one must believe in this consequence of the theory, since other consequences of the theory which are inseparably bound up with this one, such as the law of scattering of light by an electron, are confirmed by experiment.”[1]

In light of Dirac’s later comments on modern quantum theory, it is rather puzzling that he did not pursue the idea of trying to understand charged particles in terms of the motion of a pointlike charge, which is what the Zitterbewegung hypothesis seems to offer. Dirac’s views on non-leptonic fermions – which were then (1950s and 1960s) being analyzed in an effort to explain the ‘particle zoo‘ in terms of decay reactions conserving newly invented or ad hoc quantum numbers such as strangeness[2] – may be summed up by quoting the last paragraph in the last edition of his Principles of Quantum Mechanics:

“Now there are other kinds of interactions, which are revealed in high-energy physics. […] These interactions are not at present sufficiently well understood to be incorporated into a system of equations of motion.”[3]

Indeed, in light of this stated preference for kinematic models, it is somewhat baffling that Dirac did not follow up on this or any of the other implications of the Zitterbewegung hypothesis, especially because it should be noted that a reexamination of Dirac theory shows that interference between positive and negative energy states is not a necessary ingredient of Zitterbewegung theories.[4] The Zitterbewegung hypothesis also seems to offer interesting shortcuts to key results of mainstream quantum theory. For example, one can show that, for the hydrogen atom, the Zitterbewegung produces the Darwin term which plays the role in the fine structure as a small correction of the energy level of the s-orbitals.[5] This is why authors such as Hestenes refers to it as a possible alternative interpretation of mainstream quantum mechanics, which may be an exaggerated claim in light of the fact that the zbw hypothesis results from the study of electron behavior only.

Zitterbewegung models have mushroomed[6] and it is, therefore, increasingly difficult to distinguish between them. The key to understanding and distinguishing the various Zitterbewegung models may well be Wheeler‘s ‘mass without mass’ idea, which implies a distinction between the idea of (i) a pointlike electric charge (i.e. the idea of a charge only, with zero rest mass) and (ii) the idea of an electron as an elementary particle whose equivalent mass is the energy of the zbw oscillation of the pointlike charge.[7] The ‘mass without mass’ concept requires a force to act on a charge – and a charge only – to explain why a force changes the state of motion of an object – its momentum p = mγ·v(with γ referring to the Lorentz factor) – in accordance with the (relativistically correct) F = dp/dt force law.


As mentioned above, the zbw hypothesis goes back to Schrödinger’s and Dirac’s efforts to try to explain what an electron actually is. Unfortunately, both interpreted the electron as a pointlike particle with no ‘internal structure’.David Hestenes is to be credited with reviving the Zitterbewegung hypothesis in the early 1990s. While acknowledging its origin as a (trivial) solution to Dirac’s equation for electrons, Hestenes argues the Zitterbewegung should be related to the intrinsic properties of the electron (charge, spin and magnetic moment). He argues that the Zitterbewegung hypothesis amounts to a physical interpretation of the elementary wavefunction or – more boldly – to a possible physical interpretation of all of quantum mechanics: “Spin and phase [of the wavefunction] are inseparably related — spin is not simply an add-on, but an essential feature of quantum mechanics. […] A standard observable in Dirac theory is the Dirac current, which doubles as a probability current and a charge current. However, this does not account for the magnetic moment of the electron, which many investigators conjecture is due to a circulation of charge. But what is the nature of this circulation? […] Spin and phase must be kinematical features of electron motion. The charge circulation that generates the magnetic moment can then be identified with the Zitterbewegung of Schrödinger “[8] Hestenes’ interpretation amounts to an kinematic model of an electron which can be described in terms of John Wheeler‘s mass without mass concept.[9] The rest mass of the electron is analyzed as the equivalent energy of an orbital motion of a pointlike charge. This pointlike charge has no rest mass and must, therefore, move at the speed of light (which confirms Dirac’s en Schrödinger’s remarks on the nature of the Zitterbewegung). Hestenes summarizes his interpretation as follows: “The electron is nature’s most fundamental superconducting current loop. Electron spin designates the orientation of the loop in space. The electron loop is a superconducting LC circuit. The mass of the electron is the energy in the electron’s electromagnetic field. Half of it is magnetic potential energy and half is kinetic.”[10]

Hestenes‘ articles and papers on the Zitterbewegung discuss the electron only. The interpretation of an electron as a superconducting ring of current (or as a (two-dimensional) oscillator) also works for the muon electron: its theoretical Compton radius rC = ħ/mμc ≈ 1.87 fm falls within the CODATA confidence interval for the experimentally determined charge radius.[11] Hence, the theory seems to offer a remarkably and intuitive model of leptons. However, the model cannot be generalized to non-leptonic fermions (spin-1/2 particles). Its application to protons or neutrons, for example, is problematic: when inserting the energy of a proton or a neutron into the formula for the Compton radius (the rC = ħ/mc formula follows from the kinematic model), we get a radius of the order of rC = ħ/mpc ≈ 0.21 fm, which is about 1/4 of the measured value (0.84184(67) fm to 0.897(18) fm). A radius of the order of 0.2 fm is also inconsistent with the presumed radius of the pointlike charge itself. Indeed, while the pointlike charge is supposed to be pointlike, pointlike needs to be interpreted as ‘having no internal structure’: it does not imply the pointlike charge has no (small) radius itself. The classical electron radius is a likely candidate for the radius of the pointlike charge because it emerges from low-energy (Thomson) scattering experiments (elastic scattering of photons, as opposed to inelastic Compton scattering). The assumption of a pointlike charge with radius re = α·ħ/mpc) may also offer a geometric explanation of the anomalous magnetic moment.[12]

In any case, the remarks above show that a Zitterbewegung model for non-leptonic fermions is likely to be somewhat problematic: a proton, for example, cannot be explained in terms of the Zitterbewegung of a positron (or a heavier variant of it, such as the muon- or tau-positron).[13] This is why it is generally assumed the large energy (and the small size) of nucleons is to be explained by another force – a strong force which acts on a strong charge instead of an electric charge. One should note that both color and/or flavor in the standard quarkgluon model of the strong force may be thought of as zero-mass charges in ‘mass without mass’ kinematic models and, hence, the acknowledgment of this problem does not generally lead zbw theorists to abandon the quest for an alternative realist interpretation of quantum mechanics.

