# The speed of light as an angular velocity (2)

My previous post on the speed of light as an angular velocity was rather cryptic. This post will be a bit more elaborate. Not all that much, however: this stuff is and remains quite dense, unfortunately. ðŸ˜¦ But I’ll do my best to try to explain what I am thinking of. Remember the formula (orÂ definition) of theÂ elementary wavefunction:

Ïˆ =Â aÂ·eâˆ’i[EÂ·t âˆ’ pâˆ™x]/Ä§ =Â aÂ·cos(pâˆ™x/Ä§ âˆ’ Eâˆ™t/Ä§) + iÂ·aÂ·sin(pâˆ™x/Ä§ âˆ’Â Eâˆ™t/Ä§)

How should we interpret this? We know an actual particle will be represented by aÂ wave packet: a sum of wavefunctions, each with its own amplitude ak and its own argument Î¸k = (Ekâˆ™t âˆ’ pkâˆ™x)/Ä§. But… Well… Let’s see how far we get when analyzing theÂ elementaryÂ wavefunction itself only.

According to mathematicalÂ convention, the imaginary unit (i) is a 90Â°Â angle in theÂ counterclockwise direction. However, NatureÂ surely cannot be bothered about our convention of measuring phase angles – orÂ timeÂ itself – clockwiseÂ or counterclockwise. Therefore, both right- as well as left-handed polarization may be possible, as illustrated below.

The left-handed elementary wavefunction would be written as:

Ïˆ =Â aÂ·ei[EÂ·t âˆ’ pâˆ™x]/Ä§ =Â aÂ·cos(pâˆ™x/Ä§ âˆ’ Eâˆ™t/Ä§)Â âˆ’Â iÂ·aÂ·sin(pâˆ™x/Ä§ âˆ’Â Eâˆ™t/Ä§)

In my previous posts, I hypothesized that the two physical possibilities correspond to the angular momentum of our particle – say, an electron – being eitherÂ positive or negative: J = +Ä§/2 or, else,Â J = âˆ’Ä§/2. I will come back to this in a moment. Let us first further examine the functional form of the wavefunction.

We should note that both theÂ directionÂ as well as theÂ magnitudeÂ of the (linear) momentum (p) are relative: they depend on the orientation and relative velocity of our reference frame – which are, in effect, relative to the reference frame of our object. As such, the wavefunction itself is relative: another observer will obtain a different value for both the momentum (p) as well as for the energy (E). Of course, this makes us think of the relativity of the electric and magnetic field vectors (E and B) but… Well… It’s not quite the same because – as I will explain in a moment – the argument of the wavefunction, considered as a whole, is actually invariant under a Lorentz transformation.

Let me elaborate this point.Â If we consider the reference frame of the particle itself, then the idea of direction and momentum sort of vanishes, as the momentum vector shrinks to the origin itself:Â p = 0. Let us now look at howÂ the argument of the wavefunction transforms. The E and p in the argument of the wavefunction (Î¸ = Ï‰âˆ™t â€“ kâˆ™x = (E/Ä§)âˆ™t â€“ (p/Ä§)âˆ™x =Â (Eâˆ™t â€“ pâˆ™x)/Ä§) are, of course, the energy and momentum as measured in our frame of reference. Hence, we will want to write these quantities as E = Ev and p = pv = pvâˆ™v. If we then use natural time and distanceÂ units (hence, the numerical value of c is equal to 1 and, hence, the (relative) velocity is then measured as a fraction ofÂ c, with a value between 0 and 1), we can relate the energy and momentum of a moving object to its energy and momentum when at rest using the following relativistic formulas:

EvÂ =Â Î³Â·E0Â and pvÂ = Î³Â·m0âˆ™vÂ =Â Î³Â·E0âˆ™v/c2

The argument of the wavefunction can then be re-written as:

Î¸ = [Î³Â·E0/Ä§]âˆ™t â€“ [(Î³Â·E0âˆ™v/c2)/Ä§]âˆ™x = (E0/Ä§)Â·(t âˆ’ vâˆ™x/c2)Â·Î³ =Â (E0/Ä§)âˆ™t’

The Î³ in these formulas is, of course, the Lorentz factor, and t’ is theÂ properÂ time: t’Â = (t âˆ’ vâˆ™x/c2)/âˆš(1âˆ’v2/c2). Two essential points should be noted here:

1. The argument of the wavefunction is invariant. There is a primed time (t’) but there is no primedÂ Î¸ (Î¸’):Â Î¸ = (Ev/Ä§)Â·t â€“ (pv/Ä§)Â·x =Â (E0/Ä§)âˆ™t’.

