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 = m*c*^{2} and E = m·*a*^{2}·ω^{2}/2 formulas. The E = m*c*^{2} formula is very well known but also very mysterious. In contrast, the E = m·*a*^{2}·ω^{2}/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 *c *= *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 *a *in the E = m·*a*^{2}·ω^{2}/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 t_{0}, and it is easy to understand that t_{0} must be such that ω·t_{0} = 2π. Hence, t_{0} = 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·e*^{−i∙θ} = *a·e*^{−i∙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:

- K.E. = T = m·
*v*^{2}/2 = (1/2)·m·ω^{2}·*a*^{2}·sin^{2}(ω·t + Δ) - P.E. = U = k·x
^{2}/2 = (1/2)·k·*a*^{2}·cos^{2}(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·ω^{2}. Hence, the total energy—for *one *piston (or one spring) only—is equal to:

E = T + U = (1/2)· m·ω^{2}·*a*^{2}·[sin^{2}(ω·t + Δ) + cos^{2}(ω·t + Δ)] = m·*a*^{2}·ω^{2}/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·*a*^{2}·ω^{2}.

Let us now think this through. If E and m are the energy and mass of an electron, then the E = m·*a*^{2}·ω^{2} and E = m·*c*^{2} equations tell us that *c *= *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·*c*^{2}/(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·*a*^{2}·ω^{2} = m·*c*^{2} 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: *v* = *r*·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*/*a *= *c*/[ ħ/(m·*c*)] = (*c*/ħ)·(m·*c*) = m·*c*^{2}/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 *k *from m and ω. We wrote: *k = *m·ω^{2}. 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·ω^{2} = 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) = ħ^{2}/(2·E). Now, substituting E for E = m·*c*^{2} and remembering that *a *= ħ/(m·*c*), we can write this as:

I = ħ^{2}/(2·m·*c*^{2}) = (ħ^{2}·m)/(2·m^{2}·*c*^{2}) = (1/2)·m·*a*^{2}

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·*a*^{2}·ω^{2}/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·*v*^{2}/2 = (1/2)·m·ω^{2}·*a*^{2}·sin^{2}(ω·t + Δ) formula, right? 🙂 The relativistically correct definition of kinetic energy is slightly different than the T = m·*v*^{2}/2 formula. It may – or may not – make a difference. 🙂 I’ll talk about that in my next post.

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