While Hestenes‘ zbw interpretation (and the geometric calculus approach he developed) is elegant and attractive, he did not seem to have managed to convincingly explain an obvious question of critics of the model: what keeps the pointlike charge in the zbw electron in its circular orbit? To put it simply: one may think of the electron as a superconducting ring but there is no material ring to hold and guide the charge. Of course, one may argue that the electromotive force explains the motion but this raises the fine-tuning problem: the slightest deviation of the pointlike charge from its circular orbit would yield disequilibrium and, therefore, non-stability. [One should note the fine-tuning problem is also present in mainstream quantum mechanics. See, for example, the discussion in Feynman’s Lectures on Physics.] The lack of a convincing answer to these and other questions (e.g. on the distribution of (magnetic) energy within the superconducting ring) led several theorists working on electron models (e.g. Alexander Burinskii[14][15]) to move on and explore alternative geometric approaches, including Kerr-Newman geometries. Burinskii summarizes his model as follows: “The electron is a superconducting disk defined by an over-rotating black hole geometry. The charge emerges from the Möbius structure of the Kerr geometry.”[16] His advanced modelling of the electron also allows for a conceptual bridge with mainstream quantum mechanics, grand unification theories and string theory: “[…] Compatibility between gravity and quantum theory can be achieved without modifications of Einstein-Maxwell equations, by coupling to a supersymmetric Higgs model of symmetry breaking and forming a nonperturbative super-bag solution, which generates a gravity-free Compton zone necessary for consistent work of quantum theory. Super-bag is naturally upgraded to Wess-Zumino supersymmetric QED model, forming a bridge to perturbative formalism of conventional QED.”[17]

The various geometric approaches (Hestenes’ geometric calculus, Burinskii’s Kerr-Newman model, oscillator models) yield the same results (the intrinsic properties of the electron are derived from what may be referred to as kinematic equations or classical (but relativistically correct) equations) – except for a factor 2 or 1/2 or the inclusion (or not) of variable tuning parameters (Burinskii’s model, for example, allows for a variable geometry) – but the equivalence of the various models that may or may not explain the hypothetical Zitterbewegung still needs to be established.

The continued interest in zbw models may be explained because Zitterbewegung models – in particular Hestenes’ model and the oscillator model – are intuitive and, therefore, attractive. They are intuitive because they combine the Planck-Einstein relation (E = hf) and Einstein’s mass-energy equivalence (E = mc2): each cycle of the Zitterbewegung electron effectively packs (i) the unit of physical action (h) and (ii) the electron’s energy. This allows one to understand Planck’s quantum of action as the product of the electron’s energy and the cycle time: h = E·T = h·f·T = h·f/f = h. In addition, the idea of a centripetal force keeping some zero-mass pointlike charge in a circular orbit also offers a geometric explanation of Einstein’s mass-energy equivalence relation: this equation, therefore, is no longer a rather inexplicable consequence of special relativity theory.

The section below offers a general overview of the original discovery of Schrödinger and Dirac. It is followed by further analysis which may or may not help the reader to judge whether the Zitterbewegung hypothesis might, effectively, amount to what David Hestenes claims it actually is: an alternative interpretation of quantum mechanics.

Theory for a free fermion

[See the article: the author of this section does not seem to know – or does not mention, at least – that the Zitterbewegung hypothesis only applies to leptons (no strong charge).]

Experimental evidence

The Zitterbewegung may remain theoretical because, as Dirac notes, the frequency may be too high to be observable: it is the same frequency as that of a 0.511 MeV gamma-ray. However, some experiments may offer indirect evidence. Dirac’s reference to electron scattering experiments is also quite relevant because such experiments yield two radii: a radius for elastic scattering (the classical electron radius) and a radius for inelastic scattering (the Compton radius). Zittebewegung theorists think Compton scattering involves electron-photon interference: the energy of the high-energy photon (X- or gamma-ray photons) is briefly absorbed before the electron comes back to its equilibrium situation by emitting another (lower-energy) photon (the difference in the energy of the incoming and the outgoing photon gives the electron some extra momentum). Because of this presumed interference effect, Compton scattering is referred to as inelastic. In contrast, low-energy photons scatter elastically: they seem to bounce off some hard core inside of the electron (no interference).

Some experiments also claim they amount to a simulation of the Zitterbewegung of a free relativistic particle. First, with a trapped ion, by putting it in an environment such that the non-relativistic Schrödinger equation for the ion has the same mathematical form as the Dirac equation (although the physical situation is different).[18][19] Then, in 2013, it was simulated in a setup with Bose–Einstein condensates.[20]

The effective mass of the electric charge

The 2m factor in the formula for the zbw frequency and the interpretation of the Zitterbewegung in terms of a centripetal force acting on a pointlike charge with zero rest mass leads one to re-explore the concept of the effective mass of an electron. Indeed, if we write the effective mass of the pointlike charge as mγ = γm0 then we can derive its value from the angular momentum of the electron (L = ħ/2) using the general angular momentum formula L = r × p and equating r to the Compton radius:

This explains the 1/2 factor in the frequency formula for the Zitterbewegung. Substituting m for mγ in the ω = 2mc2/ħ yields an equivalence with the Planck-Einstein relation ω = mγc2/ħ. The electron can then be described as an oscillator (in two dimensions) whose natural frequency is given by the Planck-Einstein relation.[21]

Philosophy and Physics


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An intuitive interpretation of Einstein’s mass-energy equivalence relation

My dear readers – I haven’t published much lately, because I try to summarize my ideas now in short articles that might be suitable for publication in a journal. I think the latest one (on Einstein’s mass-energy relation) should be of interest. Let me just insert the summary here:

The radial velocity formula and the Planck-Einstein relation give us the Zitterbewegung (zbw) frequency (E = ħω = E/ħ) and zbw radius (a = c/ω = cħ/mc2 = ħ/mc) of  the electron. We interpret this by noting that the c = aω identity gives us the E = mc2 = ma2ω2 equation, which suggests we should combine the total energy (kinetic and potential) of two harmonic oscillators to explain the electron mass. We do so by interpreting the elementary wavefunction as a two-dimensional (harmonic) electromagnetic oscillation in real space which drives the pointlike charge along the zbw current ring. This implies a dual view of the reality of the real and imaginary part of the wavefunction:

  1. The x = acos(ωt) and y = a·sin(ωt) equations describe the motion of the pointlike charge.
  2. As an electromagnetic oscillation, we write it as E0 = E0cos(ωt+π/2) + i·E0·sin(ωt+π/2).

The magnitudes of the oscillation a and E0 are expressed in distance (m) and force per unit charge (N/C) respectively and are related because the energy of both oscillations is one and the same. The model – which implies the energy of the oscillation and, therefore, the effective mass of the electron is spread over the zbw disk – offers an equally intuitive explanation for the angular momentum, magnetic moment and the g-factor of charged spin-1/2 particles. Most importantly, the model also offers us an intuitive interpretation of Einstein’s enigmatic mass-energy equivalence relation. Going from the stationary to the moving reference frame, we argue that the plane of the zbw oscillation should be parallel to the direction of motion so as to be consistent with the results of the Stern-Gerlach experiment.

So… Well… Have fun with it ! I think I am going to sign off. 🙂 Yours – JL

The geometry of the wavefunction, electron spin and the form factor

Pre-script (dated 26 June 2020): Our ideas have evolved into a full-blown realistic (or classical) interpretation of all things quantum-mechanical. In addition, I note the dark force has amused himself by removing some material. So no use to read this. Read my recent papers instead. 🙂

Original post:

Our previous posts showed how a simple geometric interpretation of the elementary wavefunction yielded the (Compton scattering) radius of an elementary particle—for an electron, at least: for the proton, we only got the order of magnitude right—but then a proton is not an elementary particle. We got lots of other interesting equations as well… But… Well… When everything is said and done, it’s that equivalence between the E = m·a2·ω2 and E = m·c2 relations that we… Well… We need to be more specific about it.

