# Atomic magnets: precession and diagmagnetism

This and the next posts will further build on the concepts introduced in my previous post on particle spin. This post in particular will focus on some of the math we’ll need to understand what quantum mechanics is all about. The first topic is about the quantum-mechanical equivalent of the phenomenon of precession. The other topics are… Well… You’ll see… š

#### The Larmor frequency

The motion of a spinning object in a force field is quite complicated. In our post on gyroscopes, we introduced the concepts of precession and nutation. The concept of precession is illustrated below for the Earth as well as for a spinning top. In both cases, the external force is just gravity.

Nutation is an additional movement: on top of the precessional movement, a spinning object may wobble, as illustrated below.

There seems to be no analog for nutation in quantum mechanics. In fact, the terms nutation and precession seem to be used interchangeably in quantum physics, although they are very different in classical physics. But let’s not complicate things and, hence, talk about the phenomenon of precession only.

We will not re-explain the phenomenon of precession here but just remind you that the phenomenon can be described in terms of (a) the angle between the symmetry axis and the momentum vector, which we’ll denote by Īø, and (b) the angular velocity of the precession, which we’ll denote by ĻpĀ = dĻ/dt, as shown below. The J in the illustration below is theĀ angular momentum of the object. Hence, if we’d imagine it to be an electron, thenĀ JĀ would be the spinĀ angular momentum only, not its orbital angular momentumāalthough the analysis would obviously be valid for the orbital and/orĀ totalĀ angular momentum as well.

OK. Let’s look at what’s going on. The angular displacement ā which is also, rather confusingly, referred to as the angle of precessionĀ ā in the time intervalĀ Īt is, obviously, equal to ĪĻ = ĻpĀ·Īt.Ā Now, looking at the geometry of the situation, and using the small-angle approximation for the sine, one can also see that ĪJ ā (JĀ·sinĪø)Ā·(ĻpĀ·Īt). In fact, going to the limit (i.e. for infinitesimally smallĀ ĪĻ and ĪJ), we can write:

dJ/dt = ĻpĀ·JĀ·sinĪø

But the angular momentum cannot change if there’s no torque. In fact, the time rate of change of the angular momentum is equalĀ to theĀ torque. [You should look this up but, if you don’t want to do that, note that this is just the equivalent, for rotational motion, of the F = dp/dt law for linear motion.] Now, in my post on magnetic dipoles, I showed that the torqueĀ ĻĀ on a loop of current with magnetic momentĀ Ī¼Ā in an external magnetic field BĀ  is equal toĀ Ļ = Ī¼ĆB. So the magnitude of the torque is equal to |Ļ| = |Ī¼|Ā·|B|Ā·sinĪø = Ī¼Ā·BĀ·sinĪø. Therefore,Ā ĻpĀ·JĀ·sinĪø = Ī¼Ā·BĀ·sinĪø and, hence,

ĻpĀ = Ī¼Ā·B/J

However, from the generalĀ Ī¼/J = āgĀ·(qe/2m)Ā equation we derived in our previous post, we know that Ī¼/JĀ ā for an atomic magnet, that is āĀ must be equal to Ī¼/J = gĀ·qe/2m. So we get the formula we wanted to get here:

ĻpĀ = gĀ·(qe/2m)Ā·B

This equation says that the angular velocity of the precession is proportional to the magnitude of the external magnetic field, and that the constant of proportionality is equal to gĀ·(qe/2m). It’s good to do the math and actually calculateĀ the precession frequency fpĀ = Ļp/2Ļ. It’s easy.Ā We had calculated qe/2m already: it was equal to 1.6Ć10ā19Ā C divided by 2Ā·9.1Ć10ā31Ā kg, so thatās 0.0879Ć1012Ā  C/kg or 0.0879Ć1012Ā (CĀ·m)/(NĀ·s2), more or less. š Now, g is dimensionless, and B is expressed in tesla: 1 T = (NĀ·s)/(CĀ·m), so we get the sā1Ā dimension we want for a frequency. For g = 2 (so we look at the spin of the electron itself only), we get:

fpĀ =Ā Ļp/2Ļ = 2Ā·0.0879Ć1012/2Ļ ā 28Ć109 = 28 gigacycles per tesla = 28Ā GHz/T

This is a number expressed per unit of the magnetic field strength B. Note that you’ll often see this number expressed as 1.4 megacycles per gauss, using the olderĀ gaussĀ unit for magnetic field strength: 1 tesla = 10,000 gauss. For a nucleus, we get a somewhat less impressive number because the proton (or neutron) mass is so much bigger: it’s a number expressed in megacycles per tesla, indeed, and for a proton (i.e. a hydrogen nucleus), it’s about 42.58 MHz/T.

Now, you may wonder about the numbers here. Are they astronomical? Maybe. Maybe not. It’s probably good to note that theĀ strength of the magnetic field in medical MRI systemsĀ (magnetic resonance imaging systems) is only 1.5 to 3 tesla, so it’s a rather large unit. You should also note that the clock speed of the CPU in your laptop ā so that’s the speed at which it executes instructions ā is measured in GHz too, so perhaps it’s not so astronomic. I’ll let you judge. š

So… Well… That’s all nice. The key question, of course, is whether or not this classical view of the electron spinning around a proton is accurate, quantum-mechanically, that is. I’ll let Feynman answer that question provisionally:

“According to the classical theory, then, the electron orbitsāand spinsāin an atom should precess in a magnetic field. Is it also true quantum-mechanically? It is essentially true, but the meaning of the āprecessionā is different. In quantum mechanics one cannot talk about the direction of the angular momentum in the same sense as one does classically; nevertheless, there is a very close analogyāso close that we continue to call it precession.”

To distinguish classical and quantum-mechanical precession, quantum-mechanical precession is usually referred to as Larmor precession, and the frequencies above are often referred to as Larmor frequencies. However, I should note that, technically speaking, the term Larmor frequencyĀ is actually reserved for the frequency I’ll describe in the next section.Ā I should also note that theĀ ĻpĀ = gĀ·(qe/2m)Ā·B is usually written, quite simply, asĀ ĻpĀ = Ī³Ā·B. Of course, the gamma isĀ notĀ the Lorentz factor here, but the so-called gyromagnetic ratioĀ (aka as the magnetogyric ratio):Ā Ī³ =Ā gĀ·(qe/2m). Ohājust so you know: Sir Joseph Larmor was a British physicists and, yes, he developed all of the stuff we’re talking about here. š

At this point, you may wonder if and why all of the above is relevant. Well… There’s more than one answer to this question, but I’d recommend you start with reading the Wikipedia article on NMR spectroscopy. š And then you should also read Feynman’s exposĆ© on the Rabi atomic or molecular beam method for determining the precession frequency. It’s really fascinating stuff, but you are sufficiently armed now to read those things for yourself, and so I’ll just move on. Indeed, there’s something else I need to talk about here, and that’sĀ Larmor’s Theorem.

#### Larmor’s Theorem

We’ve been talking single electrons only so far. Now, you may fear that things become quite complicated when many electrons are involved and… Well… That’s true, of course. And then you may also think that things become even more complicated when external fields are involved, like that external magnetic field we introduced above, and that led our electrons to precess at extraordinary frequencies. Well… That’s notĀ true. Here we get some help: Larmor proved a theorem that basically says that, if we can work out the motions of the electronsĀ withoutĀ the external field, the solution for the motionsĀ withĀ the external field is the no-field solutionĀ with an added rotation about the axis of the field. More specifically, for an external magnetic field, the added rotation will have an angular frequency equal to:

ĻLĀ = (qe/2m)Ā·B

So that’s the same formula as we found for the angular velocity of the precession if g = 1, so that’s very easy to remember. TheĀ ĻLĀ  frequency, which is the precession frequency for g = 1, is referred to as theĀ Larmor frequency. The proof of the above is remarkably easy, but… Well… I don’t want to copy Feynman here, so I’ll just refer you to the relevant LectureĀ on it. š

#### Diamagnetism

I guess it’s about time we relate all of what we learned so far to properties of matter we can relate to, and so that’s what I’ll do here. We’re not going to talk aboutĀ ferromagnetismĀ here, i.e. the mechanism through which iron, nickel and cobalt and most of their alloys become permanentĀ magnets. That’s quite peculiar and so we willĀ notĀ discuss it here. Here we’ll talk about the very weak quantum-mechanical magnetic effectĀ ā a thousand to a million times less than the effects in ferromagnetic materialsĀ ā that occurs inĀ allĀ materials when placed in an external magnetic field.

While the effect is there inĀ allĀ materials, it’s stronger for some than for others. In fact, it’s usually so weak it is hard to detect, and so it’s usually demonstrated using elements for which the diamagnetic effect is somewhat stronger, like bismuth or antimony. The effect is demonstrated by suspending a piece of material in a non-uniform field, as illustrated below. The diamagnetic effect will cause a small displacement of the material, away from the high-field region, i.e. away from the pointed pole.

I should immediately add that some materials, like aluminium, will actually be attracted to the pointed pole, but that’s because of yet another effect that notĀ all materials share: paramagnetism. I’ll talk about that in another post, together with ferromagnetism. So… Diamagnetism: what is it?

The illustration below shows our spinning electron (q) once again. It also shows a magnetic field B but, unlike our analysis above, or the analysis in our previous post, we assume the external magnetic field is not just there. We assume it changes, because it’s been turned on or offāhopefully slowly: if not, we’d have eddy-current forces causing potentially strong impulses.

