Our previous posts showed how a simple geometric interpretation of the elementary wavefunction yielded the (Compton scattering) radius of an elementary particle—for an electron, at least: for the proton, we only got the order of magnitude right—but then a proton is not an elementary particle. We got lots of other interesting equations as well… But… Well… When everything is said and done, it’s that equivalence between the E = m·*a*^{2}·ω^{2} and E = m·*c*^{2} relations that we… Well… We need to be more *specific *about it.

Indeed, I’ve been ambiguous here and there—*oscillating *between various interpretations, so to speak. 🙂 In my own mind, I refer to my unanswered questions, or my ambiguous answers to them, as the *form factor *problem. So… Well… That explains the title of my post. But so… Well… I do want to be somewhat more *conclusive *in this post. So let’s go and see where we end up. 🙂

To help focus our mind, let us recall the metaphor of the V-2 *perpetuum mobile*, as illustrated below. With permanently closed valves, the air inside the cylinder compresses and decompresses as the pistons move up and down. It provides, therefore, a restoring force. As such, it will store potential energy, just like a spring, and the motion of the pistons will also reflect that of a mass on a spring: it is described by a sinusoidal function, with the zero point at the center of each cylinder. We can, therefore, think of the moving pistons as harmonic oscillators, just like mechanical springs. Of course, instead of two cylinders with pistons, one may also think of connecting two springs with a crankshaft, but then that’s not fancy enough for me. 🙂

At first sight, the analogy between our flywheel model of an electron and the V-twin engine seems to be complete: the 90 degree angle of our V-2 engine makes it possible to perfectly balance the pistons and we may, therefore, think of the flywheel as a (symmetric) rotating mass, whose angular momentum is given by the product of the angular frequency and the moment of inertia: L = ω·I. Of course, the moment of inertia (aka the angular mass) will depend on the *form *(or *shape*) of our flywheel:

- I = m·
*a*^{2}for a rotating*point*mass m or, what amounts to the same, for a circular*hoop*of mass m and radius*r*=*a*. - For a rotating (uniformly solid)
*disk*, we must add a 1/2 factor: I*a*^{2}/2.

How can we relate those formulas to the E = m·*a*^{2}·ω^{2} formula? The *kinetic *energy that is being stored in a flywheel is equal E* _{kinetic}* = I·ω

*/2, so that is only*

^{2}*half*of the E = m·

*a*

^{2}·ω

^{2}product if we substitute I for I = m·

*a*

^{2}. [For a disk, we get a factor 1/4, so that’s even worse!] However, our flywheel model of an electron incorporates potential energy too. In fact, the E = m·

*a*

^{2}·ω

^{2}formula just adds the (kinetic and potential) energy of two oscillators:

*we do not really consider the energy in the flywheel itself*because… Well… The essence of our flywheel model of an electron is

*not*the flywheel: the flywheel just

*transfers*energy from one oscillator to the other, but so… Well… We don’t

*include*it in our energy calculations.

**The essence of our model is that two-dimensional oscillation which**That two-dimensional oscillation—the

*drives*the electron, and which is reflected in Einstein’s E = m·*c*^{2}formula.*a*

^{2}·ω

^{2}=

*c*

^{2}equation, really—tells us that

**the**—but measured in units of

*resonant*(or*natural*)*frequency*of the fabric of spacetime is given by the speed of light*a*. [If you don’t quite get this, re-write the

*a*

^{2}·ω

^{2}=

*c*

^{2}equation as ω =

*c*/

*a*: the radius of our electron appears as a

*natural*distance unit here.]

Now, we were *extremely* happy with this interpretation not only because of the key results mentioned above, but also because it has lots of other nice consequences. Think of our probabilities as being proportional to energy densities, for example—and all of the other stuff I describe in my published paper on this. But there is even more on the horizon: a follower of this blog (a reader with an actual PhD in physics, for a change) sent me an article analyzing elementary particles as tiny black holes because… Well… If our electron is effectively spinning around, then its tangential velocity is equal to *v *= *a·*ω = *c*. Now, recent research suggest black holes are also spinning at (nearly) the speed of light. Interesting, right? However, in order to understand what she’s trying to tell me, I’ll first need to get a better grasp of general relativity, so I can relate what I’ve been writing here and in previous posts to the Schwarzschild radius and other stuff.

