The speed of light as an angular velocity

Over the weekend, I worked on a revised version of my paper on a physical interpretation of the wavefunction. However, I forgot to add the final remarks on the speed of light as an angular velocity. I know… This post is for my faithful followers only. It is dense, but let me add the missing bits here:

12

Advertisements

The geometry of the wavefunction (2)

This post further builds on the rather remarkable results we got in our previous posts. Let us start with the basics once again. The elementary wavefunction is written as:

ψ = a·ei[E·t − px]/ħa·cos(px/ħ − E∙t/ħ) + i·a·sin(px/ħ − E∙t/ħ)

Of course, Nature (or God, as Einstein would put it) does not care about our conventions for measuring an angle (i.e. the phase of our wavefunction) clockwise or counterclockwise and, therefore, the ψ = a·ei[E·t − px]/ħ function is also permitted. We know that cos(θ) = cos(−θ) and sinθ = −sin(θ), so we can write:    

ψ = a·ei[E·t − p∙x]/ħa·cos(E∙t/ħ − px/ħ) + i·a·sin(E∙t/ħ − px/ħ)

= a·cos(px/ħ − E∙t/ħ) − i·a·sin(px/ħ − E∙t/ħ)

The vectors p and x are the momentum and position vector respectively: p = (px, py, pz) and x = (x, y, z). However, if we assume there is no uncertainty about p – not about the direction, and not about the magnitude – then the direction of p can be our x-axis. In this reference frame, x = (x, y, z) reduces to (x, 0, 0), and px/ħ reduces to p∙x/ħ. This amounts to saying our particle is traveling along the x-axis or, if p = 0, that our particle is located somewhere on the x-axis. So we have an analysis in one dimension only then, which facilitates our calculations. The geometry of the wavefunction is then as illustrated below. The x-axis is the direction of propagation, and the y- and z-axes represent the real and imaginary part of the wavefunction respectively.

Note that, when applying the right-hand rule for the axes, the vertical axis is the y-axis, not the z-axis. Hence, we may associate the vertical axis with the cosine component, and the horizontal axis with the sine component. [You can check this as follows: if the origin is the (x, t) = (0, 0) point, then cos(θ) = cos(0) = 1 and sin(θ) = sin(0) = 0. This is reflected in both illustrations, which show a left- and a right-handed wave respectively.]

Now, you will remember that we speculated the two polarizations (left- versus right-handed) should correspond to the two possible values for the quantum-mechanical spin of the wave (+ħ/2 or −ħ/2). We will come back to this at the end of this post. Just focus on the essentials first: the cosine and sine components for the left-handed wave are shown below. Look at it carefully and try to understand. Needless to say, the cosine and sine function are the same, except for a phase difference of π/2: sin(θ) = cos(θ − π/2).

circular polarizaton with components

As for the wave velocity, and its direction of propagation, we know that the (phase) velocity of any waveform F(kx − ωt) is given by vp = ω/k. In our case, we find that vp = ω/k = (E/ħ)/(p/ħ) = E/p. Of course, the momentum might also be in the negative x-direction, in which case k would be equal to −p and, therefore, we would get a negative phase velocity: vp = ω/k = (E/ħ)/(−p/ħ) = −E/p.

As you know, E/ħ = ω gives the frequency in time (expressed in radians per second), while p/ħ = k gives us the wavenumber, or the frequency in space (expressed in radians per meter). [If in doubt, check my post on essential wave math.] Now, you also know that f = ω/2π  and λ = 2π/k, which gives us the two de Broglie relations:

  1. E = ħ∙ω = h∙f
  2. p = ħ∙k = h/λ

The frequency in time (oscillations or radians per second) is easy to interpret. A particle will always have some mass and, therefore, some energy, and it is easy to appreciate the fact that the wavefunction of a particle with more energy (or more mass) will have a higher density in time than a particle with less energy.

However, the second de Broglie relation is somewhat harder to interpret. Note that the wavelength is inversely proportional to the momentum: λ = h/p. Hence, if p goes to zero, then the wavelength becomes infinitely long, so we write:

If p → 0 then λ → ∞.

For the limit situation, a particle with zero rest mass (m0 = 0), the velocity may be c and, therefore, we find that p = mvv = mcc = m∙c (all of the energy is kinetic) and, therefore, p∙c = m∙c2 = E, which we may also write as: E/p = c. Hence, for a particle with zero rest mass (m0 = 0), the wavelength can be written as:

λ = h/p = hc/E = h/mc

Of course, we are talking a photon here. We get the zero rest mass for a photon. In contrast, all matter-particles should have some mass[1] and, therefore, their velocity will never equal c.[2] The question remains: how should we interpret the inverse proportionality between λ and p?

Let us first see what this wavelength λ actually represents. If we look at the ψ = a·cos(p∙x/ħ − E∙t/ħ) − i·a·sin(p∙x/ħ – E∙t/ħ) once more, and if we write p∙x/ħ as Δ, then we can look at p∙x/ħ as a phase factor, and so we will be interested to know for what x this phase factor Δ = p∙x/ħ will be equal to 2π. So we write:

Δ =p∙x/ħ = 2π ⇔ x = 2π∙ħ/p = h/p = λ

So now we get a meaningful interpretation for that wavelength. It is the distance between the crests (or the troughs) of the wave, so to speak, as illustrated below. Of course, this two-dimensional wave has no real crests or troughs: they depend on your frame of reference.

wavelength

So now we know what λ actually represent for our one-dimensional elementary wavefunction. Now, the time that is needed for one cycle is equal to T = 1/f = 2π·(ħ/E). Hence, we can now calculate the wave velocity:

v = λ/T = (h/p)/[2π·(ħ/E)] = E/p

Unsurprisingly, we just get the phase velocity that we had calculated already: v = vp = E/p. It does not answer the question: what if p is zero? What if we are looking at some particle at rest? It is an intriguing question: we get an infinitely long wavelength, and an infinite phase velocity. Now, we know phase velocities can be superluminal, but they should not be infinite. So what does the mathematical inconsistency tell us? Do these infinitely long wavelengths and infinite wave velocities tell us that our particle has to move? Do they tell us our notion of a particle at rest is mathematically inconsistent?

Maybe. But maybe not. Perhaps the inconsistency just tells us our elementary wavefunction – or the concept of a precise energy, and a precise momentum – does not make sense. This is where the Uncertainty Principle comes in: stating that p = 0, implies zero uncertainty. Hence, the σp factor in the σp∙σx ≤ ħ/2 would be zero and, therefore, σp∙σx would be zero which, according to the Uncertainty Principle, it cannot be: it can be very small, but it cannot be zero.

It is interesting to note here that σp refers to the standard deviation from the mean, as illustrated below. Of course, the distribution may be or may not be normal – we don’t know – but a normal distribution makes intuitive sense, of course. Also, if we assume the mean is zero, then the uncertainty is basically about the direction in which our particle is moving, as the momentum might then be positive or negative.

