Part IV · Quantum Mechanics — Chapter 4.1

What Classical Physics Cannot Do

Four failures, made quantitative, so that when quantisation arrives it arrives forced rather than proposed.

Tools you'll need  — Chapter 0.6 Worked example 1 above all: the Boltzmann distribution $p_{i}=\ee^{-\beta E_{i}}/Z$ derived from maximum entropy and two Lagrange multipliers, with $\beta=1/k_{B}T$ identified rather than assumed. Sections 3 and 4 both run on it and nothing else in statistical mechanics is used or needed. Chapter 0.8 §7.6 for the standing waves of a field in a box and §7.2 for the fact that the modes are independent oscillators. Section 2 here is that argument in three dimensions. Also §4.3 for the harmonic oscillator's energy as a sum of two squares, which §3.1 needs. Chapter 0.2 §5 for when an improper integral converges, which is the whole content of §3.3, and §4 for interchanging a sum and an integral. Chapter 0.3 §5 for dimensional analysis, used in §5.7 to say what $h$ is, and for $k_{e}=e^{2}/4\pi\epsilon_{0}=\alpha\hbar c$, which §7.3 needs. Chapter 0.9 §1 for Fourier series and Parseval's identity, used once in §5.6's grind box to evaluate a sum. Also §7.5 for the Cauchy distribution, whose divergence has the same structure as §3.3's. Chapter 2.5 §7.1 for the wave four-vector $k^{\mu}=(\omega/c,\vv k)$ and its null condition, §4.3 for $E=pc$ and $p\cdot p=0$, and Worked example 1, the Compton calculation, which §6.4 collects unchanged. Chapter 2.1 §3 for the electromagnetic wave equation, the dispersion relation $\omega=c\abs{\vv k}$ and the transversality of light, which supplies §2.4's factor of two. Chapter 3.9 §1.2 for the microwave background at $2.7255\ \mathrm{K}$, which Worked example 1 takes apart as a blackbody.

1 · What a classical prediction of the blackbody spectrum actually is

Announce the destination. Before anything can be predicted, there has to be something whose value does not depend on unrecorded details of the apparatus. We show that the light inside a hot cavity is such a thing: its spectrum is a universal function of frequency and temperature, identical for every cavity of every material and shape. The argument costs one page and it uses the second law of thermodynamics once.

1.1 · The object is a cavity, not a surface

A hot poker glows, and the colour of the glow depends on the poker. Iron, carbon and tungsten at the same temperature do not look the same, because each emits and absorbs different wavelengths with different efficiency, and what leaves the surface is the product of the two. So the radiation from a surface is not a universal object, and there is nothing clean to predict about it.

Change the object. Take a closed cavity with opaque walls, hold the whole thing at temperature $T$, and wait until nothing further changes. The radiation inside is then in thermal equilibrium with the walls, meaning that every part of the wall absorbs exactly as much as it emits, at every wavelength. If that were not so, the part in question would be heating or cooling, and the state would not be steady.

Let $u(\nu,T)\,\dd\nu$ be the energy per unit volume carried by radiation with frequency between $\nu$ and $\nu+\dd\nu$. To measure it, drill a hole small enough that letting radiation out does not disturb the equilibrium inside, and analyse what comes through. That is what a laboratory blackbody is, and it is why the spectrum of a furnace with a peephole is reproducible when the spectrum of a glowing surface is not.

1.2 · Kirchhoff's argument, in full

Here is the claim to be established. The function $u(\nu,T)$ depends on $\nu$ and on $T$, and on nothing else at all. Not on the material of the walls, not on the shape of the cavity, not on its volume. The argument is one Kirchhoff gave in 1859, and it is worth having complete, because it is what makes the problem well posed and because it uses exactly one thing this book has not built.

Take two cavities, $A$ and $B$, with walls of different materials and of different shapes, both held at the same temperature $T$. Now the move. Join them by a short pipe and place in the pipe a filter that transmits radiation in a narrow band of frequencies about $\nu$ and reflects everything else perfectly. Such a filter exchanges no net energy with the radiation, because what it does not transmit it returns.

Suppose $u_{A}(\nu,T)\gt u_{B}(\nu,T)$. Then more energy per unit time crosses the filter from $A$ to $B$ than from $B$ to $A$. The reason is that the flux through an aperture is proportional to the energy density on that side, and to nothing else the filter is sensitive to. So $B$ warms and $A$ cools, and two bodies that started at the same temperature have spontaneously arrived at different temperatures with nothing being done to either. That is forbidden. The same argument with $A$ and $B$ exchanged rules out $u_{B}\gt u_{A}$. Hence

and since $A$ and $B$ were arbitrary cavities, $u$ is a universal function. Let's note where each hypothesis entered. The filter's narrowness is what gives the conclusion frequency by frequency rather than merely for the total. Its perfect reflection outside the band is what stops it acting as a reservoir of its own.

1.3 · What is therefore to be predicted

The problem is now completely specified, which is a rare and valuable state of affairs. There is one function of two variables, it belongs to no particular material, and any theory of light and heat must produce it. Two things follow immediately and are worth writing down before any calculation, because they are what makes the calculation of §2 legitimate.

The cavity may be chosen for convenience. Since the answer does not depend on shape or material, take a cube with perfectly conducting walls. That is not an approximation to a real furnace. It is a different cavity with the same $u$.

The answer is an energy density, not an energy. $u$ carries no dependence on volume, so a cavity twice the size holds twice the energy at every frequency. This is what lets §2 count modes per unit volume and stop worrying about the box.

2 · Counting modes in a box

Announce the destination. We want the number of independent oscillations of the electromagnetic field per unit volume with frequency between $\nu$ and $\nu+\dd\nu$. Four steps: identify the standing waves, notice that their wavevectors form a lattice, count lattice points inside a ball, and change variable from wavenumber to frequency. The answer is $g(\nu)=8\pi\nu^{2}/c^{3}$ and every step is on this page, because this count is used three more times in the book and once here to produce a divergence.

2.1 · A field in a box is a set of independent oscillators

Chapter 0.8 §7.6 took a chain of masses and springs, let the spacing go to zero, and obtained a field obeying the wave equation. Its normal modes are the standing waves $\sin(n\pi x/L)$, and each one of them is an independent harmonic oscillator with its own frequency, its own amplitude and its own conserved energy. Chapter 2.1 §2 derived from Maxwell's equations that each Cartesian component of $\vv E$ in vacuum obeys exactly that wave equation, with $v=c$. So the argument transfers with one change: three dimensions instead of one.

In a cubical cavity of side $L$ with perfectly conducting walls, the tangential electric field must vanish on every wall. A standing wave

satisfies that condition exactly when $n_{x},n_{y},n_{z}$ are positive integers, since the sine vanishes at $0$ and at $L$ precisely then. Write the wavevector components as $k_{i}=n_{i}\pi/L$. The wave equation then forces $\omega=c\abs{\vv k}$, which is Chapter 2.1's dispersion relation, so

The reason to want the count is that each such mode is an independent oscillator. So the total energy in the cavity is a sum over modes of whatever energy each one carries, and §3 will supply the energy of one.

2.2 · The allowed wavevectors form a lattice

Read (4.1.3) as a statement about geometry rather than about waves. The allowed $\vv k$ are the points of a cubic lattice of spacing $\pi/L$, restricted to the octant where all three components are positive. Each lattice point sits at the corner of a small cube of volume $(\pi/L)^{3}$, and each such cube contains exactly one lattice point when the points are assigned to cubes consistently.

