Maxwell, Lorentz, gauges and gauge transformations

Pre-script (dated 26 June 2020): This post got severely mutilated by the removal of material by the dark force. It may, therefore, be difficult to follow the main story-line.

Original post:

I’ve done quite a few posts already on electromagnetism. They were all focused on the math one needs to understand Maxwell’s equations. Maxwell’s equations are a set of (four) differential equations, so they relate some function with its derivatives. To be specific, they relate E and B, i.e. the electric and magnetic field vector respectively, with their derivatives in space and in time. [Let me be explicit here: E and B have three components, but depend on both space as well as time, so we have three dependent and four independent variables for each function: E = (E_x, E_y, E_z) = E(x, y, z, t) and B = (B_x, B_y, B_z) = B(x, y, z, t).] That’s simple enough to understand, but the dynamics involved are quite complicated, as illustrated below.

I now want to do a series on the more interesting stuff, including an exploration of the concept of gauge in field theory, and I also want to show how one can derive the wave equation for electromagnetic radiation from Maxwell’s equations. Before I start, let’s recall the basic concept of a field.

The reality of fields

I said a couple of time already that (electromagnetic) fields are real. They’re more than just a mathematical structure. Let me show you why. Remember the formula for the electrostatic potential caused by some charge q at the origin:

We know that the (negative) gradient of this function, at any point in space, gives us the electric field vector at that point: E = –∇Φ. [The minus sign is there because of convention: we take the reference point Φ = 0 at infinity.] Now, the electric field vector gives us the force on a unit charge (i.e. the charge of a proton) at that point. If q is some positive charge, the force will be repulsive, and the unit charge will accelerate away from our q charge at the origin. Hence, energy will be expended, as force over distance implies work is being done: as the charges separate, potential energy is converted into kinetic energy. Where does the energy come from? The energy conservation law tells us that it must come from somewhere.

It does: the energy comes from the field itself. Bringing in more or bigger charges (from infinity, or just from further away) requires more energy. So the new charges change the field and, therefore, its energy. How exactly? That’s given by Gauss’ Law: the total flux out of a closed surface is equal to:

You’ll say: flux and energy are two different things. Well… Yes and no. The energy in the field depends on E. Indeed, the formula for the energy density in space (i.e. the energy per unit volume) is

Getting the energy over a larger space is just another integral, with the energy density as the integral kernel:

Feynman’s illustration below is not very sophisticated but, as usual, enlightening. 🙂

Gauss’ Theorem connects both the math as well as the physics of the situation and, as such, underscores the reality of fields: the energy is not in the electric charges. The energy is in the fields they produce. Everything else is just the principle of superposition of fields – i.e. E = E₁+ E₂– coming into play. I’ll explain Gauss’ Theorem in a moment. Let me first make some additional remarks.

First, the formulas are valid for electrostatics only (so E and B only vary in space, not in time), so they’re just a piece of the larger puzzle. 🙂 As for now, however, note that, if a field is real (or, to be precise, if its energy is real), then the flux is equally real.

Second, let me say something about the units. Field strength (E or, in this case, its normal component E_n = E·n) is measured in newton (N) per coulomb (C), so in N/C. The integral above implies that flux is measured in (N/C)·m². It’s a weird unit because one associates flux with flow and, therefore, one would expect flux is some quantity per unit time and per unit area, so we’d have the m² unit (and the second) in the denominator, not in the numerator. But so that’s true for heat transfer, for mass transfer, for fluid dynamics (e.g. the amount of water flowing through some cross-section) and many other physical phenomena. But for electric flux, it’s different. You can do a dimensional analysis of the expression above: the sum of the charges is expressed in coulomb (C), and the electric constant (i.e. the vacuum permittivity) is expressed in C²/(N·m²), so, yes, it works: C/[C²/(N·m²)] = (N/C)·m². To make sense of the units, you should think of the flux as the total flow, and of the field strength as a surface density, so that’s the flux divided by the total area, so (field strength) = (flux)/(area). Conversely, (flux) = (field strength)×(area). Hence, the unit of flux is [flux] = [field strength]×[area] = (N/C)·m².

OK. Now we’re ready for Gauss’ Theorem. 🙂 I’ll also say something about its corollary, Stokes’ Theorem. It’s a bit of a mathematical digression but necessary, I think, for a better understanding of all those operators we’re going to use.

Gauss’ Theorem

The concept of flux is related to the divergence of a vector field through Gauss’ Theorem. Gauss’s Theorem has nothing to do with Gauss’ Law, except that both are associated with the same genius. Gauss’ Theorem is:

The ∇·C in the integral on the right-hand side is the divergence of a vector field. It’s the volume density of the outward flux of a vector field from an infinitesimal volume around a given point.

Huh? What’s a volume density? Good question. Just substitute C for E in the surface and volume integral above (the integral on the left is a surface integral, and the one on the right is a volume integral), and think about the meaning of what’s written. To help you, let me also include the concept of linear density, so we have (1) linear, (2) surface and (3) volume density. Look at that representation of a vector field once again: we said the density of lines represented the magnitude of E. But what density? The representation hereunder is flat, so we can think of a linear density indeed, measured along the blue line: so the flux would be six (that’s the number of lines), and the linear density (i.e. the field strength) is six divided by the length of the blue line.

However, we defined field strength as a surface density above, so that’s the flux (i.e. the number of field lines) divided by the surface area (i.e. the area of a cross-section): think of the square of the blue line, and field lines going through that square. That’s simple enough. But what’s volume density? How do we count the number of lines inside of a box? The answer is: mathematicians actually define it for an infinitesimally small cube by adding the fluxes out of the six individual faces of an infinitesimally small cube:

So, the truth is: volume density is actually defined as a surface density, but for an infinitesimally small volume element. That, in turn, gives us the meaning of the divergence of a vector field. Indeed, the sum of the derivatives above is just ∇·C (i.e. the divergence of C), and ΔxΔyΔz is the volume of our infinitesimal cube, so the divergence of some field vector C at some point P is the flux – i.e. the outgoing ‘flow’ of C – per unit volume, in the neighborhood of P, as evidenced by writing

Indeed, just bring ΔV to the other side of the equation to check the ‘per unit volume’ aspect of what I wrote above. The whole idea is to determine whether the small volume is like a sink or like a source, and to what extent. Think of the field near a point charge, as illustrated below. Look at the black lines: they are the field lines (the dashed lines are equipotential lines) and note how the positive charge is a source of flux, obviously, while the negative charge is a sink.

Now, the next step is to acknowledge that the total flux from a volume is the sum of the fluxes out of each part. Indeed, the flux through the part of the surfaces common to two parts will cancel each other out. Feynman illustrates that with a rough drawing (below) and I’ll refer you to his Lecture on it for more detail.

So… Combining all of the gymnastics above – and integrating the divergence over an entire volume, indeed – we get Gauss’ Theorem:

Stokes’ Theorem

There is a similar theorem involving the circulation of a vector, rather than its flux. It’s referred to as Stokes’ Theorem. Let me jot it down:

We have a contour integral here (left) and a surface integral (right). The reasoning behind is quite similar: a surface bounded by some loop Γ is divided into infinitesimally small squares, and the circulation around Γ is the sum of the circulations around the little loops. We should take care though: the surface integral takes the normal component of ∇×C, so that’s (∇×C)_n= (∇×C)·n. The illustrations below should help you to understand what’s going on.

The electric versus the magnetic force

There’s more than just the electric force: we also have the magnetic force. The so-called Lorentz force is the combination of both. The formula, for some charge q in an electromagnetic field, is equal to:

Hence, if the velocity vector v is not equal to zero, we need to look at the magnetic field vector B too! The simplest situation is magnetostatics, so let’s first have a look at that.

Magnetostatics imply that that the flux of E doesn’t change, so Maxwell’s third equation reduces to c²∇×B = j/ε₀. So we just have a steady electric current (j): no accelerating charges. Maxwell’s fourth equation, ∇•B = 0, remains what is was: there’s no such thing as a magnetic charge. The Lorentz force also remains what it is, of course: F = q(E+v×B) = qE +qv×B. Also note that the v, j and the lack of a magnetic charge all point to the same: magnetism is just a relativistic effect of electricity.

What about units? Well… While the unit of E, i.e. the electric field strength, is pretty obvious from the F = qE term – hence, E = F/q, and so the unit of E must be [force]/[charge] = N/C – the unit of the magnetic field strength is more complicated. Indeed, the F = qv×B identity tells us it must be (N·s)/(m·C), because 1 N = 1C·(m/s)·(N·s)/(m·C). Phew! That’s as horrendous as it looks, and that’s why it’s usually expressed using its shorthand, i.e. the tesla: 1 T = 1 (N·s)/(m·C). Magnetic flux is the same concept as electric flux, so it’s (field strength)×(area). However, now we’re talking magnetic field strength, so its unit is T·m²= (N·s·m²)/(m·C) = (N·s·m)/C, which is referred to as the weber (Wb). Remembering that 1 volt = 1 N·m/C, it’s easy to see that a weber is also equal to 1 Wb = 1 V·s. In any case, it’s a unit that is not so easy to interpret.

Magnetostatics is a bit of a weird situation. It assumes steady fields, so the ∂E/∂t and ∂B/∂t terms in Maxwell’s equations can be dropped. In fact, c²∇×B = j/ε₀ implies that ∇·(c²∇×B ) = ∇·(j/ε₀) and, therefore, that ∇·j = 0. Now, ∇·j = –∂ρ/∂t and, therefore, magnetostatics is a situation which assumes ∂ρ/∂t = 0. So we have electric currents but no change in charge densities. To put it simply, we’re not looking at a condenser that is charging or discharging, although that condenser may act like the battery or generator that keeps the charges flowing! But let’s go along with the magnetostatics assumption. What can we say about it? Well… First, we have the equivalent of Gauss’ Law, i.e. Ampère’s Law:

We have a line integral here around a closed curve, instead of a surface integral over a closed surface (Gauss’ Law), but it’s pretty similar: instead of the sum of the charges inside the volume, we have the current through the loop, and then an extra c² factor in the denominator, of course. Combined with the ∇•B = 0 equation, this equation allows us to solve practical problems. But I am not interested in practical problems. What’s the theory behind?

The magnetic vector potential

The ∇•B = 0 equation is true, always, unlike the ∇×E = 0 expression, which is true for electrostatics only (no moving charges). It says the divergence of B is zero, always, and, hence, it means we can represent B as the curl of another vector field, always. That vector field is referred to as the magnetic vector potential, and we write:

∇·B = ∇·(∇×A) = 0 and, hence, B = ∇×A

In electrostatics, we had the other theorem: if the curl of a vector field is zero (everywhere), then the vector field can be represented as the gradient of some scalar function, so if ∇×C = 0, then there is some Ψ for which C = ∇Ψ. Substituting C for E, and taking into account our conventions on charge and the direction of flow, we get E = –∇Φ. Substituting E in Maxwell’s first equation (∇•E = ρ/ε₀) then gave us the so-called Poisson equation: ∇²Φ = ρ/ε₀, which sums up the whole subject of electrostatics really! It’s all in there!

Except magnetostatics, of course. Using the (magnetic) vector potential A, all of magnetostatics is reduced to another expression:

∇²A= −j/ε₀, with ∇·A = 0

Note the qualifier: ∇·A = 0. Why should the divergence of A be equal to zero? You’re right. It doesn’t have to be that way. We know that ∇·(∇×C) = 0, for any vector field C, and always (it’s a mathematical identity, in fact, so it’s got nothing to do with physics), but choosing A such that ∇·A = 0 is just a choice. In fact, as I’ll explain in a moment, it’s referred to as choosing a gauge. The ∇·A = 0 choice is a very convenient choice, however, as it simplifies our equations. Indeed, c²∇×B = j/ε₀ = c²∇×(∇×A), and – from our vector calculus classes – we know that ∇×(∇×C) = ∇(∇·C) – ∇²C. Combining that with our choice of A (which is such that ∇·A = 0, indeed), we get the ∇²A= −j/ε₀expression indeed, which sums up the whole subject of magnetostatics!

The point is: if the time derivatives in Maxwell’s equations, i.e. ∂E/∂t and ∂B/∂t, are zero, then Maxwell’s four equations can be nicely separated into two pairs: the electric and magnetic field are not interconnected. Hence, as long as charges and currents are static, electricity and magnetism appear as distinct phenomena, and the interdependence of E and B does not appear. So we re-write Maxwell’s set of four equations as:

Electrostatics: ∇•E = ρ/ε₀ and ∇×E = 0
Magnetostatics: ∇×B = j/c²ε₀ and ∇•B = 0

Note that electrostatics is a neat example of a vector field with zero curl and a given divergence (ρ/ε₀), while magnetostatics is a neat example of a vector field with zero divergence and a given curl (j/c²ε₀).

Electrodynamics

But reality is usually not so simple. With time-varying fields, Maxwell’s equations are what they are, and so there is interdependence, as illustrated in the introduction of this post. Note, however, that the magnetic field remains divergence-free in dynamics too! That’s because there is no such thing as a magnetic charge: we only have electric charges. So ∇·B = 0 and we can define a magnetic vector potential A and re-write B as B = ∇×A, indeed.

I am writing a vector potential field because, as I mentioned a couple of times already, we can choose A. Indeed, as long as ∇·A = 0, it’s fine, so we can add curl-free components to the magnetic potential: it won’t make a difference. This condition is referred to as gauge invariance. I’ll come back to that, and also show why this is what it is.

While we can easily get B from A because of the B = ∇×A, getting E from some potential is a different matter altogether. It turns out we can get E using the following expression, which involves both Φ (i.e. the electric or electrostatic potential) as well as A (i.e. the magnetic vector potential):

E = –∇Φ – ∂A/∂t

Likewise, one can show that Maxwell’s equations can be re-written in terms of Φ and A, rather than in terms of E and B. The expression looks rather formidable, but don’t panic:

Just look at it. We have two ‘variables’ here (Φ and A) and two equations, so the system is fully defined. [Of course, the second equation is three equations really: one for each component x, y and z.] What’s the point? Why would we want to re-write Maxwell’s equations? The first equation makes it clear that the scalar potential (i.e. the electric potential) is a time-varying quantity, so things are not, somehow, simpler. The answer is twofold. First, re-writing Maxwell’s equations in terms of the scalar and vector potential makes sense because we have (fairly) easy expressions for their value in time and in space as a function of the charges and currents. For statics, these expressions are:

So it is, effectively, easier to first calculate the scalar and vector potential, and then get E and B from them. For dynamics, the expressions are similar:

Indeed, they are like the integrals for statics, but with “a small and physically appealing modification”, as Feynman notes: when doing the integrals, we must use the so-called retarded time $t' = t - r 12 /ct’$ . The illustration below shows how it works: the influences propagate from point (2) to point (1) at the speed c, so we must use the values of ρ and j at the time $t' = t - r 12 /ct’$ indeed!

The second aspect of the answer to the question of why we’d be interested in Φ and A has to do with the topic I wanted to write about here: the concept of a gauge and a gauge transformation.

Gauges and gauge transformations in electromagnetics

Let’s see what we’re doing really. We calculate some A and then solve for B by writing: B = ∇×A. Now, I say some A because any A‘ = A + ∇Ψ, with Ψ any scalar field really. Why? Because the curl of the gradient of Ψ – i.e. curl(gradΨ) = ∇×(∇Ψ) – is equal to 0. Hence, ∇×(A + ∇Ψ) = ∇×A + ∇×∇Ψ = ∇×A.

So we have B, and now we need E. So the next step is to take Faraday’s Law, which is Maxwell’s second equation: ∇×E = –∂B/∂t. Why this one? It’s a simple one, as it does not involve currents or charges. So we combine this equation and our B = ∇×A expression and write:

∇×E = –∂(∇×A)/∂t

Now, these operators are tricky but you can verify this can be re-written as:

∇×(E + ∂A/∂t) = 0

Looking carefully, we see this expression says that E + ∂A/∂t is some vector whose curl is equal to zero. Hence, this vector must be the gradient of something. When doing electrostatics, When we worked on electrostatics, we only had E, not the ∂A/∂t bit, and we said that E tout court was the gradient of something, so we wrote $E = - \nabla Φ. We now do the same thing for E + \partial A /\partialt, so we write:$

E + ∂A/∂t = −∇Φ

So we use the same symbol Φ but it’s a bit of a different animal, obviously. However, it’s easy to see that, if the ∂A/∂t would disappear (as it does in electrostatics, where nothing changes with time), we’d get our ‘old’ −∇Φ. Now, E + ∂A/∂t = −∇Φ can be written as:

E = −∇Φ – ∂A/∂t

So, what’s the big deal? We wrote B and E as a function of Φ and A. Well, we said we could replace A by any A‘ = A + ∇Ψ but, obviously, such substitution would not yield the same E. To get the same E, we need some substitution rule for Φ as well. Now, you can verify we will get the same E if we’d substitute Φ for Φ’ = Φ – ∂Ψ/∂t. You should check it by writing it all out:

E = −∇Φ’–∂A’/∂t = −∇(Φ–∂Ψ/∂t)–∂(A+∇Ψ)/∂t

= −∇Φ+∇(∂Ψ/∂t)–∂A/∂t–∂(∇Ψ)/∂t = −∇Φ – ∂A/∂t = E

Again, the operators are a bit tricky, but the +∇(∂Ψ/∂t) and –∂(∇Ψ)/∂t terms do cancel out. Where are we heading to? When everything is said and done, we do need to relate it all to the currents and the charges, because that’s the real stuff out there. So let’s take Maxwell’s ∇•E = ρ/ε₀ equation, which has the charges in it, and let’s substitute E for E = −∇Φ – ∂A/∂t. We get:

That equation can be re-written as:

So we have one equation here relating Φ and A to the sources. We need another one, and we also need to separate Φ and A somehow. How do we do that?

Maxwell’s fourth equation, i.e. c²∇×B = j/ε₀+ ∂E/∂t can, obviously, be written as c²∇×B − ∂E/∂t = j/ε₀. Substituting both E and B yields the following monstrosity:

We can now apply the general ∇×(∇×C) = ∇(∇·C) – ∇²C identity to the first term to get:

It’s equally monstrous, obviously, but we can simplify the whole thing by choosing Φ and A in a clever way. For the magnetostatic case, we chose A such that ∇·A = 0. We could have chosen something else. Indeed, it’s not because B is divergence-free, that A has to be divergence-free too! For example, I’ll leave it to you to show that choosing ∇·A such that

also respects the general condition that any A and Φ we choose must respect the A‘ = A + ∇Ψ and Φ’ = Φ – ∂Ψ/∂t equalities. Now, if we choose ∇·A such that ∇·A = −c^–2·∂Φ/∂t indeed, then the two middle terms in our monstrosity cancel out, and we’re left with a much simpler equation for A:

In addition, doing the substitution in our other equation relating Φ and A to the sources yields an equation for Φ that has the same form:

What’s the big deal here? Well… Let’s write it all out. The equation above becomes:

That’s a wave equation in three dimensions. In case you wonder, just check one of my posts on wave equations. The one-dimensional equivalent for a wave propagating in the x direction at speed c (like a sound wave, for example) is ∂²Φ/∂x²= c^–2·∂²Φ/∂t², indeed. The equation for A yields above yields similar wave functions for A‘s components A_x, A_y, and A_z.

So, yes, it is a big deal. We’ve written Maxwell’s equations in terms of the scalar (Φ) and vector (A) potential and in a form that makes immediately apparent that we’re talking electromagnetic waves moving out at the speed c. Let me copy them again:

You may, of course, say that you’d rather have a wave equation for E and B, rather than for A and Φ. Well… That can be done. Feynman gives us two derivations that do so. The first derivation is relatively simple and assumes the source our electromagnetic wave moves in one direction only. The second derivation is much more complicated and gives an equation for E that, if you’ve read the first volume of Feynman’s Lectures, you’ll surely remember:

The links are there, and so I’ll let you have fun with those Lectures yourself. I am finished here, indeed, in terms of what I wanted to do in this post, and that is to say a few words about gauges in field theory. It’s nothing much, really, and so we’ll surely have to discuss the topic again, but at least you now know what a gauge actually is in classical electromagnetic theory. Let’s quickly go over the concepts:

Choosing the ∇·A is choosing a gauge, or a gauge potential (because we’re talking scalar and vector potential here). The particular choice is also referred to as gauge fixing.
Changing A by adding ∇ψ is called a gauge transformation, and the scalar function Ψ is referred to as a gauge function. The fact that we can add curl-free components to the magnetic potential without them making any difference is referred to as gauge invariance.
Finally, the ∇·A = −c^–2·∂Φ/∂t gauge is referred to as a Lorentz gauge.

Just to make sure you understand: why is that Lorentz gauge so special? Well… Look at the whole argument once more: isn’t it amazing we get such beautiful (wave) equations if we stick it in? Also look at the functional shape of the gauge itself: it looks like a wave equation itself! […] Well… No… It doesn’t. I am a bit too enthusiastic here. We do have the same 1/c² and a time derivative, but it’s not a wave equation. 🙂 In any case, it all confirms, once again, that physics is all about beautiful mathematical structures. But, again, it’s not math only. There’s something real out there. In this case, that ‘something’ is a traveling electromagnetic field. 🙂

But why do we call it a gauge? That should be equally obvious. It’s really like choosing a gauge in another context, such as measuring the pressure of a tyre, as shown below. 🙂

Gauges and group theory

You’ll usually see gauges mentioned with some reference to group theory. For example, you will see or hear phrases like: “The existence of arbitrary numbers of gauge functions ψ(r, t) corresponds to the U(1) gauge freedom of the electromagnetic theory.” The U(1) notation stands for a unitary group of degree n = 1. It is also known as the circle group. Let me copy the introduction to the unitary group from the Wikipedia article on it:

In mathematics, the unitary group of degree n, denoted U(n), is the group of n × n unitary matrices, with the group operation that of matrix multiplication. The unitary group is a subgroup of the general linear group GL(n, C). In the simple case n = 1, the group U(1) corresponds to the circle group, consisting of all complex numbers with absolute value 1 under multiplication. All the unitary groups contain copies of this group.

The unitary group U(n) is a real Lie group of of dimension n². The Lie algebra of U(n) consists of n × n skew-Hermitian matrices, with the Lie bracket given by the commutator. The general unitary group (also called the group of unitary similitudes) consists of all matrices A such that A*A is a nonzero multiple of the identity matrix, and is just the product of the unitary group with the group of all positive multiples of the identity matrix.

Phew! Does this make you any wiser? If anything, it makes me realize I’ve still got a long way to go. 🙂 The Wikipedia article on gauge fixing notes something that’s more interesting (if only because I more or less understand what it says):

Although classical electromagnetism is now often spoken of as a gauge theory, it was not originally conceived in these terms. The motion of a classical point charge is affected only by the electric and magnetic field strengths at that point, and the potentials can be treated as a mere mathematical device for simplifying some proofs and calculations. Not until the advent of quantum field theory could it be said that the potentials themselves are part of the physical configuration of a system. The earliest consequence to be accurately predicted and experimentally verified was the Aharonov–Bohm effect, which has no classical counterpart.

This confirms, once again, that the fields are real. In fact, what this says is that the potentials are real: they have a meaningful physical interpretation. I’ll leave it to you to expore that Aharanov-Bohm effect. In the meanwhile, I’ll study what Feynman writes on potentials and all that as used in quantum physics. It will probably take a while before I’ll get into group theory though.

Indeed, it’s probably best to study physics at a somewhat less abstract level first, before getting into the more sophisticated stuff.

The blackbody radiation problem revisited: quantum statistics

Pre-script (dated 26 June 2020): This post got mutilated by the removal of some illustrations 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:

The equipartition theorem – which states that the energy levels of the modes of any (linear) system, in classical as well as in quantum physics, are always equally spaced – is deep and fundamental in physics. In my previous post, I presented this theorem in a very general and non-technical way: I did not use any exponentials, complex numbers or integrals. Just simple arithmetic. Let’s go a little bit beyond now, and use it to analyze that blackbody radiation problem which bothered 19th century physicists, and which led Planck to ‘discover’ quantum physics. [Note that, once again, I won’t use any complex numbers or integrals in this post, so my kids should actually be able to read through it.]

Before we start, let’s quickly introduce the model again. What are we talking about? What’s the black box? The idea is that we add heat to atoms (or molecules) in a gas. The heat results in the atoms acquiring kinetic energy, and the kinetic theory of gases tells us that the mean value of the kinetic energy for each independent direction of motion will be equal to kT/2. The blackbody radiation model analyzes the atoms (or molecules) in a gas as atomic oscillators. Oscillators have both kinetic as well as potential energy and, on average, the kinetic and potential energy is the same. Hence, the energy in the oscillation is twice the kinetic energy, so its average energy is 〈E〉 = 2·kT/2 = kT. However, oscillating atoms implies oscillating electric charges. Now, electric charges going up and down radiate light and, hence, as light is emitted, energy flows away.

How exactly? It doesn’t matter. It is worth noting that 19th century physicists had no idea about the inner structure of an atom. In fact, at that time, the term electron had not yet been invented: the first atomic model involving electrons was the so-called plum pudding model, which J.J. Thompson advanced in 1904, and he called electrons “negative corpuscles“. And the Rutherford-Bohr model, which is the first model one can actually use to explain how and why excited atoms radiate light, came in 1913 only, so that’s long after Planck’s solution for the blackbody radiation problem, which he presented to the scientific community in December 1900. It’s really true: it doesn’t matter. We don’t need to know about the specifics. The general idea is all that matters. As Feynman puts it: it’s how “A hot stove cools on a cold night, by radiating the light into the sky, because the atoms are jiggling their charge and they continually radiate, and slowly, because of this radiation, the jiggling motion slows down.” 🙂

His subsequent description of the black box is equally simple: “If we enclose the whole thing in a box so that the light does not go away to infinity, then we can eventually get thermal equilibrium. We may either put the gas in a box where we can say that there are other radiators in the box walls sending light back or, to take a nicer example, we may suppose the box has mirror walls. It is easier to think about that case. Thus we assume that all the radiation that goes out from the oscillator keeps running around in the box. Then, of course, it is true that the oscillator starts to radiate, but pretty soon it can maintain its kT of energy in spite of the fact that it is radiating, because it is being illuminated, we may say, by its own light reflected from the walls of the box. That is, after a while there is a great deal of light rushing around in the box, and although the oscillator is radiating some, the light comes back and returns some of the energy that was radiated.”

So… That’s the model. Don’t you just love the simplicity of the narrative here? 🙂 Feynman then derives Rayleigh’s Law, which gives us the frequency spectrum of blackbody radiation as predicted by classical theory, i.e. the intensity (I) of the light as a function of (a) its (angular) frequency (ω) and (b) the average energy of the oscillators, which is nothing but the temperature of the gas (Boltzmann’s constant k is just what it is: a proportionality constant which makes the units come out alright). The other stuff in the formula, given hereunder, are just more constants (and, yes, the c is the speed of light!). The grand result is:

The formula looks formidable but the function is actually very simple: it’s quadratic in ω and linear in 〈E〉 = kT. The rest is just a bunch of constants which ensure all of the units we use to measures stuff come out alright. As you may suspect, the derivation of the formula is not so simple as the narrative of the black box model, and so I won’t copy it here (you can check yourself). Indeed, let’s focus on the results, not on the technicalities. Let’s have a look at the graph.

The I(ω) graphs for T = T₀ and T = 2T₀ are given by the solid black curves. They tell us how much light we should have at different frequencies. They just go up and up and up, so Rayleigh’s Law implies that, when we open our stove – and, yes, I know, some kids don’t know what a stove is – and take a look, we should burn our eyes from x-rays. We know that’s not the case, in reality, so our theory must be wrong. An even bigger problem is that the curve implies that the total energy in the box, i.e. the total of all this intensity summed up over all frequencies, is infinite: we’ve got an infinite curve here indeed, and so an infinite area under it. Therefore, as Feynman puts it: “Rayleigh’s Law is fundamentally, powerfully, and absolutely wrong.” The actual graphs, indeed, are the dashed curves. I’ll come back to them.

The blackbody radiation problem is history, of course. So it’s no longer a problem. Let’s see how the equipartition theorem solved it. We assume our oscillators can only take on equally spaced energy levels, with the space between them equal to h·f = ħ·ω. The frequency f (or ω = 2π·f) is the fundamental frequency of our oscillator, and you know h and ħ = h/2π, course: Planck’s constant. Hence, the various energy levels are given by the following formula: E_n = n·ħ·ω = n·h·f. The first five are depicted below.

Next to the energy levels, we write the probability of an oscillator occupying that energy level, which is given by Boltzmann’s Law. I wrote about Boltzmann’s Law in another post too, so I won’t repeat myself here, except for noting that Boltzmann’s Law says that the probabilities of different conditions of energy are given by e^−energy/kT = 1/e^energy/kT. Different ‘conditions of energy’ can be anything: density, molecular speeds, momenta, whatever. Here we have a probability P_n as a function of the energy E_n = n·ħ·ω, so we write: P_n = A·e^−energy/kT= A·e^{−n·ħ·ω/kT}. [Note that P₀ is equal to A, as a consequence.]

Now, we need to determine how many oscillators we have in each of the various energy states, so that’s N₀, N₁, N₂, etcetera. We’ve done that before: N₁/N₀ = P₁/P₀ = (A·e^−2ħω/kT)/(A·e^−ħω/kT) = e^−ħω/kT. Hence, N₁ = N₀·e^−ħω/kT. Likewise, it’s not difficult to see that, N₂ = N₀·e^−2ħω/kTor, more in general, that N_n = N₀·e^−nħω/kT = N₀·[e^−ħω/kT]ⁿ. To make the calculations somewhat easier, Feynman temporarily substitutes e^−ħω/kT for x. Hence, we write: N₁ = N₀·x, N₂ = N₀·x²,…, N_n = N₀·xⁿ, and the total number of oscillators is obviously N_tot = N₀+N₁+…+N_n+… = N₀·(1+x+x²+…+xⁿ+…).

What about their energy? The energy of all oscillators in state 0 is, obviously, zero. The energy of all oscillators in state 1 is N₁·ħω = ħω·N₀·x. Adding it all up for state 2 yields N₂·2·ħω = 2·ħω·N₀·x². More generally, the energy of all oscillators in state n is equal to N_n·n·ħω = n·ħω·N₀·xⁿ. So now we can write the total energy of the whole system as E_tot = E₀+E₁+…+E_n+… = 0+ħω·N₀·x+2·ħω·N₀·x²+…+n·ħω·N₀·xⁿ+… = ħω·N₀·(x+2x²+…+nxⁿ+…). The average energy of one oscillator, for the whole system, is therefore:

Now, Feynman leaves the exercise of simplifying that expression to the reader and just says it’s equal to:

I should try to figure out how he does that. It’s something like Horner’s rule but that’s not easy with infinite polynomials. Or perhaps it’s just some clever way of factoring both polynomials. I didn’t break my head over it but just checked if the result is correct. [I don’t think Feynman would dare to joke here, but one could never be sure with him it seems. :-)] Note he substituted e^−ħω/kT for x, not e^+ħω/kT, so there is a minus sign there, which we don’t have in the formula above. Hence, the denominator, e^ħω/kT–1 = (1/x)–1 = (1–x)/x, and 1/(e^ħω/kT–1) = x/(1–x). Now, if (x+2x²+…+nxⁿ+…)/(1+x+x²+…+xⁿ+…) = x/(1–x), then (x+2x²+…+nxⁿ+…)·(1–x) must be equal to x·(1+x+x²+…+xⁿ+…). Just write it out: (x+2x²+…+nxⁿ+…)·(1–x) = x+2x²+…+nxⁿ+….−x²−2x³−…−nxⁿ⁺¹+… = x+x²+…+xⁿ+… Likewise, we get x·(1+x+x²+…+xⁿ+…) = x+x²+…+xⁿ+… So, yes, done.

Now comes the Big Trick, the rabbit out of the hat, so to speak. 🙂 We’re going to substitute the classical expression for 〈E〉 (i.e. kT) in Rayleigh’s Law for it’s quantum-mechanical equivalent (i.e. 〈E〉 = ħω/[e^ħω/kT–1].

What’s the logic behind? Rayleigh’s Law gave the intensity for the various frequencies that are present as a function of (a) the frequency (of course!) and (b) the average energy of the oscillators, which is kT according to classical theory. Now, our assumption that an oscillator cannot take on just any energy value but that the energy levels are equally spaced, combined with Boltzmann’s Law, gives us a very different formula for the average energy: it’s a function of the temperature, but it’s a function of the fundamental frequency too! I copied the graph below from the Wikipedia article on the equipartition theorem. The black line is the classical value for the average energy as a function of the thermal energy. As you can see, it’s one and the same thing, really (look at the scales: they happen to be both logarithmic but that’s just to make them more ‘readable’). Its quantum-mechanical equivalent is the red curve. At higher temperatures, the two agree nearly perfectly, but at low temperatures (with low being defined as the range where kT << ħ·ω, written as h·ν in the graph), the quantum mechanical value decreases much more rapidly. [Note the energy is measured in units equivalent to h·ν: that’s a nice way to sort of ‘normalize’ things so as to compare them.]

So, without further ado, let’s take Rayleigh’s Law again and just substitute kT (i.e. the classical formula for the average energy) for the ‘quantum-mechanical’ formula for 〈E〉, i.e. ħω/[e^ħω/kT–1]. Adding the dω factor to emphasize we’re talking some continuous distribution here, we get the even grander result (Feynman calls it the first quantum-mechanical formula ever known or discussed):

So this function is the dashed I(ω) curve (I copied the graph below again): this curve does not ‘blow up’. The math behind the curve is the following: even for large ω, leading that ω³ factor in the numerator to ‘blow up’, we also have Euler’s number being raised to a tremendous power in the denominator. Therefore, the curves come down again, and so we don’t get those incredible amounts of UV light and x-rays.

So… That’s how Max Planck solved the problem and how he became the ‘reluctant father of quantum mechanics.’ The formula is not as simple as Rayleigh’s Law (we have a cubic function in the numerator, and an exponential in the denominator), but its advantage is that it’s correct. Indeed, when everything is said and done, indeed, we do want our formulas to describe something real, don’t we? 🙂

Let me conclude by looking at that ‘quantum-mechanical’ formula for the average energy once more:

〈E〉 = ħω/[e^ħω/kT–1]

It’s not a distribution function (the formula for I(ω) is the distribution function), but the –1 term in the denominator does tell us already we’re talking Bose-Einstein statistics. In my post on quantum statistics, I compared the three distribution functions. Let ‘s quickly look at them again:

Maxwell-Boltzmann (for classical particles): f(E) = 1/[A·e^E/kT]
Fermi-Dirac (for fermions): f(E) = 1/[Ae^E/kT+ 1]
Bose-Einstein (for bosons): f(E) = 1/[Ae^E/kT− 1]

So here we simply substitute ħω for E, which makes sense, as the Planck-Einstein relation tells us that the energy of the particles involved is, indeed, equal to E = ħω . Below, you’ll find the graph of these three functions, first as a function of E, so that’s f(E), and then as a function of T, so that’s f(T) (or f(kT) if you want).

The first graph, for which E is the variable, is the more usual one. As for the interpretation, you can see what’s going on: bosonic particles (or bosons, I should say) will crowd the lower energy levels (the associated probabilities are much higher indeed), while for fermions, it’s the opposite: they don’t want to crowd together and, hence, the associated probabilities are much lower. So fermions will spread themselves over the various energy levels. The distribution for ‘classical’ particles is somewhere in the middle.

In that post of mine, I gave an actual example involving nine particles and the various patterns that are possible, so you can have a look there. Here I just want to note that the math behind is easy to understand when dropping the A (that’s just another normalization constant anyway) and re-writing the formulas as follows:

Maxwell-Boltzmann (for classical particles): f(E) = e^−E/kT
Fermi-Dirac (for fermions): f(E) = e^−E/kT/[1+e^−E/kT]
Bose-Einstein (for bosons): f(E) = e^−E/kT/[1−e^−E/kT]

Just use Feynman’s substitution x = e^−ħω/kT: the Bose-Einstein distribution then becomes 1/[1/x–1] = 1/[(1–x)/x] = x/(1–x). Now it’s easy to see that the denominator of the formula of both the Fermi-Dirac as well as the Bose-Einstein distribution will approach 1 (i.e. the ‘denominator’ of the Maxwell-Boltzmann formula) if e^−E/kTapproaches zero, so that’s when E becomes larger and larger. Hence, for higher energy levels, the probability densities of the three functions approach each other indeed, as they should.

Now what’s the second graph about? Here we’re looking at one energy level only, but we let the temperature vary from 0 to infinity. The graph says that, at low temperature, the probabilities will also be more or less the same, and the three distributions only differ at higher temperatures. That makes sense too, of course!

Well… That says it all, I guess. I hope you enjoyed this post. As I’ve sort of concluded Volume I of Feynman’s Lectures with this, I’ll be silent for a while… […] Or so I think. 🙂

Strings in classical and quantum physics

Pre-scriptum (dated 26 June 2020): These posts on elementary math and physics have not suffered much from the attack by the dark force—which is good because I still like them. While my views on the true nature of light, matter and the force or forces that act on them have evolved significantly as part of my explorations of a more realist (classical) explanation of quantum mechanics, I think most (if not all) of the analysis in this post remains valid and fun to read. In fact, I find the simplest stuff is often the best. 🙂

Original post:

This post is not about string theory. The goal of this post is much more limited: it’s to give you a better understanding of why the metaphor of the string is so appealing. Let’s recapitulate the basics by see how it’s used in classical as well as in quantum physics.

In my posts on music and math, or music and physics, I described how a simple single string always vibrates in various modes at the same time: every tone is a mixture of an infinite number of elementary waves. These elementary waves, which are referred to as harmonics (or as (normal) modes, indeed) are perfectly sinusoidal, and their amplitude determines their relative contribution to the composite waveform. So we can always write the waveform F(t) as the following sum:

F(t) = a₁sin(ωt) + a₂sin(2ωt) + a₃sin(3ωt) + … + a_nsin(nωt) + …

[If this is your first reading of my post, and the formula shies you away, please try again. I am writing most of my posts with teenage kids in mind, and especially this one. So I will not use anything else than simple arithmetic in this post: no integrals, no complex numbers, no logarithms. Just a bit of geometry. That’s all. So, yes, you should go through the trouble of trying to understand this formula. The only thing that you may have some trouble with is ω, i.e. angular frequency: it’s the frequency expressed in radians per time unit, rather than oscillations per second, so ω = 2π·f = 2π/T, with f the frequency as you know it (i.e. oscillations per second) and T the period of the wave.]

I also noted that the wavelength of these component waves (λ) is determined by the length of the string (L), and by its length only: λ₁ = 2L, λ₂ = L, λ₃ = (2/3)·L. So these wavelengths do not depend on the material of the string, or its tension. At any point in time (so keeping t constant, rather than x, as we did in the equation above), the component waves look like this:

etcetera (1/8, 1/9,…,1/n,… 1/∞)

That the wavelengths of the harmonics of any actual string only depend on its length is an amazing result in light of the complexities behind: a simple wound guitar string, for example, is not simple at all (just click the link here for a quick introduction to guitar string construction). Simple piano wire isn’t simple either: it’s made of high-carbon steel, i.e. a very complex metallic alloy. In fact, you should never think any material is simple: even the simplest molecular structures are very complicated things. Hence, it’s quite amazing all these systems are actually linear systems and that, despite the underlying complexity, those wavelength ratios form a simple harmonic series, i.e. a simple reciprocal function y = 1/x, as illustrated below.

A simple harmonic series? Hmm… I can’t resist noting that the harmonic series is, in fact, a mathematical beast. While its terms approach zero as x (or n) increases, the series itself is divergent. So it’s not like 1+1/2+1/4+1/8+…+1/2ⁿ+…, which adds up to 2. Divergent series don’t add up to any specific number. Even Leonhard Euler – the most famous mathematician of all times, perhaps – struggled with this. In fact, as late as in 1826, another famous mathematician, Niels Henrik Abel (in light of the fact he died at age 26 (!), his legacy is truly amazing), exclaimed that a series like this was “an invention of the devil”, and that it should not be used in any mathematical proof. But then God intervened through Abel’s contemporary Augustin-Louis Cauchy 🙂 who finally cracked the nut by rigorously defining the mathematical concept of both convergent as well as divergent series, and equally rigorously determining their possibilities and limits in mathematical proofs. In fact, while medieval mathematicians had already grasped the essentials of modern calculus and, hence, had already given some kind of solution to Zeno’s paradox of motion, Cauchy’s work is the full and final solution to it. But I am getting distracted, so let me get back to the main story.

More remarkable than the wavelength series itself, is its implication for the respective energy levels of all these modes. The material of the string, its diameter, its tension, etc will determine the speed with which the wave travels up and down the string. [Yes, that’s what it does: you may think the string oscillates up and down, and it does, but the waveform itself travels along the string. In fact, as I explained in my previous post, we’ve got two waves traveling simultaneously: one going one way and the other going the other.] For a specific string, that speed (i.e. the wave velocity) is some constant, which we’ll denote by c. Now, c is, obviously, the product of the wavelength (i.e. the distance that the wave travels during one oscillation) and its frequency (i.e. the number of oscillations per time unit), so c = λ·f. Hence, f = c/λ and, therefore, f₁ = (1/2)·c/L, f₂ = (2/2)·c/L, f₃ = (3/2)·c/L, etcetera. More in general, we write f_n = (n/2)·c/L. In short, the frequencies are equally spaced. To be precise, they are all (1/2)·c/L apart.

Now, the energy of a wave is directly proportional to its frequency, always, in classical as well as in quantum mechanics. For example, for photons, we have the Planck-Einstein relation: E = h·f = ħ·ω. So that relation states that the energy is proportional to the (light) frequency of the photon, with h (i.e. he Planck constant) as the constant of proportionality. [Note that ħ is not some different constant. It’s just the ‘angular equivalent’ of h, so we have to use ħ = h/2π when frequencies are expressed in angular frequency, i.e. radians per second rather than hertz.] Because of that proportionality, the energy levels of our simple string are also equally spaced and, hence, inserting another proportionality constant, which I’ll denote by a instead of h (because it’s some other constant, obviously), we can write:

E_n = a·f_n = (n/2)·a·c/L

Now, if we denote the fundamental frequency f₁ = (1/2)·c/L, quite simply, by f (and, likewise, its angular frequency as ω), then we can re-write this as:

E_n = n·a·f = n·ā·ω (ā = a/2π)

This formula is exactly the same as the formula used in quantum mechanics when describing atoms as atomic oscillators, and why and how they radiate light (think of the blackbody radiation problem, for example), as illustrated below: E_n = n·ħ·ω = n·h·f. The only difference between the formulas is the proportionality constant: instead of a, we have Planck’s constant here: h, or ħ when the frequency is expressed as an angular frequency.

This grand result – that the energy levels associated with the various states or modes of a system are equally spaced – is referred to as the equipartition theorem in physics, and it is what connects classical and quantum physics in a very deep and fundamental way.

In fact, because they’re nothing but proportionality constants, the value of both a and h depends on our units. If w’d use the so-called natural units, i.e. equating ħ to 1, the energy formula becomes E_n = n·ω, and, hence, our unit of energy and our unit of frequency become one and the same. In fact, we can, of course, also re-define our time unit such that the fundamental frequency ω is one, i.e. one oscillation per (re-defined) time unit, so then we have the following remarkable formula:

E_n = n

Just think about it for a moment: what I am writing here is E₀ = 0, E₁ = 1, E₂ = 2, E₃ = 3, E₄ = 4, etcetera. Isn’t that amazing? I am describing the structure of a system here – be it an atom emitting or absorbing photons, or a macro-thing like a guitar string – in terms of its basic components (i.e. its modes), and it’s as simple as counting: 0, 1, 2, 3, 4, etc.

You may think I am not describing anything real here, but I am. We cannot do whatever we wanna do: some stuff is grounded in reality, and in reality only—not in the math. Indeed, the fundamental frequency of our guitar string – which we used as our energy unit – is a property of the string, so that’s real: it’s not just some mathematical shape out: it depends on the string’s length (which determines its wavelength), and it also depends on the propagation speed of the wave, which depends on other basic properties of the string, such as its material, its diameter, and its tension. Likewise, the fundamental frequency of our atomic oscillator is a property of the atomic oscillator or, to use a much grander term, a property of the Universe. That’s why h is a fundamental physical constant. So it’s not like π or e. [When reading physics as a freshman, it’s always useful to clearly distinguish physical constants (like Avogadro’s number, for example) from mathematical constants (like Euler’s number).]

The theme that emerges here is what I’ve been saying a couple of times already: it’s all about structure, and the structure is amazingly simple. It’s really that equipartition theorem only: all you need to know is that the energy levels of the modes of a system – any system really: an atom, a molecular system, a string, or the Universe itself – are equally spaced, and that the space between the various energy levels depends on the fundamental frequency of the system. Moreover, if we use natural units, and also re-define our time unit so the fundamental frequency is equal to 1 (so the frequencies of the other modes are 2, 3, 4 etc), then the energy levels are just 0, 1, 2, 3, 4 etc. So, yes, God kept things extremely simple. 🙂

In order to not cause too much confusion, I should add that you should read what I am writing very carefully: I am talking the modes of a system. The system itself can have any energy level, of course, so there is no discreteness at the level of the system. I am not saying that we don’t have a continuum there. We do. What I am saying is that its energy level can always be written as a (potentially infinite) sum of the energies of its components, i.e. its fundamental modes, and those energy levels are discrete. In quantum-mechanical systems, their spacing is h·f, so that’s the product of Planck’s constant and the fundamental frequency. For our guitar, the spacing is a·f (or, using angular frequency, ā·ω: it’s the same amount). But that’s it really. That’s the structure of the Universe. 🙂

Let me conclude by saying something more about a. What information does it capture? Well… All of the specificities of the string (like its material or its tension) determine the fundamental frequency f and, hence, the energy levels of the basic modes of our string. So a has nothing to do with the particularities of our string, of our system in general. However, we can, of course, pluck our string very softly or, conversely, give it a big jolt. So our a coefficient is not related to the string as such, but to the total energy of our string. In other words, a is related to those amplitudes a₁, a₂, etc in our F(t) = a₁sin(ωt) + a₂sin(2ωt) + a₃sin(3ωt) + … + a_nsin(nωt) + … wave equation.

How exactly? Well… Based on the fact that the total energy of our wave is equal to the sum of the energies of all of its components, I could give you some formula. However, that formula does use an integral. It’s an easy integral: energy is proportional to the square of the amplitude, and so we’re integrating the square of the wave function over the length of the string. But then I said I would not have any integral in this post, and so I’ll stick to that. In any case, even without the formula, you know enough now. For example, one of the things you should be able to reflect on is the relation between a and h. It’s got to do with structure, of course. 🙂 But I’ll let you think about that yourself.

[…] Let me help you. Think of the meaning of Planck’s constant h. Let’s suppose we’d have some elementary ‘wavicle’, like that elementary ‘string’ that string theorists are trying to define: the smallest ‘thing’ possible. It would have some energy, i.e. some frequency. Perhaps it’s just one full oscillation. Just enough to define some wavelength and, hence, some frequency indeed. Then that thing would define the smallest time unit that makes sense: it would the time corresponding to one oscillation. In turn, because of the E = h·f relation, it would define the smallest energy unit that makes sense. So, yes, h is the quantum (or fundamental unit) of energy. It’s very small indeed (h = 6.626070040(81)×10⁻³⁴ J·s, so the first significant digit appears only after 33 zeroes behind the decimal point) but that’s because we’re living at the macro-scale and, hence, we’re measuring stuff in huge units: the joule (J) for energy, and the second (s) for time. In natural units, h would be one. [To be precise, physicist prefer to equate ħ, rather than h, to one when talking natural units. That’s because angular frequency is more ‘natural’ as well when discussing oscillations.]

What’s the conclusion? Well… Our a will be some integer multiple of h. Some incredibly large multiple, of course, but a multiple nevertheless. 🙂

Post scriptum: I didn’t say anything about strings in this post or, let me qualify, about those elementary ‘strings’ that string theorists try to define. Do they exist? Feynman was quite skeptical about it. He was happy with the so-called Standard Model of phyics, and he would have been very happy to know that the existence Higgs field has been confirmed experimentally (that discovery is what prompted my blog!), because that confirms the Standard Model. The Standard Model distinguishes two types of wavicles: fermions and bosons. Fermions are matter particles, such as quarks and electrons. Bosons are force carriers, like photons and gluons. I don’t know anything about string theory, but my guts instinct tells me there must be more than just one mathematical description of reality. It’s the principle of duality: concepts, theorems or mathematical structures can be translated into other concepts, theorems or structures. But… Well… We’re not talking equivalent descriptions here: string theory is a different theory, it seems. For a brief but totally incomprehensible overview (for novices at least), click on the following link, provided by the C.N. Yang Institute for Theoretical Physics. If anything, it shows I’ve got a lot more to study as I am inching forward on the difficult Road to Reality. 🙂

Modes in classical and in quantum physics

Original post:

Basics

Waves are peculiar: there is one single waveform, i.e. one motion only, but that motion can always be analyzed as the sum of the motions of all the different wave modes, combined with the appropriate amplitudes and phases. Saying the same thing using different words: we can always analyze the wave function as the sum of a (possibly infinite) number of components, i.e. a so-called Fourier series:

The f(t) function can be any wave, but the simple examples in physics textbooks usually involve a string or, in two dimensions, some vibrating membrane, and I’ll stick to those examples too in this post. Feynman calls the Fourier components harmonic functions, or harmonics tout court, but the term ‘harmonic’ refers to so many different things in math that it may be better not to use it in this context. The component waves are sinusoidal functions, so sinusoidals might be a better term but it’s not in use, because a more general analysis will use complex exponentials, rather than sines and/or cosines. Complex exponentials (e.g. 10^ix) are periodic functions too, so they are totally unlike real exponential functions (e.g. (e.g. 10^x). Hence, Feynman also uses the term ‘exponentials’. At some point, he also writes that the pattern of motion (of a mode) varies ‘exponentially’ but, of course, he’s thinking of complex exponentials, and, therefore, we should substitute ‘exponentially’ for ‘sinusoidally’ when talking real-valued wave functions.

[…] I know. I am already getting into the weeds here. As I am a bit off-track anyway now, let me make another remark here. You may think that we have two types of sinusoidals, or two types of functions, in that Fourier decomposition: sines and cosines. You should not think of it that way: the sine and cosine function are essentially the same. I know your old math teacher in high school never told you that, but it’s true. They both come with the same circle (yes, I know that’s ridiculous statement but I don’t know how to phrase it otherwise): the difference between a sine and a cosines is just a phase shift: cos(ωt) = sin(ωt + π/2) and, conversely, sin(ωt) = cos(ωt − π/2). If the starting phases of all of the component waves would be the same, we’d have a Fourier decomposition involving cosines only, or sines only—whatever you prefer. Indeed, because they’re the same function except for that phase shift (π/2), we can always go from one to the other by shifting our origin of space (x) and/or time (t). However, we cannot assume that all of the component waves have the same starting phase and, therefore, we should write each component as cos(n·ωt + Φ_n), or a sine with a similar argument. Now, you’ll remember – because your math teacher in high school told you that at least 🙂 – that there’s a formula for the cosine (and sine) of the sum of two angles: we can write cos(n·ωt + Φ_n) as cos(n·ωt + Φ_n) = [cos(Φ_n)·cos(n·ωt) – sin(Φ_n)·sin(n·ωt)]. Substituting cos(Φ_n) and – sin(Φ_n) for a_n and b_n respectively gives us the a_n·cos(n·ωt) + b_n·sin(n·ωt) expressions above. In addition, the component waves may not only differ in phase, but also in amplitude, and, hence, the a_n and b_n coefficients do more than only capturing the phase differences. But let me get back on the track. 🙂

Those sinusoidals have a weird existence: they are not there, physically—or so it seems. Indeed, there is one waveform only, i.e. one motion only—and, if it’s any real wave, it’s most likely to be non-sinusoidal. At the same time, I noted, in my previous post, that, if you pluck a string or play a chord on your guitar, some string you did not pluck may still pick up one or more of its harmonics (i.e. one or more of its overtones) and, hence, start to vibrate too! It’s the resonance phenomenon. If you have a grand piano, it’s even more obvious: if you’d press the C4 key on a piano, a small hammer will strike the C4 string and it will vibrate—but the C5 string (one octave higher) will also vibrate, although nothing touched it—except for the air transmitting the sound wave (including the harmonics causing the resonance) from the C4 string, of course! So the component waves are there and, at the same time, they’re not. Whatever they are, they are more than mathematical forms: the so-called superposition principle (on which the Fourier analysis is based) is grounded in reality: it’s because we can add forces. I know that sounds extremely obvious – or ridiculous, you might say 🙂 – but it is actually not so obvious. […] I am tempted to write something about conservative forces here but… Well… I need to move on.

