Electron and photon strings

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 unit charge), then that’s kinda logical, right? 🙂 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. 🙂 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. We had the elementary wavefunction:

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

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

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

This wavefunction comes with constant probabilities |ψ|2  = a2, so we need to define a space outside of which ψ = 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 a finite number of oscillations. It will, therefore, occupy a finite amount of space. Mathematically, this corresponds to the normalization condition: all probabilities have to add up to one, as illustrated below.probability in a boxNow, all oscillations of the elementary wavefunction have the same amplitude: a. [Terminology is a bit confusing here because we use the term amplitude to refer to two very different things here: we may say a is the amplitude of the (probability) amplitude ψ. So how many oscillations do we have? What is the size of our box? Let us assume our particle is an electron, and we will reduce its motion to a one-dimensional motion only: we’re thinking of it as traveling along the x-axis. We can then use the y- and z-axes as mathematical axes only: they will show us how the magnitude and direction of the real and imaginary component of ψ. The animation below (for which I have to credit Wikipedia) shows how it looks like.wavicle animationOf course, we can have right- as well as left-handed particle waves because, while time physically goes by in one direction only (we can’t reverse time), we can count it in two directions: 1, 2, 3, etcetera or −1, −2, −3, etcetera. In the latter case, think of time ticking away. 🙂 Of course, in our physical interpretation of the wavefunction, this should explain the (spin) angular momentum of the electron, which is – for some mysterious reason that we now understand 🙂 – always equal to = ± ħ/2.

Now, because 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 8.19×10−14 N∙m, so it will pack some 1.24×1020 oscillations per second. So how long is our cylinder here? To answer that question, we need to calculate the phase 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 electronvolt only (1 eV ≈ 1.6 ×10−19 N·m) and, therefore, it will pack like 1015 oscillations per second, so that’s a density (in time) that is about 100,000 times less.

Back to the angular momentum. The classical formula for it is L = I·ω, so that’s angular frequency times angular mass. What’s the angular velocity here? That’s easy: ω = E/ħ. What’s the angular mass? If we think of our particle as a tiny cylinder, we may use the formula for its angular mass: I = m·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’s a. Of course, the mass-energy equivalence relation tells us that E = mc2, so we can write:

L = I·ω = (m·r2/2)·(E/ħ) = (1/2)·a2·m·(mc2/ħ) = (1/2)·a2·m2·c2

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 N·m·s, so that’s action or angular momentum units. Now, we know L must be equal to = ± ħ/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 = J, then we get the following equation for a:  Compton radius formulaThis is the formula for the radius of an electron. To be precise, it is the Compton scattering radius, so that’s the effective radius of an electron as determined by scattering experiments. You can calculate it: it is about 3.8616×10−13 m, so that’s the picometer scale, as we would expect.

This is a rather spectacular result. As far as I am concerned, it is spectacular enough for me to actually believe my interpretation of the wavefunction makes sense.

Let us now try to think about the length of our cylinder once again. The period of our wave is equal to T = 1/f = 1/(ω/2π) = 1/[(E/ħ)·2π] = 1/(E/h) = h/E. Now, the phase velocity (vp) will be given by:

vp = λ·= (2π/k)·(ω/2π) = ω/k = (E/ħ)/(p/ħ) = E/p = E/(m·vg) = (m·c2)/(m·vg) = c2/vg

This is very interesting, because it establishes an inverse proportionality between the group and the phase velocity of our wave, with c2 as the coefficient of inverse proportionality. In fact, this equation looks better if we write as vp·vg = c2. Of course, the group velocity (vg) is the classical velocity of our electron. This equation shows us the idea of an electron at rest doesn’t make sense: if vg = 0, then vp times zero must equal c2, which cannot be the case: electrons must 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 vp = vg = E/p = c.

How can we calculate the length of a photon or an electron? It is an interesting question. The mentioned orders or magnitude of the frequency (1015 or 1020) gives us the number of oscillations per second. But how many do we have in one photon, or in 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): it is of the order of 108, which means the wave train will last about 10–8 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 0.5×1015 oscillations per second, and the decay time is about 3.2×10–8 seconds, so that makes for (0.5×1015)·(3.2×10–8) = 16 million oscillations. Now, the wavelength is 600 nanometer (600×10–9) m), so that gives us a wavetrain with a length of (600×10–9)·(16×106) = 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, if the total energy of our sodium light photon (i.e. about 2 eV ≈ 3.3×10–19 J) are to be packed in those oscillations, then each oscillation would pack about 2×10–26 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.decay-time1

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 meter or more? Relativity theory saves us here. We need to distinguish the reference frame of the photon – riding along the wave as it is being emitted, so to speak – 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 look like a pointlike particle.

What about the electron? Can we use similar assumptions? For the photon, we can use the decay time to calculate the effective number 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? The p = m·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 = mc2 formula. Of course, v here is, obviously, the group velocity (vg), so that’s the classical velocity of our particle. Hence, we can write:

p = m·vg = (E/c2vg ⇔ vg = p/m =  p·c2/E

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

λ = vp/f = vp·T = vp⋅(h/E) = (c2/vg)·(h/E) = h/(m·vg) = h/p 

This gives us the other de Broglie relation: λ = h/p. This doesn’t help us much, although it is interesting to think about it. The = E/h relation is somewhat intuitive: higher energy, higher frequency. In contrast, what the λ = 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 second de Broglie relation like a zillion times now, and I think it’s rubbish. It’s meant to be used for the group velocity, I feel. I am saying that because we get a non-sensical energy formula out of it. Look at this:

  1. E = h·f and p = h/λ. Therefore, f = E/h and λ = p/h.
  2. v = λ = (E/h)∙(p/h) = E/p
  3. p = m·v. Therefore, E = v·p = m·v2

E = m·v2? This formula is only correct if c, in which case it becomes the E = mc2 equation. So it then describes a photon, or a massless matter-particle which… Well… That’s a contradictio in terminis. 🙂 In all other cases, we get nonsense.

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

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

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. The amplitude varies – because of the sine and cosine – but the probability remains the same: |ψ|2  = 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 wave was this one:F1

It was a physical 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/c2a2·(E22), out of which we get our formula for the scattering radius: = ħ/mc. Now how do we pack that energy in our cylinder? Assuming that energy is distributed uniformly, we’re tempted to write something like E = a2·l or, looking at the geometry of the situation:

E = π·a2·l ⇔ = E/(π·a2)

It’s just the formula for the volume of a cylinder. Using the value we got for the Compton scattering radius (= 3.8616×10−13 m), we find an l that’s equal to (8.19×10−14)/(π·14.9×10−26) =≈ 0.175×1012Meter? Yes. We get the following formula:

length formula

0.175×1012 m is 175 million kilometer. That’s – literally – astronomic. It corresponds to 583 light-seconds, or 9.7 light-minutes. 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? 🙂

Could we possibly get some less astronomic proportions? What if we impose that should equal a? We get the following condition:l over aWe find that m would have to be equal to m ≈ 1.11×10−36 kg. That’s tiny. In fact, it’s equivalent to an energy of about  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 (μm), so that’s in the infrared spectrum. It’s a funny formula: we find, basically, that the l/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): 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 B for (1/c)∙iE or for −(1/c)∙iE implies a very specific choice of reference frame. The imaginary unit is a two-dimensional concept: it only makes sense when giving it a plane view. Literally. Indeed, my formulas assume the i (or −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 physical interpretation of the wavefunction stands. 🙂


Taking the magic out of God’s number: some additional reflections

In my previous post, I explained why the fine-structure constant α is not a ‘magical’ number, even if it relates all fundamental properties of the electron: its mass, its energy, its charge, its radius, its photon scattering cross-section (i.e. the Bohr radius, or the size of the atom really) and, finally, the coupling constant for photon-electron interactions. The key to such understanding of α was the model of an electron as a tiny ball of charge. As such, we have two energy formulas for it. One is the energy that’s needed to assemble the charge from infinitely dispersed infinitesimal charges, which we denoted as Uelec. The other formula is the energy of the field of the tiny ball of charge, which we denoted as Eelec.

The formula for Eelec is calculated using the formula for the field momentum of a moving charge and, using the m = E/cmas-energy equivalence relationship, is equivalent to the electromagnetic mass. We went through the derivation in our previous post, so let me just jot down the result:

emm - 2

The second formula depends on what ball of charge we’re thinking of, because the formulas for a charged sphere and a spherical shell of charge are different: both have the same structure as the relationship above (so the energy is also proportional to the square of the electron charge and inversely proportional to the radius a), but the constant of proportionality is different. For a sphere of charge, we write:

 f sphre

For a spherical shell of charge we write:


To compare the formulas, you need to note that the square of the electron charge in the formula for the field energy is equal to e2 = qe2/4πε= ke·qe2. So we multiply the square of the actual electron charge by the Coulomb constant k= 1/4πε0. As you can see, the three formulas have exactly the same form then. It’s just the proportionality constant that’s different: it’s 2/3, 3/5 and 1/2 respectively. It’s interesting to quickly reflect on the dimensions here: [ke] ≈ 9×109 N·m2/C2, so e2 is expressed in N·m2. That makes the units come out alright, as we divide by a (so that’s in meter) and so we get the energy in joule (which is newton·meter). In fact, now that we’re here, let’s quickly calculate the value of e2: it’s that ke·qe2 product, so it’s equal to 2.3×10−28 N·m2. We can quickly check this value because we know that the classical electron radius is equal to:

classical electron radius

So we divide 2.3×10−28 N·mby mec≈ 8.2×10−14 J, so we get r≈ 2.82×10−15 m. So we’re spot on! Why did I do this check? Not really to check what I wrote. It’s more to show what’s going on. We’ve got yet another formula relating the energy and the radius of an electron here, so now we have three. In fact we have more because the formula for Uelec depends on the finer details of our model for the electron (sphere versus shell, uniform versus non-uniform distribution):

  1. Eelec = (2/3)·(e2/a): This is the formula for the energy of the field, so we may all it is external energy.
  2. Uelec = (3/5)·(e2/a), or Uelec = (1/2)·(e2/a): This is the energy needed to assemble our electron, so we might, perhaps, call it its internal energy. The first formula assumes our electron is a uniformly charged sphere. The second assumes all charges sit on the surface of the sphere. If we drop the assumption of the charge having to be uniformly distributed, we’ll find yet another formula.
  3. mece2/r0: This is the energy associated with the so-called classical electron radius (r0) and the electron’s rest mass (me).

In our previous posts, we assumed the last equation was the right one. Why? Because it’s the one that’s been verified experimentally. The discrepancies between the various proportionality coefficients – i.e. the difference between 2/3 and 1, basically – are to be explained because of the binding forces within the electron, without which the electron would just ‘explode’, as the French physicist and polymath Henri Poincaré famously put itIndeed, if the electron is a little ball of negative charge, the repulsive forces between its parts should rip it apart. So we will not say anything more about this. You can have fun yourself by googling all the various theories that try to model these binding forces. [I may do the same some day, but now I’ve got other priorities: I want to move to Feynman’s third volume of Lectures, which is devoted to quantum physics only, so I look very much forward to that.]

In this post, I just wanted to reflect once more on what constants are really fundamental and what constants are somewhat less fundamental. From all what I wrote in my previous post, I said there were three:

  1. The fine-structure constant α, which is a dimensionless number.
  2. Planck’s constant h, whose dimension is joule·second, so that’s the dimension of action.
  3. The speed of light c, whose dimension is that of a velocity.

The three are related through the following expression:

alpha re-expressed

This is an interesting expression. Let’s first check its dimension. We already explained that e2 is expressed in N·m2. That’s rather strange, because it means the dimension of e itself is N1/2·m: what’s the square root of a force of one newton? In fact, to interpret the formula above, it’s probably better to re-write eas e2 = qe2/4πε= ke·qe2. That shows you how the electron charge and Coulomb’s constant are related. Of course, they are part and parcel of one and the same force lawCoulomb’s law. We don’t need anything else, except for relativity theory, because we need to explain the magnetic force as well—and that we can do because magnetism is just a relativistic effect. Think of the field momentum indeed: the magnetic field comes into play only when we start to move our electron. The relativity effect is captured by c  in that formula for α above. As for ħ, ħ = h/2π comes with the E = h·f equation, which links us to the electron’s Compton wavelength λ through the de Broglie relation λ = h/p.

The point is: we should probably not look at α as a ‘fundamental physical constant’. It’s e2 that’s the third fundamental constant, besides h and c. Indeed, it’s from e2 that all the rest follows: the electron’s internal energy, its external energy, and its radius, and then all the rest by combining stuff with other stuff.

Now, we took the magic out of α by doing what we did in the previous posts, and that’s to combine stuff with other stuff, and so now you may think I am putting the magic back in with that formula for α, which seems to define α in terms of the three mentioned ‘fundamental’ constants. That’s not the case: this relation comes out of all of the other relationships we found, and so it’s nothing new really. It’s actually not a definition of α: it just does what it does, and that’s to relate α to the ‘fundamental’ physical constants behind.

So… No new magic. In fact, I want to close this post by taking away even more of the magic. If you read my previous post, I said that α was ‘God’s cut-off factor’ 🙂 ensuring our energy functions do not blow up, but I also said it was impossible to say why he chose 0.00729735256 as the cut-off factor. The question is actually easily answered by thinking about those two formulas we had for the internal and external energy respectively. Let’s re-write them in natural units and, temporarily, two different subscripts for α, so we write:

  1. Eelec = αe/r0: This is the formula for the energy of the field.
  2. Uelec = αu/r0: This is the energy needed to assemble our electron.

Both energies are determined by the above-mentioned laws, i.e. Coulomb’s Law and the theory of relativity, so α has got nothing to do what that. However, both energies have to be the same, and so αhas to be equal to αu. In that sense, α is, quite simply, a proportionality constant that achieves that equality. Now that explains why we can derive α from the three other constants which, as mentioned above, are probably more fundamental. In fact, we’ve got only three degrees of freedom here, so if we chose c, h and as ‘fundamental’, then α isn’t any more.

The underlying deep question behind it all is why those two energies should be equal. Why would our electron have some internal energy if it’s elementary? The answer to that question is: because it has some non-zero radius, and it has some non-zero radius because we don’t want our formula for the field energy (or the field momentum) to blow up. Now, if it has some radius, then it has to have some internal energy.

You’ll say: that makes sense, but it doesn’t answer the question. Why would it have internal energy, with or without a zero radius? If an electron is an elementary particle, then it’s really elementary, isn’t? And so then we shouldn’t try to ‘assemble’ it from an infinite number of infinitesimally small charges. You’re right, and here we can also note that the fact that the electron doesn’t blow up is firm evidence it’s very elementary indeed.

I should also note that Feynman actually doesn’t talk about the energy that’s needed to assemble a charge: he gets his Uelec = (1/2)·(e2/a) by calculating the external field energy for a spherical shell of charge, and he sticks to it—presumably because it’s the same field for a uniform or non-uniform sphere of charge. He only notes there has to be some radius because, if not, the formula he uses blows up, indeed. So – who knows? – perhaps he doesn’t quite believe that formula for the internal energy is relevant either.

So perhaps there is no internal energy indeed. Perhaps there’s just the energy of the field. So… Well… I can’t say much about this… Except… Well… Perhaps just one more thing. Let me note something that, I hope, you noticed as well: the ke·qe2 is the numerator in Coulomb’s Law itself. You also know that energy equals force times distance. So if we divide both sides by r0, we get Coulomb’s Law itself Felec = ke·qe2/r02. The only thing is: what’s the distance? It’s one charge only, and there is no distance between one charge, is there? Well… Yes and no. I have been thinking that the requirement of the internal and external energies being equal resembles the statement that the forces between two charges are equal and opposite. That ties in with the idea of the internal energy itself: remember we were basically talking forces between infinitesimally small elements of charge within the electron itself? So r0 is, perhaps, some average distance or so. There must be some way of thinking of it like that. But… Well… Which one exactly?

This kind of reflection may not make sense. Who knows? I obviously need to think all of this through and so this post is, indeed, just a bunch of reflections for which I will have more time later—hopefully. 🙂 Perhaps we’re all just pushing the matter too far. Perhaps we should just accept that the external energy has that 2/3 factor but that the actual energy of the electron should also include the equivalent energy of some binding force that holds the electron together. Well… In any case. That’s all I am going to do on this extremely complicated matter. It’s time to move indeed! So the point to take home here is probably just this:

  1. When calculating the radius of an electron using classical theory, we get in trouble: not only do we find different radii, but the radii that we find do not respect the E = meclaw. It’s only the mece2/r0 that’s relativistically correct.
  2. That suggests the electron also has some non-electric mass, which are referred to as ‘binding forces’ or ‘Poincaré stresses’, but which remain to be explained convincingly.
  3. All of this shouldn’t surprise us: for all we know, the electron is something fuzzy. 🙂

So my next posts will focus on the ‘essentials’ preparing for Feynman’s Volume on quantum mechanics. Those ‘essentials’ will still involve some classical stuff but, as you will see, even more contradictions, that – hopefully! – will then be solved in the quantum-mechanical picture of it all. 🙂

The classical explanation for the electron’s mass and radius

Feynman’s 28th Lecture in his series on electromagnetism is one of the more interesting but, at the same time, it’s one of the few Lectures that is clearly (out)dated. In essence, it talks about the difficulties involved in applying Maxwell’s equations to the elementary charges themselves, i.e. the electron and the proton. We already signaled some of these problems in previous posts. For example, in our post on the energy in electrostatic fields, we showed how our formulas for the field energy and/or the potential of a charge blow up when we use it to calculate the energy we’d need to assemble a point charge. What comes out is infinity: ∞. So our formulas tell us we’d need an infinite amount of energy to assemble a point charge.

Well… That’s no surprise, is it? The idea itself is impossible: how can one have a finite amount of charge in something that’s infinitely small? Something that has no size whatsoever? It’s pretty obvious we get some division by zero there. 🙂 The mathematical approach is often inconsistent. Indeed, a lot of blah-blah in physics is obviously just about applying formulas to situations that are clearly not within the relevant area of application of the formula. So that’s why I went through the trouble (in my previous post, that is) of explaining you how we get these energy and potential formulas, and that’s by bringing charges (note the plural) together. Now, we may assume these charges are point charges, but that assumption is not so essential. What I tried to say when being so explicit was the following: yes, a charge causes a field, but the idea of a potential makes sense only when we’re thinking of placing some other charge in that field. So point charges with ‘infinite energy’ should not be a problem. Feynman admits as much when he writes:

“If the energy can’t get out, but must stay there forever, is there any real difficulty with an infinite energy? Of course, a quantity that comes out infinite may be annoying, but what really matters is only whether there are any observable physical effects.”

So… Well… Let’s see. There’s another, more interesting, way to look at an electron: let’s have a look at the field it creates. A electron – stationary or moving – will create a field in Maxwell’s world, which we know inside out now. So let’s just calculate it. In fact, Feynman calculates it for the unit charge (+1), so that’s a positron. It eases the analysis because we don’t have to drag any minus sign along. So how does it work? Well…

We’ll have an energy flux density vector – i.e. the Poynting vector S – as well as a momentum density vector g all over space. Both are related through the g = S/c2 equation which, as I explained in my previous post, is probably best written as cg = S/c, because we’ve got units then, on both sides, that we can readily understand, like N/m2 (so that’s force per unit area) or J/m3 (so that’s energy per unit volume). On the other hand, we’ll need something that’s written as a function of the velocity of our positron, so that’s v, and so it’s probably best to just calculate g, the momentum, which is measured in N·s or kg·(m/s2)·s (both are equivalent units for the momentum p = mv, indeed) per unit volume (so we need to add a 1/ m3 to the unit). So we’ll have some integral all over space, but I won’t bother you with it. Why not? Well… Feynman uses a rather particular volume element to solve the integral, and so I want you to focus on the solution. The geometry of the situation, and the solution for g, i.e. the momentum of the field per unit volume, is what matters here.