So counting modes with $\abs{\vv k}$ below some value $k$ becomes counting lattice points inside a sphere of radius $k$, restricted to one octant. Suppose $k$ is large compared with the lattice spacing $\pi/L$. That is the same as saying the wavelength is small compared with the box, which for visible light in a laboratory cavity means a ratio of about $10^{-5}$. Then the count is the volume available divided by the volume per point:

The $\tfrac18$ is the octant and the $\tfrac43\pi k^{3}$ is the ball. Nothing else has happened.

2.3 · Two polarisations

Each allowed $\vv k$ carries not one mode but two, and the reason is a fact about Maxwell's equations rather than about counting. In vacuum $\nabla\cdot\vv E=0$, and for a wave with wavevector $\vv k$ that condition reads $\vv k\cdot\vv E_{0}=0$: the field is transverse, as Chapter 2.1 §2 derived. The vectors perpendicular to a given $\vv k$ form a two-dimensional space, so there are two independent polarisations for each $\vv k$ and no more. Hence

2.4 · From wavenumber to frequency, and the density of modes

We want the count per unit frequency. So substitute $k=2\pi\nu/c$ from (4.1.3) before differentiating. That way the chain rule is done once and visibly, rather than twice and silently:

Differentiate with respect to $\nu$ to get the number of modes in a thin band, and divide by the volume $L^{3}$ because §1.3 established that the answer is a density:

modes per unit volume per unit frequency. Check the dimensions before going on: $[\nu^{2}/c^{3}]=\mathsf{T}^{-2}\cdot\mathsf{L}^{-3}\mathsf{T}^{3}=\mathsf{T}\,\mathsf{L}^{-3}$, so $g\,\dd\nu$ is a pure number per unit volume, as a mode count must be. The box has vanished from the answer, exactly as Kirchhoff's argument said it had to.

2.5 · What was thrown away, and how big it is

Two approximations were made in (4.1.4) and both deserve a number rather than a reassurance, because a divergence is about to be blamed on this formula.

The boundary of the ball. Replacing a lattice count by a volume miscounts the points near the spherical surface and near the three coordinate planes. Both errors are surface effects, so they scale as the square of the radius against the volume's cube, and the relative error therefore falls as $1/(kL)$. Counting the lattice points exactly for spheres of radius $M$ lattice spacings gives a ratio to (4.1.4) of $0.9774$ at $M=100$ and $0.99249$ at $M=300$. That is $1-2.26/M$ to within a percent in the coefficient, which is a $1/M$ correction with a coefficient of order one, exactly as predicted. For a cavity of $1\ \mathrm{cm}$ holding visible light, $M\sim10^{5}$ and the correction is at the level of $10^{-5}$.

The choice of boundary condition. Perfectly conducting walls gave sines. Periodic boundary conditions give complex exponentials instead, with $k_{i}=2\pi n_{i}/L$ and $n_{i}$ of either sign. The lattice spacing doubles, which divides the count by $8$, but the region is now the whole ball rather than one octant, which multiplies it by $8$. The two changes cancel exactly and (4.1.7) is unchanged. It had better be, since §1.2 proved that the walls cannot matter.

What this closes. The mode density (4.1.7) is the object the book has referred to and never built. It is used again in Chapter 4.18 for occupation numbers and in Part V for field quantisation, and it is derived here once, completely, so that it never has to be quoted.

3 · Rayleigh–Jeans, and a divergent integral

Announce the destination. We give each mode of §2 the energy classical statistical mechanics assigns it, multiply, and integrate. The multiplication takes one line. The integration does not converge, and the whole of §§4 and 5 exists to repair that. Along the way we set out the two experimental facts any correct formula must reproduce, so that when one is produced you can score it against targets that were fixed in advance.

3.1 · Equipartition, derived rather than quoted

Chapter 0.6's Worked example 1 maximised the entropy $S=-\sum_{i}p_{i}\ln p_{i}$ subject to normalisation and a fixed mean energy, and obtained $p_{i}=\ee^{-\beta E_{i}}/Z$ with $Z=\sum_{i}\ee^{-\beta E_{i}}$ and $\beta=1/k_{B}T$. We want the mean energy of one oscillator, so we need that result with the sum over states replaced by an integral. The reason for the replacement is that a classical oscillator's state is a continuous pair $(q,p)$ rather than a label from a list.

Making that replacement is a step worth naming as the place where classical statistical mechanics is quietly incomplete. A sum over discrete states becomes an integral over phase space only once you decide how much phase-space area counts as one state, and classical mechanics offers no such unit. Write the unknown unit as $\varsigma$, with the dimensions of $q$ times $p$, so that

Nothing below depends on $\varsigma$. The quantity we want is $\avg{E}=-\dv{}{\beta}\ln Z$, which is the second constraint of 0.6's Worked example 1 restated, and a constant multiplying $Z$ adds a $\beta$-independent constant to $\ln Z$. Differentiation kills that constant. Keep the symbol anyway. Section 5.7 shows that the new constant $h$ has exactly the dimensions of the missing unit, and the fact that classical physics had a hole of precisely this shape is not a coincidence.

A mode is a harmonic oscillator, so by Chapter 0.8 §4.3 its energy is a sum of two squares,

Both integrals in (4.1.8) then factorise, and each is a Gaussian. Rather than evaluate them, notice that the only thing needed is how each depends on $\beta$: in $\int\ee^{-\beta a x^{2}}\dd x$, substitute $x=y/\sqrt{\beta a}$, which pulls $(\beta a)^{-1/2}$ outside and leaves a $\beta$-independent integral. So each quadratic term contributes a factor $\beta^{-1/2}$ to $Z$, and

Now that we have $\ln Z$ as a function of $\beta$, the mean energy is one differentiation away, so differentiate:

per mode. Half of that comes from the kinetic term and half from the potential term, which is the general statement of equipartition: $\tfrac12k_{B}T$ for every coordinate the energy is quadratic in. The result is independent of $m$, of $\omega$ and therefore of $\nu$, and that independence is the whole problem. A mode oscillating at $10^{18}\ \mathrm{Hz}$ is assigned exactly as much energy as one at $10\ \mathrm{Hz}$.

3.2 · The Rayleigh–Jeans law

Energy density equals modes per unit volume per unit frequency, times energy per mode. Multiply (4.1.7) by (4.1.11):

This is the Rayleigh–Jeans law, and everything in it was derived: the mode count from Maxwell's equations and a box, the energy per mode from maximum entropy. There is no fitted parameter anywhere. It is also, at low frequency, right. Right to the precision of the best measurements, in fact, which is what §5 turns into a boundary condition. Radio astronomers use it daily to convert a measured intensity into a brightness temperature, and Worked example 1 computes where it stops being safe.

3.3 · The integral, and the number

The total energy per unit volume is the area under (4.1.12):

Chapter 0.2 §5 classified exactly this: $\int^{\infty}\nu^{s}\dd\nu$ converges only for $s\lt-1$, and here $s=2$. So the failure is not a large number, an inaccuracy or a régime of validity. It is a divergence, and the phrase ultraviolet catastrophe is a literal description of where it lives. It lives in the high-frequency tail, which is exactly the region where the mode count grows.

