Read · Light quanta: from entropy to an energy scale

§1 · classical energy allocation

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A derivation

A finite window cannot cure an infinite total

What happens if every radiation mode receives classical thermal energy?

The 6 printed paragraphs this passage explains

§1, paragraph 1: The setup for section 1: a space enclosed by perfectly reflecting walls holds gas molecules and electrons that move freely and collide like the molecules of a gas, and further electrons bound to fixed points by springlike forces. These bound electrons, called resonators, emit and absorb electromagnetic waves of definite periods.
§1, paragraph 2: On the current view of how light arises, the radiation in this space at equilibrium, worked out with Maxwell's theory, should be the same as black radiation, at least if resonators of every frequency that matters are present.
§1, paragraph 3: Setting the radiation aside, collisions with the gas require each resonator vibration along one line to have the mean energy (R/N)T, where R is the gas constant, N the number of molecules in a gram-equivalent and T the temperature. With more or less energy on average, the collisions would give energy to the gas or take it from the gas, so equilibrium needs exactly this mean energy.
§1, paragraph 4: Planck had found the corresponding equilibrium between a resonator and the radiation around it, assuming the radiation is as disordered as possible: the resonator's mean energy fixes the radiation's energy density at its frequency.
§1, footnote 2: The footnote cites Planck's paper in Annalen der Physik, volume 1, page 99, 1900.
§1, paragraph 5: For the radiation at each frequency to neither gain nor lose energy on balance, both conditions must hold, and together they fix its density. The result disagrees with experience, and it also means the picture allows no definite division of energy between ether and matter: the wider the range of resonator frequencies, the larger the radiation energy, without limit.

Counting cavity modes and giving each mode the classical mean energy $k_B T$ produces a spectral energy density $\rho_\nu$ . A density must be multiplied by a frequency width to give energy per unit volume.

u_\nu = \frac{8\,\pi\,\nu^{2}}{c^{3}}\,k_B\,T

The classical energy density per unit frequency is eight pi frequency squared times thermal energy, divided by light speed cubed.

U\left(\nu_{c}\right) = \int_{0}^{\nu_{c}} u_\nu\,\mathrm{d}\nu

U\left(\nu_{c}\right) = \frac{8\,\pi\,k_B\,T}{3\,c^{3}}\,\nu_{c}^{3}

The energy density below the cutoff frequency is eight pi times Boltzmann's constant times temperature, divided by three times the speed of light cubed, times the cutoff frequency cubed.

At fixed positive temperature, doubling the cutoff multiplies this finite-range total by eight. Extending the model to arbitrarily high frequencies produces an unbounded total. The resonator laboratory keeps that failure distinct from the finite portion displayed on screen.

Open the foundation: Adding continuously

Show every step here: A finite window cannot cure an infinite total

A tool for this step

Quantities and units

Can a number of metres be compared with a number of seconds?

"The particle moved 6" says nothing until the unit is added: 6 micrometres, or 6 seconds? A physical quantity is a number times a unit, and the number alone changes when the unit does. Six micrometres is 0.006 millimetres: the same length, written with another unit.

Quantities of different kinds cannot be added or compared: 3 metres plus 2 seconds has no meaning. So in a correct equation every term, on both sides, carries the same units. Checking that is the quickest test a formula can pass or fail.

Take the Brownian diffusion coefficient, $D = k_BT/(6\pi\eta a)$ . Boltzmann's constant is in joules per kelvin, T in kelvins, the viscosity η in pascal-seconds and the radius a in metres. The units work out to square metres per second, which is what a diffusion coefficient must be. The worked example below goes through it.

Units also tell apart quantities that share a letter. The light paper's $\rho_\nu$ is an energy per unit volume per unit of frequency, joules per cubic metre per hertz, not per unit of wavelength: two different quantities with different units. And its L is the speed of light in §§1 and 2 but an absorbed energy in §9. A unit check starts from what the quantity is, not from its letter.

Worked example: Checking that D comes out in square metres per second

$k_BT$ : joules per kelvin times kelvins is joules, which are newton-metres.
$6\pi\eta a$ : 6π has no units, and pascal-seconds times metres is (newtons per square metre) × seconds × metres, which is newton-seconds per metre.
Divide: newton-metres divided by newton-seconds per metre is square metres per second.
With numbers: 1.380649 × 10⁻²³ × 290.15 ÷ (6π × 1.35 × 10⁻³ × 5 × 10⁻⁷) = 3.15 × 10⁻¹³ m²/s. That is the modern Boltzmann constant (set modern-si-2019) with the paper's water at 17 °C and its particle.

