# The quantization of magnetic moments

Pre-script (dated 26 June 2020): This post got mutilated by the removal of some material by the dark force. You should be able to follow the main story line, however. If anything, the lack of illustrations might actually help you to think things through for yourself.

Original post:

You may not have many questions after a first read of Feynman’s Lecture on the Stern-Gerlach experiment and his more general musings on the quantization of the magnetic moment of an elementary particle. [At least I didn’t have all that many after my first reading, which I summarized in a previous post.]

However, a second, third or fourth reading should trigger some, I’d think. My key question is the following: what happens to that magnetic moment of a particle – and its spin  – as it travels through a homogeneous or inhomogeneous magnetic field? We know – or, to be precise, we assume – its spin is either “up” (Jz = +ħ/2) or “down” (Jz = −ħ/2) when it enters the Stern-Gerlach apparatus, but then – when it’s moving in the field itself – we would expect that the magnetic field would, somehow, line up the magnetic moment, right?

Feynman says that it doesn’t: from all of the schematic drawings – and the subsequent discussion of Stern-Gerlach filters – it is obvious that the magnetic field – which we denote as B, and which we assume to be inhomogeneous  – should not result in a change of the magnetic moment. Feynman states it as follows: “The magnetic field produces a torque. Such a torque you would think is trying to line up the (atomic) magnet with the field, but it only causes its precession.”

[…] OK. That’s too much information already, I guess. Let’s start with the basics. The key to a good understanding of this discussion is the force formula: We should first explain this formula before discussing the obvious question: over what time – or over what distance – should we expect this force to pull the particle up or down in the magnetic field? Indeed, if the force ends up aligning the moment, then the force will disappear!

So let’s first explain the formula. We start by explaining the energy U. U is the potential energy of our particle, which it gets from its magnetic moment μ and its orientation in the magnetic field B. To be precise, we can write the following: Of course, μ and B are the magnitudes of μ and B respectively, and θ is the angle between μ and B: if the angle θ is zero, then Umag will be negative. Hence, the total energy of our particle (U) will actually be less than what it would be without the magnetic field: it is the energy when the magnetic moment of our particle is fully lined up with the magnetic field. When the angle is a right angle (θ = ±π/2), then the energy doesn’t change (Umag = 0). Finally, when θ is equal to π or −π, then its energy will be more than what it would be outside of the magnetic field. [Note that the angle θ effectively varies between –π and π – not between 0 and 2π!] Of course, we may already note that, in quantum mechanics, Umag will only take on a very limited set of values. To be precise, for a particle with spin number j = 1/2, the possible values of Umag will be limited to two values only. We will come back to that in a moment. First that force formula.

Energy is force over a distance. To be precise, when a particle is moved from point a to point b, then its change in energy can be written as the following line integral: Note that the minus sign is there because of the convention that we’re doing work against the force when increasing the (potential) energy of that what we’re moving. Also note that F∙ds product is a vector (dot) product: it is, obviously, equal to Ft times ds, with Ft the magnitude of the tangential component of the force. The equation above gives us that force formula: Feynman calls it the principle of virtual work, which sounds a bit mysterious – but so you get it by taking the derivative of both sides of the energy formula.

Let me now get back to the real mystery of quantum mechanics, which tells us that the magnetic moment – as measured along our z-axis – will only take one of two possible values. To be precise, we have the following formula for μz: This is a formula you just have to accept for the moment. It needs a bit of interpretation, and you need to watch out for the sign. The g-factor is the so-called Landé g-factor: it is equal to 1 for a so-called pure orbital moment, 2 for a so-called pure spin moment, and some number in-between in reality, which is always some mixture of the two: both the electron’s orbit around the nucleus as well as the electron’s rotation about its own axis contribute to the total angular momentum and, hence, to the total magnetic moment of our electron. As for the other factors, m and qe are, of course, the mass and the charge of our electron, and Jz is either +ħ/2 or −ħ/2. Hence, if we know g, we can easily calculate the two possible values for μz.

Now, that also means we could – theoretically – calculate the two possible values of that angle θ. For some reason, no handbook in physics ever does that. The reason is probably a good one: electron orbits, and the concept of spin itself, are not like the orbit and the spin of some planet in a planetary system. In fact, we know that we should not think of electrons like that at all: quantum physicists tell us we may only think of it as some kind of weird cloud around a center. That cloud has a density which is to be calculated by taking the absolute square of the quantum-mechanical amplitude of our electron.