Indeed, I’ve been ambiguous here and there—oscillating between various interpretations, so to speak. 🙂 In my own mind, I refer to my unanswered questions, or my ambiguous answers to them, as the form factor problem. So… Well… That explains the title of my post. But so… Well… I do want to be somewhat more conclusive in this post. So let’s go and see where we end up. 🙂

To help focus our mind, let us recall the metaphor of the V-2 perpetuum mobile, as illustrated below. With permanently closed valves, the air inside the cylinder compresses and decompresses as the pistons move up and down. It provides, therefore, a restoring force. As such, it will store potential energy, just like a spring, and the motion of the pistons will also reflect that of a mass on a spring: it is described by a sinusoidal function, with the zero point at the center of each cylinder. We can, therefore, think of the moving pistons as harmonic oscillators, just like mechanical springs. Of course, instead of two cylinders with pistons, one may also think of connecting two springs with a crankshaft, but then that’s not fancy enough for me. 🙂

V-2 engine

At first sight, the analogy between our flywheel model of an electron and the V-twin engine seems to be complete: the 90 degree angle of our V-2 engine makes it possible to perfectly balance the pistons and we may, therefore, think of the flywheel as a (symmetric) rotating mass, whose angular momentum is given by the product of the angular frequency and the moment of inertia: L = ω·I. Of course, the moment of inertia (aka the angular mass) will depend on the form (or shape) of our flywheel:

  1. I = m·a2 for a rotating point mass m or, what amounts to the same, for a circular hoop of mass m and radius a.
  2. For a rotating (uniformly solid) disk, we must add a 1/2 factor: I = m·a2/2.

How can we relate those formulas to the E = m·a2·ω2 formula? The kinetic energy that is being stored in a flywheel is equal Ekinetic = I·ω2/2, so that is only half of the E = m·a2·ω2 product if we substitute I for I = m·a2. [For a disk, we get a factor 1/4, so that’s even worse!] However, our flywheel model of an electron incorporates potential energy too. In fact, the E = m·a2·ω2 formula just adds the (kinetic and potential) energy of two oscillators: we do not really consider the energy in the flywheel itself because… Well… The essence of our flywheel model of an electron is not the flywheel: the flywheel just transfers energy from one oscillator to the other, but so… Well… We don’t include it in our energy calculations. The essence of our model is that two-dimensional oscillation which drives the electron, and which is reflected in Einstein’s E = m·c2 formula. That two-dimensional oscillation—the a2·ω2 = c2 equation, really—tells us that the resonant (or natural) frequency of the fabric of spacetime is given by the speed of light—but measured in units of a. [If you don’t quite get this, re-write the a2·ω2 = c2 equation as ω = c/a: the radius of our electron appears as a natural distance unit here.]

Now, we were extremely happy with this interpretation not only because of the key results mentioned above, but also because it has lots of other nice consequences. Think of our probabilities as being proportional to energy densities, for example—and all of the other stuff I describe in my published paper on this. But there is even more on the horizon: a follower of this blog (a reader with an actual PhD in physics, for a change) sent me an article analyzing elementary particles as tiny black holes because… Well… If our electron is effectively spinning around, then its tangential velocity is equal to ω = c. Now, recent research suggest black holes are also spinning at (nearly) the speed of light. Interesting, right? However, in order to understand what she’s trying to tell me, I’ll first need to get a better grasp of general relativity, so I can relate what I’ve been writing here and in previous posts to the Schwarzschild radius and other stuff.

Let me get back to the lesson here. In the reference frame of our particle, the wavefunction really looks like the animation below: it has two components, and the amplitude of the two-dimensional oscillation is equal to a, which we calculated as = ħ·/(m·c) = 3.8616×10−13 m, so that’s the (reduced) Compton scattering radius of an electron.


In my original article on this, I used a more complicated argument involving the angular momentum formula, but I now prefer a more straightforward calculation:

c = a·ω = a·E/ħ = a·m·c2/ħ  ⇔ = ħ/(m·c)

The question is: what is that rotating arrow? I’ve been vague and not so vague on this. The thing is: I can’t prove anything in this regard. But my hypothesis is that it is, in effect, a rotating field vector, so it’s just like the electric field vector of a (circularly polarized) electromagnetic wave (illustrated below).

There are a number of crucial differences though:

  1. The (physical) dimension of the field vector of the matter-wave is different: I associate the real and imaginary component of the wavefunction with a force per unit mass (as opposed to the force per unit charge dimension of the electric field vector). Of course, the newton/kg dimension reduces to the dimension of acceleration (m/s2), so that’s the dimension of a gravitational field.
  2. I do believe this gravitational disturbance, so to speak, does cause an electron to move about some center, and I believe it does so at the speed of light. In contrast, electromagnetic waves do not involve any mass: they’re just an oscillating field. Nothing more. Nothing less. In contrast, as Feynman puts it: “When you do find the electron some place, the entire charge is there.” (Feynman’s Lectures, III-21-4)
  3. The third difference is one that I thought of only recently: the plane of the oscillation cannot be perpendicular to the direction of motion of our electron, because then we can’t explain the direction of its magnetic moment, which is either up or down when traveling through a Stern-Gerlach apparatus.

I mentioned that in my previous post but, for your convenience, I’ll repeat what I wrote there. The basic idea here is illustrated below (credit for this illustration goes to another blogger on physics). As for the Stern-Gerlach experiment itself, let me refer you to a YouTube video from the Quantum Made Simple site.

Figure 1 BohrThe point is: the direction of the angular momentum (and the magnetic moment) of an electron—or, to be precise, its component as measured in the direction of the (inhomogeneous) magnetic field through which our electron is traveling—cannot be parallel to the direction of motion. On the contrary, it is perpendicular to the direction of motion. In other words, if we imagine our electron as spinning around some center, then the disk it circumscribes will comprise the direction of motion.

However, we need to add an interesting detail here. As you know, we don’t really have a precise direction of angular momentum in quantum physics. [If you don’t know this… Well… Just look at one of my many posts on spin and angular momentum in quantum physics.] Now, we’ve explored a number of hypotheses but, when everything is said and done, a rather classical explanation turns out to be the best: an object with an angular momentum J and a magnetic moment μ (I used bold-face because these are vector quantities) that is parallel to some magnetic field B, will not line up, as you’d expect a tiny magnet to do in a magnetic field—or not completely, at least: it will precess. I explained that in another post on quantum-mechanical spin, which I advise you to re-read if you want to appreciate the point that I am trying to make here. That post integrates some interesting formulas, and so one of the things on my ‘to do’ list is to prove that these formulas are, effectively, compatible with the electron model we’ve presented in this and previous posts.

Indeed, when one advances a hypothesis like this, it’s not enough to just sort of show that the general geometry of the situation makes sense: we also need to show the numbers come out alright. So… Well… Whatever we think our electron—or its wavefunction—might be, it needs to be compatible with stuff like the observed precession frequency of an electron in a magnetic field.

Our model also needs to be compatible with the transformation formulas for amplitudes. I’ve been talking about this for quite a while now, and so it’s about time I get going on that.