Say the number, because "infinite" is easier to shrug off than an arithmetic comparison. Cut the integral off at a frequency $\nu_{c}$ and it evaluates to $8\pi k_{B}T\nu_{c}^{3}/3c^{3}$. At $T=300\ \mathrm{K}$:

Cut off at	Rayleigh–Jeans energy density	The measured total, for comparison
$3\ \mathrm{THz}$ (far infrared)	$3.48\times10^{-8}\ \mathrm{J\,m^{-3}}$	$6.13\times10^{-6}\ \mathrm{J\,m^{-3}}$
$750\ \mathrm{THz}$ (violet)	$0.543\ \mathrm{J\,m^{-3}}$
$3\times10^{16}\ \mathrm{Hz}$ (soft X-ray)	$3.48\times10^{4}\ \mathrm{J\,m^{-3}}$

By the time the integral has been carried to violet light it has overshot the measured total by a factor of $8.9\times10^{4}$, and every further decade of frequency multiplies the answer by $10^{3}$ again with no end. Taken at its word, (4.1.13) says that a sealed box of air at room temperature contains unbounded electromagnetic energy, and that opening it would produce an unbounded flash of X-rays. Nobody has observed that.

One clarification, because the mistake is common. The divergence is not caused by the box being idealised, by the walls being perfect conductors, or by the failure of the lattice count near the origin. Those are all effects at low $k$, and §2.5 showed they are relative corrections of order $1/(kL)$ that shrink as $\nu$ grows. Everything that goes wrong goes wrong in the tail, and every correction available is at the other end.

3.4 · Two measured facts, fixed as targets before the derivation

Set the scoreboard before playing. Two features of real cavity radiation were established experimentally in the 1890s, and they are stated here so that §5's formula can be tested against numbers it was not built from.

One conversion is needed to use the second of these, and it is geometry rather than physics: the flux out of a small hole is $F=cu_{\text{tot}}/4$, derived in the grind box. The factor of four is the average of $\cos\theta$ over an outward hemisphere, weighted by solid angle.

Grind box A — why the flux through a hole is $c/4$ times the energy density

Inside the cavity the radiation is isotropic, so the energy density travelling into solid angle $\dd\Omega$ about any direction is $u_{\text{tot}}\dd\Omega/4\pi$. That energy moves at speed $c$. Through a hole of area $A$ in the wall, the component of the flow along the outward normal is $c\cos\theta$, and only directions in the outward hemisphere contribute.

$$F \;=\; \int_{\text{hemisphere}} \frac{u_{\text{tot}}}{4\pi}\,c\cos\theta\,\dd\Omega \;=\; \frac{u_{\text{tot}}c}{4\pi}\int_{0}^{2\pi}\!\!\dd\varphi\int_{0}^{\pi/2}\!\cos\theta\,\sin\theta\,\dd\theta.$$

The $\varphi$ integral gives $2\pi$. The $\theta$ integral is $\int_{0}^{1}s\,\dd s=\tfrac12$ with $s=\sin\theta$. Hence

$$F \;=\; \frac{u_{\text{tot}}c}{4\pi}\cdot 2\pi\cdot\tfrac12 \;=\; \frac{c\,u_{\text{tot}}}{4}. \;\checkmark$$

So $\sigma T^{4}=c\,u_{\text{tot}}/4$, and a formula for $u_{\text{tot}}$ delivers $\sigma$.

4 · Einstein's A and B coefficients

Announce the destination. We put atoms in the cavity and require that the radiation and the atoms be in equilibrium with each other. Writing every process that changes the atomic populations at a rate proportional to a population, imposing balance, and feeding in Chapter 0.6's Boltzmann weights produces an expression for $u$. The first attempt, with the two processes anybody would write down, fails. And it fails in a way that forces a third process nobody had seen. That is the argument's best moment, and it is why this route is worth taking rather than a shorter one.

This route is longer than the usual textbook one, and the reason for taking it belongs in front of it. The short derivation runs on Bose–Einstein statistics, which requires the symmetrisation of identical-particle states, which requires the notion of an identical particle. That last item is the final chapter of this part, so the short road would make the argument circular. Einstein's road uses nothing but rate bookkeeping and the Boltzmann distribution you already own, and it produces a physical prediction on the way that the short road hides.

4.1 · The setting, and the one identification being made

Put into the cavity a dilute gas of atoms, each of which has two energy levels $E_{1}\lt E_{2}$, and let $N_{1}$ and $N_{2}$ be the numbers of atoms in each. Dilute means the atoms interact with the radiation and not with each other, so nothing but radiation moves atoms between levels. Write the level spacing as

where $\nu$ is the frequency of the radiation this transition exchanges energy with, and $h$ is a constant.

This is an identification, not a derivation, and it is worth being precise about what is being assumed and what is not. Two things are already established. First, by §1.2 the radiation's spectrum is universal. So a relation between the atom's level spacing and the frequency it couples to cannot depend on which atom is used, because it is a property of the field. Second, by Chapter 0.3 §5's dimensional analysis, no product of powers of $\nu$ and $c$ has the dimensions of energy: $[\nu^{a}c^{b}]=\mathsf{L}^{b}\mathsf{T}^{-a-b}$, which carries no mass whatever the exponents. Some new constant is therefore unavoidable.

What is assumed is that one constant suffices, meaning that the ratio $(E_{2}-E_{1})/\nu$ is the same number at every frequency rather than a function of frequency. That assumption is tested three separate times below. Sections 5.5 and 5.6 check it against two measurements the formula was not fitted to, and §6 measures the same constant twice more in experiments involving no cavity at all. If it were false, those checks would disagree.

4.2 · Detailed balance, stated as the assumption it is

4.3 · The populations, from Chapter 0.6

Before any rates, the ratio of populations is fixed. Chapter 0.6's Worked example 1 gives $p_{i}\propto\ee^{-\beta E_{i}}$ with $\beta=1/k_{B}T$, and since the numbers of atoms in each level are proportional to those probabilities,

Two remarks before this gets used. Suppose the levels were degenerate, with $g_{1}$ and $g_{2}$ states of the same energy. Every state still carries the same Boltzmann weight, so the ratio would be $(g_{2}/g_{1})\ee^{-h\nu/k_{B}T}$, and every conclusion below acquires a $g$ in a predictable place. Section 4.6 says where. Take $g_{1}=g_{2}=1$ throughout.

And note the shape of (4.1.15) against something you compute routinely. The fraction of atoms in the upper level is $N_{2}/(N_{1}+N_{2})=1/\big(1+\ee^{h\nu/k_{B}T}\big)$, which is a logistic function of $h\nu/k_{B}T$. It is the same expression, symbol for symbol, as a two-outcome logistic model with linear predictor $-h\nu/k_{B}T$. The Boltzmann ratio is the odds of finding an atom excited, and $-h\nu/k_{B}T$ is the log-odds. That is not an analogy. Both come from normalising two exponential weights. The identity is useful because it means the intuition you have for how fast an odds ratio collapses as the linear predictor grows is exactly the right intuition for how fast the upper level empties as the frequency rises.

4.4 · Two processes, and why they do not close

Now the rates. The atoms are in a bath of radiation of spectral density $u(\nu,T)$ at the transition frequency. Consider processes whose rate is proportional to the population available to undergo them, which is what "rate constant" means and what mass action assumes.