In fact, when thinking about the two possible values for θ, we may want to remind ourselves of another peculiar consequence of the fact that the angular momentum – and, hence, the magnetic moment – is not continuous but quantized: the magnitude of the angular momentum J is not  J = √(J·J) = √J2 in quantum mechanics but J = √(J·J) = √[j·(j+1)·ħ2] = √[j·(j+1)]·ħ. For our electron, j = 1/2 and, hence, the magnitude of J is equal to J = √[(1/2)∙(3/2)]∙ ħ = √(3/4)∙ħ ≈ 0.866∙ħ. Hence, the magnitude of the angular momentum is larger than the maximum value of Jz – and not just a little bit, because the maximum value of ħ is ħ/2! That leads to that weird conclusion: in quantum mechanics, we find that the angular momentum is never completely along any one direction ! In fact, this conclusion basically undercuts the very idea of the angular momentum – and, hence, the magnetic moment – of having any precise direction at all! [This may sound spectacular, but there is actually a classical equivalent to the idea of the angular momentum having no precisely defined direction: gyroscopes may not only precess, but nutate as well. Nutation refers to a kind of wobbling around the direction of the angular momentum. For more details, see the post I wrote after my first reading of Feynman’s Lecture on the quantization of magnetic moments. :-)]

Let’s move on. So if, in quantum mechanics, we cannot associate the magnetic moment – or the angular momentum – with some specific direction, then how should we imagine it? Well… I won’t dwell on that here, but you may want to have a look at another post of mine, where I develop a metaphor for the wavefunction which may help you to sort of understand what it might be. The metaphor may help you to think of some oscillation in two directions – rather than in one only – with the two directions separated by a right angle. Hence, the whole thing obviously points in some direction but it’s not very precise. In any case, I need to move on here.

We said that the magnetic moment will take one of two values only, in any direction along which we’d want to measure it. We also said that the (maximum) value along that direction – any direction, really – will be smaller than the magnitude of the moment. [To be precise, we said that for the angular momentum, but the formulas above make it clear the conclusions also hold for the magnetic moment.] So that means that the magnetic moment is, in fact, never fully aligned with the magnetic field. Now, if it is not aligned – and, importantly, if it also does not line up – then it should precess. Now, precession is a difficult enough concept in classical mechanics, so you may think it’s going to be totally abstruse in quantum mechanics. Well… That is true – to some extent. At the same time, it is surely not unintelligible. I will not repeat Feynman’s argument here, but he uses the classical formulas once more to calculate an angular velocity and a precession frequency – although he doesn’t explain what they might actually physically represent. Let me just jot down the formula for the precession frequency: We get the same factors: g, qe and m. In addition, you should also note that the precession frequency is directly proportional  to the strength of the magnetic field, which makes sense. Now, you may wonder: what is the relevance of this? Can we actually measure any of this?

We can. In fact, you may wonder about the if I inserted above: if we can measure the Landé g-factor… Can we? We can. It’s done in a resonance experiment, which is referred to as the Rabi molecular-beam method – but then it might also be just an atomic beam, of course!

The experiment is interesting, because it shows the precession is – somehow – real. It also illustrates some other principles we have been describing above.

The set-up looks pretty complicated. We have a series of three magnets. The first magnet is just a Stern-Gerlach apparatus: a magnet with a very sharp edge on one of the pole tips so as to produce an inhomogeneous magnetic field. Indeed, a homogeneous magnetic field implies that ∂B/∂z = 0 and, hence, the force along the z-direction would be zero and our atomic magnets would not be displaced.

The second magnet is more complicated. Its magnetic field is uniform, so there are no vertical forces on the atoms and they go straight through. However, the magnet includes an extra set of coils that can produce an alternating horizontal field as well. I’ll come back to that in a moment. Finally, the third magnet is just like the first one, but with the field inverted. Have a look at it: It may not look very obvious but, after some thinking, you’ll agree that the atoms can only arrive at the detector if they follow the trajectories a and/or b. In fact, these trajectories are the only possible ones because of the slits S1 and S2.

Now what’s the idea of that horizontal field B’ in magnet 2? In a classical situation, we could change the angular momentum – and the magnetic moment – by applying some torque about the z-axis. The idea is shown in Figure (a) and (b) below. Figure (a) shows – or tries to show – some rotating field B’ – one that is always at right angles to both the angular momentum as well as to the (uniform) B field. That would be effective. However, Figure (b) shows another arrangement that is almost equally effective: an oscillating field that sort of pulls and pushes at some frequency ω. Classically, such fields would effectively change the angle of our gyroscope with respect to the z-axis. Is it also the case quantum-mechanically?

It turns out it sort of works the same in quantum mechanics. There is a big difference though. Classically, μz would change gradually, but in quantum mechanics it cannot: in quantum mechanics, it must jump suddenly from one value to the other, i.e. from +ħ/2 to −ħ/2, or the other way around. In other words, it must flip up or down. Now, if an atom flips, then it will, of course, no longer follow the (a) or (b) trajectories: it will follow some other path, like a’ or b’, which make it crash into the magnet. Now, it turns out that almost all atoms will flip if we get that frequency ω right. The graph below shows this ‘resonance’ phenomenon: there is a sharp drop in the ’current’ of atoms if ω is close or equal to ωp. What’s ωp? It’s that precession frequency for which we gave you that formula above. To make a long story short, from the experiment, we can calculate the Landé g-factor for that particular beam of atoms – say, silver atoms . So… Well… Now we know it all, don’t we?