Last but not least, those articles that relate matter-particles to (quantum) gravity—such as the one I mentioned above—are intriguing too and, hence, whatever hypotheses I advance here, I’d better check them against those more advanced theories too, right? 🙂 Unfortunately, that’s going to take me a few more years of studying… But… Well… I still have many years ahead—I hope. 🙂

Post scriptum: It’s funny how one’s brain keeps working when sleeping. When I woke up this morning, I thought: “But it is that flywheel that matters, right? That’s the energy storage mechanism and also explains how photons possibly interact with electrons. The oscillators drive the flywheel but, without the flywheel, nothing is happening. It is really the transfer of energy—through the flywheel—which explains why our flywheel goes round and round.”

It may or may not be useful to remind ourselves of the math in this regard. The motion of our first oscillator is given by the cos(ω·t) = cosθ function (θ = ω·t), and its kinetic energy will be equal to sin2θ. Hence, the (instantaneous) change in kinetic energy at any point in time (as a function of the angle θ) is equal to: d(sin2θ)/dθ = 2∙sinθ∙d(sinθ)/dθ = 2∙sinθ∙cosθ. Now, the motion of the second oscillator (just look at that second piston going up and down in the V-2 engine) is given by the sinθ function, which is equal to cos(θ − π /2). Hence, its kinetic energy is equal to sin2(θ − π /2), and how it changes (as a function of θ again) is equal to 2∙sin(θ − π /2)∙cos(θ − π /2) = = −2∙cosθ∙sinθ = −2∙sinθ∙cosθ. So here we have our energy transfer: the flywheel organizes the borrowing and returning of energy, so to speak. That’s the crux of the matter.

So… Well… What if the relevant energy formula is E = m·a2·ω2/2 instead of E = m·a2·ω2? What are the implications? Well… We get a √2 factor in our formula for the radius a, as shown below.

square 2

Now that is not so nice. For the tangential velocity, we get a·ω = √2·c. This is also not so nice. How can we save our model? I am not sure, but here I am thinking of the mentioned precession—the wobbling of our flywheel in a magnetic field. Remember we may think of Jz—the angular momentum or, to be precise, its component in the z-direction (the direction in which we measure it—as the projection of the real angular momentum J. Let me insert Feynman’s illustration here again (Feynman’s Lectures, II-34-3), so you get what I am talking about.


Now, all depends on the angle (θ) between Jz and J, of course. We did a rather obscure post on these angles, but the formulas there come in handy now. Just click the link and review it if and when you’d want to understand the following formulas for the magnitude of the presumed actual momentum:magnitude formulaIn this particular case (spin-1/2 particles), j is equal to 1/2 (in units of ħ, of course). Hence, is equal to √0.75 ≈ 0.866. Elementary geometry then tells us cos(θ) = (1/2)/√(3/4) =  = 1/√3. Hence, θ ≈ 54.73561°. That’s a big angle—larger than the 45° angle we had secretly expected because… Well… The 45° angle has that √2 factor in it: cos(45°) = sin(45°) = 1/√2.

Hmm… As you can see, there is no easy fix here. Those damn 1/2 factors! They pop up everywhere, don’t they? 🙂 We’ll solve the puzzle. One day… But not today, I am afraid. I’ll call it the form factor problem… Because… Well… It sounds better than the 1/2 or √2 problem, right? 🙂

Note: If you’re into quantum math, you’ll note ħ/(m·c) is the reduced Compton scattering radius. The standard Compton scattering radius is equal to  = (2π·ħ)/(m·c) =  h/(m·c) = h/(m·c). It doesn’t solve the √2 problem. Sorry. The form factor problem. 🙂

To be honest, I finished my published paper on all of this with a suggestion that, perhaps, we should think of two circular oscillations, as opposed to linear ones. Think of a tiny ball, whose center of mass stays where it is, as depicted below. Any rotation – around any axis – will be some combination of a rotation around the two other axes. Hence, we may want to think of our two-dimensional oscillation as an oscillation of a polar and azimuthal angle. It’s just a thought but… Well… I am sure it’s going to keep me busy for a while. 🙂polar_coordsThey are oscillations, still, so I am not thinking of two flywheels that keep going around in the same direction. No. More like a wobbling object on a spring. Something like the movement of a bobblehead on a spring perhaps. 🙂bobblehead

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Re-visiting the Complementarity Principle: the field versus the flywheel model of the matter-wave

Note: I have published a paper that is very coherent and fully explains what’s going on. There is nothing magical about it these things. Check it out: The Meaning of the Fine-Structure Constant. No ambiguity. No hocus-pocus.

Jean Louis Van Belle, 23 December 2018

Original post:

This post is a continuation of the previous one: it is just going to elaborate the questions I raised in the post scriptum of that post. Let’s first review the basics once more.

The geometry of the elementary wavefunction

In the reference frame of the particle itself, the geometry of the wavefunction simplifies to what is illustrated below: an oscillation in two dimensions which, viewed together, form a plane that would be perpendicular to the direction of motion—but then our particle doesn’t move in its own reference frame, obviously. Hence, we could be looking at our particle from any direction and we should, presumably, see a similar two-dimensional oscillation. That is interesting because… Well… If we rotate this circle around its center (in whatever direction we’d choose), we get a sphere, right? It’s only when it starts moving, that it loses its symmetry. Now, that is very intriguing, but let’s think about that later.


Let’s assume we’re looking at it from some specific direction. Then we presumably have some charge (the green dot) moving about some center, and its movement can be analyzed as the sum of two oscillations (the sine and cosine) which represent the real and imaginary component of the wavefunction respectively—as we observe it, so to speak. [Of course, you’ve been told you can’t observe wavefunctions so… Well… You should probably stop reading this. :-)] We write:

ψ = = a·ei∙θ = a·ei∙E·t/ħ = a·cos(−E∙t/ħ) + i·a·sin(−E∙t/ħ) = a·cos(E∙t/ħ) i·a·sin(E∙t/ħ) 

So that’s the wavefunction in the reference frame of the particle itself. When we think of it as moving in some direction (so relativity kicks in), we need to add the p·x term to the argument (θ = E·t − px). It is easy to show this term doesn’t change the argument (θ), because we also get a different value for the energy in the new reference frame: E= γ·E0 and so… Well… I’ll refer you to my post on this, in which I show the argument of the wavefunction is invariant under a Lorentz transformation: the way Ev and pv and, importantly, the coordinates and t relativistically transform ensures the invariance.

In fact, I’ve always wanted to read de Broglie‘s original thesis because I strongly suspect he saw that immediately. If you click this link, you’ll find an author who suggests the same. Having said that, I should immediately add this does not imply there is no need for a relativistic wave equation: the wavefunction is a solution for the wave equation and, yes, I am the first to note the Schrödinger equation has some obvious issues, which I briefly touch upon in one of my other posts—and which is why Schrödinger himself and other contemporaries came up with a relativistic wave equation (Oskar Klein and Walter Gordon got the credit but others (including Louis de Broglie) also suggested a relativistic wave equation when Schrödinger published his). In my humble opinion, the key issue is not that Schrödinger’s equation is non-relativistic. It’s that 1/2 factor again but… Well… I won’t dwell on that here. We need to move on. So let’s leave the wave equation for what it is and go back to our wavefunction.