2.Â TheÂ E0/Ä§ coefficient pops up as an angular frequency:Â E0/Ä§ =Â Ï‰0. We may refer to it asÂ theÂ frequency of the elementary wavefunction.

Now, if you don’t like the concept ofÂ angular frequency, we can also write:Â f0Â =Â Ï‰0/2Ï€ = (E0/Ä§)/2Ï€ = E0/h.Â Alternatively, and perhaps more elucidating, we get the following formula for theÂ periodÂ of the oscillation:

T0Â = 1/f0Â =Â h/E0

This is interesting, because we can look at the period as aÂ naturalÂ unit of time for our particle. This period is inverselyÂ proportional to the (rest) energy of the particle, and the constant of proportionality is h. Substituting E0Â for m0Â·c2, we may also say it’s inversely proportional to the (rest) mass of the particle, with the constant of proportionality equal to h/c2. The period of an electron, for example, would be equal to about 8Ã—10âˆ’21Â s. That’sÂ veryÂ small, and it only gets smaller for larger objects ! But what does all of this really tellÂ us? What does it actuallyÂ mean?

We can look at the sine and cosine components of the wavefunction as an oscillation inÂ twoÂ dimensions, as illustrated below.

Look at the little green dot going around. Imagine it is someÂ mass going around and around. Its circular motion is equivalent to the two-dimensional oscillation. Indeed, instead of saying it moves along a circle, we may also say it moves simultaneously (1) left and right and back again (the cosine) while also moving (2) up and down and back again (the sine).

Now, a mass that rotates about a fixed axis hasÂ angular momentum, which we can write as the vector cross-product L = rÃ—p or, alternatively, as the product of an angular velocity (Ï‰) and rotational inertia (I), aka as theÂ moment of inertia or the angular mass:Â L = IÂ·Ï‰. [Note we writeÂ L andÂ Ï‰ in boldface here because they are (axial) vectors. If we consider their magnitudes only, we write L = IÂ·Ï‰ (no boldface).]

We can now do some calculations. We already know the angular velocity (Ï‰) is equal toÂ E0/Ä§. Now, theÂ magnitude ofÂ rÂ in the L =Â rÃ—pÂ vector cross-product should equal theÂ magnitudeÂ ofÂ Ïˆ =Â aÂ·eâˆ’iâˆ™EÂ·t/Ä§, so we write:Â r = a. What’s next? Well… The momentum (p) is the product of a linear velocity (v) – in this case, theÂ tangentialÂ velocity –Â and some mass (m): p = mÂ·v. If we switch to scalarÂ instead ofÂ vector quantities, then the (tangential) velocity is given by v = rÂ·Ï‰.

So now we only need to think about what formula we should use for the angular mass. If we’re thinking, as we are doing here, of some pointÂ mass going around some center, then the formula to use isÂ I = mÂ·r2. However, we may also want to think that the two-dimensional oscillation of our point mass actually describes the surface of a disk, in which case the formula for I becomesÂ I = mÂ·r2/2. Of course, the addition of this 1/2 factor may seem arbitrary but, as you will see, it will give us a more intuitive result. This is what we get:

L = IÂ·Ï‰ = (mÂ·r2/2)Â·(E/Ä§) = (1/2)Â·a2Â·(E/c2)Â·(E/Ä§) =Â a2Â·E2/(2Â·Ä§Â·c2)

Note that our frame of reference is that of the particle itself, so we should actually write Ï‰0, m0Â and E0Â instead ofÂ Ï‰, m and E. The value of the rest energy of an electron is about 0.510 MeV, or 8.1871Ã—10âˆ’14 Nâˆ™m. Now, this momentum should equal J = Â±Ä§/2. We can, therefore, derive the (Compton scattering) radius of an electron:Substituting the various constants with their numerical values, we find that a is equal 3.8616Ã—10âˆ’13 m, which is the (reduced) Compton scattering radius of an electron.Â The (tangential) velocity (v) can now be calculated as being equal toÂ v = rÂ·Ï‰ = aÂ·Ï‰ = [Ä§Â·/(mÂ·c)]Â·(E/Ä§) =Â c. This is an amazing result. Let us think about it.