Let me get back to the lesson here. In the reference frame of our particle, the wavefunction really looks like the animation below: it has two components, and the amplitude of the two-dimensional oscillation is equal to *a*, which we calculated as *a *= ħ·/(m·*c*) = 3.8616×10^{−13} m, so that’s the (reduced) Compton scattering radius of an electron.

In my original article on this, I used a more complicated argument involving the angular momentum formula, but I now prefer a more straightforward calculation:

*c* = *a*·ω = *a*·E/ħ = *a*·m·*c*^{2}/ħ ⇔ *a *= ħ/(m·*c*)

The question is: what *is *that rotating arrow? I’ve been vague and not so vague on this. The thing is: I can’t *prove *anything in this regard. But my *hypothesis *is that it is, in effect, a *rotating **field vector*, so it’s just like the electric field vector of a (circularly polarized) electromagnetic wave (illustrated below).

There are a number of crucial differences though:

- The (physical) dimension of the field vector of the matter-wave is different: I associate the real and imaginary component of the wavefunction with a force
*per unit mass*(as opposed to the force per unit charge dimension of the electric field vector). Of course, the newton/kg dimension reduces to the dimension of acceleration (m/s^{2}), so that’s the dimension of a gravitational field. - I do believe this gravitational disturbance, so to speak, does cause an electron to move about some center, and I believe it does so at the speed of light. In contrast, electromagnetic waves do
*not*involve any mass: they’re just an oscillating*field*. Nothing more. Nothing less. In contrast, as Feynman puts it: “When you do find the electron some place, the entire charge is there.” (Feynman’s*Lectures*, III-21-4) - The third difference is one that I thought of only recently: the
*plane*of the oscillation can*not*be perpendicular to the direction of motion of our electron, because then we can’t explain the direction of its magnetic moment, which is either up or down when traveling through a Stern-Gerlach apparatus.

I mentioned that in my previous post but, for your convenience, I’ll repeat what I wrote there. The basic idea here is illustrated below (credit for this illustration goes to another blogger on physics). As for the Stern-Gerlach experiment itself, let me refer you to a YouTube video from the *Quantum Made Simple *site.

The point is: the direction of the angular momentum (and the magnetic moment) of an electron—or, to be precise, its component as measured in the direction of the (inhomogeneous) magnetic field through which our electron is *traveling*—can*not* be parallel to the direction of motion. On the contrary, it is *perpendicular* to the direction of motion. In other words, if we imagine our electron as spinning around some center, then the disk it circumscribes will *comprise *the direction of motion.

However, we need to add an interesting detail here. As you know, we don’t really have a precise direction of angular momentum in quantum physics. [If you don’t know this… Well… Just look at one of my many posts on spin and angular momentum in quantum physics.] Now, we’ve explored a number of hypotheses but, when everything is said and done, a rather classical explanation turns out to be the best: an object with an angular momentum ** J** and a magnetic moment

**(I used bold-face because these are**

*μ**vector*quantities) that is parallel to some magnetic field

**B**, will

*not*line up, as you’d expect a tiny magnet to do in a magnetic field—or not

*completely*, at least: it will

*precess*. I explained that in another post on quantum-mechanical spin, which I advise you to re-read if you want to appreciate the point that I am trying to make here. That post integrates some interesting formulas, and so one of the things on my ‘to do’ list is to prove that these formulas are, effectively, compatible with the electron model we’ve presented in this and previous posts.

Indeed, when one advances a hypothesis like this, it’s not enough to just sort of show* *that the general geometry of the situation makes sense: we also need to show the numbers come out alright. So… Well… Whatever we *think *our electron—or its wavefunction—might be, it needs to be compatible with stuff like the *observed* precession frequency* *of an electron in a magnetic field.

Our model also needs to be compatible with the transformation formulas for amplitudes. I’ve been talking about this for quite a while now, and so it’s about time I get going on that.