Let me show that diagram of the first seven harmonics of an ideal string once again. All of them, and the higher ones too, would be in our wave function. Hence, assuming there’s no phase difference between the harmonics, we’d write:

f(t) = sin(ωt) + sin(2ωt) + sin(3ωt) + … + sin(nωt) + …

The frequencies of the various modes of our ideal string are all simple multiples of the fundamental frequency ω, as evidenced from the argument in our sine functions (ω, 2ω, 3ω, etcetera). Conversely, the respective wavelengths are λ, λ/2, λ/3, etcetera. [Remember: the speed of the wave is fixed, and frequency and wavelength are inversely proportional: c = λ·f = λ/T = λ·(ω/2π).] So, yes, these frequencies and wavelengths can all be related to each other in terms of equally simple harmonic ratios: 1:2, 2:3, 3:5, 4:5 etcetera. I explained in my previous posts why that does not imply that the musical notes themselves are related in such way: the musical scale is logarithmic. So I won’t repeat myself. All of the above is just an introduction to the more serious stuff, which I’ll talk about now.

Modes in two dimensions

An analysis of waves in two dimensions is often done assuming some drum membrane. The Great Teacher played drums, as you can see from his picture in his Lectures, and there are also videos of him performing on YouTube. So that’s why the drum is used almost all textbooks now. 🙂

The illustration of one of the normal modes of a circular membrane comes from the Wikipedia article on modes. There are many other normal modes – some of them with a simpler shape, but some of them more complicated too – but this is a nice one as it also illustrates the concept of a nodal line, which is closely related to the concept of a mode. Huh? Yes. The modes of a one-dimensional string have nodes, i.e. points where the displacement is always zero. Indeed, as you can see from the illustration above (not below), the first overtone has one node, the second two, etcetera. So the equivalent of a node in two dimensions is a nodal line: for the mode shown below, we have one bisecting the disc and then another one—a circle about halfway between the edge and center. The third nodal line is the edge itself, obviously. [The author of the Wikipedia article nodes that the animation isn’t perfect, because the nodal line and the nodal circle halfway the edge and the center both move a little bit. In any case, it’s pretty good, I think. I should also learn how to make animations like that. :-)]

What’s a mode?

How do we find these modes? And how are they defined really? To explain that, I have to briefly return to the one-dimensional example. The key to solving the problem (i.e. finding the modes, and defining their characteristics) is the following fact: when a wave reaches the clamped end of a string, it will be reflected with a change in sign, as illustrated below: we’ve got that F(x+ct) wave coming in, and then it goes back indeed, but with the sign reversed.

It’s a complicated illustration because it also shows some hypothetical wave coming from the other side, where there is no string to vibrate. That hypothetical wave is the same wave, but travelling in the other direction and with the sign reversed (–F). So what’s that all about? Well… I never gave any general solution for a waveform traveling up and down a string: I just said the waveform was traveling up and down the string (now that is obvious: just look at that diagram with the seven first harmonics once again, and think about how that oscillation goes up and down with time), but so I did not really give any general solution for them (the sine and cosine functions are specific solutions). So what is the general solution?

Let’s first assume the string is not held anywhere, so that we have an infinite string along which waves can travel in either direction. In fact, the most general functional form to capture the fact that a waveform can travel in any direction is to write the displacement y as the sum of two functions: one wave traveling one way (which we’ll denote by F), and the other wave (which we’ll denote by G) traveling the other way. From the illustration above, it’s obvious that the F wave is traveling towards the negative x-direction and, hence, its argument will be x + ct. Conversely, the G wave travels in the positive x-direction, so its argument is x – ct. So we write:

y = F(x + ct) + G(x – ct)

[I’ve explained this thing about directions and why the argument in a wavefunction (x ± ct) is what it is before. You should look it up in case you don’t understand. As for the c in this equation, that’s the wave velocity once more, which is constant and which depends, as always, on the medium, so that’s the material and the diameter and the tension and whatever of the string.]

So… We know that the string is actually not infinite, but that it’s fixed to some ‘infinitely solid wall’ (as Feynman puts it). Hence, y is equal to zero there: y = 0. Now let’s choose the origin of our x-axis at the fixed end so as to simplify the analysis. Hence, where y is zero, x is also zero. Now, at x = 0, our general solution above for the infinite string becomes y = F(ct) + G(−ct) = 0, for all values of t. Of course, that means G(−ct) must be equal to –F(ct). Now, that equality is there for all values of t. So it’s there for all values of ct and −ct. In short, that equality is valid for whatever value of the argument of G and –F. As Feynman puts it: “G of anything must be –F of minus that same thing.” Now, the ‘anything’ in G is its argument: x – ct, so ‘minus that same thing’ is –(x – ct) = −x + ct. Therefore, our equation becomes:

y = F(x + ct) − F(−x + ct)

So that’s what’s depicted in the diagram above: the F(x + ct) wave ‘vanishes’ behind the wall as the − F(−x + ct) wave comes out of it. Conversely, the − F(−x + ct) is hypothetical indeed until it reaches the origin, after which it becomes the real wave. Their sum is only relevant near the origin x = 0, and on the positive side only (on the negative side of the x-axis, the F and G functions are both hypothetical). [I know, it’s not easy to follow, but textbooks are really short on this—which is why I am writing my blog: I want to help you ‘get’ it.]

Now, the results above are valid for any wave, periodic or not. Let’s now confine the analysis to periodic waves only. In fact, we’ll limit the analysis to sinusoidal wavefunctions only. So that should be easy. Yes. Too easy. I agree. 🙂

So let’s make things difficult again by introducing the complex exponential notation, so that’s Euler’s formula: e^iθ = cosθ + isinθ, with i the imaginary unit, and isinθ the imaginary component of our wave. So the only thing that is real, is cosθ.

What the heck? Just bear with me. It’s good to make the analysis somewhat more general, especially because we’ll be talking about the relevance of all of this to quantum physics, and in quantum physics the waves are complex-valued indeed! So let’s get on with it. To use Euler’s formula, we need to substitute x + ct for the phase of the wave, so that involves the angular frequency and the wavenumber. Let me just write it down:

F(x + ct) = e^iω(t+x/c) and F(−x + ct) = e^iω(t−x/c)

Huh? Yeah. Sorry. I’ll resist the temptation to go off-track here, because I really shouldn’t be copying what I wrote in other posts. Most of what I write above is really those simple relations: c = λ·f = ω/k, with k, i.e. the wavenumber, being defined as k = 2π/λ. For details, go to one of my others posts indeed, in which I explain how that works in very much detail: just click on the link here, and scroll down to the section on the phase of a wave, in which I explain why the phase of wave is equal to θ = ωt–kx = ω(t–x/c). And, yes, I know: the thing with the wave directions and the signs is quite tricky. Just remember: for a wave traveling in the positive x-direction, the signs in front of x and t are each other’s opposite but, if the wave’s traveling in the negative y-direction, they are the same. As mentioned, all the rest is usually a matter of shifting the phase, which amounts to shifting the origin of either the x- or the t-axis. I need to move on. Using the exponential notation for our sinusoidal wave, y = F(x + ct) − F(−x + ct) becomes:

y = e^iω(t+x/c) − e^iω(t−x/c)

I can hear you sigh again: Now what’s that for? What can we do with this? Just continue to bear with me for a while longer. Let’s factor the e^iωt term out. [Why? Patience, please!] So we write:

y = e^iωt[e^iωx/c) − e^−iωx/c)]

Now, you can just use Euler’s formula again to double-check that e^iθ − e^−θ = 2isinθ. [To get that result, you should remember that cos(−θ) = cosθ, but sin(−θ) = −sin(θ).] So we get:

y = e^iωt[e^iωx/c) − e^−iωx/c)] = 2ie^iωtsin(ωx/c)

Now, we’re only interested in the real component of this amplitude of course – but that’s only we’re in the classical world here, not in the real world, which is quantum-mechanical and, hence, involves the imaginary stuff also 🙂 – so we should write this out using Euler’s formula again to convert the exponential to sinusoidals again. Hence, remembering that i² = −1, we get:

y = 2ie^iωtsin(ωx/c) = 2icos(ωt)·sin(ωx/c) – 2sin(ωt)·sin(ωx/c)

!?!

OK. You need a break. So let me pause here for a while. What the hell are we doing? Is this legit? I mean… We’re talking some real wave, here, don’t we? We do. So is this conversion from/to real amplitudes to/from complex amplitudes legit? It is. And, in this case (i.e. in classical physics), it’s true that we’re interested in the real component of y only. But then it’s nice the analysis is valid for complex amplitudes as well, because we’ll be talking complex amplitudes in quantum physics.

[…] OK. I acknowledge it all looks very tricky so let’s see what we’d get using our old-fashioned sine and/or cosine function. So let’s write F(x + ct) as cos(ωt+ωx/c) and F(−x + ct) as cos(ωt−ωx/c). So we write y = cos(ωt+ωx/c) − cos(ωt−ωx/c). Now work on this using the cos(α+β) = cosα·cosβ − sinα·sinβ formula and the cos(−α) = cosα and sin(−α) = −sinα identities. You (should) get: y = −2sin(ωt)·sin(ωx/c). So that’s the real component in our y function above indeed. So, yes, we do get the same results when doing this funny business using complex exponentials as we’d get when sticking to real stuff only! Fortunately! 🙂

[Why did I get off-track again? Well… It’s true these conversions from real to complex amplitudes should not be done carelessly. It is tricky and non-intuitive, to say the least. The weird thing about it is that, if we multiply two imaginary components, we get a real component, because i² is a real number: it’s −1! So it’s fascinating indeed: we add an imaginary component to our real-valued function, do all kinds of manipulations with – including stuff that involves the use of the i² = −1 – and, when done, we just take out the real component and it’s alright: we know that the result is OK because of the ‘magic’ of complex numbers! In any case, I need to move on so I can’t dwell on this. I also explained much of the ‘magic’ in other posts already, so I shouldn’t repeat myself. If you’re interested, click on this link, for instance.]

Let’s go back to our y = – 2sin(ωt)·sin(ωx/c) function. So that’s the oscillation. Just look at the equation and think about what it tells us. Suppose we fix x, so we’re looking at one point on the string only and only let t vary: then sin(ωx/c) is some constant and it’s our sin(ωt) factor that goes up and down. So our oscillation has frequency ω, at every point x, so that’s everywhere!

Of course, this result shouldn’t surprise us, should it? That’s what we put in when we wrote F as F(x + ct) = e^iω(t+x/c) or as cos(ωt+ωx/c), isn’t it? Well… Yes and no. Yes, because you’re right: we put in that angular frequency. But then, no, because we’re talking a composite wave here: a wave traveling up and down, with the components traveling in opposite directions. Indeed, we’ve also got that G(x) = −F(–x) function here. So, no, it’s not quite the same.

Let’s fix t now, and take a snapshot of the whole wave, so now we look at x as the variable and sin(ωt) is some constant. What we see is a sine wave, and sin(ωt) is its maximum amplitude. Again, you’ll say: of course! Well… Yes. The thing is: the point where the amplitude of our oscillation is equal to zero, is always the same, regardless of t. So we have fixed nodes indeed. Where are they? The nodes are, obviously, the points where sin(ωx/c) = 0, so that’s when ωx/c is equal to 0, obviously, or – more importantly – whenever ωx/c is equal to π, 2π, 3π, 4π, etcetera. More, generally, we can say whenever ωx/c = n·π with n = 0, 1, 2,… etc. Now, that’s the same as writing x = n·π·c/ω = n·π/k = n·π·λ/2π = n·λ/2.

Now let’s remind ourselves of what λ really is: for the fundamental frequency it’s twice the length of the string, so λ = 2·L. For the next mode (i.e. the second harmonic), it’s the length itself: λ = L. For the third, it’s λ = (2/3)·L, etcetera. So, in general, it’s λ = (2/m)·L with m = 1, 2, etcetera. [We may or may not want to include a zero mode by allowing m to equal zero as well, so then there’s no oscillation and y = 0 everywhere. 🙂 But that’s a minor point.] In short, our grand result is:

x = n·λ/2 = n·(2/m)·L/2 = (n/m)·L

Of course, we have to exclude the x points lying outside of our string by imposing that n/m ≤ 1, i.e. the condition that n ≤ m. So for m = 1, n is 0 or 1, so the nodes are, effectively, both ends of the string. For m = 2, n can be 0, 1 and 2, so the nodes are the ends of the string and it’s middle point L/2. And so on and so on.

I know that, by now, you’ve given up. So no one is reading anymore and so I am basically talking to myself now. What’s the point? Well… I wanted to get here in order to define the concept of a mode: a mode is a pattern of motion, which has the property that, at any point, the object moves perfectly sinusoidally, and that all points move at the same frequency (though some will move more than others). Modes also have nodes, i.e. points that don’t move at all, and above I showed how we can find the nodes of the modes of a one-dimensional string.

Also note how remarkable that result actually is: we didn’t specify anything about that string, so we don’t care about its material or diameter or tension or whatever. Still, we know its fundamental (or normal modes), and we know their nodes: they’re a function of the length of the string, and the number of the mode only: x = (n/m)·L. While an oscillating string may seem to be the most simple thing on earth, it isn’t: think of all the forces between the molecules, for instance, as that string is vibrating. Still, we’ve got this remarkably simple formula. Don’t you find that amazing?

[…] OK… If you’re still reading, I know you want me to move on, so I’ll just do that.

Back to two dimensions

The modes are all that matters: when linear forces (i.e. linear systems) are involved, any motion can be analyzed as the sum of the motions of all the different modes, combined with appropriate amplitudes and phases. Let me reproduce the Fourier series once more (the more you see, the better you’ll understand it—I should hope!): Of course, we should generalize this also include x as a variable which, again, is easier if we’d use complex exponentials instead of the sinusoidal components. The nice illustration on Fourier analysis from Wikipedia shows how it works, in essence, that is. The red function below consists of six of those modes.

OK. Enough of this. Let’s go to the two-dimensional case now. To simplify the analysis, Feynman invented a rectangular drum. A rectangular drum is probably more difficult to play, but it’s easier to analyze—as compared to a circular drum, that is! 🙂

In two dimensions, our sinusoidal one-dimensional e^i(ωt−kx)waveform becomes e^{i(ωt−k_xx−k_yy)}. So we have a wavenumber for the x and y directions, and the sign in front is determined by the direction of the wave, so we need to check whether it moves in the positive or negative direction of the x- and y-axis respectively. Now, we can rewrite e^{i(ωt+k_xx+k_yy)} as e^iωt·e^{i(ωt+k_xx+k_yy)}, of course, which is what you see in the diagram above, except that the wave is moving in the negative y direction and, hence, we’ve got + sign in front of our k_yy term. All the rest is rather well explained in Feynman, so I’ll refer you to the textbook here.

We basically need to ensure that we have a nodal line at x = 0 and at x = a, and then we do the same for y = 0 and y = a. Then we apply exactly the same logic as for the one-dimensional string: the wave needs to be coherently reflected. The analysis is somewhat more complicated because it involves some angle of incidence now, i.e. the θ in the diagram above, so that’s another page in Feynman’s textbook. And then we have the same gymnastics for finding wavelengths in terms of the dimensions a and b, as well as in terms of n and m, where n is the number of the mode involved when fixing the nodal lines at x = 0 and x = a, and m is the number of the mode involved when fixing the nodal lines at y = 0 and y = b. Sounds difficult? Well… Yes. But I won’t copy Feynman here. Just go and check for yourself.

The grand result is that we do get some formula for a wavelength λ of what satisfies the definition of a mode: a perfectly sinusoidal motion, that has all points on the drum move at the same frequency, though some move more than others. Also, as evidenced from my illustration for the circular disk: we’ve got nodal lines, and then I mean other nodal lines, different from the edges! I’ll just give you that formula here (again, for the detail, go and check Feynman yourself):

Feynman also works out an example for a = 2b. I’ll just copy the results hereunder, which is a formula for the (angular) frequencies ω, and a table of the mode shapes in a qualitative way (I’ll leave it to you to google animations that match the illustration).

Again, we should note the amazing simplicity of the result: we don’t care about the type of membrane or whatever other material the drum is made of. It’s proportions are all that matters.

Finally, you should also note the last two columns in the table above: these just show to illustrate that, unlike our modes in the one-dimensional case, the natural frequencies here are not multiples of the fundamental frequency. As Feynman notes, we should not be led astray by the example of the one-dimensional ideal string. It’s again a departure from the Pythagorean idea, that all in Nature respects harmonic ratios. It’s just not true. Let me quote Feynman, as I have no better summary: “The idea that the natural frequencies are harmonically related is not generally true. It is not true for a system with more than one dimension, nor is it true for one-dimensional systems which are more complicated than a string with uniform density and tension.“

So… That says it all, I’d guess. Maybe I should just quote his example of a one-dimensional system that does not obey Pythagoras’ prescription: a hanging chain which, because of the weight of the chain, has higher tension at the top than at the bottom. If such chain is set in oscillation, there are various modes and frequencies, but the frequencies will not be simply multiples of each other, nor of any other number. It is also interesting to note that the mode shapes will also not be sinusoidal. However, here we’re getting into non-linear dynamics, and so I’ll you read about that elsewhere too: once again, Feynman’s analysis of non-linear systems is very accessible and an interesting read. Hence, I warmly recommend it.

Modes in three dimensions and in quantum mechanics.

Well… Unlike what you might expect, I won’t bury you under formulas this time. Let me refer you, instead, to Wikipedia’s article on the so-called Leidenfrost effect. Just do it. Don’t bother too much about the text, scroll down a bit, and play the video that comes with it. I saw it, sort of by accident, and, at first, I thought it was something very high-tech. But no: it’s just a drop of water skittering around in a hot pan. It takes on all kinds of weird forms and oscillates in the weirdest of ways, but all is nothing but an excitation of the various normal modes of it, with various amplitudes and phases, of course, as a Fourier analysis of the phenomenon dictates.

There’s plenty of other stuff around to satisfy your curiosity, all quite understandable and fun—because you now understand the basics of it for the one- and two-dimensional case.

So… Well… I’ve kept this section extremely short, because now I want to say a few words about quantum-mechanical systems. Well… In fact, I’ll simply quote Feynman on it, because he writes about in a style that’s unsurpassed. He also nicely sums up the previous conversation. Here we go:

The ideas discussed above are all aspects of what is probably the most general and wonderful principle of mathematical physics. If we have a linear system whose character is independent of the time, then the motion does not have to have any particular simplicity, and in fact may be exceedingly complex, but there are very special motions, usually a series of special motions, in which the whole pattern of motion varies exponentially with the time. For the vibrating systems that we are talking about now, the exponential is imaginary, and instead of saying “exponentially” we might prefer to say “sinusoidally” with time. However, one can be more general and say that the motions will vary exponentially with the time in very special modes, with very special shapes. The most general motion of the system can always be represented as a superposition of motions involving each of the different exponentials.

This is worth stating again for the case of sinusoidal motion: a linear system need not be moving in a purely sinusoidal motion, i.e., at a definite single frequency, but no matter how it does move, this motion can be represented as a superposition of pure sinusoidal motions. The frequency of each of these motions is a characteristic of the system, and the pattern or waveform of each motion is also a characteristic of the system. The general motion in any such system can be characterized by giving the strength and the phase of each of these modes, and adding them all together. Another way of saying this is that any linear vibrating system is equivalent to a set of independent harmonic oscillators, with the natural frequencies corresponding to the modes.

In quantum mechanics the vibrating object, or the thing that varies in space, is the amplitude of a probability function that gives the probability of finding an electron, or system of electrons, in a given configuration. This amplitude function can vary in space and time, and satisfies, in fact, a linear equation. But in quantum mechanics there is a transformation, in that what we call frequency of the probability amplitude is equal, in the classical idea, to energy. Therefore we can translate the principle stated above to this case by taking the word frequency and replacing it with energy. It becomes something like this: a quantum-mechanical system, for example an atom, need not have a definite energy, just as a simple mechanical system does not have to have a definite frequency; but no matter how the system behaves, its behavior can always be represented as a superposition of states of definite energy. The energy of each state is a characteristic of the atom, and so is the pattern of amplitude which determines the probability of finding particles in different places. The general motion can be described by giving the amplitude of each of these different energy states. This is the origin of energy levels in quantum mechanics. Since quantum mechanics is represented by waves, in the circumstance in which the electron does not have enough energy to ultimately escape from the proton, they are confined waves. Like the confined waves of a string, there are definite frequencies for the solution of the wave equation for quantum mechanics. The quantum-mechanical interpretation is that these are definite energies. Therefore a quantum-mechanical system, because it is represented by waves, can have definite states of fixed energy; examples are the energy levels of various atoms.

Isn’t that great? What a summary! It also shows a deeper understanding of classical physics makes it sooooo much better to read something about quantum mechanics. In any case, as for the examples, I should add – because that’s what you’ll often find when you google for quantum-mechanical modes – the vibrational modes of molecules. There’s tons of interesting analysis out there, and so I’ll let you now have fun with it yourself! 🙂

Music and Math

Pre-scriptum (dated 26 June 2020): These posts on elementary math and physics have not suffered much the attack by the dark force—which is good because I still like them. While my views on the true nature of light, matter and the force or forces that act on them have evolved significantly as part of my explorations of a more realist (classical) explanation of quantum mechanics, I think most (if not all) of the analysis in this post remains valid and fun to read. In fact, I find the simplest stuff is often the best. 🙂

Original post:

I ended my previous post, on Music and Physics, by emphatically making the point that music is all about structure, about mathematical relations. Let me summarize the basics:

1. The octave is the musical unit, defined as the interval between two pitches with the higher frequency being twice the frequency of the lower pitch. Let’s denote the lower and higher pitch by a and b respectively, so we say that b‘s frequency is twice that of a.

2. We then divide the [a, b] interval (whose length is unity) in twelve equal sub-intervals, which define eleven notes in-between a and b. The pitch of the notes in-between is defined by the exponential function connecting a and b. What exponential function? The exponential function with base 2, so that’s the function y = 2^x.

Why base 2? Because of the doubling of the frequencies when going from a to b, and when going from b to b + 1, and from b + 1 to b + 2, etcetera. In music, we give a, b, b + 1, b + 2, etcetera the same name, or symbol: A, for example. Or Do. Or C. Or Re. Whatever. If we have the unit and the number of sub-intervals, all the rest follows. We just add a number to distinguish the various As, or Cs, or Gs, so we write A1, A2, etcetera. Or C1, C2, etcetera. The graph below illustrates the principle for the interval between C4 and C5. Don’t think the function is linear. It’s exponential: note the logarithmic frequency scale. To make the point, I also inserted another illustration (credit for that graph goes to another blogger).

You’ll wonder: why twelve sub-intervals? Well… That’s random. Non-Western cultures use a different number. Eight instead of twelve, for example—which is more logical, at first sight at least: eight intervals amounts to dividing the interval in two equal halves, and the halves in halves again, and then once more: so the length of the sub-interval is then 1/2·1/2·1/2 = (1/2)³ = 1/8. But why wouldn’t we divide by three, so we have 9 = 3·3 sub-intervals? Or by 27 = 3·3·3? Or by 16? Or by 5?

The answer is: we don’t know. The limited sensitivity of our ear demands that the intervals be cut up somehow. [You can do tests of the sensitivity of your ear to relative frequency differences online: it’s fun. Just try them! Some of the sites may recommend a hearing aid, but don’t take that crap.] So… The bottom line is that, somehow, mankind settled on twelve sub-intervals within our musical unit—or our sound unit, I should say. So it is what it is, and the ratio of the frequencies between two successive (semi)tones (e.g. C and C#, or E and F, as E and F are also separated by one half-step only) is 2^1/12 = 1.059463… Hence, the pitch of each note is about 6% higher than the pitch of the previous note. OK. Next thing.

3. What’s the similarity between C1, C2, C3 etcetera? Or between A1, A2, A3 etcetera? The answer is: harmonics. The frequency of the first overtone of a string tuned at pitch A3 (i.e. 220 Hz) is equal to the fundamental frequency of a string tuned at pitch A4 (i.e. 440 Hz). Likewise, the frequency of the (pitch of the) C4 note above (which is the so-called middle C) is 261.626 Hz, while the frequency of the (pitch of the) next C note (C5) is twice that frequency: 523.251 Hz. [I should quickly clarify the terminology here: a tone consists of several harmonics, with frequencies f, 2·f, 3·f,… n·f,… The first harmonic is referred to as the fundamental, with frequency f. The second, third, etc harmonics are referred to as overtones, with frequency 2·f, 3·f, etc.]

To make a long story short: our ear is able to identify the individual harmonics in a tone, and if the frequency of the first harmonic of one tone (i.e. the fundamental) is the same frequency as the second harmonic of another, then we feel they are separated by one musical unit.

Isn’t that most remarkable? Why would it be that way?

My intuition tells me I should look at the energy of the components. The energy theorem tells us that the total energy in a wave is just the sum of the energies in all of the Fourier components. Surely, the fundamental must carry most of the energy, and then the first overtone, and then the second. Really? Is that so?

Well… I checked online to see if there’s anything on that, but my quick check reveals there’s nothing much out there in terms of research: if you’d google ‘energy levels of overtones’, you’ll get hundreds of links to research on the vibrational modes of molecules, but nothing that’s related to music theory. So… Well… Perhaps this is my first truly original post! 🙂 Let’s go for it. 🙂

The energy in a wave is proportional to the square of its amplitude, and we must integrate over one period (T) of the oscillation. The illustration below should help you to understand what’s going on. The fundamental mode of the wave is an oscillation with a wavelength (λ₁) that is twice the length of the string (L). For the second mode, the wavelength (λ₂) is just L. For the third mode, we find that λ₃ = (2/3)·L. More in general, the wavelength of the n^thmode is λ_n = (2/n)·L.

The illustration above shows that we’re talking sine waves here, differing in their frequency (or wavelength) only. [The speed of the wave (c), as it travels back and forth along the string, i constant, so frequency and wavelength are in that simple relationship: c = f·λ.] Simplifying and normalizing (i.e. choosing the ‘right’ units by multiplying scales with some proportionality constant), the energy of the first mode would be (proportional to):

What about the second and third modes? For the second mode, we have two oscillations per cycle, but we still need to integrate over the period of the first mode T = T₁, which is twice the period of the second mode: T₁ = 2·T₂. Hence, T₂ = (1/2)·T₁. Therefore, the argument of the sine wave (i.e. the x variable in the integral above) should go from 0 to 4π. However, we want to compare the energies of the various modes, so let’s substitute cleverly. We write:

The period of the third mode is equal to T₃ = (1/3)·T₁. Conversely, T₁ = 3·T₃. Hence, the argument of the sine wave should go from 0 to 6π. Again, we’ll substitute cleverly so as to make the energies comparable. We write:

Now that is interesting! For a so-called ideal string, whose motion is the sum of a sinusoidal oscillation at the fundamental frequency f, another at the second harmonic frequency 2·f, another at the third harmonic 3·f, etcetera, we find that the energies of the various modes are proportional to the values in the harmonic series 1, 1/2, 1/3, 1/4,… 1/n, etcetera. Again, Pythagoras’ conclusion was wrong (the ratio of frequencies of individual notes do not respect simple ratios), but his intuition was right: the harmonic series ∑n⁻¹(n = 1, 2,…,∞) is very relevant in describing natural phenomena. It gives us the respective energies of the various natural modes of a vibrating string! In the graph below, the values are represented as areas. It is all quite deep and mysterious really!

So now we know why we feel C4 and C5 have so much in common that we call them by the same name: C, or Do. It also helps us to understand why the E and A tones have so much in common: the third harmonic of the 110 Hz A2 string corresponds to the fundamental frequency of the E4 string: both are 330 Hz! Hence, E and A have ‘energy in common’, so to speak, but less ‘energy in common’ than two successive E notes, or two successive A notes, or two successive C notes (like C4 and C5).

[…] Well… Sort of… In fact, the analysis above is quite appealing but – I hate to say it – it’s wrong, as I explain in my post scriptum to this post. It’s like Pythagoras’ number theory of the Universe: the intuition behind is OK, but the conclusions aren’t quite right. 🙂

Ideality versus reality

We’ve been talking ideal strings. Actual tones coming out of actual strings have a quality, which is determined by the relative amounts of the various harmonics that are present in the tone, which is not some simple sum of sinusoidal functions. Actual tones have a waveform that may resemble something like the wavefunction I presented in my previous post, when discussing Fourier analysis. Let me insert that illustration once again (and let me also acknowledge its source once more: it’s Wikipedia). The red waveform is the sum of six sine functions, with harmonically related frequencies, but with different amplitudes. Hence, the energy levels of the various modes will not be proportional to the values in that harmonic series ∑n⁻¹, with n = 1, 2,…,∞.

Das wohltemperierte Klavier

Nothing in what I wrote above is related to questions of taste like: why do I seldomly select a classical music channel on my online radio station? Or why am I not into hip hop, even if my taste for music is quite similar to that of the common crowd (as evidenced from the fact that I like ‘Listeners’ Top’ hit lists)?

Not sure. It’s an unresolved topic, I guess—involving rhythm and other ‘structures’ I did not mention. Indeed, all of the above just tells us a nice story about the structure of the language of music: it’s a story about the tones, and how they are related to each other. That relation is, in essence, an exponential function with base 2. That’s all. Nothing more, nothing less. It’s remarkably simple and, at the same time, endlessly deep. 🙂 But so it is not a story about the structure of a musical piece itself, of a pop song of Ellie Goulding, for instance, or one of Bach’s preludes or fugues.

That brings me back to the original question I raised in my previous post. It’s a question which was triggered, long time ago, when I tried to read Douglas Hofstadter‘s Gödel, Escher and Bach, frustrated because my brother seemed to understand it, and I didn’t. So I put it down, and never ever looked at it again. So what is it really about that famous piece of Bach?

Frankly, I still amn’t sure. As I mentioned in my previous post, musicians were struggling to find a tuning system that would allow them to easily transpose musical compositions. Transposing music amounts to changing the so-called key of a musical piece, so that’s moving the whole piece up or down in pitch by some constant interval that is not equal to an octave. It’s a piece of cake now. In fact, increasing or decreasing the playback speed of a recording also amounts to transposing a piece: a increase or decrease of the playback speed by 6% will shift the pitch up or down by about one semitone. Why? Well… Go back to what I wrote above about that 12th root of 2. We’ve got the right tuning system now, and so everything is easy. Logarithms are great! 🙂

Back to Bach. Despite their admiration for the Greek ideas around aesthetics – and, most notably, their fascination with harmonic ratios! – (almost) all Renaissance musicians were struggling with the so-called Pythagorean tuning system, which was used until the 18th century and which was based on a correct observation (similar strings, under the same tension but differing in length, sound ‘pleasant’ when sounded together if – and only if – the ratio of the length of the strings is like 1:2, 2:3, 3:4, 3:5, 4:5, etcetera) but a wrong conclusion (the frequencies of musical tones should also obey the same harmonic ratios), and Bach’s so-called ‘good’ temperament tuning system was designed such that the piece could, indeed, be played in most keys without sounding… well… out of tune. 🙂

Having said that, the modern ‘equal temperament’ tuning system, which prescribes that tuning should be done such that the notes are in the above-described simple logarithmic relation to each other, had already been invented. So the true question is: why didn’t Bach embrace it? Why did he stick to ratios? Why did it take so long for the right system to be accepted?

I don’t know. If you google, you’ll find a zillion of possible explanations. As far as I can see, most are all rather mystic. More importantly, most of them do not mention many facts. My explanation is rather simple: while Bach was, obviously, a musical genius, he may not have understood what an exponential, or a logarithm, is all about. Indeed, a quick read of summary biographies reveals that Bach studied a wide range of topics, like Latin and Greek, and theology—of course! But math is not mentioned. He didn’t write about tuning and all that: all of his time went to writing musical masterpieces!

What the biographies do mention is that he always found other people’s tunings unsatisfactory, and that he tuned his harpsichords and clavichords himself. Now that is quite revealing, I’d say! In my view, Bach couldn’t care less about the ratios. He knew something was wrong with the Pythagorean system (or the variants as were then used, which are referred to as meantone temperament) and, as a musical genius, he probably ended up tuning by ear. [For those who’d wonder what I am talking about, let me quickly insert a Wikipedia graph illustrating the difference between the Pythagorean system (and two of these meantone variants) and the equal temperament tuning system in use today.]

So… What’s the point I am trying to make? Well… Frankly, I’d bet Bach’s own tuning was actually equal temperament, and so he should have named his masterpiece Das gleichtemperierte Klavier. Then we wouldn’t have all that ‘noise’ around it. 🙂

Post scriptum: Did you like the argument on the respective energy levels of the harmonics of an ideal string? Too bad. It’s wrong. I made a common mistake: when substituting variables in the integral, I ‘forgot’ to substitute the lower and upper bound of the interval over which I was integrating the function. The calculation below corrects the mistake, and so it does the required substitutions—for the first three modes at least. What’s going on here? Well… Nothing much… I just integrate over the length L taking a snapshot at t = 0 (as mentioned, we can always shift the origin of our independent variable, so here we do it for time and so it’s OK). Hence, the argument of our wave function sin(kx−ωt) reduces to kx, with k = 2π/λ, and λ= 2L, λ = L, λ= (2/3)·L for the first, second and third mode respectively. [As for solving the integral of the sine squared, you can google the formula, and please do check my substitutions. They should be OK, but… Well… We never know, do we? :-)]

[…] No… This doesn’t make all that much sense either. Those integrals yield the same energy for all three modes. Something must be wrong: shorter wavelengths (i.e. higher frequencies) are associated with higher energy levels. Full stop. So the ‘solution’ above can’t be right… […] You’re right. That’s where the time aspect comes into play. We were taking a snapshot, indeed, and the mean value of the sine squared function is 1/2 = 0.5, as should be clear from Pythagoras’ theorem: cos²x + sin²x = 1. So what I was doing is like integrating a constant function over the same-length interval. So… Well… Yes: no wonder I get the same value again and again.

[…]

We need to integrate over the same time interval. You could do that, as an exercise, but there’s a more direct approach to it: the energy of a wave is directly proportional to its frequency, so we write: E ∼ f. If the frequency doubles, triples, quadruples etcetera, then its energy doubles, triples, quadruples etcetera too. But – remember – we’re talking one string only here, with a fixed wave speed c = λ·f – so f = c/λ (read: the frequency is inversely proportional to the wavelength) – and, therefore (assuming the same (maximum) amplitude), we get that the energy level of each mode is inversely proportional to the wavelength, so we find that E ∼ 1/f.

Now, with direct or inverse proportionality relations, we can always invent some new unit that makes the relationship an identity, so let’s do that and turn it into an equation indeed. [And, yes, sorry… I apologize again to your old math teacher: he may not quite agree with the shortcut I am taking here, but he’ll justify the logic behind.] So… Remembering that λ₁ = 2L, λ₂ = L, λ₃ = (2/3)·L, etcetera, we can then write:

E₁ = (1/2)/L, E₂ = (2/2)/L, E₃ = (3/2)/L, E₄ = (4/2)/L, E₅ = (5/2)/L,…, E_n = (n/2)/L,…

That’s a really nice result, because… Well… In quantum theory, we have this so-called equipartition theorem, which says that the permitted energy levels of a harmonic oscillator are equally spaced, with the interval between them equal to h or ħ (if you use the angular frequency to describe a wave (so that’s ω = 2π·f), then Planck’s constant (h) becomes ħ = h/2π). So here we’ve got equipartition too, with the interval between the various energy levels equal to (1/2)/L.

You’ll say: So what? Frankly, if this doesn’t amaze you, stop reading—but if this doesn’t amaze you, you actually stopped reading a long time ago. 🙂 Look at what we’ve got here. We didn’t specify anything about that string, so we didn’t care about its materials or diameter or tension or how it was made (a wound guitar string is a terribly complicated thing!) or about whatever. Still, we know its fundamental (or normal) modes, and their frequency or nodes or energy or whatever depend on the length of the string only, with the ‘fundamental’ unit of energy being equal to the reciprocal length. Full stop. So all is just a matter of size and proportions. In other words, it’s all about structure. Absolute measurements don’t matter.

You may say: Bull****. What’s the conclusion? You still didn’t tell me anything about how the total energy of the wave is supposed to be distributed over its normal modes!

That’s true. I didn’t. Why? Well… I am not sure, really. I presented a lot of stuff here, but I did not present a clear and unambiguous answer as to how the total energy of a string is distributed over its modes. Not for actual strings, nor for ideal strings. Let me be honest: I don’t know. I really don’t. Having said that, my guts instinct that most of the energy – of, let’s say, a C4 note – should be in the primary mode (i.e. in the fundamental frequency) must be right: otherwise we would not call it a C4 note. So let’s try to make some assumptions. However, before doing so, let’s first briefly touch base with reality.

For actual strings (or actual musical sounds), I suspect the analysis can be quite complicated, as evidenced by the following illustration, which I took from one of the many interesting sites on this topic. Let me quote the author: “A flute is essentially a tube that is open at both ends. Air is blown across one end and sound comes out the other. The harmonics are all whole number multiples of the fundamental frequency (436 Hz, a slightly flat A₄ — a bit lower in frequency than is normally acceptable). Note how the second harmonic is nearly as intense as the fundamental. [My = blog writer’s 🙂 italics] This strong second harmonic is part of what makes a flute sound like a flute.”

Hmmm… What I see in the graph is a first harmonic that is actually more intense than its fundamental, so what’s that all about? So can we actually associate a specific frequency to that tone? Not sure. So we’re in trouble already.

If reality doesn’t match our thinking, what about ideality? Hmmm… What to say? As for ideal strings – or ideal flutes 🙂 – I’d venture to say that the most obvious distribution of energy over the various modes (or harmonics, when we’re talking sound) would is the Boltzmann distribution.

Huh? Yes. Have a look at one of my posts on statistical mechanics. It’s a weird thing: the distribution of molecular speeds in a gas, or the density of the air in the atmosphere, or whatever involving many particles and/or a great degree of complexity (so many, or such a degree of complexity, that only some kind of statistical approach to the problem works—all that involves Boltzmann’s Law, which basically says the distribution function will be a function of the energy levels involved: f = e^–energy. So… Well… Yes. It’s the logarithmic scale again. It seems to govern the Universe. 🙂

Huh? Yes. That’s why I think: the distribution of the total energy of the oscillation should be some Boltzmann function, so it should depend on the energy of the modes: most of the energy will be in the lower modes, and most of the most in the fundamental. […] Hmmm… It again begs the question: how much exactly?

Well… The Boltzmann distribution strongly resembles the ‘harmonic’ distribution shown above (1, 1/2, 1/3, 1/4 etc), but it’s not quite the same. The graph below shows how they are similar and dissimilar in shape. You can experiment yourself with coefficients and all that, but your conclusion will be the same. As they say in Asia: they are “same-same but different.” 🙂 […] It’s like the ‘good’ and ‘equal’ temperament used when tuning musical instruments: the ‘good’ temperament – which is based on harmonic ratios – is good, but not good enough. Only the ‘equal’ temperament obeys the logarithmic scale and, therefore, is perfect. So, as I mentioned already, while my assumption isn’t quite right (the distribution is not harmonic, in the Pythagorean sense), the intuition behind is OK. So it’s just like Pythagoras’ number theory of the Universe. Having said that, I’ll leave it to you to draw the correct the conclusions from it. 🙂

Music and Physics

Original post:

My first working title for this post was Music and Modes. Yes. Modes. Not moods. The relation between music and moods is an interesting research topic as well but so it’s not what I am going to write about. 🙂

It started with me thinking I should write something on modes indeed, because the concept of a mode of a wave, or any oscillator really, is quite central to physics, both in classical physics as well as in quantum physics (quantum-mechanical systems are analyzed as oscillators too!). But I wondered how to approach it, as it’s a rather boring topic if you look at the math only. But then I was flying back from Europe, to Asia, where I live and, as I am also playing a bit of guitar, I suddenly wanted to know why we like music. And then I thought that’s a question you may have asked yourself at some point of time too! And so then I thought I should write about modes as part of a more interesting story: a story about music—or, to be precise, a story about the physics behind music. So… Let’s go for it.

Philosophy versus physics

There is, of course, a very simple answer to the question of why we like music: we like music because it is music. If it would not be music, we would not like it. That’s a rather philosophical answer, and it probably satisfies most people. However, for someone studying physics, that answer can surely not be sufficient. What’s the physics behind? I reviewed Feynman’s Lecture on sound waves in the plane, combined it with some other stuff I googled when I arrived, and then I wrote this post, which gives you a much less philosophical answer. 🙂

The observation at the center of the discussion is deceptively simple: why is it that similar strings (i.e. strings made of the same material, with the same thickness, etc), under the same tension but differing in length, sound ‘pleasant’ when sounded together if – and only if – the ratio of the length of the strings is like 1:2, 2:3, 3:4, 3:5, 4:5, etc (i.e. like whatever other ratio of two small integers)?

You probably wonder: is that the question, really? It is. The question is deceptively simple indeed because, as you will see in a moment, the answer is quite complicated. So complicated, in fact, that the Pythagoreans didn’t have any answer. Nor did anyone else for that matter—until the 18th century or so, when musicians, physicists and mathematicians alike started to realize that a string (of a guitar, or a piano, or whatever instrument Pythagoras was thinking of at the time), or a column of air (in a pipe organ or a trumpet, for example), or whatever other thing that actually creates the musical tone, actually oscillates at numerous frequencies simultaneously.

The Pythagoreans did not suspect that a string, in itself, is a rather complicated thing – something which physicists refer to as a harmonic oscillator – and that its sound, therefore, is actually produced by many frequencies, instead of only one. The concept of a pure note, i.e. a tone that is free of harmonics (i.e. free of all other frequencies, except for the fundamental frequency) also didn’t exist at the time. And if it did, they would not have been able to produce a pure tone anyway: producing pure tones – or notes, as I’ll call them, somewhat inaccurately (I should say: a pure pitch) – is remarkably complicated, and they do not exist in Nature. If the Pythagoreans would have been able to produce pure tones, they would have observed that pure tones do not give any sensation of consonance or dissonance if their relative frequencies respect those simple ratios. Indeed, repeated experiments, in which such pure tones are being produced, have shown that human beings can’t really say whether it’s a musical sound or not: it’s just sound, and it’s neither pleasant (or consonant, we should say) or unpleasant (i.e. dissonant).

The Pythagorean observation is valid, however, for actual (i.e. non-pure) musical tones. In short, we need to distinguish between tones and notes (i.e. pure tones): they are two very different things, and the gist of the whole argument is that musical tones coming out of one (or more) string(s) under tension are full of harmonics and, as I’ll explain in a minute, that’s what explains the observed relation between the lengths of those strings and the phenomenon of consonance (i.e. sounding ‘pleasant’) or dissonance (i.e. sounding ‘unpleasant’).

Of course, it’s easy to say what I say above: we’re 2015 now, and so we have the benefit of hindsight. Back then – so that’s more than 2,500 years ago! – the simple but remarkable fact that the lengths of similar strings should respect some simple ratio if they are to sound ‘nice’ together, triggered a fascination with number theory (in fact, the Pythagoreans actually established the foundations of what is now known as number theory). Indeed, Pythagoras felt that similar relationships should also hold for other natural phenomena! To mention just one example, the Pythagoreans also believed that the orbits of the planets would also respect such simple numerical relationships, which is why they talked of the ‘music of the spheres’ (Musica Universalis).

We now know that the Pythagoreans were wrong. The proportions in the movements of the planets around the Sun do not respect simple ratios and, with the benefit of hindsight once again, it is regrettable that it took many courageous and brilliant people, such as Galileo Galilei and Copernicus, to convince the Church of that fact. 😦 Also, while Pythagoras’ observations in regard to the sounds coming out of whatever strings he was looking at were correct, his conclusions were wrong: the observation does not imply that the frequencies of musical notes should all be in some simple ratio one to another.

Let me repeat what I wrote above: the frequencies of musical notes are not in some simple relationship one to another. The frequency scale for all musical tones is logarithmic and, while that implies that we can, effectively, do some tricks with ratios based on the properties of the logarithmic scale (as I’ll explain in a moment), the so-called ‘Pythagorean’ tuning system, which is based on simple ratios, was plain wrong, even if it – or some variant of it (instead of the 3:2 ratio, musicians used the 5:4 ratio from about 1510 onwards) – was generally used until the 18th century! In short, Pythagoras was wrong indeed—in this regard at least: we can’t do much with those simple ratios.

Having said that, Pythagoras’ basic intuition was right, and that intuition is still very much what drives physics today: it’s the idea that Nature can be described, or explained (whatever that means), by quantitative relationships only. Let’s have a look at how it actually works for music.

Tones, noise and notes

Let’s first define and distinguish tones and notes. A musical tone is the opposite of noise, and the difference between the two is that musical tones are periodic waveforms, so they have a period T, as illustrated below. In contrast, noise is a non-periodic waveform. It’s as simple as that.

Now, from previous posts, you know we can write any period function as the sum of a potentially infinite number of simple harmonic functions, and that this sum is referred to as the Fourier series. I am just noting it here, so don’t worry about it as for now. I’ll come back to it later.

You also know we have seven musical notes: Do-Re-Mi-Fa-Sol-La-Si or, more common in the English-speaking world, A-B-C-D-E-F-G. And then it starts again with A (or Do). So we have two notes, separated by an interval which is referred to as an octave (from the Greek octo, i.e. eight), with six notes in-between, so that’s eight notes in total. However, you also know that there are notes in-between, except between E and F and between B and C. They are referred to as semitones or half-steps. I prefer the term ‘half-step’ over ‘semitone’, because we’re talking notes really, not tones.

We have, for example, F–sharp (denoted by F#), which we can also call G-flat (denoted by Gb). It’s the same thing: a sharp # raises a note by a semitone (aka half-step), and a flat b lowers it by the same amount, so F# is Gb. That’s what shown below: in an octave, we have eight notes but twelve half-steps.

Let’s now look at the frequencies. The frequency scale above (expressed in oscillations per second, so that’s the hertz unit) is a logarithmic scale: frequencies double as we go from one octave to another: the frequency of the C4 note above (the so-called middle C) is 261.626 Hz, while the frequency of the next C note (C5) is double that: 523.251 Hz. [Just in case you’d want to know: the 4 and 5 number refer to its position on a standard 88-key piano keyboard: C4 is the fourth C key on the piano.]

Now, if we equate the interval between C4 and C5 with 1 (so the octave is our musical ‘unit’), then the interval between the twelve half-steps is, obviously, 1/12. Why? Because we have 12 halve-steps in our musical unit. You can also easily verify that, because of the way logarithms work, the ratio of the frequencies of two notes that are separated by one half-step (between D# and E, for example) will be equal to 2^1/12. Likewise, the ratio of the frequencies of two notes that are separated by n half-steps is equal to 2^n/12. [In case you’d doubt, just do an example. For instance, if we’d denote the frequency of C4 as f₀, and the frequency of C# as f₁ and so on (so the frequency of D is f₂, the frequency of C5 is f₁₂, and everything else is in-between), then we can write the f₂/f₀ratio as f₂/f₀= ( f₂/f₁)(f₁/f₀) = 2^1/12·2^1/12 = 2^2/12= 2^1/6. I must assume you’re smart enough to generalize this result yourself, and that f₁₂/f₀is, obviously, equal to 2^12/12=2¹ = 2, which is what it should be!]

Now, because the frequencies of the various C notes are expressed as a number involving some decimal fraction (like 523.251 Hz, and the 0.251 is actually an approximation only), and because they are, therefore, a bit hard to read and/or work with, I’ll illustrate the next idea – i.e. the concept of harmonics – with the A instead of the C. 🙂

Harmonics

The lowest A on a piano is denoted by A0, and its frequency is 27.5 Hz. Lower A notes exist (we have one at 13.75 Hz, for instance) but we don’t use them, because they are near (or actually beyond) the limit of the lowest frequencies we can hear. So let’s stick to our grand piano and start with that 27.5 Hz frequency. The next A note is A1, and its frequency is 55 Hz. We then have A2, which is like the A on my (or your) guitar: its frequency is equal to 2×55 = 110 Hz. The next is A3, for which we double the frequency once again: we’re at 220 Hz now. The next one is the A in the illustration of the C scale above: A4, with a frequency of 440 Hz.

[Let me, just for the record, note that the A4 note is the standard tuning pitch in Western music. Why? Well… There’s no good reason really, except convention. Indeed, we can derive the frequency of any other note from that A4 note using our formula for the ratio of frequencies but, because of the properties of a logarithmic function, we could do the same using whatever other note really. It’s an important point: there’s no such thing as an absolute reference point in music: once we define our musical ‘unit’ (so that’s the so-called octave in Western music), and how many steps we want to have in-between (so that’s 12 steps—again, in Western music, that is), we get all the rest. That’s just how logarithms work. So music is all about structure, i.e. mathematical relationships. Again, Pythagoras’ conclusions were wrong, but his intuition was right.]

Now, the notes we are talking about here are all so-called pure tones. In fact, when I say that the A on our guitar is referred to as A2 and that it has a frequency of 110 Hz, then I am actually making a huge simplification. Worse, I am lying when I say that: when you play a string on a guitar, or when you strike a key on a piano, all kinds of other frequencies – so-called harmonics – will resonate as well, and that’s what gives the quality to the sound: it’s what makes it sound beautiful. So the fundamental frequency (aka as first harmonic) is 110 Hz alright but we’ll also have second, third, fourth, etc harmonics with frequency 220 Hz, 330 Hz, 440 Hz, etcetera. In music, the basic or fundamental frequency is referred to as the pitch of the tone and, as you can see, I often use the term ‘note’ (or pure tone) as a synonym for pitch—which is more or less OK, but not quite correct actually. [However, don’t worry about it: my sloppiness here does not affect the argument.]

What’s the physics behind? Look at the illustration below (I borrowed it from the Physics Classroom site). The thick black line is the string, and the wavelength of its fundamental frequency (i.e. the first harmonic) is twice its length, so we write λ₁ = 2·L or, the other way around, L = (1/2)·λ₁. Now that’s the so-called first mode of the string. [One often sees the term fundamental or natural or normal mode, but the adjective is not necessary really. In fact, I find it confusing, although I sometimes find myself using it too.]

We also have a second, third, etc mode, depicted below, and these modes correspond to the second, third, etc harmonic respectively.

For the second, third, etc mode, the relationship between the wavelength and the length of the string is, obviously, the following: L = (2/2)·λ₂= λ₂, L = L = (3/2)·λ₃, etc. More in general, for the n^th mode, L will be equal to L = (n/2)·λ_n, with n = 1, 2, etcetera. In fact, because L is supposed to be some fixed length, we should write it the other way around: λ_n = (2/n)·L.

What does it imply for the frequencies? We know that the speed of the wave – let’s denote it by c – as it travels up and down the string, is a property of the string, and it’s a property of the string only. In other words, it does not depend on the frequency. Now, the wave velocity is equal to the frequency times the wavelength, always, so we have c = f·λ. To take the example of the (classical) guitar string: its length is 650 mm, i.e. 0.65 m. Hence, the identities λ₁ = (2/1)·L, λ₂ = (2/2)·L, λ₃ = (2/3)·L etc become λ₁ = (2/1)·0.65 = 1.3 m, λ₂ = (2/2)·0.65 = 0.65 m, λ₃ = (2/3)·0.65 = 0.433.. m and so on. Now, combining these wavelengths with the above-mentioned frequencies, we get the wave velocity c = (110 Hz)·(1.3 m) = (220 Hz)·(0.65 m) = (330 Hz)·(0.433.. m) = 143 m/s.

Let me now get back to Pythagoras’ string. You should note that the frequencies of the harmonics produced by a simple guitar string are related to each other by simple whole number ratios. Indeed, the frequencies of the first and second harmonics are in a simple 2 to 1 ratio (2:1). The second and third harmonics have a 3:2 frequency ratio. The third and fourth harmonics a 4:3 ratio. The fifth and fourth harmonic 5:4, and so on and so on. They have to be. Why? Because the harmonics are simple multiples of the basic frequency. Now that is what’s really behind Pythagoras’ observation: when he was sounding similar strings with the same tension but different lengths, he was making sounds with the same harmonics. Nothing more, nothing less.

Let me be quite explicit here, because the point that I am trying to make here is somewhat subtle. Pythagoras’ string is Pythagoras’ string: he talked similar strings. So we’re not talking some actual guitar or a piano or whatever other string instrument. The strings on (modern) string instruments are not similar, and they do not have the same tension. For example, the six strings of a guitar strings do not differ in length (they’re all 650 mm) but they’re different in tension. The six strings on a classical guitar also have a different diameter, and the first three strings are plain strings, as opposed to the bottom strings, which are wound. So the strings are not similar but very different indeed. To illustrate the point, I copied the values below for just one of the many commercially available guitar string sets. It’s the same for piano strings. While they are somewhat more simple (they’re all made of piano wire, which is very high quality steel wire basically), they also differ—not only in length but in diameter as well, typically ranging from 0.85 mm for the highest treble strings to 8.5 mm (so that’s ten times 0.85 mm) for the lowest bass notes.