In our previous posts, we introduced the metaphor of twoÂ springsÂ or oscillators, whose energy was equal to E =Â mÂ·Ï‰2. Is this compatible with Einstein’s E =Â mÂ·c2Â mass-energy equivalence relation? It is. TheÂ E =Â mÂ·c2Â impliesÂ E/m =Â c2. We, therefore, can write the following:

Ï‰ = E/Ä§ =Â mÂ·c2/Ä§ = mÂ·(E/m)Â·/Ä§Â â‡” Ï‰ =Â E/Ä§

Hence, we should actually have titled this and the previous post somewhat differently: the speed of light appears as aÂ tangentialÂ velocity. Think of the following: theÂ ratioÂ ofÂ c andÂ Ï‰ is equal toÂ c/Ï‰ =Â aÂ·Ï‰/Ï‰ =Â a. Hence, the tangential and angular velocity would be the same if we’d measure distance in units ofÂ a. In other words,Â the radius of an electron appears as a natural distance unit here: if we’d measureÂ Ï‰ inÂ units ofÂ aÂ per second, rather than in radians (which are expressed in the SI unit of distance, i.e. the meter) per second, the two concepts would coincide.

More fundamentally, we may want to look at the radius of an electron as a naturalÂ unit of velocity.Â Huh?Â Yes. Just re-write theÂ c/Ï‰ =Â a asÂ Ï‰ =Â c/a. What does it say? Exactly what I said, right? As such, the radius of an electron is not only aÂ normÂ for measuring distance but also for time.Â ðŸ™‚

If you don’t quite get this, think of the following. For an electron, we get an angular frequency that is equal toÂ Ï‰ = E/Ä§ = (8.19Ã—10âˆ’14Â NÂ·m)/(1.05Ã—10âˆ’34Â NÂ·mÂ·s) â‰ˆ 7.76Ã—1020Â radiansÂ per second. That’s an incredible velocity, because radians are expressed in distance unitsâ€”so that’s inÂ meter. However, our mass is not moving along theÂ unitÂ circle, but along a much tinier orbit. TheÂ ratioÂ of the radius of the unit circle andÂ aÂ is equal to 1/a â‰ˆÂ (1 m)/(3.86Ã—10âˆ’13 m) â‰ˆ 2.59Ã—1012. Now, if we divide theÂ above-mentionedÂ velocityÂ ofÂ 7.76Ã—1020Â radiansÂ per second by this factor, we get… Right ! The speed of light: 2.998Ã—1082Â m/s. ðŸ™‚

Post scriptum: I have no clear answer to the question as to why we should use the I = mÂ·r2/2 formula, as opposed to theÂ I = mÂ·r2Â formula. It ensures we get the result we want, but this 1/2 factor is actually rather enigmatic. It makes me think of the 1/2 factor in SchrÃ¶dinger’s equation, which is also quite enigmatic. In my view, the 1/2 factor should not be there in SchrÃ¶dinger’s equation. Electron orbitals tend to be occupied byÂ twoÂ electrons with opposite spin. That’s why their energy levels should beÂ twice as much. And so I’d get rid of the 1/2 factor, solve for the energy levels, and then divide them by two again. Or something like that. ðŸ™‚ But then that’s just my personal opinion or… Well… I’ve always been intrigued by the difference between the originalÂ printedÂ edition of the Feynman Lectures and the online version, which has been edited on this point. My printed edition is the third printing, which is dated July 1966, and – on this point – it says the following:

“Donâ€™t forget thatÂ meff has nothing to do with the real mass of an electron. It may be quite differentâ€”although in commonly used metals and semiconductors it often happens to turn out to be the same general order of magnitude, about 2 to 20 timesÂ the free-space mass of the electron.”

