The flywheel model of an electron

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

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

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

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

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

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

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

The flywheel interpretation

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

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

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

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

JzN·ħ = W/ω

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

RH photon

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

electron

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

radius formula

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

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

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

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

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

dW/dt = q·εt·v

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

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

dJz/dt = q·εt·v

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

Feynman formula

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

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

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

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

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

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

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

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

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

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

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

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

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

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

planetary model

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

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

One thought on “The flywheel model of an electron

  1. Pingback: The field and flywheel model of an electron | Reading Feynman

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