# 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·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. 🙂

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.

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—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.

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:In 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. 🙂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. 🙂

# Electron spin and the geometry of the wavefunction

In our previous posts, we interpreted the elementary wavefunction ψ = a·ei∙θ = a·cosθ − i·a·sinθ as a two-dimensional oscillation in spacetime. In addition to assuming the two directions of the oscillation were perpendicular to each other, we also assumed they were perpendicular to the direction of motion. While the first assumption is essential in our interpretation, the second assumption is solely based on an analogy with a circularly polarized electromagnetic wave. We also assumed the matter wave could be right-handed as well as left-handed (as illustrated below), and that these two physical possibilities corresponded to the angular momentum being equal to plus or minus ħ/2 respectively.

This allowed us to derive the Compton scattering radius of an elementary particle. Indeed, we interpreted the rotating vector as a resultant vector, which we get by adding the sine and cosine motions, which represent the real and imaginary components of our wavefunction. The energy of this two-dimensional oscillation is twice the energy of a one-dimensional oscillator and, therefore, equal to E = m·a2·ω2. Now, the angular frequency is given by ω = E/ħ and E must, obviously, also be equal to E = m·c2. Substitition and re-arranging the terms gives us the Compton scattering radius:

The value given above is the (reduced) Compton scattering radius for an electron. For a proton, we get a value of about 2.1×10−16 m, which is about 1/4 of the radius of a proton as measured in scattering experiments. Hence, for a proton, our formula does not give us the exact (i.e. experimentally verified) value but it does give us the correct order of magnitude—which is fine because we know a proton is not an elementary particle and, hence, the motion of its constituent parts (quarks) is… Well… It complicates the picture hugely.

If we’d presume the electron charge would, effectively, be rotating about the center, then its tangential velocity is given by v = a·ω = [ħ·/(m·c)]·(E/ħ) = c, which is yet another wonderful implication of our hypothesis. Finally, the a·ω formula allowed us to interpret the speed of light as the resonant frequency of the fabric of space itself, as illustrated when re-writing this equality as follows:

This gave us a natural and forceful interpretation of Einstein’s mass-energy equivalence formula: the energy in the E = m·c2· equation is, effectively, a two-dimensional oscillation of mass.

However, while toying with this and other results (for example, we may derive a Poynting vector and show probabilities are, effectively, proportional to energy densities), I realize the plane of our two-dimensional oscillation cannot be perpendicular to the direction of motion of our particle. In fact, the direction of motion must lie in the same plane. This is a direct consequence of the direction of the angular momentum as measured by, for example, the Stern-Gerlach experiment. 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 (inhomogenous) 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 some rotating disk or a flywheel, then it will actually comprise the direction of motion.

What are the implications? I am not sure. I will definitely need to review whatever I wrote about the de Broglie wavelength in previous posts. We will also need to look at those transformations of amplitudes once again. Finally, we will also need to relate this to the quantum-mechanical formulas for the angular momentum and the magnetic moment.

Post scriptum: As in previous posts, I need to mention one particularity of our model. When playing with those formulas, we contemplated two different formulas for the angular mass: one is the formula for a rotating mass (I = m·r2/2), and the other is the one for a rotating mass (I = m·r2). The only difference between the two is a 1/2 factor, but it turns out we need it to get a sensical result. For a rotating mass, the angular momentum is equal to the radius times the momentum, so that’s the radius times the mass times the velocity: L = m·v·r. [See also Feynman, Vol. II-34-2, in this regard)] Hence, for our model, we get L = m·v·r = m·c·a = m·c·ħ/(m·c) = ħ. Now, we know it’s equal to ±ħ/2, so we need that 1/2 factor in the formula.

Can we relate this 1/2 factor to the g-factor for the electron’s magnetic moment, which is (approximately) equal to 2? Maybe. We’d need to look at the formula for a rotating charged disk. That’s for a later post, however. It’s been enough for today, right? 🙂

I would just like to signal another interesting consequence of our model. If we would interpret the radius of our disk (a)—so that’s the Compton radius of our electron, as opposed to the Thomson radius—as the uncertainty in the position of our electron, then our L = m·v·r = m·c·a = p·a = ħ/2 formula as a very particular expression of the Uncertainty Principle: p·Δx= ħ/2. Isn’t that just plain nice? 🙂

# 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.

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 viXra.org.

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:

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.

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:

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:

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.

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.

# Photons as spin-1 particles

After all of the lengthy and speculative excursions into the nature of the wavefunction for an electron, it is time to get back to Feynman’s Lectures and look at photon-electron interactions. So that’s chapter 17 and 18 of Volume III. Of all of the sections in those chapters – which are quite technical here and there – I find the one on the angular momentum of polarized light the most interesting.