In short, Pythagoras was not playing the guitar or the piano (or whatever other more sophisticated string instrument that the Greeks surely must have had too) when he was thinking of these harmonic relationships. The physical explanation behind his famous observation is, therefore, quite simple: musical tones that have the same harmonics sound pleasant, or consonant, we should say—from the Latin con-sonare, which, literally, means ‘to sound together’ (from sonare = to sound and con = with). And otherwise… Well… Then they do not sound pleasant: they are dissonant.

To drive the point home, let me emphasize that, when we’re plucking a string, we produce a sound consisting of many frequencies, all in one go. One can see it in practice: if you strike a lower A string on a piano – let’s say the 110 Hz A2 string – then its second harmonic (220 Hz) will make the A3 string vibrate too, because it’s got the same frequency! And then its fourth harmonic will make the A4 string vibrate too, because they’re both at 440 Hz. Of course, the strength of these other vibrations (or their amplitude we should say) will depend on the strength of the other harmonics and we should, of course, expect that the fundamental frequency (i.e. the first harmonic) will absorb most of the energy. So we pluck one string, and so we’ve got one sound, one tone only, but numerous notes at the same time!

In this regard, you should also note that the third harmonic of our 110 Hz A2 string corresponds to the fundamental frequency of the E4 tone: both are 330 Hz! And, of course, the harmonics of E, such as its second harmonic (2·330 Hz = 660 Hz) correspond to higher harmonics of A too! To be specific, the second harmonic of our E string is equal to the sixth harmonic of our A2 string. If your guitar is any good, and if your strings are of reasonable quality too, you’ll actually see it: the (lower) E and A strings co-vibrate if you play the A major chord, but by striking the upper four strings only. So we’ve got energy – motion really – being transferred from the four strings you do strike to the two strings you do not strike! You’ll say: so what? Well… If you’ve got any better proof of the actuality (or reality) of various frequencies being present at the same time, please tell me! 🙂

So that’s why A and E sound very well together (A, E and C#, played together, make up the so-called A major chord): our ear likes matching harmonics. And so that why we like musical tones—or why we define those tones as being musical! 🙂 Let me summarize it once more: musical tones are composite sound waves, consisting of a fundamental frequency and so-called harmonics (so we’ve got many notes or pure tones altogether in one musical tone). Now, when other musical tones have harmonics that are shared, and we sound those notes too, we get the sensation of harmony, i.e. the combination sounds consonant.

Now, i’s not difficult to see that we will always have such shared harmonics if we have similar strings, with the same tension but different lengths, being sounded together. In short, what Pythagoras observed has nothing much to do with notes, but with tones. Let’s go a bit further in the analysis now by introducing some more math. And, yes, I am very sorry: it’s the dreaded Fourier analysis indeed! 🙂

Fourier analysis

You know that we can decompose any periodic function into a sum of a (potentially infinite) series of simple sinusoidal functions, as illustrated below. I took the illustration from Wikipedia: the red function s₆(x) is the sum of six sine functions of different amplitudes and (harmonically related) frequencies. The so-called Fourier transform S(f) (in blue) relates the six frequencies with the respective amplitudes.

In light of the discussion above, it is easy to see what this means for the sound coming from a plucked string. Using the angular frequency notation (so we write everything using ω instead of f), we know that the normal or natural modes of oscillation have frequencies ω = 2π/T = 2πf (so that’s the fundamental frequency or first harmonic), 2ω (second harmonic), 3ω (third harmonic), and so on and so on.

Now, there’s no reason to assume that all of the sinusoidal functions that make up our tone should have the same phase: some phase shift Φ may be there and, hence, we should write our sinusoidal function not as cos(ωt), but as cos(ωt + Φ) in order to ensure our analysis is general enough. [Why not a sine function? It doesn’t matter: the cosine and sine function are the same, except for another phase shift of 90° = π/2.] Now, from our geometry classes, we know that we can re-write cos(ωt + Φ) as

cos(ωt + Φ) = [cos(Φ)cos(ωt) – sin(Φ)sin(ωt)]

We have a lot of these functions of course – one for each harmonic, in fact – and, hence, we should use subscripts, which is what we do in the formula below, which says that any function f(t) that is periodic with the period T can be written mathematically as:

You may wonder: what’s that period T? It’s the period of the fundamental mode, i.e. the first harmonic. Indeed, the period of the second, third, etc harmonic will only be one half, one third etcetera of the period of the first harmonic. Indeed, T₂ = (2π)/(2ω) = (1/2)·(2π)/ω = (1/2)·T₁, and T₃ = (2π)/(3ω) = (1/3)·(2π)/ω = (1/3)·T₁, and so on. However, it’s easy to see that these functions also repeat themselves after two, three, etc periods respectively. So all is alright, and the general idea behind the Fourier analysis is further illustrated below. [Note that both the formula as well as the illustration below (which I took from Feynman’s Lectures) add a ‘zero-frequency term’ a₀ to the series. That zero-frequency term will usually be zero for a musical tone, because the ‘zero’ level of our tone will be zero indeed. Also note that the a_n and b_n coefficients are, of course, equal to a_n = cos Φ_nand b_n= –sinΦ_n, so you can relate the illustration and the formula easily.]

You’ll say: What the heck! Why do we need the mathematical gymnastics here? It’s just to understand that other characteristic of a musical tone: its quality (as opposed to its pitch). A so-called rich tone will have strong harmonics, while a pure tone will only have the first harmonic. All other characteristics – the difference between a tone produced by a violin as opposed to a piano – are then related to the ‘mix’ of all those harmonics.

So we have it all now, except for loudness which is, of course, related to the magnitude of the air pressure changes as our waveform moves through the air: pitch, loudness and quality. that’s what makes a musical tone. 🙂

Dissonance

As mentioned above, if the sounds are not consonant, they’re dissonant. But what is dissonance really? What’s going on? The answer is the following: when two frequencies are near to a simple fraction, but not exact, we get so-called beats, which our ear does not like.

Huh? Relax. The illustration below, which I copied from the Wikipedia article on piano tuning, illustrates the phenomenon. The blue wave is the sum of the red and the green wave, which are originally identical. But then the frequency of the green wave is increased, and so the two waves are no longer in phase, and the interference results in a beating pattern. Of course, our musical tone involves different frequencies and, hence, different periods T₁,T₂, T₃etcetera, but you get the idea: the higher harmonics also oscillate with period T₁, and if the frequencies are not in some exact ratio, then we’ll have a similar problem: beats, and our ear will not like the sound.

Of course, you’ll wonder: why don’t we like beats in tones? We can ask that, can’t we? It’s like asking why we like music, isn’t it? […] Well… It is and it isn’t. It’s like asking why our ear (or our brain) likes harmonics. We don’t know. That’s how we are wired. The ‘physical’ explanation of what is musical and what isn’t only goes so far, I guess. 😦

Pythagoras versus Bach

From all of what I wrote above, it is obvious that the frequencies of the harmonics of a musical tone are, indeed, related by simple ratios of small integers: the frequencies of the first and second harmonics are in a simple 2 to 1 ratio (2:1); the second and third harmonics have a 3:2 frequency ratio; the third and fourth harmonics a 4:3 ratio; the fifth and fourth harmonic 5:4, etcetera. That’s it. Nothing more, nothing less.

In other words, Pythagoras was observing musical tones: he could not observe the pure tones behind, i.e. the actual notes. However, aesthetics led Pythagoras, and all musicians after him – until the mid-18th century – to also think that the ratio of the frequencies of the notes within an octave should also be simple ratios. From what I explained above, it’s obvious that it should not work that way: the ratio of the frequencies of two notes separated by n half-steps is 2^n/12, and, for most values of n, 2^n/12 is not some simple ratio. [Why? Just take your pocket calculator and calculate the value of 2^1/12: it’s 2^0.08333… = 1.0594630943… and so on… It’s an irrational number: there are no repeating decimals. Now, 2^n/12 is equal to 2^1/12·2^1/12·…·2^1/12 (n times). Why would you expect that product to be equal to some simple ratio?]

So – I said it already – Pythagoras was wrong—not only in this but also in other regards, such as when he espoused his views on the solar system, for example. Again, I am sorry to have to say that, but it is what is: the Pythagoreans did seem to prefer mathematical ideas over physical experiment. 🙂 Having said that, musicians obviously didn’t know about any alternative to Pythagoras, and they had surely never heard about logarithmic scales at the time. So… Well… They did use the so-called Pythagorean tuning system. To be precise, they tuned their instruments by equating the frequency ratio between the first and the fifth tone in the C scale (i.e. the C and G, as they did not include the C#, D# and F# semitones when counting) with the ratio 3/2, and then they used other so-called harmonic ratios for the notes in-between.

Now, the 3/2 ratio is actually almost correct, because the actual frequency ratio is 2^7/12 (we have seven tones, including the semitones—not five!), and so that’s 1.4983, approximately. Now, that’s pretty close to 3/2 = 1.5, I’d say. 🙂 Using that approximation (which, I admit, is fairly accurate indeed), the tuning of the other strings would then also be done assuming certain ratios should be respected, like the ones below.

So it was all quite good. Having said that, good musicians, and some great mathematicians, felt something was wrong—if only because there were several so-called just intonation systems around (for an overview, check out the Wikipedia article on just intonation). More importantly, they felt it was quite difficult to transpose music using the Pythagorean tuning system. Transposing music amounts to changing the so-called key of a musical piece: what one does, basically, is moving the whole piece up or down in pitch by some constant interval that is not equal to an octave. Today, transposing music is a piece of cake—Western music at least. But that’s only because all Western music is played on instruments that are tuned using that logarithmic scale (technically, it’s referred to as the 12-tone equal temperament (12-TET) system). When you’d use one of the Pythagorean systems for tuning, a transposed piece does not sound quite right.

The first mathematician who really seemed to know what was wrong (and, hence, who also knew what to do) was Simon Stevin, who wrote a manuscript based on the ’12^throot of 2 principle’ around AD 1600. It shouldn’t surprise us: the thinking of this mathematician from Bruges would inspire John Napier’s work on logarithms. Unfortunately, while that manuscript describes the basic principles behind the 12-TET system, it didn’t get published (Stevin had to run away from Bruges, to Holland, because he was protestant and the Spanish rulers at the time didn’t like that). Hence, musicians, while not quite understanding the math (or the physics, I should say) behind their own music, kept trying other tuning systems, as they felt it made their music sound better indeed.

One of these ‘other systems’ is the so-called ‘good’ temperament, which you surely heard about, as it’s referred to in Bach’s famous composition, Das Wohltemperierte Klavier, which he finalized in the first half of the 18th century. What is that ‘good’ temperament really? Well… It is what it is: it’s one of those tuning systems which made musicians feel better about their music for a number of reasons, all of which are well described in the Wikipedia article on it. But the main reason is that the tuning system that Bach recommended was a great deal better when it came to playing the same piece in another key. However, it still wasn’t quite right, as it wasn’t the equal temperament system (i.e. the 12-TET system) that’s in place now (in the West at least—the Indian music scale, for instance, is still based on simple ratios).

Why do I mention this piece of Bach? The reason is simple: you probably heard of it because it’s one of the main reference points in a rather famous book: Gödel, Escher and Bach—an Eternal Golden Braid. If not, then just forget about it. I am mentioning it because one of my brothers loves it. It’s on artificial intelligence. I haven’t read it, but I must assume Bach’s master piece is analyzed there because of its structure, not because of the tuning system that one’s supposed to use when playing it. So… Well… I’d say: don’t make that composition any more mystic than it already is. 🙂 The ‘magic’ behind it is related to what I said about A4 being the ‘reference point’ in music: since we’re using a universal logarithmic scale now, there’s no such thing as an absolute reference point any more: once we define our musical ‘unit’ (so that’s the so-called octave in Western music), and also define how many steps we want to have in-between (so that’s 12—in Western music, that is), we get all the rest. That’s just how logarithms work.

So, in short, music is all about structure, i.e. it’s all about mathematical relations, and about mathematical relations only. Again, Pythagoras’ conclusions were wrong, but his intuition was right. And, of course, it’s his intuition that gave birth to science: the simple ‘models’ he made – of how notes are supposed to be related to each other, or about our solar system – were, obviously, just the start of it all. And what a great start it was! Looking back once again, it’s rather sad conservative forces (such as the Church) often got in the way of progress. In fact, I suddenly wonder: if scientists would not have been bothered by those conservative forces, could mankind have sent people around the time that Charles V was born, i.e. around A.D. 1500 already? 🙂

Post scriptum: My example of the the (lower) E and A guitar strings co-vibrating when playing the A major chord striking the upper four strings only, is somewhat tricky. The (lower) E and A strings are associated with lower pitches, and we said overtones (i.e. the second, third, fourth, etc harmonics) are multiples of the fundamental frequency. So why is that the lower strings co-vibrate? The answer is easy: they oscillate at the higher frequencies only. If you have a guitar: just try it. The two strings you do not pluck do vibrate—and very visibly so, but the low fundamental frequencies that come out of them when you’d strike them, are not audible. In short, they resonate at the higher frequencies only. 🙂

The example that Feynman gives is much more straightforward: his example mentions the lower C (or A, B, etc) notes on a piano causing vibrations in the higher C strings (or the higher A, B, etc string respectively). For example, striking the C2 key (and, hence, the C2 string inside the piano) will make the (higher) C3 string vibrate too. But few of us have a grand piano at home, I guess. That’s why I prefer my guitar example. 🙂

Maxwell-Boltzmann, Bose-Einstein and Fermi-Dirac statistics

Pre-scriptum added much later: We have advanced much in our understanding since we wrote this post. If you are reading it because you want to understand more about the boson-fermion distinction, then you shouldn’t be here. The general distinction between bosons and fermions is a useless theoretical generalization which actually prevents you from understanding what is really going on. I am keeping this post online for documentation purposes only. It is interesting from a math point of view but you are not here to learn math, are you?

Jean Louis Van Belle, 20 May 2020

Original post:

I’ve discussed statistics, in the context of quantum mechanics, a couple of times already (see, for example, my post on amplitudes and statistics). However, I never took the time to properly explain those distribution functions which are referred to as the Maxwell-Boltzmann, Bose-Einstein and Fermi-Dirac distribution functions respectively. Let me try to do that now—without, hopefully, getting lost in too much math! It should be a nice piece, as it connects quantum mechanics with statistical mechanics, i.e. two topics I had nicely separated so far. 🙂

You know the Boltzmann Law now, which says that the probabilities of different conditions of energy are given by e^−energy/kT = 1/e^energy/kT. Different ‘conditions of energy’ can be anything: density, molecular speeds, momenta, whatever. The point is: we have some probability density function f, and it’s a function of the energy E, so we write:

f(E) = C·e^−energy/kT= C/e^energy/kT

C is just a normalization constant (all probabilities have to add up to one, so the integral of this function over its domain must be one), and k and T are also usual suspects: T is the (absolute) temperature, and k is the Boltzmann constant, which relates the temperate to the kinetic energy of the particles involved. We also know the shape of this function. For example, when we applied it to the density of the atmosphere at various heights (which are related to the potential energy, as P.E. = m·g·h), assuming constant temperature, we got the following graph. The shape of this graph is that of an exponential decay function (we’ll encounter it again, so just take a mental note of it).

graph

A more interesting application is the quantum-mechanical approach to the theory of gases, which I introduced in my previous post. To explain the behavior of gases under various conditions, we assumed that gas molecules are like oscillators but that they can only take on discrete levels of energy. [That’s what quantum theory is about!] We denoted the various energy levels, i.e. the energies of the various molecular states, by E₀, E₁, E₂,…, E_i,…, and if Boltzmann’s Law applies, then the probability of finding a molecule in the particular state E_i is proportional to e^−E_i /kT. We can then calculate the relative probabilities, i.e. the probability of being in state E_i, relative to the probability of being in state E₀, is:

P_i/P₀ = e^−E_i /kT/e^−E₀ /kT = e^{−(E_i–E₀)/kT}= 1/e^{(E_i–E₀)/kT}

Now, P_i obviously equals n_i/N, so it is the ratio of the number of molecules in state E_i (n_i) and the total number of molecules (N). Likewise, P₀ = n₀/N and, therefore, we can write:

n_i/n₀= e^{−(E_i−E₀)/kT}= 1/e^{(E_i–E₀)/kT}

This formulation is just another Boltzmann Law, but it’s nice in that it introduces the idea of a ground state, i.e. the state with the lowest energy level. We may or may not want to equate E₀ with zero. It doesn’t matter really: we can always shift all energies by some arbitrary constant because we get to choose the reference point for the potential energy.

So that’s the so-called Maxwell-Boltzmann distribution. Now, in my post on amplitudes and statistics, I had jotted down the formulas for the other distributions, i.e. the distributions when we’re not talking classical particles but fermions and/or bosons. As you know, fermions are particles governed by the Fermi exclusion principle: indistinguishable particles cannot be together in the same state. For bosons, it’s the other way around: having one in some quantum state actually increases the chance of finding another one there, and we can actually have an infinite number of them in the same state.

We also know that fermions and bosons are the real world: fermions are the matter-particles, bosons are the force-carriers, and our ‘Boltzmann particles’ are nothing but a classical approximation of the real world. Hence, even if we can’t see them in the actual world, the Fermi-Dirac and Bose-Einstein distributions are the real-world distributions. 🙂 Let me jot down the equations once again:

Fermi-Dirac (for fermions): f(E) = 1/[Ae^{(E − E_F)/kT}+ 1]

Bose-Einstein (for bosons): f(E) = 1/[Ae^E/kT− 1]

We’ve got some other normalization constant here (A), which we shouldn’t be too worried about—for the time being, that is. Now, to see how these distributions are different from the Maxwell-Boltzmann distribution (which we should re-write as f(E) = C·e^−E/kT = 1/[A·e^E/kT] so as to make all formulas directly comparable), we should just make a graph. Please go online to find a graph tool (I found a new one recently—really easy to use), and just do it. You’ll see they are all like that exponential decay function. However, in order to make a proper comparison, we would actually need to calculate the normalization coefficients and, for the Fermi energy, we would also need the Fermi energy E_F(note that, for simplicity, we did equate E₀ with zero). Now, we could give it a try, but it’s much easier to google and find an example online.

The HyperPhysics website of Georgia State University gives us one: the example assumes 6 particles and 9 energy levels, and the table and graph below compare the Maxwell-Boltzmann and Bose-Einstein distributions for the model.

Now that is an interesting example, isn’t it? In this example (but all depends on its assumptions, of course), the Maxwell-Boltzmann and Bose-Einstein distributions are almost identical. Having said that, we can clearly see that the lower energy states are, indeed, more probable with Bose-Einstein statistics than with the Maxwell-Boltzmann statistics. While the difference is not dramatic at all in this example, the difference does become very dramatic, in reality, with large numbers (i.e. high matter density) and, more importantly, at very low temperatures, at which bosons can condense into the lowest energy state. This phenomenon is referred to as Bose-Einstein condensation: it causes superfluidity and superconductivity, and it’s real indeed: it has been observed with supercooled He-4, which is not an everyday substance, but real nevertheless!

What about the Fermi-Dirac distribution for this example? The Fermi-Dirac distribution is given below: the lowest energy state is now less probable, the mid-range energies much more, and none of the six particles occupy any of the four highest energy levels. Again, while the difference is not dramatic in this example, it can become very dramatic, in reality, with large numbers (read: high matter density) and very low temperatures: at absolute zero, all of the possible energy states up to the Fermi energy level will be occupied, and all the levels above the Fermi energy will be vacant.

What can we make out of all of this? First, you may wonder why we actually have more than one particle in one state above: doesn’t that contradict the Fermi exclusion principle? No. We need to distinguish micro- and macro-states. In fact, the example assumes we’re talking electrons here, and so we can have two particles in each energy state—with opposite spin, however. At the same time, it’s true we cannot have three, or more, in any state. That results, in the example we’re looking at here, in five possible distributions only, as shown below.

The diagram is an interesting one: if the particles were to be classical particles, or bosons, then 26 combinations are possible, including the five Fermi-Dirac combinations, as shown above. Note the little numbers above the 26 possible combinations (e.g. 6, 20, 30,… 180): they are proportional to the likelihood of occurring under the Maxwell-Boltzmann assumption (so if we assume the particles are ‘classical’ particles). Let me introduce you to the math behind the example by using the diagram below, which shows three possible distributions/combinations (I know the terminology is quite confusing—sorry for that!).

If we could distinguish the particles, then we’d have 2002 micro-states, which is the total of all those little numbers on top of the combinations that are shown (6+60+180+…). However, the assumption is that we cannot distinguish the particles. Therefore, the first combination in the diagram above, with five particles in the zero energy state and one particle in state 9, occurs 6 times into 2002 and, hence, it has a probability of 6/2002 ≈ 0.003 only. In contrast, the second combination is 10 times more likely, and the third one is 30 times more likely! In any case, the point is, in the classical situation (and in the Bose-Einstein hypothesis as well), we have 26 possible macro-states, as opposed to 5 only for fermions, and so that leads to a very different density function. Capito?

No? Well, this blog is not a textbook on physics and, therefore, I should refer you to the mentioned site once again, which references a 1992 textbook on physics (Frank Blatt, Modern Physics, 1992) as the source of this example. However, I won’t do that: you’ll find the details in the Post Scriptum to this post. 🙂

Let’s first focus on the fundamental stuff, however. The most burning question is: if the real world consists of fermions and bosons, why is that that we only see the Maxwell-Boltzmann distribution in our actual (non-real?) world? 🙂 The answer is that both the Fermi-Dirac and Bose-Einstein distribution approach the Maxwell–Boltzmann distribution if higher temperatures and lower particle densities are involved. In other words, we cannot see the Fermi-Dirac distributions (all matter is fermionic, except for weird stuff like superfluid helium-4 at 1 or 2 degrees Kelvin), but they are there!

Let’s approach it mathematically: the most general formula, encompassing both Fermi-Dirac and Bose-Einstein statistics, is:

N_i(E_i) ∝ 1/[e^{(E_i − μ)/kT}± 1]

If you’d google, you’d find a formula involving an additional coefficient, g_i, which is the so-called degeneracy of the energy level E_i. I included it in the formula I used in the above-mentioned post of mine. However, I don’t want to make it any more complicated than it already is and, therefore, I omitted it this time. What you need to look at are the two terms in the denominator: e^{(E_i − μ)/kT}and ± 1.

From a math point of view, it is obvious that the values of e^{(E_i − μ)/kT}+ 1 (Fermi-Dirac) and e^{(E_i − μ)/kT}− 1 (Bose-Einstein) will approach each other if e^{(E_i − μ)/kT}is much larger than ±1, so if e^{(E_i − μ)/kT}>> 1. That’s the case, obviously, if the (E_i − μ)/kT ratio is large, so if (E_i − μ) >> kT. In fact, (E_i − μ) should, obviously, be much larger than kT for the lowest energy levels too! Now, the conditions under which that is the case are associated with the classical situation (such as a cylinder filled with gas, for example). Why?

Well… […] Again, I have to say that this blog can’t substitute for a proper textbook. Hence, I am afraid I have to leave it to you to do the necessary research to see why. 🙂 The non-mathematical approach is to simple note that quantum effects, i.e. the ±1 term, only apply if the concentration of particles is high enough. Indeed, quantum effects appear if the concentration of particles is higher than the so-called quantum concentration. Only when the quantum concentration is reached, particles will start interacting according to what they are, i.e. as bosons or as fermions. At higher temperature, that concentration will not be reached, except in massive objects such as a white dwarf (white dwarfs are stellar remnants with the mass like that of the Sun but a volume like that of the Earth). So, in general, we can say that at higher temperatures and at low concentration we will not have any quantum effects. That should settle the matter—as for now, at least.

You’ll have one last question: we derived Boltzmann’s Law from the kinetic theory of gases, but how do we derive that N_i(E_i) = 1/[Ae^{(E_i − μ)/kT}± 1] expression? Good question but, again, we’d need more than a few pages to explain that! The answer is: quantum mechanics, of course! Go check it out in Feynman’s third Volume of Lectures! 🙂

Post scriptum: combinations, permutations and multiplicity

The mentioned example from HyperPhysics is really interesting, if only because it shows you also need to master a bit of combinatorics to get into quantum mechanics. Let’s go through the basics. If we have n distinct objects, we can order hem in n! ways, with n! (read: n factorial) equal to n·(n–1)·(n–2)·…·3·2·1. Note that 0! is equal to 1, per definition. We’ll need that definition.

For example, a red, blue and green ball can be ordered in 3·2·1 = 6 ways. Each way is referred to as a permutation.

Besides permutations, we also have the concept of a k-permutation, which we can denote in a number of ways but let’s choose P(n, k). [The P stands for permutation here, not for probability.] P(n, k) is the number of ways to pick k objects out of a set of n objects. Again, the objects are supposed to be distinguishable. The formula is P(n, k) = n·(n–1)·(n–2)·…·(n–k+1) = n!/(n–k)!. That’s easy to understand intuitively: on your first pick you have n choices; on your second, n–1; on your third, n–2, etcetera. When n = k, we obviously get n! again.

There is a third concept: the k-combination (as opposed to the k-permutation), which we’ll denote by C(n, k). That’s when the order within our subset doesn’t matter: an ace, a queen and a jack taken out of some card deck are a queen, a jack, and an ace: we don’t care about the order. If we have k objects, there are k! ways of ordering them and, hence, we just have to divide P(n, k) by k! to get C(n, k). So we write: C(n, k) = P(n, k)/k! = n!/[(n–k)!k!]. You recognize C(n, k): it’s the binomial coeficient.

Now, the HyperPhysics example illustrating the three mentioned distributions (Maxwell-Boltzmann, Bose-Einstein and Fermi-Dirac) is a bit more complicated: we need to associate q energy levels with N particles. Every possible configuration is referred to as a micro-state, and the total number of possible micro-states is referred to as the multiplicity of the system, denoted by Ω(N, q). The formula for Ω(N, q) is another binomial coefficient: Ω(N, q) = (q+N–1)!/[q!(N–1)!]. Ω(N, q) = Ω(6, 9) = (9+6–1)!/[9!(6–1)!] = 2002.

In our example, however, we do not have distinct particles and, therefore, we only have 26 macro-states (as opposed to 2002 micro-states), which are also referred to, confusingly, as distributions or combinations.

Now, the number of micro-states associated with the same macro-state is given by yet another formula: it is equal to N!/[n₁!·n₂!·n₃!·…·n_q!], with n_i! the number of particles in level i. [See why we need the 0! = 1 definition? It ensures unoccupied states do not affect the calculation.] So that’s how we get those numbers 6, 60 and 180 for those three macro-states.

But how do we calculate those average numbers of particles for each energy level? In other words, how do we calculate the probability densities under the Maxwell-Boltzmann, Fermi-Dirac and Bose-Einstein hypothesis respectively?

For the Maxwell-Boltzmann distribution, we proceed as follows: for each energy level j (or E_j, I should say), we calculate n_j= ∑n_ij·P_i over all macro-states i. In this summation, we have n_ij, which is the number of particles in energy level j in micro-state i, while P_i is the probability of macro-state i as calculated by the ratio of (i) the number of micro-states associated with macro-state i and (ii) the total number of micro-states. For P_i, we gave the example of 3/2002 ≈ 0.3%. For 60 and 180, we get 60/2002 ≈ 3% and 180/2002 ≈ 9%. Calculating all the n_j‘s for j ranging from 1 to 9 should yield the numbers and the graph below indeed.

OK. That’s how it works for Maxwell-Boltzmann. Now, it is obvious that the Fermi-Dirac and the Bose-Einstein distribution should not be calculated in the same way because, if they were, they would not be different from the Maxwell-Boltzmann distribution! The trick is as follows.

For the Bose-Einstein distribution, we give all macro-states equal weight—so that’s a weight of one, as shown below. Hence, the probability P_i is, quite simply, 1/26 ≈ 3.85% for all 26 macro-states. So we use the same n_j= ∑n_ij·P_iformula but with P_i = 1/26.

Finally, I already explained how we get the Fermi-Dirac distribution: we can only have (i) one, (ii) two, or (iii) zero fermions for each energy level—not more than two! Hence, out of the 26 macro-states, only five are actually possible under the Fermi-Dirac hypothesis, as illustrated below once more. So it’s a very different distribution indeed!

Now, you’ll probably still have questions. For example, why does the assumption, for the Bose-Einstein analysis, that macro-states have equal probability favor the lower energy states? The answer is that the model also integrates other constraints: first, when associating a particle with an energy level, we do not favor one energy level over another, so all energy levels have equal probability. However, at the same time, the whole system has some fixed energy level, and so we cannot put the particles in the higher energy levels only! At the same time, we know that, if we have q particles, and the probability of a particle having some energy level j is the same for all j, then they are likely not to be all at the same energy level: they’ll be distributed, effectively, as evidenced by the very low chance (0.3% only) of having 5 particles in the ground state and 1 particle at a higher level, as opposed to the 3% and 9% chance of the other two combinations shown in that diagram with three possible Maxwell-Boltzmann (MB) combinations.

So what happens when assigning an equal probability to all 26 possible combinations (with value 1/26) is that the combinations that were previously rather unlikely – because they did have a rather heavy concentration of particles in the ground state only – are now much more likely. So that’s why the Bose-Einstein distribution, in this example at least, is skewed towards the lowest energy level—as compared to the Maxwell-Boltzmann distribution, that is.

So that’s what’s behind, and that should also answer the other question you surely have when looking at those five acceptable Fermi-Dirac configurations: why don’t we have the same five configurations starting from the top down, rather than from the bottom up? Now you know: such configuration would have much higher energy overall, and so that’s not allowed under this particular model.

There’s also this other question: we said the particles were indistinguishable, but so then we suddenly say there can be two at any energy level, because their spin is opposite. It’s obvious this is rather ad hoc as well. However, if we’d allow only one particle at any energy level, we’d have no allowable combinations and, hence, we’d have no Fermi-Dirac distribution at all in this example.

In short, the example is rather intuitive, which is actually why I like it so much: it shows how bosonic and fermionic behavior appear rather gradually, as a consequence of variables that are defined at the system level, such as density, or temperature. So, yes, you’re right if you think the HyperPhysics example lacks rigor. That’s why I think it’s such wonderful pedagogic device. 🙂

The Quantum-Mechanical Gas Law

Pre-script (dated 26 June 2020): This post has become less relevant (even irrelevant, perhaps) because my views on all things quantum-mechanical have evolved significantly as a result of my progression towards a more complete realist (classical) interpretation of quantum physics. The text also got mutilated because of the removal of material by the dark force. I keep blog posts like these mainly because I want to keep track of where I came from. I might review them one day, but I currently don’t have the time or energy for it. 🙂

Original post:

In my previous posts, it was mentioned repeatedly that the kinetic theory of gases is not quite correct: the experimentally measured values of the so-called specific heat ratio (γ) vary with temperature and, more importantly, their values differ, in general, from what classical theory would predict. It works, more or less, for noble gases, which do behave as ideal gases and for which γ is what the kinetic theory of gases would want it to be: γ = 5/3—but we get in trouble immediately, even for simple diatomic gases like oxygen or hydrogen, as illustrated below: the theoretical value is 9/7 (so that’s 1.286, more or less), but the measured value is very different.

Let me quickly remind you how we get the theoretical number. According to classical theory, a diatomic molecule like oxygen can be represented as two atoms connected by a spring. Each of the atoms absorbs kinetic energy, and for each direction of motion (x, y and z), that energy is equal to kT/2, so the kinetic energy of both atoms – added together – is 2·3·kT/2 = 3kT. However, I should immediately add that not all of that energy is to be associated with the center-of-mass motion of the whole molecule, which determines the temperature of the gas: that energy is and remains equal to the 3kT/2, always. We also have rotational and vibratory motion. The molecule can rotate in two independent directions (and any combination of these directions, of course) and, hence, rotational motion is to absorb an amount of energy equal to 2·kT/2 = kT. Finally, the vibratory motion is to be analyzed as any other oscillation, so like a spring really. There is only one dimension involved and, hence, the kinetic energy here is just kT/2. However, we know that the total energy in an oscillator is the sum of the kinetic and potential energy, which adds another kT/2 term. Putting it all together, we find that the average energy for each diatomic particle is (or should be) equal to 7·kT/2 = (7/2)kT. Now, as mentioned above, the temperature of the gas (T) is proportional to the mean molecular energy of the center-of-mass motion only (in fact, that’s how temperature is defined), with the constant of proportionality equal to 3k/2. Hence, for monatomic ideal gases, we can write: U = N·(3k/2)T and, therefore, PV = NkT = (2/3)·U. Now, γ appears as follows in the ideal gas law: PV = (γ–1)U. Therefore, γ = 2/3 + 1 = 5/3, but so that’s for monatomic ideal gases only! The total kinetic energy of our diatomic molecule is U = N·(7k/2)T and, therefore, PV = (2/7)·U. So γ must be γ = 2/7 + 1 = 9/7 ≈ 1.286 for diatomic gases, like oxygen and hydrogen.

Phew! So that’s the theory. However, as we can see from the diagram, γ approaches that value only when we heat the gas to a few thousand degrees! So what’s wrong? One assumption is that certain kinds of motions “freeze out” as the temperature falls—although it’s kinda weird to think of something ‘freezing out’ at a thousand degrees Kelvin! In any case, at the end of the 19th century, that was the assumption that was advanced, very reluctantly, by scientists such as James Jeans. However, the mystery was about to be solved then, as Max Planck, even more reluctantly, presented his quantum theory of energy at the turn of the century itself.

But the quantum theory was confirmed and so we should now see how we can apply it to the behavior of gas. In my humble view, it’s a really interesting analysis, because we’re applying quantum theory here to a phenomenon that’s usually being analyzed as a classical problem only.

Boltzmann’s Law

We derived Boltzmann’s Law in our post on the First Principles of Statistical Mechanics. To be precise, we gave Boltzmann’s Law for the density of a gas (which we denoted by n = N/V) in a force field, like a gravitational field, or in an electromagnetic field (assuming our gas particles are electrically charged, of course). We noted, however, Boltzmann’s Law was also applicable to much more complicated situations, like the one below, which shows a potential energy function for two molecules that is quite characteristic of the way molecules actually behave: when they come very close together, they repel each other but, at larger distances, there’s a force of attraction. We don’t really know the forces behind but we don’t need to: as long as these forces are conservative, they can combine in whatever way they want to combine, and Boltzmann’s Law will be applicable. [It should be obvious why. If you hesitate, just think of the definition of work and how it affects potential energy and all that. Work is force times distance, but when doing work, we’re also changing potential energy indeed! So if we’ve got a potential energy function, we can get all the rest.]

Boltzmann’s Law itself is illustrated by the graph below, which also gives the formula for it: n = n₀·e^−P.E/kT.

It’s a graph starting at n = n₀ for P.E. = 0, and it then decreases exponentially. [Funny expression, isn’t it? So as to respect mathematical terminology, I should say that it decays exponentially.] In any case, if anything, Boltzmann’s Law shows the natural exponential function is quite ‘natural’ indeed, because Boltzmann’s Law pops up in Nature everywhere! Indeed, Boltzmann’s Law is not limited to functions of potential energy only. For example, Feynman derives another Boltzmann Law for the distribution of molecular speeds or, so as to ensure the formula is also valid in relativity, the distribution of molecular momenta. In case you forgot, momentum (p) is the product of mass (m) and velocity (u), and the relevant Boltzmann Law is:

f(p)·dp = C·e^−K.E/kT·dp

The argument is not terribly complicated but somewhat lengthy, and so I’ll refer you to the link for more details. As for the f(p) function (and the dp factor on both sides of the equation), that’s because we’re not talking exact values of p but some range equal to dp and some probability of finding particles that have a momentum within that range. The principle is illustrated below for molecular speeds (denoted by u = p/m), so we have a velocity distribution below. The illustration for p would look the same: just substitute u for p.

Boltzmann’s Law can be stated, much more generally, as follows:

The probability of different conditions of energy (E), potential or kinetic, is proportional to e^−E/kT.

As Feynman notes, “This is a rather beautiful proposition, and a very easy thing to remember too!” It is, and we’ll need it for the next bit.

The quantum-mechanical theory of gases

According to quantum theory, energy comes in discrete packets, quanta, and any system, like an oscillator, will only have a discrete set of energy levels, i.e. states of different energy. An energy state is, obviously, a condition of energy and, hence, Boltzmann’s Law applies. More specifically, if we denote the various energy levels, i.e. the energies of the various molecular states, by E₀, E₁, E2,…, E_i,…, and if Boltzmann’s Law applies, then the probability of finding a molecule in the particular state E_i will be proportional to e^−E_i /kT.

Now, we know we’ve got some constant there, but we can get rid of that by calculating relative probabilities. For example, the probability of being in state E₁, relative to the probability of being in state E₀, is:

P₁/P₀ = e^−E₁ /kT/e^−E₀ /kT = e^{−(E₁–E₀)/kT}

But the relative probability P₁should, obviously, also be equal to the ratio n₁/N, i.e. the ratio of the number of molecules in state E₁ and the total number of molecules. Likewise, P₀= n₀/N. Hence, P₁/P₀ = n₁/n₀and, therefore, we can write:

n = n₀e^{−(E₁–E₀)/kT}

What can we do with that? Remember we want to explain the behavior of non-monatomic gas—like diatomic gas, for example. Now we need some other assumption, obviously. As it turns out, the assumption that we can represent a system as some kind of oscillation still makes sense! In fact, the assumption that our diatomic molecule is like a spring is equally crucial to our quantum-theoretical analysis of gases as it is to our classical kinetic theory of gases. To be precise, in both theories, we look at it as a harmonic oscillator.

Don’t panic. A harmonic oscillator is, quite simply, a system that, when displaced from its equilibrium position, experiences some kind of restoring force. Now, for it to be harmonic, the force needs to be linear. For example, when talking springs, the restoring force F will be proportional to the displacement x). It basically means we can use a linear differential equation to analyze the system, like m·(d²x/dt²) = –kx. […] I hope you recognize this equation, because you should! It’s Newton’s Law: F = m·a with F = –k·x. If you remember the equation, you’ll also remember that harmonic oscillations were sinusoidal oscillations with a constant amplitude and a constant frequency. That frequency did not depend on the amplitude: because of the sinusoidal function involved, it was easier to write that frequency as an angular frequency, which we denoted by ω₀ and which, in the case of our spring, was equal to ω₀ = (k/m)^1/2. So it’s a property of the system. Indeed, ω₀is the square root of the ratio of (1) k, which characterizes the spring (it’s its stiffness), and (2) m, i.e. the mass on the spring. Solving the differential equation yielded x = A·cos(ω₀t + Δ) as a general solution, with A the (maximum) amplitude, and Δ some phase shift determined by our t = 0 point. Let me quickly jot down too more formulas: the potential energy in the spring is kx²/2, while its kinetic energy is mv²/2, as usual (so the kinetic energy depends on the mass and its velocity, while the potential energy only depends on the displacement and the spring’s stiffness). Of course, kinetic and potential energy add up to the total energy of the system, which is constant and proportional to the square of the (maximum) amplitude: K.E. + P.E. = E ∝ A². To be precise, E = kA²/2.

That’s simple enough. Let’s get back to our molecular oscillator. While the total energy of an oscillator in classical theory can take on any value, Planck challenged that assumption: according to quantum theory, it can only take up energies equal to ħω at a time. [Note that we use the so-called reduced Planck constant here (i.e. h-bar), because we’re dealing with angular frequencies.] Hence, according to quantum theory, we have an oscillator with equally spaced energy levels, and the difference between them is ħω. Now, ħω is terribly tiny—but it’s there. Let me visualize what I just wrote:

So our expression for P₁/P₀ becomes P₁/P₀ = e^−ħω/kT/e^−0/kT = e^−ħω/kT. More generally, we have P_i/P₀ = e^{−i·ħω/kT}. So what? Well… We’ve got a function here which gives the chance of finding a molecule in state P_i relative to that of finding it in state E₀, and it’s a function of temperature. Now, the graph below illustrates the general shape of that function. It’s a bit peculiar, but you can see that the relative probability goes up and down with temperature. The graph makes it clear that, at extremely low temperatures, most particles will be in state E₀ and, of course, the internal energy of our body of gas will be close to nil.

Now, we can look at the oscillators in the bottom state (i.e. particles in the molecular energy state E₀) as being effectively ‘frozen’: they don’t contribute to the specific heat. However, as we increase the temperature, our molecules gradually begin to have an appreciable probability to be in the second state, and then in the next state, and so on, and so the internal energy of the gas increases effectively. Now, when the probability is appreciable for many states, the quantized states become nearly indistinguishable and, hence, the situation is like classical physics: it is nearly indistinguishable from a continuum of energies.

Now, while you can imagine such analysis should explain why the specific heat ratio for oxygen and hydrogen varies as it does in the very first graph of this post, you can also imagine the details of that analysis fill quite a few pages! In fact, even Feynman doesn’t include it in his Lectures. What he does include is the analysis of the blackbody radiation problem, which is remarkably similar. So… Well… For more details on that, I’ll refer you to Feynman indeed. 🙂

I hope you appreciated this little ‘lecture’, as it sort of wraps up my ‘series’ of posts on statistical mechanics, thermodynamics and, central to both, the classical theory of gases. Have fun with it all!

Entropy, energy and enthalpy

Original post:

Phew! I am quite happy I got through Feynman’s chapters on thermodynamics. Now is a good time to review the math behind it. We thoroughly understand the gas equation now:

PV = NkT = (γ–1)U

The gamma (γ) in this equation is the specific heat ratio: it’s 5/3 for ideal gases (so that’s about 1.667) and, theoretically, 4/3 ≈ 1.333 or 9/7 ≈ 1.286 for diatomic gases, depending on the degrees of freedom we associate with diatomic molecules. More complicated molecules have even more degrees of freedom and, hence, can absorb even more energy, so γ gets closer to one—according to the kinetic gas theory, that is. While we know that the kinetic gas theory is not quite accurate – an approach involving molecular energy states is a better match for reality – that doesn’t matter here. As for the term (specific heat ratio), I’ll explain that later. [I promise. 🙂 You’ll see it’s quite logical.]

The point to note is that this body of gas (or whatever substance) stores an amount of energy U that is directly proportional to the temperature (T), and Nk/(γ–1) is the constant of proportionality. We can also phrase it the other way around: the temperature is directly proportional to the energy, with (γ–1)/Nk the constant of proportionality. It means temperature and energy are in a linear relationship. [Yes, direct proportionality implies linearity.] The graph below shows the T = [(γ–1)/Nk]·U relationship for three different values of γ, ranging from 5/3 (i.e. the maximum value, which characterizes monatomic noble gases such as helium, neon or krypton) to a value close to 1, which is characteristic of more complicated molecular arrangements indeed, such as heptane (γ = 1.06) or methyl butane ((γ = 1.08). The illustration shows that, unlike monatomic gas, more complicated molecular arrangements allow the gas to absorb a lot of (heat) energy with a relatively moderate rise in temperature only.

We’ll soon encounter another variable, enthalpy (H), which is also linearly related to energy: H = γU. From a math point of view, these linear relationships don’t mean all that much: they just show these variables – temperature, energy and enthalphy – are all directly related and, hence, can be defined in terms of each other.

We can invent other variables, like the Gibbs energy, or the Helmholtz energy. In contrast, entropy, while often being mentioned as just some other state function, is something different altogether. In fact, the term ‘state function’ causes a lot of confusion: pressure and volume are state variables too. The term is used to distinguish these variables from so-called process functions, notably heat and work. Process functions describe how we go from one equilibrium state to another, as opposed to the state variables, which describe the equilibrium situation itself. Don’t worry too much about the distinction—for now, that is.

Let’s look at non-linear stuff. The PV = NkT = (γ–1)U says that pressure (P) and volume (V) are inversely proportional one to another, and so that’s a non-linear relationship. [Yes, inverse proportionality is non-linear.] To help you visualize things, I inserted a simple volume-pressure diagram below, which shows how pressure and volume are related for three different values of U (or, what amounts to the same, three different values of T).

The curves are simple hyperbolas which have the x- and y-axis as horizontal and vertical asymptote respectively. If you’ve studied social sciences (like me!) – so if you know a tiny little bit of the ‘dismal science’, i.e. economics (like me!) – you’ll note they look like indifference curves. The x- and y-axis then represent the quantity of some good X and some good Y respectively, and the curves closer to the origin are associated with lower utility. How much X and Y we will buy then, depends on (a) their price and (b) our budget, which we represented by a linear budget line tangent to the curve we can reach with our budget, and then we are a little bit happy, very happy or extremely happy, depending on our budget. Hence, our budget determines our happiness. From a math point of view, however, we can also look at it the other way around: our happiness determines our budget. [Now that‘s a nice one, isn’t it? Think about it! 🙂 And, in the process, think about hyperbolas too: the y = 1/x function holds the key to understanding both infinity and nothingness. :-)]

U is a state function but, as mentioned above, we’ve got quite a few state variables in physics. Entropy, of course, denoted by S—and enthalpy too, denoted by H. Let me remind you of the basics of the entropy concept:

The internal energy U changes because (a) we add or remove some heat from the system (ΔQ), (b) because some work is being done (by the gas on its surroundings or the other way around), or (c) because of both. Using the differential notation, we write: dU = dQ – dW, always. The (differential) work that’s being done is PdV. Hence, we have dU = dQ – PdV.
When transferring heat to a system at a certain temperature, there’s a quantity we refer to as the entropy. Remember that illustration of Feynman’s in my post on entropy: we go from one point to another on the temperature-volume diagram, taking infinitesimally small steps along the curve, and, at each step, an infinitesimal amount of work dW is done, and an infinitesimal amount of entropy dS = dQ/T is being delivered.
The total change in entropy, ΔS, is a line integral: ΔS = ∫_LdQ/T = ∫_LdS.

That’s somewhat tougher to understand than economics, and so that’s why it took me more time to come with terms with it. 🙂 Just go through Feynman’s Lecture on it, or through that post I referenced above. If you don’t want to do that, then just note that, while entropy is a very mysterious concept, it’s deceptively simple from a math point of view: ΔS = ΔQ/T, so the (infinitesimal) change in entropy is, quite simply, the ratio of (1) the (infinitesimal or incremental) amount of heat that is being added or removed as the system goes from one state to another through a reversible process and (2) the temperature at which the heat is being transferred. However, I am not writing this post to discuss entropy once again. I am writing it to give you an idea of the math behind the system.

So dS = dQ/T. Hence, we can re-write dU = dQ – dW as:

dU = TdS – PdV ⇔ dU + d(PV) = TdS – PdV + d(PV)

⇔ d(U + PV) = dH = TdS – PdV + PdV + VdP = TdS + VdP

The U + PV quantity on the left-hand side of the equation is the so-called enthalpy of the system, which I mentioned above. It’s denoted by H indeed, and it’s just another state variable, like energy: same-same but different, as they say in Asia. We encountered it in our previous post also, where we said that chemists prefer to analyze the behavior of substances using temperature and pressure as ‘independent variables’, rather than temperature and volume. Independent variables? What does that mean, exactly?

According to the PV = NkT equation, we only have two independent variables: if we assign some value to two variables, we’ve got a value for the third one. Indeed, remember that other equation we got when we took the total differential of U. We wrote U as U(V, T) and, taking the total differential, we got:

dU = (∂U/∂T)dT + (∂U/∂V)dV

We did not need to add a (∂U/∂P)dP term, because the pressure is determined by the volume and the temperature. We could also have written U = U(P, T) and, therefore, that dU = (∂U/∂T)dT + (∂U/∂P)dP. However, when working with temperature and pressure as the ‘independent’ variables, it’s easier to work with H rather than U. The point to note is that it’s all quite flexible really: we have two independent variables in the system only. The third one (and all of the other variables really, like energy or enthalpy or whatever) depend on the other two. In other words, from a math point of view, we only have two degrees of freedom in the system here: only two variables are actually free to vary. 🙂

Let’s look at that dH = TdS + VdP equation. That’s a differential equation in which not temperature and pressure, but entropy (S) and pressure (P) are ‘independent’ variables, so we write:

dH(S, P) = TdS + VdP

Now, it is not very likely that we will have some problem to solve with data on entropy and pressure. At our level of understanding, any problem that’s likely to come our way will probably come with data on more common variables, such as the heat, the pressure, the temperature, and/or the volume. So we could continue with the expression above but we don’t do that. It makes more sense to re-write the expression substituting TdS for dQ once again, so we get:

dH = dQ + VdP

That resembles our dU = dQ – PdV expression: it just substitutes V for –P. And, yes, you guessed it: it’s because the two expressions resemble each other that we like to work with H now. 🙂 Indeed, we’re talking the same system and the same infinitesimal changes and, therefore, we can use all the formulas we derived already by just substituting H for U, V for –P, and dP for dV. Huh? Yes. It’s a rather tricky substitution. If we switch V for –P (or vice versa) in a partial derivative involving T, we also need to include the minus sign. However, we do not need to include the minus sign when substituting dV and dP, and we also don’t need to change the sign of the partial derivatives of U and H when going from one expression to another! It’s a subtle and somewhat weird point, but a very important one! I’ll explain it in a moment. Just continue to read as for now. Let’s do the substitution using our rules:

dU = (∂Q/∂T)_VdT + [T(∂P/∂T)_V − P]dV becomes:

dH = (∂Q/∂T)_PdT + (∂H/∂P)_TdP = C_PdT + [–T·(∂V/∂T)_P+ V]dP

Note that, just as we referred to (∂Q/∂T)_Vas the specific heat capacity of a substance at constant volume, which we denoted by C_V, we now refer to (∂Q/∂T)_P as the specific heat capacity at constant pressure, which we’ll denote, logically, as C_P. Dropping the subscripts of the partial derivatives, we re-write the expression above as:

dH = C_PdT + [–T·(∂V/∂T)+ V]dP

So we’ve got what we wanted: we switched from an expression involving derivatives assuming constant volume to an expression involving derivatives assuming constant pressure. [In case you wondered what we wanted, this is it: we wanted an equation that helps us to solve another type of problem—another formula for a problem involving a different set of data.]

As mentioned above, it’s good to use subscripts with the partial derivatives to emphasize what changes and what is constant when calculating those partial derivatives but, strictly speaking, it’s not necessary, and you will usually not find the subscripts when googling other texts. For example, in the Wikipedia article on enthalpy, you’ll find the expression written as:

dH = C_PdT + V(1–αT)dP with α = (1/V)(∂V/∂T)

Just write it all out and you’ll find it’s the same thing, exactly. It just introduces another coefficient, α, i.e. the coefficient of (cubic) thermal expansion. If you find this formula is easier to remember, then please use this one. It doesn’t matter.

Now, let’s explain that funny business with the minus signs in the substitution. I’ll do so by going back to that infinitesimal analysis of the reversible cycle in my previous post, in which we had that formula involving ΔQ for the work done by the gas during an infinitesimally small reversible cycle: ΔW = ΔVΔP = ΔQ·(ΔT/T). Now, we can either write that as:

ΔQ = T·(ΔP/ΔT)·ΔV = dQ = T·(∂P/∂T)_V·dV – which is what we did for our analysis of (∂U/∂V)_T– or, alternatively, as
ΔQ = T·(ΔV/ΔT)·ΔP = dQ = T·(∂V/∂T)_P·dP, which is what we’ve got to do here, for our analysis of (∂H/∂P)_T.

Hence, dH = dQ + VdP becomes dH = T·(∂V/∂T)_P·dP + V·dP, and dividing all by dP gives us what we want to get: dH/dP = (∂H/∂P)_T= T·(∂V/∂T)_P+ V.

[…] Well… NO! We don’t have the minus sign in front of T·(∂V/∂T)_P, so we must have done something wrong or, else, that formula above is wrong.

The formula is right (it’s in Wikipedia, so it must be right :-)), so we are wrong. Indeed! The thing is: substituting dT, dV and dP for ΔT, ΔV and ΔP is somewhat tricky. The geometric analysis (illustrated below) makes sense but we need to watch the signs.

We’ve got a volume increase, a temperature drop and, hence, also a pressure drop over the cycle: the volume goes from V to V+ΔV (and then back to V, of course), while the pressure and the temperature go from P to P–ΔP and T to T–ΔT respectively (and then back to P and T, of course). Hence, we should write: ΔV = dV, –ΔT = dT, and –ΔP = dP. Therefore, as we replace the ratio of the infinitesimal change of pressure and temperature, ΔP/ΔT, by a proper derivative (i.e. ∂P/∂T), we should add a minus sign: ΔP/ΔT = –∂P/∂T. Now that gives us what we want: dH/dP = (∂H/∂P)_T= –T·(∂V/∂T)_P+ V, and, therefore, we can, indeed, write what we wrote above:

dU = (∂Q/∂T)_VdT + [T(∂P/∂T)_V − P]dV becomes:

dH = (∂Q/∂T)_PdT + [–T·(∂V/∂T)_P+ V]dP = C_PdT + [–T·(∂V/∂T)_P+ V]dP

Now, in case you still wonder: what’s the use of all these different expressions stating the same? The answer is simple: it depends on the problem and what information we have. Indeed, note that all derivatives we use in our expression for dH expression assume constant pressure, so if we’ve got that kind of data, we’ll use the chemists’ representation of the system. If we’ve got data describing performance at constant volume, we’ll need the physicists’ formulas, which are given in terms of derivatives assuming constant volume. It all looks complicated but, in the end, it’s the same thing: the PV = NkT equation gives us two ‘independent’ variables and one ‘dependent’ variable. Which one is which will determine our approach.