But so we’ve got some changeĀ in the magnetic flux , and so we know, because of Faraday or MaxwellĀ āĀ you choose šĀ ā that we’ll have some circulation of E, i.e. theĀ electricĀ field. The magnetic flux is B times the surface area, and the circulation is the average tangential component E times the length of the path. Because our model of the orbiting electron is so nice and symmetric, we can write Faraday’s Law here as:

EĀ·2ĻĀ·r =Ā ād(BĀ·ĻĀ·r2)/dt āĀ E =Ā ā(r/2)Ā·dB/dt

A field implies a force and, therefore, a torque on the electron. The torque is equal to the force times the lever arm, so it’s equal to (āqeĀ·E)Ā·r = āqeĀ·EĀ·r. Of course, the torque is also equal to the rate of the change of the angular momentum, so dJ/dt must equal:

dJ/dt = āqeĀ·EĀ·r = Ā qeĀ·(r/2)Ā·(dB/dt)Ā·r = (qeĀ·r2/2)Ā·(dB/dt)

Now, the assumption is that the field goes from zero to B, so ĪB = B. Therefore, ĪJ must be equal to:

ĪJ =Ā (qeĀ·r2/2)Ā·B

You should, in fact, derive this more formally, by integratingābut let’s keep things as simple as we can. š What does this formula say, really? It’s the extra angular momentum from the ‘twist’ that’s given to the electrons as the field is turned on. Now, this added angular momentum makes an extra magnetic moment which, because it is an orbital motion, is justĀ qe/2mĀ times the angular momentum that’s already there. But more angular momentum means the magnetic moment has changed, according to theĀ Ī¼ = (qe/2m)Ā·J formula we derived in our previous post, so we have:

ĪĪ¼ = ā(qe/2m)Ā·ĪJ

TheĀ minus sign is there because of Lenz’ law: theĀ addedĀ momentum is opposite to the magnetic fieldāand, yes, I know: it’s hard to keep track of all of the conventions involved here. In any case, we get the following grand equation:

So we found that theĀ inducedĀ magnetic moment is directly proportional to the magnetic fieldĀ B,Ā and opposing it. Now that is what explains why our piece of bismuth does what it does in that non-uniform magnetic field. Of course, you’ll say: why is stronger for bismuth than for other materials? And what about aluminium, or paramagnetism in general? Well… Good questions, but we’ll tackle them in the next posts. š

Let me conclude this post by copying Feynman’s little exposĆ© on why the phenomenon of diamagnetism is so particular. In fact, he notes that, because we’re talking a piece of material here that can’t spin ā so it’s held in place, so to say āĀ we should have “no magnetic effects whatsoever”. The reasoning is as follows:

This is very interesting indeed. This classical theorem basically says that the energy of a system should not be affected by the presence of a magnetic field. However, we know magnetic effects, such as the diamagnetic effect, are there, so these effects are referred to as ‘quantum-mechanical’ effects indeed: they cannotĀ be explained using classical theory only, even if all of what we wrote above used classical theory only.

I should also note another point: why do we need a non-homogeneous field? Well… The situation is comparable to what we wrote on the Stern-Gerlach experiment. If we would have a homogeneousĀ magnetic field, then we would only have a torque on all of the atomic magnets, but no net force in one or the other direction. Thereās something else here too: you may think that the forces pointing towards and away from the pointed tip should cancel each other out, so there should actually be no net movement of the material at all! Feynmanās analysis works for one atom, indeed, but does it still make sense if we look at the whole piece of material? It does, because weāre talking an induced magnetic moment thatās opposing the field, regardless of the orientation of the magnetic moment of the individual atoms in the piece of material. So, even if the individual atoms have opposite momenta, the extra induced magnetic moment will point in the same direction for all. So that solves that issue. However, it does not address Feynmanās own critical remark in regard to the supposed ‘impossibility’ of diamagnetism in classical mechanics.

But I’ll let you think about this, and sign off for today. š I hope you enjoyed this post.

# Spin and angular momentum in quantum mechanics

Note: A few years after writing the post below, I published a paper on the anomalous magnetic moment which makes (some of) what is written below irrelevant. It gives a clean classical explanation for it. Have a look by clicking on the link here !

Original blog post:

Feynman starts his Volume of LecturesĀ on quantum mechanics (so that’s Volume III of the whole series) with the rules we already know, so that’s the ‘special’ math involving probabilityĀ amplitudes, rather than probabilities. However, these introductory chapters assume theoretical zero-spinĀ particles, which means they don’t have any angularĀ momentum. While that makesĀ itĀ much easierĀ to understand the basics of quantum math, realĀ elementary particles do have angular momentum, which makes the analysis much more complicated. Therefore, Feynman makes it veryĀ clear, after his introductory chapters, that he expects all prospective readers of his third volume to first work their way through chapter 34 and 35 of the second volume, which discusses the angular momentum of elementary particles from both a classical as well as a quantum-mechanical perspective. So that’s what we will do here.Ā I have to warn you, though: while the mentioned two chapters are more generous with text than other textbooks on quantum mechanicsĀ I’ve looked at, the matter is still quite hard to digest. By way of introduction, Feynman writes the following:

“The behavior of matter on a small scaleāas we have remarked many timesāis different from anything that you are used to and is very strange indeed. Understanding of these matters comes very slowly, if at all. One never gets a comfortable feeling that these quantum-mechanical rules are ‘natural’. Of course they are, but they are not natural to our own experience at an ordinary level. The attitude that we are going to take with regard to this rule about angular momentum is quite different from many of the other things we have talked about. We are not going to try to ‘explain’ it but tell you what happens.”

I personally feel it’s not all as mysterious as Feynman claims it to be, but I’ll let you judge for yourself. So let’s just go for it and see what comes out. š

#### Atomic magnets and the g-factor

When discussing electromagnetic radiation, we introduced the concept of atomic oscillators. It was a very useful model to help us understand what’s supposed to be going on. Now we’re going to introduce atomic magnets. It is based on the classical idea of an electron orbiting around a proton. Of course, weĀ knowĀ this classical idea is wrong: we don’t have nice circular electron orbitals, and our discussion on the radius of an the electron in our previous post makes it clear that the idea of the electron itself is rather fuzzy.Ā Nevertheless, the classical concepts used to analyze rotation are also used, mutatis mutandis,Ā in quantum mechanics. Mutatis mutandis means: with necessary alterations. So…Ā Well… Let’s go for it. šĀ The basic idea is the following: an electron in a circular orbit is a circular current and, hence, it causes a magnetic field, i.e. a magnetic flux through the area of the loopāas illustrated below.

As such, we’ll have a magnetic (dipole) moment, and you may want to review my post(s) on that topicĀ so as to ensure you understand what follows. The magnetic moment (Ī¼)Ā is the product of the current (I) and the area of the loop (ĻĀ·r2), and its conventional direction is given by the Ī¼Ā vectorĀ in the illustration below, which also shows the other relevant scalar and/or vector quantities, such as the velocity v and the orbital angular momentum J. The orbital angular momentum is to be distinguished from theĀ spinĀ angular momentum, which results from the spin around its own axis. So the spin angular momentumĀ ā which is often referred to as theĀ spinĀ tout courtĀ ā isĀ notĀ depicted below, and will only be discussed in a few minutes.

Let me interject something on notation here. Feynman’s always uses J, for whatever momentum. That’s not so usual. Indeed, if you’d google a bit, you’ll see the convention is to useĀ SĀ andĀ LĀ respectively to distinguish spin and orbital angular momentum respectively. If we’d use S and L, we can write theĀ totalĀ angular momentum as J = S + L, and the illustration below shows how the S and L vectors are to be added. It looks a bit complicated, so you can skip this for now and come back to it later. But just try to visualize things:

1. The L vector is moving around, so that assumes the orbital plane is moving around too. That happens when we’d put our atomic system in a magnetic field. We’ll come back to that. In what follows, we’ll assume the orbital plane isĀ notĀ moving.
2. The S vector here is also moving, which also assumes the axis of rotation is notĀ steady. What’s going on here is referred to asĀ precession, and we discussed it when presenting the math one needs to understand gyroscopes.
3. Adding S and L yields J, the totalĀ angular momentum. Unsurprisingly, this vector wiggles around too. Don’t worry about the magnitudes of the vectors here. Also, in case you’d wonder why the axis of symmetry for the movement of theĀ J vector happens to be theĀ JzĀ axis, the answer is simple: we choseĀ the coordinate system so as to ensure that was the case.

But I am digressing. I just inserted the illustration above to give you an inkling of where we’re going with this. Indeed, what’s shown above will make it easier for you to see how we can generalize the analysis that we’ll do now, which is an analysis of the orbital angular momentum and the related magnetic moment only. Let me copy the illustration we started with once more, so you don’t have to scroll up to see what we’re talking about.

So we have a charge orbiting around some center. It’s a classical analysis, and so it’s really like a planet around the Sun, except that we should remember that likes repel, and opposites attract, so we’ve got a minus sign in the force law here.

Let’s go through the math. TheĀ magnetic moment is the current times the area of the loop. As the velocity is constant, the current is just the charge qĀ times the frequency of rotation. The frequency of rotation is, of course, the velocity (i.e. the distance traveled per second) divided by the circumference of the orbit (i.e. 2ĻĀ·r). Hence, we write: I = (qeĀ·v)/(2ĻĀ·r) and, therefore:Ā Ī¼ =Ā (qe/Ā·v)Ā·ĻĀ·r2)/(2ĻĀ·r) = qeĀ·vĀ·r/2. Note that, as per the convention, current is defined as a flow ofĀ positiveĀ charges, so the illustration above actually assumes we’re talking some protonĀ in orbit, so q = qeĀ would be the elementary charge +1. If we’d be talking an electron, then its charge is to be denoted as āqeĀ (minusĀ qe, i.e.Ā ā1), and we’d need to reverse the direction of Ī¼, which we’ll do in a moment. However, to simplify the discussion, you should just think of some positiveĀ charge orbiting the way it does in the illustration above.