Standard_deviation_diagram

The question of natural units may pop up. The Uncertainty Principle suggests a numerical value of the natural unit for momentum and distance that is equal to the square root of ħ/2, so that’s about 0.726×10−17 m for the distance unit and 0.726×10−17 N∙s for the momentum unit, as the product of both gives us ħ/2. To make this somewhat more real, we may note that 0.726×10−17 m is the attometer scale (1 am = 1×10−18 m), so that is very small but not unreasonably small.[3]

Hence, we need to superimpose a potentially infinite number of waves with energies and momenta centered on some mean value. It is only then that we get meaningful results. For example, the idea of a group velocity – which should correspond to the classical idea of the velocity of our particle – only makes sense in the context of wave packet. Indeed, the group velocity of a wave packet (vg) is calculated as follows:

vg = ∂ωi/∂ki = ∂(Ei/ħ)/∂(pi/ħ) = ∂(Ei)/∂(pi)

This assumes the existence of a dispersion relation which gives us ωi as a function of ki – what amounts to the same – Ei as a function of pi. How do we get that? Well… There are a few ways to go about it but one interesting way of doing it is to re-write Schrödinger’s equation as the following pair of equations[4]:

  1. Re(∂ψ/∂t) = −[ħ/(2meff)]·Im(∇2ψ) ⇔ ω·cos(kx − ωt) = k2·[ħ/(2meff)]·cos(kx − ωt)
  2. Im(∂ψ/∂t) = [ħ/(2meff)]·Re(∇2ψ) ⇔ ω·sin(kx − ωt) = k2·[ħ/(2meff)]·sin(kx − ωt)

These equations imply the following dispersion relation:

ω = ħ·k2/(2m)

Of course, we need to think about the subscripts now: we have ωi, ki, but… What about meff or, dropping the subscript, about m? Do we write it as mi? If so, what is it? Well… It is the equivalent mass of Ei obviously, and so we get it from the mass-energy equivalence relation: mi = Ei/c2. It is a fine point, but one most people forget about: they usually just write m. However, if there is uncertainty in the energy, then Einstein’s mass-energy relation tells us we must have some uncertainty in the (equivalent) mass too, and the two will, obviously, be related as follows: σm = σE/c2. We are tempted to do a few substitutions here. Let’s first check what we get when doing the mi = Ei/c2 substitution:

ωi = ħ·ki2/(2mi) = (1/2)∙ħ·ki2c2/Ei = (1/2)∙ħ·ki2c2/(ωi∙ħ) = (1/2)∙ħ·ki2c2i

⇔ ωi2/ki2 = c2/2 ⇔ ωi/ki = vp = c/2 !?

We get a very interesting but nonsensical condition for the dispersion relation here. I wonder what mistake I made. 😦

Let us try another substitution. The group velocity is what it is, right? It is the velocity of the group, so we can write: ki = p/ħ = mi ·vg. This gives us the following result:

ωi = ħ·(mi ·vg)2/(2mi) = ħ·mi·vg2/2

It is yet another interesting condition for the dispersion relation. Does it make any more sense? I am not so sure. That factor 1/2 troubles us. It only makes sense when we drop it. Now you will object that Schrödinger’s equation gives us the electron orbitals – and many other correct descriptions of quantum-mechanical stuff – so, surely, Schrödinger’s equation cannot be wrong. You’re right. It’s just that… Well… When we are splitting in up in two equations, as we are doing here, then we are looking at one of the two dimensions of the oscillation only and, therefore, it’s only half of the mass that counts. Complicated explanation but… Well… It should make sense, because the results that come out make sense. Think of it. So we write this:

  • Re(∂ψ/∂t) = −(ħ/meffIm(∇2ψ) ⇔ ω·cos(kx − ωt) = k2·(ħ/meff)·cos(kx − ωt)
  • Im(∂ψ/∂t) = (ħ/meffRe(∇2ψ) ⇔ ω·sin(kx − ωt) = k2·(ħ/meff)·sin(kx − ωt)

We then get the dispersion relation without that 1/2 factor:

ωi = ħ·ki2/mi

The mi = Ei/c2 substitution then gives us the result we sort of expected to see:

ωi = ħ·ki2/mi = ħ·ki2c2/Ei = ħ·ki2c2/(ωi∙ħ) ⇔ ωi/ki = vp = c

Likewise, the other calculation also looks more meaningful now:

ωi = ħ·(mi ·vg)2/mi = ħ·mi·vg2

Sweet ! 🙂

Let us put this aside for the moment and focus on something else. If you look at the illustrations above, you see we can sort of distinguish (1) a linear velocity – the speed with which those wave crests (or troughs) move – and (2) some kind of circular or tangential velocity – the velocity along the red contour line above. We’ll need the formula for a tangential velocity: vt = a∙ω.

Now, if λ is zero, then vt = a∙ω = a∙E/ħ is just all there is. We may double-check this as follows: the distance traveled in one period will be equal to 2πa, and the period of the oscillation is T = 2π·(ħ/E). Therefore, vt will, effectively, be equal to vt = 2πa/(2πħ/E) = a∙E/ħ. However, if λ is non-zero, then the distance traveled in one period will be equal to 2πa + λ. The period remains the same: T = 2π·(ħ/E). Hence, we can write:

F1

For an electron, we did this weird calculation. We had an angular momentum formula (for an electron) which we equated with the real-life +ħ/2 or −ħ/2 values of its spin, and we got a numerical value for a. It was the Compton radius: the scattering radius for an electron. Let us write it out:

F2

Using the right numbers, you’ll find the numerical value for a: 3.8616×10−13 m. But let us just substitute the formula itself here: F3

This is fascinating ! And we just calculated that vp is equal to c. For the elementary wavefunction, that is. Hence, we get this amazing result:

vt = 2c

This tangential velocity is twice the linear velocity !

Of course, the question is: what is the physical significance of this? I need to further look at this. Wave velocities are, essentially, mathematical concepts only: the wave propagates through space, but nothing else is really moving. However, the geometric implications are obviously quite interesting and, hence, need further exploration.

One conclusion stands out: all these results reinforce our interpretation of the speed of light as a property of the vacuum – or of the fabric of spacetime itself. 🙂

[1] Even neutrinos should have some (rest) mass. In fact, the mass of the known neutrino flavors was estimated to be smaller than 0.12 eV/c2. This mass combines the three known neutrino flavors.

[2] Using the Lorentz factor (γ), we can write the relativistically correct formula for the kinetic energy as KE = E − E0 = mvc2 − m0c2 = m0γc2 − m0c2 = m0c2(γ − 1). As v approaches c, γ approaches infinity and, therefore, the kinetic energy would become infinite as well.

[3] It is, of course, extremely small, but 1 am is the current sensitivity of the LIGO detector for gravitational waves. It is also thought of as the upper limit for the length of an electron, for quarks, and for fundamental strings in string theory. It is, in any case, 1,000,000,000,000,000,000 times larger than the order of magnitude of the Planck length (1.616229(38)×10−35 m).

[4] The meff is the effective mass of the particle, which depends on the medium. For example, an electron traveling in a solid (a transistor, for example) will have a different effective mass than in an atom. In free space, we can drop the subscript and just write meff = m. As for the equations, they are easily derived from noting that two complex numbers a + i∙b and c + i∙d are equal if, and only if, their real and imaginary parts are the same. Now, the ∂ψ/∂t = i∙(ħ/meff)∙∇2ψ equation amounts to writing something like this: a + i∙b = i∙(c + i∙d). Now, remembering that i2 = −1, you can easily figure out that i∙(c + i∙d) = i∙c + i2∙d = − d + i∙c.