Feynman provides an eminently readable explanation of how the electromagnetic energy of a photon may be absorbed by an electron as kinetic energy. It is entirely compatible with our physical interpretation of the wavefunction of an electron as… Well… We’ve basically been looking at the electron as a little flywheel, right? 🙂 I won’t copy Feynman here, except the illustration, which speaks for itself.

However, I do recommend you explore these two Lectures for yourself. Among other interesting passages, Feynman notes that, while photons are spin-1 particles and, therefore, are supposed to be associated with three possible values for the angular momentum (Jz = +ħ, 0 or −ħ), there are only two states: the zero case doesn’t exist. As Feynman notes: “This strange lack is related to the fact that light cannot stand still.” But I will let you explore this for yourself. 🙂

# Feynman as the Great Teacher?

While browsing for something else, I stumbled on an article which derides Feynman’s qualities as a teacher, and the Caltech Feynman Lectures themselves. It is an interesting read. Let me quote (part of) the conclusion:

“Richard Feynman constructed an “introductory” physics course at Caltech suitable primarily for perhaps imaginary extreme physics prodigies like himself or how he pictured himself as an eighteen year old. It is an open question how well the actual eighteen year old Feynman would have done in the forty-three year old Feynman’s “introductory” physics course. Like many adults had Feynman lost touch with what it had been like to be eighteen? In any case, such extreme physics prodigies made up only a small fraction of the highly qualified undergraduate students at Caltech either in the 1960’s or 1980’s. An educational system designed by extreme prodigies for extreme prodigies, often from academic families, extremely wealthy families, or other unusual backgrounds rare even among most top students as conventionally defined, is a prescription for disaster for the vast majority of students and society at large.”

The article actually reacts to a blog post from Bill Gates, who extols Feynman’s virtues as a teacher. So… Was or wasn’t he a great teacher?

It all depends on your definition of a great teacher. I respect the views in the mentioned article mentioned above—if only because the author, John F. McGowan, is not just anyone: he is a B.S. from Caltech itself, and he has a Ph.D. in physics. I don’t, so… Well… He is an authority, obviously. Frankly, I must agree I struggled with Feynman’s Lectures too, and I will probably continue to do so as I read and re-read them time after time. On the other hand, below I copy one of those typical Feynman illustrations you will not find in any other textbook. Feynman tries to give us a physical explanation of the photon-electron interaction here. Most introductory physics textbooks just don’t bother: they’ll give you the mathematical formalism and then some exercises, and that’s it. Worse, those textbooks will repeatedly tell you you can’t really ‘understand’ quantum math. Just go through the math and apply the rules. That’s the general message.

I find that very disappointing. I must admit that Feynman has racked my brain—but in a good way. I still feel I do not quite understand quantum physics “the way we would like to”. It is still “peculiar and mysterious”, but then that’s just how Richard Feynman feels about it too—and he’s humble enough to admit that in the very first paragraph of his very first Lecture on QM.

I have spent a lot of my free time over the past years thinking about a physical or geometric interpretation of the wavefunction—half of my life, in a way—and I think I found it. The article I recently published on it got downloaded for the 100th time today, and this blog – as wordy, nerdy and pedantic as it is – attracted 5,000 visitors last month alone. People like me: people who want to understand physics beyond the equations.

So… Well… Feynman himself admits he was mainly interested in the “one or two dozen students who — very surprisingly — understood almost everything in all of the lectures, and who were quite active in working with the material and worrying about the many points in an excited and interested way.” I think there are many people like those students. People like me: people who want to understand but can’t afford to study physics on a full-time basis.

For those, I think Feynman’s Lectures are truly inspirational. At the very least, they’ve provided me with many wonderful evenings of self-study—some productive, in the classical sense of the word (moving ahead) and… Some… Well… Much of what I read did—and still does—keep me awake at night. 🙂

# The speed of light as an angular velocity (2)

My previous post on the speed of light as an angular velocity was rather cryptic. This post will be a bit more elaborate. Not all that much, however: this stuff is and remains quite dense, unfortunately. 😦 But I’ll do my best to try to explain what I am thinking of. Remember the formula (or definition) of the elementary wavefunction:

ψ = a·ei[E·t − px]/ħa·cos(px/ħ − E∙t/ħ) + i·a·sin(px/ħ − E∙t/ħ)

How should we interpret this? We know an actual particle will be represented by a wave packet: a sum of wavefunctions, each with its own amplitude ak and its own argument θk = (Ek∙t − pkx)/ħ. But… Well… Let’s see how far we get when analyzing the elementary wavefunction itself only.