OK. That’s all there’s to say about the magnetic momentāfor the time being, that is. Let’s think about the angular momentum now. It’sĀ orbital angular momentum here, and so that’s the type of angular momentum we discussed in our post on gyroscopes. We denoted it as LĀ indeed ā i.e. not asĀ J, but that’s just a matter of conventionsĀ āĀ and we noted thatĀ L could be calculated as the vector cross product of the position vectorĀ r and the momentum vector p, as shown in the animation below, which also shows the torqueĀ vectorĀ Ļ.

The angular momentum LĀ changesĀ inĀ the animation above. InĀ our J case above, it doesn’t. Also, unlike what’s happening with the angular momentum of that swinging ball above, the magnitudeĀ of our J doesn’t change. It remains constant, and it’s equal to |J| = J =Ā |rĆp| =Ā |r|Ā·|p|Ā·sinĪø = rĀ·p = rĀ·mĀ·v. One should note this is a non-relativistic formula, but as the relative velocity of an electron v/cĀ is equal to the fine-structure constant, so that’s Ī±Ā ā 0.0073 (see my post on the fine-structure constantĀ if you wonder where this formula comes from), it’s OK to notĀ include the Lorentz factor in our formulas as for now.

Now, as I mentioned already, the illustration we’re using to explain Ī¼ andĀ JĀ is somewhat unreal because it assumes a positive charge q, and so Ī¼ and JĀ point in the same direction in this case, which is notĀ the case if we’d be talking an actual atomic system with an electron orbiting around a proton. But let’s go along with it as for now and so we’ll put the required minusĀ sign in later. We can combine the J =Ā rĀ·mĀ·v and Ī¼ =Ā qĀ·vĀ·r/2 formulas to write:

Ī¼ = (q/2m)Ā·J orĀ Ī¼/JĀ =Ā (q/2m)Ā (electron orbit)

In other words, theĀ ratioĀ of the magnetic moment and the angular moment depends on (1) the charge (q) and (2)Ā the mass of the charge, and on those two variablesĀ only. So the ratio does not dependĀ on the velocity v nor on the radius r. It can be noted that the q/2m factor is often referred to as the gyromagnetic factor (not to be confused with the g-factor, which we’ll introduce shortly). It’s good to do a quick dimensional check of this relation: the magnetic moment is expressed in ampĆØreĀ perĀ secondĀ times the loop area, so that’s (C/s)Ā·m2. On the right-hand side, we have the dimension of the gyromagnetic factor, which is C/kg, times the dimension of the angular momentum, which is mĀ·kgĀ·m/s, so we have the same units on both sides: CĀ·m2/s, Ā which is often written as jouleĀ perĀ teslaĀ (J/T): the jouleĀ is the energy unit (1 J = 1 NĀ·m), and the tesla measures the strength of the magnetic field (1 T = 1 (NĀ·s)/(CĀ·m). OK. So that works out.

So far, so good. TheĀ story is a little bit different for the spinĀ angular momentum and theĀ spin magnetic moment.Ā The formula is the following:

Ī¼ = (q/m)Ā·JĀ (electron spin)

This formula says that the Ī¼/J ratio is twiceĀ what it is for the orbital motion of the electron. Why is that? Feynman says “the reasons are pure quantum-mechanicalāthere is no classical explanation.” So I’d suggest we just leave that question open for the moment and see if we’re any wiser once we’ve worked ourselves through all of his LecturesĀ on quantum physics. š Let’s just go along with it as for now.

Now,Ā we can write both formulas āĀ i.e. the formula for the spin and the orbital angular momentum ā in a more general way using the format below:

Ī¼ = āgĀ·(qe/2me)Ā·J

Why the minus sign? Well… I wanted to get the sign right this time. Our model assumed someĀ positiveĀ charge in orbit, but so we want a formula for a atomic system, and so our circling charge should be anĀ electron. So the formula above is the formula for a electron, and the direction of the magnetic moment and of the angular motion will be oppositeĀ for electrons: it just replaces q by āqe. The format above also applies to anyĀ atomic system: as Feynman writes, “For an isolated atom, the direction of the magnetic moment will always be exactly opposite to the direction of the angular momentum.”Ā So the g-factor will beĀ characteristic of the state of the atom. It will beĀ 1Ā for a pure orbital moment,Ā 2 for a pure spin moment, or some other number in-between for a complicated system like an atom, indeed.

You may have one last question: whyĀ qe/2m instead of qe/m in the middle? Well… If we’d take qe/m, then g would be 1/2 for the orbitalĀ angular momentum, and the initial idea with g was that it would be some integer (we’ll quickly see that’s an idea only). So… Well… It’s just one more convention. Of course, conventions are not always respected so sometimes you’ll see the expression above written without the minus sign, so you may see it asĀ Ī¼ = gĀ·(qe/2me)Ā·J. In that case, the g-factor for our example involving the spinĀ angular momentum and the spinĀ magnetic moment, will obviously have to be written asĀ minusĀ 2.

Of course, it’s easy to see that the formula for the spin of aĀ protonĀ will look the same, except that we should take the mass of the proton in the formula, so that’sĀ mpĀ instead of me. Having said that, the elementary charge remains what it is, but so we write it without the minus sign here. To make a long story short, the formula for the proton is:Ā

Ī¼ = gĀ·(qe/2mp)Ā·J

OK. That’s clear enough. For electrons, the g-factor is referred to as the LandĆ© g-factor, while the g-factor for protons or, more generally, for any spinningĀ nucleus, is referred to as theĀ nuclearĀ g-factor. Now, you may or may not be surprised, but there’sĀ a g-factorĀ for neutrons too, despite the fact that they do not carry a netĀ charge: the explanation for it must have something to do with the quarks that make up the neutron but that’s a complicated matter which we will notĀ get into here. Finally, there is a g-factor for a whole atom, or a whole atomic system, and that’s referred to as… Well… Just the g-factor. š It’s, obviously, a number that’s characteristic of the state of the atom.

So… This was a big step forward. We’ve got all of the basics on that ‘magical’ spin number here, and so I hope it’s somewhat less ‘magical’ now. š Let me just copy the values of the g-factor for some elementary particles. It also shows how hard physicists have been trying to narrow down the uncertainty in the measurement. Quite impressive! The table comes from the Wikipedia article on it. I hope the explanations above will now enable you to read and understand that. š

Let’s now move on to the next topic.

#### Spin numbers and states

Of course, we’re talking quantum mechanics and, therefore, J can only take on a finite number of values. While that’s weird ā as weird as other quantum-mechanical things, such as the boson-fermion math, for exampleĀ ā it should not surprise us. As we will see in a moment, the values of J will determine the magneticĀ energy our system will acquire when we put in some magnetic field and, as Feynman writes: “That the energy of an atom in the magnetic field can have only certain discrete energies is really not more surprising than the fact that atoms in general have only certain discrete energy levels. Why should the same thing not hold for atoms in a magnetic field? It does. It is just correlation of this with theĀ idea of an oriented magnetic moment that brings out some of the strange implications of quantum mechanics.” Yep. That’s true. We’ll talk about that later.

Of course, you’ll probably want some ‘easier’ explanation. I am afraid I can’t give you that. All I can say is that, perhaps, you should think of ourĀ discussion on the fine-structure constant, which made it clear that the various radii of the electron, its velocity and its mass and/or energy are all related one to another and, hence, that they can only take on certain values. Indeed, of all the relations we discussed, there’s two you should alwaysĀ remember. The first relationship is theĀ UĀ = (e2/r) = Ī±/r. So that links the energy (which we can express in equivalent mass units), the electron charge and its radius. The second thing you should remember is that the Bohr radius and the classical electron radius are also related through Ī±: Ī±Ā  Ā re/r = Ī±2. So you may want to think of theĀ different values for J as being associated with different ‘orbitals’, so to speak. But that’s a very crude way of thinking about it, so I’d say: just accept the fact and see where it leads us. You’ll see, in a few moments from now, that theĀ whole thing is not unlike the quantum-mechanical explanation of the blackbody radiation problem, which assumes that the permitted energy levels (or states) are equally spaced and hĀ·f apart, with fĀ the frequency of the light that’s being absorbed and/or emitted. So the atom takes up energies onlyĀ hĀ·f at a time. Here we’ve got something similar: the energy levels that we’ll associate with the discrete values of JĀ āĀ or J‘sĀ componentsĀ , I should sayĀ ā will also be equally spaced. Let me show you how it works, as that will make things somewhat more clear.

If we have an object with a given total angular momentum J in classical mechanics, then any of its components x, y or z, could take on any value from +J toĀ āJ. That’s not the case here. The rule is that the ‘system’ ā the atom, the nucleus, or anything really ā will have a characteristic number, which is referred to as the ‘spin’ of the system and, somewhat confusingly, it’s denoted by jĀ (as you can,Ā it’s extremely important, indeed, to distinguish capital letters (like J) from small letters (like j)Ā if you want to keep track of what we’re explaining here). Now, if we have that characteristic spin number j, then any component of JĀ (think of theĀ z-direction, for example) can take on only (one of) the following values:

Note that we will always have 2j + 1 values. For example, if j = 3/2, we’ll have 2Ā·(3/2) + 1 = 4 permitted values, and in the extreme case where j is zero, we’ll still have 2Ā·0 + 1 = 1 permitted value: zero itself. So that basically says we have no angular momentum. […] OK. That should be clear enough, but let’s pause here for a moment and analyze thisājust to make sure we ‘get’ this indeed. What’s being written here? WhatĀ areĀ those numbers? Let’s do a quick dimensional analysis first. BecauseĀ j, jĀ ā 1, jĀ ā 2, etcetera are pure numbers, it’s only the dimension of Ä§ that we need to look at. We knowĀ Ä§: it’s the Planck constant h, which is expressed inĀ jouleĀ·second, i.e. JĀ·s =Ā NĀ·mĀ·s, divided by 2Ļ.Ā Ā That makes sense, because we get the same dimension for the angular momentum. Indeed, the L or J =Ā rĀ·mĀ·v formula also gives us the dimension of physical action, i.e. NĀ·mĀ·s. Just check it: [r]Ā·[m]Ā·[v] = mĀ·kgĀ·m/s = mĀ·(NĀ·s2/m)Ā·m/sĀ = NĀ·mĀ·s. Done!