You’ll note the argument (or phase) of our wavefunction moves clockwise—or counterclockwise, depending on whether you’re standing in front of behind the clock. Of course, Nature doesn’t care about where we stand or—to put it differently—whether we measure time clockwise, counterclockwise, in the positive, the negative or whatever direction. Hence, I’ve argued we can have both left- as well as right-handed wavefunctions, as illustrated below (for p ≠ 0). Our hypothesis is that these two physical possibilities correspond to the angular momentum of our electron being either positive or negative: Jz = +ħ/2 or, else, Jz = −ħ/2. [If you’ve read a thing or two about neutrinos, then… Well… They’re kinda special in this regard: they have no charge and neutrinos and antineutrinos are actually defined by their helicity. But… Well… Let’s stick to trying to describing electrons for a while.]

The line of reasoning that we followed allowed us to calculate the amplitude a. We got a result that tentatively confirms we’re on the right track with our interpretation: we found that = ħ/me·c, so that’s the Compton scattering radius of our electron. All good ! But we were still a bit stuck—or ambiguous, I should say—on what the components of our wavefunction actually are. Are we really imagining the tip of that rotating arrow is a pointlike electric charge spinning around the center? [Pointlike or… Well… Perhaps we should think of the Thomson radius of the electron here, i.e. the so-called classical electron radius, which is equal to the Compton radius times the fine-structure constant: rThomson = α·rCompton ≈ 3.86×10−13/137.]

So that would be the flywheel model.

In contrast, we may also think the whole arrow is some rotating field vector—something like the electric field vector, with the same or some other physical dimension, like newton per charge unit, or newton per mass unit? So that’s the field model. Now, these interpretations may or may not be compatible—or complementary, I should say. I sure hope they are but… Well… What can we reasonably say about it?

Let us first note that the flywheel interpretation has a very obvious advantage, because it allows us to explain the interaction between a photon and an electron, as I demonstrated in my previous post: the electromagnetic energy of the photon will drive the circulatory motion of our electron… So… Well… That’s a nice physical explanation for the transfer of energy. However, when we think about interference or diffraction, we’re stuck: flywheels don’t interfere or diffract. Only waves do. So… Well… What to say?

I am not sure, but here I want to think some more by pushing the flywheel metaphor to its logical limits. Let me remind you of what triggered it all: it was the mathematical equivalence of the energy equation for an oscillator (E = m·a2·ω2) and Einstein’s formula (E = m·c2), which tells us energy and mass are equivalent but… Well… They’re not the same. So what are they then? What is energy, and what is mass—in the context of these matter-waves that we’re looking at. To be precise, the E = m·a2·ω2 formula gives us the energy of two oscillators, so we need a two-spring model which—because I love motorbikes—I referred to as my V-twin engine model, but it’s not an engine, really: it’s two frictionless pistons (or springs) whose direction of motion is perpendicular to each other, so they are in a 90° degree angle and, therefore, their motion is, effectively, independent. In other words: they will not interfere with each other. It’s probably worth showing the illustration just one more time. And… Well… Yes. I’ll also briefly review the math one more time.

V-2 engine

If the magnitude of the oscillation is equal to a, then the motion of these piston (or the mass on a spring) will be described by x = a·cos(ω·t + Δ). Needless to say, Δ is just a phase factor which defines our t = 0 point, and ω is the natural angular frequency of our oscillator. Because of the 90° angle between the two cylinders, Δ would be 0 for one oscillator, and –π/2 for the other. Hence, the motion of one piston is given by x = a·cos(ω·t), while the motion of the other is given by x = a·cos(ω·t–π/2) = a·sin(ω·t). The kinetic and potential energy of one oscillator – think of one piston or one spring only – can then be calculated as:

  1. K.E. = T = m·v2/2 = (1/2)·m·ω2·a2·sin2(ω·t + Δ)
  2. P.E. = U = k·x2/2 = (1/2)·k·a2·cos2(ω·t + Δ)

The coefficient k in the potential energy formula characterizes the restoring force: F = −k·x. From the dynamics involved, it is obvious that k must be equal to m·ω2. Hence, the total energy—for one piston, or one spring—is equal to:

E = T + U = (1/2)· m·ω2·a2·[sin2(ω·t + Δ) + cos2(ω·t + Δ)] = m·a2·ω2/2

Hence, adding the energy of the two oscillators, we have a perpetuum mobile storing an energy that is equal to twice this amount: E = m·a2·ω2. It is a great metaphor. Somehow, in this beautiful interplay between linear and circular motion, energy is borrowed from one place and then returns to the other, cycle after cycle. However, we still have to prove this engine is, effectively, a perpetuum mobile: we need to prove the energy that is being borrowed or returned by one piston is the energy that is being returned or borrowed by the other. That is easy to do, but I won’t bother you with that proof here: you can double-check it in the referenced post or – more formally – in an article I posted on

It is all beautiful, and the key question is obvious: if we want to relate the E = m·a2·ω2 and E = m·c2 formulas, we need to explain why we could, potentially, write as a·ω = a·√(k/m). We’ve done that already—to some extent at least. The tangential velocity of a pointlike particle spinning around some axis is given by v = r·ω. Now, the radius is given by = ħ/(m·c), and ω = E/ħ = m·c2/ħ, so is equal to to v = [ħ/(m·c)]·[m·c2/ħ] = c. Another beautiful result, but what does it mean? We need to think about the meaning of the ω = √(k/m) formula here. In the mentioned article, we boldly wrote that the speed of light is to be interpreted as the resonant frequency of spacetime, but so… Well… What do we really mean by that? Think of the following.

Einstein’s E = mc2 equation implies the ratio between the energy and the mass of any particle is always the same:


This effectively reminds us of the ω2 = C1/L or ω2 = k/m formula for harmonic oscillators. The key difference is that the ω2= C1/L and ω2 = k/m formulas introduce two (or more) degrees of freedom. In contrast, c2= E/m for any particle, always. However, that is exactly the point: we can modulate the resistance, inductance and capacitance of electric circuits, and the stiffness of springs and the masses we put on them, but we live in one physical space only: our spacetime. Hence, the speed of light (c) emerges here as the defining property of spacetime: the resonant frequency, so to speak. We have no further degrees of freedom here.

Let’s think about k. [I am not trying to avoid the ω2= 1/LC formula here. It’s basically the same concept: the ω2= 1/LC formula gives us the natural or resonant frequency for a electric circuit consisting of a resistor, an inductor, and a capacitor. Writing the formula as ω2= C−1/L introduces the concept of elastance, which is the equivalent of the mechanical stiffness (k) of a spring, so… Well… You get it, right? The ω2= C1/L and ω2 = k/m sort of describe the same thing: harmonic oscillation. It’s just… Well… Unlike the ω2= C1/L, the ω2 = k/m is directly compatible with our V-twin engine metaphor, because it also involves physical distances, as I’ll show you here.] The in the ω2 = k/m is, effectively, the stiffness of the spring. It is defined by Hooke’s Law, which states that the force that is needed to extend or compress a spring by some distance  is linearly proportional to that distance, so we write: F = k·x.