According to mathematical convention, the imaginary unit (i) is a 90° angle in the counterclockwise direction. However, Nature surely cannot be bothered about our convention of measuring phase angles – or time itself – clockwise or counterclockwise. Therefore, both right- as well as left-handed polarization may be possible, as illustrated below.

The left-handed elementary wavefunction would be written as:

ψ = a·ei[E·t − px]/ħa·cos(px/ħ − E∙t/ħ) − i·a·sin(px/ħ − E∙t/ħ)

In my previous posts, I hypothesized that the two physical possibilities correspond to the angular momentum of our particle – say, an electron – being either positive or negative: J = +ħ/2 or, else, J = −ħ/2. I will come back to this in a moment. Let us first further examine the functional form of the wavefunction.

We should note that both the direction as well as the magnitude of the (linear) momentum (p) are relative: they depend on the orientation and relative velocity of our reference frame – which are, in effect, relative to the reference frame of our object. As such, the wavefunction itself is relative: another observer will obtain a different value for both the momentum (p) as well as for the energy (E). Of course, this makes us think of the relativity of the electric and magnetic field vectors (E and B) but… Well… It’s not quite the same because – as I will explain in a moment – the argument of the wavefunction, considered as a whole, is actually invariant under a Lorentz transformation.

Let me elaborate this point. If we consider the reference frame of the particle itself, then the idea of direction and momentum sort of vanishes, as the momentum vector shrinks to the origin itself: p = 0. Let us now look at how the argument of the wavefunction transforms. The E and p in the argument of the wavefunction (θ = ω∙t – kx = (E/ħ)∙t – (p/ħ)∙x = (E∙t – px)/ħ) are, of course, the energy and momentum as measured in our frame of reference. Hence, we will want to write these quantities as E = Ev and p = pv = pvv. If we then use natural time and distance units (hence, the numerical value of c is equal to 1 and, hence, the (relative) velocity is then measured as a fraction of c, with a value between 0 and 1), we can relate the energy and momentum of a moving object to its energy and momentum when at rest using the following relativistic formulas:

E= γ·E0 and p= γ·m0v = γ·E0v/c2

The argument of the wavefunction can then be re-written as:

θ = [γ·E0/ħ]∙t – [(γ·E0v/c2)/ħ]∙x = (E0/ħ)·(t − v∙x/c2)·γ = (E0/ħ)∙t’

The γ in these formulas is, of course, the Lorentz factor, and t’ is the proper time: t’ = (t − v∙x/c2)/√(1−v2/c2). Two essential points should be noted here:

1. The argument of the wavefunction is invariant. There is a primed time (t’) but there is no primed θ (θ’): θ = (Ev/ħ)·t – (pv/ħ)·x = (E0/ħ)∙t’.

2. The E0/ħ coefficient pops up as an angular frequency: E0/ħ = ω0. We may refer to it as the frequency of the elementary wavefunction.

Now, if you don’t like the concept of angular frequency, we can also write: f0 = ω0/2π = (E0/ħ)/2π = E0/h. Alternatively, and perhaps more elucidating, we get the following formula for the period of the oscillation:

T0 = 1/f0 = h/E0

This is interesting, because we can look at the period as a natural unit of time for our particle. This period is inversely proportional to the (rest) energy of the particle, and the constant of proportionality is h. Substituting Efor m0·c2, we may also say it’s inversely proportional to the (rest) mass of the particle, with the constant of proportionality equal to h/c2. The period of an electron, for example, would be equal to about 8×10−21 s. That’s very small, and it only gets smaller for larger objects ! But what does all of this really tell us? What does it actually mean?

We can look at the sine and cosine components of the wavefunction as an oscillation in two dimensions, as illustrated below.

Look at the little green dot going around. Imagine it is some mass going around and around. Its circular motion is equivalent to the two-dimensional oscillation. Indeed, instead of saying it moves along a circle, we may also say it moves simultaneously (1) left and right and back again (the cosine) while also moving (2) up and down and back again (the sine).

Now, a mass that rotates about a fixed axis has angular momentum, which we can write as the vector cross-product L = r×p or, alternatively, as the product of an angular velocity (ω) and rotational inertia (I), aka as the moment of inertia or the angular massL = I·ω. [Note we write L and ω in boldface here because they are (axial) vectors. If we consider their magnitudes only, we write L = I·ω (no boldface).]

We can now do some calculations. We already know the angular velocity (ω) is equal to E0/ħ. Now, the magnitude of r in the Lr×p vector cross-product should equal the magnitude of ψ = a·ei∙E·t/ħ, so we write: r = a. What’s next? Well… The momentum (p) is the product of a linear velocity (v) – in this case, the tangential velocity – and some mass (m): p = m·v. If we switch to scalar instead of vector quantities, then the (tangential) velocity is given by v = r·ω.