So we’ve got some kind ofĀ unit of actionĀ once more here, even if it’s not h but Ä§ = h/2Ļ. That makes it a quantum of action expressed for a radian, so that’s a unit of length, rather than for a full cycle.Ā Just so you know, Ä§ = h/2Ļ is 1Ć10ā34Ā JĀ·sĀ āĀ 6.6Ć10ā16Ā eVĀ·s, and we could chose to express the components of J in terms of h by multiplying the whole thing with 2Ļ. That would boil down to saying that our unit length is not unity but the unit circle, which is 2Ļ times unity. Huh?Ā Just think about it: h is a fundamental unit linked to one full cycle of something, so it all makes sense. Before we move on, you may want to compare the value of h orĀ Ä§ with the energy of a photon, which is 1.6 to 3.2 eV in the visible light spectrum, but you shouldĀ note that energy does not have the time dimension, and a second is an eternity in quantum physics, so the comparison is a bit tricky. So… […] Well… Let’s just move on.Ā What about those coefficients? What constraints are there?

Well… The constraint is that the difference between +j and āj must be some integer, so +jā(āj) = 2jĀ must be an integer. That implies that the spin number j is always an integer or a half-integer, depending on whether j is even or odd. Let’s do a few examples:

1. A lithium (Li-7) nucleus has spin j = 3/2 and, therefore, the permitted values for the angular momentum around any axis (the z-axis, for example) are: 3/2, 3/2ā1=1/2, 3/2ā2=ā1/2, andĀ ā3/2āall timesĀ Ä§ of course! Note that the difference between +j andĀ āj is 3, and each ‘step’ between those two levels is Ä§, as we’d like it to be.
2. The nucleus of the much rarer Lithium-6 isotope is one of the few stable nuclei that has spin j = 1, so the permitted values are 1, 0 and ā1. Again, all needs to be multiplied with Ä§ to get the actual value for the J-component that we’re looking at. So each step is ‘one’ again, and the total difference (betweenĀ +j andĀ āj) is 2.]
3. An electron is a spin-1/2 particle, and so there are only two permitted values: +Ä§/2 and āÄ§/2. So there is just one ‘step’ and it’s equal to the whole difference betweenĀ +j andĀ āj. In fact, this is the most common situation, because we’ll be talking elementary fermions most of the time.
4. Photons are an example of spin-1 ‘particles’, and ‘particles’ with integer spin are referred to as bosons. In this regard, you may heard of superfluid Helium-4, which is caused by Bose-EinsteinĀ condensationĀ near the zero temperature point, and demonstrates the integerĀ spin number ofĀ Helium-4, so it resembles Lithium-6 in this regard.

The four ‘typical’ examples makes it clear that the actual situations that we’ll be analyzing will usually be quite simple: we’ll only have 2, 3 or 4 permitted values only. As mentioned, there is this fundamental dichotomy betweenĀ fermionsĀ andĀ bosons. Fermions have half-integer spin, and all elementaryĀ fermions, such as protons, neutrons, electrons, neutrinos and quarks are spin-1/2 particles. [Note that a proton and a neutron are, strictly speaking, notĀ elementary, as their constituent parts are quarks.] Bosons have integer spin, and the bosons we know of are spin-one particles, (except for the newly discovered Higgs boson, which is an actual spin-zero particle). The photon is an example, but the helium nucleus (He-4) also has spin one, which ā as mentioned above āĀ gives rise to superfluidity when its cooled near the absolute zero point.

In any case, to make a long story short, in practice, we’ll be dealing almost exclusively with spin-1, spin-1/2 particles and, occasionally, with spin-3/2 particles. In addition, to analyze simple stuff, we’ll often pretend particles doĀ notĀ have any spin, so our ‘theoretical’ particles will often be spin zero. That’s just to simplify stuff.

We now need to learn how to do a bit of math with all of this. Before we do so, let me make some additional remarks on these permitted values. Regardless of whether or not J is ‘wobbling’ or movingĀ or not āĀ let me be clear: J is not movingĀ in the analysis above, but we’ll discuss the phenomenon of precessionĀ in the next post, and that willĀ involve a J like that J circling around theĀ JzĀ axis, so I am just preparing the terrain hereĀ ā J‘s magnitude will always be some constant, which we denoted by |J| = J.

Now there’s something really interesting here, which again distinguishes classical mechanics from quantum mechanics. As mentioned, in classical mechanics, any of J‘s components Jx,Ā JyĀ or Jz, could take on any value from +J toĀ āJ and, therefore, the maximum value of any component of J ā say JzĀ ā would be equal to J. To be precise, J would be the value of the component of JĀ in the direction of J itself. So, in classical mechanics, we’d write: |J| = +ā(JĀ·J) = +āJ2Ā =Ā J, and it would be the maximum value of any component of J. But so we said that, if the spin number of J is j, then the maximum value of any component of J was equal to jĀ·Ä§.Ā So, naturally, one would think that J = |J| = +ā(JĀ·J) = +āJ2Ā = jĀ·Ä§.

However, that’s notĀ the case in quantum mechanics: the maximum value of any component of J is notĀ J = jĀ·Ä§ but the square root of jĀ·(j+1)Ā·Ä§.

Huh?Ā Yes. Let me spell it out:Ā |J| = +ā(JĀ·J) = +āJ2Ā ā  jÄ§. Indeed, quantum math has many particularities, and this is one of them. The magnitude of JĀ isĀ not equal to the largest possible value of any component of J:

J‘s magnitude is not jÄ§ but ā(j(j+1)Ä§).

As for the proof of this, let me simplify my life and just copy Feynman here:

The formula can be easily generalized for j ā  3/2. Also note that we used a fact that we didn’t mention as yet: all possible values of the z-component (or of whatever component) of J are equally likely.

Now, the result is fascinating, but the implications are even better. Let me paraphrase Feynman as he relates them:

1. From what we have so far, we can get another interesting and somewhat surprising conclusion. In certain classical calculations the quantity that appears in the final result is the square of the magnitude of the angular momentumĀ Jāin other words, JāJĀ = J2. It turns out that it is often possible to guess at the correct quantum-mechanical formula by using the classical calculation and the following simple rule: Replace J2Ā = JāJĀ byĀ j(j+1)Ä§. This rule is commonly used, and usually gives the correct result.
2. The second implication is the one we announced already: although we would think classically that the largest possible value of the anyĀ component ofĀ JĀ is just the magnitude ofĀ J, quantum-mechanically the maximum of any component of J is always less than that, because jÄ§ is always less than ā(j(j+1)Ä§). For example, for j = 3/2 = 1.5, we haveĀ j(j+1) = (3/2)Ā·(5/2) = 15/4 = 3.75. Now, the square root of this value isĀ ā3.75 ā 1.9365, so the magnitude of J is about 30% largerĀ than the maximum value of any of J‘s components. That’s a pretty spectacular difference, obviously!Ā  Ā

The second point is quite deep: it implies that the angular momentum is ‘never completely along any direction’. Why? Well… Think of it: “any of J‘s components” also includes the component in the direction of J itself! But if the maximum value of that component is 30% less than the magnitude of J, what does that mean really? All we can say is that it implies that the concept of the direction of the magnitude itself is actually quite fuzzy in quantum mechanics! Of course, that’s got to do with the Uncertainty Principle, and so we’ll come back to this later.

In fact, ifĀ you look at the math, you may think: what’s that business with those average or expected values? A magnitude is a magnitude, isn’t it? It’s supposed to be calculated from theĀ actualĀ values of Jx, JyĀ andĀ Jz, not from some average that’s based on the (equal) likelihoods of the permitted values. You’re right. Feynman’s derivation here is quantum-mechanical from the start and, therefore, we get a quantum-mechanical result indeed: the magnitude of JĀ is calculated as the magnitude of a quantum-mechanical variable in the derivation above, not as the magnitude of a classical variable.

[…] OK. On to the next.

#### The magnetic energy of atoms

Before we start talking about this topic, we should, perhaps, relate the angular momentum to the magnetic moment once again. We can do that using theĀ Ī¼ = (q/2m)Ā·J and/orĀ Ī¼ = (q/m)Ā·JĀ formula (so that’s the simple formulas for the orbital and spin angular momentum respectively) or, else, by using the more generalĀ Ī¼ = ā gĀ·(q/2m)Ā·J formula.

Let’s use the simpler Ī¼ = (qe/2m)Ā·JĀ formula, which is the one for the orbitalĀ angular momentum.Ā What’s qe/2m? It should be equal to 1.6Ć10ā19Ā C divided by 2Ā·9.1Ć10ā31Ā kg, so that’s about 0.0879Ć1012Ā  C/kg, or 0.0879Ć1012Ā (CĀ·m)/(NĀ·s2). Now we multiply by Ä§/2 ā 0.527Ć10ā34Ā JĀ·s. We get something like 0.0463Ć10ā22Ā m2Ā·C/s or J/T. These numbers are ridiculously small, so they’re usually measured in terms of a so-called naturalĀ unit: the Bohr magneton, which I’ll explain in a moment but so here we’re interested in its value only, which is Ī¼BĀ =Ā 9.274Ć10ā24Ā J/T. Hence, Ī¼/Ī¼BĀ = 0.5 = 1/2. What a nice number!