Two to twenty times. Not 1 or 0.5 to 20 times. No. Two times. As I’ve explained a couple of times, if we’d define a new effective mass which would be twice the old concept – so meffNEWÂ = 2âˆ™meffOLDÂ – then such re-definition would not only solve a number of paradoxes and inconsistencies, but it will also justify my interpretation of energy as a two-dimensional oscillation of mass.

However, the online edition has been edited here to reflect the current knowledge about the behavior of an electron in a medium. Hence, if you click on the link above, you will read that the effective mass can be “about 0.1 to 30 times” the free-space mass of the electron. Well… This is another topic altogether, and so I’ll sign off here and let you think about it all. ðŸ™‚

# Electron and photon strings

Note: I have published a paper that is very coherent and fully explains what the idea of a photon might be. There is nothing stringy. Check it out: The Meaning of the Fine-Structure Constant. No ambiguity. No hocus-pocus.

Jean Louis Van Belle, 23 December 2018

Original post:

In my previous posts, I’ve been playing with… Well… At the very least, a new didactic approach to understanding the quantum-mechanical wavefunction. I just boldly assumed the matter-wave is a gravitational wave. I did so by associating its components with the dimension of gravitational field strength: newton per kg, which is the dimension of acceleration (N/kg = m/s2). Why? When you remember the physical dimension of the electromagnetic field is N/C (force per unitÂ charge), then that’s kinda logical, right? ðŸ™‚Â The math is beautiful. Key consequences include the following:

1. Schrodinger’s equation becomes an energy diffusion equation.
2. Energy densities give us probabilities.
3. The elementary wavefunction for the electron gives us the electron radius.
4. Spin angular momentum can be interpreted as reflecting the right- or left-handedness of the wavefunction.
5. Finally, the mysterious boson-fermion dichotomy is no longer “deep down in relativistic quantum mechanics”, as Feynman famously put it.

It’s all great. Every day brings something new. ðŸ™‚ Today I want to focus on our weird electron model and how we get God’s number (aka the fine-structure constant) out of it. Let’s recall the basics of it.Â We had the elementary wavefunction:

Ïˆ =Â aÂ·eâˆ’i[EÂ·t âˆ’ pâˆ™x]/Ä§ =Â aÂ·eâˆ’i[EÂ·t âˆ’ pâˆ™x]/Ä§ = aÂ·cos(pâˆ™x/Ä§ âˆ’ Eâˆ™t/Ä§) + iÂ·aÂ·sin(pâˆ™x/Ä§ âˆ’ Eâˆ™t/Ä§)

In one-dimensional space (think of a particle traveling along some line), the vectors (p and x) become scalars, and so we simply write:

Ïˆ =Â aÂ·eâˆ’i[EÂ·t âˆ’ pâˆ™x]/Ä§ =Â aÂ·eâˆ’i[EÂ·t âˆ’ pâˆ™x]/Ä§ = aÂ·cos(pâˆ™x/Ä§ âˆ’ Eâˆ™t/Ä§) + iÂ·aÂ·sin(pâˆ™x/Ä§ âˆ’ Eâˆ™t/Ä§)

This wavefunction comes with constantÂ probabilities |Ïˆ|2Â  = a2, so we need to define a space outside of whichÂ Ïˆ = 0. Think of the particle-in-a-box model. This is obvious oscillations pack energy, and the energy of our particle is finite. Hence, each particle – be it a photon or an electron – will pack aÂ finiteÂ number of oscillations. It will, therefore, occupy a finite amount of space. Mathematically, this corresponds to the normalization condition:Â all probabilities have to add up to one, as illustrated below.Now, allÂ oscillations of the elementary wavefunction have the same amplitude:Â a. [Terminology is a bit confusing here because we use the term amplitude to refer to two very different things here: we may sayÂ a is the amplitude of the (probability) amplitudeÂ Ïˆ. So how many oscillations do we have? What is theÂ sizeÂ of our box? Let us assume our particle is an electron, and we will reduce its motion to aÂ one-dimensionalÂ motion only: we’re thinking of it as traveling along the x-axis. We can then use the y- andÂ z-axes asÂ mathematical axes only: they will show us how the magnitude and direction of the real and imaginary component ofÂ Ïˆ. The animation below (for which I have to credit Wikipedia) shows how it looks like.Of course, we can have right- as well as left-handed particle waves because, while timeÂ physicallyÂ goes by in one direction only (we can’t reverse time), we can countÂ it in two directions: 1, 2, 3, etcetera orÂ âˆ’1,Â âˆ’2,Â âˆ’3,Â etcetera. In the latter case, think of timeÂ tickingÂ away. ðŸ™‚ Of course, in ourÂ physicalÂ interpretation of the wavefunction, this should explain the (spin) angular momentum of the electron, which is – for some mysterious reason that we now understand ðŸ™‚ – always equal toÂ JÂ =Â Â± Ä§/2.