Reality and perception

It’s quite easy to get lost in all of the math when talking quantum mechanics. In this post, I’d like to freewheel a bit. I’ll basically try to relate the wavefunction we’ve derived for the electron orbitals to the more speculative posts I wrote on how to interpret the wavefunction. So… Well… Let’s go. 🙂

If there is one thing you should remember from all of the stuff I wrote in my previous posts, then it’s that the wavefunction for an electron orbital – ψ(x, t), so that’s a complex-valued function in two variables (position and time) – can be written as the product of two functions in one variable:

ψ(x, t) = ei·(E/ħ)·t·f(x)

In fact, we wrote f(x) as ψ(x), but I told you how confusing that is: the ψ(x) and ψ(x, t) functions are, obviously, very different. To be precise, the f(x) = ψ(x) function basically provides some envelope for the two-dimensional eiθ = ei·(E/ħ)·t = cosθ + i·sinθ oscillation – as depicted below (θ = −(E/ħ)·t = ω·t with ω = −E/ħ).Circle_cos_sinWhen analyzing this animation – look at the movement of the green, red and blue dots respectively – one cannot miss the equivalence between this oscillation and the movement of a mass on a spring – as depicted below.spiral_sThe ei·(E/ħ)·t function just gives us two springs for the price of one. 🙂 Now, you may want to imagine some kind of elastic medium – Feynman’s famous drum-head, perhaps 🙂 – and you may also want to think of all of this in terms of superimposed waves but… Well… I’d need to review if that’s really relevant to what we’re discussing here, so I’d rather not make things too complicated and stick to basics.

First note that the amplitude of the two linear oscillations above is normalized: the maximum displacement of the object from equilibrium, in the positive or negative direction, which we may denote by x = ±A, is equal to one. Hence, the energy formula is just the sum of the potential and kinetic energy: T + U = (1/2)·A2·m·ω2 = (1/2)·m·ω2. But so we have two springs and, therefore, the energy in this two-dimensional oscillation is equal to E = 2·(1/2)·m·ω2 = m·ω2.

This formula is structurally similar to Einstein’s E = m·c2 formula. Hence, one may want to assume that the energy of some particle (an electron, in our case, because we’re discussing electron orbitals here) is just the two-dimensional motion of its mass. To put it differently, we might also want to think that the oscillating real and imaginary component of our wavefunction each store one half of the total energy of our particle.

However, the interpretation of this rather bold statement is not so straightforward. First, you should note that the ω in the E = m·ω2 formula is an angular velocity, as opposed to the in the E = m·c2 formula, which is a linear velocity. Angular velocities are expressed in radians per second, while linear velocities are expressed in meter per second. However, while the radian measures an angle, we know it does so by measuring a length. Hence, if our distance unit is 1 m, an angle of 2π rad will correspond to a length of 2π meter, i.e. the circumference of the unit circle. So… Well… The two velocities may not be so different after all.

There are other questions here. In fact, the other questions are probably more relevant. First, we should note that the ω in the E = m·ω2 can take on any value. For a mechanical spring, ω will be a function of (1) the stiffness of the spring (which we usually denote by k, and which is typically measured in newton (N) per meter) and (2) the mass (m) on the spring. To be precise, we write: ω2 = k/m – or, what amounts to the same, ω = √(k/m). Both k and m are variables and, therefore, ω can really be anything. In contrast, we know that c is a constant: equals 299,792,458 meter per second, to be precise. So we have this rather remarkable expression: c = √(E/m), and it is valid for any particle – our electron, or the proton at the center, or our hydrogen atom as a whole. It is also valid for more complicated atoms, of course. In fact, it is valid for any system.

Hence, we need to take another look at the energy concept that is used in our ψ(x, t) = ei·(E/ħ)·t·f(x) wavefunction. You’ll remember (if not, you should) that the E here is equal to En = −13.6 eV, −3.4 eV, −1.5 eV and so on, for = 1, 2, 3, etc. Hence, this energy concept is rather particular. As Feynman puts it: “The energies are negative because we picked our zero point as the energy of an electron located far from the proton. When it is close to the proton, its energy is less, so somewhat below zero. The energy is lowest (most negative) for n = 1, and increases toward zero with increasing n.”

Now, this is the one and only issue I have with the standard physics story. I mentioned it in one of my previous posts and, just for clarity, let me copy what I wrote at the time:

Feynman gives us a rather casual explanation [on choosing a zero point for measuring energy] in one of his very first Lectures on quantum mechanics, where he writes the following: “If we have a “condition” which is a mixture of two different states with different energies, then the amplitude for each of the two states will vary with time according to an equation like a·eiωt, with ħ·ω = E = m·c2. Hence, we can write the amplitude for the two states, for example as:

ei(E1/ħ)·t and ei(E2/ħ)·t

And if we have some combination of the two, we will have an interference. But notice that if we added a constant to both energies, it wouldn’t make any difference. If somebody else were to use a different scale of energy in which all the energies were increased (or decreased) by a constant amount—say, by the amount A—then the amplitudes in the two states would, from his point of view, be:

ei(E1+A)·t/ħ and ei(E2+A)·t/ħ

All of his amplitudes would be multiplied by the same factor ei(A/ħ)·t, and all linear combinations, or interferences, would have the same factor. When we take the absolute squares to find the probabilities, all the answers would be the same. The choice of an origin for our energy scale makes no difference; we can measure energy from any zero we want. For relativistic purposes it is nice to measure the energy so that the rest mass is included, but for many purposes that aren’t relativistic it is often nice to subtract some standard amount from all energies that appear. For instance, in the case of an atom, it is usually convenient to subtract the energy Ms·c2, where Ms is the mass of all the separate pieces—the nucleus and the electrons—which is, of course, different from the mass of the atom. For other problems, it may be useful to subtract from all energies the amount Mg·c2, where Mg is the mass of the whole atom in the ground state; then the energy that appears is just the excitation energy of the atom. So, sometimes we may shift our zero of energy by some very large constant, but it doesn’t make any difference, provided we shift all the energies in a particular calculation by the same constant.”

It’s a rather long quotation, but it’s important. The key phrase here is, obviously, the following: “For other problems, it may be useful to subtract from all energies the amount Mg·c2, where Mg is the mass of the whole atom in the ground state; then the energy that appears is just the excitation energy of the atom.” So that’s what he’s doing when solving Schrödinger’s equation. However, I should make the following point here: if we shift the origin of our energy scale, it does not make any difference in regard to the probabilities we calculate, but it obviously does make a difference in terms of our wavefunction itself. To be precise, its density in time will be very different. Hence, if we’d want to give the wavefunction some physical meaning – which is what I’ve been trying to do all along – it does make a huge difference. When we leave the rest mass of all of the pieces in our system out, we can no longer pretend we capture their energy.

So… Well… There you go. If we’d want to try to interpret our ψ(x, t) = ei·(En/ħ)·t·f(x) function as a two-dimensional oscillation of the mass of our electron, the energy concept in it – so that’s the Ein it – should include all pieces. Most notably, it should also include the electron’s rest energy, i.e. its energy when it is not in a bound state. This rest energy is equal to 0.511 MeV. […] Read this again: 0.511 mega-electronvolt (106 eV), so that’s huge as compared to the tiny energy values we mentioned so far (−13.6 eV, −3.4 eV, −1.5 eV,…).

Of course, this gives us a rather phenomenal order of magnitude for the oscillation that we’re looking at. Let’s quickly calculate it. We need to convert to SI units, of course: 0.511 MeV is about 8.2×10−14 joule (J), and so the associated frequency is equal to ν = E/h = (8.2×10−14 J)/(6.626×10−34 J·s) ≈ 1.23559×1020 cycles per second. Now, I know such number doesn’t say all that much: just note it’s the same order of magnitude as the frequency of gamma rays and… Well… No. I won’t say more. You should try to think about this for yourself. [If you do, think – for starters – about the difference between bosons and fermions: matter-particles are fermions, and photons are bosons. Their nature is very different.]