Hmm… This cannot be a coincidence… […] You’re right. It isn’t. To get the full picture, we need to include the spinĀ angular momentum, so we also need to see what the Ī¼ = (q/m)Ā·JĀ will yield. That’s easy, of course, as it’s twice the value of (q/2m)Ā·J, so Ī¼/Ī¼BĀ = 1, and so the totalĀ is equal to 3/2. So the magnetic moment of an electron has the same value (when expressed in terms of the Bohr magneton) as the spinĀ (when expressed in terms of Ä§). Now that’s justĀ sweet!

Yes, it is. All our definitions and formulas were formulated so as to make it sweet. Having said that,Ā we do have a tiny little problem. If we use the generalĀ Ī¼ = āgĀ·(q/2m)Ā·J to write the result we found for the spinĀ of the electron only (so we’re notĀ looking at the orbital momentum here), then we’d write:Ā Ī¼ = 2Ā·(q/2m)Ā·JĀ = (q/m)Ā·JĀ and, hence, the g-factor here isĀ ā2. Yes. We know that. You told me so already. What’s the issue? Well… The problem is: experiments reveal the actual value of g isĀ not exactly ā2: it’s ā2.00231930436182(52) instead, with the last two digits (in brackets) the uncertainty in the current measurements. Just check it for yourself on the NIST website. š [Please do check it: it brings some realness to this discussion.]

Hmm…. The accuracy of the measurement suggests we should take it seriously, even if we’re talking a difference of 0.1% only. We should. It can be explained, of course: it’s something quantum-mechanical. However, we’ll talk about this later. As for now, just try to understand the basics here. It’s complicated enough already, and so we’ll stay away from the nitty-gritty as long as we can.

Let’s now get back to the magneticĀ energyĀ of our atoms.Ā From our discussion on theĀ torqueĀ on a magnetic dipole in an external magnetic field, we know that our magnetic atoms will have someĀ extraĀ magnetic energy when placed in an external field. So now we have an externalĀ magnetic field B, and we derived the formula for the energy is

UmagĀ =Ā āĪ¼Ā·BĀ·cosĪø = āĪ¼Ā·B

I won’t explain the whole thing once again, but it might help to visualize the situation, which we do below. The loop here is not circular but square, and it’s a current-carrying wire instead of an electron in orbit, but I hope you get the point.

We need to chose some coordinate system to calculate stuff and so we’ll just choose our z-axis along the direction of the external magnetic field B so as to simplify those calculations. If we do that, we can just take the z-component of Ī¼ and then combine the interim result with our generalĀ Ī¼ = ā gĀ·(q/2m)Ā·J formula, so we write:

UmagĀ =Ā āĪ¼zĀ·B =Ā gĀ·(q/2m)Ā·JzĀ·B

Now, we know that theĀ maximumĀ value of JzĀ is equal to jĀ·Ä§, and so the maximum value of UmagĀ will be equalĀ g(q/2m)jÄ§B. Let’s now simplify this expression by choosing some natural unit, and that’s the unit we introduced already above: the Bohr magneton. It’s equal to (qeÄ§)/(2me) and its value is Ī¼BĀ āĀ 9.274Ć10ā24Ā J/T. So we get the result we wanted, and that is:

Let me make a few remarks here. First on that magneton: you should note there’s also something which is known as theĀ nuclearĀ magneton which, you guessed it, is calculated using theĀ proton charge and the proton mass:Ā Ī¼NĀ = (qpÄ§)/(2mp)Ā ā 5.05Ć10ā27Ā J/T. My second remark is a question: what does that formulaĀ mean, really? Well… Let me quote Feynman on that. The formula basically says the following:

“The energy of an atomic system is changed when it is put in a magnetic field by an amount that is proportional to the field, and proportional to Jz. We say that the energy of an atomic system is ‘split’ into 2jĀ + 1Ā ‘levels’ by a magnetic field. For instance, an atom whose energy is U0Ā outside a magnetic field and whose jĀ isĀ 3/2, will have four possible energies when placed in a field. We can show these energies by an energy-level diagram like that drawn below. Any particular atom can have only one of the four possible energies in any given field B. That is what quantum mechanics says about the behavior of an atomic system in a magnetic field.”

Of course, the simplest ‘atomic’ system is a single electron, which has spin 1/2 only (like most fermions really: the example in the diagram above, with spin 3/2, would be that Li-7 system or something similar). If the spin is 1/2, then there are only two energy levels, with JzĀ = Ā±Ä§/2 and, as we mentioned already, the g-factor for an electron is ā2 (again, the use of minus signs (or not) is quite confusing: I am sorry for that),Ā and soĀ our formula above becomes very simple:

UmagĀ =Ā Ā±Ā Ī¼BĀ·B

The graph above becomes the graph below, and we can now speak more loosely and say that the electron either has its spin ‘up’ (so that’s along the field), or ‘down’ (so that’s opposite the field).

By now, you’re probably tired of the math and you’ll wonder: how can weĀ proveĀ all of this permitted value business? Well… That question leads me to the last topic of my post: the Stern-Gerlach experiment.

#### The Stern-Gerlach experimentĀ

Here again, I can just copy straight of out of Feynman, and so I hope you’ll forgive me if I just do that, as I don’t think there’s any advantage to me trying to summarize what he writes on it:

“The fact that the angular momentum is quantized is such a surprising thing that we will talk a little bit about it historically. It was a shock from the moment it was discovered (although it was expected theoretically). It was first observed in an experiment done in 1922 by Stern and Gerlach. If you wish, you can consider the experiment of Stern-Gerlach as a direct justification for a belief in the quantization of angular momentum. Stern and Gerlach devised an experiment for measuring the magnetic moment of individual silver atoms. They produced a beam of silver atoms by evaporating silver in a hot oven and letting some of them come out through a series of small holes. This beam was directed between the pole tips of a special magnet, as shown in the illustration below. Their idea was the following. If the silver atom has a magnetic momentĀ Ī¼, then in a magnetic fieldĀ BĀ it has an energyĀ āĪ¼zB, where zĀ is the direction of the magnetic field. In the classical theory, Ī¼zĀ would be equal to the magnetic moment times the cosine of the angle between the moment and the magnetic field, so the extra energy in the field would be

ĪU =Ā āĪ¼Ā·BĀ·cosĪø

Of course, as the atoms come out of the oven, their magnetic moments would point in every possible direction, so there would be all values ofĀ Īø. Now if the magnetic field varies very rapidly withĀ zāif there is a strong field gradientāthen the magnetic energy will also vary with position, and there will be a force on the magnetic moments whose direction will depend on whether cosineĀ ĪøĀ is positive or negative. The atoms will be pulled up or down by a force proportional to the derivative of the magnetic energy; from the principle of virtual work,

Stern and Gerlach made their magnet with a very sharp edge on one of the pole tips in order to produce a very rapid variation of the magnetic field. The beam of silver atoms was directed right along this sharp edge, so that the atoms would feel a vertical force in the inhomogeneous field. A silver atom with its magnetic moment directed horizontally would have no force on it and would go straight past the magnet. An atom whose magnetic moment was exactly vertical would have a force pulling it up toward the sharp edge of the magnet. An atom whose magnetic moment was pointed downward would feel a downward push. Thus, as they left the magnet, the atoms would be spread out according to their vertical components of magnetic moment. In the classical theory all angles are possible, so that when the silver atoms are collected by deposition on a glass plate, one should expect a smear of silver along a vertical line. The height of the line would be proportional to the magnitude of the magnetic moment. The abject failure of classical ideas was completely revealed when Stern and Gerlach saw what actually happened. They found on the glass plate two distinct spots. The silver atoms had formed two beams.

That a beam of atoms whose spins would apparently be randomly oriented gets split up into two separate beams is most miraculous. How does the magnetic moment know that it is only allowed to take on certain components in the direction of the magnetic field? Well, that was really the beginning of the discovery of the quantization of angular momentum, and instead of trying to give you a theoretical explanation, we will just say that you are stuck with the result of this experiment just as the physicists of that day had to accept the result when the experiment was done. It is an experimental fact that the energy of an atom in a magnetic field takes on a series of individual values. For each of these values the energy is proportional to the field strength. So in a region where the field varies, the principle of virtual work tells us that the possible magnetic force on the atoms will have a set of separate values; the force is different for each state, so the beam of atoms is split into a small number of separate beams. From a measurement of the deflection of the beams, one can find the strength of the magnetic moment.”

I should note one point which Feynman hardly addresses in the analysis above: why do we need a non-homogeneous field? Wellā¦ Think of it. The individual silver atoms are not like electrons in some electric field. They are tiny little magnets, and magnets do not behave like electrons. Remember we said thereās no such thing as a magnetic charge? So that applies here. If the silver atoms are tiny magnets, with a magnetic dipole moment, then the only thing they will do is turn, so as to minimize their energy U = āĪ¼BcosĪø.

That energy is minimized when Ī¼ and B are at right angles of each other, so as to make the cosĪø factor zero, which happens when Īø = Ļ/2. Hence, in a homogeneous magnetic field, we will have a torque on the loop of current – think of our electron(s) in orbit here – but no net force pulling it in this or that direction as a whole. So the atoms would just rotate but not move in our classical analysis here.

To make the atoms themselves move towards or away one of the poles (with or without a sharp tip), the magnetic field must be non-homogeneous, so as to ensure that the force thatās pulling on one side of the loop of current is slightly different from the force thatās pulling (in the opposite direction) on the other side of the loop of current. So thatās why the field has to be non-homogeneous (or inhomogeneous as Feynman calls it), and so thatās why one pole needs to have a sharply pointed tip.