So let’s look at that geometry. It’s depicted below. We’ve got a radial electric field—a Coulomb field really, because our charge is moving at a non-relativistic speed, so v << c and we can approximate with a Coulomb field indeed. Maxwell’s equations imply that B = v×E/c2, so g = ε0E×B is what it is in the illustration below. Note that we’d have to reverse the direction of both E and B for an electron (because it’s negative), but g would be the same. It is directed obliquely toward the line of motion and its magnitude is g = (ε0v/c2)·E2·sinθ. Don’t worry about it: Feynman integrates this thing for you. 🙂 It’s not that difficult, but still… To solve it, he uses the fact that the fields are symmetric about the line of motion, which is indicated by the little arrow around the v-axis, with the Φ symbol next to it (it symbolizes the potential). [The ‘rather particular volume element’ is a ring around the v-axis, and it’s because of this symmetry that Feynman picks the ring. Feynman’s Lectures are not only great to learn physics: they’re a treasure drove of mathematical tricks too. :-)]

momentum field

As said, I don’t want to bother you with the technicalities of the integral here. This is the result:


What does this say? It says that the momentum of the field – i.e. the electromagnetic momentum, integrated over all of space – is proportional to the velocity v of our charge. That makes sense: when v = 0, we’ll have an electrostatic field all over space and, hence, some inertia, but it’s only when we try to move our charge that Newton’s Law comes into play: then we’ll need some force to overcome that inertia. It all works through the Poynting formula: S = E×B0. If nothing’s moving, then B = 0, and so we’ll have some E and, therefore, we’ll have field energy alright, but the energy flow will be zero. But when we move the charge, we’re moving the field, and so then B ≠ 0 and so it’s through B that the E in our S equation start kicking in. Does that make sense? Think about it: it’s good to try to visualize things in your mind. 🙂

The constants in the proportionality constant (2e2)/(3ac2) of our pv formula above are:

  • e2 = qe2/(4πε0), with qthe electron charge (without the minus sign) and ε0 our ubiquitous electric constant. [Note that, unlike Feynman, I prefer to not write e in italics, so as to not confuse it with Euler’s number ≈ 2.71828 etc. However, I know I am not always consistent in my notation. :-/ We don’t need Euler’s number in this post, so e or is always an expression for the electron charge, not Euler’s number. Stupid remark, perhaps, but I don’t want you to be confused.]
  • a is the radius of our charge—see we got away from the idea of a point charge? 🙂
  • c2 is just c2, i.e. our weird constant (the square of the speed of light) which seems to connect everything to everything. Indeed, think about stuff like this: S/g = c= 1/(ε0μ0).

Now, p = mv, so that formula for p basically says that our elementary charge (as mentioned, g is the same for a positron or an electron: E and B will be reversed, but g is not) has an electromagnetic mass melec equal to:

emm - 2

That’s an amazing result. We don’t need to give our electron any rest mass: just its charge and its movement will do! Super! So we don’t need any Higgs fields here! 🙂 The electromagnetic field will do!

Well… Maybe. Let’s explore what we’ve got here.

First, let’s compare that radius a in our formula to what’s found in experiments. Huh? Did someone ever try to measure the electron radius? Of course. There are all these scattering experiments in which electrons get fired at atoms. They can fly through or, else, hit something. Therefore, one can some statistical analysis and determine what is referred to as a cross-section. A cross-section is denoted by the same symbol as the standard deviation: σ (sigma). In any case… So there’s something that’s referred to as the classical electron radius, and it’s equal to the so-called Thomsom scattering length. Thomson scattering, as opposed to Compton scattering, is elastic scattering, so it preserves kinetic energy (unlike Compton scattering, where energy gets absorbed and changes frequencies). So… Well… I won’t go into too much detail but, yes, this is the electron radius we need. [I am saying this rather explicitly because there are two other numbers around: the so-called Bohr radius and, as you might imagine, the Compton scattering cross-section.]

The Thomson scattering length is 2.82 femtometer (so that’s 2.82×10−15 m), more or less that is :-), and it’s usually related to the observed electron mass mthrough the fine-structure constant α. In fact, using Planck units, we can write:  re·me= α, which is an amazing formula but, unfortunately, I can’t dwell on it here. Using ordinary m, s, C and what have you units, we can write ras:

classical electron radius

That’s good, because if we equate mand melec and switch melec and a in our formula for melec, we get:


So, frankly, we’re spot on! Well… Almost. The two numbers differ by 1/3. But who cares about a 1/3 factor indeed? We’re talking rather fuzzy stuff here – scattering cross-sections and standard deviations and all that – so… Yes. Well done! Our theory works!

Well… Maybe. Physicists don’t think so. They think the 1/3 factor is an issue. It’s sad because it really makes a lot of sense. In fact, the Dutch physicist Hendrik Lorentz – whom we know so well by now 🙂 – had also worked out that, because of the length contraction effect, our spherical charge would contract into an ellipsoid and… Well… He worked it all out, and it was not a problem: he found that the momentum was altered by the factor (1−v2/c2)−1/2, so that’s the ubiquitous Lorentz factor γ! He got this formula in the 1890s already, so that’s long before the theory of relativity had been developed. So, many years before Planck and Einstein would come up with their stuff, Hendrik Antoon Lorentz had the correct formulas already: the mass, or everything really, all should vary with that γ-factor. 🙂

Why bother about the 1/3 factor? [I should note it’s actually referred to as the 4/3 problem in physics.] Well… The critics do have a point: if we assume that (a) an electron is not a point charge – so if we allow it to have some radius a – and (b) that Maxwell’s Laws apply, then we should go all the way. The energy that’s needed to assemble an electron should then, effectively, be the same as the value we’d get out of those field energy formulas. So what do we get when we apply those formulas? Well… Let me quickly copy Feynman as he does the calculation for an electron, not looking at it as a point particle, but as a tiny shell of charge, i.e. a sphere with all charge sitting on the surface:

Feynman energy

 Let me enlarge the formula:

energy electron

Now, if we combine that with our formula for melec above, then we get:

4-3 problem

So that formula does not respect Einstein’s universal mass-energy equivalence formula E = mc2. Now, you will agree that we really want Einstein’s mass-energy equivalence relation to be respected by all, so our electron should respect it too. 🙂 So, yes, we’ve got a problem here, and it’s referred to as the 4/3 problem (yes, the ratio got turned around).

Now, you may think it got solved in the meanwhile. Well… No. It’s still a bit of a puzzle today, and the current-day explanation is not really different from what the French scientist Henri Poincaré proposed as a ‘solution’ to the problem back in the 1890s. He basically told Lorentz the following: “If the electron is some little ball of charge, then it should explode because of the repulsive forces inside. So there should be some binding forces there, and so that energy explains the ‘missing mass’ of the electron.” So these forces are effectively being referred to as Poincaré stresses, and the non-electromagnetic energy that’s associated with them – which, of course, has to be equal to 1/3 of the electromagnetic energy (I am sure you see why) 🙂 – adds to the total energy and all is alright now. We get:

U = mc2 = (melec + mPoincaré)c2

So… Yes… Pretty ad hoc. Worse, according to the Wikipedia article on electromagnetic mass, that’s still where we are. And, no, don’t read Feynman’s overview of all of the theories that were around then (so that’s in the 1960s, or earlier). As I said, it’s the one Lecture you don’t want to waste time on. So I won’t do that either.

In fact, let me try to do something else here, and that’s to de-construct the whole argument really. 🙂 Before I do so, let me highlight the essence of what was written above. It’s quite amazing really. Think of it: we say that the mass of an electron – i.e. its inertia, or the proportionality factor in Newton’s F = m·a law of motion – is the energy in the electric and magnetic field it causes. So the electron itself is just a hook for the force law, so to say. There’s nothing there, except for the charge causing the field. But so its mass is everywhere and, hence, nowhere really. Well… I should correct that: the field strength falls of as 1/rand, hence, the energy flow and momentum density that’s associated with it, falls of as 1/r4, so it falls of very rapidly and so the bulk of the energy is pretty near the charge. 🙂

[Note: You’ll remember that the field that’s associated with electromagnetic radiation falls of as 1/r, not as 1/r2, which is why there is an energy flux there which is never lost, which can travel independently through space. It’s not the same here, so don’t get confused.]

So that’s something to note: the melec = (2c−2/3)·(e2/a) has the radius in it, but that radius is only the hook, so to say. That’s fine, because it is not inconsistent with the idea of the Thomson scattering cross-section, which is the area that one can hit. Now, you’ll wonder how one can hit an electron: you can readily imagine an electron beam aimed at nuclei, but how would one hit electrons? Well… You can shoot photons at them, and see if they bounce back elastically or non-elastically. The cross-section area that bounces them off elastically must be pretty ‘hard’, and the cross-section that deflects them non-elastically somewhat less so. 🙂

OK… But… Yes? Hey! How did we get that electron radius in that formula? 

Good question! Brilliant, in fact! You’re right: it’s here that the whole argument falls apart really. We did a substitution. That radius a is the radius of a spherical shell of charge with an energy that’s equal to Uelec = (1/2)·(e2/a), so there’s another way of stating the inconsistency: the equivalent energy of melec = (2c−2/3)·(e2)/a) is equal to E = melec·c= (2/3)·(e2/a) and that’s not the same as Uelec = (1/2)·(e2/a). If we take the ratio of Uelec and melec·c=, we get the same factor: (1/2)/(2/3) = 3/4. But… Your question is superb! Look at it: putting it the way we put it reveals the inconsistency in the whole argument. We’re mixing two things here:

  1. We first calculate the momentum density, and the momentum, that’s caused by the unit charge, so we get some energy which I’ll denote as Eelec = melec·c2
  2. Now, we then assume this energy must be equal to the energy that’s needed to assemble the unit charge from an infinite number of infinitesimally small charges, thereby also assuming the unit charge is a uniformly charged sphere of charge with radius a.
  3. We then use this radius a to simplify our formula for Eelec = melec·c2

Now that is not kosher, really! First, it’s (a) a lot of assumptions, both implicit as well as explicit, and then (b) it’s, quite simply, not a legit mathematical procedure: calculating the energy in the field, or calculating the energy we need to assemble a uniformly charged sphere of radius a are two very different things.

Well… Let me put it differently. We’re using the same laws – it’s all Maxwell’s equations, really – but we should be clear about what we’re doing with them, and those two things are very different. The legitimate conclusion must be that our a is wrong. In other words, we should not assume that our electron is spherical shell of charge. So then what? Well… We could easily imagine something else, like a uniform or even a non-uniformly charged sphere. Indeed, if we’re just filling empty space with infinitesimally small charge ‘elements’, then we may want to think the density at the ‘center’ will be much higher, like what’s going on when planets form: the density of the inner core of our own planet Earth is more than four times the density of its surface material. [OK. Perhaps not very relevant here, but you get the idea.] Or, conversely, taking into account Poincaré’s objection, we may want to think all of the charge will be on the surface, just like on a perfect conductor, where all charge is surface charge!

Note that the field outside of a uniformly charged sphere and the field of a spherical shell of charge is exactly the same, so we would not find a different number for Eelec = melec·c2, but we surely would find a different number for Uelec. You may want to look up some formulas here: you’ll find that the energy of a uniformly distributed sphere of charge (so we do not assume that all of the charge sits on the surface here) is equal to (3/5)·(e2/a). So we’d already have much less of a problem, because the 3/4 factor in the Uelec = (3/4)·melec·c2 becomes a (5/3)·(2/3) = 10/9 factor. So now we have a discrepancy of some 10% only. 🙂

You’ll say: 10% is 10%. It’s huge in physics, as it’s supposed to be an exact science. Well… It is and it isn’t. Do you realize we haven’t even started to talk about stuff like spin? Indeed, in modern physics, we think of electrons as something that also spins around one or the other axis, so there’s energy there too, and we didn’t include that in our analysis.

In short, Feynman’s approach here is disappointing. Naive even, but then… Well… Who knows? Perhaps he didn’t do this Lecture himself. Perhaps it’s just an assistant or so. In fact, I should wonder why there’s still physicists wasting time on this! I should also note that naively comparing that a radius with the classical electron radius also makes little or no sense. Unlike what you’d expect, the classical electron radius re and the Thomson scattering cross-section σare not related like you might think they are, i.e. like σ= π·re2 or σ= π·(re/2)2 or σre2 or σ= π·(2·re)2 or whatever circular surface calculation rule that might make sense here. No. The Thomson scattering cross-section is equal to:

σ= (8π/3)·re2 = (2π/3)·(2·re)2 = (2/3)·π·(2·re)2 ≈ 66.5×10−30 m= 66.5 (fm)2

Why? I am not sure. I must assume it’s got to do with the standard deviation and all that. The point is, we’ve got a 2/3 factor here too, so do we have a problem really? I mean… The a we got was equal to a = (2/3)·re, wasn’t it? It was. But, unfortunately, it doesn’t mean anything. It’s just a coincidence. In fact, looking at the Thomson scattering cross-section, instead of the Thomson scattering radius, makes the ‘problem’ a little bit worse. Indeed, applying the π·r2 rule for a circular surface, we get that the radius would be equal to (8/3)1/2·re ≈ 1.633·re, so we get something that’s much larger rather than something that’s smaller here.

In any case, it doesn’t matter. The point is: this kind of comparisons should not be taken too seriously. Indeed, when everything is said and done, we’re comparing three very different things here:

  1. The radius that’s associated with the energy that’s needed to assemble our electron from infinitesimally small charges, and so that’s based on Coulomb’s law and the model we use for our electron: is it a shell or a sphere of charge? If it’s a sphere, do we want to think of it as something that’s of uniform of non-uniform density.
  2. The second radius is associated with the field of an electron, which we calculate using Poynting’s formula for the energy flow and/or the momentum density. So that’s not about the internal structure of the electron but, of course, it would be nice if we could find some model of an electron that matches this radius.
  3. Finally, there’s the radius that’s associated with elastic scattering, which is also referred to as hard scattering because it’s like the collision of two hard spheres indeed. But so that’s some value that has to be established experimentally and so it involves judicious choices because there’s probabilities and standard deviations involved.

So should we worry about the gaps between these three different concepts? In my humble opinion: no. Why? Because they’re all damn close and so we’re actually talking about the same thing. I mean: isn’t terrific that we’ve got a model that brings the first and the second radius together with a difference of 10% only? As far as I am concerned, that shows the theory works. So what Feynman’s doing in that (in)famous chapter is some kind of ‘dimensional analysis’ which confirms rather than invalidates classical electromagnetic theory. So it shows classical theory’s strength, rather than its weakness. It actually shows our formula do work where we wouldn’t expect them to work. 🙂

The thing is: when looking at the behavior of electrons themselves, we’ll need a different conceptual framework altogether. I am talking quantum mechanics here. Indeed, we’ll encounter other anomalies than the ones we presented above. There’s the issue of the anomalous magnetic moment of electrons, for example. Indeed, as I mentioned above, we’ll also want to think as electrons as spinning around their own axis, and so that implies some circulation of E that will generate a permanent magnetic dipole moment… […] OK, just think of some magnetic field if you don’t have a clue what I am saying here (but then you should check out my post on it). […] The point is: here too, the so-called ‘classical result’, so that’s its theoretical value, will differ from the experimentally measured value. Now, the difference here will be 0.0011614, so that’s about 0.1%, i.e. 100 times smaller than my 10%. 🙂

Personally, I think that’s not so bad. 🙂 But then physicists need to stay in business, of course. So, yes, it is a problem. 🙂

Post scriptum on the math versus the physics

The key to the calculation of the energy that goes into assembling a charge was the following integral:

U 4

This is a double integral which we simplified in two stages, so we’re looking at an integral within an integral really, but we can substitute the integral over the ρ(2)·dVproduct by the formula we got for the potential, so we write that as Φ(1), and so the integral above becomes:

U 5Now, this integral integrates the ρ(1)·Φ(1)·dVproduct over all of space, so that’s over all points in space, and so we just dropped the index and wrote the whole thing as the integral of ρ·Φ·dV over all of space:

U 6

We then established that this integral was mathematically equivalent to the following equation:

U 7

So this integral is actually quite simple: it just integrates EE = E2 over all of space. The illustration below shows E as a function of the distance for a sphere of radius R filled uniformly with charge.

uniform density

So the field (E) goes as for r ≤ R and as 1/rfor r ≥ R. So, for r ≥ R, the integral will have (1/r2)2 = 1/rin it. Now, you know that the integral of some function is the surface under the graph of that function. Look at the 1/r4 function below: it blows up between 1 and 0. That’s where the problem is: there needs to be some kind of cut-off, because that integral will effectively blow up when the radius of our little sphere of charge gets ‘too small’. So that makes it clear why it doesn’t make sense to use this formula to try to calculate the energy of a point charge. It just doesn’t make sense to do that.


What’s ‘too small’? Let’s look at the formula we got for our electron as a spherical shell of charge:

energy electron

So we’ve got an even simpler formula here: it’s just a 1/relation. Why is that? Well… It’s just the way the math turns it out. I copied the detail of Feynman’s calculation above, so you can double-check it. It’s quite wonderful, really. We have a very simple inversely proportional relationship between the radius of our electron and its energy as a sphere of charge. We could write it as:

Uelect  = α/, with α = e2/2

But – Hey! Wait a minute! We’ve seen something like this before, haven’t we? We did. We did when we were discussing the wonderful properties of that magical number, the fine-structure constant, which we also denoted by α. 🙂 However, because we used α already, I’ll denote the fine-structure constant as αe here, so you don’t get confused. As you can see, the fine-structure constant 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, and its mass (and, hence, its energy). So, at this stage of the argument, α can be anything, and αcannot, of course. It’s just that magical number out there, which relates everything to everything: it’s the God-given number we don’t understand. 🙂 Having said that, it seems like we’re going to get some understanding here because we know that, one the many expressions involving αe was the following one:

me = αe/re

This says that the mass of the electron is equal to the ratio of the fine-structure constant and the electron radius. [Note that we express everything in natural units here, so that’s Planck units. For the detail of the conversion, please see the relevant section on that in my one of my posts on this and other stuff.] Now, mass is equivalent to energy, of course: it’s just a matter of units, so we can equate me with Ee (this amounts to expressing the energy of the electron in a kg unit—bit weird, but OK) and so we get:

Ee = αe/re

So there we have: the fine-structure constant αe is Nature’s ‘cut-off’ factor, so to speak. Why? Only God knows. 🙂 But it’s now (fairly) easy to see why all the relations involving αe are what they are. For example, we also know that αe is the square of the electron charge expressed in Planck units, so we have:

 α = eP2 and, therefore, Ee = eP2/re

Now, you can check for yourself: it’s just a matter of re-expressing everything in standard SI units, and relating eP2 to e2, and it should all work: you should get the Uelect  = (1/2)·e2/expression. So… Well… At least this takes some of the magic out the fine-structure constant. It’s still a wonderful thing, but so you see that the fundamental relationship between (a) the energy (and, hence, the mass), (b) the radius and (c) the charge of an electron is not something God-given. What’s God-given are Maxwell’s equations, and so the Ee = αe/r= eP2/re is just one of the many wonderful things that you can get out of  them🙂

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: 🙂


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:



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.]


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. 🙂

The Uncertainty Principle and the stability of atoms

The Model of the Atom

In one of my posts, I explained the quantum-mechanical model of an atom. Feynman sums it up as follows:

“The electrostatic forces pull the electron as close to the nucleus as possible, but the electron is compelled to stay spread out in space over a distance given by the Uncertainty Principle. If it were confined in too small a space, it would have a great uncertainty in momentum. But that means it would have a high expected energy—which it would use to escape from the electrical attraction. The net result is an electrical equilibrium not too different from the idea of Thompson—only is it the negative charge that is spread out, because the mass of the electron is so much smaller than the mass of the proton.”

This explanation is a bit sloppy, so we should add the following clarification: “The wave function Ψ(r) for an electron in an atom does not describe a smeared-out electron with a smooth charge density. The electron is either here, or there, or somewhere else, but wherever it is, it is a point charge.” (Feynman’s Lectures, Vol. III, p. 21-6)

The two quotes are not incompatible: it is just a matter of defining what we really mean by ‘spread out’. Feynman’s calculation of the Bohr radius of an atom in his introduction to quantum mechanics clears all confusion in this regard:

Bohr radius

It is a nice argument. One may criticize he gets the right thing out because he puts the right things in – such as the values of e and m, for example 🙂 − but it’s nice nevertheless!