Absorption. An atom in level $1$ absorbs and goes to level $2$. Nothing happens without radiation present, so the rate is proportional to $u$ as well as to $N_{1}$:

Spontaneous emission. An excited atom drops to level $1$, emitting. An excited atom in the dark still decays, which is what a fluorescence lifetime is. So this rate is proportional to $N_{2}$ and to nothing else:

That is everything a two-process model can offer: one rate up, one rate down. So impose detailed balance on just these two, and see what spectrum comes out:

Now test it, and the test is the cheapest one available: hold $\nu$ fixed and let $T$ grow without bound. The right-hand side of (4.1.18) approaches the finite number $A_{21}/B_{12}$. But at fixed $\nu$ and large enough $T$ we are in the régime $h\nu\ll k_{B}T$, where §3.2's Rayleigh–Jeans law holds and is measurably correct, and it says $u=8\pi\nu^{2}k_{B}T/c^{3}$, which grows without bound. Note carefully how little of Rayleigh–Jeans is being used: only that $u\to\infty$ as $T\to\infty$, not the coefficient. Even that weak statement is enough.

So the two-process scheme is not merely inaccurate. It is inconsistent, for every choice of $A_{21}$ and $B_{12}$. Something is missing, and the failure says what kind of thing. The emission rate must be able to grow without bound as $u$ does, so there must be an emission process whose rate depends on $u$.

4.5 · The third process, forced

Ask what can be added. The scheme's ground rules are that every rate is proportional to a population and, if it requires radiation, to $u$. Level $2$ has one population and there is one field density, so the only term available that is not already present is

radiation already present makes an excited atom drop and emit. This is stimulated emission, and the point to hold onto is that it was not added to the model to explain anything. It was the only repair the model's own rules permitted, and it is required by consistency alone. Einstein wrote it down in 1917. No experiment had suggested it, and none would show it directly for decades.

4.6 · The balance, solved, and the first coefficient relation

With all three processes, detailed balance reads

We want $u$, so collect its terms on one side and divide:

The population ratio is the one piece of physics still missing from that expression, so bring it in now. Substituting (4.1.15) gives $N_{1}/N_{2}=\ee^{+h\nu/k_{B}T}$, and hence

Everything about the spectrum is now in two ratios of rate constants, and two conditions fix them. Here is the first, and it is one line.

Let $T\to\infty$ at fixed $\nu$. The exponential tends to $1$, so the denominator of (4.1.22) tends to $B_{12}-B_{21}$ and $u$ tends to $A_{21}/(B_{12}-B_{21})$. But $u$ must grow without bound, by the argument of §4.4. A ratio grows without bound only if its denominator goes to zero. Hence

Absorption and stimulated emission have exactly the same rate constant. Nothing was fitted, and one limit did the whole job. (With degeneracies the same line gives $g_{1}B_{12}=g_{2}B_{21}$, which is the promised predictable place.) Write $B$ for the common value, and (4.1.22) collapses to

Notice that this also closes §4.1's identification, which was left standing as an assumption. $A_{21}/B$ is a property of the atom and carries no temperature, so the entire temperature dependence of $u$ at fixed $\nu$ sits in that exponent. Two atoms coupling to the same $\nu$ with different level spacings would give the one universal $u$ two different temperature dependences, and §1.2 forbids that. So $E_{2}-E_{1}$ is a function of $\nu$ alone, and the same function for every atom. That means the only thing §4.1 assumed beyond what is now proved is that the function is a straight line through the origin. That is a narrow assumption rather than a broad one, and the three checks in §5.5, §5.6 and §6 are tests of the linearity alone.

Recap, since ten lines have gone by. Three rate constants went in, one of them belonging to a process introduced because the other two could not balance. Detailed balance plus Boltzmann weights gave (4.1.22). One limit then gave $B_{12}=B_{21}$. What is left is the complete shape of the spectrum, with a single unknown constant $A_{21}/B$ standing in front of it. That constant is a property of the atom, and the next section shows it cannot be.

5 · The Planck spectrum, and what $h$ is

Announce the destination. One constant is left in (4.1.24), and one condition remains unused: at low frequency the answer must reduce to the Rayleigh–Jeans law, which is right there. Imposing it fixes the constant and finishes the spectrum. We then differentiate the result and integrate it, and compare both against the two numbers §3.4 fixed in advance. Finally we say, carefully, what has and has not been established, and what $h$ is at this stage of the book.

5.1 · Rayleigh–Jeans as a boundary condition

Write $y=h\nu/k_{B}T$ and take $y\to0$ in (4.1.24). Since $\ee^{y}-1=y+y^{2}/2+\cdots$, the leading behaviour is

That must agree with (4.1.12), which was derived from Maxwell's equations and maximum entropy with nothing fitted, and which is measurably correct at low frequency. Setting the two equal,

The factor $k_{B}T$ cancels from both sides. It had to cancel, since $A_{21}/B$ is a ratio of rate constants and cannot depend on the temperature of a cavity the atom happens to be sitting in. That cancellation is a check, not a step. What remains is

and note what has just been established: the rate at which an excited atom decays in the dark is not independent of the rate at which it absorbs. The two are locked together, by a factor built from $h$, $c$ and the transition frequency and containing nothing about the atom at all. An atom that absorbs strongly must fluoresce quickly, and the $\nu^{3}$ says that a transition at twice the frequency decays eight times faster for the same absorption strength.

5.2 · The Planck spectrum

Substitute (4.1.27) into (4.1.24):

This is the Planck spectrum. Read it as the mode count of §2 times an average energy per mode:

The mode count is untouched. It came from Maxwell's equations and a box, and neither of those is in question. What has changed is the energy per mode, and the way it has changed is the whole content. The quantity $k_{B}T$ has been replaced by something that equals $k_{B}T$ when $h\nu\ll k_{B}T$ and falls off exponentially when $h\nu\gg k_{B}T$. High-frequency modes are not sharing the heat. That is what cures §3.3's divergence, and it is what equipartition got wrong.

5.3 · What this argument did, and what it did not do

Be exact about the logical shape here, because it is unusual and a careful reader will notice.

What was derived. That $u$ must have the form $\text{const}/(\ee^{h\nu/k_{B}T}-1)$ came from detailed balance plus Boltzmann weights plus the existence of a third process, and nothing else. That is a genuine derivation of the shape, and it is the part that could not have been guessed. Similarly, $B_{12}=B_{21}$ came from a limit and is exact.

What was supplied from outside. The constant in front came from requiring agreement with the Rayleigh–Jeans law, in the régime where that law is correct. That is the very law being replaced. So the classical theory was not merely refuted here. It was used, as a boundary condition, to fix two numbers in a formula whose form came from elsewhere. That is a legitimate move, and it is what matching asymptotics always does. But it is not the same as deriving a law from nothing, and the chapter would be dishonest to let it read as though it were.

What repairs it. Chapter 4.18 §5 derives (4.1.28) a second time and by a completely different route. That route counts the states of the field with Bose–Einstein statistics, which needs the symmetrisation postulate and therefore cannot be done here. The two routes share no step. When the second one arrives, the constant in front is computed rather than matched, and the present argument stops being a fit and becomes one of two independent confirmations. Until then, the honest statement is this. The form is derived, the normalisation is matched, and three checks follow immediately that were not used in the matching: Wien's 1896 distribution law as a limit (§5.4), the displacement constant (§5.5), and Stefan's $\sigma$ (§5.6).