The corresponding angular frequency is just the same number but multiplied by 2π (one cycle corresponds to 2π radians and, hence, ω = 2π·ν = 7.76344×1020 rad per second. Now, if our green dot would be moving around the origin, along the circumference of our unit circle, then its horizontal and/or vertical velocity would approach the same value. Think of it. We have this eiθ = ei·(E/ħ)·t = ei·ω·t = cos(ω·t) + i·sin(ω·t) function, with ω = E/ħ. So the cos(ω·t) captures the motion along the horizontal axis, while the sin(ω·t) function captures the motion along the vertical axis. Now, the velocity along the horizontal axis as a function of time is given by the following formula:

v(t) = d[x(t)]/dt = d[cos(ω·t)]/dt = −ω·sin(ω·t)

Likewise, the velocity along the vertical axis is given by v(t) = d[sin(ω·t)]/dt = ω·cos(ω·t). These are interesting formulas: they show the velocity (v) along one of the two axes is always less than the angular velocity (ω). To be precise, the velocity approaches – or, in the limit, is equal to – the angular velocity ω when ω·t is equal to ω·= 0, π/2, π or 3π/2. So… Well… 7.76344×1020 meter per second!? That’s like 2.6 trillion times the speed of light. So that’s not possible, of course!

That’s where the amplitude of our wavefunction comes in – our envelope function f(x): the green dot does not move along the unit circle. The circle is much tinier and, hence, the oscillation should not exceed the speed of light. In fact, I should probably try to prove it oscillates at the speed of light, thereby respecting Einstein’s universal formula:

c = √(E/m)

Written like this – rather than as you know it: E = m·c2 – this formula shows the speed of light is just a property of spacetime, just like the ω = √(k/m) formula (or the ω = √(1/LC) formula for a resonant AC circuit) shows that ω, the natural frequency of our oscillator, is a characteristic of the system.

Am I absolutely certain of what I am writing here? No. My level of understanding of physics is still that of an undergrad. But… Well… It all makes a lot of sense, doesn’t it? 🙂

Now, I said there were a few obvious questions, and so far I answered only one. The other obvious question is why energy would appear to us as mass in motion in two dimensions only. Why is it an oscillation in a plane? We might imagine a third spring, so to speak, moving in and out from us, right? Also, energy densities are measured per unit volume, right?

Now that‘s a clever question, and I must admit I can’t answer it right now. However, I do suspect it’s got to do with the fact that the wavefunction depends on the orientation of our reference frame. If we rotate it, it changes. So it’s like we’ve lost one degree of freedom already, so only two are left. Or think of the third direction as the direction of propagation of the wave. 🙂 Also, we should re-read what we wrote about the Poynting vector for the matter wave, or what Feynman wrote about probability currents. Let me give you some appetite for that by noting that we can re-write joule per cubic meter (J/m3) as newton per square meter: J/m3 = N·m/m3 = N/m2. [Remember: the unit of energy is force times distance. In fact, looking at Einstein’s formula, I’d say it’s kg·m2/s2 (mass times a squared velocity), but that simplifies to the same: kg·m2/s2 = [N/(m/s2)]·m2/s2.]

I should probably also remind you that there is no three-dimensional equivalent of Euler’s formula, and the way the kinetic and potential energy of those two oscillations works together is rather unique. Remember I illustrated it with the image of a V-2 engine in previous posts. There is no such thing as a V-3 engine. [Well… There actually is – but not with the third cylinder being positioned sideways.]two-timer-576-px-photo-369911-s-original

But… Then… Well… Perhaps we should think of some weird combination of two V-2 engines. The illustration below shows the superposition of two one-dimensional waves – I think – one traveling east-west and back, and the other one traveling north-south and back. So, yes, we may to think of Feynman’s drum-head again – but combining two-dimensional waves – two waves that both have an imaginary as well as a real dimension

dippArticle-14

Hmm… Not sure. If we go down this path, we’d need to add a third dimension – so w’d have a super-weird V-6 engine! As mentioned above, the wavefunction does depend on our reference frame: we’re looking at stuff from a certain direction and, therefore, we can only see what goes up and down, and what goes left or right. We can’t see what comes near and what goes away from us. Also think of the particularities involved in measuring angular momentum – or the magnetic moment of some particle. We’re measuring that along one direction only! Hence, it’s probably no use to imagine we’re looking at three waves simultaneously!

In any case… I’ll let you think about all of this. I do feel I am on to something. I am convinced that my interpretation of the wavefunction as an energy propagation mechanism, or as energy itself – as a two-dimensional oscillation of mass – makes sense. 🙂

Of course, I haven’t answered one key question here: what is mass? What is that green dot – in reality, that is? At this point, we can only waffle – probably best to just give its standard definition: mass is a measure of inertia. A resistance to acceleration or deceleration, or to changing direction. But that doesn’t say much. I hate to say that – in many ways – all that I’ve learned so far has deepened the mystery, rather than solve it. The more we understand, the less we understand? But… Well… That’s all for today, folks ! Have fun working through it for yourself. 🙂

Post scriptum: I’ve simplified the wavefunction a bit. As I noted in my post on it, the complex exponential is actually equal to ei·[(E/ħ)·− m·φ], so we’ve got a phase shift because of m, the quantum number which denotes the z-component of the angular momentum. But that’s a minor detail that shouldn’t trouble or worry you here.

Re-visiting the speed of light, Planck’s constant, and the fine-structure constant

A brother of mine sent me a link to an article he liked. Now, because we share some interest in physics and math and other stuff, I looked at it and…

Well… I was disappointed. Despite the impressive credentials of its author – a retired physics professor – it was very poorly written. It made me realize how much badly written stuff is around, and I am glad I am no longer wasting my time on it. However, I do owe my brother some explanation of (a) why I think it was bad, and of (b) what, in my humble opinion, he should be wasting his time on. 🙂 So what it is all about?

The article talks about physicists deriving the speed of light from “the electromagnetic properties of the quantum vacuum.” Now, it’s the term ‘quantum‘, in ‘quantum vacuum’, that made me read the article.

Indeed, deriving the theoretical speed of light in empty space from the properties of the classical vacuum – aka empty space – is a piece of cake: it was done by Maxwell himself as he was figuring out his equations back in the 1850s (see my post on Maxwell’s equations and the speed of light). And then he compared it to the measured value, and he saw it was right on the mark. Therefore, saying that the speed of light is a property of the vacuum, or of empty space, is like a tautology: we may just as well put it the other way around, and say that it’s the speed of light that defines the (properties of the) vacuum!

Indeed, as I’ll explain in a moment: the speed of light determines both the electric as well as the magnetic constants μand ε0, which are the (magnetic) permeability and the (electric) permittivity of the vacuum respectively. Both constants depend on the units we are working with (i.e. the units for electric charge, for distance, for time and for force – or for inertia, if you want, because force is defined in terms of overcoming inertia), but so they are just proportionality coefficients in Maxwell’s equations. So once we decide what units to use in Maxwell’s equations, then μand ε0 are just proportionality coefficients which we get from c. So they are not separate constants really – I mean, they are not separate from c – and all of the ‘properties’ of the vacuum, including these constants, are in Maxwell’s equations.

