# 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. 🙂

# Mass as a two-dimensional oscillation (2)

This post basically further develops my speculative thoughts about the real meaning of the E = m·c2 formula. However, I’ll use the relativistically correct formulas for the calculations this time, so it may look somewhat more complicated. However, I think you should be able to digest it relatively easily, as none of the math is exceedingly difficult.

My previous post explored the similarity between the formula for the energy of a harmonic oscillator and the E = m·c2 formula. Now, there is another formula that sort of resembles it: the E = m·v2/2 formula for the kinetic energy. Could we relate them somehow and – in the process – gain a better understanding of Einstein’s famous formula? I think we can, and I want to show you how. In fact, in this post, I will try to relate all three.

We should first note that the E = m·v2/2 is a non-relativistic formula. It is only correct if we assume the mass – defined as a measure of inertia, remember? – to be constant, which we know isn’t true. As an object accelerates and gains kinetic energy, its effective mass will increase. In fact, the relativistically correct formula for the kinetic energy just calculates it as the difference between (1) the total energy (which is given by the E = m·c2 formula, always) and (2) its rest energy, so we write:

K.E. = E − E0 = mv·c2 − m0·c2 = m0·γ·c2 − m0·c2 = m0·c2·(γ − 1)

The γ in this formula is, of course, the ubiquitous Lorentz factor. Hence, the correct formula for the kinetic energy is m0·c2·(γ − 1). We shouldn’t use that m·v2/2 formula. Still, the two formulas are remarkably similar: there is a squared velocity (v2 and c2) and some factor (1/2 versus γ − 1). Why the squared velocity? That’s child play, right? Yep, I effectively wrote a post on that for my kids. We have a force that acts on some object over some time and over some distance, and so that force is going to do some work. While it’s child play, we’re calculating a path or line integral here:

Child play? Perhaps, but many kids don’t know what a vector dot product is (the F·dx), and they also don’t realize we can only solve this because we assume the mass m to be constant (i.e. not a function of the velocity v). So… Well… In our flywheel model of an electron, we’ve been using a non-relativistic formula, but we’ve calculated the tangential speed as being equal to c. A recipe for disaster, right? 🙂 Can we re-do the calculations? We can. You can google a zillion publications on relativistic harmonic oscillators but I took the derivation below from a fairly simple one I’d recommend. The only correction we’ll do here is to use the relativistically correct expression of Newton’s force law: the force equals the time rate of change of the (relativistic) momentum p = mvv = γm0v. So we write:

F = dp/dt = F = –kx with p = mvv = γm0v

Multiplying both sides with = dx/dt yields the following expression: Now, when we combine two oscillators – think of the metaphor of a frictionless V-twin engine, as illustrated below 🙂 – then we know that – because of the 90° angle between the two cylinders, the motion of one piston will be equal to x = a∙cos(ω∙t), while the
motion of the other is given by y = a∙cos(ω∙t–π/2) = a∙sin(ω∙t).Now how do we calculate the total energy in this system? Should we distinguish the x– and y– components of the total momentum p? We can easily do that. Look at the animation below, and you’ll immediately understand that we can easily calculate the velocities in the x– and a y-direction: vx = dx/dt = −a·ω·sin(ω∙t) and vy = dy/dta·ω·cos(ω∙t). The sum of the square of both then gives us the tangential velocity vv2 a2∙ω2∙sin2(ω∙t) + a2∙ω2∙cos2(ω∙t) = a2∙ω2 ⇔ va∙ω.  But how do we add energies here? It’s a tricky question: we have potential energy in one oscillator, and then in the other, and these energies are being transferred from one to another through the flywheel, so to speak. So there is kinetic energy there. Can we just add it all? Let us think about our perpetuum mobile once more, noting that the equilibrium position for the piston is at the center of each cylinder. When it goes past that position, extra pressure will build up and eventually push the piston back. When it is below that position, pressure is below equilibrium and will, therefore, also want to pull the piston back. The dynamics are as follows:

• When θ is zero, the air in cylinder 1 is fully compressed, and the piston will return to the equilibrium position (x = 0) as θ goes to 90°. The flywheel will transfer energy to cylinder 2, where the piston goes from the equilibrium position to full compression. Cylinder 2 borrows energy, and will want to return to its equilibrium position.
• When θ is 90°, the air in cylinder 2 is fully compressed, and the piston will return to the equilibrium position (y = 0) as θ goes to 180°. The flywheel will transfer energy back to cylinder 1, where the piston goes past the equilibrium position to create a vacuum. The piston in cylinder 1 borrows energy, and will want to return to its equilibrium position.
• When θ is 180°, the piston in cylinder 1 is fully extended, and will want to return to equilibrium because the pressure is lower than when in equilibrium. As θ goes from 180° to 270°, the piston in cylinder 1 does effectively return to equilibrium and, through the flywheel, pushes the piston in cylinder 2 past the equilibrium to create vacuum. The piston in cylinder 2 borrows energy, and will want to return to equilibrium.
• Finally, between 270° and 360°, the piston in cylinder 2 returns to equilibrium and, through the flywheel, causes the piston in cylinder 1 to compress air. The piston in cylinder 2 borrows energy, and will want to return to equilibrium.

It is a funny thing. Where is the energy in this system? Energy is not supposed to be thought as being directional but, here, direction clearly matters! We need to think about averages here (kinetic energy is a non-directional (scalar) quantity but it’s a function of velocity, and velocity is directional. If we have two directions only (x and y), then we can write: 〈vx2〉 = 〈vy2〉 = [〈vx2〉 + 〈vy2〉]/2 = 〈v2〉/2. So this gives us a clue, but we won’t make things to complicated here. Think of it like this. While transferring energy from one piston to the other, the crankshaft will rotate with a constant angular velocity: linear motion becomes circular motion, and vice versa. So what is the total energy in the system? What if we would want to use it? What can we take from it? You’d agree we would have to take it from the flywheel, right? The usable energy is in the flywheel. Let’s have a look at that energy conservation law we derived above: The usable energy in the flywheel is the E = m·c2 term. This, and my previous post, suggests we may interpret the mass of an electron as a two-dimensional oscillation. In fact, I think my previous post is an easier read because I use the classical (non-relativistic) formulas there. This post, hopefully, demonstrates that a relativistically correct mathematical treatment doesn’t alter the picture that I’ve been trying to offer. 🙂

Of course, the more difficult thing is to go beyond this metaphor and explain how exactly this motion from borrowing and returning energy to space would actually work. 🙂 So that would be a proper ‘ring theory’ of matter. 🙂

# Mass as a two-dimensional oscillation

This post is actually not about Einstein or anything he wrote. It is just for fun. It really is because… Well… While it’s about one of the most fundamental questions in physics (what is mass?), I am not going to talk about the Higgs field or other terribly complicated stuff. In fact, I’ll be talking about relativity using formulas that I shouldn’t be using. 🙂 But you should find this very straightforward and thought-provoking. Or so I hope, at least. So see if you like it and – if you do or you don’t, whatever – please let me know by posting a comment.

In my previous post, I noted the structural similarity between the E = mc2 and E = m·a2·ω2/2 formulas. The E = mc2 formula is very well known but also very mysterious. In contrast, the E = m·a2·ω2/2 formula is not so well known but not mysterious at all! It is the formula for the energy of a harmonic oscillator: think of an oscillating spring, for example. The only difference between the two formulas is the 1/2 factor and… Well… We would also need some interpretation of the a·ω identity that comes out of this, of course – but I will get there in a moment. 🙂