5.4 · The two limits, and the two empirical laws they reproduce

Low frequency, $h\nu\ll k_{B}T$. By construction $\ee^{y}-1\to y$ and (4.1.28) becomes $8\pi\nu^{2}k_{B}T/c^{3}$, the Rayleigh–Jeans law. There is no information in that. It is the boundary condition being satisfied.

High frequency, $h\nu\gg k_{B}T$. Now $\ee^{y}-1\to\ee^{y}$ and

This is Wien's distribution law, the formula fitted to the high-frequency measurements in 1896 and known to fail in the infrared. It was not used as an input anywhere above, so this is the first real prediction. The same expression reproduces both of the empirical laws that were known before 1900, each in its own régime, and interpolates smoothly between them. How badly each fails outside its own régime is arithmetic:

$y=h\nu/k_{B}T$	Rayleigh–Jeans $/$ Planck	Wien $/$ Planck
$0.1$	$1.052$	$0.095$
$1$	$1.718$	$0.632$
$3$	$6.362$	$0.950$
$10$	$2.20\times10^{3}$	$0.99996$
$20$	$2.43\times10^{7}$	$1.000000$

The two ratios are $(\ee^{y}-1)/y$ and $1-\ee^{-y}$, both read straight off (4.1.28). The crossover sits at $y$ of order one, so the peak of the spectrum is exactly where neither classical formula works. That is why the two camps could each claim their own half of the data, and why neither could claim the middle.

5.5 · Wien's displacement, with the arithmetic

The first thing §3.4 fixed as a target. Differentiate (4.1.28) with respect to $\nu$ at fixed $T$ and set the result to zero. Substituting $y=h\nu/k_{B}T$ makes this a question about the single function $y^{3}/(\ee^{y}-1)$, because everything else is a constant multiplier. So the answer will be a pure number, and $\nu_{\max}$ will be proportional to $T$ before any arithmetic is done. Using the quotient rule,

The denominator is finite and non-zero for $y\gt0$, so set the numerator to zero. Divide by $y^{2}\ee^{y}$, which is legitimate since $y\neq0$ at a maximum:

This is transcendental and has no closed-form root, so it has to be solved numerically. The method is Chapter 0.3's first-order Taylor expansion run backwards. Near any point, $f(y+\delta)\approx f(y)+f'(y)\delta$. The step that would send $f$ to zero is therefore $\delta=-f(y)/f'(y)$, and iterating that step is Newton's method. With $f(y)=y-3+3\ee^{-y}$ and $f'(y)=1-3\ee^{-y}$, starting from $y=2.8$:

Three steps, and the root is $y_{\max}=2.8214393721$. Hence $h\nu_{\max}=2.8214393721\,k_{B}T$, or

which is $5.8789257576\times10^{10}\ \mathrm{Hz\,K^{-1}}$. Compare §3.4's measured $5.8789\times10^{10}\ \mathrm{Hz\,K^{-1}}$. Nothing was fitted to this. The only number put into the formula was matched at low frequency, and the peak sits at $y=2.82$, where that low-frequency law is wrong by a factor of $(\ee^{2.8214}-1)/2.8214=5.60$.

Problem 2 does the same calculation per unit wavelength, gets the root of $5(1-\ee^{-x})=x$, and recovers Wien's constant $b=2.8977719552\times10^{-3}\ \mathrm{m\,K}$ against the measured $2.897771955\times10^{-3}$. It also settles the trap that $\nu_{\max}\lambda_{\max}\neq c$.

5.6 · Stefan–Boltzmann, with the arithmetic

The second target. Integrate (4.1.28) over all frequencies. The substitution $y=h\nu/k_{B}T$ is the one to make, because it removes every trace of $T$ from the integrand and leaves it in the prefactor, where its power can be read off. With $\nu=k_{B}Ty/h$ and $\dd\nu=(k_{B}T/h)\dd y$,

The $T^{4}$ has appeared before the integral has been touched, which is the point of doing the substitution first: the fourth power comes from $\nu^{3}\dd\nu$ and nothing else, and would be there whatever the remaining integral turned out to be. That integral is a pure number, and the grind box evaluates it as $\pi^{4}/15=6.4939394$. So

What §3.4 asked for was the flux out of a hole rather than the energy density inside, and Grind box A supplies the conversion $F=cu_{\text{tot}}/4$. So

Put the numbers in, piece by piece, so the arithmetic is checkable. We have $\pi^{5}=306.0196848$, then $k_{B}^{4}=(1.380649\times10^{-23})^{4}=3.633566649\times10^{-92}$, then $c^{2}=8.987551787\times10^{16}$, and finally $h^{3}=(6.62607015\times10^{-34})^{3}=2.909163220\times10^{-100}$. The numerator is $2\times306.0196848\times3.633566649\times10^{-92}=2.2238858\times10^{-89}$ and the denominator is $15\times8.987551787\times10^{16}\times2.909163220\times10^{-100}=3.9219383\times10^{-82}$, so

Against §3.4's measured $5.670374419\times10^{-8}$. Every digit shown agrees. The reason it is that good rather than merely good is that since 2019 $h$, $c$ and $k_{B}$ are exact by definition, so the modern "measured" $\sigma$ is this arithmetic. Before 2019 the comparison was a genuine confrontation between an experiment and a formula, and it agreed to the precision of the experiment. Two measurements, both fixed in §3.4 before the derivation ran, both recovered from one formula containing one constant.

Grind box B — the integral $\int_{0}^{\infty}y^{3}(\ee^{y}-1)^{-1}\dd y$, and $\sum n^{-4}$ from Parseval

Step 1. Expand the denominator. For $y\gt0$ write $\dfrac{1}{\ee^{y}-1}=\dfrac{\ee^{-y}}{1-\ee^{-y}}$ and expand the geometric series in $\ee^{-y}$, which converges because $0\lt\ee^{-y}\lt1$:

$$\frac{1}{\ee^{y}-1} \;=\; \ee^{-y}\sum_{m=0}^{\infty}\ee^{-my} \;=\; \sum_{n=1}^{\infty}\ee^{-ny}.$$

Step 2. Interchange sum and integral. Every term is non-negative, so the partial sums increase monotonically and the interchange is the case Chapter 0.2 §4 discussed. (The theorem that makes it airtight is monotone convergence, proved in Chapter 4.3.)

$$\int_{0}^{\infty}\frac{y^{3}\dd y}{\ee^{y}-1} \;=\; \sum_{n=1}^{\infty}\int_{0}^{\infty}y^{3}\ee^{-ny}\,\dd y \;=\; \sum_{n=1}^{\infty}\frac{3!}{n^{4}} \;=\; 6\sum_{n=1}^{\infty}\frac{1}{n^{4}},$$

the inner integral being $\int_{0}^{\infty}y^{3}\ee^{-ny}\dd y=3!/n^{4}$ by three integrations by parts, or by differentiating $\int\ee^{-ny}\dd y=1/n$ three times with respect to $n$.