As for the force formula, itās crucial to remember that energy (or work) is force times distance. To be precise, itās the ā«Fāds integral. This integral will have a minus sign in front when weāre doing work against the force, so thatās when weāre increasing the potential energy of an object. Conversely, weāll just take the positive value when weāre converting potential energy into kinetic energy. So that explains the F = āāU/āz formula above. In fact, in the analysis above, Feynman assumes the magnetic moment doesnāt turn at all. Thatās pretty obvious from the Fz = āāU/āz = āĪ¼ācosĪøāāB/āz formula, in which Ī¼ is clearly being treated as a constant. So the Fz in this formula is a net force in the z-direction, and itās crucially dependent on the variation of the magnetic field in the z-direction. If the field would not be varying, āB/āz would be zero and, therefore, we would not have any net force in the z-direction. As mentioned above, we would only have a torque.

Well… This sort of covers all of what we wanted to cover today. š I hope you enjoyed it.

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

Note: I have published a paper that is very coherent and fully explains this so-called God-given number. There is nothing magical about it. It is just a scaling constant. Check it out: The Meaning of the Fine-Structure Constant. No ambiguity. No hocus-pocus.

Jean Louis Van Belle, 23 December 2018

Original post:

In my previous post, I explained why the fine-structure constant Ī±Ā isĀ notĀ a ‘magical’ number, even ifĀ it relates all fundamental properties of the electron: its mass, its energy, its charge, its radius, its photonĀ scattering cross-sectionĀ (i.e. the Bohr radius, or the size of the atom really) and, finally, the couplingĀ constant for photon-electron interactions.Ā The key to such understanding ofĀ Ī± 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 fieldĀ of the tiny ball of charge, which we denoted as Eelec.

The formula for EelecĀ is calculated using the formula for the field momentum of a moving charge and, using the m = E/c2Ā mas-energy equivalence relationship, is equivalentĀ to the electromagnetic mass. We went through the derivation in our previous post, so let me just jot down the result:

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

Ā

For a spherical shellĀ of charge we write:

To compare the formulas, you need to note that the square of the electron chargeĀ eĀ in the formula for the field energy isĀ equal to e2Ā = qe2/4ĻĪµ0Ā = keĀ·qe2. So we multiplyĀ the square of the actualĀ electron charge by the Coulomb constant keĀ = 1/4ĻĪµ0. As you can see, the three formulas have exactlyĀ the same form then. It’s just the proportionality constant that’s different: it’s 2/3, 3/5 and 1/2 respectively.Ā It’s interesting to quickly reflect on the dimensions here: [ke] ā 9Ć109 NĀ·m2/C2, so e2Ā is expressed in NĀ·m2. That makes the units come out alright, as we divide by a (so that’s in meter) and so we get the energy inĀ jouleĀ (which is newtonĀ·meter). In fact, now that we’re here, let’s quickly calculate the value of e2: it’s that keĀ·qe2Ā product, so it’s equal to 2.3Ć10ā28Ā NĀ·m2. We can quickly check this value because we know that the classical electron radius is equal to:

So we divide 2.3Ć10ā28Ā NĀ·m2Ā by mec2Ā ā 8.2Ć10ā14 J, so we get r0Ā ā 2.82Ć10ā15Ā 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.Ā 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 UelecĀ depends on the finer details of our model for the electron (sphere versus shell, uniform versus non-uniform distribution):

1. EelecĀ = (2/3)Ā·(e2/a): This is the formula for the energy of the field, so we may all it isĀ externalĀ energy.
2. UelecĀ = (3/5)Ā·(e2/a), or UelecĀ = (1/2)Ā·(e2/a): This is the energy needed to assembleĀ our electron, so we might, perhaps, call it itsĀ internalĀ energy. The first formula assumes our electron is a uniformly chargedĀ 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. mec2Ā =Ā e2/r0: This is the energy associated with the so-called classical electron radiusĀ (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 ā i.e. the difference between 2/3 and 1, basically āĀ are to be explained because of theĀ binding forces within the electron, without which the electron would just ‘explode’, as the French physicist and polymath Henri PoincarĆ© famously put it.Ā Indeed, 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 byĀ googlingĀ 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 ofĀ Lectures, which is devoted to quantum physicsĀ only, so I look very much forward to that.]

In this post, I just wanted to reflect once more on what constants are reallyĀ fundamentalĀ 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 Ī±, which is a dimensionless number.
2. Planck’s constant h, whose dimension is jouleĀ·second, so that’s the dimension ofĀ action.
3. The speed of light c, whose dimension is that of a velocity.

The three are related through the following expression:

This is an interesting expression. Let’s first check its dimension. We already explained thatĀ e2Ā is expressed inĀ NĀ·m2. That’s rather strange, because it means the dimension of e itself is N1/2Ā·m: what’s the square root of a force of oneĀ newton? In fact, to interpret the formula above, it’s probably better to re-writeĀ e2Ā asĀ e2Ā = qe2/4ĻĪµ0Ā = keĀ·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 law:Ā Coulomb’s law. We don’t need anything else, except for relativity theory, because we need to explain theĀ magnetic forceĀ as wellāandĀ thatĀ 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 cĀ  in that formula for Ī± above. As for Ä§, Ä§ = h/2Ļ comes with the E = hĀ·fĀ equation, which links us to the electron’s Compton wavelengthĀ Ī» through theĀ de BroglieĀ relationĀ Ī» = h/p.

The point is: we should probably not look at Ī± as a ‘fundamental physical constant’. It’s e2Ā that’s the third fundamental constant, besides h and c. Indeed, it’s fromĀ e2Ā 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 Ī± 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 Ī±, which seems to defineĀ Ī±Ā 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 notĀ a definition ofĀ Ī±: it just does what it does, and that’s toĀ relateĀ Ī± 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 thatĀ Ī± was ‘God’s cut-off factor’ š ensuring our energy functions do not blow up, but I also said it was impossible to sayĀ whyĀ he choseĀ 0.00729735256Ā as the cut-off factor. The question is actually easily answered by thinking about thoseĀ twoĀ formulas we had for the internal and external energy respectively. Let’s re-write them in natural units and, temporarily, two different subscripts forĀ Ī±, so we write:

1. EelecĀ = Ī±e/r0: This is the formula for the energy of the field.
2. UelecĀ = Ī±u/r0: This is the energy needed to assembleĀ our electron.

Both energies are determined by the above-mentioned laws, i.e. Coulomb’s Law and the theory of relativity, so Ī± has got nothing to do what that. However, both energies have to be the same, and so Ī±eĀ has to be equal toĀ Ī±u. In that sense,Ā Ī± is, quite simply, aĀ proportionality constantĀ that achieves that equality. NowĀ thatĀ explains why we canĀ deriveĀ Ī± from the three other constants which, as mentioned above, are probablyĀ moreĀ fundamental. In fact, we’ve got only three degrees of freedom here, so if we chose c,Ā h andĀ eĀ as ‘fundamental’, thenĀ Ī± isn’t any more.

The underlying deep question behind it all isĀ whyĀ those two energies should be equal.Ā 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 itĀ hasĀ to have some internal energy.

You’ll say: that makes sense, but it doesn’t answer the question. Why would it haveĀ internal energy, with or without a zero radius? If an electron is an elementaryĀ particle, then it’s reallyĀ 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 veryĀ 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 UelecĀ = (1/2)Ā·(e2/a) by calculating the externalĀ field energy for a spherical shell of charge, and he sticks to itāpresumably because it’s the same field for a uniform or non-uniformĀ sphere of charge. He only notes there has to be some radius because, if not, the formula he uses blows up, indeed. SoĀ ā who knows?Ā ā perhaps he doesn’t quite believe that formula for theĀ internalĀ 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…Ā Perhaps just one more thing. Let me note somethingĀ that, I hope, you noticed as well: theĀ keĀ·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 FelecĀ = keĀ·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 r0Ā is, perhaps, some averageĀ distance or so. There must beĀ someĀ way of thinking of it like that. But… Well… Which oneĀ 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 laterāhopefully. š 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 =Ā mec2Ā law. It’s only theĀ mec2Ā =Ā e2/r0Ā that’s relativistically correct.
2. That suggests the electron also has some non-electric mass, which are referred to as ‘binding forces’ or ‘PoincarĆ© 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. š

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, thatĀ ā hopefully!Ā ā will then be solved in the quantum-mechanicalĀ picture of itĀ all. š

# Taking the magic out of God’s number

Note: I have published a paper that is very coherent and fully explains this so-called God-given number. There is nothing magical about it. It is just a scaling constant. Check it out: The Meaning of the Fine-Structure Constant. No ambiguity. No hocus-pocus.

Jean Louis Van Belle, 23 December 2018

Original post:

I think the post scriptum to my previous post is interesting enough to separate it out as a piece of its own, so let me do that here. You’ll remember that we were trying to find some kind of a model for the electron, picturing it like a tiny little ball of charge, and then we just applied the classical energy formulas to it to see what comes out of it. TheĀ key formula is the integral that gives us the energy that goes into assembling a charge. It was the following thing:

This is a double integral which we simplified in two stages, so we’re looking at an integral within an integral really, butĀ we can substitute the integral over the Ļ(2)Ā·dV2Ā product by the formula we got for the potential, so we write that asĀ Ī¦(1), and so the integral above becomes:

Now, thisĀ integral integrates the Ļ(1)Ā·Ī¦(1)Ā·dV1Ā product over all of space, so thatās over all points in space, and so we just dropped the index and wrote the whole thing as the integral of ĻĀ·Ī¦Ā·dVĀ over all of space:

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

So this integral is actually quite simple: it just integrates Eā¢E = E2Ā over all of space. The illustration below shows E as a function of the distanceĀ rĀ for a sphere of radius R filledĀ uniformlyĀ with charge.