Maybe. As mentioned when I started this post, when going through all of this material, I always wonder why there is no magnetization effect: why would an atom remain in the same state when it crosses a magnetic field? When it’s already aligned with the magnetic field – to the maximum extent possible, that is – then it shouldn’t flip, but what if its magnetic moment is opposite? It should lower its energy by flipping, right? And it should flip just like that. Why would it need an oscillating B’ field?

In fact, Feynman does describe how the magnetization phenomenon can be analyzed – classically and quantum-mechanically, but he does that for bulk materials: solids, or liquids, or gases – anything that involves lots of atoms that are kicked around because of the thermal motions. So that involves statistical mechanics – which I am sure you’ve skipped so far. 🙂 It is a beautiful argument – which ends with an equally beautiful formula, which tells us the magnetization (M) of a material – which is defined as the net magnetic moment per unit volume – has the same direction as the magnetic field (B) and a magnitude M that is proportional the magnitude of B: The μ in this formula is the magnitude of the magnetic moment of the individual atoms and so… Well… It’s just like the formula for the electric polarization P, which we described in some other post. In fact, the formula for P and M are same-same but different, as they would say in Thailand. 🙂 But this wonderful story doesn’t answer our question. The magnetic moment of an individual particle should not stay what it is: if it doesn’t change because of all the kicking around as a result of thermal motions, then… Well… These little atomic magnets should line up. That means atoms with their spin “up” should go into the “spin-down” state.

I don’t have an answer to my own question as for now. I suspect it’s got to do with the strength of the magnetic field: a Stern-Gerlach apparatus involves a weak magnetic field. If it’s too strong, the atomic magnets must flip. Hence, a more advanced analysis should probably include that flipping effect. When quickly googling – just now – I found an MIT lab exercise on it, which also provides a historical account of the Stern-Gerlach experiment itself. I skimmed through it – and will read all of it in the coming days – but let me just quote this from the historical background section:

“Stern predicted that the effect would be be just barely observable. They had difficulty in raising support in the midst of the post war financial turmoil in Germany. The apparatus, which required extremely precise alignment and a high vacuum, kept breaking down. Finally, after a year of struggle, they obtained an exposure of sufficient length to give promise of an observable silver deposit. At first, when they examined the glass plate they saw nothing. Then, gradually, the deposit became visible, showing a beam separation of 0.2 millimeters! Apparently, Stern could only afford cheap cigars with a high sulfur content. As he breathed on the glass plate, sulfur fumes converted the invisible silver deposit into visible black silver sufide, and the splitting of the beam was discovered.”

Isn’t this funny? And great at the same time? 🙂 But… Well… The point is: the paper for that MIT lab exercise makes me realize Feynman does cut corners when explaining stuff – and some corners are more significant than others. I note, for example, that they talk about interference peaks rather than “two distinct spots on the glass plate.” Hence, the analysis is somewhat more sophisticated than Feynman pretends it to be. So, when everything is said and done, Feynman’s Lectures may indeed be reading for undergraduate students only. Is it time to move on?

 The magnetic moment – as measured in a particular coordinate system – is equal to μ = −g·[q/(2m)]·J. The factor J in this expression is the angular momentum, and the coordinate system is chosen such that its z-axis is along the direction of the magnetic field B. The component of J along the z-axis is written as Jz. This z-component of the angular momentum is what is, rather loosely, being referred to as the spin of the particle in this context. In most other contexts, spin refers to the spin number j which appears in the formula for the value of Jz, which is Jz = j∙ħ, (j−1)∙ħ, (j−2)∙ħ,…, (−j+2)∙ħ, (−j+1), −j∙ħ. Note the separation between the possible values of Jz is equal to ħ. Hence, j itself must be an integer (e.g. 1 or 2) or a half-integer (e.g. 1/2). We usually look at electrons, whose spin number j is 1/2.

 One of the pole tips of the magnet that is used in the Stern-Gerlach experiment has a sharp edge. Therefore, the magnetic field strength varies with z. We write: ∂B/∂z ≠ 0.

 The z-direction can be any direction, really.

 The original experiment was effectively done with a beam of silver atoms. The lab exercise which MIT uses to show the effect to physics students involves potassium atoms.