JavaScript is off, so the controls below cannot change anything. The construction shows its first setting, and its closing paragraph, “What it shows, in words”, describes the rest.

Try it: cancelling units

Choose a formula. Each factor is written in the base units kilogram, metre, second and kelvin, the powers are added, and the sum is compared with the units the result must have. Two of the formulas are wrong on purpose.

Formula

D = k_BT/(6πηa)

k_B, Boltzmann's constant: J/K, which is kg m² s⁻² K⁻¹, on top.
T, the temperature: K, on top.
6π, six times pi, is a pure number with no units.
η, the viscosity: Pa s, which is kg m⁻¹ s⁻¹, underneath.
a, the particle's radius: m, underneath.

Adding the powers gives m² s⁻¹. A diffusion coefficient must be m² s⁻¹: the units agree.

What it shows, in words

A formula's units can be checked without any numbers: write every factor in kilograms, metres, seconds and kelvins, add the powers of the factors on top, subtract those underneath, and halve those under a square root. The Stokes-Einstein diffusion coefficient comes out in square metres per second. Leave out the radius and a metre too many survives; leave off the square root in the spread and the result is an area, not a length. The check catches both slips, but it cannot catch a wrong pure number such as 6π.

Where this lesson stops

This lesson stops at checking units. Converting the papers' own units into SI units is a separate step, with its own conventions.

Symbols first, in the paper's letters. T is the absolute temperature. R is the gas constant and N the number of real molecules in one gram-equivalent, so R/N is the gas constant for a single molecule; this explanation also writes it $k_B$ , a later name. L in the paper is the speed of light, written c below. ν is a frequency, and $\rho_\nu\,d\nu$ is the energy per unit volume carried by the radiation whose frequency lies between ν and ν + dν.

Start from the kinetic theory of gases. A resonator, an electron bound to a point in space, exchanges energy with the gas molecules by collisions. In equilibrium the mean energy of one of its straight-line vibrations comes out as $\frac{R}{N}T$ , which is $k_BT$ (p. 134).
Planck had related such a resonator to the radiation around it. A resonator of natural frequency ν in equilibrium with radiation has, per vibration component, the mean energy $\frac{L^3}{8\pi\nu^2}\rho_\nu$ (p. 135).
If the radiation at frequency ν is neither steadily gaining nor steadily losing energy, both conditions hold at once. Set the two mean energies equal: $\frac{R}{N}T=\frac{L^3}{8\pi\nu^2}\rho_\nu$ .
Multiply both sides by $8\pi\nu^2/L^3$ to leave the density alone: $\rho_\nu=\frac{R}{N}\frac{8\pi\nu^2}{L^3}T$ (p. 136). In the explanatory letters that is the first formula below.

u_\nu = \frac{8\,\pi\,\nu^{2}}{c^{3}}\,k_B\,T

The classical energy density per unit frequency is eight pi frequency squared times thermal energy, divided by light speed cubed.

Check the units. $k_BT$ is an energy. $\nu^2/c^3$ has units of $\mathrm{s}^{-2}$ divided by $\mathrm{cm}^3\mathrm{s}^{-3}$ , which is seconds per cubic centimetre. The product is energy times seconds per volume: energy per unit volume and per unit frequency, which is what a spectral density must be.
A density becomes an energy only after it is multiplied by a frequency width. The energy per unit volume in all frequencies from 0 up to a cutoff $\nu_c$ is the integral of $\rho_\nu$ over that range.
Everything except $\nu^2$ stays constant during the integration, so it comes outside the integral. The integral of $\nu^2$ from 0 to $\nu_c$ is $\nu_c^3/3$ , which gives the second formula.

U\left(\nu_{c}\right) = \int_{0}^{\nu_{c}} u_\nu\,\mathrm{d}\nu

U\left(\nu_{c}\right) = \frac{8\,\pi\,k_B\,T}{3\,c^{3}}\,\nu_{c}^{3}

The energy density below the cutoff frequency is eight pi times Boltzmann's constant times temperature, divided by three times the speed of light cubed, times the cutoff frequency cubed.