So the field (E) goes asĀ rĀ for r ā¤ R and as 1/r2Ā for r ā„ R. So, for r ā„ R, the integral will have (1/r2)2Ā = 1/r4Ā in 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’.Ā 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.

In fact, the need for a ‘cut-off factor’ so as to ensure our energy function doesn’t ‘blow up’ is not because of the exponent in theĀ 1/r4 expression: the need is also there for any 1/r relation, as illustrated below. AllĀ 1/rnĀ function have the sameĀ pivotĀ point, as you can see from the simple illustration below. So, yes, we cannot go all the way to zero from there when integrating: we have to stopĀ somewhere.

So what’s the ‘cut-off point’? What’s ‘too small’ a radius? Let’s look at the formula we got for our electron as a shellĀ of charge (so the assumption here is that the charge isĀ uniformly distributed on the surfaceĀ of a sphere with radius a):

So we’ve got an even simpler formula here: it’s just a 1/rĀ relation (a is rĀ in this formula), not 1/r4. Why is that? Well… It’s just the way the math turns out: we’re integrating over volumes and so that involves an r3Ā factor and so it all simplifies to 1/r, and so that gives us this simple inverselyĀ proportional relationship between U and r, i.eĀ a, in this case. šĀ I copied the detail of Feynman’s calculation in my previous post, so you can double-check it. It’s quite wonderful, really. Look at it again: we have a very simpleĀ inverselyĀ proportional relationship between theĀ radiusĀ of our electron andĀ itsĀ energy as a sphere of charge. We could write it as:

UelectĀ Ā = Ī±/a, withĀ Ī± = e2/2

Still… We need the cut-off point’. Also note that, as I pointed out, we don’t necessarily need to assume that the charge in our little ball of charge (i.e. our electron) sits on the surface only: if we’d assume it’s a uniformly charged sphere of charge, we’d just get another constant of proportionality: our 1/2 factor would become a 3/5 factor, so we’d write: UelectĀ Ā = (3/5)Ā·e2/a. But we’re not interested in finding the rightĀ model here. We know theĀ UelectĀ Ā = (3/5)Ā·e2/aĀ gives us a value forĀ aĀ that differs with a 2/5 factor as the classical electron radius. That’s not so bad and so let’s go along with it. š

We’re going to look at the simpleĀ structureĀ of this relation, and all of its implications. The simple equation above says that the energy of our electron is (a) proportional to the square of its charge and (b) inversely proportional to its radius. Now,Ā thatĀ is a very remarkable result. In fact,Ā we’ve seen something like this before, and we were astonished.Ā We saw it when we were discussing the wonderful properties of that magical number, theĀ fine-structure constant, which we also denoted byĀ Ī±. However, because we usedĀ Ī± already, I’ll denote the fine-structure constant as Ī±e here, so you don’t get confused. You’ll remember thatĀ the fine-structure constant is a God-like number indeed: it links allĀ of the fundamental properties of the electron, i.e. its charge, its radius, its distance to the nucleus (i.e. the Bohr radius), its velocity, its mass (and, hence, its energy), its de BroglieĀ wavelength. Whatever: all these physical constants are all related through the fine-structure constant.Ā

In my various posts on this topic, I’ve repeatedly said that, but I never showed why it’s true, and so it was a very magical number indeed. I am going to take some of the magic out now. Not too much but… Well… You can judge for yourself how much of the magic remains after I am done here. š

So, at this stage of the argument,Ā Ī± can be anything, andĀ Ī±eĀ cannot, 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,Ā or didn’t understand, I should say. Past tense. Indeed, we’re going to get some understanding here because we know that one of theĀ many expressions involving Ī±e was the following one:

meĀ = Ī±e/re

This says that theĀ massĀ 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’sĀ 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.] In fact, the U =Ā (3/5)Ā·e2/aĀ andĀ meĀ = Ī±e/reĀ relations looks exactlyĀ the same, because one of the other equations involving the fine-structure constant was:Ā Ī±eĀ =Ā eP2. So we’ve got the square of the charge here as well! Indeed, as I’ll explain in a moment, the difference between the two formulas is just a matter of units.

Now, mass is equivalent to energy, of course: it’s just a matter of units, so we can equate meĀ with EeĀ (this amounts to expressing the energy of the electron in aĀ kg unitābit weird, but OK) and so we get:

EeĀ = Ī±e/re

So there we have: the fine-structure constantĀ Ī±eĀ is Nature’s ‘cut-off’ factor, so to speak. Why? Only God knows. š But it’s now (fairly) easy to see why all the relations involving Ī±eĀ are what they are. As I mentioned already, we also know that Ī±eĀ is the square of the electron charge expressed in Planck units, so we have:

Ā Ī±eĀ =Ā eP2Ā and, therefore,Ā EeĀ = eP2/re

Now, you can check for yourself: it’s just a matter of re-expressing everything in standard SI units, and relating eP2Ā to e2, and it should all work: you should get the EelectĀ Ā = (2/3)Ā·e2/aĀ 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 isĀ notĀ something God-given. What’s God-given are Maxwell’s equations, and so theĀ EeĀ = Ī±e/reĀ = eP2/reĀ is just one of the many wonderful things that you can get out of Ā them.

So we found God’s ‘cut-off factor’ š It’s equal to Ī±eĀ ā 0.0073 = 7.3Ć10ā3. So 7.3 thousands of… What? Well… Nothing. It’s just a pure ratio between the energy and the radius of an electron (if both are expressed in Planck units, of course). And so it determines the electron charge (again, expressed in Planck units). Indeed, we write:

ePĀ = āĪ±e

Really? Yes. JustĀ doĀ all these formulas:

ePĀ = āĪ±eĀ āĀ ā0.0073Ā·1.9Ć10ā18Ā coulombĀ ā 1.6Ā Ć10ā19 C

Just re-check itĀ withĀ all the known decimals: you’ll see it’s bang on. Let’s look at the EeĀ = meĀ = Ī±e/reĀ ratio once again. What’s theĀ meaningĀ of it? Let’s first calculate the value of reĀ and me,Ā i.e. the electron radius and electron mass expressed in Planck units. It’s equal to the classical electron radius divided by the Planck length, and then the same for the mass, so we get the following thing:

reĀ ā (2.81794Ć10ā15Ā m)/(1.6162Ć10ā35Ā m) = 1.7435Ć1020Ā

meĀ ā (9.1Ć10ā31Ā kg)/(2.17651Ć10ā8 kg) = 4.18Ć10ā23

Ī±eĀ = (4.18Ć10ā23)Ā·(1.7435Ć1020) ā 0.0073

It works like a charm, but what does it mean? Well… It’s just a ratio between two physical quantities, and the scaleĀ you use to measure those quantities matters very much. We’ve explained that the Planck mass is a rather large unit at the atomic scale and, therefore, it’s perhaps not quite appropriate to use it here. In fact, out of the many interesting expressions for Ī±e, I should highlight the following one:

Ī±eĀ = e2/(Ä§Ā·c) ā (1.60217662Ć10ā19 C)2/(4ĻĪµ0Ā·[(1.054572Ć10ā34 NĀ·mĀ·s)Ā·(2.998Ć108 m/s)])Ā ā 0.0073 once more š

Note that the elementary charge e is actually equal to qe/4ĻĪµ0, which is what I am using in the formula. I know that’s confusing, but it what it is. As for the units, it’s a bit tedious to write it all out, but you’ll get there. Note that Īµ0Ā ā 8.8542Ć10ā12Ā C2/(NĀ·m2) so… Well… All the units do cancel out, and we get a dimensionless number indeed, which is what Ī±eĀ is.

The point is: this expression links Ī±eĀ to the the de BroglieĀ relation (p =Ā h/Ī»), with Ī» the wavelengthĀ that’s associated with the electron. Of course, because of the Uncertainty Principle, we know we’re talking some wavelengthĀ rangeĀ really, so we should write the de Broglie relation asĀ Īp = hĀ·Ī(1/Ī»). Now, that, in turn, allows us to try to work out the Bohr radius, which is the other ‘dimension’ we associate with an electron. Of course, now you’ll say: why would you do that. Why would you bring in the de BroglieĀ relation here?

Well… We’re talking energy, and so we have the Planck-EinsteinĀ relation first: the energy of some particle can always be written as the product of hĀ and some frequencyĀ f: E = hĀ·f. The only thing thatĀ de Broglie relation adds is the Uncertainty Principle indeed: the frequencyĀ fĀ will be some frequency range, associated with someĀ momentumĀ range, and so that’s what the Uncertainty Principle really says. I can’t dwell too much on that here, because otherwise this post would become a book. š For more detail, you can check out one of my many posts on the Uncertainty Principle. In fact, the one I am referring to here has Feynman’s calculation of the Bohr radius, so I warmly recommend you check it out. The thrust of the argument is as follows:

1. If we assume that (a) an electron takes some space ā which Iāll denote by rĀ š ā and (b) that it has some momentum p because of its mass m and its velocity v, then the ĪxĪp =Ā Ä§Ā 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, letās just boldly write r ā Ä§/p and see what we can do with that.
2. We know that the kinetic energy of our electron equals mv2/2, which we can write as p2/2m so we get rid of the velocity factor.Wellā¦ Substituting our p ā Ä§/r conjecture, we get K.E. = Ä§2/2mr2. So thatās a formula for the kinetic energy. Next is potential.
3. TheĀ formula for the potential energy is U = q1q2/4ĻĪµ0r12. Now, weāre actually talking about the size of an atomĀ here, so one charge is the proton (+e) and the other is the electron (āe), so the potential energy is U = P.E. = āe2/4ĻĪµ0r, with rĀ the ādistanceā between the proton and the electronāso thatās the Bohr radius weāre looking for!
4. We can now write the total energy (which Iāll denote by E, but donāt confuse it with the electric field vector!) asĀ E = K.E. + P.E. =Ā Ā Ä§2/2mr2Ā āĀ e2/4ĻĪµ0r. Now,Ā the electron (whatever it is) is, obviously, in some kind of equilibrium state. Why is that obvious? Wellā¦ Otherwise our hydrogen atom wouldnāt or couldnāt exist. š Hence, itās in some kind of energy āwellā indeed, at the bottom. Such equilibrium point āat the bottomā is characterized by its derivative (in respect to whatever variable) being equal to zero. Now, the only āvariableā here isĀ rĀ (all the other symbols are physical constants), so we have to solve for dE/dr = 0. Writing it all out yields:Ā dE/dr = āÄ§2/mr3Ā + e2/4ĻĪµ0r2Ā = 0Ā ā r =Ā 4ĻĪµ0Ä§2/me2
5. We can now put the values in:Ā r =Ā 4ĻĪµ0h2/me2Ā = [(1/(9Ć109) C2/NĀ·m2)Ā·(1.055Ć10ā34Ā JĀ·s)2]/[(9.1Ć10ā31Ā kg)Ā·(1.6Ć10ā19Ā C)2]Ā = 53Ć10ā12Ā m = 53 pico-meter (pm)

Done. We’re right on the spot.Ā The Bohr radius is, effectively, about 53 trillionthsĀ of a meter indeed!