Mass as a Scale Factor for Uncertainty

Having complimented Feynman, the calculation above does raise an obvious question: why is it that we cannot confine the electron in “too small a space” but that we can do so for the nucleus (which is just one proton in the example of the hydrogen atom here). Feynman gives the answer above: because the mass of the electron is so much smaller than the mass of the proton.

Huh? What’s the mass got to do with it? The uncertainty is the same for protons and electrons, isn’t it?

Well… It is, and it isn’t. 🙂 The Uncertainty Principle – usually written in its more accurate σxσp ≥ ħ/2 expression – applies to both the electron and the proton – of course! – but the momentum is the product of mass and velocity (p = m·v), and so it’s the proton’s mass that makes the difference here. To be specific, the mass of a proton is about 1836 times that of an electron. Now, as long as the velocities involved are non-relativistic—and they are non-relativistic in this case: the (relative) speed of electrons in atoms is given by the fine-structure constant α = v/c ≈ 0.0073, so the Lorentz factor is very close to 1—we can treat the m in the p = m·v identity as a constant and, hence, we can also write: Δp = Δ(m·v) = m·Δv. So all of the uncertainty of the momentum goes into the uncertainty of the velocity. Hence, the mass acts likes a reverse scale factor for the uncertainty. To appreciate what that means, let me write ΔxΔp = ħ as:

ΔxΔv = ħ/m

It is an interesting point, so let me expand the argument somewhat. We actually use a more general mathematical property of the standard deviation here: the standard deviation of a variable scales directly with the scale of the variable. Hence, we can write: σ(k·x) = k·σ(x), with k > 0. So the uncertainty is, indeed, smaller for larger masses. Larger masses are associated with smaller uncertainties in their position x. To be precise, the uncertainty is inversely proportional to the mass and, hence, the mass number effectively acts like a reverse scale factor for the uncertainty.

Of course, you’ll say that the uncertainty still applies to both factors on the left-hand side of the equation, and so you’ll wonder: why can’t we keep Δx the same and multiply Δv with m, so its product yields ħ again? In other words, why can’t we have a uncertainty in velocity for the proton that is 1836 times larger than the uncertainty in velocity for the electron? The answer to that question should be obvious: the uncertainty should not be greater than the expected value. When everything is said and done, we’re talking a distribution of some variable here (the velocity variable, to be precise) and, hence, that distribution is likely to be the Maxwell-Boltzmann distribution we introduced in previous posts. Its formula and graph are given below:

Formula M-B distribution428px-Maxwell-Boltzmann_distribution_pdf

In statistics (and in probability theory), they call this a chi distribution with three degrees of freedom and a scale parameter which is equal to a = (kT/m)1/2. The formula for the scale parameter shows how the mass of a particle indeed acts as a reverse scale parameter. The graph above shows three graphs for a = 1, 2 and 5 respectively. Note the square root though: quadrupling the mass (keeping kT the same) amounts to going from a = 2 to a = 1, so that’s halving a. Indeed, [kT/(4m)]1/2 = (1/2)(kT/m)1/2. So we can’t just do what we want with Δv (like multiplying it with 1836, as suggested). In fact, the graph and the formulas show that Feynman’s assumption that we can equate p with Δp (i.e. his assumption that “the momenta must be of the order p = ħ/Δx, with Δx the spread in position”), more or less at least, is quite reasonable.

Of course, you are very smart and so you’ll have yet another objection: why can’t we associate a much higher momentum with the proton, as that would allow us to associate higher velocities with the proton? Good question. My answer to that is the following (and it might be original, as I didn’t find this anywhere else). When everything is said and done, we’re talking two particles in some box here: an electron and a proton. Hence, we should assume that the average kinetic energy of our electron and our proton is the same (if not, they would be exchanging kinetic energy until it’s more or less equal), so we write <melectron·v2electron/2> = <mproton·v2proton/2>. We can re-write this as mp/m= 1/1836 = <v2e>/<v2p> and, therefore, <v2e> = 1836·<v2p>. Now, <v2> ≠ <v>2 and, hence, <v> ≠ √<v2>. So the equality does not imply that the expected velocity of the electron is √1836 ≈ 43 times the expected velocity of the proton. Indeed, because of the particularities of the distribution, there is a difference between (a) the most probable speed, which is equal to √2·a ≈ 1.414·a, (b) the root mean square speed, which is equal to √<v2> = √3·a ≈ 1.732·a, and, finally, (c) the mean or expected speed, which is equal to <v> = 2·(2/π)1/2·a ≈ 1.596·a.

However, we are not far off. We could use any of these three values to roughly approximate Δv, as well as the scale parameter a itself: our answers would all be of the same order. However, to keep the calculations simple, let’s use the most probable speed. Let’s equate our electron mass with unity, so the mass of our proton is 1836. Now, such mass implies a scale factor (i.e. a) that’s √1836 ≈ 43 times smaller. So the most probable speed of the proton and, therefore, its spread, would be about √2/√1836 = √(2/1836) ≈ 0.033 that of the electron, so we write: Δvp ≈ 0.033·Δve. Now we can insert this in our ΔxΔv = ħ/m = ħ/1836 identity. We get: ΔxpΔvp = Δxp·√(2/1836)·Δve = ħ/1836. That, in turn, implies that √(2·1836)·Δxp = ħ/Δve, which we can re-write as: Δx= Δxe/√(2·1836) ≈ Δxe/60. In other words, the expected spread in the position of the proton is about 60 times smaller than the expected spread of the electron. More in general, we can say that the spread in position of a particle, keeping all else equal, is inversely proportional to (2m)1/2. Indeed, in this case, we multiplied the mass with about 1800, and we found that the uncertainty in position went down with a factor 1/60 = 1/√3600. Not bad as a result ! Is it precise? Well… It could be like √3·√m or 2·(2/π)1/2··√m depending on our definition of ‘uncertainty’, but it’s all of the same order. So… Yes. Not bad at all… 🙂

You’ll raise a third objection now: the radius of a proton is measured using the femtometer scale, so that’s expressed in 10−15 m, which is not 60 but a million times smaller than the nanometer (i.e. 10−9 m) scale used to express the Bohr radius as calculated by Feynman above. You’re right, but the 10−15 m number is the charge radius, not the uncertainty in position. Indeed, the so-called classical electron radius is also measured in femtometer and, hence, the Bohr radius is also like a million times that number. OK. That should settle the matter. I need to move on.

Before I do move on, let me relate the observation (i.e. the fact that the uncertainty in regard to position decreases as the mass of a particle increases) to another phenomenon. As you know, the interference of light beams is easy to observe. Hence, the interference of photons is easy to observe: Young’s experiment involved a slit of 0.85 mm (so almost 1 mm) only. In contrast, the 2012 double-slit experiment with electrons involved slits that were 62 nanometer wide, i.e. 62 billionths of a meter! That’s because the associated frequencies are so much higher and, hence, the wave zone is much smaller. So much, in fact, that Feynman could not imagine technology would ever be sufficiently advanced so as to actually carry out the double slit experiment with electrons. It’s an aspect of the same: the uncertainty in position is much smaller for electrons than it is for photons. Who knows: perhaps one day, we’ll be able to do the experiment with protons. 🙂 For further detail, I’ll refer you one of my posts on this.

What’s Explained, and What’s Left Unexplained?

There is another obvious question: if the electron is still some point charge, and going around as it does, why doesn’t it radiate energy? Indeed, the Rutherford-Bohr model had to be discarded because this ‘planetary’ model involved circular (or elliptical) motion and, therefore, some acceleration. According to classical theory, the electron should thus emit electromagnetic radiation, as a result of which it would radiate its kinetic energy away and, therefore, spiral in toward the nucleus. The quantum-mechanical model doesn’t explain this either, does it?

I can’t answer this question as yet, as I still need to go through all Feynman’s Lectures on quantum mechanics. You’re right. There’s something odd about the quantum-mechanical idea: it still involves a electron moving in some kind of orbital − although I hasten to add that the wavefunction is a complex-valued function, not some real function − but it does not involve any loss of kinetic energy due to circular motion apparently!

There are other unexplained questions as well. For example, the idea of an electrical point charge still needs to be re-conciliated with the mathematical inconsistencies it implies, as Feynman points out himself in yet another of his Lectures. :-/

Finally, you’ll wonder as to the difference between a proton and a positron: if a positron and an electron annihilate each other in a flash, why do we have a hydrogen atom at all? Well… The proton is not the electron’s anti-particle. For starters, it’s made of quarks, while the positron is made of… Well… A positron is a positron: it’s elementary. But, yes, interesting question, and the ‘mechanics’ behind the mutual destruction are quite interesting and, hence, surely worth looking into—but not here. 🙂

Having mentioned a few things that remain unexplained, the model does have the advantage of solving plenty of other questions. It explains, for example, why the electron and the proton are actually right on top of each other, as they should be according to classical electrostatic theory, and why they are not at the same time: the electron is still a sort of ‘cloud’ indeed, with the proton at its center.

The quantum-mechanical ‘cloud’ model of the electron also explains why “the terrific electrical forces balance themselves out, almost perfectly, by forming tight, fine mixtures of the positive and the negative, so there is almost no attraction or repulsion at all between two separate bunches of such mixtures” (Richard Feynman, Introduction to Electromagnetism, p. 1-1) or, to quote from one of his other writings, why we do not fall through the floor as we walk:

“As we walk, our shoes with their masses of atoms push against the floor with its mass of atoms. In order to squash the atoms closer together, the electrons would be confined to a smaller space and, by the uncertainty principle, their momenta would have to be higher on the average, and that means high energy; the resistance to atomic compression is a quantum-mechanical effect and not a classical effect. Classically, we would expect that if we were to draw all the electrons and protons closer together, the energy would be reduced still further, and the best arrangement of positive and negative charges in classical physics is all on top of each other. This was well known in classical physics and was a puzzle because of the existence of the atom. Of course, the early scientists invented some ways out of the trouble—but never mind, we have the right way out, now!”

So that’s it, then. Except… Well…

The Fine-Structure Constant

When talking about the stability of atoms, one cannot escape a short discussion of the so-called fine-structure constant, denoted by α (alpha). I discussed it another post of mine, so I’ll refer you there for a more comprehensive overview. I’ll just remind you of the basics:

(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) Finally, 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. [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.]

Note that, from (2) and (4), we also 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.

In addition, in the mentioned post, I also related α to the so-called coupling constant determining the strength of the interaction between electrons and photons. So… What a magical number indeed ! It suggests some unity that our little model of the atom above doesn’t quite capture. As far as I am concerned, it’s one of the many other ‘unexplained questions’, and one of my key objectives, as I struggle through Feynman’s Lectures, is to understand it all. 🙂 One of the issues is, of course, how to relate this coupling constant to the concept of a gauge, which I briefly discussed in my previous post. In short, I’ve still got a long way to go… 😦

Post Scriptum: The de Broglie relations and the Uncertainty Principle

My little exposé on mass being nothing but a scale factor in the Uncertainty Principle is a good occasion to reflect on the Uncertainty Principle once more. Indeed, what’s the uncertainty about, if it’s not about the mass? It’s about the position in space and velocity, i.e. it’s movement and time. Velocity or speed (i.e. the magnitude of the velocity vector) is, in turn, defined as the distance traveled divided by the time of travel, so the uncertainty is about time as well, as evidenced from the ΔEΔt = h expression of the Uncertainty Principle. But how does it work exactly?

Hmm… Not sure. Let me try to remember the context. We know that the de Broglie relation, λ = h/p, which associates a wavelength (λ) with the momentum (p) of a particle, is somewhat misleading, because we’re actually associating a (possibly infinite) bunch of component waves with a particle. So we’re talking some range of wavelengths (Δλ) and, hence, assuming all these component waves travel at the same speed, we’re also talking a frequency range (Δf). The bottom line is that we’ve got a wave packet and we need to distinguish the velocity of its phase (vp) versus the group velocity (vg), which corresponds to the classical velocity of our particle.

I think I explained that pretty well in one of my previous posts on the Uncertainty Principle, so I’d suggest you have a look there. The mentioned post explains how the Uncertainty Principle relates position (x) and momentum (p) as a Fourier pair, and it also explains that general mathematical property of Fourier pairs: the more ‘concentrated’ one distribution is, the more ‘spread out’ its Fourier transform will be. In other words, it is not possible to arbitrarily ‘concentrate’ both distributions, i.e. both the distribution of x (which I denoted as Ψ(x) as well as its Fourier transform, i.e. the distribution of p (which I denoted by Φ(p)). So, if we’d ‘squeeze’ Ψ(x), then its Fourier transform Φ(p) will ‘stretch out’.

That was clear enough—I hope! But how do we go from ΔxΔp = h to ΔEΔt = h? Why are energy and time another Fourier pair? To answer that question, we need to clearly define what energy and what time we are talking about. The argument revolves around the second de Broglie relation: E = h·f. How do we go from the momentum p to the energy E? And how do we go from the wavelength λ to the frequency f?

The answer to the first question is the energy-mass equivalence: E = mc2, always. This formula is relativistic, as m is the relativistic mass, so it includes the rest mass m0 as well as the equivalent mass of its kinetic energy m0v2/2 + … [Note, indeed, that the kinetic energy – defined as the excess energy over its rest energy – is a rapidly converging series of terms, so only the m0v2/2 term is mentioned.] Likewise, momentum is defined as p = mv, always, with m the relativistic mass, i.e. m = (1−v2/c2)−1/2·m0 = γ·m0, with γ the Lorentz factor. The E = mc2 and p = mv relations combined give us the E/c = m·c = p·c/v or E·v/c = p·c relationship, which we can also write as E/p = c2/v. However, we’ll need to write E as a function of p for the purpose of a derivation. You can verify that E− p2c= m02c4) and, hence, that E = (p2c+ m02c4)1/2.

Now, to go from a wavelength to a frequency, we need the wave velocity, and we’re obviously talking the phase velocity here, so we write: vp = λ·f. That’s where the de Broglie hypothesis comes in: de Broglie just assumed the Planck-Einstein relation E = h·ν, in which ν is the frequency of a massless photon, would also be valid for massive particles, so he wrote: E = h·f. It’s just a hypothesis, of course, but it makes everything come out alright. More in particular, the phase velocity vp = λ·f can now be re-written, using both de Broglie relations (i.e. h/p = λ and E/h = f) as vp = (E/h)·(p/h) = E/p = c2/v. Now, because v is always smaller than c for massive particles (and usually very much smaller), we’re talking a superluminal phase velocity here! However, because it doesn’t carry any signal, it’s not inconsistent with relativity theory.

Now what about the group velocity? To calculate the group velocity, we need the frequencies and wavelengths of the component waves. The dispersion relation assumes the frequency of each component wave can be expressed as a function of its wavelength, so f = f(λ). Now, it takes a bit of wave mechanics (which I won’t elaborate on here) to show that the group velocity is the derivative of f with respect to λ, so we write vg = ∂f/∂λ. Using the two de Broglie relations, we get: vg = ∂f/∂λ = ∂(E/h)/∂(p/h) = ∂E/∂p = ∂[p2c+ m02c4)1/2]/∂p. Now, when you write it all out, you should find that vg = ∂f/∂λ = pc2/E = c2/vp = v, so that’s the classical velocity of our particle once again.

Phew! Complicated! Yes. But so we still don’t have our ΔEΔt = h expression! All of the above tells us how we can associate a range of momenta (Δp) with a range of wavelengths (Δλ) and, in turn, with a frequency range (Δf) which then gives us some energy range (ΔE), so the logic is like:

Δp ⇒ Δλ ⇒ Δf ⇒ ΔE

Somehow, the same sequence must also ‘transform’ our Δx into Δt. I googled a bit, but I couldn’t find any clear explanation. Feynman doesn’t seem to have one in his Lectures either so, frankly, I gave up. What I did do in one of my previous posts, is to give some interpretation. However, I am not quite sure if it’s really the interpretation: there are probably several ones. It must have something to do with the period of a wave, but I’ll let you break your head over it. 🙂 As far as I am concerned, it’s just one of the other unexplained questions I have as I sort of close my study of ‘classical’ physics. So I’ll just make a mental note of it. [Of course, please don’t hesitate to send me your answer, if you’d have one!] Now it’s time to really dig into quantum mechanics, so I should really stay silent for quite a while now! 🙂

The Strange Theory of Light and Matter (III)

This is my third and final comments on Feynman’s popular little booklet: The Strange Theory of Light and Matter, also known as Feynman’s Lectures on Quantum Electrodynamics (QED).

The origin of this short lecture series is quite moving: the death of Alix G. Mautner, a good friend of Feynman’s. She was always curious about physics but her career was in English literature and so she did not manage the math. Hence, Feynman introduces this 1985 publication by writing: “Here are the lectures I really prepared for Alix, but unfortunately I can’t tell them to her directly, now.”

Alix Mautner died from a brain tumor, and it is her husband, Leonard Mautner, who sponsored the QED lectures series at the UCLA, which Ralph Leigton transcribed and published as the booklet that we’re talking about here. Feynman himself died a few years later, at the relatively young age of 69. Tragic coincidence: he died of cancer too. Despite all this weirdness, Feynman’s QED never quite got the same iconic status of, let’s say, Stephen Hawking’s Brief History of Time. I wonder why, but the answer to that question is probably in the realm of chaos theory. 🙂 I actually just saw the movie on Stephen Hawking’s life (The Theory of Everything), and I noted another strange coincidence: Jane Wilde, Hawking’s first wife, also has a PhD in literature. It strikes me that, while the movie documents that Jane Wilde gave Hawking three children, after which he divorced her to marry his nurse, Elaine, the movie does not mention that he separated from Elaine too, and that he has some kind of ‘working relationship’ with Jane again.

Hmm… What to say? I should get back to quantum mechanics here or, to be precise, to quantum electrodynamics.

One reason why Feynman’s Strange Theory of Light and Matter did not sell like Hawking’s Brief History of Time, might well be that, in some places, the text is not entirely accurate. Why? Who knows? It would make for an interesting PhD thesis in History of Science. Unfortunately, I have no time for such PhD thesis. Hence, I must assume that Richard Feynman simply didn’t have much time or energy left to correct some of the writing of Ralph Leighton, who transcribed and edited these four short lectures a few years before Feynman’s death. Indeed, when everything is said and done, Ralph Leighton is not a physicist and, hence, I think he did compromise – just a little bit – on accuracy for the sake of readability. Ralph Leighton’s father, Robert Leighton, an eminent physicist who worked with Feynman, would probably have done a much better job.

I feel that one should not compromise on accuracy, even when trying to write something reader-friendly. That’s why I am writing this blog, and why I am writing three posts specifically on this little booklet. Indeed, while I’d warmly recommend that little book on QED as an excellent non-mathematical introduction to the weird world of quantum mechanics, I’d also say that, while Ralph Leighton’s story is great, it’s also, in some places, not entirely accurate indeed.

So… Well… I want to do better than Ralph Leighton here. Nothing more. Nothing less. 🙂 Let’s go for it.

I. Probability amplitudes: what are they?

The greatest achievement of that little QED publication is that it manages to avoid any reference to wave functions and other complicated mathematical constructs: all of the complexity of quantum mechanics is reduced to three basic events or actions and, hence, three basic amplitudes which are represented as ‘arrows’—literally.

Now… Well… You may or may not know that a (probability) amplitude is actually a complex number, but it’s not so easy to intuitively understand the concept of a complex number. In contrast, everyone easily ‘gets’ the concept of an ‘arrow’. Hence, from a pedagogical point of view, representing complex numbers by some ‘arrow’ is truly a stroke of genius.

Whatever we call it, a complex number or an ‘arrow’, a probability amplitude is something with (a) a magnitude and (b) a phase. As such, it resembles a vector, but it’s not quite the same, if only because we’ll impose some restrictions on the magnitude. But I shouldn’t get ahead of myself. Let’s start with the basics.

A magnitude is some real positive number, like a length, but you should not associate it with some spatial dimension in physical space: it’s just a number. As for the phase, we could associate that concept with some direction but, again, you should just think of it as a direction in a mathematical space, not in the real (physical) space.