So now we only need to think about what formula we should use for the angular mass. If we’re thinking, as we are doing here, of some point mass going around some center, then the formula to use is I = m·r2. However, we may also want to think that the two-dimensional oscillation of our point mass actually describes the surface of a disk, in which case the formula for I becomes I = m·r2/2. Of course, the addition of this 1/2 factor may seem arbitrary but, as you will see, it will give us a more intuitive result. This is what we get:

L = I·ω = (m·r2/2)·(E/ħ) = (1/2)·a2·(E/c2)·(E/ħ) = a2·E2/(2·ħ·c2)

Note that our frame of reference is that of the particle itself, so we should actually write ω0, m0 and E0 instead of ω, m and E. The value of the rest energy of an electron is about 0.510 MeV, or 8.1871×10−14 N∙m. Now, this momentum should equal J = ±ħ/2. We can, therefore, derive the (Compton scattering) radius of an electron:Substituting the various constants with their numerical values, we find that a is equal 3.8616×10−13 m, which is the (reduced) Compton scattering radius of an electron. The (tangential) velocity (v) can now be calculated as being equal to v = r·ω = a·ω = [ħ·/(m·c)]·(E/ħ) = c. This is an amazing result. Let us think about it.

In our previous posts, we introduced the metaphor of two springs or oscillators, whose energy was equal to E = m·ω2. Is this compatible with Einstein’s E = m·c2 mass-energy equivalence relation? It is. The E = m·c2 implies E/m = c2. We, therefore, can write the following:

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

Hence, we should actually have titled this and the previous post somewhat differently: the speed of light appears as a tangential velocity. Think of the following: the ratio of c and ω is equal to c/ω = a·ω/ω = a. Hence, the tangential and angular velocity would be the same if we’d measure distance in units of a. In other words, the radius of an electron appears as a natural distance unit here: if we’d measure ω in units of per second, rather than in radians (which are expressed in the SI unit of distance, i.e. the meter) per second, the two concepts would coincide.

More fundamentally, we may want to look at the radius of an electron as a natural unit of velocityHuh? Yes. Just re-write the c/ω = a as ω = c/a. What does it say? Exactly what I said, right? As such, the radius of an electron is not only a norm for measuring distance but also for time. 🙂

If you don’t quite get this, think of the following. For an electron, we get an angular frequency that is equal to ω = E/ħ = (8.19×10−14 N·m)/(1.05×10−34 N·m·s) ≈ 7.76×1020 radians per second. That’s an incredible velocity, because radians are expressed in distance units—so that’s in meter. However, our mass is not moving along the unit circle, but along a much tinier orbit. The ratio of the radius of the unit circle and is equal to 1/a ≈ (1 m)/(3.86×10−13 m) ≈ 2.59×1012. Now, if we divide the above-mentioned velocity of 7.76×1020 radians per second by this factor, we get… Right ! The speed of light: 2.998×1082 m/s. 🙂

Post scriptum: I have no clear answer to the question as to why we should use the I = m·r2/2 formula, as opposed to the I = m·r2 formula. It ensures we get the result we want, but this 1/2 factor is actually rather enigmatic. It makes me think of the 1/2 factor in Schrödinger’s equation, which is also quite enigmatic. In my view, the 1/2 factor should not be there in Schrödinger’s equation. Electron orbitals tend to be occupied by two electrons with opposite spin. That’s why their energy levels should be twice as much. And so I’d get rid of the 1/2 factor, solve for the energy levels, and then divide them by two again. Or something like that. 🙂 But then that’s just my personal opinion or… Well… I’ve always been intrigued by the difference between the original printed edition of the Feynman Lectures and the online version, which has been edited on this point. My printed edition is the third printing, which is dated July 1966, and – on this point – it says the following:

“Don’t forget that meff has nothing to do with the real mass of an electron. It may be quite different—although in commonly used metals and semiconductors it often happens to turn out to be the same general order of magnitude, about 2 to 20 times the free-space mass of the electron.”

Two to twenty times. Not 1 or 0.5 to 20 times. No. Two times. As I’ve explained a couple of times, if we’d define a new effective mass which would be twice the old concept – so meffNEW = 2∙meffOLD – then such re-definition would not only solve a number of paradoxes and inconsistencies, but it will also justify my interpretation of energy as a two-dimensional oscillation of mass.

However, the online edition has been edited here to reflect the current knowledge about the behavior of an electron in a medium. Hence, if you click on the link above, you will read that the effective mass can be “about 0.1 to 30 times” the free-space mass of the electron. Well… This is another topic altogether, and so I’ll sign off here and let you think about it all. 🙂