# Tag Archives: zitterbewegung

# 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 = *c*ħ/m*c*^{2} = ħ/m*c*) of the electron. We interpret this by noting that the *c* = *a*ω identity gives us the E = m*c*^{2} = m*a*^{2}ω^{2} 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:

- The
*x*=*a*cos(ωt) and*y*=*a*·sin(ωt) equations describe the motion of the pointlike charge. - As an electromagnetic oscillation, we write it as
*E*_{0}=*E*_{0}cos(ωt+π/2) +*i*·*E*_{0}·sin(ωt+π/2).

The magnitudes of the oscillation *a* and *E*_{0} 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

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·*a*^{2}·ω^{2} and E = m·*c*^{2} 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. 🙂

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:

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

How can we relate those formulas to the E = m·*a*^{2}·ω^{2} formula? The *kinetic *energy that is being stored in a flywheel is equal E* _{kinetic}* = I·ω

*/2, so that is only*

^{2}*half*of the E = m·

*a*

^{2}·ω

^{2}product if we substitute I for I = m·

*a*

^{2}. [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·

*a*

^{2}·ω

^{2}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**That two-dimensional oscillation—the

*drives*the electron, and which is reflected in Einstein’s E = m·*c*^{2}formula.*a*

^{2}·ω

^{2}=

*c*

^{2}equation, really—tells us that

**the**—but measured in units of

*resonant*(or*natural*)*frequency*of the fabric of spacetime is given by the speed of light*a*. [If you don’t quite get this, re-write the

*a*

^{2}·ω

^{2}=

*c*

^{2}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 *v *= *a·*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 *a *= 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·*c*^{2}/ħ ⇔ *a *= 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:

- 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/s^{2}), so that’s the dimension of a gravitational field. - 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) - The third difference is one that I thought of only recently: the
*plane*of the oscillation can*not*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.

The 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*—can*not* 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

**(I used bold-face because these are**

*μ**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 sin^{2}θ. Hence, the (instantaneous) *change *in kinetic energy at any point in time (as a function of the angle θ) is equal to: d(sin^{2}θ)/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 sin^{2}(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·*a*^{2}·ω^{2}/2 instead of E = m·*a*^{2}·ω^{2}? What are the implications? Well… We get a √2 factor in our formula for the radius *a*, as shown below.

Now that is *not *so nice. For the tangential velocity, we get *v *= *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 * J_{z}*—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

*. Let me insert Feynman’s illustration here again (Feynman’s*

**J***Lectures*, II-34-3), so you get what I am talking about.

Now, all depends on the angle (θ) between * J_{z}* and

**, 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**

*J**magnitude*of the presumed

*actual*momentum:In this particular case (spin-1/2 particles),

*j*is equal to 1/2 (in units of ħ, of course). Hence,

*J*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 *a *= *ħ*/(m·*c*) is the *reduced *Compton scattering radius. The standard Compton scattering radius is equal to *a·*2π* *= (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. 🙂They 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. 🙂

# 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·e*^{−i∙θ} = *a·e*^{−i∙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*·

*term to the argument (θ = E·t −*

**x****p**∙

**x**). 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·E*

_{v }_{0}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 E

*and p*

_{v}*and, importantly, the coordinates*

_{v}*x*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 wave

*function*.

You’ll note the argument (or *phase*) of our wavefunction moves clockwise—or *counter*clockwise, 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: *J _{z}* = +ħ/2 or, else,

*J*= −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

_{z}*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 *a *= ħ/m_{e}·*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:* r _{Thomson} = α·r_{Compton}* ≈ 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·*a*^{2}·ω^{2}) and Einstein’s formula (E = m·*c*^{2}), 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·*a*^{2}·ω^{2} 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.

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:

- K.E. = T = m·
*v*^{2}/2 = (1/2)·m·ω^{2}·*a*^{2}·sin^{2}(ω·t + Δ) - P.E. = U = k·x
^{2}/2 = (1/2)·k·*a*^{2}·cos^{2}(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·ω^{2}. Hence, the total energy—for *one *piston, or one spring—is equal to:

E = T + U = (1/2)· m·ω^{2}·*a*^{2}·[sin^{2}(ω·t + Δ) + cos^{2}(ω·t + Δ)] = m·*a*^{2}·ω^{2}/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·*a*^{2}·ω^{2}. 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 viXra.org.

