# The flywheel model of an electron

In one of my previous posts, I developed a flywheel model for an electron – or for charged matter-particles in general. You can review the basics in the mentioned post or, else, download a more technical paper from this link. Here I just want to present some more general remarks about the conceptual difficulties. Let’s recall the basics of the model. We think of an electron as a point charge moving about some center, so that’s the green dot in the animation below. We can then analyze its movement in terms of two perpendicular oscillations, i.e. the sine (blue) and cosine (red) functions.

The idea is that space itself – by some mechanism we’d need to elaborate – provides a restoring force when our charge moves away from the center. So it wants to go back to the r = (x, y) = 0, but it never actually does because of these restoring forces, which are perpendicular to each other and just make it spin around. This may sound crazy but the original inspiration for this – the mathematical similarity between the E = m·c2 and E = m·a2·ω2 formulas – is intriguing enough to pursue the idea. Let me show you.

Think of an electron with energy E (the equivalent mass of this electron is, of course, m = E/c2). Now look at the following. If we substitute ω for E/ħ (the frequency of the matter-wave) and for ħ/m·c (i.e. the Compton scattering radius of an electron), we can write:

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

We get the same when equating the E = m·a2·ω2 and E = m·c2 formulas:

a2·ω2 = m·c2  ⇔ a·ω

Wow! Did we just prove something? No. Not really. We only showed that our E = m·a2·ω2 = m·c2 equation might make sense. Note that the a·ω = c product is the tangential velocity vtangential = r·ω = a·ω = c.

Our rotating charge should also have some angular momentum, right? In fact, the Stern-Gerlach experiment tells us the angular momentum should be equal to ± ħ/2. Is it? We will need a formula for the angular mass, aka the moment of inertia. Let’s try the one for a rotating disk: I = (1/2)·m·a2. The formula for the angular momentum itself is just L = I·ω. Combining both gives us the following result:

L = ħ/2 = I·ω = (1/2)·m·a2·ω = (1/2)·(E/c2a2·(E/ħ) ⇔ = ħ/(m·c)

Did we just get the formula for the Compton scattering radius out of our model? Well… Yes. It is a rather Grand Result, I think. So… Well… It might make sense, but… Well… Does it, really?

First, we should note that the math is not so easy as it might seem at first: we would need to use relativistically correct formulas everywhere. I made a start at that, and it’s quite encouraging. For example, I showed in one of my previous posts, that the classical E = m·v2/2 formula for the kinetic energy of a mass on a spring effectively morphs into the relativistically correct E = mv·c2 = m0·γ·c2 formula. So it looks like the annoying 1/2 factors in the classical formulas effectively disappear in a relativistically correct analysis. However, there is a lot more to be checked than just the math for one oscillator, and I haven’t done that yet.

The key issue is the interpretation of the mass factor. Our pointlike charge can only travel at light speed if its rest mass is equal to zero. But that is a problem for the model, because the idea of accelerations or decelerations as a result of a restoring force is no longer consistent: if there is no rest mass, even the tiniest force on it will give it an infinite acceleration or deceleration. Hence, it is tempting to assume the charge must have some mass, and the concepts of the electromagnetic and/or the effective mass spring to mind. However, I don’t immediately see how the specifics of this could be worked out. So we’re a bit stuck here. Still, it is tempting to try to think of it like a photon, for which we can effectively state that all of its energy – and the equivalent mass – is in the oscillation: E = ħ·ω and m = E/c2 = ħ·ω/c2.

Let’s think of a few other questions. What if the energy of our electron increases? Our a·ω = c tells that = c/ω = ħ/E should decrease. Does that make any sense? We may think of it like this. The E = m·c2 equation tells us that the ratio between the energy and the mass of a particle is constant: E/m = c2. So we can write:

E/m = c2 = a2·ω2 = a2·E22 ⇔ a·E/ħ = c

This tells us the same: is inversely proportional to E, and the constant of inverse proportionality is ħ. Hence, our disk becomes smaller but spins faster. What happens to the angular mass and the angular momentum? The angular momentum must L = ħ/2 = I·ω must remain the same, and you can check it does. Hence, the angular mass I effectively diminishes as ω = E/ħ goes up.

Hence, it all makes sense – intuitively, that is. In fact, our model is basically an adaptation of the more general idea of electromagnetic mass: we’re saying the energy in the (two-dimensional) oscillation gives our electron the equivalent mass that we measure in real-life experiments. Having said that, the idea of a pointlike charge with zero rest mass – accelerating and decelerating in two perpendicular directions – remains a strange one.

Let us think about the nature of the restoring force. The oscillator model is, effectively, based on the assumption we have a linear force here. To be precise, the restoring force is thought of as being directly proportional with the distance from the equilibrium: F = −k·x. The proportionality constant is the stiffness of our spring. It just comes with the system. So what’s the equivalent of in our model here? For a non-relativistic spring, it is easy to show that – while a constant – will always be equal to k = m·ω2. Hence, if we put a larger mass on the spring, then the frequency of the oscillation must go down, and vice versa. But what mass concept should we apply here? Before we further explore this question, let us look at those relativistic corrections. Why? Well… I think it’s about time I show you how they affect the analysis, right? Let me gratefully acknowledge an obviously brilliant B.A. student who did an excellent paper on relativistic springs (link and references here), from which I copy the illustration below.

The graphs show what happens to the trajectory and the velocity when speeds go up all the way to the relativistic limit. The light grey is the classical (non-relativistic) case: we recognize the sinusoidal motion and velocity function. See how these shapes change when velocities go to the relativistic limit (see the grey and (for velocities reaching the speed of light) the black curves). The sinusoidal shape of the functions disappears: the velocity very rapidly changes between −c and +c. In fact, the velocity function starts resembling a square wave function.

This is quite interesting, but… Well… It doesn’t make us much wiser, and it surely does not answer our question: how should we interpret the stiffness coefficient here? The relativistically correct force law is given below, but that doesn’t tell us anything new: tells us that the restoring force is going to be directly proportional to the distance from the equilibrium (zero) point:

F = dp/dt = d(mv)/dt = mvdv/dt = mva = −k·x

However, we may think of this, perhaps: when we analyze a mechanical (or any other typical) oscillator, we think of being given. But for our flywheel model, we find that the value of is going to depend on the energy and the amplitude of the oscillation. To be precise, if we just take the E = m·a2·ω2 = m·c2 and k = m·ω2 formulas for granted, we can write k as a function of the energy:

E = k·a2 ⇔ = E/a2 = (E·E2)/(c2ħ2) ⇔ = E3/(c2ħ2)

It is a funny formula: it tells us the stiffness – or whatever the equivalent is in this model of ours – increases exponentially with the energy. It is a weird formula, but it is consistent with the other conclusion above: if the energy goes up, the radius becomes smaller. It’s because space becomes stickier. 🙂

But space is space, right? And its defining characteristic – the constant speed of light – is… Well… It is what it is: it just tells us that the ratio between the energy and mass for any particle is always the same:

What else does it tell us? Well… If our model is correct, it also tells us the tangential velocity of our charge will always be equal to c. In that sense, then, it tells us space is space. There is one more relation that we might mention here:

a·ω ⇔ ω = c/a

The energy defines the radius in our model (remember the = ħ/E relation). Now, the relationship above tells us that, in turn, defines the frequency, and in a rather particular way: the frequency is inversely proportional to the radius, and the proportionality coefficient is, once again, c. 🙂