Step 3. The sum, from Parseval. Take $f(x)=x^{2}$ on $[-\pi,\pi]$ and expand it in Chapter 0.9 §1's orthonormal basis $\big\{(2\pi)^{-1/2},\ \pi^{-1/2}\cos nx,\ \pi^{-1/2}\sin nx\big\}$. The sine coefficients vanish because $x^{2}$ is even, and

$$c_{0}=\frac{1}{\sqrt{2\pi}}\int_{-\pi}^{\pi}\!x^{2}\dd x=\frac{2\pi^{3}}{3\sqrt{2\pi}}, \qquad c_{n}=\frac{1}{\sqrt{\pi}}\int_{-\pi}^{\pi}\!x^{2}\cos nx\,\dd x=\frac{4\pi(-1)^{n}}{n^{2}\sqrt{\pi}},$$

the second by two integrations by parts, the boundary terms surviving only in the second one because $\sin n\pi=0$ and $\cos n\pi=(-1)^{n}$. Parseval's identity says $\norm{f}^{2}=\sum_{k}\abs{c_{k}}^{2}$. That is Chapter 0.5 §2, transferred to this basis by the completeness that Chapter 0.9 §1.3 flagged as quoted and Chapter 4.3 proves. Here $\norm{f}^{2}=\int_{-\pi}^{\pi}x^{4}\dd x=\tfrac25\pi^{5}$. So

$$\frac{2\pi^{5}}{5} \;=\; \frac{4\pi^{6}}{9\cdot 2\pi} \;+\; \sum_{n=1}^{\infty}\frac{16\pi^{2}}{n^{4}\pi} \;=\; \frac{2\pi^{5}}{9} \;+\; 16\pi\sum_{n=1}^{\infty}\frac{1}{n^{4}}.$$

Rearranged, $16\pi\sum n^{-4}=2\pi^{5}\big(\tfrac15-\tfrac19\big)=\tfrac{8\pi^{5}}{45}$, so $\sum n^{-4}=\pi^{4}/90$. Therefore

$$\int_{0}^{\infty}\frac{y^{3}\dd y}{\ee^{y}-1} \;=\; 6\cdot\frac{\pi^{4}}{90} \;=\; \frac{\pi^{4}}{15} \;=\; 6.4939394023. \;\checkmark$$

(Checked numerically: the same integral evaluated by quadrature gives $6.4939394023$.)

5.7 · What $h$ is, at this stage

Say plainly what has and has not been claimed about the new constant, because everything later in this part depends on not over-reading it now.

Its dimensions. From (4.1.14), $h$ is an energy divided by a frequency, so $[h]=\mathsf{ML}^{2}\mathsf{T}^{-2}\cdot\mathsf{T}=\mathsf{ML}^{2}\mathsf{T}^{-1}$. Those are the dimensions of action, which is energy times time, and which is also momentum times length and angular momentum. Chapter 1.2 built a whole subject out of a quantity with those dimensions, and Chapter 1.3 gave phase space its area. So the coincidence is not one. It is the subject of Chapter 4.10.

Its value. Fitted. The formula's shape was derived and its normalisation matched, and the one remaining number is whatever makes (4.1.28) agree with the measured spectrum: $h=6.62607015\times10^{-34}\ \mathrm{J\,s}$. Since 2019 that value is exact by definition, because the kilogram is now defined by fixing it. But that is a statement about the SI, not about physics, and in 1900 it was a fit to a curve.

What it is not, yet. Nothing above says that energy comes in indivisible units, that the oscillator's energy levels are discrete, or that light is made of particles. All that has been shown is that the average energy of a cavity mode is $h\nu/(\ee^{h\nu/k_{B}T}-1)$ rather than $k_{B}T$. Planck himself resisted the stronger readings for years and was right to want more evidence. Section 6 supplies it, from two experiments that involve no cavity.

One structural remark, and it is the reason to have kept $\varsigma$ in §3.1. The classical partition function (4.1.8) needed a unit of phase-space area and classical mechanics had none to offer. The constant that has just appeared has exactly the dimensions of phase-space area, meaning momentum times length, which is Chapter 1.3's $\dd q\,\dd p$. It is the missing unit, and Chapter 4.10 makes the statement precise by showing that a classical orbit enclosing area $\mathcal{A}$ in phase space corresponds to about $\mathcal{A}/h$ quantum states.

temperature T / K

5772 K

cut-off ν_c / ν_max

10^1.00

microwave background, 2.7255 K

room, 300 K

filament, 2800 K

solar surface, 5772 K

peak of the plotted Planck curve: nu_max = 3.39332e+14 Hz nu_max / T = 5.878926e+10 (formula: 5.878926e+10)

integral to cut-off: Planck 8.397622e-1 J/m^3 Rayleigh-Jeans 9.681505e+2 J/m^3 ratio 1.15e+3

a T^4 = 8.397622e-1 J/m^3 implied sigma = c(Planck integral)/4T^4 = 5.670374e-8 (measured 5.670374e-08) [SETTLED]

6 · Light carries momentum in parcels: the photoelectric effect and Compton

Announce the destination. Two experiments, neither involving a cavity. The first measures $h$ a second time, from a straight line whose slope has nothing to do with §5, and shows that light delivers energy in amounts fixed by frequency. The second shows that the parcel carries momentum too, and does so by a calculation Chapter 2.5 has already done in full. Together they promote the relation $p^{\mu}=\hbar k^{\mu}$, which Chapter 2.5 could only note as a structural resemblance, into a law.

6.1 · The classical prediction, stated before it is contradicted

Shine light on a clean metal surface in vacuum and electrons come off. Classically light is a wave carrying energy density $\propto E^{2}$ spread continuously over the surface, and an electron bound in the metal is a charge in that oscillating field, driven and gradually shaken loose. Three predictions follow, and it is worth writing them down before the data, because each is definite.

(i) The maximum kinetic energy should rise with intensity. A stronger field shakes harder, so an electron freed by a bright beam should leave faster than one freed by a dim beam of the same colour.

(ii) There should be no threshold frequency. Any colour should work eventually: energy accumulates, and enough of it accumulated is enough of it accumulated.

(iii) There should be a measurable delay at low intensity. Estimate it. A beam of intensity $I$ falling on an atom of radius $r\approx10^{-10}\ \mathrm{m}$ delivers power $I\pi r^{2}$ to that atom. With $I=1\ \mathrm{W\,m^{-2}}$, which is bright, the area is $\pi(10^{-10})^{2}=3.14\times10^{-20}\ \mathrm{m^{2}}$ and the power is $3.14\times10^{-20}\ \mathrm{W}$. To accumulate the $2.28\ \mathrm{eV}=3.65\times10^{-19}\ \mathrm{J}$ needed to free an electron from sodium takes

That estimate is generous to the classical picture in every direction. It assumes the atom collects the whole beam falling on its geometric cross-section and loses none of it. Reduce the intensity by a factor of a million, which is still an easily visible beam, and the delay becomes $1.16\times10^{7}\ \mathrm{s}$, or four and a half months.

6.2 · What is measured

Every one of (i), (ii) and (iii) contradicts a prediction of §6.1, and (iv) is not a prediction of §6.1 at all. A wave picture supplies no reason for a universal slope, because it supplies no constant with the right dimensions.

6.3 · $K_{\max}=h\nu-\phi$, and a second measurement of $h$

The reading that accounts for all four facts at once is that light of frequency $\nu$ delivers its energy to one electron in an indivisible amount $h\nu$, and it is the same $h$, on the hypothesis being tested. Then an electron that absorbs one such amount and pays the minimum energy $\phi$ required to escape the metal emerges with at most

Check each fact against it.