Now, we left one thing unexplained. Why do we refer to γ as the specific heat ratio? The answer is: it is the ratio of the specific heat capacities indeed, so we can write:

γ = C_P/C_V

However, it is important to note that that’s valid for ideal gases only. In that case, we know that the (∂U/∂V)_Pderivative in our dU = (∂U/∂T)_VdT + (∂U/∂V)_TdV expression is zero: we can change the volume, but if the temperature remains the same, the internal energy remains the same. Hence, dU = (∂U/∂T)_VdT = C_VdT, and dU/dT = C_V. Likewise, the (∂H/∂P)_Tderivative in our dH = (∂H/∂T)_PdT + (∂H/∂P)_TdP expression is zero—for ideal gases, that is. Hence, dH = (∂H/∂T)_PdT = C_PdT, and dH/dT = C_P. Hence,

C_P/C_V = (dH/dT)/(dU/dT) = dH/dU

Does that make sense? If dH/dU = γ, then H must be some linear function of U. More specifically, H must be some function H = γU + c, with c some constant (it’s the so-called constant of integration). Now, γ is supposed to be constant too, of course. That’s all perfectly fine: indeed, combining the definition of H (H = U + PV), and using the PV = (γ–1)U relation, we have H = U + (γ–1)U = γU (hence, c = 0). So, yes, dH/dU = γ, and γ = C_P/C_V.

Note the qualifier, however: we’re assuming γ is constant (which does not imply the gas has to be ideal, so the interpretation is less restrictive than you might think it is). If γ is not a constant, it’s a different ballgame. […] So… Is γ actually constant? The illustration below shows γ is not constant for common diatomic gases like hydrogen or (somewhat less common) oxygen. It’s the same for other gases: when mentioning γ, we need to state the temperate at which we measured it too. 😦 However, the illustration also shows the assumption of γ being constant holds fairly well if temperature varies only slightly (like plus or minus 100° C), so that’s OK. 🙂

I told you so: the kinetic gas theory is not quite accurate. An approach involving molecular energy states works much better (and is actually correct, as it’s consistent with quantum theory). But so we are where we are and I’ll save the quantum-theoretical approach for later. 🙂

So… What’s left? Well… If you’d google the Wikipedia article on enthalphy in order to check if I am not writing nonsense, you’ll find it gives γ as the ratio of H and U itself: γ = H/U. That’s not wrong, obviously (γ = H/U = γU/U = γ), but that formula doesn’t really explain why γ is referred to as the specific heat ratio, which is what I wanted to do here.

OK. We’ve covered a lot of ground, but let’s reflect some more. We did not say a lot about entropy, and/or the relation between energy and entropy. Too bad… The relationship between entropy and energy is obviously not so simple as between enthalpy and energy. Indeed, because of that easy H = γU relationship, enthalpy emerges as just some auxiliary variable: some temporary variable we need to calculate something. Entropy is, obviously, something different. Unlike enthalpy, entropy involves very complicated thinking, involving (ir)reversibility and all that. So it’s quite deep, I’d say – but I’ll write more about that later. I think this post has gone as far as it should. 🙂

Is gas a reversible engine?

Original post:

We’ve worked on very complicated matters in the previous posts. In this post, I am going to tie up a few loose ends, not only about the question in the title but also other things. Let me first review a few concepts and constructs.

Temperature

We’ve talked a lot about temperature, but what it is really? You have an answer ready of course: it is the mean kinetic energy of the molecules of a gas or a substance. You’re right. To be precise, it is the mean kinetic energy of the center-of-mass (CM) motions of the gas molecules.

The added precision in the definition above already points out temperature is not just mean kinetic energy or, to put it differently, that the concept of mean kinetic energy itself is not so simple when we are not talking ideal gases. So let’s be precise indeed. First, let me jot down the formula for the mean kinetic energy of the CM motions of the gas particles:

(K.E.)_CM= <(1/2)·mv²>

Now let’s recall the most fundamental law in the kinetic theory of gases, which states that the mean value of the kinetic energy for each independent direction of motion will be equal to kT/2. [I know you know the kinetic theory of gases itself is not accurate – we should be talking about molecular energy states – but let’s go along with it.] Now, because we have only three independent directions of motions (the x, y and z directions) for ideal gas molecules (or atoms, I should say), the mean kinetic energy of the gas particles is kT/2 + kT/2 + kT/2 = 3kT/2.

What’s going on here is that we are actually defining temperature here: we basically say that the kinetic energy is linearly proportional to something that we define as the temperature. For practical reasons, that constant of proportionality is written as 3k/2, with k the Boltzmann constant. So we write our definition of temperature as:

(K.E.)_CM = 3kT/2 ⇔ T = (3k/2)^–1<(1/2)·mv²> = [2/(3k)]·(K.E.)_CM

What happens with temperature when considering more complex gases, such as diatomic gases? Nothing. The temperature will still be proportional to the kinetic energy of the center-of-mass motions, but we should just note it’s the (K.E.)_CMof the whole diatomic molecule, not of the individual atoms. The thing with more complicated arrangements is that, when adding or removing heat, we’ve got something else going on too: part of the energy will go into the rotational and vibratory motions inside the molecule, which is why we’ll need to add a lot more heat in order to achieve the same change in temperature or, vice versa, we’ll be able to extract a lot more heat out of the gas – as compared to an ideal gas, that is – for the same drop in temperature. [When talking molecular energy states, rather than independent directions of motions, we’re saying the same thing: energy does not only go in center-of-mass motion but somewhere else too.]

You know the ideal gas law is based on the reasoning above and the PV = NkT equation, which is always valid. For ideal gases, we write:

PV = NkT = Nk(3k/2)^–1<(1/2)mv²> = (2/3)N<(1/2)·mv²> = (2/3)·U

For diatomic gases, we have to use another coefficient. According to our theory above, which distinguishes 6 independent directions of motions, the mean kinetic energy is twice 3kT/2 now, so that’s 3kT, and, hence, we write: T = (3k)^–1<K.E.> =

PV = NkT = Nk(3k)^–1<K.E.> = (1/3)·U

The two equations above will usually be written as PV = (γ–1)U, so γ, which is referred to as the specific heat ratio, would be equal 5/3 ≈ 1.67 for ideal gases and 4/3 ≈ 1.33 for diatomic gases. [If you read my previous posts, you’ll note I used 9/7 ≈ 1.286, but that’s because Feynman suddenly decides to add the potential energy of the oscillator as another ‘independent direction of motion’.]

Now, if we’re not adding or removing heat to/from the gas, we can do a differential analysis yielding a differential equation (what did you expect?), which we can then integrate to find that P = C/V^γ relationship. You’ve surely seen it before. The C is some constant related to the energy and/or the state of the gas. It is actually interesting to plot the pressure-volume relationship using that P = C/V^γ relationship for various values of γ. The blue graph below assumes γ = 5/3 ≈ 1.667, which is the theoretical value for ideal gases (γ for helium or krypton comes pretty close to that), while the red graph gives the same relationship for γ = 4/3 ≈ 1.33, which is the theoretical value for diatomic gases (gases like bromine and iodine have a γ that’s close to that).

Let me repeat that this P = C/V^γ relationship is only valid for adiabatic expansion or compression: we do not add or remove heat and, hence, this P = C/V^γ function gives us the adiabatic segments only in a Carnot cycle (i.e. the adiabatic lines in a pressure-volume diagram). Now, it is interesting to observe that the slope of the adiabatic line for the ideal gas is more negative than the slope of the adiabatic line for the diatomic gas: the blue curve is the steeper one. That’s logical: for the same volume change, we should get a bigger drop in pressure for the ideal gas, as compared to the diatomic gas, because… Well… You see the logic, don’t you?

Let’s freewheel a bit and see what it implies for our Carnot cycle.

Carnot engines with ideal and non-ideal gas

We know that, if we could build an ideal frictionless gas engine (using a cylinder with a piston or whatever other device we can think of), its efficiency will be determined by the amount of work it can do over a so-called Carnot cycle, which consists of four steps: (1) isothermal expansion (gas absorbs heat and the volume expands at constant temperature), (2) adiabatic expansion (the volume expands while the temperature drops), (3) isothermal compression (the volume decreases at constant temperature, so heat is taken out), and (4) isothermal compression (the volume decreases as we bring the gas back to the same temperature).

It is important to note that work is being done, by the gas on its surroundings, or by the surroundings on the gas, during each step of the cycle: work is being done by the gas as it expands, always, and work is done on the gas as it is being compressed, always.

You also know that there is only one Carnot efficiency, which is defined as the ratio of (a) the net amount of work we get out of our machine in one such cycle, which we’ll denote by W, and (b) the amount of heat we have to put in to get it (Q₁). We’ve also shown that W is equal to the difference between the heat we put during the first step (isothermal expansion) and the heat that’s taken out in the third step (isothermal compression): W = Q₁ − Q₂, which basically means that all heat is converted into useful work—which is why it’s an efficient engine! We also know that the formula for the efficiency is given by:

W/Q₁ = (T₁ − T₂)/T₁.

Where’s Q₂ in this formula? It’s there, implicitly, as the efficiency of the engine depends on T₂. In fact, that’s the crux of the matter: for efficient engines, we also have the same Q₁/T₁= Q₂/T₂ratio, which we define as the entropy S = Q₁/T₁= Q₂/T₂. We’ll come back to this.

Now how does it work for non-ideal gases? Can we build an equally efficient engine with actual gases? This was, in fact, Carnot’s original question, and we haven’t really answered it in our previous posts, because we weren’t quite ready for it. Let’s consider the various elements to the answer:

Because we defined temperature the way we defined it, it is obvious that the gas law PV = NkT still holds for diatomic gases, or whatever gas (such as steam vapor, for example, the stuff which was used in Carnot’s time). Hence, the isothermal lines in our pressure-volume diagrams don’t change. For a given temperature T, we’ll have the same green and red isothermal line in the diagram above.
However, the adiabatic lines (i.e .the blue and purple lines in the diagram above) for the non-ideal gas are much flatter than the one for an ideal gas. Now, just take that diagram and draw two flatter curves through point a and c indeed—but not as flat as the isothermal segments, of course! What you’ll notice is that the area of useful work becomes much smaller.

What does that imply in terms of efficiency? Well… Also consider the areas under the graph which, as you know, represent the amount of work done during each step (and you really need to draw the graph here, otherwise you won’t be able to follow my argument):

The phase of isothermal expansion will be associated with a smaller volume change, because our adiabatic line for the diatomic gas intersects the T = T₁ isothermal line at a smaller value for V. Hence, less work is being done during that stage.
However, more work will be done during adiabatic expansion, and the associated volume change is also larger.
The isothermal compression phase is also associated with a smaller volume change, because our adiabatic line for the diatomic gas intersects the T = T₂ isothermal line at a larger value for V.
Finally, adiabatic compression requires more work to be done to get from T₂ to T₁again, and the associated volume change is also larger.

The net result is clear from the graph: the net amount of work that’s being done over the complete cycle is less for our non-ideal gas than as compared to our engine working with ideal gas. But, again, the question here is what it implies in terms of efficiency? What about the W/Q₁ ratio?

The problem is that we cannot see how much heat is being put in (Q₁) and how much heat is being taken out (Q₂) from the graph. The only thing we know is that we have an engine working here between the same temperature T₁ to T₂. Hence, if we use subscript A for the ideal gas engine and subscript B for the one working with ordinary (i.e. non-ideal) gas, and if both engines are to have the same efficiency W/Q₁= W_B/Q_{1_B}= W_A/Q_{1_A}, then it’s obvious that,

if W_A> W_B, then Q_{1_A} > Q_{1_B}.

Is that consistent with what we wrote above for each of the four steps? It is. Heat energy is taken in during the first step only, as the gas expands isothermally. Now, because the temperature stays the same, there is no change in internal energy, and that includes no change in the internal vibrational and rotational energy. All of the heat energy is converted into work. Now, because the volume change is less, the work will be less and, hence, the heat that’s taken in must also be less. The same goes for the heat that’s being taken out during the third step, i.e. the isothermal compression stage: we’ve got a smaller volume change here and, hence, the surroundings of the gas do less work, and a lesser amount of heat energy is taken out.

So what’s the grand conclusion? It’s that we can build an ideal gas engine working between the same temperature T₁ and T₁, and with exactly the same efficiency and W/Q₁ = (T₁ − T₂)/T₁using non-ideal gas. Of course, there must be some difference! You’re right: there is. While the ordinary gas machine will be as efficient as the ideal gas machine, it will not do the same amount of work. The key to understanding this is to remember that efficiency is a ratio, not some absolute number. Let’s go through it. Because their efficiency is the same, we know that the W/Q₁ ratios for both engines (A and B) is the same and, hence, we can write:

W_A/W_B = Q_{1_A}/Q_{1_B}

What about the entropy? The entropy S = Q_{1_A}/T₁ = Q_{2_A}/T₂ is not the same for both machines. For example, if the engine with ideal gas (A) does twice the work of the engine with ordinary gas (B), then Q_{1_A} will also be twice the amount Q_{1_B}. Indeed, S_A = Q_{1_A} /T₁ and S_B = Q_{1_B}/T₁. Hence, S_A/S_B = Q_{1_A}/Q_{1_B}. For example, if Q_{1_A} = 2·Q_{1_B}, then engine A’s entropy will also be twice that of engine B. [Now that we’re here, I should also note you’ll have the same ratio for Q_{2_A}. Indeed, we know that, for an efficient machine, we have: Q₁/T₁= Q₂/T₂. Hence, Q_{1_A}/Q_{2_A} = T₁/T₂ and Q_{1_B}/Q_{2_B} = T₁/T₂. So Q_{1_A}/Q_{2_A}= Q_{1_B}/Q_{2_B}and, therefore, So Q_{1_A}/Q_{1_B}= Q_{2_A}/Q_{2_B}.]

Why would the entropy be any different? We’ve got the same number of particles, the same volume and the same working temperatures, and so the only difference is that the particles in engine B are diatomic: the molecules consist of two atoms, rather than one only. An intuitive answer to the question as to why the entropy is different can be given by comparing it to another example, which I mentioned in a previous post, for which the entropy is also different fro some non-obvious reason. Indeed, we can think of the two atoms as the equivalent of the white and black particles in the box (see my previous post on entropy): if we allow the white and black particles to mix in the same volume, rather than separate them in two compartments, then the entropy goes up (we calculated the increase as equal to k·ln2). Likewise, the entropy is much lower if all particles have to come in pairs, which is the case for a diatomic gas. Indeed, if they have to come in pairs, we significantly reduce the number of ways all particles can be arranged, or the ‘disorder’, so to say. As the entropy is a measure of that number (one can loosely define entropy as the logarithm of the number of ways), the entropy must go down as well. Can we illustrate that using the ΔS = Nkln(V₂/V₁) formula we introduced in our previous post, or our more general S(V, T) = Nk[lnV + (1/γ-1)lnT] + a formula? Maybe. Let’s give it a try.

We know that our diatomic molecules have an average kinetic energy equal to 3kT/2. Well… Sorry. I should be precise: that’s the kinetic energy of their center-of-mass motion only! Now, let us suppose all our diatomic molecules spit up. We know the average kinetic energy of the constituent parts will also equal 3kT/2. Indeed, if a gas molecule consists of two atoms (let’s just call them atom A and B respectively), and if their combined mass is M = m_A+ m_B, we know that:

<m_Av_A²/2> = <m_Bv_B²/2> = <Mv_CM²/2> = 3kT/2

Hence, if they split, we’ll have twice the number of particles (2N) in the same volume with the same average kinetic energy: 3kT/2. Hence, we double the energy, but the average kinetic energy of the particles is the same, so the temperature should be the same. Hmm… You already feel something is wrong here… What about the energy that we associated with the internal motions within the molecule, i.e. the internal rotational and vibratory motions of the atoms, when they were still part of the same molecule? That was also equal to 3kT/2, wasn’t it? It was. Yes. In case you forgot why, let me remind you: the total energy is the sum of the (average) kinetic energy of the two atoms, so that’s <m_Av_A²/2> + <m_Bv_B²/2> = 3kT/2 + 3kT/2 = 3kT. Now, that sum is also equal to the sum of the center-of-mass motion (which is 3 kT/2) and the average kinetic energy of the rotational and vibratory motions. Hence, the average kinetic energy of the rotational and vibratory motions is 3kT – 3 kT/2 = 3 kT/2. It’s all part of the same theorem: the average kinetic energy for each independent direction of motion is kT/2, and the number of degrees of freedom for a molecule consisting of r atoms is 3, because each atom can move in three directions. Rotation involves another two independent motions (in three dimensions, we’ve got two axes of rotation only), and vibration another one. So the kinetic energy going into rotation is kT/2 + kT/2 = kT and for vibration it’s kT/2. Adding all yields 3kT/2 + kT + kT/2 = 3kT.

The arithmetic is quite tricky. Indeed, you may think that, if we split the molecule, that the rotational and vibratory energy has to go somewhere, and that it is only natural to assume that, when we spit the diatomic molecule, the individual atoms have to absorb it. Hence, you may think that the temperature of the gas will be higher. How much higher? We had an average energy of 3kT per molecule in the diatomic situation, but so now we have twice as many particles, and hence, the average energy per particle now is… Re-read what I wrote above: it’s just 3kT/2 again. The energy that’s associated with the center-of-mass motions and the rotational and vibratory motions is not something extra: it’s part of the average kinetic energy of the atoms themselves. So no rise in temperature!

Having said that, our PV = NkT = (2/3)U equation obviously doesn’t make any sense anymore, as we’ve got twice as many particles now. While the temperature has not gone up, both the internal energy and the pressure have doubled, as we’ve got twice as many particles hitting the walls of our cylinder now. To restore the pressure to its ex ante value, we need to increase the volume. Remember, however, that pressure is force per unit surface area, not per volume unit: P = F/A. So we don’t have to double the volume: we only have to double the surface area. Now, it all depends on the shape of the volume: are we thinking of a box or of some sphere? One thing we know though: if we calculate the volume using some radius r, which may also be the length of the edge of a cube, then we know the volume is going to be proportional to r³, while the surface area is going to be proportional to r². Hence, the ratio between the surface area and the volume is going to be proportional to r²/r³ = r^2/3. So that’s another 2/3 ratio which pops us here, as an exponent this time. It’s not a coincidence, obviously.

Hmm… Interesting exercise. I’ll let you work it out. I am sure you’ll find some sensible value for the new volume, so you should able to use that ΔS = Nkln(V₂/V₁) formula. However, you also need to think about the comparability of the two situations. We wanted to compare two equal volumes with an equal number of particles (diatomic molecules versus atoms), and so you’ll need to move back in that direction to get a final answer to your question. Please do mail me the answer: I hope it makes sense. 🙂

Inefficient engines

When trying to understand efficient engines, it’s interesting to also imagine how inefficient engines work, so as to see what they imply for our Carnot diagram. Suppose we’ve tried to build a Carnot engine in our kitchen, and we end up with one that is fairly frictionless, and fairly well isolated, so there is little heat loss during the heat transfer steps. We also have good contact surfaces so we think the the heat transfer processes will also be fairly frictionless, so to speak. So we did our calculations and built the engine using the best kitchen design and engineering practices. Now it’s the time for the test. Will it work?

What might happen is the following: while we’ve designed the engine to get some net amount of work out of it (in each and every cycle) that is given by the isothermic and adiabatic lines below, we may find that we’re not able to keep the temperature constant. So we try to follow the green isothermic line alright, but we can’t. We may also find that, when our heat counter tells us we’ve put Q₁ in already, that our piston hasn’t moved out quite as far we thought it would. So… Damn, we’re never going to get to c. What’s the reason? Some heat loss, because our isolation wasn’t perfect, and friction.

So we’re likely to have followed an actual path that’s closer to the red arrow, which brings us near point d. So we’ve missed point c. We have no choice, however: the temperature has dropped to T₂ and, hence, we need to start with the next step. Which one? The second? The third? It’s not quite clear, because our actual path on the pressure-volume diagram doesn’t follow any of our ideal isothermal or adiabatic lines. What to do? Let’s just take some heat out and start compressing to see what happens. If we’ve followed a path like the red arrow, we’re likely to be on something like the black arrow now. Indeed, if we’ve got a problem with friction or heat loss, we’ll continue to have that problem, and so the temperature will drop much faster than we think it should, and so we will not have the expected volume decrease. In fact, we’re not able to maintain the temperature even at T₂. What horror! We can’t repeat our process and, hence, it is surely not reversible! All our work for nothing! We have to start all over and re-examine our design.

So our kitchen machine goes nowhere. But then how do actual engines work? The answer is: they put much more heat in, and they also take much more heat out. More importantly, they’re also working much below the theoretical efficiency of an ideal engine, just like our kitchen machine. So that’s why we’ve got the valves and all that in a steam engine. Also note that a car engine works entirely different: it converts chemical energy into heat energy by burning fuel inside of the cylinder. Do we get any useful work out? Of course! My Lamborghini is fantastic. 🙂 Is it efficient? Nope. We’re converting huge amounts of heat energy into a very limited amount of useful work, i.e. the type of energy we need to drive the wheels of my car, or a dynamo. Actual engines are a shadow only of ideal engines. So what’s the Carnot cycle really? What does it mean in practice? Does the mathematical model have any relevance at all?

The Carnot cycle revisited

Let’s look at those differential equations once again. [Don’t be scared by the concept of a differential equation. I’ll come back to it. Just keep reading.] Let’s start with the ΔU = (∂U/∂T)ΔT + (∂U/∂V)ΔV equation, which mathematical purists would probably prefer to write as:

dU = (∂U/∂T)dT + (∂U/∂V)dV

I find Feynman’s use of the Δ symbol more appropriate, because, when dividing by dV or dT, we get dU/dV and dU/dt, which makes us think we’re dealing with ordinary derivatives here, and we are not: it’s partial derivatives that matter here. [I’ll illustrate the usefulness of distinguishing the Δ and d symbol in a moment.] Feynman is even more explicit about that as he uses subscripts for the partial derivatives, so he writes the equation above as:

ΔU = (∂U/∂T)_VΔT+ (∂U/∂V)_TΔV

However, partial derivatives always assume the other variables are kept constant and, hence, the subscript is not needed. It makes the notation rather cumbersome and, hence, I think it makes the analysis even more unreadable than it already is. In any case, it is obvious that we’re looking at a situation in which all changes: the volume, the temperature and the pressure. However, in the PV = NkT equation (which, I repeat, is valid for all gases, ideal or not, and in all situations, be it adiabatic or isothermal expansion or compression), we have only two independent variables for a given number of particles. We can choose: volume and temperature, or pressure and temperature, or volume and pressure. The third variable depends on the two other variables and, hence, is referred to as dependent. Now, one should not attach too much importance to the terms (dependent or independent does not mean more or less fundamental) but, when everything is said and done, we need to make a choice when approaching the problem. In physics, we usually look at the volume and the temperature as the ‘independent’ variables but the partial derivative notation makes it clear it doesn’t matter. With three variables, we’ll have three partial derivatives: ∂P/∂T, ∂V/∂T and ∂P/∂V, and their reciprocals ∂T/∂P, ∂T/∂V and ∂V/∂P too, of course!

Having said that, when calculating the value of derived variables like energy, or entropy, or enthalpy (which is a state variable used in chemistry), we’ll use two out of the three mentioned variables only, because the third one is redundant, so to speak. So we’ll have some formula for the internal energy of a gas that depends on temperature and volume only, so we write:

U = U(V, T)

Now, in physics, one will often only have a so-called differential equation for a variable, i.e. something that is written in terms of differentials and derivatives, so we’ll do that here too. But let me give some other example first. You may or may not remember that we had this differential equation telling us how the density (n = N/V) of the atmosphere changes with the height (h), as a function of the molecular mass (m), the temperature (T) and the density (n) itself: dn/dh = –(mg/kT)·n, with g the gravitational constant and k the Boltzmann constant. Now, it is not always easy to go from a differential equation to a proper formula, but this one can be solved rather easily. Indeed, a function which has a derivative that is proportional to itself (that’s what this differential equation says really) is an exponential, and the solution was n = n₀e^–mgh/kT, with n₀ some other constant (the density at h = 0, which can be chosen anywhere). This explicit formula for n says that the density goes down exponentially with height, which is what we would expect.

Let’s get back to our gas though. We also have differentials here, which are infinitesimally small changes in variables. As mentioned above, we prefer to write them with a Δ in front (rather than using the d symbol)—i.e. we write ΔT, ΔU, ΔU, or ΔQ. When we have two variables only, say x and y, we can use the d symbol itself and, hence, write Δx and Δy as dx and dy. However, it’s still useful to distinguish, in order to write something like this:

Δy = (dy/dx)Δx

This says we can approximate the change in y at some point x when we know the derivative there. For a function in two variables, we can write the same, which is what we did at the very start of this analysis:

ΔU = (∂U/∂T)ΔT + (∂U/∂V)ΔV

Note that the first term assumes constant volume (because of the ∂U/∂T derivative), while the second assumes constant temperature (because of the ∂U/∂V derivative).

Now, we also have a second equation for ΔU, expressed in differentials only (so no partial derivatives here):

ΔU = ΔQ – PΔV

This equation basically states that the internal energy of a gas can change because (a) some heat is added or removed or (b) some work is being done by or on the gas as its volume gets bigger or smaller. Note the minus sign in front of PΔV: it’s there to ensure the signs come out alright. For example, when compressing the gas (so ΔV is negative), ΔU = – PΔV will be positive. Conversely, when letting the gas expand (so ΔV is positive), ΔU = – PΔV will be negative, as it should be.

What’s the relation between these two equations? Both are valid, but you should surely not think that, just because we have a ΔV in the second term of each equation, we can write –P = ∂U/∂V. No.

Having said that, let’s look at the first term of the ΔU = (∂U/∂T)ΔT + (∂U/∂V)ΔV equation and analyze it using the ΔU = ΔQ – PΔV equation. We know (∂U/∂T)ΔT assumes we keep the volume constant, so ΔV = 0 and, hence, ΔU = ΔQ: all the heat goes into changing the internal energy; none goes into doing some work. Therefore, we can write:

(∂U/∂T)ΔT = (∂Q/∂T)ΔT = C_VΔT

You already know that we’ve got a name for that C_V function (remember: a derivative is a function too!): it’s the (specific) heat capacity of the gas (or whatever substance) at constant volume. For ideal gases, C_V is some constant but, remember, we’re not limiting ourselves to analyzing ideal gases only here!

So we’re done with the first term in that ΔU = (∂U/∂T)ΔT + (∂U/∂V)ΔV. Now it’s time for the second one: (∂U/∂V)ΔV. Now both ΔQ and –PΔV are relevant: the internal energy changes because (a) some heat is being added and (b) because the volume changes and, hence, some work is being done. You know what we need to find. It’s that weird formula:

∂U/∂V = T(∂P/∂T) – P

But how do we get there? We can visualize what’s going on as a tiny Carnot cycle. So we think of gas as an ideal engine itself: we put some heat in (ΔQ) which gets an isothermal expansion at temperature T going, during a tiny little instant, doing a little bit of work. But then we stop adding heat and, hence, we’ll have some tiny little adiabatic expansion, during which the gas keeps going and also does a tiny amount of work as it pushes against the surrounding gas molecules. However, this step involves an infinitesimally small temperature drop—just a little bit, to T–ΔT. And then the surrounding gas will start pushing back and, hence, we’ve got some isothermal compression going, at temperature T–ΔT, which is then followed, once again, by adiabatic compression as the temperature goes back to T. The last two steps involve the surroundings of the tiny little volume of gas we’re looking at, doing work on the gas, instead of the other way around.

You’ll say this sounds very fishy. It does, but it is Feynman’s analysis, so who am I to doubt it? You’ll ask: where does the heat go, and where does the work go? Indeed, if ΔQ is Q₁, what about Q₂? Also, we can sort of imagine that the gas can sort of store the energy of the work that’s being done during step 1 and 2, to then give (most of it) back during step 3 and 4, but what about the net work that’s being done in this cycle, which is (see the diagram) equal to W = Q₁– Q₂ = ΔPΔV? Where does that go? In some kind of flywheel or something? Obviously not! Hmm… Not sure. In any case, Q₁ is infinitesimally small and, hence, nearing zero. Q₂ is even smaller, so perhaps we should equate it to zero and just forget about it. As for the net work done by the cycle, perhaps this may just go into moving the gas molecules in the equally tiny volume of gas we’re looking at. Hence, perhaps there’s nothing left to be transferred to the surrounding gas. In short, perhaps we should look at ΔQ as the energy that’s needed to do just one cycle.

Well… No. If gas is an ideal engine, we’re talking elastic collisions and, hence, it’s not like a transient, like something that peters out. The energy has to go somewhere—and it will. The tiny little volume we’re looking at will come back to its original state, as it should, because we’re looking at (∂U/∂V)ΔV, which implies we’re doing an analysis at constant temperature, but the energy we put in has got to go somewhere: even if Q₂ is zero, and all of ΔQ goes into work, it’s still energy that has to go somewhere!

It does go somewhere, of course! It goes into the internal energy of the gas we’re looking at. It adds to the kinetic energy of the surrounding gas molecules. The thing is: when doing such infinitesimal analysis, it becomes difficult to imagine the physics behind. All is blurred. Indeed, if we’re talking a very small volume of gas, we’re talking a limited number of particles also and, hence, these particles doing work on other gas particles, or these particles getting warmer or colder as they collide with the surrounding body of gas, it all becomes more or less the same. To put it simply: they’re more likely to follow the direction of the red and black arrows in our diagram above. So, yes, the theoretical analysis is what it is: a mathematical idealization, and so we shouldn’t think that’s what actually going on in a gas—even if Feynman tries to think of it in that way. So, yes, I agree with some critics, but to a very limited extent only, who say that Feynman’s Lectures on thermodynamics aren’t the best in the Volume: it may be simpler to just derive the equation we need from some Hamiltonian or whatever other mathematical relationship involving state variables like entropy or what have you. However, I do appreciate Feynman’s attempt to connect the math with the physics, which is what he’s doing here. If anything, it’s sure got me thinking!

In any case, we need to get on with the analysis, so let’s wrap it up. We know the net amount of work that’s being done is equal to W = Q₁(T₁ – T₂)/ T₁ = ΔQ(ΔT/T). So that’s equal to ΔPΔV and, hence, we can write:

net work done by the gas = ΔPΔV = ΔQ(ΔT/T)

This implies ΔQ = T(ΔP/ΔT)ΔV. Now, looking at the diagram, we can appreciate ΔP/ΔT is equal to ∂P/∂T (ΔP is the change in pressure at constant volume). Hence, ΔQ = T(∂P/∂T)ΔV. Now we have to add the work, so that’s −PΔV. We get:

ΔU = ΔQ − PΔV = T(∂P/∂T)ΔV − PΔV ⇔ ΔU/ΔV = ∂U/∂V = T(∂P/∂T) − P

So… We are where we wanted to be. 🙂 It’s a rather surprising analysis, though. Is the Q₂ = 0 assumption essential? It is, as part of the analysis of the analysis of the second term in the ΔU = (∂U/∂T)ΔT + (∂U/∂V)ΔV expression, that is. Make no mistake: the W = Q₁(T₁−T₂)/ T₁ = ΔQ(ΔT/T) formula is valid, always, and the Q₂ is taken into account in it implicitly, because of the ΔT (which is defined using T₂). However, if Q₂ would not be zero, it would add to the internal energy without doing any work and, as such, it would be part of the first term in the ΔU = (∂U/∂T)ΔT + (∂U/∂V)ΔV expression: we’d have heat that is not changing the volume (and, hence, that is not doing any work) but that’s just… Well… Heat that’s just adding heat to the gas. 🙂

To wrap everything up, let me jot down the whole thing now:

ΔU = (∂Q/∂T)·ΔT + [T(∂P/∂T) − P]·ΔV

Now, strangely enough, while we started off saying the second term in our ΔU expression assumed constant temperature (because of the ∂U/∂V derivative), we now re-write that second term using the ∂P/∂T derivative, which assumes constant volume! Now, our first term assumes constant volume too, and so we end up with an expression which assumes constant volume throughout! At the same time, we do have that ΔV factor of course, which implies we do not really assume volume is constant. On the contrary: the question we started off with was about how the internal energy changes with temperature and volume. Hence, the assumptions of constant temperature and volume only concern the partial derivatives that we are using to calculate that change!

Now, as for the model itself, let me repeat: when doing such analysis, it is very difficult to imagine the physics behind. All is blurred. When talking infinitesimally small volumes of gas, one cannot really distinguish between particles doing work on other gas particles, or these particles getting warmer or colder as they collide with them. It’s all the same. So, in reality, the actual paths are more like the red and black arrows in our diagram above. Even for larger volumes of gas, we’ve got a problem: one volume of gas is not thermally isolated from another and, hence, ideal gas is not some Carnot engine. A Carnot engine is this theoretical construct, which assumes we can nicely separate isothermal from adiabatic expansion/compression. In reality, we can’t. Even to get the isothermal expansion started, we need a temperature difference in order to get the energy flow going, which is why the assumption of frictionless heat transfer is so important. But what’s frictionless, and what’s an infinitesimal temperature difference? In the end, it’s a difference, right? So we already have some entropy increase: some heat (let’s say ΔQ) leaves the reservoir, which has temperature T, and enters the cylinder, which has to have a temperature that’s just-a-wee bit lower, let’s say T – ΔT. Hence, the entropy of the reservoir is reduced by ΔQ/T, and the entropy of the cylinder is increased by ΔQ/(T – ΔT). Hence, ΔS = ΔQ/(T–ΔT) – ΔQ/T = ΔQΔT/[T(T–ΔT)].

You’ll say: sure, but then the temperature in the cylinder must go up to T and… No. Why? We don’t have any information on the volume of the cylinder here. We should also involve the time derivatives, so we should start asking questions like: how much power goes into the cylinder, so what’s the energy exchange per unit time here? The analysis will become endlessly more complicated of course – it may have played a role in Sadi Carnot suffering from “mania” and “general delirium” when he got older 🙂 – but you should arrive at the same conclusion: when everything is said and done, the model is what it is, and that’s a mathematical model of some ideal engine – i.e. an idea of a device we don’t find in Nature, and which we’ll never be able to actually build – that shows how we could, potentially, get some energy out of a gas when using some device build to do just that. As mentioned above, thinking in terms of actual engines – like steam engines or, worse, combustion engines – does not help. Not at all really: just try to understand the Carnot cycle as it’s being presented, and that’s usual a mathematical presentation, which is why textbooks always remind the reader to not take the cylinder and piston thing too literally.

Let me note one more thing. Apart from the heat or energy loss question, there’s another unanswered question: from what source do we take the energy to move our cylinder from one heat reservoir to the other? We may imagine it all happens in space so there’s no gravity and all that (so we do not really have to spend some force just holding it) but even then: we have to move it from one place to another, and so that involves some acceleration and deceleration and, hence, some force times a distance. In short, the conclusion is all the same: the reversible Carnot cycle does not really exist and entropy increases, always.

With this, you should be able to solve some practical problems, which should help you to get the logic of it all. Let’s start with one.

Feynman’s rubber band engine

Feynman’s rubber band engine shows the model is quite general indeed, so it’s not limited to some Carnot engine only. A rubber band engine? Yes. When we heat a rubber band, it does not expand: it contracts, as shown below.

Why? It’s not very intuitive: heating a metal bar makes it longer, not shorter. It’s got to do with the fact that rubber consists of an enormous tangle of long chains of molecules: think of molecular spaghetti. But don’t worry about the details: just accept we could build an engine using the fact, as shown above. It’s not a very efficient machine (Feynman thinks he’d need heating lamps delivering 400 watts of power to lift a fly with it), but let’s apply our thermodynamic relations:

When we heat the rubber band, it will pull itself in, thereby doing some work. We can write that amount of work as FD So that’s like -PΔV in our ΔU = ΔQ – PΔV equation, but not that F has a direction that’s opposite to the direction of the pressure, so we don’t have the minus sign.
So here we can write: ΔU = ΔQ + FΔL.

So what? Well… We can re-write all of our gas equations by substituting –F for P and L for V, and they’ll apply! For example, when analyzing that infinitesimal Carnot cycle above, we found that ΔQ = T(∂P/∂T)ΔV, with ΔQ the heat that’s needed to change the volume by ΔV at constant temperature. So now we can use the above-mentioned substitution (P becomes –F and V becomes L) to calculate the heat that’s needed to change the length of the rubber band by ΔL at constant temperature: it is equal to ΔQ = –T(∂F/∂T)ΔL. The result may not be what we like (if we want the length to change significantly, we’re likely to need a lot of heat and, hence, we’re likely to end up melting the rubber), but it is what it is. 🙂

As Feynman notes: the power of these thermodynamic equations is that we can apply them to very different situations than gas. Another example is a reversible electric cell, like a rechargeable storage battery. Having said that, the assumption that these devices are all efficient is a rather theoretical one and, hence, that constrains the usefulness of our equations significantly. Having said that, engineers have to start somewhere, and the efficient Carnot cycle is the obvious point of departure. It is also a theoretical reference point to calculate actual efficiencies of actual engines, of course.

Post scriptum: Thermodynamic temperature

Let me quickly say something about an alternative definition of temperature: it’s what Feynman refers to as the thermodynamic definition. It’s an equivalent to the kinetic definition really, but let me quickly show why. As we think about efficient engines, it would be good to have some reference temperature T₂, so we can drop the subscripts and have our engines run between T and that reference temperature, which we’ll simply call ‘one degree’ (1°). The amount of heat that an ideal engine will deliver at that reference temperature is denoted by Q_S, so we can drop the subscript for Q₁ and denote it, quite simply, as Q.

We’ve defined entropy as S = Q/T, so Q = ST and Q_S = S·1°. So what? Nothing much. Just note we can use the S = Q/T and Q_S = S×1° equations to define temperature in terms of entropy. This definition is referred to as the thermodynamic definition, and it is fully equivalent with our kinetic energy. It’s just a different approach. Feynman makes kind of a big deal out of this but, frankly, there’s nothing more to it.

Just note that the definition also works for our ideal engine with non-ideal gas: the amounts of heat involved for the engine with non-ideal gas, i.e. Q and Q_S, will be proportionally less than the Q and Q_S amounts for the reversible engine with ideal gas. Remember that Q_{1_A}/Q_{1_B}= Q_{2_A}/Q_{2_B} equation, in case you’d have doubts.] Hence, we do not get some other thermodynamical temperature! All makes sense again, as it should! 🙂

The Ideal versus the Actual Gas Law

Original post:

In previous posts, we referred, repeatedly, to the so-called ideal gas law, for which we have various expressions. The expression we derived from analyzing the kinetics involved when individual gas particles (atoms or molecules) move and collide was P·V = N·k·T, in which the variables are P (pressure), V (volume), N (the number of particles in the given volume), T (temperature) and k (the Boltzmann constant). We also wrote it as P·V = (2/3)·U, in which U represents the total energy, i.e. the sum of the energies of all gas particles. We also said the P·V = (2/3)·U formula was only valid for monatomic gases, in which case U is the kinetic energy of the center-of-mass motion of the atoms.

In order to provide some more generality, the equation is often written as P·V = (γ–1)·U. Hence, for monatomic gases, we have γ = 5/3. For a diatomic gas, we’ll also have vibrational and rotational kinetic energy. As we pointed out in a previous post, each independent direction of motion, i.e. each degree of freedom in the system, will absorb an amount of energy equal to k·T/2. For monatomic gases, we have three independent directions of motion (x, y, z) and, hence, the total energy U = 3·k·T/2 = (2/3)·U.

Finally, when we’re considering adiabatic expansion/compression only – so when we do not add or remove any heat to/from to the gas – we can also write the ideal gas law as PV^γ= C, with C some constant. [It is important to note that this PV^γ= C relation can be derived from the more general P·V = (γ–1)·U expression, but that the two expressions are not equivalent. Please have a look at the P.S. to this post on this, which shows how we get that PV^γ= constant expression, and talks a bit about its meaning.]

So what’s the gas law for diatomic gas, like O₂, i.e. oxygen? The key to the analysis of diatomic gases is, basically, a model which represents the oxygen molecule as two atoms connected by a spring, but with a force law that’s not as simplistic as Hooke’s law: we’re not looking at some linear force, but a force that’s referred to as a van der Waals force. The image below gives a vague idea of what that might imply. Remember: when moving an object in a force field, we change its potential energy, and the work done, as we move with or against the force, is equal to the change in potential energy. The graph below shows the force is anything but linear.

The illustration above is a graph of potential energy for two molecules, but we can also apply it for the ‘spring’ model for two atoms within a single molecule. For the detail, I’ll refer you to Feynman’s Lecture on this. It’s not that the full story is too complicated: it’s just too lengthy to reproduce it in this post. Just note the key point of the whole story: one arrives at a theoretical value for γ that is equal to γ = 9/7 ≈ 1.286. Wonderful! Yes. Except for the fact that value does not correspond to what is measured in reality: the experimentally confirmed value for γ for oxygen (O₂) is about 1.40.

What about other gases? When measuring the value for other diatomic gases, like iodine (I₂) or bromine (Br₂), we get a value closer to the theoretical value (1.30 and 1.32 respectively) but, still, there’s a variation to be explained here. The value for hydrogen H₂ is about 1.4, so that’s like oxygen again. For other gases, we again get different values. Why? What’s the problem?

It cannot be explained using classical theory. In addition, doing the measurements for oxygen and hydrogen at various temperatures also reveals that γ is a function of temperature, as shown below. Now that’s another experimental fact that does not line up with our kinetic theory of gases!

Reality is right, always. Hence, our theory must be wrong. Our analysis of the independent direction of motions inside of a molecule doesn’t work—even for the simple case of a diatomic molecule. Great minds such as James Clerk Maxwell couldn’t solve the puzzle in the 19th century and, hence, had to admit classical theory was in trouble. Indeed, popular belief has it that the black-body radiation problem was the only thing classical theory couldn’t explain in the late 19th century but that’s not true: there were many more problems keeping physicists awake. But so we’ve got a problem here. As Feynman writes: “We might try some force law other than a spring but it turns out that anything else will only make γ higher. If we include more forms of energy, γ approaches unity more closely, contradicting the facts. All the classical theoretical things that one can think of will only make it worse. The fact is that there are electrons in each atom, and we know from their spectra that there are internal motions; each of the electrons should have at least kT/2 of kinetic energy, and something for the potential energy, so when these are added in, γ gets still smaller. It is ridiculous. It is wrong.”

So what’s the answer? The answer is to be found in quantum mechanics. Indeed, one can develop a model distinguishing various molecular states with various energy levels E₀, E₁, E₂,…, E_i,…, and then associate a probability distribution which gives us the probability of finding a molecule in a particular state. Some more assumptions, all quite similar to the assumptions used by Planck when he solved the black-body radiation problem, then give us what we want: to put it simply, it is like some of the motions ‘freeze out’ at lower temperatures. As a result, γ goes up as we go down in temperature.

Hence, quantum mechanics saves the day, again. However, that’s not what I want to write about here. What I want to do here is to give you an equation for the internal energy of a gas which is based on what we can actually measure, so that’s pressure, volume and temperature. I’ll refer to it as the Actual Gas Law, because it takes into account that γ is not some fixed value (so it’s not some natural constant, like Planck’s or Boltzmann’s constant), and it also takes into account that we’re not always gas—ideal or actual gas—but also liquids and solids.

Now, we have many inter-connected variables here, and so the analysis is quite complicated. In fact, it’s a great opportunity to learn more about partial derivatives and how we can use them. So the lesson is as much about math as it about physics. In fact, it’s probably more about math. 🙂 Let’s see what we can make out of it.

Energy, work, force, pressure and volume

First, I should remind you that work is something that is done by a force on some object in the direction of the displacement of that object. Hence, work is force times distance. Now, because the force may actually vary as our object is being displaced and while the work is being done, we represent work as a line integral:

W = ∫F·ds

We write F and s in bold-face and, hence, we’ve got a vector dot product here, which ensures we only consider the component of the force in the direction of the displacement: F·Δs = |F|·|Δs|·cosθ, with θ the angle between the force and the displacement.

As for the relationship between energy and work, you know that one: as we do work on an object, we change its energy, and that’s what we are looking at here: the (internal) energy of our substance. Indeed, when we have a volume of gas exerting pressure, it’s the same thing: some force is involved (pressure is the force per unit area, so we write: P = F/A) and, using the model of the box with the frictionless piston (illustrated below), we write:

dW = F(–dx) = – PAdx = – PdV

The dW = – PdV formula is the one we use when looking at infinitesimal changes. When going through the full thing, we should integrate, as the volume (and the pressure) changes over the trajectory, so we write:

W = ∫PdV

Now, it is very important to note that the formulas above (dW = – PdV and W = ∫PdV) are always valid. Always? Yes. We don’t care whether or not the compression (or expansion) is adiabatic or isothermal. [To put it differently, we don’t care whether or not heat is added to (or removed from) the gas as it expands (or decreases in volume).] We also don’t keep track of the temperature here. It doesn’t matter. Work is work.

Now, as you know, an integral is some area under a graph so I can rephrase our result as follows: the work that is being done by a gas, as it expands (or the work that we need to put in in order to compress it), is the area under the pressure-volume graph, always.

Of course, as we go through a so-called reversible cycle, getting work out of it, and then putting some work back in, we’ll have some overlapping areas cancelling each other. That’s how we derived the amount of useful (i.e. net) work that can be done by an ideal gas engine (illustrated below) as it goes through a Carnot cycle, taking in some amount of heat Q₁ from one reservoir (which is usually referred to as the boiler) and delivering some other amount of heat (Q₁) to another reservoir (usually referred to as the condenser). As I don’t want to repeat myself too much, I’ll refer you to one of my previous posts for more details. Hereunder, I just present the diagram once again. If you want to understand anything of what follows, you need to understand it—thoroughly.

It’s important to note that work is being done in each of the four steps of the cycle, and that the work done by the gas is positive when it expands, and negative when its volume is being reduced. So, let me repeat: the W = ∫PdV formula is valid for both adiabatic as well as isothermal expansion/compression. We just need to be careful about the sign and see in which direction it goes. Having said that, it’s obvious adiabatic and isothermal expansion/compression are two very different things and, hence, their impact on the (internal) energy of the gas is quite different:

Adiabatic compression/expansion assumes that no (external) heat energy (Q) is added or removed and, hence, all the work done goes into changing the internal energy (U). Hence, we can write: W = PΔV = –ΔU and, therefore, ΔU = –PΔV. Of course, adiabatic compression/expansion must involve a change in temperature, as the kinetic energy of the gas molecules is being transferred from/to the piston. Hence, the temperature (which is nothing but the average kinetic energy of the molecules) changes.
In contrast, isothermal compression/expansion (i.e. a volume change without any change in temperature) must involve an exchange of heat energy with the surroundings so to allow the temperature to remain constant. So ΔQ ≠ 0 in this case.

The grand but simple formula capturing all is, obviously:

ΔU = ΔQ – PΔV

It says what we’ve said already: the internal energy of a substance (a gas) changes because some work is being done as its volume changes and/or because some heat is added or removed.

Now we have to get serious about partial derivatives, which relate one variable (the so-called ‘dependent’ variable) to another (the ‘independent’ variable). Of course, in reality, all depends on all and, hence, the distinction is quite artificial. Physicists tend to treat temperature and volume as the ‘independent’ variables, while chemists seem to prefer to think in terms of pressure and temperature. In math, it doesn’t matter all that much: we simply take the reciprocal and there you go: dy/dx = 1/(dx/dy). We go from one to another. Well… OK… We’ve got a lot of variables here, so… Yes. You’re right. It’s not going to be that simple, obviously! 🙂

Differential analysis

If we have some function f in two variables, x and y, then we can write: Δf = f(x + Δx, y + Δy) – f(x, y). We can then write the following clever thing:

What’s being said here is that we can approximate Δf using the partial derivatives ∂f/∂x and ∂f/∂y. Note that the formula above actually implies that we’re evaluating the (partial) ∂f/∂x derivative at point (x, y+Δy), rather than the point (x, y) itself. It’s a minor detail, but I think it’s good to signal it: this ‘clever thing’ is just pedagogical. [Feynman is the greatest teacher of all times! :-)] The mathematically correct approach is to simply give the formal definition of partial derivatives, and then just get on with it:

Now, let us apply that Δf formula to what we’re interested in, and that’s the change in the (internal) energy U. So we write:

Now, we can’t do anything with this, in practice, because we cannot directly measure the two partial derivatives. So, while this is an actual gas law (which is what we want), it’s not a practical one, because we can’t use it. 🙂 Let’s see what we can do about that. We need to find some formula for those partial derivatives. Let’s have a look at the (∂U/∂T)_Vfactor first. That factor is defined and referred to as the specific heat capacity at constant volume, and it’s usually denoted by C_V. Hence, we write:

C_V = specific heat capacity at constant volume = (∂U/∂T)_V

Heat capacity? But we’re talking internal energy here? It’s the same. Remember that ΔU = ΔQ – PΔV formula: if we keep the volume constant, then ΔV = 0 and, hence, ΔU = ΔQ. Hence, all of the change in internal energy (and I really mean all of the change) is the heat energy we’re adding or removing from the gas. Hence, we can also write C_V in its more usual definitional form:

C_V= (∂Q/∂T)_V

As for its interpretation, you should look at it as a ratio: C_Vis the amount of heat one must put into (or remove from) a substance in order to change its temperature by one degree with the volume held constant. Note that the term ‘specific heat capacity’ is usually referred to as the ‘specific heat’, as that’s shorter and simpler. However, you can see it’s some kind of ‘capacity’ indeed. More specifically, it’s a capacity of a substance to absorb heat. Now that’s stuff we can actually measure and, hence, we’re done with the first term in that ΔU = ΔT·(∂U/∂T)_V+ ΔV·(∂U/∂V)_Texpression, which we can now write as:

ΔT·(∂U/∂T)_V= ΔT·(∂Q/∂T)_V= ΔT·C_V

OK. So we’re done with the first term. Just to make sure we’re on the right track here, let’s have a quick look at the units here: the unit in which we should measure C_Vis, obviously, joule per degree (Kelvin), i.e. J/K. And then we multiply with ΔT, which is measured in degrees Kelvin, and we get some amount in Joule. Fine. We’re done, indeed. 🙂

Let’s look at the second term now, i.e. the ΔV·(∂U/∂V)_T term. Now, you may think that we could define C_T = (∂U/∂V)_Tas the specific heat capacity at constant temperature because… Well… Hmm… It is the amount of heat one must put into (or remove from) a substance in order to change its volume by one unit with the temperature held constant, isn’t it? So we write C_T = (∂U/∂V)_T = (∂Q/∂V)_T and we’re done here too, aren’t we?

NO! HUGE MISTAKE!

It’s not that simple. Two very different things are happening here. Indeed, the change in (internal) energy ΔU, as the volume changes by ΔV while keeping the temperature constant (we’re looking at that (∂U/∂V)_T factor here, and I’ll remind you of that subscript T a couple of times), consists of two parts:

First, the volume is not being kept constant and, hence, the internal energy (U) changes because work is being done.
Second, the internal energy (U) also changes because heat is being put in, so the temperature can be kept constant indeed.

So we cannot simplify. We’re stuck with the full thing: ΔU = ΔQ – PΔV, in which – PΔV is the (infinitesimal amount of) work that’s being done on the substance, and ΔQ is the (infinitesimal amount of) heat that’s being put in. What can we do? How can we relate this to actual measurables?

Now, the logic is quite abstruse, so please be patient and bear with me. The key to the analysis is that diagram of the reversible Carnot cycle, with the shaded area representing the net work that’s being done, except that we’re now talking infinitesimally small changes in volume, temperature and pressure. So we redraw the diagram and get something like this:

Now, you can easily see the equivalence between the shaded area and the ΔPΔV rectangle below:

So the work done by the gas is the shaded area, whose surface is equal to ΔPΔV. […] But… Hey, wait a minute! You should object: we are not talking ideal engines here and, hence, we are not going through a full Carnot cycle, are we? We’re calculating the change in internal energy when the temperature changes with ΔT, the volume changes with ΔV, and the pressure changes with ΔP. Full stop. So we’re not going back to where we came from and, hence, we should not be analyzing this thing using the Carnot cycle, should we? Well… Yes and no. More yes than no. Remember we’re looking at the second term only here: ΔV·(∂U/∂V)_T. So we are changing the volume (and, hence, the internal energy) but the subscript in the (∂U/∂V)_Tterm makes it clear we’re doing so at constant temperature. In practice, that means we’re looking at a theoretical situation here that assumes a complete and fully reversible cycle indeed. Hence, the conceptual idea is, indeed, that we put some heat in, that the gas does some work as it expands, and that we then are actually putting some work back in to bring the gas back to its original temperature T. So, in short, yes, the reversible cycle idea applies.