Let me insert a parenthesis here. If I say the ‘real’ or ‘physical’ space, I mean the space in which the electrons and photons and all other real-life objects that we’re looking at exist and move. That’s a non-mathematical definition. In fact, in math, the real space is defined as a coordinate space, with sets of real numbers (vectors) as coordinates, so… Well… That’s a mathematical space only, not the ‘real’ (physical) space. So the real (vector) space is not real. 🙂 The mathematical real space may, or may not, accurately describe the real (physical) space. Indeed, you may have heard that physical space is curved because of the presence of massive objects, which means that the real coordinate space will actually not describe it very accurately. I know that’s a bit confusing but I hope you understand what I mean: if mathematicians talk about the real space, they do not mean the real space. They refer to a vector space, i.e. a mathematical construct. To avoid confusion, I’ll use the term ‘physical space’ rather than ‘real’ space in the future. So I’ll let the mathematicians get away with using the term ‘real space’ for something that isn’t real actually. 🙂

End of digression. Let’s discuss these two mathematical concepts – magnitude and phase – somewhat more in detail.

A. The magnitude

Let’s start with the magnitude or ‘length’ of our arrow. We know that we have to square these lengths to find some probability, i.e. some real number between 0 and 1. Hence, the length of our arrows cannot be larger than one. That’s the restriction I mentioned already, and this ‘normalization’ condition reinforces the point that these ‘arrows’ do not have any spatial dimension (not in any real space anyway): they represent a function. To be specific, they represent a wavefunction.

If we’d be talking complex numbers instead of ‘arrows’, we’d say the absolute value of the complex number cannot be larger than one. We’d also say that, to find the probability, we should take the absolute square of the complex number, so that’s the square of the magnitude or absolute value of the complex number indeed. We cannot just square the complex number: it has to be the square of the absolute value.

Why? Well… Just write it out. [You can skip this section if you’re not interested in complex numbers, but I would recommend you try to understand. It’s not that difficult. Indeed, if you’re reading this, you’re most likely to understand something of complex numbers and, hence, you should be able to work your way through it. Just remember that a complex number is like a two-dimensional number, which is why it’s sometimes written using bold-face (z), rather than regular font (z). However, I should immediately add this convention is usually not followed. I like the boldface though, and so I’ll try to use it in this post.] The square of a complex number z = a + bi is equal to z= a+ 2abi – b2, while the square of its absolute value (i.e. the absolute square) is |z|= [√(a+ b2)]2 = a+ b2. So you can immediately see that the square and the absolute square of a complex numbers are two very different things indeed: it’s not only the 2abi term, but there’s also the minus sign in the first expression, because of the i= –1 factor. In case of doubt, always remember that the square of a complex number may actually yield a negative number, as evidenced by the definition of the imaginary unit itself: i= –1.

End of digression. Feynman and Leighton manage to avoid any reference to complex numbers in that short series of four lectures and, hence, all they need to do is explain how one squares a length. Kids learn how to do that when making a square out of rectangular paper: they’ll fold one corner of the paper until it meets the opposite edge, forming a triangle first. They’ll then cut or tear off the extra paper, and then unfold. Done. [I could note that the folding is a 90 degree rotation of the original length (or width, I should say) which, in mathematical terms, is equivalent to multiplying that length with the imaginary unit (i). But I am sure the kids involved would think I am crazy if I’d say this. 🙂 So let me get back to Feynman’s arrows.

B. The phase

Feynman and Leighton’s second pedagogical stroke of genius is the metaphor of the ‘stopwatch’ and the ‘stopwatch hand’ for the variable phase. Indeed, although I think it’s worth explaining why z = a + bi = rcosφ + irsinφ in the illustration below can be written as z = reiφ = |z|eiφ, understanding Euler’s representation of complex number as a complex exponential requires swallowing a very substantial piece of math and, if you’d want to do that, I’ll refer you to one of my posts on complex numbers).


The metaphor of the stopwatch represents a periodic function. To be precise, it represents a sinusoid, i.e. a smooth repetitive oscillation. Now, the stopwatch hand represents the phase of that function, i.e. the φ angle in the illustration above. That angle is a function of time: the speed with which the stopwatch turns is related to some frequency, i.e. the number of oscillations per unit of time (i.e. per second).

You should now wonder: what frequency? What oscillations are we talking about here? Well… As we’re talking photons and electrons here, we should distinguish the two:

  1. For photons, the frequency is given by Planck’s energy-frequency relation, which relates the energy (E) of a photon (1.5 to 3.5 eV for visible light) to its frequency (ν). It’s a simple proportional relation, with Planck’s constant (h) as the proportionality constant: E = hν, or ν = E/h.
  2. For electrons, we have the de Broglie relation, which looks similar to the Planck relation (E = hf, or f = E/h) but, as you know, it’s something different. Indeed, these so-called matter waves are not so easy to interpret because there actually is no precise frequency f. In fact, the matter wave representing some particle in space will consist of a potentially infinite number of waves, all superimposed one over another, as illustrated below.


For the sake of accuracy, I should mention that the animation above has its limitations: the wavetrain is complex-valued and, hence, has a real as well as an imaginary part, so it’s something like the blob underneath. Two functions in one, so to speak: the imaginary part follows the real part with a phase difference of 90 degrees (or π/2 radians). Indeed, if the wavefunction is a regular complex exponential reiθ, then rsin(φ–π/2) = rcos(φ), which proves the point: we have two functions in one here. 🙂 I am actually just repeating what I said before already: the probability amplitude, or the wavefunction, is a complex number. You’ll usually see it written as Ψ (psi) or Φ (phi). Here also, using boldface (Ψ or Φ instead of Ψ or Φ) would usefully remind the reader that we’re talking something ‘two-dimensional’ (in mathematical space, that is), but this convention is usually not followed.

Photon wave

In any case… Back to frequencies. The point to note is that, when it comes to analyzing electrons (or any other matter-particle), we’re dealing with a range of frequencies f really (or, what amounts to the same, a range of wavelengths λ) and, hence, we should write Δf = ΔE/h, which is just one of the many expressions of the Uncertainty Principle in quantum mechanics.

Now, that’s just one of the complications. Another difficulty is that matter-particles, such as electrons, have some rest mass, and so that enters the energy equation as well (literally). Last but not least, one should distinguish between the group velocity and the phase velocity of matter waves. As you can imagine, that makes for a very complicated relationship between ‘the’ wavelength and ‘the’ frequency. In fact, what I write above should make it abundantly clear that there’s no such thing as the wavelength, or the frequency: it’s a range really, related to the fundamental uncertainty in quantum physics. I’ll come back to that, and so you shouldn’t worry about it here. Just note that the stopwatch metaphor doesn’t work very well for an electron!

In his postmortem lectures for Alix Mautner, Feynman avoids all these complications. Frankly, I think that’s a missed opportunity because I do not think it’s all that incomprehensible. In fact, I write all that follows because I do want you to understand the basics of waves. It’s not difficult. High-school math is enough here. Let’s go for it.

One turn of the stopwatch corresponds to one cycle. One cycle, or 1 Hz (i.e. one oscillation per second) covers 360 degrees or, to use a more natural unit, 2π radians. [Why is radian a more natural unit? Because it measures an angle in terms of the distance unit itself, rather than in arbitrary 1/360 cuts of a full circle. Indeed, remember that the circumference of the unit circle is 2π.] So our frequency ν (expressed in cycles per second) corresponds to a so-called angular frequency ω = 2πν. From this formula, it should be obvious that ω is measured in radians per second.

We can also link this formula to the period of the oscillation, T, i.e. the duration of one cycle. T = 1/ν and, hence, ω = 2π/T. It’s all nicely illustrated below. [And, yes, it’s an animation from Wikipedia: nice and simple.]


The easy math above now allows us to formally write the phase of a wavefunction – let’s denote the wavefunction as φ (phi), and the phase as θ (theta) – as a function of time (t) using the angular frequency ω. So we can write: θ = ωt = 2π·ν·t. Now, the wave travels through space, and the two illustrations above (i.e. the one with the super-imposed waves, and the one with the complex wave train) would usually represent a wave shape at some fixed point in time. Hence, the horizontal axis is not t but x. Hence, we can and should write the phase not only as a function of time but also of space. So how do we do that? Well… If the hypothesis is that the wave travels through space at some fixed speed c, then its frequency ν will also determine its wavelength λ. It’s a simple relationship: c = λν (the number of oscillations per second times the length of one wavelength should give you the distance traveled per second, so that’s, effectively, the wave’s speed).

Now that we’ve expressed the frequency in radians per second, we can also express the wavelength in radians per unit distance too. That’s what the wavenumber does: think of it as the spatial frequency of the wave. We denote the wavenumber by k, and write: k = 2π/λ. [Just do a numerical example when you have difficulty following. For example, if you’d assume the wavelength is 5 units distance (i.e. 5 meter) – that’s a typical VHF radio frequency: ν = (3×10m/s)/(5 m) = 0.6×108 Hz = 60 MHz – then that would correspond to (2π radians)/(5 m) ≈ 1.2566 radians per meter. Of course, we can also express the wave number in oscillations per unit distance. In that case, we’d have to divide k by 2π, because one cycle corresponds to 2π radians. So we get the reciprocal of the wavelength: 1/λ. In our example, 1/λ is, of course, 1/5 = 0.2, so that’s a fifth of a full cycle. You can also think of it as the number of waves (or wavelengths) per meter: if the wavelength is λ, then one can fit 1/λ waves in a meter.


Now, from the ω = 2πν, c = λν and k = 2π/λ relations, it’s obvious that k = 2π/λ = 2π/(c/ν) = (2πν)/c = ω/c. To sum it all up, frequencies and wavelengths, in time and in space, are all related through the speed of propagation of the wave c. More specifically, they’re related as follows:

c = λν = ω/k

From that, it’s easy to see that k = ω/c, which we’ll use in a moment. Now, it’s obvious that the periodicity of the wave implies that we can find the same phase by going one oscillation (or a multiple number of oscillations back or forward in time, or in space. In fact, we can also find the same phase by letting both time and space vary. However, if we want to do that, it should be obvious that we should either (a) go forward in space and back in time or, alternatively, (b) go back in space and forward in time. In other words, if we want to get the same phase, then time and space sort of substitute for each other. Let me quote Feynman on this: “This is easily seen by considering the mathematical behavior of a(tr/c). Evidently, if we add a little time Δt, we get the same value for a(tr/c) as we would have if we had subtracted a little distance: ΔcΔt.” The variable a stands for the acceleration of an electric charge here, causing an electromagnetic wave, but the same logic is valid for the phase, with a minor twist though: we’re talking a nice periodic function here, and so we need to put the angular frequency in front. Hence, the rate of change of the phase in respect to time is measured by the angular frequency ω. In short, we write:

θ = ω(t–x/c) = ωt–kx

Hence, we can re-write the wavefunction, in terms of its phase, as follows:

φ(θ) = φ[θ(x, t)] = φ[ωt–kx]

Note that, if the wave would be traveling in the ‘other’ direction (i.e. in the negative x-direction), we’d write φ(θ) = φ[kx+ωt]. Time travels in one direction only, of course, but so one minus sign has to be there because of the logic involved in adding time and subtracting distance. You can work out an example (with a sine or cosine wave, for example) for yourself.

So what, you’ll say? Well… Nothing. I just hope you agree that all of this isn’t rocket science: it’s just high-school math. But so it shows you what that stopwatch really is and, hence, – but who am I? – would have put at least one or two footnotes on this in a text like Feynman’s QED.

Now, let me make a much longer and more serious digression:

Digression 1: on relativity and spacetime

As you can see from the argument (or phase) of that wave function φ(θ) = φ[θ(x, t)] = φ[ωt–kx] = φ[–k(x–ct)], any wave equation establishes a deep relation between the wave itself (i.e. the ‘thing’ we’re describing) and space and time. In fact, that’s what the whole wave equation is all about! So let me say a few things more about that.

Because you know a thing or two about physics, you may ask: when we’re talking time, whose time are we talking about? Indeed, if we’re talking photons going from A to B, these photons will be traveling at or near the speed of light and, hence, their clock, as seen from our (inertial) frame of reference, doesn’t move. Likewise, according to the photon, our clock seems to be standing still.

Let me put the issue to bed immediately: we’re looking at things from our point of view. Hence, we’re obviously using our clock, not theirs. Having said that, the analysis is actually fully consistent with relativity theory. Why? Well… What do you expect? If it wasn’t, the analysis would obviously not be valid. 🙂 To illustrate that it’s consistent with relativity theory, I can mention, for example, that the (probability) amplitude for a photon to travel from point A to B depends on the spacetime interval, which is invariant. Hence, A and B are four-dimensional points in spacetime, involving both spatial as well as time coordinates: A = (xA, yA, zA, tA) and B = (xB, yB, zB, tB). And so the ‘distance’ – as measured through the spacetime interval – is invariant.

Now, having said that, we should draw some attention to the intimate relationship between space and time which, let me remind you, results from the absoluteness of the speed of light. Indeed, one will always measure the speed of light c as being equal to 299,792,458 m/s, always and everywhere. It does not depend on your reference frame (inertial or moving). That’s why the constant c anchors all laws in physics, and why we can write what we write above, i.e. include both distance (x) as well as time (t) in the wave function φ = φ(x, t) = φ[ωt–kx] = φ[–k(x–ct)]. The k and ω are related through the ω/k = c relationship: the speed of light links the frequency in time (ν = ω/2π = 1/T) with the frequency in space (i.e. the wavenumber or spatial frequency k). There is only degree of freedom here: the frequency—in space or in time, it doesn’t matter: ν and ω are not independent.  [As noted above, the relationship between the frequency in time and in space is not so obvious for electrons, or for matter waves in general: for those matter-waves, we need to distinguish group and phase velocity, and so we don’t have a unique frequency.]

Let me make another small digression within the digression here. Thinking about travel at the speed of light invariably leads to paradoxes. In previous posts, I explained the mechanism of light emission: a photon is emitted – one photon only – when an electron jumps back to its ground state after being excited. Hence, we may imagine a photon as a transient electromagnetic wave–something like what’s pictured below. Now, the decay time of this transient oscillation (τ) is measured in nanoseconds, i.e. billionths of a second (1 ns = 1×10–9 s): the decay time for sodium light, for example, is some 30 ns only.

decay time

However, because of the tremendous speed of light, that still makes for a wavetrain that’s like ten meter long, at least (30×10–9 s times 3×10m/s is nine meter, but you should note that the decay time measures the time for the oscillation to die out by a factor 1/e, so the oscillation itself lasts longer than that). Those nine or ten meters cover like 16 to 17 million oscillations (the wavelength of sodium light is about 600 nm and, hence, 10 meter fits almost 17 million oscillations indeed). Now, how can we reconcile the image of a photon as a ten-meter long wavetrain with the image of a photon as a point particle?

The answer to that question is paradoxical: from our perspective, anything traveling at the speed of light – including this nine or ten meter ‘long’ photon – will have zero length because of the relativistic length contraction effect. Length contraction? Yes. I’ll let you look it up, because… Well… It’s not easy to grasp. Indeed, from the three measurable effects on objects moving at relativistic speeds – i.e. (1) an increase of the mass (the energy needed to further accelerate particles in particle accelerators increases dramatically at speeds nearer to c), (2) time dilation, i.e. a slowing down of the (internal) clock (because of their relativistic speeds when entering the Earth’s atmosphere, the measured half-life of muons is five times that when at rest), and (3) length contraction – length contraction is probably the most paradoxical of all.

Let me end this digression with yet another short note. I said that one will always measure the speed of light c as being equal to 299,792,458 m/s, always and everywhere and, hence, that it does not depend on your reference frame (inertial or moving). Well… That’s true and not true at the same time. I actually need to nuance that statement a bit in light of what follows: an individual photon does have an amplitude to travel faster or slower than c, and when discussing matter waves (such as the wavefunction that’s associated with an electron), we can have phase velocities that are faster than light! However, when calculating those amplitudes, is a constant.

That doesn’t make sense, you’ll say. Well… What can I say? That’s how it is unfortunately. I need to move on and, hence, I’ll end this digression and get back to the main story line. Part I explained what probability amplitudes are—or at least tried to do so. Now it’s time for part II: the building blocks of all of quantum electrodynamics (QED).

II. The building blocks: P(A to B), E(A to B) and j

The three basic ‘events’ (and, hence, amplitudes) in QED are the following:

1. P(A to B)

P(A to B) is the (probability) amplitude for a photon to travel from point A to B. However, I should immediately note that A and B are points in spacetime. Therefore, we associate them not only with some specific (x, y, z) position in space, but also with a some specific time t. Now, quantum-mechanical theory gives us an easy formula for P(A to B): it depends on the so-called (spacetime) interval between the two points A and B, i.e. I = Δr– Δt= (x2–x1)2+(y2–y1)2+(z2–z1)– (t2–t1)2. The point to note is that the spacetime interval takes both the distance in space as well as the ‘distance’ in time into account. As I mentioned already, this spacetime interval does not depend on our reference frame and, hence, it’s invariant (as long as we’re talking reference frames that move with constant speed relative to each other). Also note that we should measure time and distance in equivalent units when using that Δr– Δtformula for I. So we either measure distance in light-seconds or, else, we measure time in units that correspond to the time that’s needed for light to travel one meter. If no equivalent units are adopted, the formula is I = Δrc·Δt2.

Now, in quantum theory, anything is possible and, hence, not only do we allow for crooked paths, but we also allow for the difference in time to differ from  the time you’d expect a photon to need to travel along some curve (whose length we’ll denote by l), i.e. l/c. Hence, our photon may actually travel slower or faster than the speed of light c! There is one lucky break, however, that makes all come out alright: it’s easy to show that the amplitudes associated with the odd paths and strange timings generally cancel each other out. [That’s what the QED booklet shows.] Hence, what remains, are the paths that are equal or, importantly, those that very near to the so-called ‘light-like’ intervals in spacetime only. The net result is that light – even one single photon – effectively uses a (very) small core of space as it travels, as evidenced by the fact that even one single photon interferes with itself when traveling through a slit or a small hole!

[If you now wonder what it means for a photon to interfere for itself, let me just give you the easy explanation: it may change its path. We assume it was traveling in a straight line – if only because it left the source at some point in time and then arrived at the slit obviously – but so it no longer travels in a straight line after going through the slit. So that’s what we mean here.]

2. E(A to B)

E(A to B) is the (probability) amplitude for an electron to travel from point A to B. The formula for E(A to B) is much more complicated, and it’s the one I want to discuss somewhat more in detail in this post. It depends on some complex number j (see the next remark) and some real number n.

3. j

Finally, an electron could emit or absorb a photon, and the amplitude associated with this event is denoted by j, for junction number. It’s the same number j as the one mentioned when discussing E(A to B) above.

Now, this junction number is often referred to as the coupling constant or the fine-structure constant. However, the truth is, as I pointed out in my previous post, that these numbers are related, but they are not quite the same: α is the square of j, so we have α = j2. There is also one more, related, number: the gauge parameter, which is denoted by g (despite the g notation, it has nothing to do with gravitation). The value of g is the square root of 4πε0α, so g= 4πε0α. I’ll come back to this. Let me first make an awfully long digression on the fine-structure constant. It will be awfully long. So long that it’s actually part of the ‘core’ of this post actually.

Digression 2: on the fine-structure constant, Planck units and the Bohr radius

The value for j is approximately –0.08542454.

How do we know that?

The easy answer to that question is: physicists measured it. In fact, they usually publish the measured value as the square root of the (absolute value) of j, which is that fine-structure constant α. Its value is published (and updated) by the US National Institute on Standards and Technology. To be precise, the currently accepted value of α is 7.29735257×10−3. In case you doubt, just check that square root:

j = –0.08542454 ≈ –√0.00729735257 = –√α

As noted in Feynman’s (or Leighton’s) QED, older and/or more popular books will usually mention 1/α as the ‘magical’ number, so the ‘special’ number you may have seen is the inverse fine-structure constant, which is about 137, but not quite:

1/α = 137.035999074 ± 0.000000044

I am adding the standard uncertainty just to give you an idea of how precise these measurements are. 🙂 About 0.32 parts per billion (just divide the 137.035999074 number by the uncertainty). So that‘s the number that excites popular writers, including Leighton. Indeed, as Leighton puts it:

“Where does this number come from? Nobody knows. It’s one of the greatest damn mysteries of physics: a magic number that comes to us with no understanding by man. You might say the “hand of God” wrote that number, and “we don’t know how He pushed his pencil.” We know what kind of a dance to do experimentally to measure this number very accurately, but we don’t know what kind of dance to do on the computer to make this number come out, without putting it in secretly!”