• Intensity sets how many amounts arrive per second and not how large each is, so $K_{\max}$ is independent of intensity. That is fact (i).
• Below $\nu_{0}=\phi/h$ no single amount is large enough, and no waiting helps, because energy is not being accumulated. Those are facts (ii) and (iii).
• And (4.1.40) is a straight line in $\nu$ with slope $h$ and intercept $-\phi$, so the slope is universal and the intercept is the material's. That is fact (iv).

Measuring the slope measures $h$. In practice one measures the stopping potential $V_{\text{stop}}$, the retarding voltage that just prevents any electron from reaching the collector, so that $eV_{\text{stop}}=K_{\max}$ and

Take the two extreme mercury lines from a sodium surface. At $253.7\ \mathrm{nm}$, $\nu=1.181681\times10^{15}\ \mathrm{Hz}$ and $V_{\text{stop}}=2.60704\ \mathrm{V}$. At $435.8\ \mathrm{nm}$, $\nu=6.87913\times10^{14}\ \mathrm{Hz}$ and $V_{\text{stop}}=0.56498\ \mathrm{V}$. The slope is

so $h=e\times4.135667\times10^{-15}=1.602176634\times10^{-19}\times4.135667\times10^{-15} =6.626069\times10^{-34}\ \mathrm{J\,s}$. That is §5.7's value to six figures, obtained from a photocell and a voltmeter with no furnace, no spectrometer and no thermodynamics anywhere in the apparatus. Two measurements of the same constant by methods sharing no equipment is what turns §4.1's identification from an assumption into a result. Problem 3 does the full five-point fit.

6.4 · Compton scattering, collected from Chapter 2.5

Chapter 2.5's Worked example 1 did this calculation in full and this chapter reuses it unchanged, exactly as that chapter said it would. Here is the setup it used. A particle of energy $E$ and momentum $E/c$, massless so that $p\cdot p=0$ by 2.5 §4.3, strikes an electron at rest. Conservation of four-momentum is imposed. The unknown electron four-momentum is isolated on one side and squared, which replaces it by $m_{e}^{2}c^{2}$ and removes the recoil direction from the problem entirely. The result is

Two things about that formula matter here and did not matter in Chapter 2.5.

The one quoted input was $h$, and it is no longer quoted. Chapter 2.5 flagged $E=hc/\lambda$ as an import and named this chapter as where it would be argued for. It has now been argued for twice, in §5 and §6.3, so the flag is discharged and the Compton formula stands on this book's own results.

The independence of $\lambda$ is the whole evidential weight. A classical wave shakes an electron at the driving frequency, and an oscillating charge re-radiates at the frequency it is oscillating at. So the scattered light has the same wavelength as the incident light, and the predicted shift is exactly zero, at every angle, for every material. The measured shift is not small. At $90^{\circ}$ on $71\ \mathrm{pm}$ X-rays it is $2.43\ \mathrm{pm}$, a $3.4\%$ change, and at $180^{\circ}$ it is $4.85\ \mathrm{pm}$, or $6.8\%$. This is not a discrepancy in a coefficient. It is a measured effect where the classical prediction is zero, and it is reproduced to the precision of the experiment by treating the encounter as a two-body collision with the four-momentum bookkeeping of Part II applied without modification.

6.5 · $p^{\mu}=\hbar k^{\mu}$, promoted from a suspicion to a law

Chapter 2.5 §7.1 ended with a deliberately unfinished observation. It had built the wave four-vector $k^{\mu}=(\omega/c,\vv k)$ and proved it a four-vector by counting crests. It then noted that for light in vacuum $\omega=c\abs{\vv k}$, so $k\cdot k=0$. Setting that beside $p\cdot p=0$ from its §4.3, it remarked that the two objects have the same transformation law, the same null condition and the same relation between their time and space components. It closed with "Chapter 4.1 supplies the missing constant." Here it is.

A plane light wave travelling along the unit vector $\hat n$ has, from Maxwell's equations and Chapter 2.5 §4.3, energy and momentum in the ratio $E=\abs{\vv p}c$ with $\vv p$ along $\hat n$. So whatever amount of energy $E$ the wave delivers, it delivers with it four-momentum

The two are proportional, with the same unit vector, because both are built from the same propagation direction, so $p^{\mu}=(E/\omega)\,k^{\mu}$ and the entire question is what $E/\omega$ is. Sections 5 and 6.3 established that the energy is delivered in amounts $E=h\nu$, so $E/\omega=h\nu/2\pi\nu=h/2\pi$. Give that combination its own name, since every equation in the rest of this part carries it:

which is a change of units and nothing more. Here $h$ goes with $\nu$ and $\hbar$ goes with $\omega=2\pi\nu$, so that $E=h\nu=\hbar\omega$. Then

Read off the two components separately, because each is a statement that was measured independently. The time component is $E=\hbar\omega=h\nu$, which is §6.3's photoelectric line. The space component is $\vv p=\hbar\vv k$, so $\abs{\vv p}=\hbar\cdot2\pi/\lambda=h/\lambda$. That is precisely the momentum Chapter 2.5's Compton calculation assigned the incident radiation, and the calculation reproduced the measurement. So each half of (4.1.46) has its own experiment, and the two experiments are unrelated. That is what promotes the relation from a structural suspicion about two null four-vectors into a law.

One consequence is worth having now, since Part V leans on it. Because (4.1.46) is an equation between four-vectors, it holds in every inertial frame with the same $\hbar$: Lorentz-transform both sides and the relation is unchanged. So $\hbar$ is a Lorentz invariant, which a relation between a frame-dependent energy and a frame-dependent frequency had no obligation to be.

6.6 · What has not been shown

7 · Spectral lines, and why an orbiting electron is a catastrophe

Announce the destination. The last two failures are about atoms, and they are of a different kind from the first two. Those were numerical: a formula gave the wrong number, or infinity. These are structural: classical mechanics has no way to produce the form of what is observed, and the one classical model of an atom that reproduces the right energy scale destroys itself in sixteen picoseconds. We do the integral and give the number.

7.1 · The lines, and the arithmetic that fits them

Balmer noticed in 1885 that those five numbers are fitted by one formula. Take the reciprocal of each wavelength and divide by $\tfrac14-1/n^{2}$ for $n=3,4,5,6,7$ in turn:

$n$	$\lambda$ / nm	$1/\lambda$ / $\mathrm{m^{-1}}$	$\tfrac14-1/n^{2}$	ratio / $\mathrm{m^{-1}}$
$3$	$656.4614$	$1.523319\times10^{6}$	$0.138889$	$1.096790\times10^{7}$
$4$	$486.2721$	$2.056462\times10^{6}$	$0.187500$	$1.096780\times10^{7}$
$5$	$434.1692$	$2.303250\times10^{6}$	$0.210000$	$1.096785\times10^{7}$
$6$	$410.2892$	$2.437305\times10^{6}$	$0.222222$	$1.096787\times10^{7}$
$7$	$397.1198$	$2.518132\times10^{6}$	$0.229592$	$1.096786\times10^{7}$

The five ratios agree to nine parts in a million. That is not a fit with five parameters. It is one number, $R=1.096786\times10^{7}\ \mathrm{m^{-1}}$, reproducing five measurements. Rydberg then generalised it, and the generalisation predicts series nobody had looked for:

Setting $n_{1}=1$ predicts an ultraviolet series starting at $121.57\ \mathrm{nm}$, found by Lyman in 1906. Setting $n_{1}=3$ predicts an infrared series starting at $1875\ \mathrm{nm}$, found by Paschen in 1908. So (4.1.47) is not a curve-fit but a formula with predictive content, and it is left entirely unexplained here. Chapter 4.13 derives it, including the value of $R$, and that derivation is one of the three or four things quantum mechanics is believed for.

