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. 🙂

E = mc2

Isn’t it remarkable that the best-known formula in physics, E = mc2, is actually one that we cannot really prove? As such, this formula is like a physical law: we think such laws are universally true, but we cannot be sure. Why? Because the experiments and observations since Einstein came up with this formula 115 years ago (for the context, you can check the Wikipedia history section on it) only suggest it is true. Of course, these experiments and observations very strongly suggest the E = mc2 formula is true but… Well… Karl Popper (and – more importantly – Albert Einstein himself) told us that we should always be skeptical: some more advanced alien might visit Earth one day and demonstrate – through some spectacular experiment – that our E = mc2 formula (and all of the other associated laws and lemmas) is, in fact, not quite correct.

That would be even more dramatic, I guess, then the 1887 Michelson-Morley experiment – that ‘most famous failed experiment in history’ which, instead of confirming what everyone thought to be true at the time, told us that Galilean (or Newtonian) relativity was, in fact, not true. It took Einstein and a whole lot of other bright guys (the likes of James Clerk Maxwell and Hendrik Antoon Lorentz) almost 20 years to fix the problem (Einstein published his special relativity theory in 1905), so… Well… We can only hope that this alien will be friendly and, immediately after blowing up our cherished beliefs, will also show us how to fix our formulas. 🙂

Of course, you will say one can find lots of proofs of E = mc2 when surfing the web, but these ‘proofs’ are actually just simple illustrations of the law: they only ‘prove’ that the E = mc2 formula is consistent with other statements and laws. For example, a lot of these so-called proofs will show you that the E = mc2 is consistent with special relativity. They will show you, for example, that the E = mc2 formula is consistent with the relativistic mass formula m = mv = γm0, or that radiation has, effectively, some equivalent mass. But then… Well… That consistency doesn’t prove the E = mc2 formula.

The E = mc2 formula is just something like Newton’s law of gravitation, or Maxwell’s equations: we can’t prove those either. We can just work out all of their implications and check if they are consistent with experiments and observations, and we accept them because they are. In fact, most of these so-called proofs don’t even help to understand what Einstein’s mass-energy equivalence formula actually means. They just talk about its manifestations or consequences.

For example, we all know that the equivalent mass of the binding energy between matter-particles in a nuclear fission reaction is converted into destructive heat and radiation energy. However, we also know that a nuclear explosion does not actually annihilate any elementary particles. So you might say it doesn’t really convert (rest) mass to energy. It is just binding energy that gets released – or converted into some other form of energy. As such, this oft-quoted example just illustrates that energy has an equivalent mass. So it just says what it says. Nothing more. Nothing less. This example does not tell us, for example, if and how it might work the other way around. Can we convert radiation energy back into mass? Probably not, right? Why? Because of entropy and what have you. In other words, we can surely not say that mass and energy are the same. Equivalent. Yes. But not the same.

Now, you might still be inclined to think they are, because there is actually a much better illustration of mass to energy conversion than the classical nuclear bomb: think of a positron and an electron coming together. [Just to make sure, anti-matter is just matter with an opposite electric charge. There is no such thing as negative matter.] Let’s say their rest mass is m0. So… Well… The positron and the electron will effectively annihilate each other in a flash and the  resulting radiation will have an energy that’s equal to E = m0c2. So that’s a much better illustration of how the rest mass of an elementary particle can be converted directly into energy.

Still, this equivalence between mass and energy does not imply the energy and mass concepts are, effectively, the same. For starters, their physical dimension is different. Equivalent (1 kg = 1 N·s2/m), yes, but not the same. Or “same-same but different”, as they’d say in Asia. 🙂 More importantly, this illustration of this so-called equivalence between mass and energy still doesn’t prove the formula: this experiment just adds to a zillion other observations and experiments which have turned this formula into a generally accepted statement – something that is thought of as being true. In fact, in physics, we cannot really prove something is true: everything we know is true only until someone comes along and shows us why it is not true. Experiments can only confirm what we believe is true or – if they don’t work out – they prove us that what we believed is wrong. Hence, strictly speaking, experiments and observations can only tell us what might be true, or confirm our beliefs by showing us what is definitely not true. Of course, that is more than good enough for most of us. I, for one, am convinced that the E = mc2 formula is true. So it’s my truth, for sure! [Just to make sure you know where I stand: I fully accept science! Creationism and other nonsense is definitely not my truth!]

The point is: I want to understand the formula, and that’s where most of these proofs fail miserably too: not only don’t they prove anything, but they also don’t really tell us what the E = mc2 formula really means. How should we think of the annihilation of matter and anti-matter? What happens there, really?