[…] I know, it’s very confusing. I am actually struggling with the analysis myself, so don’t be too hard on yourself. Think about it, but don’t lose sleep over it. 🙂 I added a note on it in the P.S. to this post on it so you can check that out too. However, I need to get back to the analysis itself here. From our discussion of the Carnot cycle and ideal engines, we know that the work done is equal to the difference between the heat that’s being put in and the heat that’s being delivered: W = Q₁ – Q₂. Now, because we’re talking reversible processes here, we also know that Q₁/T₁ = Q₂/T₂. Hence, Q₂ = (T ₂/T₁)Q₁ and, therefore, the work done is also equal to W = Q₁– (T ₂/T₁)Q₁ = Q₁(1 – T ₂/T₁) = Q₁[(T₁– T₂)/T₁]= Q₁(ΔT/T₁). Let’s now drop the subscripts by equating Q₁ with ΔQ, so we have:

W = ΔQ(ΔT/T)

You should note that ΔQ is not the difference between Q₁ and Q₂. It is not. ΔQ is the heat we put in as it expands isothermally from volume V to volume V + ΔV. I am explicit about it because the Δ symbol usually denotes some difference between two values. In case you wonder how we can do away with Q₂, think about it. […] The answer is that we did not really get away with it: the information is captured in the ΔT factor, as T–ΔT is the final temperature reached by the gas as it expands adiabatically on the second leg of the cycle, and the change in temperature obviously depends on Q₂! Again, it’s all quite confusing because we’re looking at infinitesimal changes only, but the analysis is valid. [Again, go through the P.S. of this post if you want more remarks on this, although I am not sure they’re going to help you much. The logic is really very deep.]

[…] OK… I know you’re getting tired, but we’re almost done. Hang in there. So what do we have now? The work done by the gas as it goes through this infinitesimally small cycle is the shaded area in the diagram above, and it is equal to:

W = ΔPΔV = ΔQ(ΔT/T)

From this, it follows that ΔQ = T·ΔV·ΔP/ΔT. Now, you should look at the diagram once again to check what ΔP actually stands for: it’s the change in pressure when the temperature changes at constant volume. Hence, using our partial derivative notation, we write:

ΔP/ΔT = (∂P/∂T)_V

We can now write ΔQ = T·ΔV·(∂P/∂T)_Vand, therefore, we can re-write ΔU = ΔQ – PΔV as:

ΔU = T·ΔV·(∂P/∂T)_V– PΔV

Now, dividing both sides by ΔV, and writing all using the partial derivative notation, we get:

ΔU/ΔV = (∂U/∂V)_T= T·(∂P/∂T)_V– P

So now we know how to calculate the (∂U/∂V)_Tfactor, from measurable stuff, in that ΔU = ΔT·(∂U/∂T)_V+ ΔV·(∂U/∂V)_Texpression, and so we’re done. Let’s write it all out:

ΔU = ΔT·(∂U/∂T)_V+ ΔV·(∂U/∂V)_T= ΔT·C_V+ ΔV·[T·(∂P/∂T)_V– P]

Phew! That was tough, wasn’t it? It was. Very tough. As far as I am concerned, this is probably the toughest of all I’ve written so far.

Dependent and independent variables

Let’s pause to take stock of what we’ve done here. The expressions above should make it clear we’re actually treating temperature and volume as the independent variables, and pressure and energy as the dependent variables, or as functions of (other) variables, I should say. Let’s jot down the key equations once more:

ΔU = ΔQ – PΔV
ΔU = ΔT·(∂U/∂T)_V+ ΔV·(∂U/∂V)_T
(∂U/∂T)_V= (∂Q/∂T)_V = C_V
(∂U/∂V)_T= T·(∂P/∂T)_V– P

It looks like Chinese, doesn’t it? 🙂 What can we do with this? Plenty. Especially the first equation is really handy for analyzing and solving various practical problems. The second equation is much more difficult and, hence, less practical. But let’s try to apply this equation for actual gases to an ideal gas—just to see if we’re getting our ideal gas law once again. 🙂 We know that, for an ideal gas, the internal energy depends on temperature, not on V. Indeed, if we change the volume but we keep the temperature constant, the internal energy should be the same, as it only depends on the motion of the molecules and their number. Hence, (∂U/∂V)_Tmust equal zero and, hence, T·(∂P/∂T)_V– P = 0. Replacing the partial derivative with an ordinary one (not forgetting that the volume is kept constant), we get:

T·(dP/dT)– P = 0 (constant volume)

⇔ (1/P)·(dP/dT) = 1/T (constant volume)

Integrating both sides yields: lnP = lnT + constant. This, in turn, implies that P = T × constant. [Just re-write the first constant as the (natural) logarithm of some other constant, i.e. the second constant, obviously).] Now that’s consistent with our ideal gas P = NkT/V, because N, k and V are all constant. So, yes, the ideal gas law is a special case of our more general thermodynamical expression. Fortunately! 🙂

That’s not very exciting, you’ll say—and you’re right. You may be interested – although I doubt it 🙂 – in the chemists’ world view: they usually have performance data (read: values for derivatives) measured under constant pressure. The equations above then transform into:

ΔH = Δ(U + P·V) = ΔQ + VΔP
ΔH = ΔT·(∂H/∂T)_P+ ΔP·(∂H/∂P)_T
(∂H/∂P)_T= –T·(∂V/∂T)_P+ V

H? Yes. H is another so-called state variable, so it’s like entropy or internal energy but different. As they say in Asia: “Same-same but different.” 🙂 It’s defined as H = U + PV and its name is enthalpy. Why do we need it? Because some clever man noted that, if you take the total differential of P·V, i.e. Δ(P·V) = P·ΔV + V·ΔP, and our ΔU = ΔQ – P·ΔV expression, and you add both sides of both expressions, you get Δ(U + P·V) = ΔQ + VΔP. So we’ve substituted –P for V – so as to please the chemists – and all our equations hold provided we substitute U for H and, importantly, –P for V. [Note the sign switch is to be applied to derivatives as well: if we substitute P for –V, then ∂P/∂T becomes ∂(–V)/∂T = –(∂V/∂T)!

So that’s the chemists’ model of the world, and they’ll usually measure the specific heat capacity at constant pressure, rather than at constant volume. Indeed, one can show the following:

(∂H/∂T)_P= (∂Q/∂T)_P= C_P = the specific heat capacity at constant pressure

In short, while we referred to γ as the specific heat ratio in our previous posts, assuming we’re talking ideal gases only, we can now appreciate the fact there is actually no such thing as the specific heat: there are various variables and, hence, various definitions. Indeed, it’s not only pressure or volume: the specific heat capacity of some substance will usually also be expressed as a function of its mass (i.e. per kg), the number of particles involved (i.e. per mole), or its volume (i.e. per m³). In that case, we talk about the molar or volumetric heat capacity respectively. The name for the same thing expressed in joule per degree Kelvin and per kg (J/kg·K) is the same: specific heat capacity. So we’ve got three different concepts here, and two ways of measuring them: at constant pressure or at constant volume. No wonder one gets quite confused when googling tables listing the actual values! 🙂

Now, there’s one question left: why is γ being referred to as the specific heat ratio? The answer is simple: it actually is the ratio of the specific heat capacities C_P and C_V. Hence, γ is equal to:

γ = C_P/C_V

I could show you how that works. However, I would just be copying the Wikipedia article on it, so I won’t do that: you’re sufficiently knowledgeable now to check it out yourself, and verify it’s actually true. Good luck with it ! In the process, please also do check why C_Pis always larger than C_Vso you can explain why γ is always larger than one. 🙂

Post scriptum: As usual, Feynman’s Lectures, were the inspiration here—once more. Now, Feynman has a habit of ‘integrating’ expressions and, frankly, I never found a satisfactory answer to a pretty elementary question: integration in regard to what variable? His exposé on both the ideal as well as the actual gas law has enlightened me. The answer is simple: it doesn’t matter. 🙂 Let me show that by analyzing the following argument of Feynman:

So… What is that ‘integration’ that ‘yields’ that γlnV + lnP = lnC expression? Are we solving some differential equation here? Well… Yes. But let’s be practical and take the derivative of the expression in regard to V, P and T respectively. Let’s first see where we come from. The fundamental equation is PV = (γ–1)U. That means we’ve got two ‘independent’ variables, and one that ‘depends’ on the others: if we fix P and V, we have U, or if we fix U, then P and V are in an inversely proportional relationship. That’s easy enough. We’ve got three ‘variables’ here: U, P and V—or, in differential form, dU, dP and dV. However, Feynman eliminates one by noting that dU = –PdV. He rightly notes we can only do that because we’re talking adiabatic expansion/compression here: all the work done while expanding/compressing the gas goes into changing the internal energy: no heat is added or removed. Hence, there is no dQ term here.

So we are left with two ‘variables’ only now: P and V, or dP and dV when talking differentials. So we can choose: P depends on V, or V depends on P. If we think of V as the independent variable, we can write:

d[γ·lnV + lnP]/dV = γ·(1/V)·(dV/dV) + (1/P)·(dP/dV), while d[lnC]/dV = 0

So we have γ·(1/V)·(dV/dV) + (1/P)·(dP/dV) = 0, and we can then multiply sides by dV to get:

(γ·dV/V) + (dP/P) = 0,

which is the core equation in this argument, so that’s the one we started off with. Picking P as the ‘independent’ variable and, hence, integrating with respect to P yields the same:

d[γ·lnV + lnP]/dP = γ·(1/V)·(dV/dP) + (1/P)·(dP/dP), while d[lnC]/dP = 0

Multiplying both sides by dP yields the same thing: (γ·dV/V) + (dP/P) = 0. So it doesn’t matter, indeed. But let’s be smart and assume both P and V, or dP and dV, depend on some implicit variable—a parameter really. The obvious candidate is temperature (T). So we’ll now integrate and differentiate in regard to T. We get:

d[γ·lnV + lnP]/dT = γ·(1/V)·(dV/dT) + (1/P)·(dP/dT), while d[lnC]/dT = 0

We can, once again, multiply both sides with dT and – surprise, surprise! – we get the same result:

(γ·dV/V) + (dP/P) = 0

The point is that the γlnV + lnP = lnC expression is damn valid, and C or lnC or whatever is ‘the constant of integration’ indeed, in regard to whatever variable: it doesn’t matter. So then we can, indeed, take the exponential of both sides (which is much more straightforward than ‘integrating both sides’), so we get:

e^{γlnV + lnP}= e^lnC= C

It then doesn’t take too much intelligence to see that e^{γlnV + lnP}= e^{(lnV)^γ}^+lnP= e^{(lnV)^γ}·e^lnP= V^γ·P = P·V^γ. So we’ve got the grand result that what we wanted: PV^γ= C, with C some constant determined by the situation we’re in (think of the size of the box, or the density of the gas).

So, yes, we’ve got a ‘law’ here. We should just remind ourselves, always, that it’s only valid when we’re talking adiabatic compression or expansion: so we we do not add or remove heat energy or, as Feynman puts it, much more succinctly, “no heat is being lost“. And, of course, we’re also talking ideal gases only—which excludes a number of real substances. 🙂 In addition, we’re talking adiabatic processes only: we’re not adding nor removing heat.

It’s a weird formula: the pressure times the volume to the 5/3 power is a constant for monatomic gas. But it works: as long as individual atoms are not bound to each other, the law holds. As mentioned above, when various molecular states, with associated energy levels are at play, it becomes an entirely different ballgame. 🙂

I should add one final note as to the functional form of PV^γ= C. We can re-write it as P = C/V^γ. Because The shape of that graph is similar to the P = NkT/V relationship we started off with. Putting the two equations side by side, makes it clear our constant and temperature are obviously related one to another, but they are not directly proportional to each other. In fact, as the graphs below clearly show, the P = NkT/V gives us these isothermal lines on the pressure-volume graph (i.e. they show P and V are related at constant temperature), while the P = C/V^γ equation gives us the adiabatic lines. Just google an online function graph tool, and you can now draw your own diagrams of the Carnot cycle! Just change the denominator (i.e. the constants C and T in both equations). 🙂

Now, I promised I would say something more about that infinitesimal Carnot cycle: why is it there? Why don’t we limit the analysis to just the first two steps? In fact, the shortest and best explanation I can give is something like this: think of the whole cycle as the first step in a reversible process really. We put some heat in (ΔQ) and the gas does some work, but so that heat has to go through the whole body of gas, and the energy has to go somewhere too. In short, the heat and the work is not being absorbed by the surroundings but it all stays in the ‘system’ that we’re analyzing, so to speak, and that’s why we’re going through the full cycle, not the first two steps only. Now, this ‘answer’ may or may not satisfy you, but I can’t do better. You may want to check Feynman’s explanation itself, but he’s very short on this and, hence, I think it won’t help you much either. 😦

Entropy

Pre-script (dated 26 June 2020): This post has become less relevant (even irrelevant, perhaps) because my views on all things quantum-mechanical have evolved significantly as a result of my progression towards a more complete realist (classical) interpretation of quantum physics. It also got mutilated because of an attack by dark forces. I keep blog posts like these mainly because I want to keep track of where I came from. I might review them one day, but I currently don’t have the time or energy for it. 🙂

Original post:

The two previous posts were quite substantial. Still, they were only the groundwork for what we really want to talk about: entropy, and the second law of thermodynamics, which you probably know as follows: all of the energy in the universe is constant, but its entropy is always increasing. But what is entropy really? And what’s the nature of this so-called law?

Let’s first answer the second question: Wikipedia notes that this law is more like an empirical finding that has been accepted as an axiom. That probably sums it up best. That description does not downplay its significance. In fact, Newton’s laws of motion, or Einstein’s relatively principle, have the same status: axioms in physics – as opposed to those in math – are grounded in reality. At the same time, and just like in math, one can often choose alternative sets of axioms. In other words, we can derive the law of ever-increasing entropy from other principles, notably the Carnot postulate, which basically says that, if the whole world were at the same temperature, it would impossible to reversibly extract and convert heat energy into work. I talked about that in my previous post, and so I won’t go into more detail here. The bottom line is that we need two separate heat reservoirs at different temperatures, denoted by T₁and T₂, to convert heat into useful work.

Let’s go to the first question: what is entropy, really?

Defining entropy

Feynman, the Great Teacher, defines entropy as part of his discussion on Carnot’s ideal reversible heat engine, so let’s have a look at it once more. Carnot’s ideal engine can do some work by taking an amount of heat equal to Q₁out of one heat reservoir and putting an amount of heat equal to Q₂ into the other one (or, because it’s reversible, it can also go the other way around, i.e. it can absorb Q₂ and put Q₁ back in, provided we do the same amount of work W on the engine).

The work done by such machine, or the work that has to be done on the machine when reversing the cycle, is equal W = Q₁ – Q₂ (the equation shows the machine is as efficient as it can be, indeed: all of the difference in heat energy is converted into useful work, and vice versa—nothing gets ‘lost’ in frictional energy or whatever else!). Now, because it’s a reversible thermodynamic process, one can show that the following relationship must hold:

Q₁/T₁= Q₂/T₂

This law is valid, always, for any reversible engine and/or for any reversible thermodynamic process, for any Q₁, Q₂, T₁and T₂. [Ergo, it is not valid for non-reversible processes and/or non-reversible engines, i.e. real machines.] Hence, we can look at Q/T as some quantity that remains unchanged: an equal ‘amount’ of Q/T is absorbed and given back, and so there is no gain or loss of Q/T (again, if we’re talking reversible processes, of course). [I need to be precise here: there is no net gain or loss in the Q/T of the substance of the gas. The first reservoir obviously looses Q₁/T₁, and the second reservoir gains Q₂/T₂. The whole environment only remains unchanged if we’d reverse the cycle.]

In fact, this Q/T ratio is the entropy, which we’ll denote by S, so we write:

S = Q₁/T₁= Q₂/T₂

What the above says, is basically the following: whenever the engine is reversible, this relationship between the heats must follow: if the engine absorbs Q₁at T₁and delivers Q₂at T₂, then Q₁is to T₁as Q₂is to T₂ and, therefore, we can define the entropy S as S = Q/T. That implies, obviously:

Q = S·T

From these relations (S = Q/T and Q = S·T), it is obvious that the unit of entropy has to be joule per degree (Kelvin), i.e. J/K. As such, it has the same dimension as the Boltzmann constant, k_B≈ 1.38×10⁻²³J/K, which we encountered in the ideal gas formula PV = NkT, and which relates the mean kinetic energy of atoms or molecules in an ideal gas to the temperature. However, while k_Bis, quite simply, a constant of proportionality, S is obviously not a constant: its value depends on the system or, to continue with the mathematical model we’re using, the heat engine we’re looking at.

Still, this definition and relationships do not really answer the question: what is entropy, really? Let’s further explore the relationships so as to try to arrive at a better understanding.

I’ll continue to follow Feynman’s exposé here, so let me use his illustrations and arguments. The first argument revolves around the following set-up, involving three reversible engines (1, 2 and 3), and three temperatures (T₁ > T₂> T₃):

Engine 1 runs between T₁ and T₃and delivers W₁₃by taking in Q₁ at T₁ and delivering Q₃ at T₃. Similarly, engine 2 and 3 deliver or absorb W₃₂ and W₁₂respectively by running between T₃ and T₂ and between T₂ and T₁respectively. Now, if we let engine 1 and 2 work in tandem, so engine 1 produces W₁₃and delivers Q₃, which is then taken in by engine 2, using an amount of work W₃₂, the net result is the same as what engine 3 is doing: it runs between T₁ and T₂and delivers W₁₂, so we can write:

W₁₂= W₁₃– W₃₂

This result illustrates that there is only one Carnot efficiency, which Carnot’s Theorem expresses as follows:

All reversible engines operating between the same heat reservoirs are equally efficient.
No actual engine operating between two heat reservoirs can be more efficient than a Carnot engine operating between the same reservoirs.

Now, it’s obvious that it would be nice to have some kind of gauge – or a standard, let’s say – to describe the properties of ideal reversible engines in order to compare them. We can define a very simple gauge by assuming T₁in the diagram above is one degree. One degree what? Whatever: we’re working in Kelvin for the moment, but any absolute temperature scale will do. [An absolute temperature scale uses an absolute zero. The Kelvin scale does that, but the Rankine scale does so too: it just uses different units than the Kelvin scale (the Rankine units correspond to Fahrenheit units, while the Kelvin units correspond to Celsius degrees).] So what we do is to let our ideal engines run between some temperature T – at which it absorbs or delivers a certain heat Q – and 1° (one degree), at which it delivers or absorbs an amount of heat which we’ll denote by Q_S. [Of course, I note this assumes that ideal engines are able to run between one degree Kelvin (i.e. minus 272.15 degrees Celsius) and whatever other temperature. Real (man-made) engines are obviously likely to not have such tolerance. :-)] Then we can apply the Q = S·T equation and write:

Q_S= S·1°

Like that we solve the gauge problem when measuring the efficiency of ideal engines, for which the formula is W/Q₁= (T₁ – T₂)/T₁. In my previous post, I illustrated that equation with some graphs for various values of T₂(e.g. T₂= 4, 1, or 0.3). [In case you wonder why these values are so small, it doesn’t matter: we can scale the units, or assume 1 unit corresponds to 100 degrees, for example.] These graphs all look the same but cross the x-axis (i.e. the T₁-axis) at different points (at T₁= 4, 1, and 0.3 respectively, obviously). But let us now use our gauge and, hence, standardize the measurement by setting T₂to 1. Hence, the blue graph below is now the efficiency graph for our engine: it shows how the efficiency (W/Q₁) depends on its working temperature T₁only. In fact, if we drop the subscripts, and define Q as the heat that’s taken in (or delivered when we reverse the machine), we can simply write:

W/Q = (T – 1)/T = 1 – 1/T

Note the formula allows for negative values of the efficiency W/Q: if T₁would be lower than one degree, we’d have to put work in and, hence, our ideal engine would have negative efficiency indeed. Hence, the formula is consistent over the whole temperature domain T > 0. Also note that, coincidentally, the three-engine set-up and the W/Q formula also illustrate the scalability of our theoretical reversible heat engines: we can think of one machine substituting for two or three others, or any combination really: we can have several machines of equal efficiency working in parallel, thereby doubling, tripling, quadruping, etcetera, the output as well as the heat that’s being taken in. Indeed, W/Q = 2W/2Q= 3W/3Q₁= 4W/4Qand so on.

Also, looking at that three-engine model once again, we can set T $3$ to one degree and re-state the result in terms of our standard temperature:

If one engine, absorbing heat Q₁at T $1$ , delivers the heat Q_S at one degree, and if another engine absorbing heat Q₂at T $2$ , will also deliver the same heat Q_S at one degree, then it follows that an engine which absorbs heat Q₁at the temperature T $1$ will deliver heat Q₂ $if it runs between T 1 and T 2 .$

That’s just stating what we showed, but it’s an important result. All these machines are equivalent, so to say, and, as Feynman notes, all we really have to do is to find how much heat (Q) we need to put in at the temperature Tin order to deliver a certain amount of heat Q_Sat the unit temperature (i.e. one degree). If we can do that, then we have everything. So let’s go for it.

Measuring entropy

We already mentioned that we can look at the entropy S = Q/T as some quantity that remains unchanged as long as we’re talking reversible thermodynamic processes. Indeed, as much Q/T is absorbed as is given back in a reversible cycle or, in other words: there is no net change in entropy in a reversible cycle. But what does it mean really?

Well… Feynman defines the entropy of a system, or a substance really (think of that body of gas in the cylinder of our ideal gas engine), as a function of its condition, so it is a quantity which is similar to pressure (which is a function of density, volume and temperature: P = NkT/V), or internal energy (which is a function of pressure and volume (U = (3/2)·PV) or, substituting the pressure function, of density and temperature: U = (3/2)·NkT). That doesn’t bring much clarification, however. What does it mean? We need to go through the full argument and the illustrations here.

Suppose we have a body of gas, i.e. our substance, at some volume V $a$ and some temperature $T$ $a$ $(i.e. condition a),$ $and we bring it into some other condition (b), so it now has volume V b and temperature T b, as shown below. [Don’t worry about the ΔS = S b - S a and ΔS = S a - S b formulas as for now. I’ll explain them in a minute.]$

$You may think that a and b are, once again, steps in the reversible cycle of a Carnot engine, but no! What we’re doing here is something different altogether: we’ve got the same body of gas at point b but in a completely different condition : indeed, both the volume and temperature (and, hence, its pressure) of the gas is different in b as compared to a . What we do assume, however, is that the gas went from condition a to condition b$ $through a completely reversible$ $process . Cycle, process? What’s the difference? What do we mean with that?$

$As Feynman notes, we can think of going from a to b through a series of steps, during which tiny reversible heat engines take out an infinitesimal amount of heat dQ in tiny little reservoirs at the temperature corresponding to that point on the path. [Of course, depending on the path, we may have to add heat (and, hence, do work rather than getting work out). However, in this case, we see a temperature rise but also an expansion of volume, the net result of which is that the substance actually does some (net) work from a to b, rather than us having to put (net) work in.] So the process consists, in principle, of a (potentially infinite) number of tiny little cycles . The thinking is$ $illustrated below.$

Don’t panic. It’s one of the most beautiful illustrations in all of Feynman’s Lectures, IMHO. Just analyze it. We’ve got the same horizontal and vertical axis here, showing volume and temperature respectively, and the same points a and b showing the condition of the gas before and after and, importantly, also the same path from condition a to condition b, as in the previous illustration. It takes a pedagogic genius like Feynman to think of this: he just draws all those tiny little reservoirs and tiny engines on a mathematical graph to illustrate what’s going on: at each step, an infinitesimal amount of work dW is done, and an infinitesimal amount of entropy dS = dQ/T is being delivered at the unit temperature.

As mentioned, depending on the path, some steps may involve doing some work on those tiny engines, rather than getting work out of them, but that doesn’t change the analysis. Now, we can write the total entropy that is taken out of the substance (or the little reservoirs, as Feynman puts it), as we go from condition a to b, as:

ΔS = S_b – S_a

Now, in light of all the above, it’s easy to see that this ΔS can be calculated using the following integral:

So we have a function S here which depends on the ‘condition’ indeed—i.e. the volume and the temperature (and, hence, the pressure) of the substance. Now, you may or may not notice that it’s a function that is similar to our internal energy formula (i.e. the formula for U). At the same time, it’s not internal energy. It’s something different. We write:

S = S(V, T)

So now we can rewrite our integral formula for change in S as we go from a to b as:

Now, a similar argument as the one we used when discussing Carnot’s postulate (all ideal reversible engines operating between two temperatures are essentially equivalent) can be used to demonstrate that the change in entropy does not depend on the path: only the start and end point (i.e. point a and b) matter. In fact, the whole discussion is very similar to the discussion of potential energy when conservative force fields are involved (e.g. gravity or electromagnetism): the difference between the values for our potential energy function at different points was absolute. The paths we used to go from one point to another didn’t matter. The only thing we had to agree on was some reference point, i.e. a zero point. For potential energy, that zero point is usually infinity. In other words, we defined zero potential energy as the potential energy of a charge or a mass at an infinite distance away from the charge or mass that’s causing the field.

Here we need to do the same: we need to agree on a zero point for S, because the formula above only gives the difference of entropy between two conditions. Now, that’s where the third law of thermodynamics comes in, which simply states that the entropy of any substance at the absolute zero temperature (T = 0) is zero, so we write:

S = 0 at T = 0

That’s easy enough, isn’t it?

Now, you’ll wonder whether we can actually calculate something with that. We can. Let me simply reproduce Feynman’s calculation of the entropy function for an ideal gas. You’ll need to pull all that I wrote in this and my previous posts together, but you should be able to follow his line of reasoning:

Huh? I know. At this point, you’re probably suffering from formula overkill. However, please try again. Just go over the text and the formulas above, and try to understand what they really mean. [In case you wonder about the formula with the ln[V_b/V_a] factor (i.e. the reference to section 44.4), you can check it in my previous post.] So just try to read the S(V, T) formula: it says that a substance (a gas, liquid or solid) consisting of N atoms or molecules, at some temperature T and with some volume V, is associated with some exact value for its entropy S(V, T). The constant, a, should, of course, ensure that S(V, T) = 0 at T = 0.

The first thing you can note is that S is an increasing function of V at constant temperature T. Conversely, decreasing the volume results in a decrease of entropy. To be precise, using the formula for S, we can derive the following formula for the difference in entropy when keeping the temperature constant at some value T:

S_b – S_a = S(V_b, T) – S(V_a, T)

= ΔS = N·k·ln[V_b/V_a]

What this formula says, for example, is that we’d do nothing but double the volume (while keeping the temperature constant) of a gas when going from to a to b (hence, V_b/V_a= 2), the entropy will change by N·k·ln(2) ≈ 0.7·N·k. Conversely, if we would halve the volume (again, assuming the temperature remains constant), then the change in entropy will be N·k·ln(0.5) ≈ –0.7·N·k.

The graph below shows how it works. It’s quite simple really: it’s just the ln(x) function, and I just inserted it here so you have an idea of how the entropy changes with volume. [In case you would think it looks the same like that efficiency graph, i.e. the graph of the W/Q = (T – 1)/T = 1 – 1/T function, think again: the efficiency graph has a horizontal asymptote (y = 1), while the logarithmic function does not have any horizontal asymptote.]

Now, you may think entropy changes only marginally as we keep increasing the volume, but you should also think twice here. It’s just the nature of the logarithmic scale. Indeed, when we double the volume, going from V = 1 to V = 2, for example, the change in entropy will be equal to N·k·ln(2) ≈ 0.7·N·k. Now, that’s the same change as going from V = 2 to V = 4, and the same as going from V = 4 to V = 8. So, if we double the volume three times in a row, the total change in entropy will be that of going from V = 1 to V = 8, which is equal to N·k·ln(8) = N·k·ln(2³) = 3·ln(2). So, yes, looking at the intervals here that are associated with the same ln(2) increase in entropy, i.e. [1, 2], [2, 4] and [4, 8] respectively, you may think that the increase in entropy is marginal only, as it’s the same increase but the length of each interval is double that of the previous one. However, when reducing the volume, the logic works the other way around, and so the logarithmic function ensures the change is anything but marginal. Indeed, if we halve the volume, going from V = 1 to V = 1/2, and then halve it again, to V = 1/4, and the again, to V = 1/8, we get the same change in entropy once more—but with a minus sign in front, of course: N·k·ln(2^–3) = –3·ln(2)—but the same ln(2) change is now associated with intervals on the x-axis (between 1 and 0.5, 0.5 and 0.25, and 0.25 and 0.125 respectively) that are getting smaller and smaller as we further reduce the volume. In fact, the length of each interval is now half of that of the previous interval. Hence, the change in entropy is anything but marginal now!

[In light of the fact that the (negative) change in entropy becomes larger and larger as we further reduce the volume, and in a way that’s anything but marginal, you may now wonder, for a very brief moment, whether or not the entropy might actually take on a negative value. The answer is obviously no. The change in entropy can take on a large negative volume but the S(V, T) = N·k·[ln(V) + ln(T)/(γ–1)] + a formula, with a ensuring that the entropy is zero at T = 0, ensures things come out alright—as it should, of course!]

Now, as we’re continue to try to understand what entropy really means, it’s quite interesting to think of what this formula implies at the level of the atoms or molecules that make up the gas: the entropy change per molecule is k·ln2 – or k·ln(1/2) when compressing the gas at the same temperature. Now, its kinetic energy remains the same – because – don’t forget! – we’re changing the volume at constant temperature here. So what causes the entropy change here really? Think about it: the only thing that changed, physically, is how much room the molecule has to run around in—as Feynman puts it aptly. Hence, while everything stays the same (atoms or molecules with the same temperature and energy), we still have an entropy increase (or decrease) when the distribution of the molecules changes.

This remark brings us to the connection between order and entropy, which you vaguely know, for sure, but probably never quite understood because, if you did, you wouldn’t be reading this post. 🙂 So I’ll talk about in a moment. I first need to wrap up this section, however, by showing why all of the above is, somehow, related to that ever-increasing entropy law. 🙂

However, before doing that, I want to quickly note something about that assumption of constant temperature here. How can it remain constant? When a body of gas expands, its temperature should drop, right? Well… Yes. But only if it is pushing against something, like in cylinder with a piston indeed, or as air escapes from a tyre and pushes against the (lower-pressure) air outside of the tyre. What happens here is that the kinetic energy of the gas molecules is being transferred (to the piston, or to the gas molecules outside of the tyre) and, hence, temperature decreases indeed. In such case, the assumption is that we add (or remove) heat from our body of gas as we expand (or decrease) its volume. Having said that, in a more abstract analysis, we could envisage a body of gas that has nothing to push against, except for the walls of its container, which have the same temperature. In such more abstract analysis, we need not worry about how we keep temperature constant: the point here is just to compare the ex post and ex ante entropy of the volume. That’s all.

The Law of Ever-Increasing Entropy

With all of the above, we’re finally armed to ‘prove’ the second law of thermodynamics which we can also state as follows indeed: while the energy of the universe is constant, its entropy is always increasing. Why is this so? Out of respect, I’ll just quote Feynman once more, as I can’t see how I could possibly summarize it better:

So… That should sum it all up. You should re-read the above a couple of times, so you’re sure you grasp it. I’ll also let Feynman summarize all of those ‘laws’ of thermodynamics that we have just learned as, once more, I can’t see how I could possibly write more clearly or succinctly. His statement is much more precise that the statement we started out with: the energy of the universe is always constant but its entropy is always increasing. As Feynman notes, this version of the two laws of thermodynamics don’t say that entropy stays the same in a reversible cycle, and also doesn’t say what entropy actually is. So Feynman’s summary is much more precise and, hence, much better indeed:

Entropy and order

What I wrote or reproduced above may not have satisfied you. So we’ve got this funny number, S, describing some condition or state of a substance, but you may still feel you don’t really know what it means. Unfortunately, I cannot do all that much about that. Indeed, technically speaking, a quantity like entropy (S) is a state function, just like internal energy (U), or like enthalpy (usually denoted by H), a related concept which you may remember from chemistry and which is defined H = U + PV. As such, you may just think of S as some number that pops up in a thermodynamical equations. It’s perfectly fine to think of it like that. However, if you’re reading this post, then it’s likely you do so because some popular science book mentioned entropy and related it to order and/or disorder indeed. However, I need to disappoint you here: that relationship is not as straightforward as you may think it is. To get some idea, let’s go through another example, which I’ll also borrow from Feynman.

Let’s go back to that relationship between volume and entropy, keeping temperature constant:

ΔS = N·k·ln[V_b/V_a]

We discussed, rather at length, how entropy increases as we allow a body of gas to expand. As the formula shows, it increases logarithmically with the ratio of the ex ante and ex post volume. Now, let us think about two gases, which we can think of as ‘white’ and ‘black’ respectively. Or neon or argon. Whatever. Two different gases. Let’s suppose we’ve kept them into two separate compartments of a box, with some barrier in-between them.

Now, you know that, if we’d take out the barrier, they’ll mix it. That’s just a fact of life. As Feynman puts it: somehow, the whites will worm their way across in the space of blacks, and the blacks will worm their way, by accident, into the space of whites. [There’s a bit of a racist undertone in this, isn’t there? But then I am sure Feynman did not intend it that way.] Also, as he notes correctly: we’ve got a very simple example here of an irreversible process which is completely composed of reversible events. We know this mixing will not affect the kinetic (or internal) energy of the gas. Having said that, both the white and the black molecules now have ‘much more room to run around in’. So is there a change in entropy? You bet.

If we take away that barrier, it’s just similar to moving that piston out when we were discussing one volume of gas only. Indeed, we effectively double the volume for the whites, and we double the volume for the blacks, while keeping all at the same temperature. Hence, both the entropy of the white and black gas increases. By how much? Look at the formula: the amount is given by the product of the number of molecules (N), the Boltzman constant (k), and ln(2), i.e. the natural logarithm of the ratio of the ex post and ex ante volumes: ΔS = N·k·ln[V_b/V_a].

So, yes, entropy increases as the molecules are now distributed over a much larger space. Now, if we stretch our mind a bit, we could define as a measure of order, or disorder, especially when considering the process going the other way: suppose the gases were mixed up to begin with and, somehow, we manage to neatly separate them in two separate volumes, each half of the original. You’d agree that amounts to an increase in order and, hence, you’d also agree that, if entropy is, somehow, some measure for disorder, entropy should decrease–which it obviously does using that ΔS = N·k·ln[V_b/V_a] formula. Indeed, we calculated ΔS as –0.7·N·k.

However, the interpretation is quite peculiar and, hence, not as straightforward as popular science books suggest. Indeed, from that S(V, T) = Nk[lnV + (1/γ−1)lnT] + a formula, it’s obvious we can also decrease entropy by decreasing the number of molecules, or by decreasing the temperature. You’ll have to admit that in both cases (decrease in N, or decrease in T), you’ll have to be somewhat creative in interpreting such decrease as a decrease in disorder.

So… What more can we say? Nothing much. However, in order to be complete, I should add a final note on this discussion of entropy measuring order (or, to be more precise, measuring disorder). It’s about another concept of entropy, the so-called Shannon entropy. It’s a concept from information theory, and our entropy and the Shannon entropy do have something in common: in both, we see that logarithm pop up. It’s quite interesting but, as you might expect, complicated. Hence, I should just refer you to the Wikipedia article on it, from which I took the illustration and text below.

We’ve got two coins with two faces here. They can, obviously, be arranged in 2²= 4 ways. Now, back in 1948, the so-called father of information theory, Claude Shannon, thought it was nonsensical to just use that number (4) to represent the complexity of the situation. Indeed, if we’d take three coins, or four, or five, respectively, then we’d have 2³= 8, 2⁴= 16, and 2⁵= 32 ways, respectively, of combining them. Now, you’ll agree that, as a measure of the complexity of the situation, the exponents 1, 2, 3, 4 etcetera describe the situation much better than 2, 4, 8, 16 etcetera.

Hence, Shannon defined the so-called information entropy as, in this case, the base 2 logarithm of the number of possibilities. To be precise, the information entropy of the situation which we’re describing here (i.e. the ways a set of coins can be arranged) is equal to S = N = log₂(2^N) = 1, 2, 3, 4 etcetera for N = 1, 2, 3, 4 etcetera. In honor of Shannon, the unit is shannons. [I am not joking.] However, information theorists usually talk about bits, rather than shannons. [We’re not talking a computer bit here, although the two are obviously related, as computer bits are binary too.]

Now, one of the many nice things of logarithmic functions is that it’s easy to switch bases. Hence, instead of expressing information entropy in bits, we can also express it in trits (for base 3 logarithms), nats (for base e logarithms, so that’s the natural logarithmic function ln), or dits (for base 10 logarithms). So… Well… Feynman is right in noting that “the logarithm of the number of ways we can arrange the molecules is (the) entropy”, but that statement needs to be qualified: the concepts of information entropy and entropy tout court, as used in the context of thermodynamical analysis, are related but, as usual, they’re also different. 🙂 Bridging the two concepts involves probability distributions and other stuff. One extremely simple popular account illustrates the principle behind as follows:

Suppose that you put a marble in a large box, and shook the box around, and you didn’t look inside afterwards. Then the marble could be anywhere in the box. Because the box is large, there are many possible places inside the box that the marble could be, so the marble in the box has a high entropy. Now suppose you put the marble in a tiny box and shook up the box. Now, even though you shook the box, you pretty much know where the marble is, because the box is small. In this case we say that the marble in the box has low entropy.

Frankly, examples like this make only very limited sense. They may, perhaps, help us imagine, to some extent, how probability distributions of atoms or molecules might change as the atoms or molecules get more space to move around in. Having said that, I should add that examples like this are, at the same time, also so simplistic they may confuse us more than they enlighten us. In any case, while all of this discussion is highly relevant to statistical mechanics and thermodynamics, I am afraid I have to leave it at this one or two remarks. Otherwise this post risks becoming a course! 🙂

Now, there is one more thing we should talk about here. As you’ve read a lot of popular science books, you probably know that the temperature of the Universe is decreasing because it is expanding. However, from what you’ve learnt so far, it is hard to see why that should be the case. Indeed, it is easy to see why the temperature should drop/increase when there’s adiabatic expansion/compression: momentum and, hence, kinetic energy, is being transferred from/to the piston indeed, as it moves out or into the cylinder while the gas expands or is being compressed. But the expanding universe has nothing to push against, does it? So why should its temperature drop? It’s only the volume that changes here, right? And so its entropy (S) should increase, in line with the ΔS = S_b – S_a = S(V_b, T) – S(V_a, T) = ΔS = N·k·ln[V_b/V_a] formula, but not its temperature (T), which is nothing but the (average) kinetic energy of all of the particles it contains. Right? Maybe.

[By the way, in case you wonder why we believe the Universe is expanding, that’s because we see it expanding: an analysis of the redshifts and blueshifts of the light we get from other galaxies reveals the distance between galaxies is increasing. The expansion model is often referred to as the raisin bread model: one doesn’t need to be at the center of the Universe to see all others move away: each raisin in a rising loaf of raisin bread will see all other raisins moving away from it as the loaf expands.]

Why is the Universe cooling down?

This is a complicated question and, hence, the answer is also somewhat tricky. Let’s look at the entropy formula for an increasing volume of gas at constant temperature once more. Its entropy must change as follows:

ΔS = S_b – S_a = S(V_b, T) – S(V_a, T) = ΔS = N·k·ln[V_b/V_a]

Now, the analysis usually assumes we have to add some heat to the gas as it expands in order to keep the temperature (T) and, hence, its internal energy (U) constant. Indeed, you may or may not remember that the internal energy is nothing but the product of the number of gas particles and their average kinetic energy, so we can write:

U = N<mv²/2>

In my previous post, I also showed that, for an ideal gas (i.e. no internal motion inside of the gas molecules), the following equality holds: PV = (2/3)U. For a non-ideal gas, we’ve got a similar formula, but with a different coefficient: PV = (γ−1)U. However, all these formulas were based on the assumption that ‘something’ is containing the gas, and that ‘something’ involves the external environment exerting a force on the gas, as illustrated below.

As Feynman writes: “Suppose there is nothing, a vacuum, on the outside of the piston. What of it? If the piston were left alone, and nobody held onto it, each time it got banged it would pick up a little momentum and it would gradually get pushed out of the box. So in order to keep it from being pushed out of the box, we have to hold it with a force F.” We know that the pressure is the force per unit area: P = F/A. So can we analyze the Universe using these formulas?

Maybe. The problem is that we’re analyzing limiting situations here, and that we need to re-examine our concepts when applying them to the Universe. 🙂

The first question, obviously, is about the density of the Universe. You know it’s close to a vacuum out there. Close. Yes. But how close? If you google a bit, you’ll find lots of hard-to-read articles on the density of the Universe. If there’s one thing you need to pick up from them, is that, in order for the Universe to expand forever, it should have some critical density (denoted by ρ_c), which is like a watershed point between an expanding and a contracting Universe.

So what about it? According to Wikipedia, the critical density is estimated to be approximately five atoms (of monatomic hydrogen) per cubic metre, whereas the average density of (ordinary) matter in the Universe is believed to be 0.2 atoms per cubic metre. So that’s OK, isn’t it?

Well… Yes and no. We also have non-ordinary matter in the Universe, which is usually referred to as dark matter in the Universe. The existence of dark matter, and its properties, are inferred from its gravitational effects on visible matter and radiation. In addition, we’ve got dark energy as well. I don’t know much about it, but it seems the dark energy and the dark matter bring the actual density (ρ) of the Universe much closer to the critical density. In fact, cosmologists seem to agree thatρ ≈ ρ_c and, according to a very recent scientific research mission involving an ESA space observatory doing very precise measurements of the Universe’s cosmic background radiation, the Universe should consist of 4.82 ± 0.05% ordinary matter,25.8 ± 0.4% dark matter and 69 ± 1% dark energy. I’ll leave it to you to challenge that. 🙂

OK. Very low density. So that means very low pressure obviously. But what’s the temperature? I checked on the Physics Stack Exchange site, and the best answer is pretty nuanced: it depends on what you want to average. To be precise, the quoted answer is:

If one averages by volume, then one is basically talking about the ‘temperature’ of the photons that reach us as cosmic background radiation—which is the temperature of the Universe that those popular science books refer to. In that case, we get an average temperature of 2.72 degrees Kelvin. So that’s pretty damn cold!
If we average by observable mass, then our measurement is focused mainly on the temperature of all of the hydrogen gas (most matter in the Universe is hydrogen), which has a temperature of a few 10s of Kelvin. Only one tenth of that mass is in stars, but their temperatures are far higher: in the range of 10⁴to 10⁵ degrees. Averaging gives a range of 10³to 10⁴ degrees Kelvin. So that’s pretty damn hot!
Finally, including dark matter and dark energy, which is supposed to have even higher temperature, we’d get an average by total mass in the range of 10⁷ Kelvin. That’s incredibly hot!

This is enlightening, especially the first point: we’re not measuring the average kinetic energy of matter particles here but some average energy of (heat) radiation per unit volume. This ‘cosmological’ definition of temperature is quite different from the ‘physical’ definition that we have been using and the observation that this ‘temperature’ must decrease is quite logical: if the energy of the Universe is a constant, but its volume becomes larger and larger as the Universe expands, then the energy per unit volume must obviously decrease.

So let’s go along with this definition of ‘temperature’ and look at an interesting study of how the Universe is supposed to have cooled down in the past. It basically measures the temperature of that cosmic background radiation, i.e. a remnant of the Big Bang, a few billion years ago, which was a few degrees warmer then than it is now. To be precise, it was measured as 5.08 ± 0.1 degrees Kelvin, and this decrease has nothing to do with our simple ideal gas laws but with the Big Bang theory, according to which the temperature of the cosmic background radiation should, indeed, drop smoothly as the universe expands.

Going through the same logic but the other way around, if the Universe had the same energy at the time of the Big Bang, it was all focused in a very small volume. Now, very small volumes are associated with very small entropy according to that S(V, T) = N·k·[ln(V) + ln(T)/(γ–1)] + a formula, but then temperature was not the same obviously: all that energy has to go somewhere, and a lot of it was obviously concentrated in the kinetic energy of its constituent particles (whatever they were) and, hence, a lot of it was in their temperature.

So it all makes sense now. It was good to check out it out, as it reminds us that we should not try to analyze the Universe as a simple of body of gas that’s not contained in anything in order to then apply our equally simple ideal gas formulas. Our approach needs to be much more sophisticated. Cosmologists need to understand physics (and thoroughly so), but there’s a reason why it’s a separate discipline altogether. 🙂

First Principles of Thermodynamics

Pre-script (dated 26 June 2020): This post has become less relevant (even irrelevant, perhaps) because my views on all things quantum-mechanical have evolved significantly as a result of my progression towards a more complete realist (classical) interpretation of quantum physics. I keep blog posts like these mainly because I want to keep track of where I came from. I might review them one day, but I currently don’t have the time or energy for it. 🙂

Original post:

Thermodynamics is not an easy topic, but one can’t avoid it in physics. The main obstacle, probably, is that we very much like to think in terms of dependent and independent variables. While that approach is still valid in thermodynamics, it is more complicated, because it is often not quite clear what the dependent and independent variables are. We’ve got a lot of quantities in thermodynamics indeed: volume, pressure, internal energy, temperature and – soon to be defined – entropy, which are all some function of each other. Hence, the math involves partial derivatives and other subtleties. Let’s try to get through the basics.

Volume, pressure, temperature and the ideal gas law

We all know what a volume is. That’s an unambiguous quantity. Pressure and temperature are not so unambiguous. In fact, as far as I am concerned, the key to understanding thermodynamics is to be able to not only distinguish but also relate pressure and temperature.

The pressure of a gas or a liquid (P) is the force, per unit area, exerted by the atoms or molecules in that gas or liquid as they hit a surface, such as a piston, or the wall of the body that contains it. Hence, pressure is expressed in newton per square meter: 1 pascal (Pa) = 1 N/m². It’s a small unit for daily use: the standard atmospheric pressure is 1 atm = 101,325 Pa = 1.01325×10⁵ Pa = 1.01325 bar. We derived the formula for pressure in the previous post:

P = F/A = (2/3)·n·〈m·v²/2〉

This formula shows that the pressure depends on two variables:

The density of the gas or the liquid (i.e. the number of particles per unit volume, so it’s two variables really: a number and a volume), and
Their average kinetic energy.

Now, this average kinetic energy of the particles is nothing but the temperature (T), except that, because of historical reasons, we define temperature (expressed in degrees Kelvin) using a constant of proportionality—the Boltzmann constant k = k_B. In addition, in order to get rid of that ugly 2/3 factor in our next formula, we’ll also throw in a 3/2 factor. Hence, we re-write the average kinetic energy 〈m·v²/2〉 as:

〈m·v²/2〉 = (3/2)·k·T

Now we substitute that definition into the first equation (while also noting that, if n is the number of particles in a unit volume, we will have N = n·V atoms in a volume V) to get what we want: the so-called ideal gas law, which you should remember from your high-school days:

PV = NkT

The equation implies that, for a given number of particles (for some given substance, that is), and for some given temperature, pressure and volume are inversely proportional one to another: P = NkT/V. The curve representing that relationship between P and V has the same shape as the reciprocal function y = 1/x. To be precise, it has the same shape as a rectangular hyperbola with the center at the origin, i.e. the shape of an y = m/x curve, assuming non-negative values for x and y only. The illustration below shows that graph for m = 1, 3 and 0.3 respectively. We’ll need that graph later when looking at more complicated graphs depicting processes during which we will not keep temperature constant—so that’s why I quickly throw it in here.

Of course, n·〈m·v²/2〉 is the number of atoms times the average kinetic energy of each and, therefore, it is also the internal energy of the gas. Hence, we can also write the PV = NkT equation as:

PV = (2/3)·U

We should immediately note that we’re considering an ideal gas here, so we disregard any possibility of excitation or motion inside the atoms or molecules. It matters because, if we’re decreasing the volume and, hence, increasing the pressure, we’ll be doing work, and the energy needs to go somewhere. The equation above assumes it all goes into that 〈m·v²/2〉 factor and, hence, into the temperature. Hence, it is obvious that, if were to allow for all kinds of rotational and vibratory motions inside of the atoms or molecules motions also, then the analysis would become more complicated. Having said, in my previous post I showed that the complications are limited: we can account for all kinds of internal motion by inserting another coefficient—i.e. other than 2/3. For example, Feynman calculates it as 2/7, rather than 2/3, for the diatomic oxygen molecule. That is why we usually see a much more general expression of the equation above. We will write:

PV = (γ – 1)·U

The gamma (γ) in the equation above is the rather infamous specific heat ratio, and so it’s equal to 5/3 for the ideal gas (5/3 – 1 = 2/3). I call γ infamous because its theoretical value does not match the experimental value for most gases. For example, while I just noted γ’s theoretical value for O₂(i.e. he diatomic oxygen molecule) – it’s 9/7 ≈ 1.286, because 9/7 – 1 = 2/7), the experimentally measured value for O₂is 1.399. The difference can only be explained using quantum mechanics, which is obviously not the topic of this post, and so we won’t write much about γ. However, I need to say one or two things about it—which I’ll do by showing how we could possibly measure it. Let me reproduce the illustration in my previous post here.

The pressure is the force per unit area (P = F/A and, hence, F = P·A), and compressing the gas amounts to applying a force over some (infinitesimal) distance dx. Hence, the (differential) work done is equal to dW = F·(−dx) = – P·A·dx = – P·dV, as A·dx = dV, obviously (the area A times the distance dx is the volume change). Now, all the work done goes into changing the internal energy U: there is no heat energy that’s being added or removed here, and no other losses of energy. That’s why it’s referred to as a so-called adiabatic compression, from the Greek a (not), dia (through) and bainein (to go): no heat is going through. The cylinder is thermally insulated. Hence, we write:

dU = – P·dV

This is a very simple differential equation. Note the minus sign: the volume is going to decrease while we do work by compressing the piston, thereby increasing the internal energy. [If you are clever (which, of course, you are), you’ll immediately say that, with increasing internal energy, we should also have an increase in pressure and, hence, we shouldn’t treat P as some constant. You’re right, but so we’re doing a marginal analysis only here: we’ll deal with the full thing later. As mentioned above, the complete picture involves partial derivatives and other mathematical tricks.]

Taking the total differential of U = PV/(γ – 1), we also have another equation:

dU = (P·dV + V·dP)/(γ – 1)

Hence, we have – P·dV = (P·dV + V·dP)/(γ – 1) or, rearranging the terms:

γdV/V + dP/P = 0

Assuming that γ is constant (which is true in theory but not in practice—another reason why this γ is rather infamous), we can integrate this. It gives γlnV + lnP = lnC, with lnC the constant of integration. Now we take the exponential of both sides to get that other formulation of the gas law, which you also may or may not remember from your high-school days:

PV^γ= C (a constant)

So here you have the answer to the question as to how we can measure γ: the pressure times the volume to the γth power must be some constant. To be precise, for monatomic gases the pressure times the volume to the 5/3 ≈ 1.67 power must be a constant. The formula works for gases like helium, krypton and argon. However, the issue is more complicated when looking at more complex molecules. You should also note the serious limitation in this analysis: we should not think of P as a constant in the dU = – P·dV equation! But I’ll come back to this. As for now, just take note of it and move on to the next topic.

The Carnot heat engine

The definitions above should help us to understand and distinguish isothermal expansion and compression versus adiabatic expansion and compression which, in turn, should help us to understand what the Carnot cycle is all about. We’re looking at a so-called reversible engine here: there is no friction, and we also assume heat flows ‘frictionless’. The cycle is illustrated below: this so-called heat engine takes an amount of heat (Q₁) from a high temperature (T₁) heat pad (often referred to as the furnace or the boiler or, more generally, the heat source) and uses it to make some body (i.e. a piston in a cylinder in Carnot’s example) do some work, with some other amount of heat (Q₂) goes back into some cold sink (usually referred to as the condenser), which is nothing but a second pad at much lower temperature (T₂).

Carnot cycle

The four steps involved are the following:

(1) Isothermal expansion: The gas absorbs heat and expands while keeping the same temperature (T₁). As the number of gas atoms or molecules, and their temperature, stays the same, the heat does work, as the gas expands and pushes the piston upwards. So that’s isothermal expansion. The next is different.

(2) Adiabatic expansion: The cylinder and piston are now removed from the heat pad, and the gas continues to expand, thereby doing even more work by pushing the piston further upwards. However, as the piston and cylinder are assumed to be thermally insulated, they neither gain nor lose heat. So it is the gas that loses internal energy: its temperature drops. So the gas cools. How much? It depends on the temperature of the condenser, i.e. T₂, or – if there’s no condenser – the temperature of the surroundings. Whatever, the temperature cannot fall below T₂.