Is it Leighton, or did Feynman really say this? Not sure. While the fine-structure constant is a very special number, it’s not the only ‘special’ number. In fact, we derive it from other ‘magical’ numbers. To be specific, I’ll show you how we derive it from the fundamental properties – as measured, of course – of the electron. So, in fact, I should say that we do know how to make this number come out, which makes me doubt whether Feynman really said what Leighton said he said. 🙂

So we can derive α from some other numbers. That brings me to the more complicated answer to the question as to what the value of j really is: j‘s value is the electron charge expressed in Planck units, which I’ll denote by –eP:

j = –eP

[You may want to reflect on this, and quickly verify on the Web. The Planck unit of electric charge, expressed in Coulomb, is about 1.87555×10–18 C. If you multiply that j = –eP, so with –0.08542454, you get the right answer: the electron charge is about –0.160217×10–18 C.]

Now that is strange.

Why? Well… For starters, when doing all those quantum-mechanical calculations, we like to think of j as a dimensionless number: a coupling constant. But so here we do have a dimension: electric charge.

Let’s look at the basics. If is –√α, and it’s also equal to –eP, then the fine-structure constant must also be equal to the square of the electron charge eP, so we can write:

α = eP2

You’ll say: yes, so what? Well… I am pretty sure that, if you’ve ever seen a formula for α, it’s surely not this simple j = –eP or α = eP2 formula. What you’ve seen, most likely, is one or more of the following expressions below :

Fine-structure constant formula

That’s a pretty impressive collection of physical constants, isn’t it? 🙂 They’re all different but, somehow, when we combine them in one or the other ratio (we have not less than five different expressions here (each identity is a separate expression), and I could give you a few more!), we get the very same number: α. Now that is what I call strange. Truly strange. Incomprehensibly weird!

You’ll say… Well… Those constants must all be related… Of course! That’s exactly the point I am making here. They are, but look how different they are: mmeasures mass, rmeasures distance, e is a charge, and so these are all very different numbers with very different dimensions. Yet, somehow, they are all related through this α number. Frankly, I do not know of any other expression that better illustrates some kind of underlying unity in Nature than the one with those five identities above.

Let’s have a closer look at those constants. You know most of them already. The only constants you may not have seen before are μ0Rand, perhaps, ras well as m. However, these can easily be defined as some easy function of the constants that you did see before, so let me quickly do that:

  1. The μ0 constant is the so-called magnetic constant. It’s something similar as ε0 and it’s referred to as the magnetic permeability of the vacuum. So it’s just like the (electric) permittivity of the vacuum (i.e. the electric constant ε0) and the only reason why this blog hasn’t mentioned this constant before is because I haven’t really discussed magnetic fields so far. I only talked about the electric field vector. In any case, you know that the electric and magnetic force are part and parcel of the same phenomenon (i.e. the electromagnetic interaction between charged particles) and, hence, they are closely related. To be precise, μ0ε0 = 1/c= c–2. So that shows the first and second expression for α are, effectively, fully equivalent. [Just in case you’d doubt that μ0ε0 = 1/c2, let me give you the values: μ0 = 4π·10–7 N/A2, and ε0 = (1/4π·c2)·10C2/N·m2. Just plug them in, and you’ll see it’s bang on. Moreover, note that the ampere (A) unit is equal to the coulomb per second unit (C/s), so even the units come out alright. 🙂 Of course they do!]
  2. The ke constant is the Coulomb constant and, from its definition ke = 1/4πε0, it’s easy to see how those two expressions are, in turn, equivalent with the third expression for α.
  3. The Rconstant is the so-called von Klitzing constant. Huh? Yes. I know. I am pretty sure you’ve never ever heard of that one before. Don’t worry about it. It’s, quite simply, equal to Rh/e2. Hence, substituting (and don’t forget that h = 2πħ) will demonstrate the equivalence of the fourth expression for α.
  4. Finally, the re factor is the classical electron radius, which is usually written as a function of me, i.e. the electron mass: re = e2/4πε0mec2. Also note that this also implies that reme = e2/4πε0c2. In words: the product of the electron mass and the electron radius is equal to some constant involving the electron (e), the electric constant (ε0), and c (the speed of light).

I am sure you’re under some kind of ‘formula shock’ now. But you should just take a deep breath and read on. The point to note is that all these very different things are all related through α.

So, again, what is that α really? Well… A strange number indeed. It’s dimensionless (so we don’t measure in kg, m/s, eV·s or whatever) and it pops up everywhere. [Of course, you’ll say: “What’s everywhere? This is the first time I‘ve heard of it!” :-)]

Well… Let me start by explaining the term itself. The fine structure in the name refers to the splitting of the spectral lines of atoms. That’s a very fine structure indeed. 🙂 We also have a so-called hyperfine structure. Both are illustrated below for the hydrogen atom. The numbers n, JI, and are quantum numbers used in the quantum-mechanical explanation of the emission spectrum, which is  also depicted below, but note that the illustration gives you the so-called Balmer series only, i.e. the colors in the visible light spectrum (there are many more ‘colors’ in the high-energy ultraviolet and the low-energy infrared range).



To be precise: (1) n is the principal quantum number: here it takes the values 1 or 2, and we could say these are the principal shells; (2) the S, P, D,… orbitals (which are usually written in lower case: s, p, d, f, g, h and i) correspond to the (orbital) angular momentum quantum number l = 0, 1, 2,…, so we could say it’s the subshell; (3) the J values correspond to the so-called magnetic quantum number m, which goes from –l to +l; (4) the fourth quantum number is the spin angular momentum s. I’ve copied another diagram below so you see how it works, more or less, that is.

hydrogen spectrum

Now, our fine-structure constant is related to these quantum numbers. How exactly is a bit of a long story, and so I’ll just copy Wikipedia’s summary on this: ” The gross structure of line spectra is the line spectra predicted by the quantum mechanics of non-relativistic electrons with no spin. For a hydrogenic atom, the gross structure energy levels only depend on the principal quantum number n. However, a more accurate model takes into account relativistic and spin effects, which break the degeneracy of the the energy levels and split the spectral lines. The scale of the fine structure splitting relative to the gross structure energies is on the order of ()2, where Z is the atomic number and α is the fine-structure constant.” There you go. You’ll say: so what? Well… Nothing. If you aren’t amazed by that, you should stop reading this.

It is an ‘amazing’ number, indeed, and, hence, it does quality for being “one of the greatest damn mysteries of physics”, as Feynman and/or Leighton put it. Having said that, I would not go as far as to write that it’s “a magic number that comes to us with no understanding by man.” In fact, I think Feynman/Leighton could have done a much better job when explaining what it’s all about. So, yes, I hope to do better than Leighton here and, as he’s still alive, I actually hope he reads this. 🙂

The point is: α is not the only weird number. What’s particular about it, as a physical constant, is that it’s dimensionless, because it relates a number of other physical constants in such a way that the units fall away. Having said that, the Planck or Boltzmann constant are at least as weird.

So… What is this all about? Well… You’ve probably heard about the so-called fine-tuning problem in physics and, if you’re like me, your first reaction will be to associate fine-tuning with fine-structure. However, the two terms have nothing in common, except for four letters. 🙂 OK. Well… I am exaggerating here. The two terms are actually related, to some extent at least, but let me explain how.

The term fine-tuning refers to the fact that all the parameters or constants in the so-called Standard Model of physics are, indeed, all related to each other in the way they are. We can’t sort of just turn the knob of one and change it, because everything falls apart then. So, in essence, the fine-tuning problem in physics is more like a philosophical question: why is the value of all these physical constants and parameters exactly what it is? So it’s like asking: could we change some of the ‘constants’ and still end up with the world we’re living in? Or, if it would be some different world, how would it look like? What if was some other number? What if ke or ε0 was some other number? In short, and in light of those expressions for α, we may rephrase the question as: why is α what is is?

Of course, that’s a question one shouldn’t try to answer before answering some other, more fundamental, question: how many degrees of freedom are there really? Indeed, we just saw that ke and εare intimately related through some equation, and other constants and parameters are related too. So the question is like: what are the ‘dependent’ and the ‘independent’ variables in this so-called Standard Model?

There is no easy answer to that question. In fact, one of the reasons why I find physics so fascinating is that one cannot easily answer such questions. There are the obvious relationships, of course. For example, the ke = 1/4πεrelationship, and the context in which they are used (Coulomb’s Law) does, indeed, strongly suggest that both constants are actually part and parcel of the same thing. Identical, I’d say. Likewise, the μ0ε0 = 1/crelation also suggests there’s only one degree of freedom here, just like there’s only one degree of freedom in that ω/k = relationship (if we set a value for ω, we have k, and vice versa). But… Well… I am not quite sure how to phrase this, but… What physical constants could be ‘variables’ indeed?

It’s pretty obvious that the various formulas for α cannot answer that question: you could stare at them for days and weeks and months and years really, but I’d suggest you use your time to read more of Feynman’s real Lectures instead. 🙂 One point that may help to come to terms with this question – to some extent, at least – is what I casually mentioned above already: the fine-structure constant is equal to the square of the electron charge expressed in Planck units: α = eP2.

Now, that’s very remarkable because Planck units are some kind of ‘natural units’ indeed (for the detail, see my previous post: among other things, it explains what these Planck units really are) and, therefore, it is quite tempting to think that we’ve actually got only one degree of freedom here: α itself. All the rest should follow from it.


It should… But… Does it?

The answer is: yes and no. To be frank, it’s more no than yes because, as I noted a couple of times already, the fine-structure constant relates a lot of stuff but it’s surely not the only significant number in the Universe. For starters, I said that our E(A to B) formula has two ‘variables’:

  1. We have that complex number j, which, as mentioned, is equal to the electron charge expressed in Planck units. [In case you wonder why –eP ≈ –0.08542455 is said to be an amplitude, i.e. a complex number or an ‘arrow’… Well… Complex numbers include the real numbers and, hence, –0.08542455 is both real and complex. When combining ‘arrows’ or, to be precise, when multiplying some complex number with –0.08542455, we will (a) shrink the original arrow to about 8.5% of its original value (8.542455% to be precise) and (b) rotate it over an angle of plus or minus 180 degrees. In other words, we’ll reverse its direction. Hence, using Euler’s notation for complex numbers, we can write: –1 = eiπ eiπ and, hence, –0.085 = 0.085·eiπ = 0.085·eiπ. So, in short, yes, j is a complex number, or an ‘arrow’, if you prefer that term.]
  2. We also have some some real number n in the E(A to B) formula. So what’s the n? Well… Believe it or not, it’s the electron mass! Isn’t that amazing?

You’ll say: “Well… Hmm… I suppose so.” But then you may – and actually should – also wonder: the electron mass? In what units? Planck units again? And are we talking relativistic mass (i.e. its total mass, including the equivalent mass of its kinetic energy) or its rest mass only? And we were talking α here, so can we relate it to α too, just like the electron charge?

These are all very good questions. Let’s start with the second one. We’re talking rather slow-moving electrons here, so the relativistic mass (m) and its rest mass (m0) is more or less the same. Indeed, the Lorentz factor γ in the m = γm0 equation is very close to 1 for electrons moving at their typical speed. So… Well… That question doesn’t matter very much. Really? Yes. OK. Because you’re doubting, I’ll quickly show it to you. What is their ‘typical’ speed?

We know we shouldn’t attach too much importance to the concept of an electron in orbit around some nucleus (we know it’s not like some planet orbiting around some star) and, hence, to the concept of speed or velocity (velocity is speed with direction) when discussing an electron in an atom. The concept of momentum (i.e. velocity combined with mass or energy) is much more relevant. There’s a very easy mathematical relationship that gives us some clue here: the Uncertainty Principle. In fact, we’ll use the Uncertainty Principle to relate the momentum of an electron (p) to the so-called Bohr radius r (think of it as the size of a hydrogen atom) as follows: p ≈ ħ/r. [I’ll come back on this in a moment, and show you why this makes sense.]

Now we also know its kinetic energy (K.E.) is mv2/2, which we can write as p2/2m. Substituting our p ≈ ħ/r conjecture, we get K.E. = mv2/2 = ħ2/2mr2. This is equivalent to m2v2 = ħ2/r(just multiply both sides with m). From that, we get v = ħ/mr. Now, one of the many relations we can derive from the formulas for the fine-structure constant is re = α2r. [I haven’t showed you that yet, but I will shortly. It’s a really amazing expression. However, as for now, just accept it as a simple formula for interim use in this digression.] Hence, r = re2. The rfactor in this expression is the so-called classical electron radius. So we can now write v = ħα2/mre. Let’s now throw c in: v/c = α2ħ/mcre. However, from that fifth expression for α, we know that ħ/mcre = α, so we get v/c = α. We have another amazing result here: the v/c ratio for an electron (i.e. its speed expressed as a fraction of the speed of light) is equal to that fine-structure constant α. So that’s about 1/137, so that’s less than 1% of the speed of light. Now… I’ll leave it to you to calculate the Lorentz factor γ but… Well… It’s obvious that it will be very close to 1. 🙂 Hence, the electron’s speed – however we want to visualize that – doesn’t matter much indeed, so we should not worry about relativistic corrections in the formulas.

Let’s now look at the question in regard to the Planck units. If you know nothing at all about them, I would advise you to read what I wrote about them in my previous post. 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.]

Now, you may or may not remember that the Planck time and length units are unimaginably small, but that the Planck mass unit is actually quite sizable—at the atomic scale, that is. Indeed, the Planck mass is something huge, like the mass of an eyebrow hair, or a flea egg. Is that huge? Yes. Because if you’d want to pack it in a Planck-sized particle, it would make for a tiny black hole. 🙂 No kidding. That’s the physical significance of the Planck mass and the Planck length and, yes, it’s weird. 🙂

Let me give you some values. First, the Planck mass itself: it’s about 2.1765×10−8 kg. Again, if you think that’s tiny, think again. From the E = mc2 equivalence relationship, we get that this is equivalent to 2 giga-joule, approximately. Just to give an idea, that’s like the monthly electricity consumption of an average American family. So that’s huge indeed! 🙂 [Many people think that nuclear energy involves the conversion of mass into energy, but the story is actually more complicated than that. In any case… I need to move on.]

Let me now give you the electron mass expressed in the Planck mass unit:

  1. Measured in our old-fashioned super-sized SI kilogram unit, the electron mass is me = 9.1×10–31 kg.
  2. The Planck mass is mP = 2.1765×10−8 kg.
  3. Hence, the electron mass expressed in Planck units is meP = me/mP = (9.1×10–31 kg)/(2.1765×10−8 kg) = 4.181×10−23.

We can, once again, write that as some function of the fine-structure constant. More specifically, we can write:

meP = α/reP = α/α2rP  = 1/αrP

So… Well… Yes: yet another amazing formula involving α.

In this formula, we have reP and rP, which are the (classical) electron radius and the Bohr radius expressed in Planck (length) units respectively. So you can see what’s going on here: we have all kinds of numbers here expressed in Planck units: a charge, a radius, a mass,… And we can relate all of them to the fine-structure constant

Why? Who knows? I don’t. As Leighton puts it: that’s just the way “God pushed His pencil.” 🙂

Note that the beauty of natural units ensures that we get the same number for the (equivalent) energy of an electron. Indeed, from the E = mc2 relation, we know the mass of an electron can also be written as 0.511 MeV/c2. Hence, the equivalent energy is 0.511 MeV (so that’s, quite simply, the same number but without the 1/cfactor). Now, the Planck energy EP (in eV) is 1.22×1028 eV, so we get EeP = Ee/EP = (0.511×10eV)/(1.22×1028 eV) = 4.181×10−23. So it’s exactly the same as the electron mass expressed in Planck units. Isn’t that nice? 🙂

Now, are all these numbers dimensionless, just like α? The answer to that question is complicated. Yes, and… Well… No:

  1. Yes. They’re dimensionless because they measure something in natural units, i.e. Planck units, and, hence, that’s some kind of relative measure indeed so… Well… Yes, dimensionless.
  2. No. They’re not dimensionless because they do measure something, like a charge, a length, or a mass, and when you chose some kind of relative measure, you still need to define some gauge, i.e. some kind of standard measure. So there’s some ‘dimension’ involved there.

So what’s the final answer? Well… The Planck units are not dimensionless. All we can say is that they are closely related, physically. I should also add that we’ll use the electron charge and mass (expressed in Planck units) in our amplitude calculations as a simple (dimensionless) number between zero and one. So the correct answer to the question as to whether these numbers have any dimension is: expressing some quantities in Planck units sort of normalizes them, so we can use them directly in dimensionless calculations, like when we multiply and add amplitudes.

Hmm… Well… I can imagine you’re not very happy with this answer but it’s the best I can do. Sorry. I’ll let you further ponder that question. I need to move on.  

Note that that 4.181×10−23 is still a very small number (23 zeroes after the decimal point!), even if it’s like 46 million times larger than the electron mass measured in our conventional SI unit (i.e. 9.1×10–31 kg). Does such small number make any sense? The answer is: yes, it does. When we’ll finally start discussing that E(A to B) formula (I’ll give it to you in a moment), you’ll see that a very small number for n makes a lot of sense.

Before diving into it all, let’s first see if that formula for that alpha, that fine-structure constant, still makes sense with me expressed in Planck units. Just to make sure. 🙂 To do that, we need to use the fifth (last) expression for a, i.e. the one with re in it. Now, in my previous post, I also gave some formula for re: re = e2/4πε0mec2, which we can re-write as reme = e2/4πε0c2. If we substitute that expression for reme  in the formula for α, we can calculate α from the electron charge, which indicates both the electron radius and its mass are not some random God-given variable, or “some magic number that comes to us with no understanding by man“, as Feynman – well… Leighton, I guess – puts it. No. They are magic numbers alright, one related to another through the equally ‘magic’ number α, but so I do feel we actually can create some understanding here.

At this point, I’ll digress once again, and insert some quick back-of-the-envelope argument from Feynman’s very serious Caltech Lectures on Physics, in which, as part of the introduction to quantum mechanics, he calculates the so-called Bohr radius from Planck’s constant h. Let me quickly explain: the Bohr radius is, roughly speaking, the size of the simplest atom, i.e. an atom with one electron (so that’s hydrogen really). So it’s not the classical electron radius re. However, both are also related to that ‘magical number’ α. To be precise, if we write the Bohr radius as r, then re = α2r ≈ 0.000053… times r, which we can re-write as:

α = √(re /r) = (re /r)1/2

So that’s yet another amazing formula involving the fine-structure constant. In fact, it’s the formula I used as an ‘interim’ expression to calculate the relative speed of electrons. I just used it without any explanation there, but I am coming back to it here. Alpha again…

Just think about it for a while. In case you’d still doubt the magic of that number, let me write what we’ve discovered so far:

(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) Finally – I’ll show you in a moment – 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. 🙂

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),… In short,


Now that should answer the question in regard to the degrees of freedom we have here, doesn’t it? It looks like we’ve got only one degree of freedom here. Indeed, if we’ve got some value for α, then we’ve have the electron charge, and from the electron charge, we can calculate the Bohr radius r (as I will show below), and if we have r, we have mand re. And then we can also calculate v, which gives us its momentum (mv) and its kinetic energy (mv2/2). In short,


Isn’t that amazing? Hmm… You should reserve your judgment as for now, and carefully go over all of the formulas above and verify my statement. If you do that, you’ll probably struggle to find the Bohr radius from the charge (i.e. from α). So let me show you how you do that, because it will also show you why you should, indeed, reserve your judgment. In other words, I’ll show you why alpha does NOT give us everything! The argument below will, finally, prove some of the formulas that I didn’t prove above. Let’s go for it:

1. If we assume that (a) an electron takes some space – which I’ll denote by r 🙂 – and (b) that it has some momentum p because of its mass m and its velocity v, then the ΔxΔp = ħ relation (i.e. the Uncertainty Principle in its roughest form) suggests that the order of magnitude of r and p should be related in the very same way. Hence, let’s just boldly write r ≈ ħ/p and see what we can do with that. So we equate Δx with r and Δp with p. As Feynman notes, this is really more like a ‘dimensional analysis’ (he obviously means something very ‘rough’ with that) and so we don’t care about factors like 2 or 1/2. [Indeed, note that the more precise formulation of the Uncertainty Principle is σxσ≥ ħ/2.] In fact, we didn’t even bother to define r very rigorously. We just don’t care about precise statements at this point. We’re only concerned about orders of magnitude. [If you’re appalled by the rather rude approach, I am sorry for that, but just try to go along with it.]