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# 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… 🙂

#### 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 θ, and (b) the angular velocity of the precession, which we’ll denote by ω= dφ/dt, as shown below. The J in the illustration below is the angular momentum of the object. Hence, if we’d imagine it to be an electron, then J would be the spin angular momentum only, not its orbital angular momentum—although the analysis would obviously be valid for the orbital and/or total angular momentum as well. OK. Let’s look at what’s going on. The angular displacement – which is also, rather confusingly, referred to as the angle of precession – in the time interval Δt is, obviously, equal to Δφ = ωp·Δt. Now, looking at the geometry of the situation, and using the small-angle approximation for the sine, one can also see that ΔJ ≈ (J·sinθ)·(ωp·Δt). In fact, going to the limit (i.e. for infinitesimally small Δφ and ΔJ), we can write:

dJ/dt = ωp·J·sinθ

But the angular momentum cannot change if there’s no torque. In fact, the time rate of change of the angular momentum is equal to the 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 torque τ on a loop of current with magnetic moment μ in an external magnetic field B  is equal to τ = μ×B. So the magnitude of the torque is equal to |τ| = |μ|·|B|·sinθ = μ·B·sinθ. Therefore, ωp·J·sinθ = μ·B·sinθ and, hence,

ω= μ·B/J

However, from the general μ/J = –g·(qe/2m) equation we derived in our previous post, we know that μ/J – for an atomic magnet, that is – must be equal to μ/J = g·qe/2m. So we get the formula we wanted to get here:

ω= g·(qe/2m)·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 g·(qe/2m). It’s good to do the math and actually calculate the precession frequency fp = ωp/2π. It’s easy. We had calculated qe/2m already: it was equal to 1.6×10−19 C divided by 2·9.1×10−31 kg, so that’s 0.0879×1012  C/kg or 0.0879×1012 (C·m)/(N·s2), more or less. 🙂 Now, g is dimensionless, and B is expressed in tesla: 1 T = (N·s)/(C·m), so we get the s−1 dimension we want for a frequency. For g = 2 (so we look at the spin of the electron itself only), we get:

fp = ωp/2π = 2·0.0879×1012/2π ≈ 28×109 = 28 gigacycles per tesla = 28 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 older gauss 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 the strength of the magnetic field in medical MRI systems (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 – so that’s the speed at which it executes instructions – is measured in GHz too, so perhaps it’s not so astronomic. I’ll let you judge. 🙂

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 orbits—and spins—in an atom should precess in a magnetic field. Is it also true quantum-mechanically? It is essentially true, but the meaning of the “precession” 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 analogy—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 frequency is actually reserved for the frequency I’ll describe in the next section. I should also note that the ω= g·(qe/2m)·B is usually written, quite simply, as ω= γ·B. Of course, the gamma is not the Lorentz factor here, but the so-called gyromagnetic ratio (aka as the magnetogyric ratio): γ = g·(qe/2m). Oh—just so you know: Sir Joseph Larmor was a British physicists and, yes, he developed all of the stuff we’re talking about here. 🙂

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. 🙂 And then you should also read Feynman’s exposé 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’s 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 not true. Here we get some help: Larmor proved a theorem that basically says that, if we can work out the motions of the electrons without the external field, the solution for the motions with the external field is the no-field solution 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:

ω= (qe/2m)·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. The ωL  frequency, which is the precession frequency for g = 1, is referred to as the 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 Lecture on it. 🙂

#### 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 about ferromagnetism here, i.e. the mechanism through which iron, nickel and cobalt and most of their alloys become permanent magnets. That’s quite peculiar and so we will not discuss it here. Here we’ll talk about the very weak quantum-mechanical magnetic effect – a thousand to a million times less than the effects in ferromagnetic materials – that occurs in all materials when placed in an external magnetic field.

While the effect is there in all 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 not 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 off—hopefully slowly: if not, we’d have eddy-current forces causing potentially strong impulses. But so we’ve got some change in the magnetic flux , and so we know, because of Faraday or Maxwell – you choose 🙂 – that we’ll have some circulation of E, i.e. the electric 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:

E·2π·r = −d(B·π·r2)/dt ⇔ E = −(r/2)·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 (−qe·E)·r = −qe·E·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 = −qe·E·r =  qe·(r/2)·(dB/dt)·r = (qe·r2/2)·(dB/dt)

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

ΔJ = (qe·r2/2)·B

You should, in fact, derive this more formally, by integrating—but let’s keep things as simple as we can. 🙂 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 just qe/2m times the angular momentum that’s already there. But more angular momentum means the magnetic moment has changed, according to the μ = (qe/2m)·J formula we derived in our previous post, so we have:

Δμ = –(qe/2m)·ΔJ

The minus sign is there because of Lenz’ law: the added momentum is opposite to the magnetic field—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 the induced magnetic moment is directly proportional to the magnetic field B, 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. 🙂

Let me conclude this post by copying Feynman’s little exposé 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 – so it’s held in place, so to say – 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 cannot 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 homogeneous 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. There’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! Feynman’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 we’re talking an induced magnetic moment that’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 Feynman’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. 🙂 I hope you enjoyed this post.

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