In fact, when Maxwell compared the theoretical value of c with its presumed actual value, he didn’t compare c‘s theoretical value with the speed of light as measured by astronomers (like that 17th century Ole Roemer, to which our professor refers: he had a first go at it by suggesting some specific value for it based on his observations of the timing of the eclipses of one of Jupiter’s moons), but with c‘s value as calculated from the experimental values of μand ε0! So he knew very well what he was looking at. In fact, to drive home the point, it may also be useful to note that the Michelson-Morley experiment – which accurately measured the speed of light – was done some thirty years later. So Maxwell had already left this world by then—very much in peace, because he had solved the mystery all 19th century physicists wanted to solve through his great unification: his set of equations covers it all, indeed: electricity, magnetism, light, and even relativity!

I think the article my brother liked so much does a very lousy job in pointing all of that out, but that’s not why I wouldn’t recommend it. It got my attention because I wondered why one would try to derive the speed of light from the properties of the quantum vacuum. In fact, to be precise, I hoped the article would tell me what the quantum vacuum actually is. Indeed, as far as I know, there’s only one vacuum—one ’empty space’: empty is empty, isn’t it? 🙂 So I wondered: do we have a ‘quantum’ vacuum? And, if so, what is it, really?

Now, that is where the article is really disappointing, I think. The professor drops a few names (like the Max Planck Institute, the University of Paris-Sud, etcetera), and then, promisingly, mentions ‘fleeting excitations of the quantum vacuum’ and ‘virtual pairs of particles’, but then he basically stops talking about quantum physics. Instead, he wanders off to share some philosophical thoughts on the fundamental physical constants. What makes it all worse is that even those thoughts on the ‘essential’ constants are quite off the mark.

So… This post is just a ‘quick and dirty’ thing for my brother which, I hope, will be somewhat more thought-provoking than that article. More importantly, I hope that my thoughts will encourage him to try to grind through better stuff.

On Maxwell’s equations and the properties of empty space

Let me first say something about the speed of light indeed. Maxwell’s four equations may look fairly simple, but that’s only until one starts unpacking all those differential vector equations, and it’s only when going through all of their consequences that one starts appreciating their deep mathematical structure. Let me quickly copy how another blogger jotted them down: 🙂

god-said-maxwell-equation

As I showed in my above-mentioned post, the speed of light (i.e. the speed with which an electromagnetic pulse or wave travels through space) is just one of the many consequences of the mathematical structure of Maxwell’s set of equations. As such, the speed of light is a direct consequence of the ‘condition’, or the properties, of the vacuum indeed, as Maxwell suggested when he wrote that “we can scarcely avoid the inference that light consists in the transverse undulations of the same medium which is the cause of electric and magnetic phenomena”.

Of course, while Maxwell still suggests light needs some ‘medium’ here – so that’s a reference to the infamous aether theory – we now know that’s because he was a 19th century scientist, and so we’ve done away with the aether concept (because it’s a redundant hypothesis), and so now we also know there’s absolutely no reason whatsoever to try to “avoid the inference.” 🙂 It’s all OK, indeed: light is some kind of “transverse undulation” of… Well… Of what?

We analyze light as traveling fields, represented by two vectors, E and B, whose direction and magnitude varies both in space as well as in time. E and B are field vectors, and represent the electric and magnetic field respectively. An equivalent formulation – more or less, that is (see my post on the Liénard-Wiechert potentials) – for Maxwell’s equations when only one (moving) charge is involved is:

E

B

This re-formulation, which is Feynman’s preferred formula for electromagnetic radiation, is interesting in a number of ways. It clearly shows that, while we analyze the electric and magnetic field as separate mathematical entities, they’re one and the same phenomenon really, as evidenced by the B = –er×E/c equation, which tells us the magnetic field from a single moving charge is always normal (i.e. perpendicular) to the electric field vector, and also that B‘s magnitude is 1/times the magnitude of E, so |B| = B = |E|/c = E/c. In short, B is fully determined by E, or vice versa: if we have one of the two fields, we have the other, so they’re ‘one and the same thing’ really—not in a mathematical sense, but in a real sense.

Also note that E and B‘s magnitude is just the same if we’re using natural units, so if we equate c with 1. Finally, as I pointed out in my post on the relativity of electromagnetic fields, if we would switch from one reference frame to another, we’ll have a different mix of E and B, but that different mix obviously describes the same physical reality. More in particular, if we’d be moving with the charges, the magnetic field sort of disappears to re-appear as an electric field. So the Lorentz force F = Felectric + Fmagnetic = qE + qv×B is one force really, and its ‘electric’ and ‘magnetic’ component appear the way they appear in our reference frame only. In some other reference frame, we’d have the same force, but its components would look different, even if they, obviously, would and should add up to the same. [Well… Yes and no… You know there’s relativistic corrections to be made to the forces to, but that’s a minor point, really. The force surely doesn’t disappear!]

All of this reinforces what you know already: electricity and magnetism are part and parcel of one and the same phenomenon, the electromagnetic force field, and Maxwell’s equations are the most elegant way of ‘cutting it up’. Why elegant? Well… Click the Occam tab. 🙂

Now, after having praised Maxwell once more, I must say that Feynman’s equations above have another advantage. In Maxwell’s equations, we see two constants, the electric and magnetic constant (denoted by μand ε0 respectively), and Maxwell’s equations imply that the product of the electric and magnetic constant is the reciprocal of c2: μ0·ε= 1/c2. So here we see εand only, so no μ0, so that makes it even more obvious that the magnetic and electric constant are related one to another through c.

[…] Let me digress briefly: why do we have c2 in μ0·ε= 1/c2, instead of just c? That’s related to the relativistic nature of the magnetic force: think about that B = E/c relation. Or, better still, think about the Lorentz equation F = Felectric + Fmagnetic = qE + qv×B = q[E + (v/c)×(E×er)]: the 1/c factor is there because the magnetic force involves some velocity, and any velocity is always relative—and here I don’t mean relative to the frame of reference but relative to the (absolute) speed of light! Indeed, it’s the v/c ratio (usually denoted by β = v/c) that enters all relativistic formulas. So the left-hand side of the μ0·ε= 1/c2 equation is best written as (1/c)·(1/c), with one of the two 1/c factors accounting for the fact that the ‘magnetic’ force is a relativistic effect of the ‘electric’ force, really, and the other 1/c factor giving us the proper relationship between the magnetic and the electric constant. To drive home the point, I invite you to think about the following:

  • μ0 is expressed in (V·s)/(A·m), while εis expressed in (A·s)/(V·m), so the dimension in which the μ0·εproduct is expressed is [(V·s)/(A·m)]·[(A·s)/(V·m)] = s2/m2, so that’s the dimension of 1/c2.
  • Now, this dimensional analysis makes it clear that we can sort of distribute 1/c2 over the two constants. All it takes is re-defining the fundamental units we use to calculate stuff, i.e. the units for electric charge, for distance, for time and for force – or for inertia, as explained above. But so we could, if we wanted, equate both μ0 as well as εwith 1/c.
  • Now, if we would then equate c with 1, we’d have μ0 = ε= c = 1. We’d have to define our units for electric charge, for distance, for time and for force accordingly, but it could be done, and then we could re-write Maxwell’s set of equations using these ‘natural’ units.

In any case, the nitty-gritty here is less important: the point is that μand εare also related through the speed of light and, hence, they are ‘properties’ of the vacuum as well. [I may add that this is quite obvious if you look at their definition, but we’re approaching the matter from another angle here.]

In any case, we’re done with this. On to the next!