(3) Isothermal compression: Now we (or the surroundings) will be doing work on the gas (as opposed to the gas doing work on its surroundings). The piston is being pushed back, and so the gas is slowly being compressed while, importantly, keeping it at the same temperature T₂. Therefore, it delivers, through the head pad, a heat amount Q₂ to the second heat reservoir (i.e. the condenser).

(4) Adiabatic compression: We take the cylinder off the heat pad and continue to compress it, without letting any heat flow out this time around. Hence, the temperature must rise, back to T₁. At that point, we can put it back on the first heat pad, and start the Carnot cycle all over again.

The graph below shows the relationship between P and V, and temperature (T), as we move through this cycle. For each cycle, we put in Q₁ at temperature T₁, and take out Q₂ at temperature T₂, and then the gas does some work, some net work, or useful work as it’s labeled below.

Let’s go step by step once again:

Isothermal expansion: Our engine takes in Q₁ at temperature T₁ from the heat source (isothermal expansion), as we move along line segment (1) from point a to point b on the graph above: the pressure drops, the volume increases, but the temperature stays the same.
Adiabatic expansion: We take the cylinder off the heat path and continue to let the gas expand. Hence, it continues to push the piston, and we move along line segment (2) from point b to c: the pressure further drops, and the volume further increases, but the temperature drops too—from T₁ to T₂ to be precise.
Isothermal compression: Now we bring the cylinder in touch with the T₂ reservoir (the condenser or cold sink) and we now compress the gas (so we do work on the gas, instead of letting the gas do work on its surroundings). As we compress the gas, we reduce the volume and increase the pressure, moving along line segment (3) from c to d, while the temperature of the gas stays at T₂.
Adiabatic compression: Finally, we take the cylinder of the cold sink, but we further compress the gas. As its volume further decreases, its pressure and, importantly, its temperature too rises, from T₂ to T₁ – so we move along line segment 4 from d to a – and then we put it back on the heat source to start another cycle.

We could also reverse the cycle. In that case, the steps would be the following:

Our engine would first take in Q₂ at temperature T₂ (isothermal expansion). We move along line segment (3) here but in the opposite direction: from d to c.
Then we would push the piston to compress the gas (so we’d be doing some work on the gas, rather than have the gas do work on its surroundings) so as to increase the temperature from T₂ to T₁ (adiabatic compression). On the graph, we go from c to b along line segment (2).
Then we would bring the cylinder in touch with the T₁ reservoir and further compress the gas so an amount of heat equal to Q₁ is being delivered to the boiler at (the higher) temperature T₁ (isothermal compression). So we move along line segment (1) from b to a.
Finally, we would let the gas expand, adiabatically, so the temperature drops, back to T₂(line segment (4), from a to d), so we can put it back on the T₂ reservoir, on which we will let it further expand to take in Q₂again.

It’s interesting to note that the only reason why we can get the machine to do some net work (or why, in the reverse case, we are able to transfer heat by putting some work into some machine) is that there is some mechanism here that allows the machine to take in and transfer heat through isothermal expansion and compression. If we would only have adiabatic expansion and compression, then we’d just be going back and forth between temperature T₁and T₂ without getting any net work out of the engine. The shaded area in the graph above then collapses into a line. That is why actual steam engines are very complicated and involve valves and other engineering tricks, such as multiple expansion. Also note that we need two heat reservoirs: we can imagine isothermal expansion and compression using one heat reservoir only but then the engine would also not be doing any net work that is useful to us.

Let’s analyze the work that’s being doing during such Carnot cycle somewhat more in detail.

The work done when compressing a gas, or the work done by a gas as it expands, is an integral. I won’t explain in too much detail here but just remind you of that dW = F·(−dx) = – P·A·dx = – P·dV formula. From this, it’s easy to see that the integral is ∫ PdV.

An integral is an area under a curve: just substitute P for y = f(x) and V for x, and think of ∫ f(x)dx = ∫ y dx. So the area under each of the numbered curves is the work done by or on the gas in the corresponding step. Hence, the net work done (i.e. the so-called useful work) is the shaded area of the picture.

So what is it exactly?

Well… Assuming there are no other losses, the work done should, of course, be equal to the difference in the heat that was put in, and the heat that was taken out, so we write:

W = Q₁– Q₂

So that’s key to understanding it all: an efficient (Carnot) heat engine is one that converts all of the heat energy (i.e. Q₁– Q₂) into useful work or, conversely, which converts all of the work done on the gas into heat energy.

Schematically, Carnot’s reversible heat engine is represented as follows:

So what? You may we’ve got it all now, and that there’s nothing to add to the topic. But that’s not the case. No. We will want to know more about the exact relationship between Q₁, Q₂, T₁and T₂. Why? Because we want to be able to answer the very same questions Sadi Carnot wanted to answer, like whether or not the engine could be made more efficient by using another liquid or gas. Indeed, as a young military engineer, fascinated by the steam engines that had – by then – become quite common, Carnot wanted to find an unambiguous answer to two questions:

How much work can we get out of a heat source? Can all heat be used to do useful work?
Could we improve heat engines by replacing the steam with some other working fluid or gas?

These questions obviously make sense, especially in regard to the relatively limited efficiency of steam engines. Indeed, the actual efficiency of the best steam engines at the time was only 10 to 20 percent, and that’s under favorable conditions!

Sadi Carnot attempted to answer these in a memoir, published as a popular work in 1824 when he was only 28 years old. It was entitled Réflexions sur la Puissance Motrice du Feu (Reflections on the Motive Power of Fire). Let’s see if we can make sense of it using more modern and common language. [As for Carnot’s young age, like so many, he was not destined to live long: he was interned in a private asylum in 1832 suffering from ‘mania’ and ‘general delirium’, and died of cholera shortly after, aged 36.]

Carnot’s Theorem

You may think that both questions have easy answers. The first question is, obviously, related to the principle of conservation of energy. So… Well… If we’d be able to build a frictionless Carnot engine, including a ‘frictionless’ heat transfer mechanism, then, yes, we’d be able to convert all heat energy into useful work. But that’s an ideal only.

The second question is more difficult. The formal answer is the following: if an engine is reversible, then it makes no difference how it is designed. In other words, the amount of work that we’ll get out of a reversible Carnot heat engine as it absorbs a given amount of heat (Q₁) at temperature T₁and delivers some other amount of heat (Q₂) at some other temperature T₂ does not depend on the design of the machine. More formally, Carnot’s Theorem can be expressed as follows:

All reversible engines operating between the same heat reservoirs are equally efficient.
No actual engine operating between two heat reservoirs can be more efficient than a Carnot engine operating between the same reservoirs.

Feynman sort of ‘proves’ this Theorem from what he refers to as Carnot’s postulate. However, I feel his ‘proof’ is not a real proof, because Carnot’s postulate is too closely related to the Theorem, and so I feel he’s basically proving something using the result of the proof! However, in order to be complete, I did reproduce Feynman’s ‘proof’ of Carnot’s Theorem in the post scriptum to this post.

So… That’s it. What’s left to do is to actually calculate the efficiency of an ideal reversible Carnot heat engine, so let’s do that now. In fact, the calculation below is much more of a real proof of Carnot’s Theorem and, hence, I’d recommend you go through it.

The efficiency of an ideal engine

Above, I said I would need the result that PV^γis equal to some constant. We do, in the following proof that, for an ideal engine, the following relationship holds, always, for any Q₁, Q₂, T₁and T₂:

Q₁/T₁= Q₂/T₂

Now, we still don’t have the efficiency with this. The efficiency of an ideal engine is the ratio of the amount of work done and the amount of heat it takes in:

Efficiency = W/Q₁

But W is equal to Q₁– Q₂. Hence, re-writing the equation with the two heat/temperature ratios above as Q₂= (T₂/T₁)·Q₁, we get: W = Q₁(1 – T₂/T₁) = Q₁(T₁ – T₂)/T₁. The grand result is:

Efficiency = W/Q₁= (T₁ – T₂)/T₁

Let me help you to interpret this result by inserting a graph for T₁ going from zero to 20 degrees, and for T₂ set at 0.3, 1 and 4 degrees respectively.

The graph makes it clear we need some kind of gauge so as to be able to actually compare the efficiency of ideal engines. I’ll come back to that in my next post. However, in the meanwhile, please note that the result makes sense: T₁needs to be higher than T₂for the efficiency to be positive (of course, we can interpret negative values for the efficiency just as well, as they imply we need to do work on the engine, rather than the engine doing work for us), and the efficiency is always less than unity, getting closer to one as the working temperature of the engine goes up.

Where does the power go?

So we have an engine that does useful work – so it works, literally – and we know where it gets its energy for that: it takes in more heat than it returns. But where is the work going? It is used to do something else, of course—like moving a car. Now how does that work, exactly? The gas exerts a force on the piston, thereby giving it an acceleration a = F/m, in accordance with Newton’s Law: F = m·a.

That’s all great. But then we need to re-compress the gas and, therefore, we need to (a) decelerate the piston, (b) reverse its direction and (c) push it back in. So that should cancel all of the work, shouldn’t it?

Well… No.

Let’s look at the Carnot cycle once more to show why. The illustrations below reproduce the basic steps in the cycle and the diagram relating pressure, volume and temperature for each of the four steps once more.

Above, I wrote that the only reason why we can get the machine to do some net work (or why, in the reverse case, we are able to transfer heat from lower to higher temperature by doing some (net) work on it) is that there is some mechanism here that allows the machine to take in and transfer heat through isothermal expansion and compression and that, if we would only have adiabatic expansion and compression, then we’d just be going back and forth between temperature T₁and T₂ without getting any net work out of the engine.

Now, that’s correct and incorrect at the same time. Just imagine a cylinder and a piston in equilibrium, i.e. the pressure on the inside and the outside of the piston are the same. Then we could push it in a bit but, as soon as we release, it would come back to its equilibrium situation. In fact, as we assume the piston can move in and out without any friction whatsoever, we’d probably have a transient response before the piston settles back into the steady state position (see below). Hence, we’d be moving back and forth on segment (2), or segment (4), in that P-V-T diagram above.

The point is: segment (2) and segment (4) are not the same: points a and b, and points c and d, are marked by the same temperature (T₁and T₂respectively) butpressure and volume is very different. Why? Because we had a step in-between step (2) and (4): isothermal compression, which reduced the volume, i.e. step (3). Hence, the area underneath these two segments is different too. Indeed, you’ll remember we can write dW = F·(−dx) = – P·A·dx = – P·dV and, hence, the work done (or put in) during each step of the cycle is equal to the integral ∫ PdV, so that’s the area under each of the line segments. So it’s not like these two steps do not contribute to the net work that’s being done through the cycle. They do. Likewise, step (1) and (3) are not each other’s mirror image: they too take place at different volume and pressure, but that’s easier to see because they take place at different temperature and involve different amounts of heat (Q₁and Q₂respectively).

But, again, what happens to the work? When everything is said and done, the piston does move up and down over the same distance in each cycle, and we know that work is force times distance. Hence, if the distance is the same… Yes. You’re right: the piston must exert some net force on something or, to put it differently, the energy W = Q₁− Q₂must go somewhere. Now that’s where the time variable comes in, which we’ve neglected so far.

Let’s assume we connect the piston to a flywheel, as illustrated below, there had better be some friction on it because, if not, the flywheel would spin faster and faster and, eventually, spin out of control and all would break down. Indeed, each cycle would transfer additional kinetic energy to the flywheel. When talking work and kinetic energy, one usually applies the following formula: W = Q₁and Q₂= Δ[mv²/2] = [mv²/2]_after− [mv²/2]_before. However, we’re talking rotational kinetic energy so we should use the rotational equivalent for mv²/2, which is Iω²/2, in which I is the moment of inertia of the mass about the center of rotation and ω is the angular velocity.

You get the point. As we’re talking time now, we should also remind you of the concept of power. Power is the amount of work or energy being delivered, or consumed, per unit of time (i.e. per second). So we can write it as P(t) = dW/dt. For linear motion, P(t) can be written as the vector product (I mean the scalar, inner or dot product here) of the force and velocity vectors, so P(t) = F·v. Again, when rotation is involved, we’ve got an equivalent formula: P(t) = τ·ω, in which τ represents the torque and ω is, once again, the angular velocity of the flywheel. Again, we’d better ensure some load is placed on the engine, otherwise it will spin out of control as vand/or ω get higher and higher and, hence, the power involved gets higher and higher too, until all breaks down.

So… Now you know it all. 🙂

Post scriptum: The analysis of the Carnot cycle involves some subtleties which I left out. For example, you may wonder why the gas would actually expand isothermically in the first step of the Carnot cycle. Indeed, if it’s at the same temperature T₁as the heat source, there should be no heat flow between the heat pad and the gas and, hence, no gas expansion, no? Well… No. 🙂 The gas particles pound on every wall, but only the piston can move. As the piston moves out, frictionless, inside of the cylinder, kinetic energy is being transferred from the gas particles to the piston and, hence, the gas temperature will want to drop—but then that temperature drop will immediately cause a heat transfer. That’s why the description of a Carnot engine also postulates ‘frictionless’ heat transfer.

In fact, I note that Feynman himself struggles a bit to correctly describe what’s going on here, as his description of the Carnot cycle suggests some active involvement is needed to make the piston move and ensure the temperature does not drop too fast. Indeed, he actually writes following: “If we pull the piston out too fast, the temperature of the gas will fall too much below T and then the process will not quite be reversible.” This sounds, and actually is, a bit nonsensical: no pulling is needed, as the gas does all of the work while pushing the piston and, while it does, its temperature tends to drop, so it will suck it heat in order to equalize its temperature with its surroundings (i.e. the heat source). The situation is, effectively, similar to that of a can with compressed air: we can let the air expand, and thereby we let it do some work. However, the air will not re-compress itself by itself. To re-compress the air, you’ll need to apply the same force (or pressure I should say) but in the reverse direction.

Finally, I promised I would reproduce Feynman’s ‘proof’ of Carnot’s Theorem. This ‘proof’ involves the following imaginary set-up (see below): we’ve got engines, A and B. We assume A is an ideal reversible engine, while B may or may not be reversible. We don’t care about its design. We just assume that both can do work by taking a certain amount of heat out of one reservoir and putting another amount of heat back into another reservoir. In fact, in this set-up, we assume both engines share a large enough reservoir so as to be able to transfer heat through that reservoir.

Engine A can take an amount of heat equal to Q₁at temperature T₁ from the first reservoir, do an amount of work equal to W, and then deliver an amount of heat equal to Q₂= Q₁– W at temperature T₂ to the second reservoir. However, because it’s a reversible machine, it can also go the other way around, i.e. it can take Q₂= Q₁– W from the second reservoir, have the surroundings do an amount of work W on it, and then deliver Q₁= Q₂+ W at temperature T₁. We know that engine B can do the same, except that, because it’s different, the work might be different as well, so we’ll denote it by W’.

Now, let us suppose that the design of engine B is, somehow, more efficient, so we can get more work out of B for the same Q₁and the same temperatures T₁and T₂. What we’re saying, then, is that W’ – W is some positive non-zero amount. If that would be true, we could combine both machines. Indeed, we could have engine B take Q₁from the reservoir at temperature T₁, do an amount of work equal to W on engine A so it delivers the same amount Q₁back to the reservoir at the same temperature T₁, and we’d still be left with some positive amount of useful work W’ – W. In fact, because the amount of heat in the first reservoir is restored (in each cycle, we take Q₁out but we also put the same amount of heat back in), we could include it as part of the machine. It would no longer need to be some huge external thing with unlimited heat capacity.

So it’s great! Each cycle gives us an amount of useful work equal to W’ – W. What about the energy conservation law? Well… engine A takes Q₁– W from the reservoir at temperature T₂, and engine B gives Q₁– W’ back to it, so we’re taking a net amount of heat equal to (Q₁– W) – (Q₁– W’) = W’ – W out of the T₂ reservoir. So that works out too! So we’ve got a combined machine converting thermal energy into useful work. It looks like a nice set-up, doesn’t it?

Yes. The problem is that, according to Feynman, it cannot work. Why not? Because it violates Carnot’s postulate. The reasoning here is not easy. Let’s me try to do my best to present the argument correctly. What’s the problem? The problem is that we’ve got an engine here that operates at one temperature only. Now, according to Carnot’s postulate, it is not possible to extract the energy of heat at a single temperature with no other change in the system or the surroundings. Why not? Feynman gives the example of the can with compressed air. Imagine a can of compressed air indeed, and imagine we let the air expand, to drive a piston, for example. Now, we can imagine that our can with compressed air was in touch with a large heat reservoir at the same temperature, so its temperature doesn’t drop. So we’ve done work with that can at a single temperature. However, this doesn’t violate Carnot’s postulate because we’ve also changed the system: the air has expanded. It would only violate Carnot’s postulate if we’d find a way to put the air back in using exactly the same amount of work, so the process would be fully reversible. Now, Carnot’s postulate says that’s not possible at the same temperature. If the whole world is at the same temperature, then it is not possible to reversibly extract and convert heat energy into work.

I am not sure the example of the can with compressed air helps, but Feynman obviously thinks it should. He then phrases Carnot’s postulate as follows: “It is not possible to obtain useful work from a reservoir at a single temperature with no other changes.” He therefore claims that the combined machine as described above cannot exist. Ergo, W’ cannot be greater than W. Switching the role of A and B (so B becomes reversible too now), he concludes that W can also not be greater than W’. Hence, W and W’ have to be equal.

Hmm… I know that both philosophers and engineers have worked tirelessly to try to disprove Carnot’s postulate, and that they all failed. Hence, I don’t want to try to disprove Carnot’s postulate. In fact, I don’t doubt its truth at all. All that I am saying here is that I do have my doubts on the logical rigor of Feynman’s ‘proof’. It’s like… Well… It’s just a bit too tautological I’d say.

First Principles of Statistical Mechanics

Pre-script (dated 26 June 2020): This post has become less relevant (even irrelevant, perhaps) because my views on all things quantum-mechanical have evolved significantly as a result of my progression towards a more complete realist (classical) interpretation of quantum physics. I keep blog posts like these mainly because I want to keep track of where I came from. I might review them one day, but I currently don’t have the time or energy for it. 🙂

Original post:

Feynman seems to mix statistical mechanics and thermodynamics in his chapters on it. At first, I thought all was rather messy but, as usual, after re-reading it a couple of times, it all makes sense. Let’s have a look at the basics. We’ll start by talking about gases first.

The ideal gas law

The pressure P is the force we have to apply to the piston containing the gas (see below)—per unit area, that is. So we write: P = F/A. Compressing the gas amounts to applying a force over some (infinitesimal) distance dx. This will change the internal energy (U) of the gas by an infinitesimal amount dU. Hence, we can write:

dU = F·(−dx) = – P·A·dx = – P·dV

However, before looking at the dynamics, let’s first look at the stationary situation: let’s assume the volume of the gas does not change, and so we just have the gas atoms bouncing of the piston and, hence, exerting pressure on it. Every gas atom or particle delivers a momentum 2mv_xto the piston (the factor 2 is there because the piston does not bounce back, so there is no transfer of momentum). If there are N atoms in the volume N, then there are n = N/V in each unit volume. Of course, only the atoms within a distance v_x·t are going to hit the piston within the time t and, hence, the number of atoms hitting the piston within that time is n·A·v_x·t. Per unit time (i.e. per second), it’s n·A·v_x·t/t = n·A·v_x. Hence, the total momentum that’s being transferred per second is n·A·v_x·2mv_x.

So far, so good. Indeed, we know that the force is equal to the amount of momentum that’s being transferred per second. If you forget, just check the definitions and units: a force of 1 newton gives an mass of 1 kg an acceleration of 1 m/s per second, so 1 N = 1 kg·m/s²= 1 kg·(m/s)/s. [The kg·(m/s) unit is the unit of momentum (mass times velocity), obviously. So there we are.] Hence,

P = F/A = n·A·v_x·2mv_x/A = 2nmv_x²

Of course, we need to take an average 〈v_x²〉 here, and we should drop the factor 2 because half of the atoms/particles move away from the piston, rather than towards it. In short, we get:

P = F/A = nm〈v_x²〉

Now, the average velocity in the x-, y- and z-direction are all the same and uncorrelated, so 〈v_x²〉 = 〈v_y²〉 = 〈v_z²〉 = [〈v_x²〉 + 〈v_y²〉 + 〈v_z²〉]/3 = 〈v²〉/3. So we don’t worry about any direction and simply write:

P = F/A = (2/3)·n·〈m·v²/2〉

[As Feynman notes, the math behind this is not difficult but, at the same time, it is also less straightforward than one might expect.] The last factor is, obviously, the kinetic energy of the (center-of-mass) motion of the atom or particle. Multiplying by V gives:

P·V = (2/3)·N·〈m·v²/2〉 = (2/3)·U

[If this confuses you, note that n = N/V, so V = N/n.] Now, that’s not a law you’ll remember from your high school days because… Well… This U – the internal energy of a gas – how do you measure that? We should link it to a measure we do know, and that’s temperature. The atoms or molecules in a gas will have an average kinetic energy which we could define as… Well… That average should have been defined as the temperature but, for historical reasons, the scale of what we know as the ‘temperature’ variable (T) is different. We need to apply a conversion factor, which is usually written as k. In fact, the conversion factor will be (3/2)·k. The 3/2 factor has been thrown in here to get rid of it later (in a few seconds, that is). To make a long story short, we write the mean atomic or molecular energy as (3/2)·k·T = 3kT/2.

Now, you should also remember that we have three independent directions of motion. Hence, the kinetic energy associated with the component of motion in any of the three directions x, y or z is only 1/2 kT = (3kT/2)/3 = kT/2. [This seems trivial, but the idea of associating energy with some direction is actually quite fundamental.] Now, I said we’d get rid of that 3/2 factor. Indeed, applying the above-mentioned definition of temperature, we get:

P·V = (2/3)·N·〈m·v²/2〉 = (2/3)·N·3kT/2 = N·k·T

Now that is a formula you may or may not remember from your high school days! 🙂 The k factor is a constant of proportionality, which makes the units come out alright. The P·V = (2/3)·U formula tells us both sides of the equation must be expressed in joule (J), i.e. the dimension of energy. Now, N is a pure number, so our k in that N·k·T expression must be expressed in joule per degree (Kelvin). To be precise, k is (about) 1.38×10⁻²³joule for every degree Kelvin, so it’s a very tiny constant: it’s referred to as the Boltzmann constant and it’s usually denoted with a capital B as subscript (k_B). As for how the product of pressure and volume can (also) yield something in joule, you can work that out for yourself, remembering the definition of a joule. […] Well… OK. Let me do it for you: [P]·[V] = (N/m²)·m³ = N·m = J. 🙂

One immediate implication of the formula above is that gases at the same temperature and pressure, in the same volume, must consist of an equal number of atoms/molecules. You’ll say: of course – because you remember that from your high school classes. However, thinking about it some more – and also in light of what we’ll be learning a bit later on gases composed of more complex molecules (diatomic molecules, for example) – you’ll have to admit it’s not all that obvious as a result.

Now, the number of atoms/molecules is usually measured in moles: one mole (or mol) is 6.02×10²³units (more or less, that is). To be somewhat more precise, its CODATA value is 6.02214129(27)×10²³. That number is Avogadro’s number (or constant), after the Italian mathematical physicist Amedeo Avogadro – who stated that law above, which is referred to as Avogradro’s Law: gases at the same temperature and pressure, in the same volume, must consist of an equal number of atoms/molecules. Avogadro’s number is defined as the amount of any substance that contains as many elementary entities (e.g. atoms, molecules, ions or electrons) as there are atoms in 12 grams of pure carbon-12 (¹²C), the isotope of carbon with relative atomic mass of exactly 12 (also by definition). Avogadro’s constant is one of the base units in the International Systems of Units, usually denoted by N_A or – as Feynman does – N₀.

Now, if we reinterpret N as the number of moles, rather than the number of atoms, ions or molecules in a gas, we can re-write the same equation using the so-called universal or ideal gas constant, which is equal to R = (1.38×10⁻²³joule)×(6.02×10²³/mol) per degree Kelvin = 8.314 J·K⁻¹·mol⁻¹. In short, the ideal gas constant is the product of two other constants: the Boltzmann constant (k_B) and the Avogadro number (N₀). So we get:

P·V = N·R·T with N = no. of moles and R = k_B·N₀

As you can see, you need to watch out with all those different constants and notations in use.

The ideal gas law and internal motion

There’s an interesting and essential remark to be made in regard to complex molecules in a gas. A complex molecule is any molecule that is not mono-atomic. The simplest example of a complex molecule is a diatomic molecule, consisting of two atoms, which we’ll denote by A and B, with mass m_Aand m_Brespectively. A and B are together but are able to oscillate or move relative to one another. In short, we also have some internal motion here, in addition to the motion of the whole thing, which will also has some kinetic energy. Hence, the kinetic energy of the gas consists of two parts:

The kinetic energy of the so-called center-of-mass motion of the whole thing (i.e. the molecule), which we’ll denote by M = m_A+ m_B, and
The kinetic energy of the rotational and vibratory motions of the two atoms (A and B) inside the molecule.

We noted that for single atoms the mean value of the kinetic energy in one direction is kT/2 and that the total kinetic energy is 3kT/2, i.e. three times as much. So what do we have here? Well… The reasoning we followed for the single atoms is also valid for the diatomic molecule considered as a single body of total mass M and with some center-of-mass velocity v_CM. Hence, we can write that

M·v_CM²/2 = (3/2)·kT

So that’s the same, regardless of whether or not we’re considering the separate pieces or the whole thing. But let’s look at the separate pieces now. We need some vector analysis here, because A and B can move in separate directions, so we have v_Aand v_B(note the boldface used for vectors). So what’s the relation between v_Aand v_Bon the one hand, and v_CM on the other? The analysis is somewhat tricky here but – assuming that the v_Aand v_Brepresentations themselves are some idealization of the actual rotational and vibratory movements of the A and B atoms – we can write:

v_CM = (m_Av_A+ m_Bv_B)/M

Now we need to calculate 〈v_CM²〉, of course, i.e. the average velocity squared. I’ll refer you to Feynman for the details which, in the end, do lead to that M·v_CM²/2 = (3/2)·kT equation. The whole calculation depends on the assumption that the relative velocity w = v_A– v_Bis not any more likely to point in one direction than another, so its average component in any direction is zero. Indeed, the interim result is that

M·v_CM²/2 = (3/2)·kT + 2m_Am_B〈v_A·v_B〉/M

Hence, one needs to prove, somehow, that 〈v_A·v_B〉 is zero in order to get the result we want, which is what that assumption about the relative velocity w ensures. Now, we still don’t have the kinetic energy of the A and B parts of the molecule. Because A and B can move in all three directions in space, their average kinetic energy 〈m_A·v_A²/2〉 and 〈m_B·v_B²/2〉 is also 3·k·T/2. Now, adding 3·k·T/2 and 3·k·T/2 yields 3kT. So now we have what we wanted:

The kinetic energy of the center-of-mass motion of the diatomic molecule is (3/2)·k·T.
The total energy of the diatomic molecule is the sum of the energies of A and B, and so that’s 3·k·T/2 + 3·k·T/2 = 3 k·T.
The kinetic energy of the internal rotational and vibratory motions of the two atoms (A and B) inside the molecule is the difference, so that’s 3·k·T – (3/2)·k·T = (3/2)·k·T.

The more general result can be stated as follows:

A r-atom molecule in a gas will have a kinetic energy of (3/2)·r·k·T, on average, of which:
3/2·k·T is kinetic energy of the center-of-mass motion of the entire molecule,
The rest, (3/2)·(r−1)·k·T, is internal vibrational and rotational kinetic energy.

Another way to state is that, for an r-atom molecule, we find that the average energy for each ‘independent direction of motion’, i.e. for each degree of freedom in the system, is kT/2, with the number of degrees of freedom being equal to 3r.

So in this particular case (example of a diatomic molecule), we have 6 degrees of freedom (two times three), because we have three directions in space for each of the two atoms. A common error is to consider the center-of-mass energy as something separate, rather than including it as a part of the total energy. So always remember: the total kinetic energy is, quite simply, the sum of the kinetic energies of the separate atoms, which can be separated into (1) the kinetic energy associated with the center-of-mass motion and (2) the kinetic energy of the internal motions.

You see? It is not that difficult, is it? Let’s move on to the next topic.

The exponential atmosphere

Feynman uses this rather intriguing title to introduce Boltzmann’s Law, which is a law about densities. Let’s jot it down first:

n = n₀·e^−P.E/kT

In this equation, P.E. is the potential energy, k is our Boltzmann constant, and T is the temperature expressed in Kelvin. As for n₀, that’s just a constant which depends on the reference point (P.E. = 0). What are we calculating here? Densities, so that’s the relative or absolute number of molecules per unit volume, so we look for a formula for a variable like n = N/V.

Let’s do an example: the ‘exponential’ atmosphere. 🙂 Feynman models our ‘atmosphere’ as a huge column of gas (see below). To simplify the analysis, we make silly assumptions. For example, we assume the temperature is the same at all heights. That’s assured by the mechanism for equalizing temperature: if the molecules on top would have less energy than those at the bottom, the molecules at the bottom would shake the molecules at the top, via the rod and the balls. That’s a very theoretical set-up, of course, but let’s just go along with it. The idea is that – when thermal equilibrium is reached – the average kinetic energy of all molecules is the same.

So, if the temperature is the same, then what’s different? The pressure, of course, which is determined by the number of molecules per unit volume. The pressure must increase with lower altitude because it has to hold, so to speak, the weight of all the gas above it. Conversely, as we go higher, the atmosphere becomes more tenuous. So what’s the ‘law’ or formula here?

We’ll use our gas law: PV = NkT, which we can re-write as P = nkT with n = N/V, so n is the number of molecules per unit volume indeed. What’s stated here is that the pressure (P) and the number of molecules per unit volume (n) are directly proportional, with kT the proportionality factor. So we have gravity (the g force) and we can do a differential analysis: what happens when we go from h to h + dh? If m is the mass of each molecule, and if we assume we’re looking at unit areas (both at h as well as h + dh), then the gravitational force on each molecule will be mg, and ndh will be the total number of molecules in that ‘unit section’.

Now, we can write dP as dP = P_h+dh− P_h and, of course, we know that the difference in pressure must be sufficient to hold, so to speak, the molecules in that small unit section dh. So we can write the following:

dP = P_h+dh− P_h = − m·g·n·dh

Now, P is P = nkT and, hence, because we assume T to be constant, we can write the whole equation as dP = k·T·dn = − m·g·n·dh. From that, we get a differential equation:

dn/dh = −(m·g)/(k·T)·n

We all hate differential equations, of course, but this one has an easy solution: the equation basically states we should find a function for n which has a derivative which is proportional to itself. Of course, we know that the exponential function is such function, so the solution of the differential equation is:

n = n₀·e^−mgh/kT

The n₀ factor is the constant of integration and is, as mentioned above, the density at h = 0. Also note that mgh is, indeed, the potential energy of the molecules, increasing with height. So we have a Boltzmann Law indeed here, which we can write as n = n₀·e^−P.E/kT. Done ! The illustration below was also taken from Feynman, and illustrates the ‘exponential atmosphere’ for two gases: oxygen and hydrogen. Because their mass is very different, the curve is different too: it shows how, in theory and in practice, lighter gases will dominate at great heights, because the exponentials for the heavier stuff have all died out.

Generalization

It is easy to show that we’ll have a Boltzmann Law in any situation where the force comes from a potential. In other words, we’ll have a Boltzmann Law in any situation for which the work done when taking a molecule from x to x + dx can be represented as potential energy. An example would be molecules that are electrically charged and attracted by some electric field or another charge that attracts them. In that case, we have an electric force of attraction which varies with position and acts on all molecules. So we could take two parallel planes in the gas, separated by a distance dx indeed, and we’d have a similar situation: the force on each atom, times the number of atoms in the unit section that’s delineated by dx, would have to be balanced by the pressure change, and we’d find a similar ‘law’: n = n₀·e^−P.E/kT.

Let’s quickly show it. The key variable is, once again, the density n: n = N/V. If we assume volume and temperature remain constant, then we can use our gas law to write the pressure as P = NkT/V = kT·n, which implies that any change in pressure must involve a density change. To be precise, dP = d(kT·n) = kT·dn. Now, we’ve got a force, and moving a molecule from x to x + dx involves work, which is the force times the distance, so the work is F·dx. The force can be anything, but we assume it’s conservative, like the electromagnetic force or gravity. Hence, the force field can be represented by a potential and the work done is equal to the change in potential energy. Hence, we can write: Fdx = –d(P.E.). Why the minus sign? If the force is doing work, we’re moving with the force and, hence, we’ll have a decrease in potential energy. Conversely, if the surroundings are doing work against the force, we’ll increase potential energy.

Now, we said the force must be balanced by the pressure. What does that mean, exactly? It’s the same analysis as the one we did for our ‘exponential’ atmosphere: we’ve got a small slice, given by dx, and the difference in pressure when going from x to x + dx must be sufficient to hold, so to speak, the molecules in that small unit section dx. [Note we assume we’re talking unit areas once again.] So, instead of writing dP = P_h+dh− P_h = − m·g·n·dh, we now write dP = F·n·dx. So, when it’s a gravitational field, the magnitude of the force involved is, obviously, F = m·g.

The minus sign business is confusing, as usual: it’s obvious that dP must be negative for positive dh, and vice versa, but here we are moving with the force, so no minus sign is needed. If you find that confusing, let me give you another way of getting that dP = F·n·dx expression. The pressure is, quite simply, the force times the number of particles, so P = F·N. Dividing both sides by V yields P/V = F·N/V = F·n. Therefore, P = F·n·V and, hence, dP must be equal to dP = d(F·n·V) = F·n·dV = F·n·dx. [Again, the assumption is that our unit of analysis is the unit area.] […] OK. I need to move on. Combining (1) dP = d(kT·n) = kT·dn, (2) dP = F·n·dx and (3) Fdx = –d(P.E.), we get:

kT·dn = –d(P.E.)·n ⇔ dn/d(P.E.) = −[1/(kT)]·n

That’s, once again, a differential equation that’s easy to solve. Indeed, we’ve repeated it ad nauseam: a function which has a derivative proportional to itself is an exponential. Hence, we have our grand equation:

n = n₀·e^−P.E/kT

If the whole thing troubles you, just remember that the key to solving problems like this is to clearly identify and separate the so-called ‘dependent’ and ‘independent’ variables. In this case, we want a formula for n and, hence, it’s potential energy that’s the ‘independent’ variable. That’s all. In case of doubt: just do the derivation: d(n₀·e^−P.E./kT)/d(P.E.) = −n₀·e^−P.E/kT·1/(kT) = −n/(kT).

The graph looks the same, of course: the density is greatest at P.E. = 0. To be precise, the density there will be equal to n = n₀·e⁰= n₀ (don’t think it’s infinity there!). And for higher (potential) energy values, we get lower density values. It’s a simple but powerful graph, and so you should always remember it.

Boltzmann’s Law is a very simple law but it can be applied to very complicated situations. Indeed, while the law is simple, the potential energy curve can be very complicated. So our Law can be applied to other situations than gravity or the electric force. The potential can combine a number of forces (as long as they’re all conservative), as shown in the graph below, which shows a situation in which molecules will attract each other at a distance r > r₀(and, hence, their potential energy decreases as they come closer together), but repel each other strongly as r becomes smaller than r₀(so potential energy increases, and very much so as we try to force them on top of each other).

Again, despite the complicated shape of the curve, the density function will – in essence – follow Boltzmann’s Law: in a given volume, the density will be highest at the distance of minimum energy, and the density will be much less at other distances. So, yes, Boltzmann’s Law is pretty powerful !

The Uncertainty Principle revisited

Pre-script (dated 26 June 2020): This post has become less relevant (even irrelevant, perhaps) because my views on all things quantum-mechanical have evolved significantly as a result of my progression towards a more complete realist (classical) interpretation of quantum physics. I keep blog posts like these mainly because I want to keep track of where I came from. I might review them one day, but I currently don’t have the time or energy for it. 🙂

Original post:

I’ve written a few posts on the Uncertainty Principle already. See, for example, my post on the energy-time expression for it (ΔE·Δt ≥ h). So why am I coming back to it once more? Not sure. I felt I left some stuff out. So I am writing this post to just complement what I wrote before. I’ll do so by explaining, and commenting on, the ‘semi-formal’ derivation of the so-called Kennard formulation of the Principle in the Wikipedia article on it.

The Kennard inequalities, σ_xσ_p ≥ ħ/2 and σ_Eσ_t ≥ ħ/2, are more accurate than the more general Δx·Δp ≥ h and ΔE·Δt ≥ h expressions one often sees, which are an early formulation of the Principle by Niels Bohr, and which Heisenberg himself used when explaining the Principle in a thought experiment picturing a gamma-ray microscope. I presented Heisenberg’s thought experiment in another post, and so I won’t repeat myself here. I just want to mention that it ‘proves’ the Uncertainty Principle using the Planck-Einstein relations for the energy and momentum of a photon:

E = hf and p = h/λ

Heisenberg’s thought experiment is not a real proof, of course. But then what’s a real proof? The mentioned ‘semi-formal’ derivation looks more impressive, because more mathematical, but it’s not a ‘proof’ either (I hope you’ll understand why I am saying that after reading my post). The main difference between Heisenberg’s thought experiment and the mathematical derivation in the mentioned Wikipedia article is that the ‘mathematical’ approach is based on the de Broglie relation. That de Broglie relation looks the same as the Planck-Einstein relation (p = h/λ) but it’s fundamentally different.

Indeed, the momentum of a photon (i.e. the p we use in the Planck-Einstein relation) is not the momentum one associates with a proper particle, such as an electron or a proton, for example (so that’s the p we use in the de Broglie relation). The momentum of a particle is defined as the product of its mass (m) and velocity (v). Photons don’t have a (rest) mass, and their velocity is absolute (c), so how do we define momentum for a photon? There are a couple of ways to go about it, but the two most obvious ones are probably the following:

We can use the classical theory of electromagnetic radiation and show that the momentum of a photon is related to the magnetic field (we usually only analyze the electric field), and the so-called radiation pressure that results from it. It yields the p = E/c formula which we need to go from E = hf to p = h/λ, using the ubiquitous relation between the frequency, the wavelength and the wave velocity (c = λf). In case you’re interested in the detail, just click on the radiation pressure link).
We can also use the mass-energy equivalence E = mc². Hence, the equivalent mass of the photon is E/c², which is relativistic mass only. However, we can multiply that mass with the photon’s velocity, which is c, thereby getting the very same value for its momentum p = c·E/c²= E/c.

So Heisenberg’s ‘proof’ uses the Planck-Einstein relations, as it analyzes the Uncertainty Principle more as an observer effect: probing matter with light, so to say. In contrast, the mentioned derivation takes the de Broglie relation itself as the point of departure. As mentioned, the de Broglie relations look exactly the same as the Planck-Einstein relationship (E = hf and p = h/λ) but the model behind is very different. In fact, that’s what the Uncertainty Principle is all about: it says that the de Broglie frequency and/or wavelength cannot be determined exactly: if we want to localize a particle, somewhat at least, we’ll be dealing with a frequency range Δf. As such, the de Broglie relation is actually somewhat misleading at first. Let’s talk about the model behind.

A particle, like an electron or a proton, traveling through space, is described by a complex-valued wavefunction, usually denoted by the Greek letter psi (Ψ) or phi (Φ). This wavefunction has a phase, usually denoted as θ (theta) which – because we assume the wavefunction is a nice periodic function – varies as a function of time and space. To be precise, we write θ as θ = ωt – kx or, if the wave is traveling in the other direction, as θ = kx – ωt.

I’ve explained this in a couple of posts already, including my previous post, so I won’t repeat myself here. Let me just note that ω is the angular frequency, which we express in radians per second, rather than cycles per second, so ω = 2πf (one cycle covers 2π rad). As for k, that’s the wavenumber, which is often described as the spatial frequency, because it’s expressed in cycles per meter or, more often (and surely in this case), in radians per meter. Hence, if we freeze time, this number is the rate of change of the phase in space. Because one cycle is, again, 2π rad, and one cycle corresponds to the wave traveling one wavelength (i.e. λ meter), it’s easy to see that k = 2π/λ. We can use these definitions to re-write the de Broglie relations E = hf and p = h/λ as:

E = ħω and p = ħk with h = h/2π

What about the wave velocity? For a photon, we have c = λf and, hence, c = (2π/k)(ω/2π) = ω/k. For ‘particle waves’ (or matter waves, if you prefer that term), it’s much more complicated, because we need to distinguish between the so-called phase velocity (v_p) and the group velocity (v_g). The phase velocity is what we’re used to: it’s the product of the frequency (the number of cycles per second) and the wavelength (the distance traveled by the wave over one cycle), or the ratio of the angular frequency and the wavenumber, so we have, once again, λf = ω/k = v_p. However, this phase velocity is not the classical velocity of the particle that we are looking at. That’s the so-called group velocity, which corresponds to the velocity of the wave packet representing the particle (or ‘wavicle’, if your prefer that term), as illustrated below.

The animation below illustrates the difference between the phase and the group velocity even more clearly: the green dot travels with the ‘wavicles’, while the red dot travels with the phase. As mentioned above, the group velocity corresponds to the classical velocity of the particle (v). However, the phase velocity is a mathematical point that actually travels faster than light. It is a mathematical point only, which does not carry a signal (unlike the modulation of the wave itself, i.e. the traveling ‘groups’) and, hence, it does not contradict the fundamental principle of relativity theory: the speed of light is absolute, and nothing travels faster than light (except mathematical points, as you can, hopefully, appreciate now).

The two animations above do not represent the quantum-mechanical wavefunction, because the functions that are shown are real-valued, not complex-valued. To imagine a complex-valued wave, you should think of something like the ‘wavicle’ below or, if you prefer animations, the standing waves underneath (i.e. C to H: A and B just present the mathematical model behind, which is that of a mechanical oscillator, like a mass on a spring indeed). These representations clearly show the real as well as the imaginary part of complex-valued wave-functions.

With this general introduction, we are now ready for the more formal treatment that follows. So our wavefunction Ψ is a complex-valued function in space and time. A very general shape for it is one we used in a couple of posts already:

Ψ(x, t) ∝ e^{i(kx – ωt)}= cos(kx – ωt) + isin(kx – ωt)

If you don’t know anything about complex numbers, I’d suggest you read my short crash course on it in the essentials page of this blog, because I don’t have the space nor the time to repeat all of that. Now, we can use the de Broglie relationship relating the momentum of a particle with a wavenumber (p = ħk) to re-write our psi function as:

Ψ(x, t) ∝ e^{i(kx – ωt)}= e^{i(px/ħ – ωt)}

Note that I am using the ‘proportional to’ symbol (∝) because I don’t worry about normalization right now. Indeed, from all of my other posts on this topic, you know that we have to take the absolute square of all these probability amplitudes to arrive at a probability density function, describing the probability of the particle effectively being at point x in space at point t in time, and that all those probabilities, over the function’s domain, have to add up to 1. So we should insert some normalization factor.

Having said that, the problem with the wavefunction above is not normalization really, but the fact that it yields a uniform probability density function. In other words, the particle position is extremely uncertain in the sense that it could be anywhere. Let’s calculate it using a little trick: the absolute square of a complex number equals the product of itself with its (complex) conjugate. Hence, if z = re^iθ, then │z│² = zz* = re^iθ·re^–iθ= r²e^iθ^–iθ= r²e⁰= r². Now, in this case, assuming unique values for k, ω, p, which we’ll note as k₀, ω₀, p₀ (and, because we’re freezing time, we can also write t = t₀), we should write:

│Ψ(x)│² = │a₀e^{i(p₀x/ħ – ω₀t₀)}│² = │a₀e^ip₀x/ħe^{–iω₀t₀}│² = │a₀e^ip₀x/ħ│² │e^–i^ω^₀t₀│² = a₀²

Note that, this time around, I did insert some normalization constant a₀ as well, so that’s OK. But so the problem is that this very general shape of the wavefunction gives us a constant as the probability for the particle being somewhere between some point a and another point b in space. More formally, we get the surface for a rectangle when we calculate the probability P[a ≤ X ≤ b] as we should calculate it, which is as follows:

More specifically, because we’re talking one-dimensional space here, we get P[a ≤ X ≤ b] = (b–a)·a₀². Now, you may think that such uniform probability makes sense. For example, an electron may be in some orbital around a nucleus, and so you may think that all ‘points’ on the orbital (or within the ‘sphere’, or whatever volume it is) may be equally likely. Or, in another example, we may know an electron is going through some slit and, hence, we may think that all points in that slit should be equally likely positions. However, we know that it is not the case. Measurements show that not all points are equally likely. For an orbital, we get complicated patterns, such as the one shown below, and please note that the different colors represent different complex numbers and, hence, different probabilities.

Also, we know that electrons going through a slit will produce an interference pattern—even if they go through it one by one! Hence, we cannot associate some flat line with them: it has to be a proper wavefunction which implies, once again, that we can’t accept a uniform distribution.

In short, uniform probability density functions are not what we see in Nature. They’re non-uniform, like the (very simple) non-uniform distributions shown below. [The left-hand side shows the wavefunction, while the right-hand side shows the associated probability density function: the first two are static (i.e. they do not vary in time), while the third one shows a probability distribution that does vary with time.]

I should also note that, even if you would dare to think that a uniform distribution might be acceptable in some cases (which, let me emphasize this, it is not), an electron can surely not be ‘anywhere’. Indeed, the normalization condition implies that, if we’d have a uniform distribution and if we’d consider all of space, i.e. if we let a go to –∞ and b to +∞, then a₀²would tend to zero, which means we’d have a particle that is, literally, everywhere and nowhere at the same time.

In short, a uniform probability distribution does not make sense: we’ll generally have some idea of where the particle is most likely to be, within some range at least. I hope I made myself clear here.

Now, before I continue, I should make some other point as well. You know that the Planck constant (h or ħ) is unimaginably small: about 1×10⁻³⁴J·s (joule-second). In fact, I’ve repeatedly made that point in various posts. However, having said that, I should add that, while it’s unimaginably small, the uncertainties involved are quite significant. Let us indeed look at the value of ħ by relating it to that σ_xσ_p ≥ ħ/2 relation.

Let’s first look at the units. The uncertainty in the position should obviously be expressed in distance units, while momentum is expressed in kg·m/s units. So that works out, because 1 joule is the energy transferred (or work done) when applying a force of 1 newton (N) over a distance of 1 meter (m). In turn, one newton is the force needed to accelerate a mass of one kg at the rate of 1 meter per second per second (this is not a typing mistake: it’s an acceleration of 1 m/s per second, so the unit is m/s²: meter per second squared). Hence, 1 J·s = 1 N·m·s = 1 kg·m/s²·m·s = kg·m²/s. Now, that’s the same dimension as the ‘dimensional product’ for momentum and distance: m·kg·m/s = kg·m²/s.

Now, these units (kg, m and s) are all rather astronomical at the atomic scale and, hence, h and ħ are usually expressed in other dimensions, notably eV·s (electronvolt-second). However, using the standard SI units gives us a better idea of what we’re talking about. If we split the ħ = 1×10⁻³⁴J·s value (let’s forget about the 1/2 factor for now) ‘evenly’ over σ_xand σ_p – whatever that means: all depends on the units, of course! – then both factors will have magnitudes of the order of 1×10⁻¹⁷: 1×10⁻¹⁷m times 1×10⁻¹⁷kg·m/s gives us 1×10⁻³⁴J·s.

You may wonder how this 1×10⁻¹⁷m compares to, let’s say, the classical electron radius, for example. The classical electron radius is, roughly speaking, the ‘space’ an electron seems to occupy as it scatters incoming light. The idea is illustrated below (credit for the image goes to Wikipedia, as usual). The classical electron radius – or Thompson scattering length – is about 2.818×10⁻¹⁵m, so that’s almost 300 times our ‘uncertainty’ (1×10⁻¹⁷m). Not bad: it means that we can effectively relate our ‘uncertainty’ in regard to the position to some actual dimension in space. In this case, we’re talking the femtometer scale (1 fm = 10⁻¹⁵m), and so you’ve surely heard of this before.

What about the other ‘uncertainty’, the one for the momentum (1×10⁻¹⁷kg·m/s)? What’s the typical (linear) momentum of an electron? Its mass, expressed in kg, is about 9.1×10⁻³¹ kg. We also know its relative velocity in an electron: it’s that magical number α = v/c, about which I wrote in some other posts already, so v = αc ≈ 0.0073·3×10⁸m/s ≈ 2.2×10⁶m/s. Now, 9.1×10⁻³¹ kg times 2.2×10⁶m/s is about 2×10^–26kg·m/s, so our proposed ‘uncertainty’ in regard to the momentum (1×10⁻¹⁷kg·m/s) is half a billion times larger than the typical value for it. Now that is, obviously, not so good. [Note that calculations like this are extremely rough. In fact, when one talks electron momentum, it’s usual angular momentum, which is ‘analogous’ to linear momentum, but angular momentum involves very different formulas. If you want to know more about this, check my post on it.]

Of course, now you may feel that we didn’t ‘split’ the uncertainty in a way that makes sense: those –17 exponents don’t work, obviously. So let’s take 1×10^–26kg·m/s for σ_p, which is half of that ‘typical’ value we calculated. Then we’d have 1×10⁻⁸m for σ_x (1×10⁻⁸m times 1×10^–26kg·m/s is, once again, 1×10^–34J·s). But then that uncertainty suddenly becomes a huge number: 1×10⁻⁸m is 100 angstrom. That’s not the atomic scale but the molecular scale! So it’s huge as compared to the pico- or femto-meter scale (1 pm = 1×10⁻¹² m, 1 fm = 1×10⁻¹⁵ m) which we’d sort of expect to see when we’re talking electrons.

OK. Let me get back to the lesson. Why this digression? Not sure. I think I just wanted to show that the Uncertainty Principle involves ‘uncertainties’ that are extremely relevant: despite the unimaginable smallness of the Planck constant, these uncertainties are quite significant at the atomic scale. But back to the ‘proof’ of Kennard’s formulation. Here we need to discuss the ‘model’ we’re using. The rather simple animation below (again, credit for it has to go to Wikipedia) illustrates it wonderfully.

Look at it carefully: we start with a ‘wave packet’ that looks a bit like a normal distribution, but it isn’t, of course. We have negative and positive values, and normal distributions don’t have that. So it’s a wave alright. Of course, you should, once more, remember that we’re only seeing one part of the complex-valued wave here (the real or imaginary part—it could be either). But so then we’re superimposing waves on it. Note the increasing frequency of these waves, and also note how the wave packet becomes increasingly localized with the addition of these waves. In fact, the so-called Fourier analysis, of which you’ve surely heard before, is a mathematical operation that does the reverse: it separates a wave packet into its individual component waves.

So now we know the ‘trick’ for reducing the uncertainty in regard to the position: we just add waves with different frequencies. Of course, different frequencies imply different wavenumbers and, through the de Broglie relationship, we’ll also have different values for the ‘momentum’ associated with these component waves. Let’s write these various values as k_n, ω_n, and p_n respectively, with n going from 0 to N. Of course, our point in time remains frozen at t₀. So we get a wavefunction that’s, quite simply, the sum of N component waves and so we write:

Ψ(x) = ∑ a_ne^{i(p_nx/ħ – ω_nt₀)}= ∑ a_ne^ip_nx/ħe^–iω_nt₀= ∑ A_ne^ip_nx/ħ

Note that, because of the e^–iω_nt₀, we now have complex-valued coefficients A_n = a_ne^–iω_nt₀ in front. More formally, we say that A_n represents the relative contribution of the mode p_n to the overall Ψ(x) wave. Hence, we can write these coefficients A as a function of p. Because Greek letters always make more of an impression, we’ll use the Greek letter Φ (phi) for it. 🙂 Now, we can go to the continuum limit and, hence, transform that sum above into an infinite sum, i.e. an integral. So our wave function then becomes an integral over all possible modes, which we write as:

Don’t worry about that new 1/√2πħ factor in front. That’s, once again, something that has to do with normalization and scales. It’s the integral itself you need to understand. We’ve got that Φ(p) function there, which is nothing but our A_n coefficient, but for the continuum case. In fact, these relative contributions Φ(p) are now referred to as the amplitude of all modes p, and so Φ(p) is actually another wave function: it’s the wave function in the so-called momentum space.

You’ll probably be very confused now, and wonder where I want to go with an integral like this. The point to note is simple: if we have that Φ(p) function, we can calculate (or derive, if you prefer that word) the Ψ(x) from it using that integral above. Indeed, the integral above is referred to as the Fourier transform, and it’s obviously closely related to that Fourier analysis we introduced above.

Of course, there is also an inverse transform, which looks exactly the same: it just switches the wave functions (Ψ and Φ) and variables (x and p), and then (it’s an important detail!), it has a minus sign in the exponent. Together, the two functions – as defined by each other through these two integrals – form a so-called Fourier integral pair, also known as a Fourier transform pair, and the variables involved are referred to as conjugate variables. So momentum (p) and position (x) are conjugate variables and, likewise, energy and time are also conjugate variables (but so I won’t expand on the time-energy relation here: please have a look at one of my others posts on that).