Now, becauseÂ a is some constant here, we may think of our box as a cylinder along the x-axis. Now, the rest mass of an electron is about 0.510 MeV, so that’s around 8.19Ã—10âˆ’14 Nâˆ™m, so it will pack some 1.24Ã—1020Â oscillations per second. So how long is our cylinder here? To answer that question, we need to calculate theÂ phaseÂ velocity of our wave. We’ll come back to that in a moment. Just note how this compares to a photon: the energy of a photon will typically be a few electronvoltÂ only (1 eVÂ â‰ˆ 1.6 Ã—10âˆ’19Â NÂ·m) and, therefore, it will pack like 1015Â oscillations per second, so that’s a density (in time) that is about 100,000 timesÂ less.

Back to the angular momentum. The classical formula for it isÂ L = IÂ·Ï‰, so that’s angular frequency times angular mass. What’s the angular velocity here? That’s easy:Â Ï‰ =Â E/Ä§. What’s the angular mass? If we think of our particle as a tiny cylinder,Â we may use the formula for its angular mass: I = mÂ·r2/2. We have m: that’s the electron mass, right? Right? So what is r? That should be the magnitude of the rotating vector, right? So that’sÂ a. Of course, the mass-energyÂ equivalence relation tells us that E = mc2, so we can write:

L = IÂ·Ï‰ = (mÂ·r2/2)Â·(E/Ä§) = (1/2)Â·a2Â·mÂ·(mc2/Ä§) = (1/2)Â·a2Â·m2Â·c2/Ä§

Does it make sense? Maybe. Maybe not. You can check the physical dimensions on both sides of the equation, and that works out: we do get something that is expressed in NÂ·mÂ·s, so that’s actionÂ orÂ angular momentumÂ units. Now, weÂ knowÂ L must be equal toÂ JÂ =Â Â± Ä§/2. [As mentioned above, the plus or minus sign depends on the left- or right-handedness of our wavefunction, so don’t worry about that.] How do we know that? Because of the Stern-Gerlach experiment, which has been repeated a zillion times, if not more. Now, if L =Â J, then we get the following equation for a:Â Â This is the formula for the radius of an electron. To be precise, it is theÂ Compton scattering radius, so that’s theÂ effectiveÂ radius of an electron as determined by scattering experiments. You can calculate it:Â it is about 3.8616Ã—10âˆ’13 m, so that’s theÂ picometerÂ scale, as we would expect.

This isÂ a rather spectacular result. As far as I am concerned, it is spectacular enough for me to actuallyÂ believeÂ myÂ interpretation of the wavefunction makes sense.

Let us now try to think about theÂ lengthÂ of our cylinder once again. The period of our wave is equal to T = 1/f = 1/(Ï‰/2Ï€) = 1/[(E/Ä§)Â·2Ï€] =Â 1/(E/h) = h/E. Now, theÂ phaseÂ velocity (vp) will be given by:

vpÂ =Â Î»Â·fÂ = (2Ï€/k)Â·(Ï‰/2Ï€) =Â Ï‰/k =Â (E/Ä§)/(p/Ä§) = E/p = E/(mÂ·vg) = (mÂ·c2)/(mÂ·vg) = c2/vg

This isÂ veryÂ interesting, because it establishes anÂ inverseÂ proportionality betweenÂ the group and the phase velocity of our wave, withÂ c2Â as the coefficient ofÂ inverseÂ proportionality.Â In fact, this equation looks better if we write asÂ vpÂ·vgÂ =Â c2. Of course, theÂ groupÂ velocityÂ (vg) is theÂ classicalÂ velocity of our electron. This equation shows us the idea of an electron at rest doesn’t make sense: ifÂ vgÂ = 0, thenÂ vpÂ times zero must equalÂ c2, which cannot be the case: electronsÂ mustÂ move in space. More generally, speaking, matter-particles must move in space, with the photon as our limiting case: it moves at the speed of light. Hence, for a photon, we find that vpÂ =Â vgÂ = E/p =Â c.