On quantum oscillations, Planck’s constant, and Planck units 

The second thought I want to develop is about the mentioned quantum oscillation. What is it? Or what could it be? An electromagnetic wave is caused by a moving electric charge. What kind of movement? Whatever: the charge could move up or down, or it could just spin around some axis—whatever, really. For example, if it spins around some axis, it will have a magnetic moment and, hence, the field is essentially magnetic, but then, again, E and B are related and so it doesn’t really matter if the first cause is magnetic or electric: that’s just our way of looking at the world: in another reference frame, one that’s moving with the charges, the field would essential be electric. So the motion can be anything: linear, rotational, or non-linear in some irregular way. It doesn’t matter: any motion can always be analyzed as the sum of a number of ‘ideal’ motions. So let’s assume we have some elementary charge in space, and it moves and so it emits some electromagnetic radiation.

So now we need to think about that oscillation. The key question is: how small can it be? Indeed, in one of my previous posts, I tried to explain some of the thinking behind the idea of the ‘Great Desert’, as physicists call it. The whole idea is based on our thinking about the limit: what is the smallest wavelength that still makes sense? So let’s pick up that conversation once again.

The Great Desert lies between the 1032 and 1043 Hz scale. 1032 Hz corresponds to a photon energy of Eγ = h·f = (4×10−15 eV·s)·(1032 Hz) = 4×1017 eV = 400,000 tera-electronvolt (1 TeV = 1012 eV). I use the γ (gamma) subscript in my Eγ symbol for two reasons: (1) to make it clear that I am not talking the electric field E here but energy, and (2) to make it clear we are talking ultra-high-energy gamma-rays here.

In fact, γ-rays of this frequency and energy are theoretical only. Ultra-high-energy gamma-rays are defined as rays with photon energies higher than 100 TeV, which is the upper limit for very-high-energy gamma-rays, which have been observed as part of the radiation emitted by so-called gamma-ray bursts (GRBs): flashes associated with extremely energetic explosions in distant galaxies. Wikipedia refers to them as the ‘brightest’ electromagnetic events know to occur in the Universe. These rays are not to be confused with cosmic rays, which consist of high-energy protons and atomic nuclei stripped of their electron shells. Cosmic rays aren’t rays really and, because they consist of particles with a considerable rest mass, their energy is even higher. The so-called Oh-My-God particle, for example, which is the most energetic particle ever detected, had an energy of 3×1020 eV, i.e. 300 million TeV. But it’s not a photon: its energy is largely kinetic energy, with the rest mass m0 counting for a lot in the m in the E = m·c2 formula. To be precise: the mentioned particle was thought to be an iron nucleus, and it packed the equivalent energy of a baseball traveling at 100 km/h! 

But let me refer you to another source for a good discussion on these high-energy particles, so I can get get back to the energy of electromagnetic radiation. When I talked about the Great Desert in that post, I did so using the Planck-Einstein relation (E = h·f), which embodies the idea of the photon being valid always and everywhere and, importantly, at every scale. I also discussed the Great Desert using real-life light being emitted by real-life atomic oscillators. Hence, I may have given the (wrong) impression that the idea of a photon as a ‘wave train’ is inextricably linked with these real-life atomic oscillators, i.e. to electrons going from one energy level to the next in some atom. Let’s explore these assumptions somewhat more.

Let’s start with the second point. Electromagnetic radiation is emitted by any accelerating electric charge, so the atomic oscillator model is an assumption that should not be essential. And it isn’t. For example, whatever is left of the nucleus after alpha or beta decay (i.e. a nuclear decay process resulting in the emission of an α- or β-particle) it likely to be in an excited state, and likely to emit a gamma-ray for about 10−12 seconds, so that’s a burst that’s about 10,000 times shorter than the 10–8 seconds it takes for the energy of a radiating atom to die out. [As for the calculation of that 10–8 sec decay time – so that’s like 10 nanoseconds – I’ve talked about this before but it’s probably better to refer you to the source, i.e. one of Feynman’s Lectures.]

However, what we’re interested in is not the energy of the photon, but the energy of one cycle. In other words, we’re not thinking of the photon as some wave train here, but what we’re thinking about is the energy that’s packed into a space corresponding to one wavelength. What can we say about that?

As you know, that energy will depend both on the amplitude of the electromagnetic wave as well as its frequency. To be precise, the energy is (1) proportional to the square of the amplitude, and (2) proportional to the frequency. Let’s look at the first proportionality relation. It can be written in a number of ways, but one way of doing it is stating the following: if we know the electric field, then the amount of energy that passes per square meter per second through a surface that is normal to the direction in which the radiation is going (which we’ll denote by S – the s from surface – in the formula below), must be proportional to the average of the square of the field. So we write S ∝ 〈E2〉, and so we should think about the constant of proportionality now. Now, let’s not get into the nitty-gritty, and so I’ll just refer to Feynman for the derivation of the formula below:

S = ε0c·〈E2

So the constant of proportionality is ε0c. [Note that, in light of what we wrote above, we can also write this as S = (1/μ0·c)·〈(c·B)2〉 = (c0)·〈B2〉, so that underlines once again that we’re talking one electromagnetic phenomenon only really.] So that’s a nice and rather intuitive result in light of all of the other formulas we’ve been jotting down. However, it is a ‘wave’ perspective. The ‘photon’ perspective assumes that, somehow, the amplitude is given and, therefore, the Planck-Einstein relation only captures the frequency variable: Eγ = h·f.

Indeed, ‘more energy’ in the ‘wave’ perspective basically means ‘more photons’, but photons are photons: they have a definite frequency and a definite energy, and both are given by that Planck-Einstein relation. So let’s look at that relation by doing a bit of dimensional analysis:

  • Energy is measured in electronvolt or, using SI units, joule: 1 eV ≈ 1.6×10−19 J. Energy is force times distance: 1 joule = 1 newton·meter, which means that a larger force over a shorter distance yields the same energy as a smaller force over a longer distance. The oscillations we’re talking about here involve very tiny distances obviously. But the principle is the same: we’re talking some moving charge q, and the power – which is the time rate of change of the energy – that goes in or out at any point of time is equal to dW/dt = F·v, with W the work that’s being done by the charge as it emits radiation.
  • I would also like to add that, as you know, forces are related to the inertia of things. Newton’s Law basically defines a force as that what causes a mass to accelerate: F = m·a = m·(dv/dt) = d(m·v)/dt = dp/dt, with p the momentum of the object that’s involved. When charges are involved, we’ve got the same thing: a potential difference will cause some current to change, and one of the equivalents of Newton’s Law F = m·a = m·(dv/dt) in electromagnetism is V = L·(dI/dt). [I am just saying this so you get a better ‘feel’ for what’s going on.]
  • Planck’s constant is measured in electronvolt·seconds (eV·s) or in, using SI units, in joule·seconds (J·s), so its dimension is that of (physical) action, which is energy times time: [energy]·[time]. Again, a lot of energy during a short time yields the same energy as less energy over a longer time. [Again, I am just saying this so you get a better ‘feel’ for these dimensions.]
  • The frequency f is the number of cycles per time unit, so that’s expressed per second, i.e. in herz (Hz) = 1/second = s−1.

So… Well… It all makes sense: [x joule] = [6.626×10−34 joule]·[1 second]×[f cycles]/[1 second]. But let’s try to deepen our understanding even more: what’s the Planck-Einstein relation really about?