Now, I thought of copying and explaining the proof of Kennard’s inequality from Wikipedia’s article on the Uncertainty Principle (you need to click on the show button in the relevant section to see it), but then that’s pretty boring math, and simply copying stuff is not my objective with this blog. More importantly, the proof has nothing to do with physics. Nothing at all. Indeed, it just proves a general mathematical property of Fourier pairs. More specifically, it proves that, the more concentrated one function is, the more spread out its Fourier transform must be. In other words, it is not possible to arbitrarily concentrate both a function and its Fourier transform.

So, in this case, if we’d ‘squeeze’ Ψ(x), then its Fourier transform Φ(p) will ‘stretch out’, and so that’s what the proof in that Wikipedia article basically shows. In other words, there is some ‘trade-off’ between the ‘compaction’ of Ψ(x), on the one hand, and Φ(p), on the other, and so that is what the Uncertainty Principle is all about. Nothing more, nothing less.

But… Yes? What’s all this talk about ‘squeezing’ and ‘compaction’? We can’t change reality, can we? Well… Here we’re entering the philosophical field, of course. How do we interpret the Uncertainty Principle? It surely does look like us trying to measure something has some impact on the wavefunction. In fact, usually, our measurement – of either position or momentum – usually makes the wavefunctions collapse: we suddenly know where the particle is and, hence, ψ(x) seems to collapse into one point. Alternatively, we measure its momentum and, hence, Φ(p) collapses.

That’s intriguing. In fact, even more intriguing is the possibility we may only partially affect those wavefunctions with measurements that are somewhat less ‘drastic’. It seems a lot of research is focused on that (just Google for partial collapse of the wavefunction, and you’ll finds tons of references, including presentations like this one).

Hmm… I need to further study the topic. The decomposition of a wave into its component waves is obviously something that works well in physics—and not only in quantum mechanics but also in much more mundane examples. Its most general application is signal processing, in which we decompose a signal (which is a function of time) into the frequencies that make it up. Hence, our wavefunction model makes a lot of sense, as it mirrors the physics involved in oscillators and harmonics obviously.

Still… I feel it doesn’t answer the fundamental question: what is our electron really? What do those wave packets represent? Physicists will say questions like this don’t matter: as long as our mathematical models ‘work’, it’s fine. In fact, if even Feynman said that nobody – including himself – truly understands quantum mechanics, then I should just be happy and move on. However, for some reason, I can’t quite accept that. I should probably focus some more on that de Broglie relationship, p = h/λ, as it’s obviously as fundamental to my understanding of the ‘model’ of reality in physics as that Fourier analysis of the wave packet. So I need to do some more thinking on that.

The de Broglie relationship is not intuitive. In fact, I am not ashamed to admit that it actually took me quite some time to understand why we can’t just re-write the de Broglie relationship (λ = h/p) as an uncertainty relation itself: Δλ = h/Δp. Hence, let me be very clear on this:

Δx = h/Δp (that’s the Uncertainty Principle) but Δλ ≠ h/Δp !

Let me quickly explain why.

If the Δ symbol expresses a standard deviation (or some other measurement of uncertainty), we can write the following:

p = h/λ ⇒ Δp = Δ(h/λ) = hΔ(1/λ) ≠ h/Δp

So I can take h out of the brackets after the Δ symbol, because that’s one of the things that’s allowed when working with standard deviations. More in particular, one can prove the following:

The standard deviation of some constant function is 0: Δ(k) = 0
The standard deviation is invariant under changes of location: Δ(x + k) = Δ(x + k)
Finally, the standard deviation scales directly with the scale of the variable: Δ(kx) = |k |Δ(x).

However, it is not the case that Δ(1/x) = 1/Δx. However, let’s not focus on what we cannot do with Δx: let’s see what we can do with it. Δx equals h/Δp according to the Uncertainty Principle—if we take it as an equality, rather than as an inequality, that is. And then we have the de Broglie relationship: p = h/λ. Hence, Δx must equal:

Δx = h/Δp = h/[Δ(h/λ)] =h/[hΔ(1/λ)] = 1/Δ(1/λ)

That’s obvious, but so what? As mentioned, we cannot write Δx = Δλ, because there’s no rule that says that Δ(1/λ) = 1/Δλ and, therefore, h/Δp ≠ Δλ. However, what we can do is define Δλ as an interval, or a length, defined by the difference between its lower and upper bound (let’s denote those two values by λ_a and λ_b respectively. Hence, we write Δλ = λ_b – λ_a. Note that this does not assume we have a continuous range of values for λ: we can have any number of frequencies λ_nbetween λ_a and λ_b, but so you see the point: we’ve got a range of values λ, discrete or continuous, defined by some lower and upper bound.

Now, the de Broglie relation associates two values p_a and p_b with λ_a and λ_b respectively: p_a = h/λ_a and p_b = h/λ_b. Hence, we can similarly define the corresponding Δp interval as p_a – p_b, with p_a = h/λ_a and p_b= h/λ_b. Note that, because we’re taking the reciprocal, we have to reverse the order of the values here: if λ_b > λ_a, then p_a = h/λ_a > p_b= h/λ_b. Hence, we can write Δp = Δ(h/λ) = p_a – p_b = h/λ₁ – h/λ₂= h(1/λ₁ – 1/λ₂) = h[λ₂ – λ₁]/λ₁λ₂. In case you have a bit of difficulty, just draw some reciprocal functions (like the ones below), and have fun connecting intervals on the horizontal axis with intervals on the vertical axis using these functions.

Now, h[λ₂ – λ₁]/λ₁λ₂) is obviously something very different than h/Δλ = h/(λ₂ – λ₁). So we can surely not equate the two and, hence, we cannot write that Δp = h/Δλ.

Having said that, the Δx = 1/Δ(1/λ) = λ₁λ₂/(λ₂ – λ₁) that emerges here is quite interesting. We’ve got a ratio here, λ₁λ₂/(λ₂ – λ₁, which shows that Δx depends only on the upper and lower bounds of the Δλ range. It does not depend on whether or not the interval is discrete or continuous.

The second thing that is interesting to note is Δx depends not only on the difference between those two values (i.e. the length of the interval) but also on their value: if the length of the interval, i.e. the difference between the two frequencies is the same, but their values as such are higher, then we get a higher value for Δx, i.e. a greater uncertainty in the position. Again, this shows that the relation between Δλ and Δx is not straightforward. But so we knew that already, and so I’ll end this post right here and right now. 🙂

The Strange Theory of Light and Matter (III)

Pre-script (dated 26 June 2020): This post has become less relevant (even irrelevant, perhaps) because my views on all things quantum-mechanical have evolved significantly as a result of my progression towards a more complete realist (classical) interpretation of quantum physics. I keep blog posts like these mainly because I want to keep track of where I came from. I might review them one day, but I currently don’t have the time or energy for it. 🙂

Original post:

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

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

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

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

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

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

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

I. Probability amplitudes: what are they?

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

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

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

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

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

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

A. The magnitude

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

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

Why? Well… Just write it out. [You can skip this section if you’re not interested in complex numbers, but I would recommend you try to understand. It’s not that difficult. Indeed, if you’re reading this, you’re most likely to understand something of complex numbers and, hence, you should be able to work your way through it. Just remember that a complex number is like a two-dimensional number, which is why it’s sometimes written using bold-face (z), rather than regular font (z). However, I should immediately add this convention is usually not followed. I like the boldface though, and so I’ll try to use it in this post.] The square of a complex number z = a + bi is equal to z²= a²+ 2abi – b², while the square of its absolute value (i.e. the absolute square) is |z|²= [√(a²+ b²)]² = a²+ b². So you can immediately see that the square and the absolute square of a complex numbers are two very different things indeed: it’s not only the 2abi term, but there’s also the minus sign in the first expression, because of the i²= –1 factor. In case of doubt, always remember that the square of a complex number may actually yield a negative number, as evidenced by the definition of the imaginary unit itself: i²= –1.

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

B. The phase

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

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

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

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

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

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

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

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

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

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

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

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

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

c = λν = ω/k

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

θ = ω(t–x/c) = ωt–kx

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

φ(θ) = φ[θ(x, t)] = φ[ωt–kx]

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

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

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

Digression 1: on relativity and spacetime

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

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

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

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

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

However, because of the tremendous speed of light, that still makes for a wavetrain that’s like ten meter long, at least (30×10^–9s times 3×10⁸m/s is nine meter, but you should note that the decay time measures the time for the oscillation to die out by a factor 1/e, so the oscillation itself lasts longer than that). Those nine or ten meters cover like 16 to 17 million oscillations (the wavelength of sodium light is about 600 nm and, hence, 10 meter fits almost 17 million oscillations indeed). Now, how can we reconcile the image of a photon as a ten-meter long wavetrain with the image of a photon as a point particle?

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

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

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

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

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

1. P(A to B)

P(A to B) is the (probability) amplitude for a photon to travel from point A to B. However, I should immediately note that A and B are points in spacetime. Therefore, we associate them not only with some specific (x, y, z) position in space, but also with a some specific time t. Now, quantum-mechanical theory gives us an easy formula for P(A to B): it depends on the so-called (spacetime) interval between the two points A and B, i.e. I = Δr²– Δt²= (x₂–x₁)²+(y₂–y₁)²+(z₂–z₁)²– (t₂–t₁)². The point to note is that the spacetime interval takes both the distance in space as well as the ‘distance’ in time into account. As I mentioned already, this spacetime interval does not depend on our reference frame and, hence, it’s invariant (as long as we’re talking reference frames that move with constant speed relative to each other). Also note that we should measure time and distance in equivalent units when using that Δr²– Δt²formula for I. So we either measure distance in light-seconds or, else, we measure time in units that correspond to the time that’s needed for light to travel one meter. If no equivalent units are adopted, the formula is I = Δr²– c·Δt².

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

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

2. E(A to B)

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

3. j

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

Now, this junction number is often referred to as the coupling constant or the fine-structure constant. However, the truth is, as I pointed out in my previous post, that these numbers are related, but they are not quite the same: α is the square of j, so we have α = j². There is also one more, related, number: the gauge parameter, which is denoted by g (despite the g notation, it has nothing to do with gravitation). The value of g is the square root of 4πε₀α, so g²= 4πε₀α. I’ll come back to this. Let me first make an awfully long digression on the fine-structure constant. It will be awfully long. So long that it’s actually part of the ‘core’ of this post actually.

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

The value for j is approximately –0.08542454.

How do we know that?

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

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

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

1/α = 137.035999074 ± 0.000000044

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

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

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

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

j = –e_P

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

Now that is strange.

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

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

α = e_P²

You’ll say: yes, so what? Well… I am pretty sure that, if you’ve ever seen a formula for α, it’s surely not this simple j = –e_P or α = e_P² formula. What you’ve seen, most likely, is one or more of the following expressions below :

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

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

Let’s have a closer look at those constants. You know most of them already. The only constants you may not have seen before are μ₀, R_Kand, perhaps, r_eas well as m_e. However, these can easily be defined as some easy function of the constants that you did see before, so let me quickly do that:

The μ₀ constant is the so-called magnetic constant. It’s something similar as ε₀ and it’s referred to as the magnetic permeability of the vacuum. So it’s just like the (electric) permittivity of the vacuum (i.e. the electric constant ε₀) and the only reason why this blog hasn’t mentioned this constant before is because I haven’t really discussed magnetic fields so far. I only talked about the electric field vector. In any case, you know that the electric and magnetic force are part and parcel of the same phenomenon (i.e. the electromagnetic interaction between charged particles) and, hence, they are closely related. To be precise, μ₀ε₀ = 1/c²= c^–2. So that shows the first and second expression for α are, effectively, fully equivalent. [Just in case you’d doubt that μ₀ε₀ = 1/c², let me give you the values: μ₀ = 4π·10^–7N/A², and ε₀ = (1/4π·c²)·10⁷C²/N·m². Just plug them in, and you’ll see it’s bang on. Moreover, note that the ampere (A) unit is equal to the coulomb per second unit (C/s), so even the units come out alright. 🙂 Of course they do!]
The k_e constant is the Coulomb constant and, from its definition k_e = 1/4πε₀, it’s easy to see how those two expressions are, in turn, equivalent with the third expression for α.
The R_Kconstant is the so-called von Klitzing constant. Huh? Yes. I know. I am pretty sure you’ve never ever heard of that one before. Don’t worry about it. It’s, quite simply, equal to R_K= h/e². Hence, substituting (and don’t forget that h = 2πħ) will demonstrate the equivalence of the fourth expression for α.
Finally, the r_e factor is the classical electron radius, which is usually written as a function of m_e, i.e. the electron mass: r_e = e²/4πε₀m_ec². Also note that this also implies that r_em_e = e²/4πε₀c². In words: the product of the electron mass and the electron radius is equal to some constant involving the electron (e), the electric constant (ε₀), and c (the speed of light).

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

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

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

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

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

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

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

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

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

Of course, that’s a question one shouldn’t try to answer before answering some other, more fundamental, question: how many degrees of freedom are there really? Indeed, we just saw that k_e and ε₀are intimately related through some equation, and other constants and parameters are related too. So the question is like: what are the ‘dependent’ and the ‘independent’ variables in this so-called Standard Model?

There is no easy answer to that question. In fact, one of the reasons why I find physics so fascinating is that one cannot easily answer such questions. There are the obvious relationships, of course. For example, the k_e = 1/4πε₀relationship, and the context in which they are used (Coulomb’s Law) does, indeed, strongly suggest that both constants are actually part and parcel of the same thing. Identical, I’d say. Likewise, the μ₀ε₀ = 1/c²relation also suggests there’s only one degree of freedom here, just like there’s only one degree of freedom in that ω/k = c relationship (if we set a value for ω, we have k, and vice versa). But… Well… I am not quite sure how to phrase this, but… What physical constants could be ‘variables’ indeed?

It’s pretty obvious that the various formulas for α cannot answer that question: you could stare at them for days and weeks and months and years really, but I’d suggest you use your time to read more of Feynman’s real Lectures instead. 🙂 One point that may help to come to terms with this question – to some extent, at least – is what I casually mentioned above already: the fine-structure constant is equal to the square of the electron charge expressed in Planck units: α = e_P².

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

[…]

It should… But… Does it?

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

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

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

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

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

Now we also know its kinetic energy (K.E.) is mv²/2, which we can write as p²/2m. Substituting our p ≈ ħ/r conjecture, we get K.E. = mv²/2 = ħ²/2mr². This is equivalent to m²v² = ħ²/r²(just multiply both sides with m). From that, we get v = ħ/mr. Now, one of the many relations we can derive from the formulas for the fine-structure constant is r_e = α²r. [I haven’t showed you that yet, but I will shortly. It’s a really amazing expression. However, as for now, just accept it as a simple formula for interim use in this digression.] Hence, r = r_e/α². The r_efactor in this expression is the so-called classical electron radius. So we can now write v = ħα²/mr_e. Let’s now throw c in: v/c = α²ħ/mcr_e. However, from that fifth expression for α, we know that ħ/mcr_e = α, so we get v/c = α. We have another amazing result here: the v/c ratio for an electron (i.e. its speed expressed as a fraction of the speed of light) is equal to that fine-structure constant α. So that’s about 1/137, so that’s less than 1% of the speed of light. Now… I’ll leave it to you to calculate the Lorentz factor γ but… Well… It’s obvious that it will be very close to 1. 🙂 Hence, the electron’s speed – however we want to visualize that – doesn’t matter much indeed, so we should not worry about relativistic corrections in the formulas.

Let’s now look at the question in regard to the Planck units. If you know nothing at all about them, I would advise you to read what I wrote about them in my previous post. Let me just note we get those Planck units by equating not less than five fundamental physical constants to 1, notably (1) the speed of light, (2) Planck’s (reduced) constant, (3) Boltzmann’s constant, (4) Coulomb’s constant and (5) Newton’s constant (i.e. the gravitational constant). Hence, we have a set of five equations here (c = ħ = k_B = k_e = G = 1), and so we can solve that to get the five Planck units, i.e. the Planck length unit, the Planck time unit, the Planck mass unit, the Planck energy unit, the Planck charge unit and, finally (oft forgotten), the Planck temperature unit. Of course, you should note that all mass and energy units are directly related because of the mass-energy equivalence relation E = mc², which simplifies to E = m if c is equated to 1. [I could also say something about the relation between temperature and (kinetic) energy, but I won’t, as it would only further confuse you.]

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

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

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

Measured in our old-fashioned super-sized SI kilogram unit, the electron mass is m_e = 9.1×10^–31kg.
The Planck mass is m_P = 2.1765×10⁻⁸kg.
Hence, the electron mass expressed in Planck units is m_{e_P} = m_e/m_P = (9.1×10^–31kg)/(2.1765×10⁻⁸kg) = 4.181×10⁻²³.

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

m_{e_P} = α/r_{e_P} = α/α²r_P = 1/αr_P

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

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

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

Note that the beauty of natural units ensures that we get the same number for the (equivalent) energy of an electron. Indeed, from the E = mc² relation, we know the mass of an electron can also be written as 0.511 MeV/c². Hence, the equivalent energy is 0.511 MeV (so that’s, quite simply, the same number but without the 1/c²factor). Now, the Planck energy E_P (in eV) is 1.22×10²⁸eV, so we get E_{e_P} = E_e/E_P= (0.511×10⁶eV)/(1.22×10²⁸eV) = 4.181×10⁻²³. So it’s exactly the same as the electron mass expressed in Planck units. Isn’t that nice? 🙂

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

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

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

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

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

Before diving into it all, let’s first see if that formula for that alpha, that fine-structure constant, still makes sense with m_e expressed in Planck units. Just to make sure. 🙂 To do that, we need to use the fifth (last) expression for a, i.e. the one with r_e in it. Now, in my previous post, I also gave some formula for r_e: r_e = e²/4πε₀m_ec², which we can re-write as r_em_e = e²/4πε₀c². If we substitute that expression for r_em_e in the formula for α, we can calculate α from the electron charge, which indicates both the electron radius and its mass are not some random God-given variable, or “some magic number that comes to us with no understanding by man“, as Feynman – well… Leighton, I guess – puts it. No. They are magic numbers alright, one related to another through the equally ‘magic’ number α, but so I do feel we actually can create some understanding here.

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

α = √(r_e /r) = (r_e /r)^1/2

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

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

(1) α is the square of the electron charge expressed in Planck units: α = e_P².

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

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

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

Note that, from (2) and (4), we find that:

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

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

(6) The electron mass (in Planck units) is equal to m_e = α/r_e = e_P²/r_e. Using the Bohr radius, we get m_e = 1/αr = 1/e_P²r.

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

IT IS ALL IN ALPHA!

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

ALPHA GIVES US EVERYTHING!

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

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

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

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

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

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

E = K.E. + P.E. = ħ²/2mr²– e²/4πε₀r

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

dE/dr = –ħ²/mr³+ e²/4πε₀r²= 0 ⇔ r = 4πε₀ħ²/me²

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

r = 4πε₀h²/me²

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

= 53×10^–12m = 53 pico-meter (pm)

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

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

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

Waw ! Isn’t it great that such simple calculations yield such great results? 🙂 [Of course, you’ll note that the omission of the 1/2 factor in the Uncertainty Principle was quite strategic. :-)] Using the r = 4πε₀ħ²/me²formula for the Bohr radius, you can now easily check the r_e = α²r formula. You should find what we jotted down already: the classical electron radius is equal to r_e = e²/4πε₀m_ec². To be precise, r_e = (53×10^–6)·(53×10^–12m) = 2.8×10^–15m. Now that’s again something you should check on the Internet. Guess what? […] It’s right on the spot again. 🙂

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

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

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

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

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

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

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

But let’s look at a few constants indeed. For example, if we have c, h and α, then we can calculate the electric charge e and, hence, the electric constant ε₀= e²/2αhc. From that, we get Coulomb’s constant k_e, because k_e is defined as 1/4πε₀… But…

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

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

?!?

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

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

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

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

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

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

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

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

The E(A to B) formula

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

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

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

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

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

So… Well… How do we do it?

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

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

Well… No! Not at all! Because an electron, unlike the photon, has a rest mass indeed – measured as not less than 0.511 MeV/c², to be precise (note the rather particular MeV/c²unit: it’s from the E = mc²formula) – one should use a different energy value! Indeed, we should include the rest mass energy, which is 0.511 MeV. So, almost all of the energy here is rest mass energy! There’s also another complication. For the photon, there is an easy relationship between the wavelength and the frequency: it has no mass and, hence, all its energy is kinetic, or movement so to say, and so we can use that ν = E/h relationship to calculate its frequency ν: it’s equal to ν = E/h = (1 eV)/(4.13567×10^–15eV·s) ≈ 0.242×10¹⁵Hz = 242 tera-hertz (1 THz = 10¹²oscillations per second). Now, knowing that light travels at the speed of light, we can check the result by calculating the wavelength using the λ = c/ν relation. Let’s do it: (2.998×10⁸m/s)/(242×10¹²Hz) ≈ 1240 nm. So… Yes, done!

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

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

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

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

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

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

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

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

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

P(A to C)∗n²∗P(C to B)

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

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

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

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

P(A to E)∗n²∗P(E to D)∗n²∗P(D to B)

So we have an (n²)²= n⁴factor here, which is good, because two hops should be much less likely than one hop. So what do we get? Well… (4.181×10⁻²³)⁴≈ 305×10⁻⁹². Pretty tiny, huh? 🙂 Of course, any point in space is a potential hop for the electron’s flight from point A to B and, hence, there’s a lot of paths and a lot of amplitudes (or ‘arrows’ if you want), which, again, is consistent with a very tiny value for n indeed.

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

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

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

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

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

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

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

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

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

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

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

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

α = 2πr_e/λ_e= r_ek_e

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

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

Post scriptum (5 October 2015):

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

Fields and charges (II)

Pre-script (dated 26 June 2020): This post has become less relevant (even irrelevant, perhaps) because my views on all things quantum-mechanical have evolved significantly as a result of my progression towards a more complete realist (classical) interpretation of quantum physics. In addition, some of the material was removed by a dark force (that also created problems with the layout, I see now). In any case, we recommend you read our recent papers. I keep blog posts like these mainly because I want to keep track of where I came from. I might review them one day, but I currently don’t have the time or energy for it. 🙂

Original post:

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

Note: In case you’re not interested in units, you can skip the first part of this post. However, please do look at the section on the electric constant ε₀and, most importantly, the section on natural units—especially Planck units, as I will touch upon the topic of gauge coupling parameters there and, hence, on quantum mechanics. Also, the third and last part, on the theoretical contradictions inherent in the idea of point charges, may be of interest to you.]

The field energy integrals

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

Now, we also said that that ε₀factor in the second integral (i.e. the one with the vector dot product E•E =|E||E|cos(0) = E²) is there to make the units come out alright. Now, when I say that, what does it mean really? I’ll explain. Let me first make a few obvious remarks:

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

The volt

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

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

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

Also note the following:

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

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

The farad

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

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

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

C = Q/V

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

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

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

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

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

The capacity of a parallel-plate condenser is C = ε₀A/d. In this formula, we have, once again, that ubiquitous electric constant ε₀(think of it as just another coefficient of proportionality), and then A, i.e. the area of the plates, and d, i.e. the separation between the two plates.
The capacity of a charged sphere of radius r (so we’re talking the capacity of a single conductor here) is C = 4πε₀r. This may remind you of the formula for the surface of a sphere (A = 4πr²), but note we’re not squaring the radius. It’s just a linear relationship with r.

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

Other electrical units

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

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

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

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

The concept of impedance

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

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

Z = |Z|eⁱ^θ

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

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

V = I∗Z = I∗|Z|eⁱ^θ

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

V = |V|eⁱ^{(ωt +}^θ_V⁾
I = |I|eⁱ^{(ωt +}^θ_I⁾

This works out as follows:

V = I∗Z = |I|eⁱ^{(ωt +}^θ_I⁾∗|Z|eⁱ^θ= |I||Z|eⁱ^{(ωt +}^θ_I^+ θ)= |V|eⁱ^{(ωt +}^θ_V⁾

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

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

The electric constant ε₀

Let’s now look at that energy density formula once again. When looking at that u = ε₀E²/2 formula, you may think that its unit should be the square of the unit in which we measure field strength. How do we measure field strength? It’s defined as the force on a unit charge (E = F/q), so it should be newton per coulomb (N/C). Because the coulomb can also be expressed in newton·meter/volt (1 V = 1 J/C = N·m/C and, hence, 1 C = 1 N·m/V), we can express field strength not only in newton/coulomb but also in volt per meter: 1 N/C = 1 N·V/N·m = 1 V/m. How do we get from N²/C² and/or V²/m²to J/m³?

Well… Let me first note there’s no issue in terms of units with that ρΦ formula in the first integral for U: [ρ]·[Φ] = (C/m³)·V = [(N·m/V)/m³)·V = (N·m)/m³ = J/m³. No problem whatsoever. It’s only that second expression for U, with the u = ε₀E²/2 in the integrand, that triggers the question. Here, we just need to accept that we need that ε₀factor to make the units come out alright. Indeed, just like other physical constants (such as c, G, or h, for example), it has a dimension: its unit is either C²/N·m² or, what amounts to the same, C/V·m. So the units come out alright indeed if, and only if, we multiply the N²/C² and/or V²/m²units with the dimension of ε₀:

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

Done!

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

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

This numerical value of ε₀is as important as its unit to ensure both expressions for U yield the same result. Indeed, as you may or may not remember from the second of my two posts on vector calculus, if we have a curl-free field C (that means ∇×C = 0 everywhere, which is the case when talking electrostatics only, as we are doing here), then we can always find some scalar field ψ such that C = ∇ψ. But so here we have E = –ε₀∇Φ, and so it’s not the minus sign that distinguishes the expression from the C = ∇ψ expression, but the ε₀factor in front.

It’s just like the vector equation for heat flow: h = –κ∇T. Indeed, we also have a constant of proportionality here, which is referred to as the thermal conductivity. Likewise, the electric constant ε₀is also referred to as the permittivity of the vacuum (or of free space), for similar reasons obviously!

Natural units

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

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

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

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

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

The fine-structure constant

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

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

0.08542455? That’s a bit of a weird number, isn’t it? You’re right. And trying to write it in terms of the charge of a u or d quark doesn’t make it any better. Also, note that the first four significant digits (8542) correspond to the first four significant digits after the decimal point of our ε₀constant. So what’s the physical significance here? Some other limit of quantum theory?

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

The only constants you have not seen before are μ₀, R_Kand, perhaps, r_eas well as m_e. However, these can be defined as a function of the constants that you did see before:

The μ₀ constant is the so-called magnetic constant. It’s something similar as ε₀ and it’s referred to as the magnetic permeability of the vacuum. So it’s just like the (electric) permittivity of the vacuum (i.e. the electric constant ε₀) and the only reason why you haven’t heard of this before is because we haven’t discussed magnetic fields so far. In any case, you know that the electric and magnetic force are part and parcel of the same phenomenon (i.e. the electromagnetic interaction between charged particles) and, hence, they are closely related. To be precise, μ₀= 1/ε₀c². That shows the first and second expression for α are, effectively, fully equivalent.
Now, from the definition of k_e = 1/4πε₀, it’s easy to see how those two expressions are, in turn, equivalent with the third expression for α.
The R_Kconstant is the so-called von Klitzing constant, but don’t worry about it: it’s, quite simply, equal to R_K= h/e². Hene, substituting (and don’t forget that h = 2πħ) will demonstrate the equivalence of the fourth expression for α.
Finally, the r_e factor is the classical electron radius, which is usually written as a function of m_e, i.e. the electron mass: r_e = e²/4πε₀m_ec². This very same equation implies that r_em_e = e²/4πε₀c². So… Yes. It’s all the same really.

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

The e²constant is (roughly) equal to (1.6022×10^–19C)²= 2.5670×10^–38C². Coulomb’s constant k_e= 1/4πε₀is about 8.9876×10⁹N·m²/C². Hence, the numerator e²k_e≈ 23.0715×10^–29N·m².
The (rounded) denominator is ħc = (1.05457×10^–34 N·m·s)(2.998×10⁸ m/s) = 3.162×10^–26 N·m².
Hence, we get α = k_ee²/ħc ≈ 7.297×10^–3 = 0.007297.

Note that this number is, effectively, dimensionless. Now, the interesting thing is that if we calculate α using Planck units, we get an e²constant that is (roughly) equal to 0.08542455²= … 0.007297! Now, because all of the other constants are equal to 1 in Planck’s system of units, that’s equal to α itself. So… Yes ! The two values for α are one and the same in the two systems of units and, of course, as you might have guessed, the fine-structure constant is effectively dimensionless because it does not depend on our units of measurement. So what does it correspond to?

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

A photon travels from point A to B. To keep things simple and stupid, Feynman denoted this amplitude by P(A to B), and please note that the P stands for photon, not for probability. I should also note that we have an easy formula for P(A to B): it depends on the so-called space-time interval between the two points A and B, i.e. I = Δr²– Δt²= (x₂–x₁)²+(y₂–y₁)²+(z₂–z₁)²– (t₂–t₁)². Hence, the space-time interval takes both the distance in space as well as the ‘distance’ in time into account.
An electron travels from point A to B: this was denoted by E(A to B) because… Well… You guessed it: the E of electron. The formula for E(A to B) was much more complicated, but the two key elements in the formula was some complex number j (see below), and some other (real) number n.
Finally, an electron could emit or absorb a photon, and the amplitude associated with this event was denoted by j, for junction.

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

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

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

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

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

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

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

Do point charges exist?

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

Huh?

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

E = q/4πε₀r²

Hence, the energy density u is ε₀E²/2 = q²/32πε₀r⁴. Now, we have that volume integral U = (ε₀/2)∫E•EdV = ∫(ε₀E²/2)dV integral. As Feynman notes, nothing prevents us from taking a spherical shell for the volume element dV, instead of an infinitesimal cube. This spherical shell would have the charge q at its center, an inner radius equal to r, an infinitesimal thickness dr, and, finally, a surface area 4πr²(that’s just the general formula for the surface area of a spherical shell, which I also noted above). Hence, its (infinitesimally small) volume is 4πr²dr, and our integral becomes:

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

So… Where did we do wrong?

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

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

U = Q²/8πε₀r

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

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

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

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

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

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

Now that explains the physical significance of that Planck mass ! We said Nature has some kind of maximum allowable mass for point-like particles, or the mass capable of holding a single elementary charge. What’s going on is, as we try to pile more charge on top of the charge that’s already there, we add energy. Now, energy has an equivalent mass. Indeed, the Planck charge (q_P≈ 1.8755×10⁻¹⁸ C), the Planck length (l_P= 1.616×10⁻³⁵ m), the Planck energy (1.956×10⁹ J), and the Planck mass (2.1765×10⁻⁸kg) are all related. Now things start making sense. Indeed, we said that the Planck mass is tiny but, still, it’s something we can imagine, like a flea’s egg or the mass of a hair of a eyebrow. The associated energy (E = mc², so that’s (2.1765×10⁻⁸kg)·(2.998×10⁸ m/s)² ≈ 19.56×10⁸ kg·m²/s²= 1.956×10⁹ joule indeed.

Now, how much energy is that? Well… That’s about 2 giga-joule, obviously, but so what’s that in daily life? It’s about the energy you would get when burning 40 liter of fuel. It’s also likely to amount, more or less, to your home electricity consumption over a month. So it’s sizable, and so we’re packing all that energy into a Planck volume (l_P³≈ 4×10⁻¹⁰⁵m³). If we’d manage that, we’d be able to create tiny black holes, because that’s what that little Planck volume would become if we’d pack so much energy in it. So… Well… Here I just have to refer you to more learned writers than I am. As Wikipedia notes dryly: “The physical significance of the Planck length is a topic of theoretical research. Since the Planck length is so many orders of magnitude smaller than any current instrument could possibly measure, there is no way of examining it directly.”

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

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

Fields and charges (I)

Pre-script (dated 26 June 2020): This post has become less relevant (even irrelevant, perhaps) because my views on all things quantum-mechanical have evolved significantly as a result of my progression towards a more complete realist (classical) interpretation of quantum physics. In addition, some of the material was removed by a dark force (that also created problems with the layout, I see now). In any case, we recommend you read our recent papers. I keep blog posts like these mainly because I want to keep track of where I came from. I might review them one day, but I currently don’t have the time or energy for it. 🙂

Original post:

My previous posts focused mainly on photons, so this one should be focused more on matter-particles, things that have a mass and a charge. However, I will use it more as an opportunity to talk about fields and present some results from electrostatics using our new vector differential operators (see my posts on vector analysis).

Before I do so, let me note something that is obvious but… Well… Think about it: photons carry the electromagnetic force, but have no electric charge themselves. Likewise, electromagnetic fields have energy and are caused by charges, but so they also carry no charge. So… Fields act on a charge, and photons interact with electrons, but it’s only matter-particles (notably the electron and the proton, which is made of quarks) that actually carry electric charge. Does that make sense? It should. 🙂

Another thing I want to remind you of, before jumping into it all head first, are the basic units and relations that are valid always, regardless of what we are talking about. They are represented below:

Let me recapitulate the main points:

The speed of light is always the same, regardless of the reference frame (inertial or moving), and nothing can travel faster than light (except mathematical points, such as the phase velocity of a wavefunction).
This universal rule is the basis of relativity theory and the mass-energy equivalence relation E = mc².
The constant speed of light also allows us to redefine the units of time and/or distance such that c = 1. For example, if we re-define the unit of distance as the distance traveled by light in one second, or the unit of time as the time light needs to travel one meter, then c = 1.
Newton’s laws of motion define a force as the product of a mass and its acceleration: F = m·a. Hence, mass is a measure of inertia, and the unit of force is 1 newton (N) = 1 kg·m/s².
The momentum of an object is the product of its mass and its velocity: p = m·v. Hence, its unit is 1 kg·m/s = 1 N·s. Therefore, the concept of momentum combines force (N) as well as time (s).
Energy is defined in terms of work: 1 Joule (J) is the work done when applying a force of one newton over a distance of one meter: 1 J = 1 N·m. Hence, the concept of energy combines force (N) and distance (m).
Relativity theory establishes the relativistic energy-momentum relation pc = Ev/c, which can also be written as E² = p²c²+ m₀²c⁴, with m₀the rest mass of an object (i.e. its mass when the object would be at rest, relative to the observer, of course). These equations reduce to m = E and E²= p²+ m₀²when choosing time and/or distance units such that c = 1. The mass m is the total mass of the object, including its inertial mass as well as the equivalent mass of its kinetic energy.
The relationships above establish (a) energy and time and (b) momentum and position as complementary variables and, hence, the Uncertainty Principle can be expressed in terms of both. The Uncertainty Principle, as well as the Planck-Einstein relation and the de Broglie relation (not shown on the diagram), establish a quantum of action, h, whose dimension combines force, distance and time (h ≈ 6.626×10⁻³⁴ N·m·s). This quantum of action (Wirkung) can be defined in various ways, as it pops up in more than one fundamental relation, but one of the more obvious approaches is to define h as the proportionality constant between the energy of a photon (i.e. the ‘light particle’) and its frequency: h = E/ν.

Note that we talked about forces and energy above, but we didn’t say anything about the origin of these forces. That’s what we are going to do now, even if we’ll limit ourselves to the electromagnetic force only.

Electrostatics

According to Wikipedia, electrostatics deals with the phenomena and properties of stationary or slow-moving electric charges with no acceleration. Feynman usually uses the term when talking about stationary charges only. If a current is involved (i.e. slow-moving charges with no acceleration), the term magnetostatics is preferred. However, the distinction does not matter all that much because – remarkably! – with stationary charges and steady currents, the electric and magnetic fields (E and B) can be analyzed as separate fields: there is no interconnection whatsoever! That shows, mathematically, as a neat separation between (1) Maxwell’s first and second equation and (2) Maxwell’s third and fourth equation:

Electrostatics: (i) ∇•E = ρ/ε₀ and (ii) ∇×E = 0.
Magnetostatics: (iii) c²∇×B = j/ε₀ and (iv) ∇•B = 0.

Electrostatics: The ρ in equation (i) is the so-called charge density, which describes the distribution of electric charges in space: ρ = ρ(x, y, z). To put it simply: ρ is the ‘amount of charge’ (which we’ll denote by Δq) per unit volume at a given point. As for ε₀, that’s a constant which ensures all units are ‘compatible’. Equation (i) basically says we have some flux of E, the exact amount of which is determined by the charge density ρ or, more in general, by the charge distribution in space. As for equation (ii), i.e. ∇×E = 0, we can sort of forget about that. It means the curl of E is zero: everywhere, and always. So there’s no circulation of E. Hence, E is a so-called curl-free field, in this case at least, i.e. when only stationary charges and steady currents are involved.

Magnetostatics: The j in (iii) represents a steady current indeed, causing some circulation of B. The c²factor is related to the fact that magnetism is actually only a relativistic effect of electricity, but I can’t dwell on that here. I’ll just refer you to what Feynman writes about this in his Lectures, and warmly recommend to read it. Oh… Equation (iv), ∇•B = 0, means that the divergence of B is zero: everywhere, and always. So there’s no flux of B. None. So B is a divergence-free field.

Because of the neat separation, we’ll just forget about B and talk about E only.

The electric potential

OK. Let’s try to go through the motions as quickly as we can. As mentioned in my introduction, energy is defined in terms of work done. So we should just multiply the force and the distance, right? 1 Joule = 1 newton × 1 meter, right? Well… Yes and no. In discussions like this, we talk potential energy, i.e. energy stored in the system, so to say. That means that we’re looking at work done against the force, like when we carry a bucket of water up to the third floor or, to use a somewhat more scientific description of what’s going on, when we are separating two masses. Because we’re doing work against the force, we put a minus sign in front of our integral:

Now, the electromagnetic force works pretty much like gravity, except that, when discussing gravity, we only have positive ‘charges’ (the mass of some object is always positive). In electromagnetics, we have positive as well as negative charge, and please note that two like charges repel (that’s not the case with gravity). Hence, doing work against the electromagnetic force may involve bringing like charges together or, alternatively, separating opposite charges. We can’t say. Fortunately, when it comes to the math of it, it doesn’t matter: we will have the same minus sign in front of our integral. The point is: we’re doing work against the force, and so that’s what the minus sign stands for. So it has nothing to do with the specifics of the law of attraction and repulsion in this case (electromagnetism as opposed to gravity) and/or the fact that electrons carry negative charge. No.

Let’s get back to the integral. Just in case you forgot, the integral sign ∫ stands for an S: the S of summa, i.e. sum in Latin, and we’re using these integrals because we’re adding an infinite number of infinitesimally small contributions to the total effort here indeed. You should recognize it, because it’s a general formula for energy or work. It is, once again, a so-called line integral, so it’s a bit different than the ∫f(x)dx stuff you learned from high school. Not very different, but different nevertheless. What’s different is that we have a vector dot product F•ds after the integral sign here, so that’s not like f(x)dx. In case you forgot, that f(x)dx product represents the surface of an infinitesimally rectangle, as shown below: we make the base of the rectangle smaller and smaller, so dx becomes an infinitesimal indeed. And then we add them all up and get the area under the curve. If f(x) is negative, then the contributions will be negative.

But so we don’t have little rectangles here. We have two vectors, F and ds, and their vector dot product, F•ds, which will give you… Well… I am tempted to write: the tangential component of the force along the path, but that’s not quite correct: if ds was a unit vector, it would be true—because then it’s just like that h•n product I introduced in our first vector calculus class. However, ds is not a unit vector: it’s an infinitesimal vector, and, hence, if we write the tangential component of the force along the path as F_t, then F•ds = |F||ds|cosθ = F·cosθ·ds = F_t·ds. So this F•ds is a tangential component over an infinitesimally small segment of the curve. In short, it’s an infinitesimally small contribution to the total amount of work done indeed. You can make sense of this by looking at the geometrical representation of the situation below.

I am just saying this so you know what that integral stands for. Note that we’re not adding arrows once again, like we did when calculating amplitudes or so. It’s all much more straightforward really: a vector dot product is a scalar, so it’s just some real number—just like any component of a vector (tangential, normal, in the direction of one of the coordinates axes, or in whatever direction) is not a vector but a real number. Hence, W is also just some real number. It can be positive or negative because… Well… When we’d be going down the stairs with our bucket of water, our minus sign doesn’t disappear. Indeed, our convention to put that minus sign there should obviously not depend on what point a and b we’re talking about, so we may actually be going along the direction of the force when going from a to b.

As a matter of fact, you should note that’s actually the situation which is depicted above. So then we get a negative number for W. Does that make sense? Of course it does: we’re obviously not doing any work here as we’re moving along the direction, so we’re surely not adding any (potential) energy to the system. On the contrary, we’re taking energy out of the system. Hence, we are reducing its (potential) energy and, hence, we should have a negative value for W indeed. So, just think of the minus sign being there to ensure we add potential energy to the system when going against the force, and reducing it when going with the force.

OK. You get this. You probably also know we’ll re-define W as a difference in potential between two points, which we’ll write as Φ(b) – Φ(a). Now that should remind you of your high school integral ∫f(x)dx once again. For a definite integral over a line segment [a, b], you’d have to find the antiderivative of f(x), which you’d write as F(x), and then you’d take the difference F(b) – F(a) too. Now, you may or may not remember that this antiderivative was actually a family of functions F(x) + k, and k could be any constant – 5/9, 6π, 3.6×10¹²⁴, 0.86, whatever! – because such constant vanishes when taking the derivative.

Here we have the same, we can define an infinite number of functions Φ(r) + k, of which the gradient will yield… Stop! I am going too fast here. First, we need to re-write that W function above in order to ensure we’re calculating stuff in terms of the unit charge, so we write:

Huh? Well… Yes. I am using the definition of the field E here really: E is the force (F) when putting a unit charge in the field. Hence, if we want the work done per unit charge, i.e. W(unit), then we have to integrate the vector dot product E·ds over the path from a to b. But so now you see what I want to do. It makes the comparison with our high school integral complete. Instead of taking a derivative in regard to one variable only, i.e. dF(x)/dx) = f(x), we have a function Φ here not in one but in three variables: Φ = Φ(x, y, z) = Φ(r) and, therefore, we have to take the vector derivative (or gradient as it’s called) of Φ to get E:

∇Φ(x, y, z) = (∂Φ/∂x, ∂Φ/∂y, ∂Φ/∂z) = –E(x, y, z)

But so it’s the same principle as what you learned how to use to solve your high school integral. Now, you’ll usually see the expression above written as:

E = –∇Φ

Why so short? Well… We all just love these mysterious abbreviations, don’t we? 🙂 Jokes aside, it’s true some of those vector equations pack an awful lot of information. Just take Feynman’s advice here: “If it helps to write out the components to be sure you understand what’s going on, just do it. There is nothing inelegant about that. In fact, there is often a certain cleverness in doing just that.” So… Let’s move on.

I should mention that we can only apply this more sophisticated version of the ‘high school trick’ because Φ and E are like temperature (T) and heat flow (h): they are fields. T is a scalar field and h is a vector field, and so that’s why we can and should apply our new trick: if we have the scalar field, we can derive the vector field. In case you want more details, I’ll just refer you to our first vector calculus class. Indeed, our so-called First Theorem in vector calculus was just about the more sophisticated version of the ‘high school trick’: if we have some scalar field ψ (like temperature or potential, for example: just substitute the ψ in the equation below for T or Φ), then we’ll always find that:

The Γ here is the curve between point 1 and 2, so that’s the path along which we’re going, and ∇ψ must represent some vector field.

Let’s go back to our W integral. I should mention that it doesn’t matter what path we take: we’ll always get the same value for W, regardless of what path we take. That’s why the illustration above showed two possible paths: it doesn’t matter which one we take. Again, that’s only because E is a vector field. To be precise, the electrostatic field is a so-called conservative vector field, which means that we can’t get energy out of the field by first carrying some charge along one path, and then carrying it back along another. You’ll probably find that’s obvious, and it is. Just note it somewhere in the back of your mind.

So we’re done. We should just substitute E for ∇Φ, shouldn’t we? Well… Yes. For minus ∇Φ, that is. Another minus sign. Why? Well… It makes that W(unit) integral come out alright. Indeed, we want a formula like W = Φ(b) – Φ(a), not like Φ(a) – Φ(b). Look at it. We could, indeed, define E as the (positive) gradient of some scalar field ψ = –Φ, and so we could write E = ∇ψ, but then we’d find that W = –[ψ(b) – ψ(a)] = ψ(a) – ψ(b).

You’ll say: so what? Well… Nothing much. It’s just that our field vectors would point from lower to higher values of ψ, so they would be flowing uphill, so to say. Now, we don’t want that in physics. Why? It just doesn’t look good. We want our field vectors to be directed from higher potential to lower potential, always. Just think of it: heat (h) flows from higher temperature (T) to lower, and Newton’s apple falls from greater to lower height. Likewise, when putting a unit charge in the field, we want to see it move from higher to lower electric potential. Now, we can’t change the direction of E, because that’s the direction of the force and Nature doesn’t care about our conventions and so we can’t choose the direction of the force. But we can choose our convention. So that’s why we put a minus sign in front of ∇Φ when writing E = –∇Φ. It makes everything come out alright. 🙂 That’s why we also have a minus sign in the differential heat flow equation: h = –κ∇T.

So now we have the easy W(unit) = Φ(b) – Φ(a) formula that we wanted all along. Now, note that, when we say a unit charge, we mean a plus one charge. Yes: +1. So that’s the charge of the proton (it’s denoted by e) so you should stop thinking about moving electrons around! [I am saying this because I used to confuse myself by doing that. You end up with the same formulas for W and Φ but it just takes you longer to get there, so let me save you some time here. :-)]

But… Yes? In reality, it’s electrons going through a wire, isn’t? Not protons. Yes. But it doesn’t matter. Units are units in physics, and they’re always +1, for whatever (time, distance, charge, mass, spin, etcetera). Always. For whatever. Also note that in laboratory experiments, or particle accelerators, we often use protons instead of electrons, so there’s nothing weird about it. Finally, and most fundamentally, if we have a –e charge moving through a neutral wire in one direction, then that’s exactly the same as a +e charge moving in the other way.

Just to make sure you get the point, let’s look at that illustration once again. We already said that we have F and, hence, E pointing from a to b and we’ll be reducing the potential energy of the system when moving our unit charge from a to b, so W was some negative value. Now, taking into account we want field lines to point from higher to lower potential, Φ(a) should be larger than Φ(b), and so… Well.. Yes. It all makes sense: we have a negative difference Φ(b) – Φ(a) = W(unit), which amounts, of course, to the reduction in potential energy.

The last thing we need to take care of now, is the reference point. Indeed, any Φ(r) + k function will do, so which one do we take? The approach here is to take a reference point P₀at infinity. What’s infinity? Well… Hard to say. It’s a place that’s very far away from all of the charges we’ve got lying around here. Very far away indeed. So far away we can say there is nothing there really. No charges whatsoever. 🙂 Something like that. 🙂 In any case. I need to move on. So Φ(P₀) is zero and so we can finally jot down the grand result for the electric potential Φ(P) (aka as the electrostatic or electric field potential):

So now we can calculate all potentials, i.e. when we know where the charges are at least. I’ve shown an example below. As you can see, besides having zero potential at infinity, we will usually also have one or more equipotential surfaces with zero potential. One could say these zero potential lines sort of ‘separate’ the positive and negative space. That’s not a very scientifically accurate description but you know what I mean.

Let me make a few final notes about the units. First, let me, once again, note that our unit charge is plus one, and it will flow from positive to negative potential indeed, as shown below, even if we know that, in an actual electric circuit, and so now I am talking about a copper wire or something similar, that means the (free) electrons will move in the other direction.

If you’re smart (and you are), you’ll say: what about the right-hand rule for the magnetic force? Well… We’re not discussing the magnetic force here but, because you insist, rest assured it comes out alright. Look at the illustration below of the magnetic force on a wire with a current, which is a pretty standard one.

So we have a given B, because of the bar magnet, and then v, the velocity vector for the… Electrons? No. You need to be consistent. It’s the velocity vector for the unit charges, which are positive (+e). Now just calculate the force F = qv×B = ev×B using the right-hand rule for the vector cross product, as illustrated below. So v is the thumb and B is the index finger in this case. All you need to do is tilt your hand, and it comes out alright.

But… We know it’s electrons going the other way. Well… If you insist. But then you have to put a minus sign in front of the q, because we’re talking minus e (–e). So now v is in the other direction and so v×B is in the other direction indeed, but our force F = qv×B = –ev×B is not. Fortunately not, because physical reality should not depend on our conventions. 🙂 So… What’s the conclusion. Nothing. You may or may not want to remember that, when we say that our current j current flows in this or that direction, we actually might be talking electrons (with charge minus one) flowing in the opposite direction, but then it doesn’t matter. In addition, as mentioned above, in laboratory experiments or accelerators, we may actually be talking protons instead of electrons, so don’t assume electromagnetism is the business of electrons only.

To conclude this disproportionately long introduction (we’re finally ready to talk more difficult stuff), I should just make a note on the units. Electric potential is measured in volts, as you know. However, it’s obvious from all that I wrote above that it’s the difference in potential that matters really. From the definition above, it should be measured in the same unit as our unit for energy, or for work, so that’s the joule. To be precise, it should be measured in joule per unit charge. But here we have one of the very few inconsistencies in physics when it comes to units. The proton is said to be the unit charge (e), but its actual value is measured in coulomb (C). To be precise: +1 e = 1.602176565(35)×10⁻¹⁹C. So we do not measure voltage – sorry, potential difference 🙂 – in joule but in joule per coulomb (J/C).

Now, we usually use another term for the joule/coulomb unit. You guessed it (because I said it): it’s the volt (V). One volt is one joule/coulomb: 1 V = 1 J/C. That’s not fair, you’ll say. You’re right, but so the proton charge e is not a so-called SI unit. Is the Coulomb an SI unit? Yes. It’s derived from the ampere (A) which, believe it or not, is actually an SI base unit. One ampere is 6.241×10¹⁸ electrons (i.e. one coulomb) per second. You may wonder how the ampere (or the coulomb) can be a base unit. Can they be expressed in terms of kilogram, meter and second, like all other base units. The answer is yes but, as you can imagine, it’s a bit of a complex description and so I’ll refer you to the Web for that.

The Poisson equation

I started this post by saying that I’d talk about fields and present some results from electrostatics using our ‘new’ vector differential operators, so it’s about time I do that. The first equation is a simple one. Using our E = –∇Φ formula, we can re-write the ∇•E = ρ/ε₀ equation as:

∇•E = ∇•∇Φ = ∇²Φ = –ρ/ε₀

This is a so-called Poisson equation. The ∇² operator is referred to as the Laplacian and is sometimes also written as Δ, but I don’t like that because it’s also the symbol for the total differential, and that’s definitely not the same thing. The formula for the Laplacian is given below. Note that it acts on a scalar field (i.e. the potential function Φ in this case).

As Feynman notes: “The entire subject of electrostatics is merely the study of the solutions of this one equation.” However, I should note that this doesn’t prevent Feynman from devoting at least a dozen of his Lectures on it, and they’re not the easiest ones to read. [In case you’d doubt this statement, just have a look at his lecture on electric dipoles, for example.] In short: don’t think the ‘study of this one equation’ is easy. All I’ll do is just note some of the most fundamental results of this ‘study’.

Also note that ∇•E is one of our ‘new’ vector differential operators indeed: it’s the vector dot product of our del operator (∇) with E. That’s something very different than, let’s say, ∇Φ. A little dot and some bold-face type make an enormous difference here. 🙂 You may or may remember that we referred to the ∇• operator as the divergence (div) operator (see my post on that).

Gauss’ Law

Gauss’ Law is not to be confused with Gauss’ Theorem, about which I wrote elsewhere. It gives the flux of E through a closed surface S, any closed surface S really, as the sum of all charges inside the surface divided by the electric constant ε₀(but then you know that constant is just there to make the units come out alright).

The derivation of Gauss’ Law is a bit lengthy, which is why I won’t reproduce it here, but you should note its derivation is based, mainly, on the fact that (a) surface areas are proportional to r² (so if we double the distance from the source, the surface area will quadruple), and (b) the magnitude of E is given by an inverse-square law, so it decreases as 1/r². That explains why, if the surface S describes a sphere, the number we get from Gauss’ Law is independent of the radius of the sphere. The diagram below (credit goes to Wikipedia) illustrates the idea.

The diagram can be used to show how a field and its flux can be represented. Indeed, the lines represent the flux of E emanating from a charge. Now, the total number of flux lines depends on the charge but is constant with increasing distance because the force is radial and spherically symmetric. A greater density of flux lines (lines per unit area) means a stronger field, with the density of flux lines (i.e. the magnitude of E) following an inverse-square law indeed, because the surface area of a sphere increases with the square of the radius. Hence, in Gauss’ Law, the two effect cancel out: the two factors vary with distance, but their product is a constant.