7.2 · Nothing classical produces whole numbers

Look at what (4.1.47) is made of: two integers and one constant. Now ask where a classical atom could get an integer from.

A charge moving in a Coulomb field has, by Chapter 1.3's Kepler problem, a continuous family of bound orbits: pick any energy $E\lt0$ and any angular momentum consistent with it, and there is an orbit. The orbital frequency is a continuous function of the energy, and a charge circling at frequency $f$ radiates at $f$ and its harmonics, by Chapter 2.1's wave equation with a periodic source. So a gas of classical atoms with energies spread over any range at all should emit a continuum. What is measured is a set of isolated lines whose positions are fixed to six figures and are the same for every hydrogen atom in the universe.

This is a different kind of failure from §3.3's, and worth naming as such. There, a calculation produced a number that was wrong. Here, the classical framework has no room for the observed structure: there is no continuous parameter to tune, no coefficient to adjust, and no limit in which integers emerge from (4.1.47)'s left-hand side. Integers appear in classical physics in exactly one place, as mode numbers of a wave confined to a region, which is §2.1's $n_{x},n_{y},n_{z}$. Following that hint is what the rest of Part IV does, with Chapter 4.13 supplying the two integers in (4.1.47) specifically.

7.3 · Larmor's formula, and the size of an atom

7.4 · The integral, and sixteen picoseconds

Model the atom as an electron in a circular orbit of radius $r$ about a proton. Two things follow before any radiation is considered, from Newton's second law and the Coulomb force alone. Balancing the centripetal acceleration against the Coulomb attraction gives $m_{e}v^{2}/r=k_{e}/r^{2}$, so $m_{e}v^{2}=k_{e}/r$. That makes the kinetic energy $k_{e}/2r$, and the total energy

Pause on the first of these before continuing, because it is the one thing the classical model gets right. At the measured radius, $E=-k_{e}/2r_{0}=-2.17987\times10^{-18}\ \mathrm{J}=-13.606\ \mathrm{eV}$, against a measured ionisation energy for hydrogen of $13.598\ \mathrm{eV}$. The classical picture has the energy scale of an atom right to a part in a thousand. Everything else about it is about to fail completely. That combination of right scale and wrong structure is what makes the failure interesting rather than merely wrong.

Now let it radiate. Substituting the acceleration into Larmor's formula,

At $r=r_{0}$ this is $4.669\times10^{-8}\ \mathrm{W}$, which for one atom is enormous: it is $13.6\ \mathrm{eV}$ of binding energy divided by $47\ \mathrm{ps}$. That crude estimate is the right order and the exact answer is three times smaller, because the power grows as the orbit shrinks.

Do it properly. The orbit's energy falls at the rate it radiates, so $\dv{E}{t}=-P$. Differentiate (4.1.48) with respect to $r$ and use the chain rule, since $r$ is what is changing:

That is separable, in the sense of Chapter 0.8 §2.1. So put every $r$ on one side and every $t$ on the other, and integrate from the initial radius to zero:

Both sides carry a minus sign and a factor of three, and both of those cancel. Solving that last equality for the time gives

The numbers, piece by piece. We have $c^{3}=2.694400\times10^{25}$, then $m_{e}^{2}=(9.1093837139\times10^{-31})^{2}=8.298087\times10^{-61}$, then $r_{0}^{3}=(5.29177\times10^{-11})^{3}=1.481845\times10^{-31}$, and finally $k_{e}^{2}=(2.307078\times10^{-28})^{2}=5.322607\times10^{-56}$. The numerator is $2.694400\times10^{25}\times8.298087\times10^{-61}\times1.481845\times10^{-31}=3.313168\times10^{-66}$ and the denominator is $4\times5.322607\times10^{-56}=2.129043\times10^{-55}$, so

Two derived checks on that number, both exact rather than numerical. The crude estimate $\abs{E(r_{0})}/P(r_{0})$ equals $3c^{3}m_{e}^{2}r_{0}^{3}/4k_{e}^{2}$, which is exactly $3t_{\text{fall}}$, so the ratio is three, as the arithmetic above showed. And the number of orbits completed is $\int\dd t/T(r)$ with $T=2\pi\sqrt{m_{e}r^{3}/k_{e}}$. Substituting $s=r^{3}$ turns that into an elementary integral and gives exactly $2t_{\text{fall}}/T(r_{0})$, which with $T(r_{0})=1.520\times10^{-16}\ \mathrm{s}$ is

So the classical hydrogen atom completes about two hundred thousand orbits and then does not exist, sixteen picoseconds after it is assembled. And the light it emits on the way is a chirp: the orbital frequency starts at $1/T(r_{0})=6.58\times10^{15}\ \mathrm{Hz}$, in the ultraviolet, and rises without bound as $r\to0$, sweeping continuously through every frequency above that. Compare with §7.1's five sharp lines whose positions are stable to six figures.

Both features are wrong, and they are wrong in different directions. The atom is not merely unstable on some long timescale. It is unstable on a timescale shorter than the period of any molecular vibration and shorter than the fastest chemical process anyone can name, so that no sample of hydrogen could ever be assembled. And the spectrum is not merely shifted. It is continuous where the measurement is discrete. Nothing about the model can be adjusted to repair either, because both come from the same two things: that a bound charge can have any energy, and that an accelerating charge radiates.

7.5 · The four failures, collected

Line them up with their numbers, because the list is the chapter's product.

What classical physics predicts	What is measured	Discrepancy
Cavity energy density $\int8\pi\nu^{2}k_{B}T/c^{3}\,\dd\nu$	$aT^{4}$, finite	Infinite against $6.13\times10^{-6}\ \mathrm{J\,m^{-3}}$ at $300\ \mathrm{K}$
Photoemission delay $\approx11.6\ \mathrm{s}$ at $1\ \mathrm{W\,m^{-2}}$, $K_{\max}$ rising with intensity, no threshold	Delay under $10^{-9}\ \mathrm{s}$, $K_{\max}$ independent of intensity, sharp threshold	$10^{10}$ in time, and two qualitative facts reversed
Compton shift $=0$ at every angle	$2.43\ \mathrm{pm}$ at $90^{\circ}$, independent of $\lambda$ and material	A measured effect against a prediction of zero
Continuous emission, and the atom collapses in $1.56\times10^{-11}\ \mathrm{s}$	Five sharp lines fitted by two integers, and atoms are stable	Structural: no continuous parameter can produce integers

Notice what the four have in common, because it is the thing Chapter 4.2 changes. In every case the classical theory assigns a continuous range of values to something the measurement finds restricted. The list is the energy of a cavity mode, the energy transferred to one electron, the momentum carried by a light beam, and the energy of a bound electron. And in every case the restriction is set by one constant with the dimensions of action, measured twice here by unrelated means and agreeing.

8 · Worked examples

9 · Your turn