Now that is interesting, isn’t it? We’re talking exactly the same thing here: spacetime is, presumably, isotropic, so it should oscillate the same in any direction—I am talking those sine and cosine oscillations now, but in physical space—so there is nothing imaginary here: all is real or… Well… As real as we can imagine it to be. 🙂

We can elaborate the point as follows. The F = k·x equation implies k is a force per unit distance: k = F/x. Hence, its physical dimension is newton per meter (N/m). Now, the in this equation may be equated to the maximum extension of our spring, or the amplitude of the oscillation, so that’s the radius in the metaphor we’re analyzing here. Now look at how we can re-write the a·ω = a·√(k/m) equation:


In case you wonder about the E = F·a substitution: just remember that energy is force times distance. [Just do a dimensional analysis: you’ll see it works out.] So we have a spectacular result here, for several reasons. The first, and perhaps most obvious reason, is that we can actually derive Einstein’s E = m·c2 formula from our flywheel model. Now, that is truly glorious, I think. However, even more importantly, this equation suggests we do not necessarily need to think of some actual mass oscillating up and down and sideways at the same time: the energy in the oscillation can be thought of a force acting over some distance, regardless of whether or not it is actually acting on a particle. Now, that energy will have an equivalent mass which is—or should be, I’d say… Well… The mass of our electron or, generalizing, the mass of the particle we’re looking at.

Huh? Yes. In case you wonder what I am trying to get at, I am trying to convey the idea that the two interpretations—the field versus the flywheel model—are actually fully equivalent, or compatible, if you prefer that term. In Asia, they would say: they are the “same-same but different” 🙂 but, using the language that’s used when discussing the Copenhagen interpretation of quantum physics, we should actually say the two models are complementary.

You may shrug your shoulders but… Well… It is a very deep philosophical point, really. 🙂 As far as I am concerned, I’ve never seen a better illustration of the (in)famous Complementarity Principle in quantum physics because… Well… It goes much beyond complementarity. This is about equivalence. 🙂 So it’s just like Einstein’s equation. 🙂

Post scriptum: If you read my posts carefully, you’ll remember I struggle with those 1/2 factors here and there. Textbooks don’t care about them. For example, when deriving the size of an atom, or the Rydberg energy, even Feynman casually writes that “we need not trust our answer [to questions like this] within factors like 2, π, etcetera.” Frankly, that’s disappointing. Factors like 2, 1/2, π or 2π are pretty fundamental numbers, and so they need an explanation. So… Well… I do loose sleep over them. :-/ Let me advance some possible explanation here.

As for Feynman’s model, and the derivation of electron orbitals in general, I think it’s got to do with the fact that electrons do want to pair up when thermal motion does not come into play: think of the Cooper pairs we use to explain superconductivity (so that’s the BCS theory). The 1/2 factor in Schrödinger’s equation also has weird consequences (when you plug in the elementary wavefunction and do the derivatives, you get a weird energy concept: E = m·v2, to be precise). This problem may also be solved when assuming we’re actually calculating orbitals for a pair of electrons, rather than orbitals for just one electron only. [We’d get twice the mass (and, presumably, the charge, so… Well… It might work—but I haven’t done it yet. It’s on my agenda—as so many other things, but I’ll get there… One day. :-)]

So… Well… Let’s get back to the lesson here. In this particular context (i.e. in the context of trying to find some reasonable physical interpretation of the wavefunction), you may or may not remember (if not, check my post on it) ‘ll remember I had to use the I = m·r2/2 formula for the angular momentum, as opposed to the I = m·r2 formula. I = m·r2/2 (with the 1/2 factor) gives us the angular momentum of a disk with radius r, as opposed to a point mass going around some circle with radius r. I noted that “the addition of this 1/2 factor may seem arbitrary”—and it totally is, of course—but so it gave us the result we wanted: the exact (Compton scattering) radius of our electron.

Now, the arbitrary 1/2 factor may or may be explained as follows. In the field model of our electron, the force is linearly proportional to the extension or compression. Hence, to calculate the energy involved in stretching it from x = 0 to a, we need to calculate it as the following integral:

half factor

So… Well… That will give you some food for thought, I’d guess. 🙂 If it racks your brain too much—or if you’re too exhausted by this point (which is OK, because it racks my brain too!)—just note we’ve also shown that the energy is proportional to the square of the amplitude here, so that’s a nice result as well… 🙂

Talking food for thought, let me make one final point here. The c2 = a2·k/m relation implies a value for k which is equal to k = m·c2/a = E/a. What does this tell us? In one of our previous posts, we wrote that the radius of our electron appeared as a natural distance unit. We wrote that because of another reason: the remark was triggered by the fact that we can write the cratio as c/ω = a·ω/ω = a. This implies the tangential and angular velocity in our flywheel model of an electron would be the same if we’d measure distance in units of a. Now, the E = a·k = a·F/(just re-writing…) implies that the force is proportional to the energy— F = (x/a)·E — and the proportionality coefficient is… Well… x/a. So that’s the distance measured in units of a. So… Well… Isn’t that great? The radius of our atom appearing as a natural distance unit does fit in nicely with our geometric interpretation of the wavefunction, doesn’t it? I mean… Do I need to say more?

I hope not because… Well… I can’t explain any better for the time being. I hope I sort of managed to convey the message. Just to make sure, in case you wonder what I was trying to do here, it’s the following: I told you appears as a resonant frequency of spacetime and, in this post, I tried to explain what that really means. I’d appreciate if you could let me know if you got it. If not, I’ll try again. 🙂 When everything is said and done, one only truly understands stuff when one is able to explain it to someone else, right? 🙂 Please do think of more innovative or creative ways if you can! 🙂

OK. That’s it but… Well… I should, perhaps, talk about one other thing here. It’s what I mentioned in the beginning of this post: this analysis assumes we’re looking at our particle from some specific direction. It could be any direction but… Well… It’s some direction. We have no depth in our line of sight, so to speak. That’s really interesting, and I should do some more thinking about it. Because the direction could be any direction, our analysis is valid for any direction. Hence, if our interpretation would happen to be some true—and that’s a big if, of course—then our particle has to be spherical, right? Why? Well… Because we see this circular thing from any direction, so it has to be a sphere, right?

Well… Yes. But then… Well… While that logic seems to be incontournable, as they say in French, I am somewhat reluctant to accept it at face value. Why? I am not sure. Something inside of me says I should look at the symmetries involved… I mean the transformation formulas for wavefunction when doing rotations and stuff. So… Well… I’ll be busy with that for a while, I guess. 😦

Post scriptum 2: You may wonder whether this line of reasoning would also work for a proton. Well… Let’s try it. Because its mass is so much larger than that of an electron (about 1835 times), the = ħ/(m·c) formula gives a much smaller radius: 1835 times smaller, to be precise, so that’s around 2.1×10−16 m, which is about 1/4 of the so-called charge radius of a proton, as measured by scattering experiments. So… Well… We’re not that far off, but… Well… We clearly need some more theory here. Having said that, a proton is not an elementary particle, so its mass incorporates other factors than what we’re considering here (two-dimensional oscillations).