2. From our discussions on energy, we know that the kinetic energy is mv2/2, which we can write as p2/2m so we get rid of the velocity factor. [Why? Because we can’t really imagine what it is anyway. As I said a couple of times already, we shouldn’t think of electrons as planets orbiting around some star. That model doesn’t work.] So… What’s next? Well… Substituting our p ≈ ħ/r conjecture, we get K.E. = ħ2/2mr2. So that’s a formula for the kinetic energy. Next is potential.

3. Unfortunately, the discussion on potential energy is a bit more complicated. You’ll probably remember that we had an easy and very comprehensible formula for the energy that’s needed (i.e. the work that needs to be done) to bring two charges together from a large distance (i.e. infinity). Indeed, we derived that formula directly from Coulomb’s Law (and Newton’s law of force) and it’s U = q1q2/4πε0r12. [If you think I am going too fast, sorry, please check for yourself by reading my other posts.] Now, we’re actually talking about the size of an atom here in my previous post, so one charge is the proton (+e) and the other is the electron (–e), so the potential energy is U = P.E. = –e2/4πε0r, with r the ‘distance’ between the proton and the electron—so that’s the Bohr radius we’re looking for!

[In case you’re struggling a bit with those minus signs when talking potential energy  – I am not ashamed to admit I did! – let me quickly help you here. It has to do with our reference point: the reference point for measuring potential energy is at infinity, and it’s zero there (that’s just our convention). Now, to separate the proton and the electron, we’d have to do quite a lot of work. To use an analogy: imagine we’re somewhere deep down in a cave, and we have to climb back to the zero level. You’ll agree that’s likely to involve some sweat, don’t you? Hence, the potential energy associated with us being down in the cave is negative. Likewise, if we write the potential energy between the proton and the electron as U(r), and the potential energy at the reference point as U(∞) = 0, then the work to be done to separate the charges, i.e. the potential difference U(∞) – U(r), will be positive. So U(∞) – U(r) = 0 – U(r) > 0 and, hence, U(r) < 0. If you still don’t ‘get’ this, think of the electron being in some (potential) well, i.e. below the zero level, and so it’s potential energy is less than zero. Huh? Sorry. I have to move on. :-)]

4. We can now write the total energy (which I’ll denote by E, but don’t confuse it with the electric field vector!) as

E = K.E. + P.E. =  ħ2/2mr– e2/4πε0r

Now, the electron (whatever it is) is, obviously, in some kind of equilibrium state. Why is that obvious? Well… Otherwise our hydrogen atom wouldn’t or couldn’t exist. 🙂 Hence, it’s in some kind of energy ‘well’ indeed, at the bottom. Such equilibrium point ‘at the bottom’ is characterized by its derivative (in respect to whatever variable) being equal to zero. Now, the only ‘variable’ here is r (all the other symbols are physical constants), so we have to solve for dE/dr = 0. Writing it all out yields:

dE/dr = –ħ2/mr+ e2/4πε0r= 0 ⇔ r = 4πε0ħ2/me2

You’ll say: so what? Well… We’ve got a nice formula for the Bohr radius here, and we got it in no time! 🙂 But the analysis was rough, so let’s check if it’s any good by putting the values in:

r = 4πε0h2/me2

= [(1/(9×109) C2/N·m2)·(1.055×10–34 J·s)2]/[(9.1×10–31 kg)·(1.6×10–19 C)2]

= 53×10–12 m = 53 pico-meter (pm)

So what? Well… Double-check it on the Internet: the Bohr radius is, effectively, about 53 trillionths of a meter indeed! So we’re right on the spot! 

[In case you wonder about the units, note that mass is a measure of inertia: one kg is the mass of an object which, subject to a force of 1 newton, will accelerate at the rate of 1 m/s per second. Hence, we write F = m·a, which is equivalent to m = F/a. Hence, the kg, as a unit, is equivalent to 1 N/(m/s2). If you make this substitution, we get r in the unit we want to see: [(C2/N·m2)·(N2·m2·s2)/[(N·s2/m)·C2] = m.]

Moreover, if we take that value for r and put it in the (total) energy formula above, we’d find that the energy of the electron is –13.6 eV. [Don’t forget to convert from joule to electronvolt when doing the calculation!] Now you can check that on the Internet too: 13.6 eV is exactly the amount of energy that’s needed to ionize a hydrogen atom (i.e. the energy that’s needed to kick the electron out of that energy well)!

Waw ! Isn’t it great that such simple calculations yield such great results? 🙂 [Of course, you’ll note that the omission of the 1/2 factor in the Uncertainty Principle was quite strategic. :-)] Using the r = 4πε0ħ2/meformula for the Bohr radius, you can now easily check the re = α2r formula. You should find what we jotted down already: the classical electron radius is equal to re = e2/4πε0mec2. To be precise, re = (53×10–6)·(53×10–12m) = 2.8×10–15 m. Now that’s again something you should check on the Internet. Guess what? […] It’s right on the spot again. 🙂

We can now also check that α = m·re formula: α = m·r= 4.181×10−23 times… Hey! Wait! We have to express re in Planck units as well, of course! Now, (2.81794×10–15 m)/(1.616×10–35 m) ≈ 1.7438 ×1020. So now we get 4.181×10−23 times 1.7438×1020 = 7.29×10–3 = 0.00729 ≈ 1/137. Bingo! We got the magic number once again. 🙂

So… Well… Doesn’t that confirm we actually do have it all with α?

Well… Yes and no… First, you should note that I had to use h in that calculation of the Bohr radius. Moreover, the other physical constants (most notably c and the Coulomb constant) were actually there as well, ‘in the background’ so to speak, because one needs them to derive the formulas we used above. And then we have the equations themselves, of course, most notably that Uncertainty Principle… So… Well…

It’s not like God gave us one number only (α) and that all the rest flows out of it. We have a whole bunch of ‘fundamental’ relations and ‘fundamental’ constants here.

Having said that, it’s true that statement still does not diminish the magic of alpha.

Hmm… Now you’ll wonder: how many? How many constants do we need in all of physics?

Well… I’d say, you should not only ask about the constants: you should also ask about the equations: how many equations do we need in all of physics? [Just for the record, I had to smile when the Hawking of the movie says that he’s actually looking for one formula that sums up all of physics. Frankly, that’s a nonsensical statement. Hence, I think the real Hawking never said anything like that. Or, if he did, that it was one of those statements one needs to interpret very carefully.]

But let’s look at a few constants indeed. For example, if we have c, h and α, then we can calculate the electric charge e and, hence, the electric constant ε= e2/2αhc. From that, we get Coulomb’s constant ke, because ke is defined as 1/4πε0… But…

Hey! Wait a minute! How do we know that ke = 1/4πε0? Well… From experiment. But… Yes? That means 1/4π is some fundamental proportionality coefficient too, isn’t it?

Wow! You’re smart. That’s a good and valid remark. In fact, we use the so-called reduced Planck constant ħ in a number of calculations, and so that involves a 2π factor too (ħ = h/2π). Hence… Well… Yes, perhaps we should consider 2π as some fundamental constant too! And, then, well… Now that I think of it, there’s a few other mathematical constants out there, like Euler’s number e, for example, which we use in complex exponentials.


I am joking, right? I am not saying that 2π and Euler’s number are fundamental ‘physical’ constants, am I? [Note that it’s a bit of a nuisance we’re also using the symbol for Euler’s number, but so we’re not talking the electron charge here: we’re talking that 2.71828…etc number that’s used in so-called ‘natural’ exponentials and logarithms.]

Well… Yes and no. They’re mathematical constants indeed, rather than physical, but… Well… I hope you get my point. What I want to show here, is that it’s quite hard to say what’s fundamental and what isn’t. We can actually pick and choose a bit among all those constants and all those equations. As one physicist puts its: it depends on how we slice it. The one thing we know for sure is that a great many things are related, in a physical way (α connects all of the fundamental properties of the electron, for example) and/or in a mathematical way (2π connects not only the circumference of the unit circle with the radius but quite a few other constants as well!), but… Well… What to say? It’s a tough discussion and I am not smart enough to give you an unambiguous answer. From what I gather on the Internet, when looking at the whole Standard Model (including the strong force, the weak force and the Higgs field), we’ve got a few dozen physical ‘fundamental’ constants, and then a few mathematical ones as well.

That’s a lot, you’ll say. Yes. At the same time, it’s not an awful lot. Whatever number it is, it does raise a very fundamental question: why are they what they are? That brings us back to that ‘fine-tuning’ problem. Now, I can’t make this post too long (it’s way too long already), so let me just conclude this discussion by copying Wikipedia on that question, because what it has on this topic is not so bad:

“Some physicists have explored the notion that if the physical constants had sufficiently different values, our Universe would be so radically different that intelligent life would probably not have emerged, and that our Universe therefore seems to be fine-tuned for intelligent life. The anthropic principle states a logical truism: the fact of our existence as intelligent beings who can measure physical constants requires those constants to be such that beings like us can exist.

I like this. But the article then adds the following, which I do not like so much, because I think it’s a bit too ‘frivolous’:

“There are a variety of interpretations of the constants’ values, including that of a divine creator (the apparent fine-tuning is actual and intentional), or that ours is one universe of many in a multiverse (e.g. the many-worlds interpretation of quantum mechanics), or even that, if information is an innate property of the universe and logically inseparable from consciousness, a universe without the capacity for conscious beings cannot exist.”

Hmm… As said, I am quite happy with the logical truism: we are there because alpha (and a whole range of other stuff) is what it is, and we can measure alpha (and a whole range of other stuff) as what it is, because… Well… Because we’re here. Full stop. As for the ‘interpretations’, I’ll let you think about that for yourself. 🙂

I need to get back to the lesson. Indeed, this was just a ‘digression’. My post was about the three fundamental events or actions in quantum electrodynamics, and so I was talking about that E(A to B) formula. However, I had to do that digression on alpha to ensure you understand what I want to write about that. So let me now get back to it. End of digression. 🙂

The E(A to B) formula

Indeed, I must assume that, with all these digressions, you are truly despairing now. Don’t. We’re there! We’re finally ready for the E(A to B) formula! Let’s go for it.

We’ve now got those two numbers measuring the electron charge and the electron mass in Planck units respectively. They’re fundamental indeed and so let’s loosen up on notation and just write them as e and m respectively. Let me recap:

1. The value of e is approximately –0.08542455, and it corresponds to the so-called junction number j, which is the amplitude for an electron-photon coupling. When multiplying it with another amplitude (to find the amplitude for an event consisting of two sub-events, for example), it corresponds to a ‘shrink’ to less than one-tenth (something like 8.5% indeed, corresponding to the magnitude of e) and a ‘rotation’ (or a ‘turn’) over 180 degrees, as mentioned above.

Please note what’s going on here: we have a physical quantity, the electron charge (expressed in Planck units), and we use it in a quantum-mechanical calculation as a dimensionless (complex) number, i.e. as an amplitude. So… Well… That’s what physicists mean when they say that the charge of some particle (usually the electric charge but, in quantum chromodynamics, it will be the ‘color’ charge of a quark) is a ‘coupling constant’.

2. We also have m, the electron mass, and we’ll use in the same way, i.e. as some dimensionless amplitude. As compared to j, it’s is a very tiny number: approximately 4.181×10−23. So if you look at it as an amplitude, indeed, then it corresponds to an enormous ‘shrink’ (but no turn) of the amplitude(s) that we’ll be combining it with.

So… Well… How do we do it?

Well… At this point, Leighton goes a bit off-track. Just a little bit. 🙂 From what he writes, it’s obvious that he assumes the frequency (or, what amounts to the same, the de Broglie wavelength) of an electron is just like the frequency of a photon. Frankly, I just can’t imagine why and how Feynman let this happen. It’s wrong. Plain wrong. As I mentioned in my introduction already, an electron traveling through space is not like a photon traveling through space.

For starters, an electron is much slower (because it’s a matter-particle: hence, it’s got mass). Secondly, the de Broglie wavelength and/or frequency of an electron is not like that of a photon. For example, if we take an electron and a photon having the same energy, let’s say 1 eV (that corresponds to infrared light), then the de Broglie wavelength of the electron will be 1.23 nano-meter (i.e. 1.23 billionths of a meter). Now that’s about one thousand times smaller than the wavelength of our 1 eV photon, which is about 1240 nm. You’ll say: how is that possible? If they have the same energy, then the f = E/h and ν = E/h should give the same frequency and, hence, the same wavelength, no?

Well… No! Not at all! Because an electron, unlike the photon, has a rest mass indeed – measured as not less than 0.511 MeV/c2, to be precise (note the rather particular MeV/c2 unit: it’s from the E = mc2 formula) – one should use a different energy value! Indeed, we should include the rest mass energy, which is 0.511 MeV. So, almost all of the energy here is rest mass energy! There’s also another complication. For the photon, there is an easy relationship between the wavelength and the frequency: it has no mass and, hence, all its energy is kinetic, or movement so to say, and so we can use that ν = E/h relationship to calculate its frequency ν: it’s equal to ν = E/h = (1 eV)/(4.13567×10–15 eV·s) ≈ 0.242×1015 Hz = 242 tera-hertz (1 THz = 1012 oscillations per second). Now, knowing that light travels at the speed of light, we can check the result by calculating the wavelength using the λ = c/ν relation. Let’s do it: (2.998×10m/s)/(242×1012 Hz) ≈ 1240 nm. So… Yes, done!

But so we’re talking photons here. For the electron, the story is much more complicated. That wavelength I mentioned was calculated using the other of the two de Broglie relations: λ = h/p. So that uses the momentum of the electron which, as you know, is the product of its mass (m) and its velocity (v): p = mv. You can amuse yourself and check if you find the same wavelength (1.23 nm): you should! From the other de Broglie relation, f = E/h, you can also calculate its frequency: for an electron moving at non-relativistic speeds, it’s about 0.123×1021 Hz, so that’s like 500,000 times the frequency of the photon we we’re looking at! When multiplying the frequency and the wavelength, we should get its speed. However, that’s where we get in trouble. Here’s the problem with matter waves: they have a so-called group velocity and a so-called phase velocity. The idea is illustrated below: the green dot travels with the wave packet – and, hence, its velocity corresponds to the group velocity – while the red dot travels with the oscillation itself, and so that’s the phase velocity. [You should also remember, of course, that the matter wave is some complex-valued wavefunction, so we have both a real as well as an imaginary part oscillating and traveling through space.]

Wave_group (1)

To be precise, the phase velocity will be superluminal. Indeed, using the usual relativistic formula, we can write that p = γm0v and E = γm0c2, with v the (classical) velocity of the electron and what it always is, i.e. the speed of light. Hence, λ = h/γm0v and = γm0c2/h, and so λf = c2/v. Because v is (much) smaller than c, we get a superluminal velocity. However, that’s the phase velocity indeed, not the group velocity, which corresponds to v. OK… I need to end this digression.

So what? Well, to make a long story short, the ‘amplitude framework’ for electrons is differerent. Hence, the story that I’ll be telling here is different from what you’ll read in Feynman’s QED. I will use his drawings, though, and his concepts. Indeed, despite my misgivings above, the conceptual framework is sound, and so the corrections to be made are relatively minor.

So… We’re looking at E(A to B), i.e. the amplitude for an electron to go from point A to B in spacetime, and I said the conceptual framework is exactly the same as that for a photon. Hence, the electron can follow any path really. It may go in a straight line and travel at a speed that’s consistent with what we know of its momentum (p), but it may also follow other paths. So, just like the photon, we’ll have some so-called propagator function, which gives you amplitudes based on the distance in space as well as in the distance in ‘time’ between two points. Now, Ralph Leighton identifies that propagator function with the propagator function for the photon, i.e. P(A to B), but that’s wrong: it’s not the same.

The propagator function for an electron depends on its mass and its velocity, and/or on the combination of both (like it momentum p = mv and/or its kinetic energy: K.E. = mv2 = p2/2m). So we have a different propagator function here. However, I’ll use the same symbol for it: P(A to B).

So, the bottom line is that, because of the electron’s mass (which, remember, is a measure for inertia), momentum and/or kinetic energy (which, remember, are conserved in physics), the straight line is definitely the most likely path, but (big but!), just like the photon, the electron may follow some other path as well.

So how do we formalize that? Let’s first associate an amplitude P(A to B) with an electron traveling from point A to B in a straight line and in a time that’s consistent with its velocity. Now, as mentioned above, the P here stands for propagator function, not for photon, so we’re talking a different P(A to B) here than that P(A to B) function we used for the photon. Sorry for the confusion. 🙂 The left-hand diagram below then shows what we’re talking about: it’s the so-called ‘one-hop flight’, and so that’s what the P(A to B) amplitude is associated with.

Diagram 1Now, the electron can follow other paths. For photons, we said the amplitude depended on the spacetime interval I: when negative or positive (i.e. paths that are not associated with the photon traveling in a straight line and/or at the speed of light), the contribution of those paths to the final amplitudes (or ‘final arrow’, as it was called) was smaller.

For an electron, we have something similar, but it’s modeled differently. We say the electron could take a ‘two-hop flight’ (via point C or C’), or a ‘three-hop flight’ (via D and E) from point A to B. Now, it makes sense that these paths should be associated with amplitudes that are much smaller. Now that’s where that n-factor comes in. We just put some real number n in the formula for the amplitude for an electron to go from A to B via C, which we write as:

P(A to C)∗n2∗P(C to B)

Note what’s going on here. We multiply two amplitudes, P(A to C) and P(C to B), which is OK, because that’s what the rules of quantum mechanics tell us: if an ‘event’ consists of two sub-events, we need to multiply the amplitudes (not the probabilities) in order to get the amplitude that’s associated with both sub-events happening. However, we add an extra factor: n2. Note that it must be some very small number because we have lots of alternative paths and, hence, they should not be very likely! So what’s the n? And why n2 instead of just n?

Well… Frankly, I don’t know. Ralph Leighton boldly equates n to the mass of the electron. Now, because he obviously means the mass expressed in Planck units, that’s the same as saying n is the electron’s energy (again, expressed in Planck’s ‘natural’ units), so n should be that number m = meP = EeP = 4.181×10−23. However, I couldn’t find any confirmation on the Internet, or elsewhere, of the suggested n = m identity, so I’ll assume n = m indeed, but… Well… Please check for yourself. It seems the answer is to be found in a mathematical theory that helps physicists to actually calculate j and n from experiment. It’s referred to as perturbation theory, and it’s the next thing on my study list. As for now, however, I can’t help you much. I can only note that the equation makes sense.

Of course, it does: inserting a tiny little number n, close to zero, ensures that those other amplitudes don’t contribute too much to the final ‘arrow’. And it also makes a lot of sense to associate it with the electron’s mass: if mass is a measure of inertia, then it should be some factor reducing the amplitude that’s associated with the electron following such crooked path. So let’s go along with it, and see what comes out of it.

A three-hop flight is even weirder and uses that n2 factor two times:

P(A to E)∗n2∗P(E to D)∗n2∗P(D to B)

So we have an (n2)= nfactor here, which is good, because two hops should be much less likely than one hop. So what do we get? Well… (4.181×10−23)≈ 305×10−92. Pretty tiny, huh? 🙂 Of course, any point in space is a potential hop for the electron’s flight from point A to B and, hence, there’s a lot of paths and a lot of amplitudes (or ‘arrows’ if you want), which, again, is consistent with a very tiny value for n indeed.