We can get rid of that 1/2 factor by combining two oscillators. Think of two frictionless pistons (or springs) in a 90° degree angle, as shown below. Because their motion is perpendicular to each other, their motion is independent, and so we can effectively add the power (and energy) of both. In fact, the 90° degree angle explains why a Ducati is more efficient than a Harley-Davidson, whose cylinders are at an angle of 45°. The 45° angle makes for great sound but… Well… Not so efficient. 🙂

The in the E = m·a2·ω2/2 formula is the magnitude of the oscillation, and the motion of these pistons (or of a mass on a spring) will be described by an x = a·cos(ω·t + Δ) function. The x is just the displacement from the center, and the Δ is just a phase factor which defines our t = 0 point. The ω is the angular frequency of our oscillator: it is defined by the time that is needed for a complete cycle, which is referred to as the period of the oscillation. We denote the period as t0, and it is easy to understand that t0 must be such that ω·t0 = 2π. Hence, t0 = 2π/ω. Note that, because of the 90° angle between the two cylinders, the phase factor Δ would be zero 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 animation below abstracts away from pistons, springs or whatever other physical oscillation we might think of – but it represents the same. We just denote the phase itself as θ = ω·t.

Now think of an electron as a charge moving about some center, so that’s the green dot in the animation above. We can then also analyze its movement in terms of two perpendicular oscillations, i.e. the sine and cosine functions shown above. Now, you may or may not know that an elementary wavefunction consists of the same: a sine and as cosine. Indeed, using Euler’s notation, we write:

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

Remembering that multiplication by the imaginary unit (i) amounts to a rotation by 90°. To be precise, because of the minus sign in front of the sine, we have a rotation by minus 90° here, but that doesn’t change the analysis: Nature doesn’t care about our convention for i, or for our convention for measuring angles clockwise or counter-clockwise, and it does allow the angular momentum to be either positive or negative (it is ± ħ/2 for an electron). But let’s further develop our analogy by getting back to our oscillators. The kinetic and potential energy of one oscillator – think of one piston or one spring only – can 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) only—is equal to:

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

That is the formula we started out with and, yes, if we would add the energy of the two oscillators, we’d effectively have a perpetuum mobile storing an energy that is equal to twice this amount: E = m·a2·ω2.

Let us now think this through. If E and m are the energy and mass of an electron, then the E = m·a2·ω2 and E = m·c2 equations tell us that a·ω. What are a and ω here? Well… The de Broglie relations suggest we should equate ω to E/ħ. As for a, we could take the Compton scattering radius of the electron, which is equal to ħ/(m·c). So we write:

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

Did we prove anything here? No. We don’t prove anything in this post. We’re just having fun. We only showed that our E = m·a2·ω2 = m·c2 equation might (let me put in italics: might) make sense. 🙂

Let me show you something else. If this flywheel model makes sense, then we can, obviously, also calculate a tangential velocity for our charge. The tangential velocity is the product of the radius and the angular velocity: vr·ω = a·ω = c. In our previous post, we wrote that we should think of the speed of light as the resonant frequency of the spacetime fabric, but we should probably take that back. The speed of light emerges as the speed of the charge in what I’ll now officially refer to as my flywheel model of an electron.

Is there a resonant frequency here? If so, how should we interpret it? Well… From our a·ω = c, we get that:

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

So the answer is: no. No resonant frequency of spacetime. The frequency is the frequency of our electron – not of the fabric of spacetime. However, we can, perhaps, think of another analogy. The natural frequency of a spring (ω) depends on (1) the mass on that spring (m) and (2) the restoring force, which is equal to F = −k·x. The k factor here is the stiffness of the spring. Could we, likewise, talk about the stiffness of the spacetime fabric? We know that, for a spring, we can calculate from m and ω. We wrote: k = m·ω2. Can we do anything with that? Probably not. The mass in our flywheel model is the equivalent mass of the energy in the (two-dimensional) oscillation. It is not some actual mass going up and down and back and forth. In fact, if the tangential velocity of our charge would be equal to c – which it is in our model – then the charge itself should have zero (rest) mass ! Hence, the stiffness would be equal to k = 0·ω2 = 0!