To answer that question, let’s think some more about the wave function. As you know, it’s customary to express the frequency as an angular frequency ω, as used in the wave function A(x, t) = A0·sin(kx − ωt). The angular frequency is the frequency expressed in radians per second. That’s because we need an angle in our wave function, and so we need to relate x and t to some angle. The way to think about this is as follows: one cycle takes a time T (i.e. the period of the wave) which is equal to T = 1/f. Yes: one second divided by the number of cycles per second gives you the time that’s needed for one cycle. One cycle is also equivalent to our argument ωt going around the full circle (i.e. 2π), so we write:  ω·T = 2π and, therefore:

ω = 2π/T = 2π·f

Now we’re ready to play with the Planck-Einstein relation. We know it gives us the energy of one photon really, but what if we re-write our equation Eγ = h·f as Eγ/f = h? The dimensions in this equation are:

[x joule]·[1 second]/[cyles] = [6.626×10−34 joule]·[1 second]

⇔ = 6.626×10−34 joule per cycle

So that means that the energy per cycle is equal to 6.626×10−34 joule, i.e. the value of Planck’s constant.

Let me rephrase truly amazing result, so you appreciate it—perhaps: regardless of the frequency of the light (or our electromagnetic wave, in general) involved, the energy per cycle, i.e. per wavelength or per period, is always equal to 6.626×10−34 joule or, using the electronvolt as the unit, 4.135667662×10−15 eV. So, in case you wondered, that is the true meaning of Planck’s constant!

Now, if we have the frequency f, we also have the wavelength λ, because the velocity of the wave is the frequency times the wavelength: = λ·f and, therefore, λ = c/f. So if we increase the frequency, the wavelength becomes smaller and smaller, and so we’re packing the same amount of energy – admittedly, 4.135667662×10−15 eV is a very tiny amount of energy – into a space that becomes smaller and smaller. Well… What’s tiny, and what’s small? All is relative, of course. 🙂 So that’s where the Planck scale comes in. If we pack that amount of energy into some tiny little space of the Planck dimension, i.e. a ‘length’ of 1.6162×10−35 m, then it becomes a tiny black hole, and it’s hard to think about how that would work.

[…] Let me make a small digression here. I said it’s hard to think about black holes but, of course, it’s not because it’s ‘hard’ that we shouldn’t try it. So let me just mention a few basic facts. For starters, black holes do emit radiation! So they swallow stuff, but they also spit stuff out. More in particular, there is the so-called Hawking radiation, as Roger Penrose and Stephen Hawking discovered.

Let me quickly make a few remarks on that: Hawking radiation is basically a form of blackbody radiation, so all frequencies are there, as shown below: the distribution of the various frequencies depends on the temperature of the black body, i.e. the black hole in this case. [The black curve is the curve that Lord Rayleigh and Sir James Jeans derived in the late 19th century, using classical theory only, so that’s the one that does not correspond to experimental fact, and which led Max Planck to become the ‘reluctant’ father of quantum mechanics. In any case, that’s history and so I shouldn’t dwell on this.]

600px-Black_body

The interesting thing about blackbody radiation, including Hawking radiation, is that it reduces energy and, hence, the equivalent mass of our blackbody. So Hawking radiation reduces the mass and energy of black holes and is therefore also known as black hole evaporation. So black holes that lose more mass than they gain through other means are expected to shrink and ultimately vanish. Therefore, there’s all kind of theories that say why micro black holes, like that Planck scale black hole we’re thinking of right now, should be much larger net emitters of radiation than large black holes and, hence, whey they should shrink and dissipate faster.

Hmm… Interesting… What do we do with all of this information? Well… Let’s think about it as we continue our trek on this long journey to reality over the next year or, more probably, years (plural). 🙂

The key lesson here is that space and time are intimately related because of the idea of movement, i.e. the idea of something having some velocity, and that it’s not so easy to separate the dimensions of time and distance in any hard and fast way. As energy scales become larger and, therefore, our natural time and distance units become smaller and smaller, it’s the energy concept that comes to the fore. It sort of ‘swallows’ all other dimensions, and it does lead to limiting situations which are hard to imagine. Of course, that just underscores the underlying unity of Nature, and the mysteries involved.

So… To relate all of this back to the story that our professor is trying to tell, it’s a simple story really. He’s talking about two fundamental constants basically, c and h, pointing out that c is a property of empty space, and h is related to something doing something. Well… OK. That’s really nothing new, and surely not ground-breaking research. 🙂

Now, let me finish my thoughts on all of the above by making one more remark. If you’ve read a thing or two about this – which you surely have – you’ll probably say: this is not how people usually explain it. That’s true, they don’t. Anything I’ve seen about this just associates the 1043 Hz scale with the 1028 eV energy scale, using the same Planck-Einstein relation. For example, the Wikipedia article on micro black holes writes that “the minimum energy of a microscopic black hole is 1019 GeV [i.e. 1028 eV], which would have to be condensed into a region on the order of the Planck length.” So that’s wrong. I want to emphasize this point because I’ve been led astray by it for years. It’s not the total photon energy, but the energy per cycle that counts. Having said that, it is correct, however, and easy to verify, that the 1043 Hz scale corresponds to a wavelength of the Planck scale: λ = c/= (3×10m/s)/(1043 s−1) = 3×10−35 m. The confusion between the photon energy and the energy per wavelength arises because of the idea of a photon: it travels at the speed of light and, hence, because of the relativistic length contraction effect, it is said to be point-like, to have no dimension whatsoever. So that’s why we think of packing all of its energy in some infinitesimally small place. But you shouldn’t think like that. The photon is dimensionless in our reference frame: in its own ‘world’, it is spread out, so it is a wave train. And it’s in its ‘own world’ that the contradictions start… 🙂

OK. Done!

My third and final point is about what our professor writes on the fundamental physical constants, and more in particular on what he writes on the fine-structure constant. In fact, I could just refer you to my own post on it, but that’s probably a bit too easy for me and a bit difficult for you 🙂 so let me summarize that post and tell you what you need to know about it.

The fine-structure constant

The fine-structure constant α is a dimensionless constant which also illustrates the underlying unity of Nature, but in a way that’s much more fascinating than the two or three things the professor mentions. Indeed, it’s quite incredible how this number (α = 0.00729735…, but you’ll usually see it written as its reciprocal, which is a number that’s close to 137.036…) links charge with the relative speeds, radii, and the mass of fundamental particles and, therefore, how this number also these concepts with each other. And, yes, the fact that it is, effectively, dimensionless, unlike h or c, makes it even more special. Let me quickly sum up what the very same number α all stands for:

(1) α is the square of the electron charge expressed in Planck units: α = eP2.

(2) α is the square root of the ratio of (a) the classical electron radius and (b) the Bohr radius: α = √(re /r). You’ll see this more often written as re = α2r. Also note that this is an equation that does not depend on the units, in contrast to equation 1 (above), and 4 and 5 (below), which require you to switch to Planck units. It’s the square of a ratio and, hence, the units don’t matter. They fall away.

(3) α is the (relative) speed of an electron: α = v/c. [The relative speed is the speed as measured against the speed of light. Note that the ‘natural’ unit of speed in the Planck system of units is equal to c. Indeed, if you divide one Planck length by one Planck time unit, you get (1.616×10−35 m)/(5.391×10−44 s) = m/s. However, this is another equation, just like (2), that does not depend on the units: we can express v and c in whatever unit we want, as long we’re consistent and express both in the same units.]

(4) α is also equal to the product of (a) the electron mass (which I’ll simply write as me here) and (b) the classical electron radius re (if both are expressed in Planck units): α = me·re. Now think that’s, perhaps, the most amazing of all of the expressions for α. [If you don’t think that’s amazing, I’d really suggest you stop trying to study physics. :-)]

Also note that, from (2) and (4), we find that:

(5) The electron mass (in Planck units) is equal me = α/r= α/α2r = 1/αr. So that gives us an expression, using α once again, for the electron mass as a function of the Bohr radius r expressed in Planck units.