Phew!

Yes… I know… Relax. We’re almost done. You should now be able to figure out why the classical electron radius and the Bohr radius can also be related to each other through the fine-structure constant. We write:

meĀ = Ī±/reĀ = Ī±/Ī±2rĀ = 1/Ī±r

So we get that Ī±/reĀ =Ā 1/Ī±r and, therefore, we get re/r = Ī±2, which explains why Ī± is also equal to the so-called junction number, or the coupling constant, for an electron-photon coupling (see my post on the quantum-mechanical aspects of the photon-electron interaction). It gives a physical meaning to the probability (which, as you know, is the absolute square of the probability amplitude) in terms of the chance of a photon actually ‘hitting’ the electron as it goes through the atom. Indeed, the ratio of the Thomson scattering cross-section and the Bohr size of the atom should be of the same order asĀ re/r, and so that’s Ī±2.

[Note: To be fully correct and complete, I should add that the coupling constant itself is notĀ Ī±2Ā but āĪ± = eP. Why do we have this square root? You’re right: the fact that the probability is the absolute squareĀ of the amplitude explains one square root (āĪ±2Ā = Ī±), but not two. The thing is: the photon-electron interaction consists of twoĀ things. First, the electron sort of ‘absorbs’ the photon, and then it emits another one, that has the same or a different frequency depending on whether or not the ‘collision’ was elastic or not. So if we denote the coupling constant as j, then the whole interaction will have a probability amplitude equal to j2. In fact, theĀ value which Feynman uses in his wonderful popular presentation of quantum mechanics (The Strange Theory of Light and Matter), is āĪ± ā ā0.0073. I am not quite sure why the minus sign is there. It must be something with the angles involved (the emitted photon will not be following the trajectory of the incoming photon) or, else, with the special arithmetic involved in boson-fermion interactions (we add amplitudes when bosons are involved, butĀ subtractĀ amplitudes when it’s fermions interacting.Ā I’ll probably find out once I am true through Feynman’s third volume ofĀ Lectures, which focus on quantum mechanics only.]

Finally, the last bit of unexplained ‘magic’ in the fine-structure constant is that the fine-structure constant (which I’ve started to write as Ī± again, instead of Ī±e) also gives us the (classical) relative speed of an electron, so that’s its speed as it orbits around the nucleus (according to the classical theory, that is), so we write

Ī± = v/cĀ =Ā Ī²

I should go through the motions hereĀ ā I’ll probably do so in the coming daysĀ ā but you can see we must be able to get it out somehow from all what we wrote above. See how powerfulĀ ourĀ UelectĀ Ā ā¼ e2/a relation really is? It links the electron, charge, its radius and its energy, and it’s all we need to all the rest out of it: its mass, its momentum, its speed and ā through the Uncertainty Principle ā the Bohr radius, which is the size of the atom.

We’ve come a long way. This is truly a milestone. We’ve taken the magic out of God’s numberāto some extent at least. š

You’ll have one last question, of course: if proportionality constants are all about theĀ scaleĀ in which weĀ measureĀ the physical quantities on either side of an equation, is there some way the fine-structure constant would come out differently? That’s the same as asking: what if we’d measure energy in units that are equivalent to the energy of an electron, and the radius of our electron just as… Well… What if we’d equate our unit of distance with the radius of the electron, so we’d write re = 1? What would happen toĀ Ī±? Well…Ā I’ll let you figure that one out yourself. I am tired and so I should go to bed now. š

[…] OK. OK. Let me tell you. It’s not that simple here. All those relationships involving Ī±, in one form or the other, are very deep. They relate a lot of stuff to a lot of stuff, and we can appreciate that only when doing a dimensional analysis. A dimensional analysis of the EeĀ = Ī±e/reĀ = eP2/rĀ yields [eP2/r] = C2/m on the right-hand side and [Ee] = J = NĀ·mĀ on the left-hand side. How can we reconcile both? The coulomb is an SI base unitĀ , so we can’t ‘translate’ it into something with N and m. [To be fully correct, for some reason, theĀ ampĆØreĀ (i.e. coulombĀ per second) was chosen as an SI base unit, but they’re interchangeable in regard to their place in the international system of units: they can’t be reduced.] So we’ve got a problem. Yes. That’s where we sort of ‘smuggled’ the 4ĻĪµ0 factor in when doing our calculations above. ThatĀ Īµ0 constant is, obviously, not ‘as fundamental’ as cĀ orĀ Ī± (just think of the cā2Ā =Ā Īµ0Ī¼0 relationship to understand what I mean here) but, still, it was necessary to make the dimensions come out alright: we need the reciprocalĀ dimensionĀ of Īµ0, i.e. (NĀ·m2)/C2, to make the dimensional analysis work.Ā We get: (C2/m)Ā·(NĀ·m2)/C2Ā = NĀ·m = J, i.e.Ā joule, so that’s the unit in which we measure energy or ā using the E = mc2Ā equivalenceĀ ā mass, which is the aspect of energy emphasizing itsĀ inertia.

So the answer is: no. Changing units won’t change alpha. So all that’s left is to play with it now. Let’s try to do that. Let me first plot that EeĀ = meĀ = Ī±e/re = 0.00729735256/re:

Unsurprisingly, we find theĀ pivotĀ point of this curve is at the intersection of the diagonal and the curve itself, so that’s at the (0.00729735256,Ā 0.00729735256) point, where slopes are Ā± 1, i.e. plus or minus unity.Ā What does this show? Nothing much. What?Ā I can hear you: I should be excited because… Well… Yes! Think of it. If youĀ would have to chose a cut-off point, you’d chose this one, wouldn’t you? š Sure, you’re right. How exciting! Let me show you. Look at it! It proves that God thinks in terms of logarithms. He has chosenĀ Ī± such that ln(E) = ln(Ī±/r) = lnĪ± ā lnrĀ = lnĪ± ā lnrĀ = 0, so ln Ī± = lnr and, therefore, Ī± = r. š

Huh? Excuse me?

I am sorry. […] Well… I am not, of course… š I just wanted to illustrate the kind of exercise some people are tempted to do. It’s no use. The fine-structure constant is what it is: it sort ofĀ summarizesĀ an awful lot of formulas. It basically shows what Maxwell’s equation imply in terms of theĀ structureĀ of an atom defined as a negativeĀ charge orbiting around some positive charge. It shows we can get calculate everything as a function of something else, and that’s what the fine-structure constant tells us: it relates everything to everything. However, when everything is said and done, the fine-structure constant shows us two things:

1. Maxwell’s equations are complete: we canĀ construct a complete model of the electron and the atom, which includes: the electron’s energy and mass, its velocity, its own radius, and the radius of the atom. [I might have forgotten one of the dimensions here, but you’ll add it. :-)]
2. God doesn’t want our equations to blow up. Our equations are all correct but, in reality, there’s a cut-off factor that ensures we don’t go to the limit with them.

So the fine-structure constant anchors our world, so to speak. In other words: of all the worlds that are possible, we live in this one.

[…] It’s pretty good as far as I am concerned. Isn’t it amazing that our mind is able to justĀ graspĀ things like that? I know my approach here is pretty intuitive, and with ‘intuitive’, I mean ‘not scientific’ here. š Frankly, I don’t like the talk about physicists “looking into God’s mind.” I don’t think that’s what they’re trying to do. I think they’re just trying to understand the fundamentalĀ unityĀ behind it all. And that’s religion enough for me. š

So… What’s the conclusion? Nothing much. We’ve sort of concluded our description of the classical world… Well… Of its ‘electromagnetic sector’ at least. š That sector can be summarized in Maxwell’s equations, which describe an infinite world of possible worlds. However, God fixed three constants:Ā h,Ā cĀ andĀ Ī±. So we live in a world that’s defined by this Trinity of fundamental physical constants. Why is it not two, or four?

My guts instinct tells me it’s because we live in three dimensions, and so there’s three degrees of freedom really. But what about time? Time is the fourth dimension, isn’t it? Yes. But time is symmetric in the ‘electromagnetic’ sector: we can reverse the arrow of time in our equations and everything still works. TheĀ arrow of timeĀ involves other theories: statistics (physicists refer to it as ‘statistical mechanics‘) and the ‘weak force’ sector, which I discussed when talking about symmetries in physics. So… Well… We’re not done. God gave us plenty of other stuff to try to understand. š