To answer that question, we need to answer a much more fundamental question: what is mass? And what is energy? It is not easy to define energy. It comes in many forms (e.g. chemical versus nuclear energy), and various other distinctions – such as the distinction between potential and kinetic energy – may cause even more confusion. Is it any easier to define mass? Maybe. Maybe not. Let’s try it. Newton’s laws associate two very different things with mass: it is, first, a measure of inertia (resistance to a change in motion) and, second, it is the cause of the gravitational force.

Let us briefly discuss the second aspect first: Einstein’s general relativity theory sort of explains gravitation away, by pointing out that a mass causes spacetime to curve. We no longer think of spacetime as an abstract mathematical space now, but as a physical space: it is our space now, and it’s bent. Think of Einstein’s famous remark: “Physical objects are not in space, but these objects are spatially extended. In this way, the concept “empty space” loses its meaning.” Hence, general relativity theory is not just another equivalent representation of the same thing (gravitation): the metaphysics are very different.

Let us turn to the first aspect: mass as a measure of inertia. When one or more forces act on an object with some mass, some power is being delivered to that object. I hope you’ve learned enough about physics to vaguely remember we can write that using vectors and a time derivative. Don’t worry if you can’t quite follow the mathematical argument. Just try to get the basic idea of it. If T is the kinetic energy of some object with some mass, we can write the following:The dT/dt is the time rate of change of the kinetic energy, and we use bold letters (e.g. F, v or s) to denote vectors, so they are directional numbers, so to speak: they do not only have a magnitude but a direction as well. The product between two vectors (e.g. F·v) is a vector dot product (so it’s commutative, unlike a vector cross product). OK. Onwards. You should note that the formula above is fully relativistically correct. Why? Because the formula for the momentum p = mv = mvv uses the relativistic mass concept, so it recognizes the mass of an object increases as it gains speed according to the Lorentz correction:Onwards! T is the kinetic energy. However, if kinetic energy is all that changes (the potential energy is just the equivalent energy of the rest mass here), then the time rate of change of the total energy E will be equal to the time rate of change of the kinetic energy T. If we then assume that the E = mc2 formula is correct, we can write the following:Note that we substituted F for dp/dt = d(mv)/dt. This too is relativistically correct: the force is the time rate of change of the momentum of an object. In fact, to correct Newton’s law for relativistic effects, we only need to re-write it like this: F = dp/dt = d(mv)/dt = mvdv/dt = mva. All we do is substitute the mass factor m for the relativistic mass m = mv. Now, it takes a few tricks (e.g. multiply both sides by 2m) to check that this equation is equivalent to this:In case you don’t see it, you may want to check the original story, which I got from Feynman here. Now, if the derivatives of two quantities are the same, then the quantities themselves differ by a constant only, so we write: m2c2 = m2v2 + C. What is the constant C? The formula must be valid for all v, so let us choose v = 0. We get: m02c2 = 0 + C = C. Substitution then gives us this: m2c2 = m2v2 + m02c2. Finally, dividing by c2 and rearranging the terms gives us the relativistic mass formula:Isn’t this amazing? We cannot prove the E = mc2 formula, but if we use it as an axiom – so if we assume it to be true – then it gives us the relativistic mass formula. So the logic is the following: if the E = mc2 formula is true, then the m = mv = γm0 is true as well. The logic does not go the other way. Why? Because the proof above uses this arrow at some place: ⇒. One way. Not an arrow like this: ⇔. 🙂

Still, the question I started out with remains: what is mass? I haven’t said anything about that so far. The truth: it is a bit complicated. In fact, I have my own little fun theory on this. It is based on the remarkable structural similarity between the relativistic energy formula and the formula for the total energy of an oscillator:

1. E = mc2
2. E = mω2/2

In fact, I should write the second formula as E = m·a2·ω2/2: the is the amplitude of the oscillation, which may or (more likely) may not be equal to one. The point is: the and the ω in these two formulas both describe a velocity – linear or, in the case of E = mω2/2 – angular. Of course, there is the 1/2 factor in the E = mω2/2 formula, but that is exactly the point that inspired me to explore the following question: what if we’d think of mass as some oscillation in two dimensions, so it stores an amount of energy that is equal to E = 2∙ mω2/2 = mω2. Indeed, Einstein’s E = mc2 equation implies the ratio between the energy and the mass of any particle is always the same:If you have ever read anything about oscillators – mechanical or electrical – this should remind you of the ω2 = C−1/L or ω2 = k/m formulas for electric and mechanical oscillators respectively. The key difference is that the ω2 = C−1/L (electric circuit) and ω2 = k/m (mechanical spring) formulas introduce two (or more) degrees of freedom. In contrast, c2 = E/m for any particle, always. But 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.

This gives rise to what I refer to as a flywheel model for elementary particles. More about that later. 😊 Or… Well… If you don’t want to wait, here are the links to my two papers on this:

Have fun ! 🙂