The flywheel model of an electron

One of my readers sent me the following question on the geometric (or even physical) interpretation of the wavefunction that I’ve been offering in recent posts:

Does this mean that the wave function is merely describing excitations in a matter field; or is this unsupported?

My reply was very short: “Yes. In fact, we can think of a matter-particle as a tiny flywheel that stores energy.”

However, I realize this answer answers the question only partially. Moreover, I now feel I’ve been quite ambiguous in my description. When looking at the geometry of the elementary wavefunction (see the animation below, which shows us a left- and right-handed wave respectively), two obvious but somewhat conflicting interpretations readily come to mind:

(1) One is that the components of the elementary wavefunction represent an oscillation (in two dimensions) of a field. We may call it a matter field (yes, think of the scalar Higgs field here), but we could also think of it as an oscillation of the spacetime fabric itself: a tiny gravitational wave, in effect. All we need to do here is to associate the sine and cosine component with a physical dimension. The analogy here is the electromagnetic field vector, whose dimension is force per unit charge (newton/coulomb). So we may associate the sine and cosine components of the wavefunction with, say, the force per unit mass dimension (newton/kg) which, using Newton’s Law (F = m·a) reduces to the dimension of acceleration (m/s2), which is the dimension of gravitational fields. I’ll refer to this interpretation as the field interpretation of the matter wave (or wavefunction).

(2) The other interpretation is what I refer to as the flywheel interpretation of the electron. If you google this, you won’t find anything. However, you will probably stumble upon the so-called Zitterbewegung interpretation of quantum mechanics, which is a more elaborate theory based on the same basic intuition. The Zitterbewegung (a term which was coined by Erwin Schrödinger himself, and which you’ll see abbreviated as zbw) is, effectively, a local circulatory motion of the electron, which is presumed to be the basis of the electron’s spin and magnetic moment. All that I am doing, is… Well… I think I do push the envelope of this interpretation quite a bit. 🙂

The first interpretation implies our rotating arrow is, effectively, some field vector. In contrast, the second interpretation implies it’s only the tip of the rotating arrow that, literally, matters: we should look at it as a pointlike charge moving around a central axis, which is the direction of propagation. Let’s look at both.

The flywheel interpretation

The flywheel interpretation has an advantage over the field interpretation, because it also gives us a wonderfully simple physical interpretation of the interaction between electrons and photons—or, further speculating, between matter-particles (fermions) and force-carrier particles (bosons) in general. In fact, Feynman shows how this might work—but in a rather theoretical Lecture on symmetries and conservation principles, and he doesn’t elaborate much, so let me do that for him. The argument goes as follows.

A light beam—an electromagnetic wave—consists of a large number of photons. These photons are thought of as being circularly polarized: look at those animations above again. The Planck-Einstein equation tells us the energy of each photon is equal to E = ħ·ω = h·f. [I should, perhaps, quickly note that the frequency is, obviously, the frequency of the electromagnetic wave. It, therefore, is not to be associated with a matter wave: the de Broglie wavelength and the wavelength of light are very different concepts, even if the Planck-Einstein equation looks the same for both.]

Now, if our beam consists of photons, the total energy of our beam will be equal to W = N·E = N·ħ·ω. It is crucially important to note that this energy is to be interpreted as the energy that is carried by the beam in a certain time: we should think of the beam as being finite, somehow, in time and in space. Otherwise, our reasoning doesn’t make sense.

The photons carry angular momentum. Just look at those animations (above) once more. It doesn’t matter much whether or not we think of light as particles or as a wave: you can see there is angular momentum there. Photons are spin-1 particles, so the angular momentum will be equal to ± ħ. Hence, then the total angular momentum Jz (the direction of propagation is supposed to be the z-axis here) will be equal to JzN·ħ. [This, of course, assumes all photons are polarized in the same way, which may or may not be the case. You should just go along with the argument right now.] Combining the W = N·ħ·ω and JzN·ħ equations, we get:

JzN·ħ = W/ω

For a photon, we do accept the field interpretation, as illustrated below. As mentioned above, the z-axis here is the direction of propagation (so that’s the line of sight when looking at the diagram). So we have an electric field vector, which we write as ε (epsilon) so as to not cause any confusion with the Ε we used for the energy. [You may wonder if we shouldn’t also consider the magnetic field vector, but then we know the magnetic field vector is, basically, a relativistic effect which vanishes in the reference frame of the charge itself.] The phase of the electric field vector is φ = ω·t.

RH photon

Now, a charge (so that’s our electron now) will experience a force which is equal to F = q·ε. We use bold letters here because F and ε are vectors. We now need to look at our electron which, in our interpretation of the elementary wavefunction, we think of as rotating about some axis. So that’s what’s represented below. [Both illustrations are Feynman’s, not mine. As for the animations above, I borrowed them from Wikipedia.]


Now, in previous posts, we calculated the radius based on a similar argument as the one Feynman used to get that JzN·ħ = W/ω equation. I’ll refer you those posts and just mention the result here: r is the Compton scattering radius for an electron, which is equal to:

radius formula

An equally spectacular implication of our flywheel model of the electron was the following: we found that the angular velocity v was equal to vr·ω = [ħ·/(m·c)]·(E/ħ) = c. Hence, in our flywheel model of an electron, it is effectively spinning around at the speed of light. Note that the angular frequency (ω) in the vr·ω equation is not the angular frequency of our photon: it’s the frequency of our electron. So we use the same Planck-Einstein equation (ω = E/ħ) but the energy E is the (rest) energy of our electron, so that’s about 0.511 MeV (so that’s an order of magnitude which is 100,000 to 300,000 times that of photons in the visible spectrum). Hence, the angular frequencies of our electron and our photon are very different. Feynman casually reflects this difference by noting the phases of our electron and our photon will differ by a phase factor, which he writes as φ0.

Just to be clear here, at this point, our analysis here diverges from Feynman’s. Feynman had no intention whatsoever to talk about Schrödinger’s Zitterbewegung hypothesis when he wrote what he wrote back in the 1960s. In fact, Feynman is very reluctant to venture into physical interpretations of the wavefunction in all his Lectures on quantum mechanics—which is surprising. Because he comes so tantalizing close at many occasions—as he does here: he describes the motion of the electron here as that of a harmonic oscillator which can be driven by an external electric field. Now that is a physical interpretation, and it is totally consistent with the one I’ve advanced in my recent posts. Indeed, Feynman also describes it as an oscillation in two dimensions—perpendicular to each other and to the direction of motion, as we do— in both the flywheel as well as the field interpretation of the wavefunction!

This point is important enough to quote Feynman himself in this regard:

“We have often described the motion of the electron in the atom as a harmonic oscillator which can be driven into oscillation by an external electric field. We’ll suppose that the atom is isotropic, so that it can oscillate equally well in the x– or y- directions. Then in the circularly polarized light, the x displacement and the displacement are the same, but one is 90° behind the other. The net result is that the electron moves in a circle.”

