# 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™‚