So, to make a long story short, E(A to B) will be a giant sum (i.e. some kind of integral indeed) of a lot of different ways an electron can go from point A to B. It will be a series of terms P(A to E) + P(A to C)∗n2∗P(C to B) + P(A to E)∗n2∗P(E to D)∗n2∗P(D to B) + … for all possible intermediate points C, D, E, and so on.

What about the j? The junction number of coupling constant. How does that show up in the E(A to B) formula? Well… Those alternative paths with hops here and there are actually the easiest bit of the whole calculation. Apart from taking some strange path, electrons can also emit and/or absorb photons during the trip. In fact, they’re doing that constantly actually. Indeed, the image of an electron ‘in orbit’ around the nucleus is that of an electron exchanging so-called ‘virtual’ photons constantly, as illustrated below. So our image of an electron absorbing and then emitting a photon (see the diagram on the right-hand side) is really like the tiny tip of a giant iceberg: most of what’s going on is underneath! So that’s where our junction number j comes in, i.e. the charge (e) of the electron.

So, when you hear that a coupling constant is actually equal to the charge, then this is what it means: you should just note it’s the charge expressed in Planck units. But it’s a deep connection, isn’t? When everything is said and done, a charge is something physical, but so here, in these amplitude calculations, it just shows up as some dimensionless negative number, used in multiplications and additions of amplitudes. Isn’t that remarkable?

d2 d3

The situation becomes even more complicated when more than one electron is involved. For example, two electrons can go in a straight line from point 1 and 2 to point 3 and 4 respectively, but there’s two ways in which this can happen, and they might exchange photons along the way, as shown below. If there’s two alternative ways in which one event can happen, you know we have to add amplitudes, rather than multiply them. Hence, the formula for E(A to B) becomes even more complicated.


Moreover, a single electron may first emit and then absorb a photon itself, so there’s no need for other particles to be there to have lots of j factors in our calculation. In addition, that photon may briefly disintegrate into an electron and a positron, which then annihilate each other to again produce a photon: in case you wondered, that’s what those little loops in those diagrams depicting the exchange of virtual photons is supposed to represent. So, every single junction (i.e. every emission and/or absorption of a photon) involves a multiplication with that junction number j, so if there are two couplings involved, we have a j2 factor, and so that’s 0.085424552 = α ≈ 0.0073. Four couplings implies a factor of 0.085424554 ≈ 0.000053.

Just as an example, I copy two diagrams involving four, five or six couplings indeed. They all have some ‘incoming’ photon, because Feynman uses them to explain something else (the so-called magnetic moment of a photon), but it doesn’t matter: the same illustrations can serve multiple purposes.

d6 d7

Now, it’s obvious that the contributions of the alternatives with many couplings add almost nothing to the final amplitude – just like the ‘many-hop’ flights add almost nothing – but… Well… As tiny as these contributions are, they are all there, and so they all have to be accounted for. So… Yes. You can easily appreciate how messy it all gets, especially in light of the fact that there are so many points that can serve as a ‘hop’ or a ‘coupling’ point!

So… Well… Nothing. That’s it! I am done! I realize this has been another long and difficult story, but I hope you appreciated and that it shed some light on what’s really behind those simplified stories of what quantum mechanics is all about. It’s all weird and, admittedly, not so easy to understand, but I wouldn’t say an understanding is really beyond the reach of us, common mortals. 🙂

Post scriptum: When you’ve reached here, you may wonder: so where’s the final formula then for E(A to B)? Well… I have no easy formula for you. From what I wrote above, it should be obvious that we’re talking some really awful-looking integral and, because it’s so awful, I’ll let you find it yourself. 🙂

I should also note another reason why I am reluctant to identify n with m. The formulas in Feynman’s QED are definitely not the standard ones. The more standard formulations will use the gauge coupling parameter about which I talked already. I sort of discussed it, indirectly, in my first comments on Feynman’s QED, when I criticized some other part of the book, notably its explanation of the phenomenon of diffraction of light, which basically boiled down to: “When you try to squeeze light too much [by forcing it to go through a small hole], it refuses to cooperate and begins to spread out”, because “there are not enough arrows representing alternative paths.”

Now that raises a lot of questions, and very sensible ones, because that simplification is nonsensical. Not enough arrows? That statement doesn’t make sense. We can subdivide space in as many paths as we want, and probability amplitudes don’t take up any physical space. We can cut up space in smaller and smaller pieces (so we analyze more paths within the same space). The consequence – in terms of arrows – is that directions of our arrows won’t change but their length will be much and much smaller as we’re analyzing many more paths. That’s because of the normalization constraint. However, when adding them all up – a lot of very tiny ones, or a smaller bunch of bigger ones – we’ll still get the same ‘final’ arrow. That’s because the direction of those arrows depends on the length of the path, and the length of the path doesn’t change simply because we suddenly decide to use some other ‘gauge’.

Indeed, the real question is: what’s a ‘small’ hole? What’s ‘small’ and what’s ‘large’ in quantum electrodynamics? Now, I gave an intuitive answer to that question in that post of mine, but it’s much more accurate than Feynman’s, or Leighton’s. The answer to that question is: there’s some kind of natural ‘gauge’, and it’s related to the wavelength. So the wavelength of a photon, or an electron, in this case, comes with some kind of scale indeed. That’s why the fine-structure constant is often written in yet another form:

α = 2πree = rek

λe and kare the Compton wavelength and wavenumber of the electron (so kis not the Coulomb constant here). The Compton wavelength is the de Broglie wavelength of the electron. [You’ll find that Wikipedia defines it as “the wavelength that’s equivalent to the wavelength of a photon whose energy is the same as the rest-mass energy of the electron”, but that’s a very confusing definition, I think.]

The point to note is that the spatial dimension in both the analysis of photons as well as of matter waves, especially in regard to studying diffraction and/or interference phenomena, is related to the frequencies, wavelengths and/or wavenumbers of the wavefunctions involved. There’s a certain ‘gauge’ involved indeed, i.e. some measure that is relative, like the gauge pressure illustrated below. So that’s where that gauge parameter g comes in. And the fact that it’s yet another number that’s closely related to that fine-structure constant is… Well… Again… That alpha number is a very magic number indeed… 🙂


Post scriptum (5 October 2015):

Much stuff is physics is quite ‘magical’, but it’s never ‘too magical’. I mean: there’s always an explanation. So there is a very logical explanation for the above-mentioned deep connection between the charge of an electron, its energy and/or mass, its various radii (or physical dimensions) and the coupling constant too. I wrote a piece about that, much later than when I wrote the piece above. I would recommend you read that piece too. It’s a piece in which I do take the magic out of ‘God’s number’. Understanding it involves a deep understanding of electromagnetism, however, and that requires some effort. It’s surely worth the effort, though.

Fields and charges (II)

My previous posts was, perhaps, too full of formulas, without offering much reflection. Let me try to correct that here by tying up a few loose ends. The first loose end is about units. Indeed, I haven’t been very clear about that and so let me somewhat more precise on that now.

Note: In case you’re not interested in units, you can skip the first part of this post. However, please do look at the section on the electric constant εand, most importantly, the section on natural units—especially Planck units, as I will touch upon the topic of gauge coupling parameters there and, hence, on quantum mechanics. Also, the third and last part, on the theoretical contradictions inherent in the idea of point charges, may be of interest to you.]

The field energy integrals

When we wrote that down that u = ε0E2/2 formula for the energy density of an electric field (see my previous post on fields and charges for more details), we noted that the 1/2 factor was there to avoid double-counting. Indeed, those volume integrals we use to calculate the energy over all space (i.e. U = ∫(u)dV) count the energy that’s associated with a pair of charges (or, to be precise, charge elements) twice and, hence, they have a 1/2 factor in front. Indeed, as Feynman notes, there is no convenient way, unfortunately, of writing an integral that keeps track of the pairs so that each pair is counted just once. In fact, I’ll have to come back to that assumption of there being ‘pairs’ of charges later, as that’s another loose end in the theory.

U 6

U 7

Now, we also said that that εfactor in the second integral (i.e. the one with the vector dot product EE =|E||E|cos(0) = E2) is there to make the units come out alright. Now, when I say that, what does it mean really? I’ll explain. Let me first make a few obvious remarks:

  1. Densities are always measures in terms per unit volume, so that’s the cubic meter (m3). That’s, obviously, an astronomical unit at the atomic or molecular scale.
  2. Because of historical reasons, the conventional unit of charge is not the so-called elementary charge +e (i.e. the charge of a proton), but the coulomb. Hence, the charge density ρ is expressed in Coulomb per cubic meter (C/m3). The coulomb is a rather astronomical unit too—at the atomic or molecular scale at least: 1 e ≈ 1.6022×10−19 C. [I am rounding here to four digits after the decimal point.]
  3. Energy is in joule (J) and that’s, once again, a rather astronomical unit at the lower end of the scales. Indeed, theoretical physicists prefer to use the electronvolt (eV), which is the energy gained (or lost) when an electron (so that’s a charge of –e, i.e. minus e) moves across a potential difference of one volt. But so we’ll stick to the joule as for now, not the eV, because the joule is the SI unit that’s used when defining most electrical units, such as the ampere, the watt and… Yes. The volt. Let’s start with that one.

The volt

The volt unit (V) measures both potential (energy) as well as potential difference (in both cases, we mean electric potential only, of course). Now, from all that you’ve read so far, it should be obvious that potential (energy) can only be measured with respect to some reference point. In physics, the reference point is infinity, which is so far away from all charges that there is no influence there. Hence, any charge we’d bring there (i.e. at infinity) will just stay where it is and not be attracted or repelled by anything. We say the potential there is zero: Φ(∞) = 0. The choice of that reference point allows us, then, to define positive or negative potential: the potential near positive charges will be positive and, vice versa, the potential near negative charges will be negative. Likewise, the potential difference between the positive and negative terminal of a battery will be positive.

So you should just note that we measure both potential as well as potential difference in volt and, hence, let’s now answer the question of what a volt really is. The answer is quite straightforward: the potential at some point r = (x, y, z) measures the work done when bringing one unit charge (i.e. +e) from infinity to that point. Hence, it’s only natural that we define one volt as one joule per unit charge:

1 volt = 1 joule/coulomb (1 V = 1 J/C).

Also note the following:

  1. One joule is the energy energy transferred (or work done) when applying a force of one newton over a distance of one meter, so one volt can also be measured in newton·meter per coulomb: 1 V = 1 J/C = N·m/C.
  2. One joule can also be written as 1 J = 1 V·C.

It’s quite easy to see why that energy = volt-coulomb product makes sense: higher voltage will be associated with higher energy, and the same goes for higher charge. Indeed, the so-called ‘static’ on our body is usually associated with potential differences of thousands of volts (I am not kidding), but the charges involved are extremely small, because the ability of our body to store electric charge is minimal (i.e. the capacitance (aka capacity) of our body). Hence, the shock involved in the discharge is usually quite small: it is measured in milli-joules (mJ), indeed.

The farad

The remark on ‘static’ brings me to another unit which I should mention in passing: the farad. It measures the capacitance (formerly known as the capacity) of a capacitor (formerly known as a condenser). A condenser consists, quite simply, of two separated conductors: it’s usually illustrated as consisting of two plates or of thin foils (e.g. aluminum foil) separated by an insulating film (e.g. waxed paper), but one can also discuss the capacity of a single body, like our human body, or a charged sphere. In both cases, however, the idea is the same: we have a ‘positive’ charge on one side (+q), and a ‘negative’ charge on the other (–q). In case of a single object, we imagine the ‘other’ charge to be some other large object (the Earth, for instance, but it can also be a car or whatever object that could potentially absorb the charge on our body) or, in case of the charged sphere, we could imagine some other sphere of much larger radius. The farad will then measure the capacity of one or both conductors to store charge.

Now, you may think we don’t need another unit here if that’s the definition: we could just express the capacity of a condensor in terms of its maximum ‘load’, couldn’t we? So that’s so many coulomb before the thing breaks down, when the waxed paper fails to separate the two opposite charges on the aluminium foil, for example. No. It’s not like that. It’s true we can not continue to increase the charge without consequences. However, what we want to measure with the farad is another relationship. Because of the opposite charges on both sides, there will be a potential difference, i.e. a voltage difference. Indeed, a capacitor is like a little battery in many ways: it will have two terminals. Now, it is fairly easy to show that the potential difference (i.e. the voltage) between the two plates will be proportional to the charge. Think of it as follows: if we double the charges, we’re doubling the fields, right? So then we need to do twice the amount of work to carry the unit charge (against the field) from one plate to the other. Now, because the distance is the same, that means the potential difference must be twice what it was.

Now, while we have a simple proportionality here between the voltage and the charge, the coefficient of proportionality will depend on the type of conductors, their shape, the distance and the type of insulator (aka dielectric) between them, and so on and so on. Now, what’s being measured in farad is that coefficient of proportionality, which we’ll denote by C(the proportionality coefficient for the charge), CV ((the proportionality coefficient for the voltage) or, because we should make a choice between the two, quite simply, as C. Indeed, we can either write (1) Q = CQV or, alternatively, V = CVQ, with C= 1/CV. As Feynman notes, “someone originally wrote the equation of proportionality as Q = CV”, so that’s what it will be: the capacitance (aka capacity) of a capacitor (aka condenser) is the ratio of the electric charge Q (on each conductor) to the potential difference V between the two conductors. So we know that’s a constant typical of the type of condenser we’re talking about. Indeed, the capacitance is the constant of proportionality defining the linear relationship between the charge and the voltage means doubling the voltage, and so we can write:

C = Q/V

Now, the charge is measured in coulomb, and the voltage is measured in volt, so the unit in which we will measure C is coulomb per volt (C/V), which is also known as the farad (F):

1 farad = 1 coulomb/volt (1 F = 1 C/V)

[Note the confusing use of the same symbol C for both the unit of charge (coulomb) as well as for the proportionality coefficient! I am sorrry about that, but so that’s convention!].

To be precise, I should add that the proportionality is generally there, but there are exceptions. More specifically, the way the charge builds up (and the way the field builds up, at the edges of the capacitor, for instance) may cause the capacitance to vary a little bit as it is being charged (or discharged). In that case, capacitance will be defined in terms of incremental changes: C = dQ/dV.

Let me conclude this section by giving you two formulas, which are also easily proved but so I will just give you the result:

  1. The capacity of a parallel-plate condenser is C = ε0A/d. In this formula, we have, once again, that ubiquitous electric constant ε(think of it as just another coefficient of proportionality), and then A, i.e. the area of the plates, and d, i.e. the separation between the two plates.
  2. The capacity of a charged sphere of radius r (so we’re talking the capacity of a single conductor here) is C = 4πε0r. This may remind you of the formula for the surface of a sphere (A = 4πr2), but note we’re not squaring the radius. It’s just a linear relationship with r.

I am not giving you these two formulas to show off or fill the pages, but because they’re so ubiquitous and you’ll need them. In fact, I’ll need the second formula in this post when talking about the other ‘loose end’ that I want to discuss.

Other electrical units

From your high school physics classes, you know the ampere and the watt, of course:

  1. The ampere is the unit of current, so it measures the quantity of charge moving or circulating per second. Hence, one ampere is one coulomb per second: 1 A = 1 C/s.
  2. The watt measures power. Power is the rate of energy conversion or transfer with respect to time. One watt is one joule per second: 1 W = 1 J/s = 1 N·m/s. Also note that we can write power as the product of current and voltage: 1 W = (1 A)·(1 V) = (1 C/s)·(1 J/C) = 1 J/s.

Now, because electromagnetism is such well-developed theory and, more importantly, because it has so many engineering and household applications, there are many other units out there, such as:

  • The ohm (Ω): that’s the unit of electrical resistance. Let me quickly define it: the ohm is defined as the resistance between two points of a conductor when a (constant) potential difference (V) of one volt, applied to these points, produces a current (I) of one ampere. So resistance (R) is another proportionality coefficient: R = V/I, and 1 ohm (Ω) = 1 volt/ampere (V/A). [Again, note the (potential) confusion caused by the use of the same symbol (V) for voltage (i.e. the difference in potential) as well as its unit (volt).] Now, note that it’s often useful to write the relationship as V = R·I, so that gives the potential difference as the product of the resistance and the current.
  • The weber (Wb) and the tesla (T): that’s the unit of magnetic flux (i.e. the strength of the magnetic field) and magnetic flux density (i.e. one tesla = one weber per square meter) respectively. So these have to do with the field vector B, rather than E. So we won’t talk about it here.
  • The henry (H): that’s the unit of electromagnetic inductance. It’s also linked to the magnetic effect. Indeed, from Maxwell’s equations, we know that a changing electric current will cause the magnetic field to change. Now, a changing magnetic field causes circulation of E. Hence, we can make the unit charge go around in some loop (we’re talking circulation of E indeed, not flux). The related energy, or the work that’s done by a unit of charge as it travels (once) around that loop, is – quite confusingly! – referred to as electromotive force (emf). [The term is quite confusing because we’re not talking force but energy, i.e. work, and, as you know by now, energy is force times distance, so energy and force are related but not the same.] To ensure you know what we’re talking about, let me note that emf is measured in volts, so that’s in joule per coulomb: 1 V = 1 J/C. Back to the henry now. If the rate of change of current in a circuit (e.g. the armature winding of an electric motor) is one ampere per second, and the resulting electromotive force (remember: emf is energy per coulomb) is one volt, then the inductance of the circuit is one henry. Hence, 1 H = 1 V/(1 A/s) = 1 V·s/A.     

The concept of impedance

You’ve probably heard about the so-called impedance of a circuit. That’s a complex concept, literally, because it’s a complex-valued ratio. I should refer you to the Web for more details, but let me try to summarize it because, while it’s complex, that doesn’t mean it’s complicated. 🙂 In fact, I think it’s rather easy to grasp after all you’ve gone through already. 🙂 So let’s give it a try.

When we have a simple direct current (DC), then we have a very straightforward definition of resistance (R), as mentioned above: it’s a simple ratio between the voltage (as measured in volt) and the current (as measured in ampere). Now, with alternating current (AC) circuits, it becomes more complicated, and so then it’s the concept of impedance that kicks in. Just like resistance, impedance also sort of measures the ‘opposition’ that a circuit presents to a current when a voltage is applied, but we have a complex ratio—literally: it’s a ratio with a magnitude and a direction, or a phase as it’s usually referred to. Hence, one will often write the impedance (denoted by Z) using Euler’s formula:

Z = |Z|eiθ

Now, if you don’t know anything about complex numbers, you should just skip all of what follows and go straight to the next section. However, if you do know what a complex number is (it’s an ‘arrow’, basically, and if θ is a variable, then it’s a rotating arrow, or a ‘stopwatch hand’, as Feynman calls it in his more popular Lectures on QED), then you may want to carry on reading.

The illustration below (credit goes to Wikipedia, once again) is, probably, the most generic view of an AC circuit that one can jot down. If we apply an alternating current, both the current as well as the voltage will go up and down. However, the current signal will lag the voltage signal, and the phase factor θ tells us by how much. Hence, using complex-number notation, we write:

V = IZ = I∗|Z|eiθ


Now, while that resembles the V = R·I formula I mentioned when discussing resistance, you should note the bold-face type for V and I, and the ∗ symbol I am using here for multiplication. First the ∗ symbol: that’s a convention Feynman adopts in the above-mentioned popular account of quantum mechanics. I like it, because it makes it very clear we’re not talking a vector cross product A×B here, but a product of two complex numbers. Now, that’s also why I write V and I in bold-face: they have a phase too and, hence, we can write them as:

  • = |V|ei(ωt + θV)
  • = |I|ei(ωt + θI)

This works out as follows:

IZ = |I|ei(ωt + θI)∗|Z|eiθ = |I||Z|ei(ωt + θ+ θ) = |V|ei(ωt + θV)

Indeed, because the equation must hold for all t, we can equate the magnitudes and phases and, hence, we get: |V| = |I||Z| and θ= θI + θ. But voltage and current is something real, isn’t it? Not some complex number? You’re right. The complex notation is used mainly to simplify the calculus, but it’s only the real part of those complex-valued functions that count. [In any case, because we limit ourselves to complex exponentials here, the imaginary part (which is the sine, as opposed to the real part, which is the cosine) is the same as the real part, but with a lag of its own (π/2 or 90 degrees, to be precise). Indeed: when writing Euler’s formula out (eiθ = cos(θ) + isin(θ), you should always remember that the sine and cosine function are basically the same function: they differ only in the phase, as is evident from the trigonometric identity sin(θ+π/) = cos(θ).]