Finally, we can also substitute (1) in (5) to get:

(6) The electron mass (in Planck units) is equal to me = α/r = eP2/re. Using the Bohr radius, we get me = 1/αr = 1/eP2r.

So… As you can see, this fine-structure constant really links all of the fundamental properties of the electron: its charge, its radius, its distance to the nucleus (i.e. the Bohr radius), its velocity, its mass (and, hence, its energy),…

So… Why is what it is?

Well… We all marvel at this, but what can we say about it, really? I struggle how to interpret this, just as much – or probably much more 🙂 – as the professor who wrote the article I don’t like (because it’s so imprecise, and that’s what made me write all what I am writing here).

Having said that, it’s obvious that it points to a unity beyond these numbers and constants that I am only beginning to appreciate for what it is: deep, mysterious, and very beautiful. But so I don’t think that professor does a good job at showing how deep, mysterious and beautiful it all is. But then that’s up to you, my brother and you, my imaginary reader, to judge, of course. 🙂

[…] I forgot to mention what I mean with ‘Planck units’. Well… Once again, I should refer you to one of my other posts. But, yes, that’s too easy for me and a bit difficult for you. 🙂 So let me just note we get those Planck units by equating not less than five fundamental physical constants to 1, notably (1) the speed of light, (2) Planck’s (reduced) constant, (3) Boltzmann’s constant, (4) Coulomb’s constant and (5) Newton’s constant (i.e. the gravitational constant). Hence, we have a set of five equations here (ħ = kB = ke = G = 1), and so we can solve that to get the five Planck units, i.e. the Planck length unit, the Planck time unit, the Planck mass unit, the Planck energy unit, the Planck charge unit and, finally (oft forgotten), the Planck temperature unit. Of course, you should note that all mass and energy units are directly related because of the mass-energy equivalence relation E = mc2, which simplifies to E = m if c is equated to 1. [I could also say something about the relation between temperature and (kinetic) energy, but I won’t, as it would only further confuse you.]

OK. Done! 🙂

Addendum: How to think about space and time?

If you read the argument on the Planck scale and constant carefully, then you’ll note that it does not depend on the idea of an indivisible photon. However, it does depend on that Planck-Einstein relation being valid always and everywhere. Now, the Planck-Einstein relation is, in its essence, a fairly basic result from classical electromagnetic theory: it incorporates quantum theory – remember: it’s the equation that allowed Planck to solve the black-body radiation problem, and so it’s why they call Planck the (reluctant) ‘Father of Quantum Theory’ – but it’s not quantum theory.

So the obvious question is: can we make this reflection somewhat more general, so we can think of the electromagnetic force as an example only. In other words: can we apply the thoughts above to any force and any movement really?

The truth is: I haven’t advanced enough in my little study to give the equations for the other forces. Of course, we could think of gravity, and I developed some thoughts on how gravity waves might look like, but nothing specific really. And then we have the shorter-range nuclear forces, of course: the strong force, and the weak force. The laws involved are very different. The strong force involves color charges, and the way distances work is entirely different. So it would surely be some different analysis. However, the results should be the same. Let me offer some thoughts though:

  • We know that the relative strength of the nuclear force is much larger, because it pulls like charges (protons) together, despite the strong electromagnetic force that wants to push them apart! So the mentioned problem of trying to ‘pack’ some oscillation in some tiny little space should be worse with the strong force. And the strong force is there, obviously, at tiny little distances!
  • Even gravity should become important, because if we’ve got a lot of energy packed into some tiny space, its equivalent mass will ensure the gravitational forces also become important. In fact, that’s what the whole argument was all about!
  • There’s also all this talk about the fundamental forces becoming one at the Planck scale. I must, again, admit my knowledge is not advanced enough to explain how that would be possible, but I must assume that, if physicists are making such statements, the argument must be fairly robust.

So… Whatever charge or whatever force we are talking about, we’ll be thinking of waves or oscillations—or simply movement, but it’s always a movement in a force field, and so there’s power and energy involved (energy is force times distance, and power is the time rate of change of energy). So, yes, we should expect the same issues in regard to scale. And so that’s what’s captured by h.

As we’re talking the smallest things possible, I should also mention that there are also other inconsistencies in the electromagnetic theory, which should (also) have their parallel for other forces. For example, the idea of a point charge is mathematically inconsistent, as I show in my post on fields and charges. Charge, any charge really, must occupy some space. It cannot all be squeezed into one dimensionless point. So the reasoning behind the Planck time and distance scale is surely valid.

In short, the whole argument about the Planck scale and those limits is very valid. However, does it imply our thinking about the Planck scale is actually relevant? I mean: it’s not because we can imagine how things might look like  – they may look like those tiny little black holes, for example – that these things actually exist. GUT or string theorists obviously think they are thinking about something real. But, frankly, Feynman had a point when he said what he said about string theory, shortly before his untimely death in 1988: “I don’t like that they’re not calculating anything. I don’t like that they don’t check their ideas. I don’t like that for anything that disagrees with an experiment, they cook up an explanation—a fix-up to say, ‘Well, it still might be true.'”

It’s true that the so-called Standard Model does not look very nice. It’s not like Maxwell’s equations. It’s complicated. It’s got various ‘sectors’: the electroweak sector, the QCD sector, the Higgs sector,… So ‘it looks like it’s got too much going on’, as a friend of mine said when he looked at a new design for mountainbike suspension. 🙂 But, unlike mountainbike designs, there’s no real alternative for the Standard Model. So perhaps we should just accept it is what it is and, hence, in a way, accept Nature as we can see it. So perhaps we should just continue to focus on what’s here, before we reach the Great Desert, rather than wasting time on trying to figure out how things might look like on the other side, especially because we’ll never be able to test our theories about ‘the other side.’

On the other hand, we can see where the Great Desert sort of starts (somewhere near the 1032 Hz scale), and so it’s only natural to think it should also stop somewhere. In fact, we know where it stops: it stops at the 1043 Hz scale, because everything beyond that doesn’t make sense. The question is: is there actually there? Like fundamental strings or whatever you want to call it. Perhaps we should just stop where the Great Desert begins. And what’s the Great Desert anyway? Perhaps it’s a desert indeed, and so then there is absolutely nothing there. 🙂

Hmm… There’s not all that much one can say about it. However, when looking at the history of physics, there’s one thing that’s really striking. Most of what physicists can think of, in the sense that it made physical sense, turned out to exist. Think of anti-matter, for instance. Paul Dirac thought it might exist, that it made sense to exist, and so everyone started looking for it, and Carl Anderson found in a few years later (in 1932). In fact, it had been observed before, but people just didn’t pay attention, so they didn’t want to see it, in a way. […] OK. I am exaggerating a bit, but you know what I mean. The 1930s are full of examples like that. There was a burst of scientific creativity, as the formalism of quantum physics was being developed, and the experimental confirmations of the theory just followed suit.

In the field of astronomy, or astrophysics I should say, it was the same with black holes. No one could really imagine the existence of black holes until the 1960s or so: they were thought of a mathematical curiosity only, a logical possibility. However, the circumstantial evidence now is quite large and so… Well… It seems a lot of what we can think of actually has some existence somewhere. 🙂

So… Who knows? […] I surely don’t. And so I need to get back to the grind and work my way through the rest of Feynman’s Lectures and the related math. However, this was a nice digression, and so I am grateful to my brother he initiated it. 🙂