It is all beautiful, and the key question is obvious: if we want to relate the E = m·*a*^{2}·ω^{2} and E = m·*c*^{2} formulas, we need to explain why we could, potentially, write *c *as *c *= *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 *r *is given by *a *= ħ/(m·*c*), and ω = E/ħ = m·*c*^{2}/ħ, so *v *is equal to to *v *= [ħ/(m·*c*)]·[m·*c*^{2}/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 = m*c*^{2} equation implies the *ratio* between the energy and the mass of *any *particle is always the same:

This effectively reminds us of the ω^{2} = *C*^{–}^{1}/*L* or ω^{2} = k/m formula for harmonic oscillators. The key difference is that the ω^{2}= *C*^{–}^{1}/*L* and ω^{2} = k/m formulas introduce *two *(or more) degrees of freedom. In contrast, *c*^{2}= 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 ω^{2}= 1/*LC* formula here. It’s basically the same concept: the ω^{2}= 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 ω^{2}= *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 ω^{2}= *C*^{–}^{1}/*L* and ω^{2} = k/m sort of describe the same thing: harmonic oscillation. It’s just… Well… Unlike the ω^{2}= *C*^{–}^{1}/*L*, the ω^{2} = k/m is *directly *compatible with our V-twin engine metaphor, because it also involves *physical distances*, as I’ll show you here.] The *k *in the ω^{2} = 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 *x * 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 *x *in this equation may be equated to the *maximum *extension of our spring, or the *amplitude *of the oscillation, so that’s the radius *r *= *a *in the metaphor we’re analyzing here. Now look at how we can re-write the *c *= *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·*c*^{2} 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*Now,*

**on a particle.***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·*v*^{2}, 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·*r*^{2}/2 formula for the angular momentum, as opposed to the I = m·*r*^{2} formula. I = m·*r*^{2}/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 *x *= *a*, we need to calculate it as the following integral:

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 *c*^{2}* *= *a*^{2}·k/m relation implies a value for k which is equal to k = m·*c*^{2}/*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 *c*/ω *ratio *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/*x *(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 *c *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 *a *= 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/s

^{2}), 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. 🙂

*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

*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 *f *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 *N *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 *J _{z}* (the direction of propagation is supposed to be the

*z*-axis here) will be equal to

*J*=

_{z}*N*·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

*J*=

_{z}*N*·ħ equations, we get:

*J _{z}* =

*N*·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.

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 *r *based on a similar argument as the one Feynman used to get that *J _{z}* =

*N*·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:

An equally spectacular implication of our flywheel model of the electron was the following: we found that the angular *velocity v* was equal to *v* = *r*·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 *v* = *r*·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 φ_{0}.

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 *y *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

**τ**=

**×**

*r***F**. Needless to say, we have two

*vector*cross-products here. Hence, if we use the

**τ**=

**×**

*r***F**formula, we need to find the

*tangential*component of the force (

**F**

_{t}), whose magnitude will be equal to F

_{t}= q·ε

_{t}

*.*Now, 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:

*d*W/*d*t = q·ε_{t}·*v*

[If you have trouble, remember *v *is equal to *d*s/*d*t = Δs/Δt for Δt → 0, and re-write the equation above as *d*W = q·ε_{t}·*v*·*d*t = q·ε_{t}·*d*s = F_{t}·*d*s. *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 = F_{t}·*r* = q·ε_{t}·*r*, so we get:

*d**J _{z}*/

*d*t = q·ε

_{t}·

*v*

The *ratio *of *d*W/*d*t and *d**J _{z}*/

*d*t gives us the following interesting equation:

Now, Feynman tries to relate this to the *J _{z}* =

*N*·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

*d*

*J*and

_{z}*d*W 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

*d*W over this time interval, we get W. Likewise, if we integrate

*d*

*J*over the

_{z}*same*time interval, we should get the

*total*angular momentum that our electron is

*absorbing*from the light beam. Now, because

*d*

*J*=

_{z}*d*W/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 *v* = *r*·ω = [ħ·/(m·*c*)]·(E/ħ) = *c*, right? Right. So… If ω increases, but *r*·ω must equal the speed of light, then *r *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 *J _{z}*, namely +1, 0, −1 (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

*j*which is standing still, there must be the 2

*j*+1 possible states with values of

*J*going in steps of 1 from −

_{z}*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 *dis*advantage 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.

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.

The 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.