Right on! But so what happens really? As our light beam—the photons, really—are being absorbed by our electron (or our atom), it absorbs angular momentum. In other words, there is a torque about the central axis. Let me remind you of the formulas for the angular momentum and for torque respectively: L = r×p and τr×F. Needless to say, we have two vector cross-products here. Hence, if we use the τr×F formula, we need to find the tangential component of the force (Ft), whose magnitude will be equal to Ft = q·εtNow, energy is force over some distance so… Well… You may need to think about it for a while but, if you’ve understood all of the above, you should also be able to understand the following formula:

dW/dt = q·εt·v

[If you have trouble, remember is equal to ds/dt = Δs/Δt for Δt → 0, and re-write the equation above as dW = q·εt·v·dt = q·εt·ds = Ft·ds. Capito?]

Now, you may or may not remember that the time rate of change of angular momentum must be equal to the torque that is being applied. Now, the torque is equal to τ = Ft·r = q·εt·r, so we get:

dJz/dt = q·εt·v

The ratio of dW/dt and dJz/dt gives us the following interesting equation:

Feynman formula

Now, Feynman tries to relate this to the JzN·ħ = W/ω formula but… Well… We should remind ourselves that the angular frequency of these photons is not the angular frequency of our electron. So… Well… What can we say about this equation? Feynman suggests to integrate dJz and dW over some time interval, which makes sense: as mentioned, we interpreted W as the energy that is carried by the beam in a certain time. So if we integrate dW over this time interval, we get W. Likewise, if we integrate dJz over the same time interval, we should get the total angular momentum that our electron is absorbing from the light beam. Now, because dJz = dW/ω, we do concur with Feynman’s conclusion: the total angular momentum which is being absorbed by the electron is proportional to the total energy of the beam, and the constant of proportionality is equal to 1/ω.

It’s just… Well… The ω here is the angular frequency of the electron. It’s not the angular frequency of the beam. Not in our flywheel model of the electron which, admittedly, is not the model which Feynman used in his analysis. Feynman’s analysis is simpler: he assumes an electron at rest, so to speak, and then the beam drives it so it goes around in a circle with a velocity that is, effectively, given by the angular frequency of the beam itself. So… Well… Fine. Makes sense. As said, I just pushed the analysis a bit further along here. Both analyses raise an interesting question: how and where is the absorbed energy being stored? What is the mechanism here?

In Feynman’s analysis, the answer is quite simple: the electron did not have any motion before but does spin around after the beam hit it. So it has more energy now: it wasn’t a tiny flywheel before, but it is now!

In contrast, in my interpretation of the matter wave, the electron was spinning around already, so where does the extra energy go now? As its energy increases, ω = E/ħ must increase, right? Right. At the same time, the velocity v = r·ω must still be equal to vr·ω = [ħ·/(m·c)]·(E/ħ) = c, right? Right. So… If ω increases, but r·ω must equal the speed of light, then must actually decrease somewhat, right?

Right. It’s a weird but inevitable conclusion, it seems. I’ll let you think about it. 🙂

To conclude this post—which, I hope, the reader who triggered it will find interesting—I would like to quote Feynman on an issue on which most textbooks remain silent: the two-state nature of photons. I will just quote him without trying to comment or alter what he writes, because what he writes is clear enough, I think:

“Now let’s ask the following question: If light is linearly polarized in the x-direction, what is its angular momentum? Light polarized in the x-direction can be represented as the superposition of RHC and LHC polarized light. […] The interference of these two amplitudes produces the linear polarization, but it has equal probabilities to appear with plus or minus one unit of angular momentum. [Macroscopic measurements made on a beam of linearly polarized light will show that it carries zero angular momentum, because in a large number of photons there are nearly equal numbers of RHC and LHC photons contributing opposite amounts of angular momentum—the average angular momentum is zero.]

Now, we have said that any spin-one particle can have three values of Jz, namely +101 (the three states we saw in the Stern-Gerlach experiment). But light is screwy; it has only two states. It does not have the zero case. This strange lack is related to the fact that light cannot stand still. For a particle of spin which is standing still, there must be the 2j+1 possible states with values of Jz going in steps of from j to +j. But it turns out that for something of spin j with zero mass only the states with the components +j and j along the direction of motion exist. For example, light does not have three states, but only two—although a photon is still an object of spin one.”

In his typical style and frankness—for which he is revered by some (like me) but disliked by others—he admits this is very puzzling, and not obvious at all! Let me quote him once more:

“How is this consistent with our earlier proofs—based on what happens under rotations in space—that for spin-one particles three states are necessary? For a particle at rest, rotations can be made about any axis without changing the momentum state. Particles with zero rest mass (like photons and neutrinos) cannot be at rest; only rotations about the axis along the direction of motion do not change the momentum state. Arguments about rotations around one axis only are insufficient to prove that three states are required. We have tried to find at least a proof that the component of angular momentum along the direction of motion must for a zero mass particle be an integral multiple of ħ/2—and not something like ħ/3. Even using all sorts of properties of the Lorentz transformation and what not, we failed. Maybe it’s not true. We’ll have to talk about it with Prof. Wigner, who knows all about such things.”

The reference to Eugene Wigner is historically interesting. Feynman probably knew him very well—if only because they had both worked together on the Manhattan Project—and it’s true Wigner was not only a great physicist but a mathematical genius as well. However, Feynman probably quotes him here for the 1963 Nobel Prize he got for… Well… Wigner’s “contributions to the theory of the atomic nucleus and elementary particles, particularly through the discovery and application of fundamental symmetry principles.” 🙂 I’ll let you figure out how what I write about in this post, and symmetry arguments, might be related. 🙂

That’s it for today, folks! I hope you enjoyed this. 🙂

Post scriptum: The main disadvantage of the flywheel interpretation is that it doesn’t explain interference: waves interfere—some rotating mass doesn’t. Ultimately, the wave and flywheel interpretation must, somehow, be compatible. One way to think about it is that the electron can only move as it does—in a “local circulatory motion”—if there is a force on it that makes it move the way it does. That force must be gravitational because… Well… There is no other candidate, is there? [We’re not talking some electron orbital here—some negative charge orbiting around a positive nucleus. We’re just considering the electron itself.] So we just need to prove that our rotating arrow will also represent a force, whose components will make our electron move the way it does. That should not be difficult. The analogy of the V-twin engine should do the trick. I’ll deal with that in my next post. If we’re able to provide such proof (which, as mentioned, should not be difficult), it will be a wonderful illustration of the complementarity principle. 🙂

However, just thinking about it does raise some questions already. Circular motion like this can be explained in two equivalent ways. The most obvious way to think about it is to assume some central field. It’s the planetary model (illustrated below). However, that doesn’t suit our purposes because it’s hard – if possible at all – to relate it to the wavefunction oscillation.

planetary model

The second model is our two-spring or V-twin engine model (illustrated below), but then what is the mass here? One hypothesis that comes to mind is that we’re constantly accelerating and decelerating an electric charge (the electron charge)—against all other charges in the Universe, so to speak. So that’s a force over a distance—energy. And energy has an equivalent mass.

V-2 engineThe question which remains open, then, is the following: what is the nature of this force? In previous posts, I suggested it might be gravitational, but so here we’re back to the drawing board: we’re talking an electrical force, but applied to some mass which acquires mass because of… Well… Because of the force—because of the oscillation (the moving charge) itself. Hmm…. I need to think about this.

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