Now, if we describe the location of charges in terms of charge densities (ρ), then we can write Q_int as:

Now, Gauss’ Law also applies to an infinitesimal cubical surface and, in one of my posts on vector calculus, I showed that the flux of E out of such cube is given by ∇•E·dV. At this point, it’s probably a good idea to remind you of what this ‘new’ vector differential operator ∇•, i.e. our ‘divergence’ operator, stands for: the divergence of E (i.e. ∇• applied to E, so that’s ∇•E) represents the volume density of the flux of E out of an infinitesimal volume around a given point. Hence, it’s the flux per unit volume, as opposed to the flux out of the infinitesimal cube itself, which is the product of ∇•E and dV, i.e. ∇•E·dV.

So what? Well… Gauss’ Law applied to our infinitesimal volume gives us the following equality:

That, in turn, simplifies to:

So that’s Maxwell’s first equation once again, which is equivalent to our Poisson equation: ∇•E = ∇²Φ = –ρ/ε₀. So what are we doing here? Just listing equivalent formulas? Yes. I should also note they can be derived from Coulomb’s law of force, which is probably the one you learned in high school. So… Yes. It’s all consistent. But then that’s what we should expect, of course. 🙂

The energy in a field

All these formulas look very abstract. It’s about time we use them for something. A lot of what’s written in Feynman’s Lectures on electrostatics is applied stuff indeed: it focuses, among other things, on calculating the potential in various circumstances and for various distributions of charge. Now, funnily enough, while that ∇•E = –ρ/ε₀ equation is equivalent to Coulomb’s law and, obviously, much more compact to write down, Coulomb’s law is easier to start with for basic calculations. Let me first write Coulomb’s law. You’ll probably recognize it from your high school days:

F₁is the force on charge q₁, and F₂is the force on charge q₂. Now, q₁and q₂. may attract or repel each other but, in both cases, the forces will be equal and opposite. [In case you wonder, yes, that’s basically the law of action and reaction.] The e₁₂ vector is the unit vector from q₂to q₁, not from q₁to q₂, as one might expect. That’s because we’re not talking gravity here: like charges do not attract but repel and, hence, we have to switch the order here. Having said that, that’s basically the only peculiar thing about the equation. All the rest is standard:

The force is inversely proportional to the square of the distance and so we have an inverse-square law here indeed.
The force is proportional to the charge(s).
Finally, we have a proportionality constant, 1/4πε₀, which makes the units come out alright. You may wonder why it’s written the way it’s written, i.e. with that 4π factor, but that factor (4π or 2π) actually disappears in a number of calculations, so then we will be left with just a 1/ε₀ or a 1/2ε₀ factor. So don’t worry about it.

We want to calculate potentials and all that, so the first thing we’ll do is calculate the force on a unit charge. So we’ll divide that equation by q₁, to calculate E(1) = F₁/q₁:

Piece of cake. But… What’s E(1) really? Well… It’s the force on the unit charge (+e), but so it doesn’t matter whether or not that unit charge is actually there, so it’s the field E caused by a charge q₂. [If that doesn’t make sense to you, think again.] So we can drop the subscripts and just write:

What a relief, isn’t it? The simplest formula ever: the (magnitude) of the field as a simple function of the charge q and its distance (r) from the point that we’re looking at, which we’ll write as P = (x, y, z). But what origin are we using to measure x, y and z. Don’t be surprised: the origin is q.

Now that’s a formula we can use in the Φ(P) integral. Indeed, the antiderivative is ∫(q/4πε₀r²)dr. Now, we can bring q/4πε₀out and so we’re left with ∫(1/r²)dr. Now ∫(1/r²)dr is equal to –1/r + k, and so the whole antiderivative is –q/4πε₀r + k. However, the minus sign cancels out with the minus sign in front of the Φ(P) = Φ(x, y, z) integral, and so we get:

You should just do the integral to check this result. It’s the same integral but with P₀(infinity) as point a and P as point b in the integral, so we have ∞ as start value and r as end value. The integral then yields Φ(P) – Φ(P₀) = –q/4πε₀[1/r – 1/∞). [The k constant falls away when subtracting Φ(P₀) from Φ(P).] But 1/∞ = 0, and we had a minus sign in front of the integral, which cancels the sign of –q/4πε₀. So, yes, we get the wonderfully simple result above. Also please do quickly check if it makes sense in terms of sign: the unit charge is +e, so that’s a positive charge. Hence, Φ(x, y, z) will be positive if the sign of q is also positive, but negative if q would happen to be negative. So that’s OK.

Also note that the potential – which, remember, represents the amount of work to be done when bringing a unit charge (e) from infinity to some distance r from a charge q – is proportional to the charge of q. We also know that the force and, hence, the work is proportional to the charge that we are bringing in (that’s how we calculated the work per unit in the first place: by dividing the total amount of work by the charge). Hence, if we’d not bring some unit charge but some other charge q₂, the work done would also be proportional to q₂. Now, we need to make sure we understand what we’re writing and so let’s tidy up and re-label our first charge once again as q₁, and the distance r as r₁₂, because that’s what r is: the distance between the two charges. We then have another obvious but nice result: the work done in bringing two charges together from a large distance (infinity) is

Now, one of the many nice properties of fields (scalar or vector fields) and the associated energies (because that’s what we are talking about here) is that we can simply add up contributions. For example, if we’d have many charges and we’d want to calculate the potential Φ at a point which we call 1, we can use the same Φ(r) = q/4πε₀r formula which we had derived for one charge only, for all charges, and then we simply add the contributions of each to get the total potential:

Now that we’re here, I should, of course, also give the continuum version of this formula, i.e. the formula used when we’re talking charge densities rather than individual charges. The sum then becomes an infinite sum (i.e. an integral), and q_j (note that j goes from 2 to n) becomes a variable which we write as ρ(2). We get:

Going back to the discrete situation, we get the same type of sum when bringing multiple pairs of charges q_i and q_j together. Hence, the total electrostatic energy U is the sum of the energies of all possible pairs of charges:

It’s been a while since you’ve seen any diagram or so, so let me insert one just to reassure you it’s as simple as that indeed:

Now, we have to be aware of the risk of double-counting, of course. We should not be adding q_iq_j/4πε₀r_ijtwice. That’s why we write ‘all pairs’ under the ∑ summation sign, instead of the usual i, j subscripts. The continuum version of this equation below makes that 1/2 factor explicit:

Hmm… What kind of integral is that? It’s a so-called double integral because we have two variables here. Not easy. However, there’s a lucky break. We can use the continuum version of our formula for Φ(1) to get rid of the ρ(2) and dV₂ variables and reduce the whole thing to a more standard ‘single’ integral. Indeed, we can write:

Now, because our point (2) no longer appears, we can actually write that more elegantly as:

That looks nice, doesn’t it? But do we understand it? Just to make sure. Let me explain it. The potential energy of the charge ρdV is the product of this charge and the potential at the same point. The total energy is therefore the integral over ϕρdV, but then we are counting energies twice, so that’s why we need the 1/2 factor. Now, we can write this even more beautifully as:

Isn’t this wonderful? We have an expression for the energy of a field, not in terms of the charges or the charge distribution, but in terms of the field they produce.

I am pretty sure that, by now, you must be suffering from ‘formula overload’, so you probably are just gazing at this without even bothering to try to understand. Too bad, and you should take a break then or just go do something else, like biking or so. 🙂

First, you should note that you know this E•E expression already: E•E is just the square of the magnitude of the field vector E, so E•E = E². That makes sense because we know, from what we know about waves, that the energy is always proportional to the square of an amplitude, and so we’re just writing the same here but with a little proportionality constant (ε₀).

OK, you’ll say. But you probably still wonder what use this formula could possibly have. What is that number we get from some integration over all space? So we associate the Universe with some number U and then what? Well… Isn’t that just nice? 🙂 Jokes aside, we’re actually looking at that E•E = E²product inside of the integral as representing an energy density (i.e. the energy per unit volume). We’ll denote that with a lower-case u symbol and so we write:

Just to make sure you ‘get’ what we’re talking about here: u is the energy density in the little cube dV in the rather simplistic (and, therefore, extremely useful) illustration below (which, just like most of what I write above, I got from Feynman).

Now that should make sense to you—I hope. 🙂 In any case, if you’re still with me, and if you’re not all formula-ed out you may wonder how we get that ε₀E•E = ε₀E² expression from that ρΦ expression. Of course, you know that E = –∇Φ, and we also have the Poisson equation ∇²Φ = –ρ/ε₀, but that doesn’t get you very far. It’s one of those examples where an easy-looking formula requires a lot of gymnastics. However, as the objective of this post is to do some of that, let me take you through the derivation.

Let’s do something with that Poisson equation first, so we’ll re-write it as ρ = –ε₀∇²Φ, and then we can substitute ρ in the integral with the ρΦ product. So we get:

Now, you should check out those fancy formulas with our new vector differential operators which we listed in our second class on vector calculus, but, unfortunately, none of them apply. So we have to write it all out and see what we get:

Now that looks horrendous and so you’ll surely think we won’t get anywhere with that. Well… Physicists don’t despair as easily as we do, it seems, and so they do substitute it in the integral which, of course, becomes an even more monstrous expression, because we now have two volume integrals instead of one! Indeed, we get:

But if ∇Φ is a vector field (it’s minus E, remember!), then Φ∇Φ is a vector field too, and we can then apply Gauss’ Theorem, which we mentioned in our first class on vector calculus, and which – mind you! – has nothing to do with Gauss’ Law. Indeed, Gauss produced so much it’s difficult to keep track of it all. 🙂 So let me remind you of this theorem. [I should also show why Φ∇Φ still yields a field, but I’ll assume you believe me.] Gauss’ Theorem basically shows how we can go from a volume integral to a surface integral:

If we apply this to the second integral in our U expression, we get:

So what? Where are we going with this? Relax. Be patient. What volume and surface are we talking about here? To make sure we have all charges and influences, we should integrate over all space and, hence, the surface goes to infinity. So we’re talking a (spherical) surface of enormous radius R whose center is the origin of our coordinate system. I know that sounds ridiculous but, from a math point of view, it is just the same like bringing a charge in from infinity, which is what we did to calculate the potential. So if we don’t difficulty with infinite line integrals, we should not have difficulty with infinite surface and infinite volumes. That’s all I can, so… Well… Let’s do it.

Let’s look at that product Φ∇Φ•n in the surface integral. Φ is a scalar and ∇Φ is a vector, and so… Well… ∇Φ•n is a scalar too: it’s the normal component of ∇Φ = –E. [Just to make sure, you should note that the way we define the normal unit vector n is such that ∇Φ•n is some positive number indeed! So n will point in the same direction, more or less, as ∇Φ = –E. So the θ angle between ∇Φ = –E and n is surely less than ± 90° and, hence, the cosine factor in the ∇Φ•n = |∇Φ||n|cosθ = |∇Φ|cosθ is positive, and so the whole vector dot product is positive.]

So, we have a product of two scalars here. What happens with them if R goes to infinity? Well… The potential varies as 1/r as we’re going to infinity. That’s obvious from that Φ = (q/4πε₀)(1/r) formula: just think of q as some kind of average now, which works because we assume all charges are located within some finite distance, while we’re going to infinity. What about ∇Φ•n? Well… Again assuming that we’re reasonably far away from the charges, we’re talking the density of flux lines here (i.e. the magnitude of E) which, as shown above, follows an inverse-square law, because the surface area of a sphere increases with the square of the radius. So ∇Φ•n varies not as 1/r but as 1/r². To make a long story short, the whole product Φ∇Φ•n falls of as 1/r goes to infinity. Now, we shouldn’t forget we’re integrating a surface integral here, with r = R, and so it’s R going to infinity. So that surface integral has to go to zero when we include all space. The volume integral still stands however, so our formula for U now consists of one term only, i.e. the volume integral, and so we now have:

Done !

What’s left?

In electrostatics? Lots. Electric dipoles (like polar molecules), electrolytes, plasma oscillations, ionic crystals, electricity in the atmosphere (like lightning!), dielectrics and polarization (including condensers), ferroelectricity,… As soon as we try to apply our theory to matter, things become hugely complicated. But the theory works. Fortunately! 🙂 I have to refer you to textbooks, though, in case you’d want to know more about it. [I am sure you don’t, but then one never knows.]

What I wanted to do is to give you some feel for those vector and field equations in the electrostatic case. We now need to bring magnetic field back into the picture and, most importantly, move to electrodynamics, in which the electric and magnetic field do not appear as completely separate things. No! In electrodynamics, they are fully interconnected through the time derivatives ∂E/∂t and ∂B/∂t. That shows they’re part and parcel of the same thing really: electromagnetism.

But we’ll try to tackle that in future posts. Goodbye for now!

The wave-particle duality revisited

Pre-script (dated 26 June 2020): This post has become less relevant (even irrelevant, perhaps) because my views on all things quantum-mechanical have evolved significantly as a result of my progression towards a more complete realist (classical) interpretation of quantum physics. I keep blog posts like these mainly because I want to keep track of where I came from. I might review them one day, but I currently don’t have the time or energy for it. 🙂

Original post:

As an economist, having some knowledge of what’s around in my field (social science), I think I am well-placed to say that physics is not an easy science. Its ‘first principles’ are complicated, and I am not ashamed to say that, after more than a year of study now, I haven’t reached what I would call a ‘true understanding’ of it.

Sometimes, the teachers are to be blamed. For example, I just found out that, in regard to the question of the wave function of a photon, the answer of two nuclear scientists was plain wrong. Photons do have a de Broglie wave, and there is a fair amount of research and actual experimenting going on trying to measure it. One scientific article which I liked in particular, and I hope to fully understand a year from now or so, is on such ‘direct measurement of the (quantum) wavefunction‘. For me, it drove home the message that these idealized ‘thought experiments’ that are supposed to make autodidacts like me understand things better, are surely instructive in regard to the key point, but confusing in other respects.

A typical example of such idealized thought experiment is the double-slit experiment with ‘special detectors’ near the slits, which may or may not detect a photon, depending on whether or not they’re switched on as well as on their accuracy. Depending on whether or not the detectors are switched on, and their accuracy, we get full interference (a), no interference (b), or a mixture of (a) and (b), as shown in (c) and (d).

I took the illustrations from Feynman’s lovely little book, QED – The Strange Theory of Light and Matter, and he surely knows what he’s talking about. Having said that, the set-up raises a key question in regard to these detectors: how do they work, exactly? More importantly, how do they disturb the photons?

I googled for actual double-slit experiments with such ‘special detectors’ near the slits, but only found such experiments for electrons. One of these, a 2010 experiment of an Italian team, suggests that it’s the interaction between the detector and the electron wave that may cause the interference pattern to disappear. The idea is shown below. The electron is depicted as an incoming plane wave, which breaks up as it goes through the slits. The slit on the left has no ‘filter’ (which you may think of as a detector) and, hence, the plane wave goes through as a cylindrical wave. The slit on the right-hand side is covered by a ‘filter’ made of several layers of ‘low atomic number material’, so the electron goes through but, at the same time, the barrier creates a spherical wave as it goes through. The researchers note that “the spherical and cylindrical wave do not have any phase correlation, and so even if an electron passed through both slits, the two different waves that come out cannot create an interference pattern on the wall behind them.” [Needless to say, while being represented as ‘real’ waves here, the ‘waves’ are, in fact, complex-valued psi functions.]

In fact, to be precise, there actually still was an interference effect if the filter was thin enough. Let me quote the reason for that: “The thicker the filter, the greater the probability for inelastic scattering. When the electron suffers inelastic scattering, it is localized. This means that its wavefunction collapses and, after the measurement act, it propagates roughly as a spherical wave from the region of interaction, with no phase relation at all with other elastically or inelastically scattered electrons. If the filter is made thick enough, the interference effects cancels out almost completely.”

This, of course, doesn’t solve the mystery. The mystery, in such experiments, is that, when we put detectors, it is either the detector at A or the detector at B that goes off. They should never go off together—”at half strength, perhaps?”, as Feynman puts it. That’s why I used italics when writing “even if an electron passed through both slits.” The electron, or the photon in a similar set-up, is not supposed to do that. As mentioned above, the wavefunction collapses or reduces. Now that’s where these so-called ‘weak measurement’ experiments come in: they indicate the interaction doesn’t have to be that way. It’s not all or nothing: our observations should not necessarily destroy the wavefunction. So, who knows, perhaps we will be able, one day, to show that the wavefunction does go through both slits, as it should (otherwise the interference pattern cannot be explained), and then we will have resolved the paradox.

I am pretty sure that, when that’s done, physicists will also be able to relate the image of a photon as a transient electromagnetic wave (first diagram below), being emitted by an atomic oscillator for a few nanoseconds only (we gave the example for sodium light, for which the decay time was 3.2×10^–8seconds) with the image of a photon as a de Broglie wave (second diagram below). I look forward to that day. I think it will come soon.

Spin

Pre-script (dated 26 June 2020): This post has become less relevant (even irrelevant, perhaps) because my views on all things quantum-mechanical have evolved significantly as a result of my progression towards a more complete realist (classical) interpretation of quantum physics. In addition, some of the material was removed by a dark force (that also created problems with the layout, I see now). In any case, we recommend you read our recent papers. I keep blog posts like these mainly because I want to keep track of where I came from. I might review them one day, but I currently don’t have the time or energy for it. 🙂

Original post:

In the previous posts, I showed how the ‘real-world’ properties of photons and electrons emerge out of very simple mathematical notions and shapes. The basic notions are time and space. The shape is the wavefunction.

Let’s recall the story once again. Space is an infinite number of three-dimensional points (x, y, z), and time is a stopwatch hand going round and round—a cyclical thing. All points in space are connected by an infinite number of paths – straight or crooked, whatever – of which we measure the length. And then we have ‘photons’ that move from A to B, but so we don’t know what is actually moving in space here. We just associate each and every possible path (in spacetime) between A and B with an amplitude: an ‘arrow‘ whose length and direction depends on (1) the length of the path l (i.e. the ‘distance’ in space measured along the path, be it straight or crooked), and (2) the difference in time between the departure (at point A) and the arrival (at point B) of our photon (i.e. the ‘distance in time’ as measured by that stopwatch hand).

Now, in quantum theory, anything is possible and, hence, not only do we allow for crooked paths, but we also allow for the difference in time to differ from l/c. Hence, our photon may actually travel slower or faster than the speed of light c! There is one lucky break, however, that makes all come out alright: the arrows associated with the odd paths and strange timings cancel each other out. Hence, what remains, are the nearby paths in spacetime only—the ‘light-like’ intervals only: a small core of space which our photon effectively uses as it travels through empty space. And when it encounters an obstacle, like a sheet of glass, it may or may not interact with the other elementary particle–the electron. And then we multiply and add the arrows – or amplitudes as we call them – to arrive at a final arrow, whose square is what physicists want to find, i.e. the likelihood of the event that we are analyzing (such a photon going from point A to B, in empty space, through two slits, or through as sheet of glass, for example) effectively happening.

The combining of arrows leads to diffraction, refraction or – to use the more general description of what’s going on – interference patterns:

Adding two identical arrows that are ‘lined up’ yields a final arrow with twice the length of either arrow alone and, hence, a square (i.e. a probability) that is four times as large. This is referred to as ‘positive’ or ‘constructive’ interference.
Two arrows of the same length but with opposite direction cancel each other out and, hence, yield zero: that’s ‘negative’ or ‘destructive’ interference.

Both photons and electrons are represented by wavefunctions, whose argument is the position in space (x, y, z) and time (t), and whose value is an amplitude or ‘arrow’ indeed, with a specific direction and length. But here we get a bifurcation. When photons interact with other, their wavefunctions interact just like amplitudes: we simply add them. However, when electrons interact with each other, we have to apply a different rule: we’ll take a difference. Indeed, anything is possible in quantum mechanics and so we combine arrows (or amplitudes, or wavefunctions) in two different ways: we can either add them or, as shown below, subtract one from the other.

There are actually four distinct logical possibilities, because we may also change the order of A and B in the operation, but when calculating probabilities, all we need is the square of the final arrow, so we’re interested in its final length only, not in its direction (unless we want to use that arrow in yet another calculation). And so… Well… The fundamental duality in Nature between light and matter is based on this dichotomy only: identical (elementary) particles behave in one of two ways: their wavefunctions interfere either constructively or destructively, and that’s what distinguishes bosons (i.e. force-carrying particles, such as photons) from fermions (i.e. matter-particles, such as electrons). The mathematical description is complete and respects Occam’s Razor. There is no redundancy. One cannot further simplify: every logical possibility in the mathematical description reflects a physical possibility in the real world.

Having said that, there is more to an electron than just Fermi-Dirac statistics, of course. What about its charge, and this weird number, its spin?,

Well… That’s what’s this post is about. As Feynman puts it: “So far we have been considering only spin-zero electrons and photons, fake electrons and fake photons.”

I wouldn’t call them ‘fake’, because they do behave like real photons and electrons already but… Yes. We can make them more ‘real’ by including charge and spin in the discussion. Let’s go for it.

Charge and spin

From what I wrote above, it’s clear that the dichotomy between bosons and fermions (i.e. between ‘matter-particles’ and ‘force-carriers’ or, to put it simply, between light and matter) is not based on the (electric) charge. It’s true we cannot pile atoms or molecules on top of each other because of the repulsive forces between the electron clouds—but it’s not impossible, as nuclear fusion proves: nuclear fusion is possible because the electrostatic repulsive force can be overcome, and then the nuclear force is much stronger (and, remember, no quarks are being destroyed or created: all nuclear energy that’s being released or used is nuclear binding energy).

It’s also true that the force-carriers we know best, notably photons and gluons, do not carry any (electric) charge, as shown in the table below. So that’s another reason why we might, mistakenly, think that charge somehow defines matter-particles. However, we can see that matter-particles, first carry very different charges (positive or negative, and with very different values: 1/3, 2/3 or 1), and even be neutral, like the neutrinos. So, if there’s a relation, it’s very complex. In addition, one of the two force-carrier for the weak force, the W boson, can have positive or negative charge too, so that doesn’t make sense, does it? [I admit the weak force is a bit of a ‘special’ case, and so I should leave it out of the analysis.] The point is: the electric charge is what it is, but it’s not what defines matter. It’s just one of the possible charges that matter-particles can carry. [The other charge, as you know, is the color charge but, to confuse the picture once again, that’s a charge that can also be carried by gluons, i.e. the carriers of the strong force.]

So what is it, then? Well… From the table above, you can see that the property of ‘spin’ (i.e. the third number in the top left-hand corner) matches the above-mentioned dichotomy in behavior, i.e. the two different types of interference (bosons versus fermions or, to use a heavier term, Bose-Einstein statistics versus Fermi-Dirac statistics): all matter-particles are so-called spin-1/2 particles, while all force-carriers (gauge bosons) all have spin one. [Never mind the Higgs particle: that’s ‘just’ a mechanism to give (most) elementary particles some mass.]

So why is that? Why are matter-particles spin-1/2 particles and force-carries spin-1 particles? To answer that question, we need to answer the question: what’s this spin number? And to answer that question, we first need to answer the question: what’s spin?

Spin in the classical world

In the classical world, it’s, quite simply, the momentum associated with a spinning or rotating object, which is referred to as the angular momentum. We’ve analyzed the math involved in another post, and so I won’t dwell on that here, but you should note that, in classical mechanics, we distinguish two types of angular momentum:

Orbital angular momentum: that’s the angular momentum an object gets from circling in an orbit, like the Earth around the Sun.
Spin angular momentum: that’s the angular momentum an object gets from spinning around its own axis., just like the Earth, in addition to rotating around the Sun, is rotating around its own axis (which is what causes day and night, as you know).

The math involved in both is pretty similar, but it’s still useful to distinguish the two, if only because we’ll distinguish them in quantum mechanics too! Indeed, when I analyzed the math in the above-mentioned post, I showed how we represent angular momentum by a vector that’s perpendicular to the direction of rotation, with its direction given by the ubiquitous right-hand rule—as in the illustration below, which shows both the angular momentum (L) as well as the torque (τ) that’s produced by a rotating mass. The formulas are given too: the angular momentum L is the vector cross product of the position vector r and the linear momentum p, while the magnitude of the torque τ is given by the product of the length of the lever arm and the applied force. An alternative approach is to define the angular velocity ω and the moment of inertia I, and we get the same result: L = Iω.

Of course, the illustration above shows orbital angular momentum only and, as you know, we no longer have a ‘planetary model’ (aka the Rutherford model) of an atom. So should we be looking at spin angular momentum only?

Well… Yes and no. More yes than no, actually. But it’s ambiguous. In addition, the analogy between the concept of spin in quantum mechanics, and the concept of spin in classical mechanics, is somewhat less than straightforward. Well… It’s not straightforward at all actually. But let’s get on with it and use more precise language. Let’s first explore it for light, not because it’s easier (it isn’t) but… Well… Just because. 🙂

The spin of a photon

I talked about the polarization of light in previous posts (see, for example, my post on vector analysis): when we analyze light as a traveling electromagnetic wave (so we’re still in the classical analysis here, not talking about photons as ‘light particles’), we know that the electric field vector oscillates up and down and is, in fact, likely to rotate in the xy-plane (with z being the direction of propagation). The illustration below shows the idealized (aka limiting) case of perfectly circular polarization: if there’s polarization, it is more likely to be elliptical. The other limiting case is plane polarization: in that case, the electric field vector just goes up and down in one direction only. [In case you wonder whether ‘real’ light is polarized, it often is: there’s an easy article on that on the Physics Classroom site.]

The illustration above uses Dirac’s bra-ket notation |L〉 and |R〉 to distinguish the two possible ‘states’, which are left- or right-handed polarization respectively. In case you forgot about bra-ket notations, let me quickly remind you: an amplitude is usually denoted by 〈x|s〉, in which 〈x| is the so-called ‘bra’, i.e. the final condition, and |s〉 is the so-called ‘ket’, i.e. the starting condition, so 〈x|s〉 could mean: a photon leaves at s (from source) and arrives at x. It doesn’t matter much here. We could have used any notation, as we’re just describing some state, which is either |L〉 (left-handed polarization) or |R〉 (right-handed polarization). The more intriguing extras in the illustration above, besides the formulas, are the values: ± ħ = ±h/2π. So that’s plus or minus the (reduced) Planck constant which, as you know, is a very tiny constant. I’ll come back to that. So what exactly is being represented here?

At first, you’ll agree it looks very much like the momentum of light (p) which, in a previous post, we calculated from the (average) energy (E) as p = E/c. Now, we know that E is related to the (angular) frequency of the light through the Planck-Einstein relation E = hν = ħω. Now, ω is the speed of light (c) times the wave number (k), so we can write: p = ħω = ħck/c = ħk. The wave number is the ‘spatial frequency’, expressed either in cycles per unit distance (1/λ) or, more usually, in radians per unit distance (k = 2π/λ), so we can also write p = ħk = h/λ. Whatever way we write it, we find that this momentum (p) depends on the energy and/or, what amounts to saying the same, the frequency and/or the wavelength of the light.

So… Well… The momentum of light is not just h or ħ, i.e. what’s written in that illustration above. So it must be something different. In addition, I should remind you this momentum was calculated from the magnetic field vector, as shown below (for more details, see my post on vector calculus once again), so it had nothing to do with polarization really.

Finally, last but not least, the dimensions of ħ and p = h/λ are also different (when one is confused, it’s always good to do a dimensional analysis in physics):

The dimension of Planck’s constant (both h as well as ħ = h/2π) is energy multiplied by time (J·s or eV·s) or, equivalently, momentum multiplied by distance. It’s referred to as the dimension of action in physics, and h is effectively, the so-called quantum of action.
The dimension of (linear) momentum is… Well… Let me think… Mass times velocity (mv)… But what’s the mass in this case? Light doesn’t have any mass. However, we can use the mass-energy equivalence: 1 eV = 1.7826×10⁻³⁶ kg. [10⁻³⁶? Well… Yes. An electronvolt is a very tiny measure of energy.] So we can express p in eV·m/s units.

Hmm… We can check: momentum times distance gives us the dimension of Planck’s constant again – (eV·m/s)·m = eV·s. OK. That’s good… […] But… Well… All of this nonsense doesn’t make us much smarter, does it? 🙂 Well… It may or may not be more useful to note that the dimension of action is, effectively, the same as the dimension of angular momentum. Huh? Why? Well… From our classical L = r×p formula, we find L should be expressed in m·(eV·m/s) = eV·m²/s units, so that’s… What? Well… Here we need to use a little trick and re-express energy in mass units. We can then write L in kg·m²/s units and, because 1 Newton (N) is 1 kg⋅m/s², the kg·m²/s unit is equivalent to the N·m·s = J·s unit. Done!

Having said that, all of this still doesn’t answer the question: are the linear momentum of light, i.e. our p, and those two angular momentum ‘states’, |L〉 and |R〉, related? Can we relate |L〉 and |R〉 to that L = r×p formula?

The answer is simple: no. The |L〉 and |R〉 states represent spin angular momentum indeed, while the angular momentum we would derive from the linear momentum of light using that L = r×p is orbital angular momentum. Let’s introduce the proper symbols: orbital angular momentum is denoted by L, while spin angular momentum is denoted by S. And then the total angular momentum is, quite simply, J = L + S.

L and S can both be calculated using either a vector cross product r × p (but using different values for r and p, of course) or, alternatively, using the moment of inertia tensor I and the angular velocity ω. The illustrations below (which I took from Wikipedia) show how, and also shows how L and S are added to yield J = L + S.

So what? Well… Nothing much. The illustration above show that the analysis – which is entirely classical, so far – is pretty complicated. [You should note, for example, that in the S = Iω and L = Iω formulas, we don’t use the simple (scalar) moment of inertia but the moment of inertia tensor (so that’s a matrix denoted by I, instead of the scalar I), because S (or L) and ω are not necessarily pointing in the same direction.

By now, you’re probably very confused and wondering what’s wiggling really. The answer for the orbital angular momentum is: it’s the linear momentum vector p. Now…

Hey! Stop! Why would that vector wiggle?

You’re right. Perhaps it doesn’t. The linear momentum p is supposed to be directed in the direction of travel of the wave, isn’t it? It is. In vector notation, we have p = ħk, and that k vector (i.e. the wavevector) points in the direction of travel of the wave indeed and so… Well… No. It’s not that simple. The wave vector is perpendicular to the surfaces of constant phase, i.e. the so-called wave fronts, as show in the illustration below (see the direction of e_k, which is a unit vector in the direction of k).

So, yes, if we’re analyzing light moving in a straight one-dimensional line only, or we’re talking a plane wave, as illustrated below, then the orbital angular momentum vanishes.

But the orbital angular momentum L does not vanish when we’re looking at a real light beam, like the ones below. Real waves? Well… OK… The ones below are idealized wave shapes as well, but let’s say they are somewhat more real than a plane wave. 🙂

So what do we have here? We have wavefronts that are shaped as helices, except for the one in the middle (marked by m = 0) which is, once again, an example of plane wave—so for that one (m = 0), we have zero orbital angular momentum indeed. But look, very carefully, at the m = ± 1 and m = ± 2 situations. For m = ± 1, we have one helical surface with a step length equal to the wavelength λ. For m = ± 2, we have two intertwined helical surfaces with the step length of each helix surface equal to 2λ. [Don’t worry too much about the second and third column: they show a beam cross-section (so that’s not a wave front but a so-called phase front) and the (averaged) light intensity, again of a beam cross-section.] Now, we can further generalize and analyze waves composed of m helices with the step length of each helix surface equal to |m|λ. The Wikipedia article on OAM (orbital angular momentum of light), from which I got this illustration, gives the following formula to calculate the OAM:

The same article also notes that the quantum-mechanical equivalent of this formula, i.e. the orbital angular momentum of the photons one would associate with the not-cylindrically-symmetric waves above (i.e. all those for which m ≠ 0), is equal to:

L_z = mħ

So what? Well… I guess we should just accept that as a very interesting result. For example, I duly note that L_zis along the direction of propagation of the wave (as indicated by the z subscript), and I also note the very interesting fact that, apparently, L_zcan be either positive or negative. Now, I am not quite sure how such result is consistent with the idea of radiation pressure, but I am sure there must be some logical explanation to that. The other point you should note is that, once again, any reference to the energy (or to the frequency or wavelength) of our photon has disappeared. Hmm… I’ll come back to this, as I promised above already.

The thing is that this rather long digression on orbital angular momentum doesn’t help us much in trying to understand what that spin angular momentum (SAM) is all about. So, let me just copy the final conclusion of the Wikipedia article on the orbital angular momentum of light: the OAM is the component of angular momentum of light that is dependent on the field spatial distribution, not on the polarization of light.

So, again, what’s the spin angular momentum? Well… The only guidance we have is that same little drawing again and, perhaps, another illustration that’s supposed to compare SAM with OAM (underneath).

Now, the Wikipedia article on SAM (spin angular momentum), from which I took the illustrations above, gives a similar-looking formula for it:

When I say ‘similar-looking’, I don’t mean it’s the same. [Of course not! Spin and orbital angular momentum are two different things!]. So what’s different in the two formulas? Well… We don’t have any del operator (∇) in the SAM formula, and we also don’t have any position vector (r) in the integral kernel (or integrand, if you prefer that term). However, we do find both the electric field vector (E) as well as the (magnetic) vector potential (A) in the equation again. Hence, the SAM (also) takes both the electric as well as the magnetic field into account, just like the OAM. [According to the author of the article, the expression also shows that the SAM is nonzero when the light polarization is elliptical or circular, and that it vanishes if the light polarization is linear, but I think that’s much more obvious from the illustration than from the formula… However, I realize I really need to move on here, because this post is, once again, becoming way too long. So…]

OK. What’s the equivalent of that formula in quantum mechanics?

Well… In quantum mechanics, the SAM becomes a ‘quantum observable’, described by a corresponding operator which has only two eigenvalues:

S_z = ± ħ

So that corresponds to the two possible values for J_z, as mentioned in the illustration, and we can understand, intuitively, that these two values correspond to two ‘idealized’ photons which describe a left- and right-handed circularly polarized wave respectively.

So… Well… There we are. That’s basically all there is to say about it. So… OK. So far, so good.

But… Yes? Why do we call a photon a spin-one particle?

That has to do with convention. A so-called spin-zero particle has no degrees of freedom in regard to polarization. The implied ‘geometry’ is that a spin-zero particle is completely symmetric: no matter in what direction you turn it, it will always look the same. In short, it really behaves like a (zero-dimensional) mathematical point. As you can see from the overview of all elementary particles, it is only the Higgs boson which has spin zero. That’s why the Higgs field is referred to as a scalar field: it has no direction. In contrast, spin-one particles, like photons, are also ‘point particles’, but they do come with one or the other kind of polarization, as evident from all that I wrote above. To be specific, they are polarized in the xy-plane, and can have one of two directions. So, when rotating them, you need a full rotation of 360° if you want them to look the same again.

Now that I am here, let me exhaust the topic (to a limited extent only, of course, as I don’t want to write a book here) and mention that, in theory, we could also imagine spin-2 particles, which would look the same after half a rotation (180°). However, as you can see from the overview, none of the elementary particles has spin-2. A spin-2 particle could be like some straight stick, as that looks the same even after it is rotated 180 degrees. I am mentioning the theoretical possibility because the graviton, if it would effectively exist, is expected to be a massless spin-2 boson. [Now why do I mention this? Not sure. I guess I am just noting this to remind you of the fact that the Higgs boson is definitely not the (theoretical) graviton, and/or that we have no quantum theory for gravity.]

Oh… That’s great, you’ll say. But what about all those spin-1/2 particles in the table? You said that all matter-particles are spin 1/2 particles, and that it’s this particular property that actually makes them matter-particles. So what’s the geometry here? What kind of ‘symmetries’ do they respect?

Well… As strange as it sounds, a spin-1/2 particle needs two full rotations (2×360°=720°) until it is again in the same state. Now, in regard to that particularity, you’ll often read something like: “There is nothing in our macroscopic world which has a symmetry like that.” Or, worse, “Common sense tells us that something like that cannot exist, that it simply is impossible.” [I won’t quote the site from which I took this quotes, because it is, in fact, the site of a very respectable research center!] Bollocks! The Wikipedia article on spin has this wonderful animation: look at how the spirals flip between clockwise and counterclockwise orientations, and note that it’s only after spinning a full 720 degrees that this ‘point’ returns to its original configuration after spinning a full 720 degrees.

So, yes, we can actually imagine spin-1/2 particles, and with not all that much imagination, I’d say. But… OK… This is all great fun, but we have to move on. So what’s the ‘spin’ of these spin-1/2 particles and, more in particular, what’s the concept of ‘spin’ of an electron?

The spin of an electron

When starting to read about it, I thought that the angular momentum of an electron would be easier to analyze than that of a photon. Indeed, while a photon has no mass and no electric charge, that analysis with those E and B vectors is damn complicated, even when sticking to a strictly classical analysis. For an electron, the classical picture seems to be much more straightforward—but only at first indeed. It quickly becomes equally weird, if not more.

We can look at an electron in orbit as a rotating electrically charged ‘cloud’ indeed. Now, from Maxwell’s equations (or from your high school classes even), you know that a rotating electric charged body creates a magnetic dipole. So an electron should behave just like a tiny bar magnet. Of course, we have to make certain assumptions about the distribution of the charge in space but, in general, we can write that the magnetic dipole moment μ is equal to:

In case you want to know, in detail, where this formula comes from, let me refer you to Feynman once again, but trust me – for once 🙂 – it’s quite straightforward indeed: the L in this formula is the angular momentum, which may be the spin angular momentum, the orbital angular momentum, or the total angular momentum. The e and m are, of course, the charge and mass of the electron respectively.

So that’s a good and nice-looking formula, and it’s actually even correct except for the spin angular momentum as measured in experiments. [You’ll wonder how we can measure orbital and spin angular momentum respectively, but I’ll talk about an 1921 experiment in a few minutes, and so that will give you some clue as to that mystery. :-)] To be precise, it turns out that one has to multiply the above formula for μ with a factor which is referred to as the g-factor. [And, no, it’s got nothing to do with the gravitational constant or… Well… Nothing. :-)] So, for the spin angular momentum, the formula should be:

Experimental physicists are constantly checking that value and they know measure it to be something like g = is 2.00231930419922 ± 1.5×10⁻¹². So what’s the explanation for that g? Where does it come from? There is, in fact, a classical explanation for it, which I’ll copy hereunder (yes, from Wikipedia). This classical explanation is based on assuming that the distribution of the electric charge of the electron and its mass does not coincide:

Why do I mention this classical explanation? Well… Because, in most popular books on quantum mechanics (including Feynman’s delightful QED), you’ll read that (a) the value for g can be derived from a quantum-theoretical equation known as Dirac’s equation (or ‘Dirac theory’, as it’s referred to above) and, more importantly, that (b) physicists call the “accurate prediction of the electron g-factor” from quantum theory (i.e. ‘Dirac’s theory’ in this case) “one of the greatest triumphs” of the theory.

So what about it? Well… Whatever the merits of both explanations (classical or quantum-mechanical), they are surely not the reason why physicists abandoned the classical theory. So what was the reason then? What a stupid question! You know that already! The Rutherford model was, quite simply, not consistent: according to classical theory, electrons should just radiate their energy away and spiral into the nucleus. More in particular, there was yet another experiment that wasn’t consistent with classical theory, and it’s one that’s very relevant for the discussion at hand: it’s the so-called Stern-Gerlach experiment.

It was just as ‘revolutionary’ as the Michelson-Morley experiment (which couldn’t measure the speed of light), or the discovery of the positron in 1932. The Stern-Gerlach experiment was done in 1921, so that’s many years before quantum theory replaced classical theory and, hence, it’s not one of those experiments confirming quantum theory. No. Quite the contrary. It was, in fact, one of the experiments that triggered the so-called quantum revolution. Let me insert the experimental set-up and summarize it (below).

sterngerlach

The German scientists Otto Stern and Walther Gerlach produced a beam of electrically-neutral silver atoms and let it pass through a (non-uniform) magnetic field. Why silver atoms? Well… Silver atoms are easy to handle (in a lab, that is) and easy to detect with a photoplate.
These atoms came out of an oven (literally), in which the silver was being evaporated (yes, one can evaporate silver), so they had no special orientation in space and, so Stern and Gerlach thought, the magnetic moment (or spin) of the outer electrons in these atoms would point into all possible directions in space.
As expected, the magnetic field did deflect the silver atoms, just like it would deflect little dipole magnets if you would shoot them through the field. However, the pattern of deflection was not the one which they expected. Instead of hitting the plate all over the place, within some contour, of course, only the contour itself was hit by the atoms. There was nothing in the middle!
And… Well… It’s a long story, but I’ll make it short. There was only one possible explanation for that behavior, and that’s that the magnetic moments – and, therefore the spins – had only two orientations in space, and two possible values only which – Surprise, surprise! – are ±ħ/2 (so that’s half the value of the spin angular momentum of photons, which explains the spin-1/2 terminology).

The spin angular momentum of an electron is more popularly known as ‘up’ or ‘down’.

So… What about it? Well… It explains why a atomic orbital can have two electrons, rather than one only and, as such, the behavior of the electron here is the basis of the so-called periodic table, which explains all properties of the chemical elements. So… Yes. Quantum theory is relevant, I’d say. 🙂

Conclusion

This has been a terribly long post, and you may no longer remember what I promised to do. What I promised to do, is to write some more about the difference between a photon and an electron and, more in particular, I said I’d write more about their charge, and that “weird number”: their spin. I think I lived up to that promise. The summary is simple:

Photons have no (electric) charge, but they do have spin. Their spin is linked to their polarization in the xy-plane (if z is the direction of propagation) and, because of the strangeness of quantum mechanics (i.e. the quantization of (quantum) observables), the value for this spin is either +ħ or –ħ, which explains why they are referred to as spin-one particles (because either value is one unit of the Planck constant).
Electrons have both electric charge as well as spin. Their spin is different and is, in fact, related to their electric charge. It can be interpreted as the magnetic dipole moment, which results from the fact we have a spinning charge here. However, again, because of the strangeness of quantum mechanics, their dipole moment is quantized and can take only one of two values: ±ħ/2, which is why they are referred to as spin-1/2 particles.

So now you know everything you need to know about photons and electrons, and then I mean real photons and electrons, including their properties of charge and spin. So they’re no longer ‘fake’ spin-zero photons and electrons now. Isn’t that great? You’ve just discovered the real world! 🙂

So… I am done—for the moment, that is… 🙂 If anything, I hope this post shows that even those ‘weird’ quantum numbers are rooted in ‘physical reality’ (or in physical ‘geometry’ at least), and that quantum theory may be ‘crazy’ indeed, but that it ‘explains’ experimental results. Again, as Feynman says:

“We understand how Nature works, but not why Nature works that way. Nobody understands that. I can’t explain why Nature behave in this particular way. You may not like quantum theory and, hence, you may not accept it. But physicists have learned to realize that whether they like a theory or not is not the essential question. Rather, it is whether or not the theory gives predictions that agree with experiment. The theory of quantum electrodynamics describes Nature as absurd from the point of view of common sense. But it agrees fully with experiment. So I hope you can accept Nature as She is—absurd.”

Frankly speaking, I am not quite prepared to accept Nature as absurd: I hope that some more familiarization with the underlying mathematical forms and shapes will make it look somewhat less absurd. More, I hope that such familiarization will, in the end, make everything look just as ‘logical’, or ‘natural’ as the two ways in which amplitudes can ‘interfere’.

Post scriptum: I said I would come back to the fact that, in the analysis of orbital and spin angular momentum of a photon (OAM and SAM), the frequency or energy variable sort of ‘disappears’. So why’s that? Let’s look at those expressions for |L〉 and |R〉 once again:

Formula L spin

What’s written here really? If |L〉 and |R〉 are supposed to be equal to either +ħ or –ħ, then that product of e^i(kz–ωt)with the 3×1 matrix (which is a ‘column vector’ in this case) does not seem to make much sense, does it? Indeed, you’ll remember that e^i(kz–ωt)just a regular wave function. To be precise, its phase is φ = kz–ωt (with z the direction of propagation of the wave), and its real and imaginary part can be written as e^iφ = cos(φ) + isin(φ) = a + bi. Multiplying it with that 3×1 column vector (1, i, 0) or (1, –i, 0) just yields another 3×1 column vector. To be specific, we get:

The 3×1 ‘vector’ (a + bi, –b+ai, 0) for |L〉, and
The 3×1 ‘vector’ (a + bi, b–ai, 0) for |R〉.

So we have two new ‘vectors’ whose components are complex numbers. Furthermore, we can note that their ‘x’-component is the same, their ‘y’-component is each other’s opposite –b+ai = –(b–ai), and their ‘z’-component is 0.

So… Well… In regard to their ‘y’-component, I should note that’s just the result of the multiplication with i and/or –i: multiplying a complex number with i amounts to a 90° degree counterclockwise rotation, while multiplication with –i amounts to the same but clockwise. Hence, we must arrive at two complex numbers that are each other’s opposite. [Indeed, in complex analysis, the value –1 = e^iπ= e^–iπ is a 180° rotation, both clockwise (φ = –π) or counterclockwise (φ = +π), of course!.]

Hmm… Still… What does it all mean really? The truth is that it takes some more advanced math to interpret the result. To be precise, pure quantum states, such |L〉 and |R〉 here, are represented by so-called ‘state vectors’ in a Hilbert space over complex numbers. So that’s what we’ve got here. So… Well… I can’t say much more about this right now: we’ll just need to study some more before we’ll ‘understand’ those expressions for |L〉 and |R〉. So let’s not worry about it right now. We’ll get there.

Just for the record, I should note that, initially, I thought 1/√2 factor in front gave some clue as to what’s going on here: 1/√2 ≈ 0.707 is a factor that’s used to calculate the root mean square (RMS) value for a sine wave. It’s illustrated below. The RMS value is a ‘special average’ one can use to calculate the energy or power (i.e. energy per time unit) of a wave. [Using the term ‘average’ is misleading, because the average of a sine wave is 1/2 over half a cycle, and 0 over a fully cycle, as you can easily see from the shape of the function. But I guess you know what I mean.]

Indeed, you’ll remember that the energy (E) of a wave is proportional to the square of its amplitude (A): E ∼ A². For example, when we have a constant current I, the power P will be proportional to its square: P ∼ I². With a varying current (I) and voltage (V), the formula is more complicated but we can simply it using the rms values: P_avg = V_RMS·I_RMS.

So… Looking at that formula, should we think of h and/or ħ as some kind of ‘average’ energy, like the energy of a photon per cycle or per radian? That’s an interesting idea so let’s explore it. If the energy of a photon is equal to E = h·ν = h·ω/2π = ħω, then we can also write:

h = E/ν and/or ħ = E/ω

So, yes: h is the energy of a photon per cycle obviously and, because the phase covers 2π radians during each cycle, and ħ must be the energy of the photon per radian! That’s a great result, isn’t it? It also gives a wonderfully simple interpretation to Planck’s quantum of action!

Well… No. We made at least two mistakes here. The first mistake is that if we think of a photon as wave train being radiated by an atom – which, as we calculated in another post, lasts about 3.2×10^–8 seconds – the graph for its energy is going to resemble the graph of its amplitude, so it’s going to die out and each oscillation will carry less and less energy. Indeed, the decay time given here (τ = 3.2×10^–8 seconds) was the time it takes for the radiation (we assumed sodium light with a wavelength of 600 nanometer) to die out by a factor 1/e. To be precise, the shape of the energy curve is E = E₀e^−t/τ, and so it’s an envelope resembling the A(t) curve below.

Indeed, remember, the energy of a wave is determined not only by its frequency (or wavelength) but also by its amplitude, and so we cannot assume the amplitude of a ‘photon wave’ is going to be the same everywhere. Just for the record: note that the energy of a wave is proportional to the frequency (doubling the frequency doubles the energy) but, when linking it to the amplitude, we should remember that the energy is proportional to the square of the amplitude, so we write E ∼ A².

The second mistake is that both ν and ω are the light frequency (expressed in cycles or radians respectively) of the light per second, i.e per time unit. So that’s not the number of cycles or radians that we should associate with the wavetrain! We should use the number of cycles (or radians) packed into one photon. We can calculate that easily from the value for the decay time τ. Indeed, for sodium light, which which has a frequency of 500 THz (500×10¹²oscillations per second) and a wavelength of 600 nm (600×10^–9meter), we said the radiation lasts about 3.2×10^–8seconds (that’s actually the time it takes for the radiation’s energy to die out by a factor 1/e, so the wavetrain will actually last (much) longer, but so the amplitude becomes quite small after that time), and so that makes for some 16 million oscillations, and a ‘length’ of the wavetrain of about 9.6 meter! Now, the energy of a sodium light photon is about 2eV (h·ν ≈ 4×10⁻¹⁵electronvolt·second times 0.5×10¹⁵cycles/sec = 2eV) and so we could say the average energy of each of those 16 million oscillations would be 2/(16×10⁶) eV = 0.125×10^–6 eV. But, from all that I wrote above, it’s obvious that this number doesn’t mean all that much, because the wavetrain is not likely to be shaped very regularly.

So, in short, we cannot say that h is the photon energy per cycle or that ħ is the photon energy per radian! That’s not only simplistic but, worse, false. Planck’s constant is what is is: a factor of proportionality for which there is no simple ‘arithmetic’ and/or ‘geometric’ explanation. It’s just there, and we’ll need to study some more math to truly understand the meaning of those two expressions for |L〉 and |R〉.

Having said that, and having thought about it all some more, I find there’s, perhaps, a more interesting way to re-write E = h·ν:

h = E/ν = (λ/c)E = T·E

T? Yes. T is the period, so that’s the time needed for one oscillation: T is just the reciprocal of the frequency (T = 1/ν = λ/c). It’s a very tiny number, because we divide (1) a very small number (the wavelength of light measured in meter) by (2) a very large number (the distance (in meter) traveled by light). For sodium light, T is equal to 2×10^–15seconds, so that’s two femtoseconds, i.e. two quadrillionths of a second.

Now, we can think of the period as a fraction of a second, and smaller fractions are, obviously, associated with higher frequencies and, what amounts to the same, shorter wavelengths (and, hence, higher energies). However, when writing T = λ/c, we can also think of T being another kind of fraction: λ/c can also be written as the ratio of the wavelength and the distance traveled by light in one second, i.e. a light-second (remember that light-seconds are measures of length, not of distance). The two fractions are the same when we express time and distance in equivalent units indeed (i.e. distance in light-second, or time in sec/c units).

So that links h to both time as well as distance and we may look at h as some kind of fraction or energy ‘density’ even (although the term ‘density’ in this context is not quite accurate). In the same vein, I should note that, if there’s anything that should make you think about h, is the fact that its value depends on how we measure time and distance. For example, if w’d measure time in other units (for example, the more ‘natural’ unit defined by the time light needs to travel one meter), then we get a different unit for h. And, of course, you also know we can relate energy to distance (1 J = 1 N·m). But that’s something that’s obvious from h‘s dimension (J·s), and so I shouldn’t dwell on that.

Hmm… Interesting thoughts. I think I’ll develop these things a bit further in one of my next posts. As for now, however, I’ll leave you with your own thoughts on it.

Note 1: As you’re trying to follow what I am writing above, you may have wondered whether or not the duration of the wavetrain that’s emitted by an atom is a constant, or whether or not it packs some constant number of oscillations. I’ve thought about that myself as I wrote down the following formula at some point of time:

h = (the duration of the wave)·(the energy of the photon)/(the number of oscillations in the wave)

As mentioned above, interpreting h as some kind of average energy per oscillation is not a great idea but, having said that, it would be a good exercise for you to try to answer that question in regard to the duration of these wavetrains, and/or the number of oscillations packed into them, yourself. There are various formulas for the Q of an atomic oscillator, but the simplest one is the one expressed in terms of the so-called classical electron radius r₀:

Q = 3λ/4πr₀

As you can see, the Q depends on λ: higher wavelengths (so lower energy) are associated with higher Q. In fact, the relationship is directly proportional: twice the wavelength will give you twice the Q. Now, the formula for the decay time τ is also dependent on the wavelength. Indeed, τ = 2Q/ω = Qλ/πc. Combining the two formulas yields (if I am not mistaken):

τ = 3λ²/4π²r₀c.

Hence, the decay time is proportional to the square of the wavelength. Hmm… That’s an interesting result. But I really need to learn how to be a bit shorter, and so I’ll really let you think now about what all this means or could mean.

Note 2: If that 1/√2 factor has nothing to do with some kind of rms calculation, where does it come from? I am not sure. It’s related to state vector math, it seems, and I haven’t started that as yet. I just copy a formula from Wikipedia here, which shows the same factor in front:

The formula above is said to represent the “superposition of joint spin states for two particles”. My gut instinct tells me 1/√2 factor has to do with the normalization condition and/or with the fact that we have to take the (absolute) square of the (complex-valued) amplitudes to get the probability.