Now, that should be more than enough in terms of an introduction to the units used in electromagnetic theory. Hence, let’s move on.

The electric constant ε0

Let’s now look at  that energy density formula once again. When looking at that u = ε0E2/2 formula, you may think that its unit should be the square of the unit in which we measure field strength. How do we measure field strength? It’s defined as the force on a unit charge (E = F/q), so it should be newton per coulomb (N/C). Because the coulomb can also be expressed in newton·meter/volt (1 V = 1 J/C = N·m/C and, hence, 1 C = 1 N·m/V), we can express field strength not only in newton/coulomb but also in volt per meter: 1 N/C = 1 N·V/N·m = 1 V/m. How do we get from N2/C2 and/or V2/mto J/m3?

Well… Let me first note there’s no issue in terms of units with that ρΦ formula in the first integral for U: [ρ]·[Φ] = (C/m3)·V = [(N·m/V)/m3)·V = (N·m)/m3 = J/m3. No problem whatsoever. It’s only that second expression for U, with the u = ε0E2/2 in the integrand, that triggers the question. Here, we just need to accept that we need that εfactor to make the units come out alright. Indeed, just like other physical constants (such as c, G, or h, for example), it has a dimension: its unit is either C2/N·m2 or, what amounts to the same, C/V·m. So the units come out alright indeed if, and only if, we multiply the N2/C2 and/or V2/m2 units with the dimension of ε0:

  1. (N2/C2)·(C2/N·m2) = (N2·m)·(1/m3) = J/m3
  2. (V2/m2)·(C/V·m) = V·C/m3 = (V·N·m/V)/m= N·m/m3 = J/m3


But so that’s the units only. The electric constant also has a numerical value:

ε0 = 8.854187817…×10−12 C/V·m ≈ 8.8542×10−12 C/V·m

This numerical value of εis as important as its unit to ensure both expressions for U yield the same result. Indeed, as you may or may not remember from the second of my two posts on vector calculus, if we have a curl-free field C (that means ×= 0 everywhere, which is the case when talking electrostatics only, as we are doing here), then we can always find some scalar field ψ such that C = ψ. But so here we have E = –ε0Φ, and so it’s not the minus sign that distinguishes the expression from the C = ψ expression, but the εfactor in front.

It’s just like the vector equation for heat flow: h = –κT. Indeed, we also have a constant of proportionality here, which is referred to as the thermal conductivity. Likewise, the electric constant εis also referred to as the permittivity of the vacuum (or of free space), for similar reasons obviously!

Natural units

You may wonder whether we can’t find some better units, so we don’t need the rather horrendous 8.8542×10−12 C/V·m factor (I am rounding to four digits after the decimal point). The answer is: yes, it’s possible. In fact, there are several systems in which the electric constant (and the magnetic constant, which we’ll introduce later) reduce to 1. The best-known are the so-called Gaussian and Lorentz-Heaviside units respectively.

Gauss defined the unit of charge in what is now referred to as the statcoulomb (statC), which is also referred to as the franklin (Fr) and/or the electrostatic unit of charge (esu), but I’ll refer you to the Wikipedia article on it in case you’d want to find out more about it. You should just note the definition of this unit is problematic in other ways. Indeed, it’s not so easy to try to define ‘natural units’ in physics, because there are quite a few ‘fundamental’ relations and/or laws in physics and, hence, equating this or that constant to one usually has implications on other constants. In addition, one should note that many choices that made sense as ‘natural’ units in the 19th century seem to be arbitrary now. For example:

  1. Why would we select the charge of the electron or the proton as the unit charge (+1 or –1) if we now assume that protons (and neutrons) consists of quarks, which have +2/3 or –1/3?
  2. What unit would we choose as the unit for mass, knowing that, despite all of the simplification that took place as a result of the generalized acceptance of the quark model, we’re still stuck with quite a few elementary particles whose mass would be a ‘candidate’ for the unit mass? Do we chose the electron, the u quark, or the d quark?

Therefore, the approach to ‘natural units’ has not been to redefine mass or charge or temperature, but the physical constants themselves. Obvious candidates are, of course, c and ħ, i.e. the speed of light and Planck’s constant. [You may wonder why physicists would select ħ, rather than h, as a ‘natural’ unit, but I’ll let you think about that. The answer is not so difficult.] That can be done without too much difficulty indeed, and so one can equate some more physical constants with one. The next candidate is the so-called Boltzmann constant (kB). While this constant is not so well known, it does pop up in a great many equations, including those that led Planck to propose his quantum of action, i.e.(see my post on Planck’s constant). When we do that – so when we equate c, ħ and kB with one (ħ = kB = 1), we still have a great many choices, so we need to impose further constraints. The next is to equate the gravitational constant with one, so then we have ħ = kB = G = 1.

Now, it turns out that the ‘solution’ of this ‘set’ of four equations (ħ = kB = G = 1) does, effectively, lead to ‘new’ values for most of our SI base units, most notably length, time, mass and temperature. These ‘new’ units are referred to as Planck units. You can look up their values yourself, and I’ll let you appreciate the ‘naturalness’ of the new units yourself. They are rather weird. The Planck length and time are usually referred to as the smallest possible measurable units of length and time and, hence, they are related to the so-called limits of quantum theory. Likewise, the Planck temperature is a related limit in quantum theory: it’s the largest possible measurable unit of temperature. To be frank, it’s hard to imagine what the scale of the Planck length, time and temperature really means. In contrast, the scale of the Planck mass is something we actually can imagine – it is said to correspond to the mass of an eyebrow hair, or a flea egg – but, again, its physical significance is not so obvious: Nature’s maximum allowed mass for point-like particles, or the mass capable of holding a single elementary charge. That triggers the question: do point-like charges really exist? I’ll come back to that question. But first I’ll conclude this little digression on units by introducing the so-called fine-structure constant, of which you’ve surely heard before.

The fine-structure constant

I wrote that the ‘set’ of equations ħ = kB = G = 1 gave us Planck units for most of our SI base units. It turns out that these four equations do not lead to a ‘natural’ unit for electric charge. We need to equate a fifth constant with one to get that. That fifth constant is Coulomb’s constant (often denoted as ke) and, yes, it’s the constant that appears in Coulomb’s Law indeed, as well as in some other pretty fundamental equations in electromagnetics, such as the field caused by a point charge q: E = q/4πε0r2. Hence, ke = 1/4πε0. So if we equate kwith one, then ε0 will, obviously, be equal to ε= 1/4π.

To make a long story short, adding this fifth equation to our set of five also gives us a Planck charge, and I’ll give you its value: it’s about 1.8755×10−18 C. As I mentioned that the elementary charge is 1 e ≈ 1.6022×10−19 C, it’s easy to that the Planck charge corresponds to some 11.7 times the charge of the proton. In fact, let’s be somewhat more precise and round, once again, to four digits after the decimal point: the qP/e ratio is about 11.7062. Conversely, we can also say that the elementary charge as expressed in Planck units, is about 1/11.7062 ≈ 0.08542455. In fact, we’ll use that ratio in a moment in some other calculation, so please jot it down.

0.08542455? That’s a bit of a weird number, isn’t it? You’re right. And trying to write it in terms of the charge of a u or d quark doesn’t make it any better. Also, note that the first four significant digits (8542) correspond to the first four significant digits after the decimal point of our εconstant. So what’s the physical significance here? Some other limit of quantum theory?

Frankly, I did not find anything on that, but the obvious thing to do is to relate is to what is referred to as the fine-structure constant, which is denoted by α. This physical constant is dimensionless, and can be defined in various ways, but all of them are some kind of ratio of a bunch of these physical constants we’ve been talking about:

Fine-structure constant formula

The only constants you have not seen before are μ0Rand, perhaps, ras well as m. However, these can be defined as a function of the constants that you did see before:

  1. The μ0 constant is the so-called magnetic constant. It’s something similar as ε0 and it’s referred to as the magnetic permeability of the vacuum. So it’s just like the (electric) permittivity of the vacuum (i.e. the electric constant ε0) and the only reason why you haven’t heard of this before is because we haven’t discussed magnetic fields so far. In any case, you know that the electric and magnetic force are part and parcel of the same phenomenon (i.e. the electromagnetic interaction between charged particles) and, hence, they are closely related. To be precise, μ= 1/ε0c2. That shows the first and second expression for α are, effectively, fully equivalent.
  2. Now, from the definition of ke = 1/4πε0, it’s easy to see how those two expressions are, in turn, equivalent with the third expression for α.
  3. The Rconstant is the so-called von Klitzing constant, but don’t worry about it: it’s, quite simply, equal to Rh/e2. Hene, substituting (and don’t forget that h = 2πħ) will demonstrate the equivalence of the fourth expression for α.
  4. Finally, the re factor is the classical electron radius, which is usually written as a function of me, i.e. the electron mass: re = e2/4πε0mec2. This very same equation implies that reme = e2/4πε0c2. So… Yes. It’s all the same really.

Let’s calculate its (rounded) value in the old units first, using the third expression:

  • The econstant is (roughly) equal to (1.6022×10–19 C)= 2.5670×10–38 C2. Coulomb’s constant k= 1/4πεis about 8.9876×10N·m2/C2. Hence, the numerator e2k≈ 23.0715×10–29 N·m2.
  • The (rounded) denominator is ħc = (1.05457×10–34 N·m·s)(2.998×108 m/s) = 3.162×10–26 N·m2.
  • Hence, we get α = kee2/ħc ≈ 7.297×10–3 = 0.007297.

Note that this number is, effectively, dimensionless. Now, the interesting thing is that if we calculate α using Planck units, we get an econstant that is (roughly) equal to 0.08542455= … 0.007297! Now, because all of the other constants are equal to 1 in Planck’s system of units, that’s equal to α itself. So… Yes ! The two values for α are one and the same in the two systems of units and, of course, as you might have guessed, the fine-structure constant is effectively dimensionless because it does not depend on our units of measurement. So what does it correspond to?

Now that would take me a very long time to explain, but let me try to summarize what it’s all about. In my post on quantum electrodynamics (QED) – so that’s the theory of light and matter basically and, most importantly, how they interact – I wrote about the three basic events in that theory, and how they are associated with a probability amplitude, so that’s a complex number, or an ‘arrow’, as Feynman puts it: something with (a) a magnitude and (b) a direction. We had to take the absolute square of these amplitudes in order to calculate the probability (i.e. some real number between 0 and 1) of the event actually happening. These three basic events or actions were:

  1. A photon travels from point A to B. To keep things simple and stupid, Feynman denoted this amplitude by P(A to B), and please note that the P stands for photon, not for probability. I should also note that we have an easy formula for P(A to B): it depends on the so-called space-time interval between the two points A and B, i.e. I = Δr– Δt= (x2–x1)2+(y2–y1)2+(z2–z1)– (t2–t1)2. Hence, the space-time interval takes both the distance in space as well as the ‘distance’ in time into account.
  2. An electron travels from point A to B: this was denoted by E(A to B) because… Well… You guessed it: the of electron. The formula for E(A to B) was much more complicated, but the two key elements in the formula was some complex number j (see below), and some other (real) number n.
  3. Finally, an electron could emit or absorb a photon, and the amplitude associated with this event was denoted by j, for junction.

Now, that junction number j is about –0.1. To be somewhat more precise, I should say it’s about –0.08542455.

–0.08542455? That’s a bit of a weird number, isn’t it? Hey ! Didn’t we see this number somewhere else? We did, but before you scroll up, let’s first interpret this number. It looks like an ordinary (real) number, but it’s an amplitude alright, so you should interpret it as an arrow. Hence, it can be ‘combined’ (i.e. ‘added’ or ‘multiplied’) with other arrows. More in particular, when you multiply it with another arrow, it amounts to a shrink to a bit less than one-tenth (because its magnitude is about 0.085 = 8.5%), and half a turn (the minus sign amounts to a rotation of 180°). Now, in that post of mine, I wrote that I wouldn’t entertain you on the difficulties of calculating this number but… Well… We did see this number before indeed. Just scroll up to check it. We’ve got a very remarkable result here:

j ≈ –0.08542455 = –√0.007297 = –√α = –e expressed in Planck units

So we find that our junction number j or – as it’s better known – our coupling constant in quantum electrodynamics (aka as the gauge coupling parameter g) is equal to the (negative) square root of that fine-structure constant which, in turn, is equal to the charge of the electron expressed in the Planck unit for electric charge. Now that is a very deep and fundamental result which no one seems to be able to ‘explain’—in an ‘intuitive’ way, that is.

I should immediately add that, while we can’t explain it, intuitively, it does make sense. A lot of sense actually. Photons carry the electromagnetic force, and the electromagnetic field is caused by stationary and moving electric charges, so one would expect to find some relation between that junction number j, describing the amplitude to emit or absorb a photon, and the electric charge itself, but… An equality? Really?

Well… Yes. That’s what it is, and I look forward to trying to understand all of this better. For now, however, I should proceed with what I set out to do, and that is to tie up a few loose ends. This was one, and so let’s move to the next, which is about the assumption of point charges.

Note: More popular accounts of quantum theory say α itself is ‘the’ coupling constant, rather than its (negative) square –√α = j = –e (expressed in Planck units). That’s correct: g or j are, technically speaking, the (gauge) coupling parameter, not the coupling constant. But that’s a little technical detail which shouldn’t bother you. The result is still what it is: very remarkable! I should also note that it’s often the value of the reciprocal (1/α) that is specified, i.e. 1/0.007297 ≈ 137.036. But so now you know what this number actually stands for. 🙂

Do point charges exist?

Feynman’s Lectures on electrostatics are interesting, among other things, because, besides highlighting the precision and successes of the theory, he also doesn’t hesitate to point out the contradictions. He notes, for example, that “the idea of locating energy in the field is inconsistent with the assumption of the existence of point charges.”


Yes. Let’s explore the point. We do assume point charges in classical physics indeed. The electric field caused by a point charge is, quite simply:

E = q/4πε0r2

Hence, the energy density u is ε0E2/2 = q2/32πε0r4. Now, we have that volume integral U = (ε0/2)∫EEdV = ∫(ε0E2/2)dV integral. As Feynman notes, nothing prevents us from taking a spherical shell for the volume element dV, instead of an infinitesimal cube. This spherical shell would have the charge q at its center, an inner radius equal to r, an infinitesimal thickness dr, and, finally, a surface area 4πr(that’s just the general formula for the surface area of a spherical shell, which I also noted above). Hence, its (infinitesimally small) volume is 4πr2dr, and our integral becomes:


To calculate this integral, we need to take the limit of –q2/8πε0r for (a) r tending to zero (r→0) and for (b) r tending to infinity (r→∞). The limit for r = ∞ is zero. That’s OK and consistent with the choice of our reference point for calculating the potential of a field. However, the limit for r = 0 is zero is infinity! Hence, that U = (ε0/2)∫EEdV basically says there’s an infinite amount of energy in the field of a point charge! How is that possible? It cannot be true, obviously.

So… Where did we do wrong?

Your first reaction may well be that this very particular approach (i.e. replacing our infinitesimal cubes by infinitesimal shells) to calculating our integral is fishy and, hence, not allowed. Maybe you’re right. Maybe not. It’s interesting to note that we run into similar problems when calculating the energy of a charged sphere. Indeed, we mentioned the formula for the capacity of a charged sphere: C = 4πε0r. Now, there’s a similarly easy formula for the energy of a charged sphere. Let’s look at how we charge a condenser:

  • We know that the potential difference between two plates of a condenser represents the work we have to do, per unit charge, to transfer a charge (Q) from one plate to the other. Hence, we can write V = ΔU/ΔQ.
  • We will, of course, want to do a differential analysis. Hence, we’ll transfer charges incrementally, one infinitesimal little charge dQ at the time, and re-write V as V = dU/dQ or, what amounts to the same: dU = V·dQ.
  • Now, we’ve defined the capacitance of a condenser as C = Q/V. [Again, don’t be confused: C stands for capacity here, measured in coulomb per volt, not for the coulomb unit.] Hence, we can re-write dU as dU = Q·dQ/C.
  • Now we have to integrate dU going from zero charge to the final charge Q. Just do a little bit of effort here and try it. You should get the following result: U = Q2/2C. [We could re-write this as U = (C2/V2)/2C =  C·V2/2, which is a form that may be more useful in some other context but not here.]
  • Using that C = 4πε0r formula, we get our grand result. The energy of a charged sphere is:

U = Q2/8πε0r

From that formula, it’s obvious that, if the radius of our sphere goes to zero, its energy should also go to infinity! So it seems we can’t really pack a finite charge Q in one single point. Indeed, to do that, our formula says we need an infinite amount of energy. So what’s going on here?

Nothing much. You should, first of all, remember how we got that integral: see my previous post for the full derivation indeed. It’s not that difficult. We first assumed we had pairs of charges qi and qfor which we calculated the total electrostatic energy U as the sum of the energies of all possible pairs of charges:

U 3

And, then, we looked at a continuous distribution of charge. However, in essence, we still did the same: we counted the energy of interaction between infinitesimal charges situated at two different points (referred to as point 1 and 2 respectively), with a 1/2 factor in front so as to ensure we didn’t double-count (there’s no way to write an integral that keeps track of the pairs so that each pair is counted only once):

U 4

Now, we reduced this double integral by a clever substitution to something that looked a bit better:

U 6

Finally, some more mathematical tricks gave us that U = (ε0/2)∫EEdV integral.

In essence, what’s wrong in that integral above is that it actually includes the energy that’s needed to assemble the finite point charge q itself from an infinite number of infinitesimal parts. Now that energy is infinitely large. We just can’t do it: the energy required to construct a point charge is ∞.

Now that explains the physical significance of that Planck mass ! We said Nature has some kind of maximum allowable mass for point-like particles, or the mass capable of holding a single elementary charge. What’s going on is, as we try to pile more charge on top of the charge that’s already there, we add energy. Now, energy has an equivalent mass. Indeed, the Planck charge (q≈ 1.8755×10−18 C), the Planck length (l= 1.616×10−35 m), the Planck energy (1.956×109 J), and the Planck mass (2.1765×10−8 kg) are all related. Now things start making sense. Indeed, we said that the Planck mass is tiny but, still, it’s something we can imagine, like a flea’s egg or the mass of a hair of a eyebrow. The associated energy (E = mc2, so that’s (2.1765×10−8 kg)·(2.998×108 m/s)2 ≈ 19.56×108 kg·m2/s= 1.956×109 joule indeed.

Now, how much energy is that? Well… That’s about 2 giga-joule, obviously, but so what’s that in daily life? It’s about the energy you would get when burning 40 liter of fuel. It’s also likely to amount, more or less, to your home electricity consumption over a month. So it’s sizable, and so we’re packing all that energy into a Planck volume (lP≈ 4×10−105 m3). If we’d manage that, we’d be able to create tiny black holes, because that’s what that little Planck volume would become if we’d pack so much energy in it. So… Well… Here I just have to refer you to more learned writers than I am. As Wikipedia notes dryly: “The physical significance of the Planck length is a topic of theoretical research. Since the Planck length is so many orders of magnitude smaller than any current instrument could possibly measure, there is no way of examining it directly.”

So… Well… That’s it for now. The point to note is that we would not have any theoretical problems if we’d assume our ‘point charge’ is actually not a point charge but some small distribution of charge itself. You’ll say: Great! Problem solved! 

Well… For now, yes. But Feynman rightly notes that assuming that our elementary charges do take up some space results in other difficulties of explanation. As we know, these difficulties are solved in quantum mechanics, but so we’re not supposed to know that when doing these classical analyses. 🙂