Last but not least, those articles that relate matter-particles to (quantum) gravity—such as the one I mentioned above—are intriguing too and, hence, whatever hypotheses I advance here, I’d better check them against those more advanced theories too, right? 🙂 Unfortunately, that’s going to take me a few more years of studying… But… Well… I still have many years ahead—I hope. 🙂

**Post scriptum**: It’s funny how one’s brain keeps working when sleeping. When I woke up this morning, I thought: “But it *is *that flywheel that matters, right? That’s the energy storage mechanism and also explains how photons possibly interact with electrons. The oscillators *drive *the flywheel but, without the flywheel, nothing is happening. It is really the *transfer *of energy—through the flywheel—which explains why our flywheel goes round and round.”

It may or may not be useful to remind ourselves of the math in this regard. The *motion *of our first oscillator is given by the cos(ω·t) = cosθ function (θ = ω·t), and its kinetic energy will be equal to sin^{2}θ. Hence, the (instantaneous) *change *in kinetic energy at any point in time (as a function of the angle θ) is equal to: d(sin^{2}θ)/dθ = 2∙sinθ∙d(sinθ)/dθ = 2∙sinθ∙cosθ. Now, the motion of the second oscillator (just look at that second piston going up and down in the V-2 engine) is given by the sinθ function, which is equal to cos(θ − π /2). Hence, its kinetic energy is equal to sin^{2}(θ − π /2), and how it *changes *(as a function of θ again) is equal to 2∙sin(θ − π /2)∙cos(θ − π /2) = = −2∙cosθ∙sinθ = −2∙sinθ∙cosθ. So here we have our energy transfer: the flywheel organizes the borrowing and returning of energy, so to speak. That’s the crux of the matter.

So… Well… What *if *the relevant energy formula is E = m·*a*^{2}·ω^{2}/2 instead of E = m·*a*^{2}·ω^{2}? What are the implications? Well… We get a √2 factor in our formula for the radius *a*, as shown below.

Now that is *not *so nice. For the tangential velocity, we get *v *= *a*·ω = √2·*c*. This is also *not *so nice. How can we save our model? I am not sure, but here I am thinking of the mentioned precession—the *wobbling *of our flywheel in a magnetic field. Remember we may think of * J_{z}*—the angular momentum or, to be precise, its component in the

*z*-direction (the direction in which we

*measure*it—as the projection of the

*real*angular momentum

*. Let me insert Feynman’s illustration here again (Feynman’s*

**J***Lectures*, II-34-3), so you get what I am talking about.

Now, all depends on the angle (θ) between * J_{z}* and

**, of course. We did a rather obscure post on these angles, but the formulas there come in handy now. Just click the link and review it if and when you’d want to understand the following formulas for the**

*J**magnitude*of the presumed

*actual*momentum:In this particular case (spin-1/2 particles),

*j*is equal to 1/2 (in units of ħ, of course). Hence,

*J*is equal to √0.75 ≈ 0.866. Elementary geometry then tells us cos(θ) = (1/2)/√(3/4) = = 1/√3. Hence, θ ≈ 54.73561°. That’s a big angle—larger than the 45° angle we had secretly expected because… Well… The 45° angle has that √2 factor in it: cos(45°) = sin(45°) = 1/√2.

Hmm… As you can see, there is no easy fix here. Those damn 1/2 factors! They pop up everywhere, don’t they? 🙂 We’ll solve the puzzle. One day… But not today, I am afraid. I’ll call it the form factor problem… Because… Well… It sounds better than the 1/2 or √2 problem, right?* *🙂

**Note**: If you’re into quantum math, you’ll note *a *= *ħ*/(m·*c*) is the *reduced *Compton scattering radius. The standard Compton scattering radius is equal to *a·*2π* *= (2π·*ħ*)/(m·*c*) = *h*/(m·*c*) = *h*/(m·*c*). It doesn’t solve the √2 problem. Sorry. The form factor problem. 🙂

To be honest, I finished my published paper on all of this with a suggestion that, perhaps, we should think of two *circular *oscillations, as opposed to linear ones. Think of a tiny ball, whose center of mass stays where it is, as depicted below. Any rotation – around any axis – will be some combination of a rotation around the two other axes. Hence, we may want to think of our two-dimensional oscillation as an oscillation of a polar and azimuthal angle. It’s just a thought but… Well… I am sure it’s going to keep me busy for a while. 🙂They are oscillations, still, so I am not thinking of *two *flywheels that keep going around in the same direction. No. More like a wobbling object on a spring. Something like the movement of a bobblehead on a spring perhaps. 🙂