Let me offer another calculation instead. If this flywheel model makes sense, then the electron will have some angular momentum, right? The angular momentum is equal to L = ω·I, so that’s the product of the angular velocity (ω) and the moment of inertia (I), aka the angular mass. Now, from the Stern-Gerlach experiment, we know that the angular momentum of an electron is equal to ± ħ/2. So now we can calculate the moment of inertia as I = L/ω = (ħ/2)/(E/ħ) = ħ2/(2·E). Now, substituting E for E = m·c2 and remembering that = ħ/(m·c), we can write this as:

I = ħ2/(2·m·c2) = (ħ2·m)/(2·m2·c2) = (1/2)·m·a2

Do we recognize that formula? Yes. It’s the formula for the angular mass of a solid disc, or a hoop about the diameter, as shown below. Which of the two makes most sense? I am not sure. I’ll let you think about that. 🙂

So… Well… That’s it! 🙂

[…]

🙂

Why am I smiling? Well… I hope this post makes you think about stuff yourself because… Well… That was my only objective: have fun by thinking about stuff yourself! 🙂

Do these calculations – and the analogy itself – make any sense? My truthful answer is: I am not sure. I really don’t know. Of course, I would very much like to think that this analogy may represent something real. Why? Because it would allow us to associate the wavefunction with something real and, therefore (see my paper on this), it would also allow us to think of Schrödinger’s equation as representing something real. To be precise, it would allow us to think of Schrödinger’s equation as an energy diffusion equation, but… Well… That is somewhat more difficult to explain than what I explained above. 😦

The essential question, of course, is: what gives that pointlike charge that circular motion? What is the origin of what Schrödinger himself referred to (I admit: he did so in a very different context) as a Zitterbewegung?

All that I’ve written above, assumes space is not just some abstract mathematical space. It is real, somehow, and perfectly elastic. So that’s why we can advance a model that assumes an an electron is nothing but a charge (with zero rest mass) that bounces around in it. All of its mass is the equivalent mass of the energy in the oscillation itself. It is, of course, a crazy hypothesis that we cannot prove but, as far as I can see, while crazy, the hypothesis is consistent with what we know about the weird wonderful world of quantum mechanics.

The main weakness in the argument is the following: if the charge itself has zero rest mass, then our E = m·a2·ω2/2 equation reduces to E = 0. Is the analogy still valid? And how can we possibly associate some angular mass with something that is going around but has zero rest mass? This is, effectively, a flywheel model without a flywheel. I may have explained matter as a two-dimensional oscillation, but I haven’t told you what is oscillating. Or… Well… I did. What is oscillating is the charge, with zero rest mass.

The analogy with a photon is obvious. A photon has zero rest mass too! Now, the wavefunction of a photon is the electromagnetic wave. Can we say what is oscillating there? Yes, we can! Of course: it is the electromagnetic field! But what’s the field? You will say the electromagnetic field has a physical dimension: newton per coulomb (N/C). This begs the next question: what’s the physical dimension of the wavefunction? Could it be the same?

My answer is: yes, or maybe. 🙂 Our model assumes it is, effectively, the charge of the electron that is oscillating. Hence, why wouldn’t it be the same?

In any case, I will let you think about that for yourself. As you can see, in physics, an answer to one question may trigger many more. 🙂 If you have any good feedback, please comment! 🙂

Post scriptum: The mentioned weakness in the argument should also be related to the fact that we are using classical non-relativistic formulas. If our charge is really moving at a speed at or near light speed around some center, we should probably have another look at our K.E. = T = m·v2/2 = (1/2)·m·ω2·a2·sin2(ω·t + Δ) formula, right? 🙂 The relativistically correct definition of kinetic energy is slightly different than the T = m·v2/2 formula. It may – or may not – make a difference. 🙂 I’ll talk about that in my next post.