How can we calculate theÂ lengthÂ of a photon or an electron? It is an interesting question. The mentioned orders or magnitude of the frequency (1015Â or 1020) gives us the number of oscillations per second. But how many do we have inÂ oneÂ photon, or inÂ one electron?

Let’s first think about photons, because we have more clues here. Photons are emitted by atomic oscillators: atoms going from one state (energy level) to another. We know how to calculate to calculate the Q of these atomic oscillators (see, for example, Feynman I-32-3):Â it is of the order of 108, which means the wave train will last about 10â€“8Â seconds (to be precise, that is the time it takes for the radiation to die out by a factor 1/e). Now, the frequency of sodium light, for example, is 0.5Ã—1015Â oscillations per second, and the decay time is about 3.2Ã—10â€“8Â seconds, so that makes for (0.5Ã—1015)Â·(3.2Ã—10â€“8) = 16 million oscillations. Now, the wavelength is 600 nanometer (600Ã—10â€“9) m), so that gives us a wavetrain with a length of (600Ã—10â€“9)Â·(16Ã—106) = 9.6 m.

These oscillations may or may not have the same amplitude and, hence, each of these oscillations may pack a different amount of energies. However,Â if the total energy of our sodium light photon (i.e. about 2 eVÂ â‰ˆÂ 3.3Ã—10â€“19Â J) are to be packed in those oscillations, then each oscillation would pack about 2Ã—10â€“26Â J, on average, that is. We speculated in other posts on how we might imagine the actual wave pulse that atoms emit when going from one energy state to another, so we don’t do that again here. However, the following illustration of the decay of a transient signal dies out may be useful.

This calculation is interesting. It also gives us an interesting paradox: if a photon is a pointlike particle, how can we say its length is like 10 meterÂ or more? Relativity theory saves us here. We need to distinguish the reference frame of the photon â€“ riding along the wave as it is being emitted, so to speak â€“ and our stationary reference frame, which is that of the emitting atom. Now, because the photon travels at the speed of light, relativistic length contraction will make it lookÂ like a pointlike particle.

What about the electron? Can we use similarÂ assumptions? For the photon, we can use the decay time to calculate the effective numberÂ of oscillations. What can we use for an electron? We will need to make some assumption about the phase velocity or, what amounts to the same, the group velocity of the particle. What formulas can we use? TheÂ p = mÂ·v is the relativistically correct formula for the momentum of an object if m = mv, so that’s the same m we use in the E =Â mc2Â formula. Of course,Â vÂ here is, obviously, the group velocity (vg), so that’s the classical velocity of our particle. Hence, we can write:

p = mÂ·vgÂ = (E/c2)Â·vgÂ â‡”Â vgÂ = p/m =Â Â pÂ·c2/E

This is just another way of writing thatÂ vgÂ =Â c2/vpÂ or vpÂ =Â c2/vgÂ so it doesn’t help, does it? Maybe. Maybe not. Let us substitute in our formula for the wavelength:

Î» =Â vp/fÂ =Â vpÂ·TÂ =Â vpâ‹…(h/E) = (c2/vg)Â·(h/E) = h/(mÂ·vg) = h/pÂ

This gives us the otherÂ de BroglieÂ relation:Â Î» =Â h/p. This doesn’t help us much, although it is interesting to think about it. TheÂ fÂ = E/h relation is somewhat intuitive: higher energy, higher frequency. In contrast, what the Î» =Â h/p relation tells us that we get an infinite wavelength if the momentum becomes really small. What does this tell us? I am not sure. Frankly, I’ve look at the secondÂ de BroglieÂ relation like a zillion times now, and I think it’s rubbish. It’s meant to be used for the groupÂ velocity, I feel. I am saying that because we get a non-sensical energy formula out of it. Look at this:

1. E = hÂ·f and p = h/Î». Therefore, f = E/h and Î» = p/h.
2. vÂ =Â fÂ·Î» = (E/h)âˆ™(p/h) = E/p
3. p = mÂ·v. Therefore, E = vÂ·p = mÂ·v2

E = mÂ·v2? This formula is only correct ifÂ vÂ =Â c, in which case it becomes theÂ E = mc2 equation. So it then describes a photon, or a massless matter-particle which… Well… That’s a contradictio in terminis. ðŸ™‚ In all other cases, we get nonsense.

Let’s try something differently.Â  If our particle is at rest, then p = 0 and theÂ pÂ·x/Ä§ term in our wavefunction vanishes, so it’s just:

Ïˆ =Â aÂ·eâˆ’iÂ·EÂ·t/Ä§ =Â aÂ·cos(Eâˆ™t/Ä§) âˆ’ iÂ·aÂ·sin(Eâˆ™t/Ä§)

Hence, our wave doesn’t travel. It has the same amplitude at every point in space at any point in time. Both the phase and group velocity become meaningless concepts. TheÂ amplitude variesÂ – because of the sine and cosine – but the probability remains the same:Â |Ïˆ|2Â  = a2. Hmm… So we need to find another way to define the size of our box. One of the formulas I jotted down in my paper in which I analyze the wavefunction as a gravitational waveÂ was this one:

It was a physicalÂ normalization condition: the energy contributions of the waves that make up a wave packet need to add up to the total energy of our wave. Of course, for our elementary wavefunction here, the subscripts vanish and so the formula reduces to E = (E/c2)Â·a2Â·(E2/Ä§2), out of which we get our formula for the scattering radius: aÂ =Â Ä§/mc. Now how do we pack that energy in our cylinder?Â Assuming that energy is distributed uniformly, we’re tempted to write something like E =Â a2Â·l or, looking at the geometry of the situation:

E = Ï€Â·a2Â·l â‡”Â lÂ = E/(Ï€Â·a2)

It’s just the formula for the volume of a cylinder.Â Using the value we got for the Compton scattering radius (aÂ =Â 3.8616Ã—10âˆ’13 m), we find anÂ l that’sÂ equal to (8.19Ã—10âˆ’14)/(Ï€Â·14.9Ã—10âˆ’26) =â‰ˆ 0.175Ã—1012Meter?Â Yes. We get the following formula:

0.175Ã—1012Â m is 175 millionÂ kilometer. That’s – literally – astronomic. It corresponds to 583 light-seconds, or 9.7 light-minutes.Â So that’s about 1.17 times the (average) distance between the Sun and the Earth. You can see that we do need to build a wave packet: that space is a bit too large to look for an electron, right? ðŸ™‚

Could we possibly get some less astronomic proportions? What if weÂ imposeÂ thatÂ lÂ should equalÂ a? We get the following condition:We find that m would have to be equal to m â‰ˆ 1.11Ã—10âˆ’36Â kg. That’s tiny. In fact, it’s equivalent to an energy of aboutÂ  equivalent to 0.623 eV (which you’ll see written as 623 milli-eV. This corresponds to light with a wavelength of about 2 micro-meter (Î¼m), so that’s in the infrared spectrum. It’s a funny formula: we find, basically, that theÂ l/aÂ ratio is proportional to m4. Hmm… What should we think of this? If you have any ideas, let me know !

Post scriptum (3 October 2017):Â The paper is going well. Getting lots of downloads, and the views on my blog are picking up too. But I have been vicious. Substituting BÂ for (1/c)âˆ™iâˆ™EÂ or for âˆ’(1/c)âˆ™iâˆ™EÂ implies a very specific choice of reference frame. The imaginary unit is a two-dimensional concept: it only makes sense when giving it a planeÂ view. Literally. Indeed, myÂ formulas assume the iÂ (or âˆ’i) plane is perpendicular to the direction of propagation of the elementary quantum-mechanical wavefunction. So… Yes. The need for rotation matrices is obvious. But my physicalÂ interpretation of the wavefunction stands. ðŸ™‚