Worked comparison at one fixed temperature: a cutoff twice as high gives $2^3=8$ times the energy, and a cutoff ten times as high gives $10^3=1000$ times. The total keeps growing as the cutoff grows.
Remove the cutoff, as the paper does at the end of §1: the integral of $\nu^2\,d\nu$ from 0 to infinity is infinite, so the radiation energy in the space would be infinite.
The paper draws two conclusions from this relation: it disagrees with experience, and in this picture there can be no definite distribution of energy between the ether and matter.
Keep names with their dates. A radiation law of this $\nu^2T$ form had been proposed by Rayleigh in June 1900; Jeans's correction of its coefficient came in 1905, parallel to this paper rather than behind it. The phrase 'ultraviolet catastrophe' for the divergence is Ehrenfest's, from 1911, and the paper does not use it.
A plotted curve that stops at some frequency shows where the calculation stopped, not a property of light. The resonator laboratory reports the finite part it draws, and refuses to give the unbounded total as a number, rather than showing it as zero.

A tool for this step

Adding continuously

How do you add up something that varies continuously?

Chance spread along a line is measured per unit length: a density. Where the density is the same all along a stretch, the chance of the stretch is the density times its length. A particle equally likely to be anywhere in a 4 µm stretch has density 0.25 per micrometre, and the chance of finding it in the first 1.5 µm is 0.25 × 1.5 = 0.375.

Where the density changes from place to place, cut the line into narrow strips. Each strip contributes its height times its width, as if the density were flat across it. Add the strips, then cut them narrower and add again. For a smooth curve the totals settle on one value, and that value is the integral:

\int_a^b p(x)\,dx = \lim_{\Delta x \to 0} \sum_i p(x_i)\,\Delta x

The integral of p of x from a to b is the value that the sum of p of x i times delta x settles on as delta x shrinks.

The units multiply. A probability density is a probability per metre, so a strip of it, times a width in metres, is a plain probability. For a density the whole area must be 1, because the particle has to be somewhere. The Brownian paper's §4 writes exactly that for the chances of a jump Δ, ∫φ(Δ) dΔ = 1, and the light-quanta paper's §3 adds entropy over every frequency, S = v∫φ dν.

One more tool recurs in the papers: integration by parts. It is the product rule for derivatives added up over an interval, and it moves a derivative from one factor to the other at the price of an endpoint term:

\int_a^b u\frac{dv}{dx}\,dx = [uv]_a^b - \int_a^b v\frac{du}{dx}\,dx

The integral of u times the derivative of v equals u v at b minus u v at a, minus the integral of v times the derivative of u.

The endpoint term, uv at b minus uv at a, is part of the answer. It disappears only when uv is zero at both ends, as it is for a density that falls to zero far out on both sides.

Worked example: A rising density, added up in narrower strips

A particle's density rises steadily across a 4 µm stretch, from 0 at the left end to 0.5 per micrometre at the right. The whole area is a triangle, ½ × 4 × 0.5 = 1.
Four strips 1 µm wide, each taking the height at its left edge: 0 + 0.125 + 0.25 + 0.375 = 0.75.
Eight strips 0.5 µm wide: the heights add to 1.75, and times 0.5 that is 0.875.
Sixteen strips 0.25 µm wide: 0.9375. The shortfall halves each time the strips do: 0.25, then 0.125, then 0.0625.
The totals settle on 1, the area of the triangle. That settled value is the integral.

A flat density gives the same total for every cut, which is why the uniform stretch above needed no strips. A curved one, such as the Brownian bell curve, does, and the integral is the value the totals settle on.

The same triangle, cut two ways

A density rising from 0 to 0.5 per micrometre across 4 μm. Each strip is as tall as the curve at its own left edge, which is why every strip falls short and why the shortfall shrinks when the strips do.

4 strips add to 0.75; 8 strips add to 0.875. The triangle itself is 1. Cut narrower and the total climbs toward it without ever passing it, and the value it climbs toward is the integral.

What it shows, in words

Both drawings are the same density, rising in a straight line from 0 at the left end to 0.5 per micrometre at the right, across 4 micrometres. The first is cut into 4 strips of 1 micrometre each and the second into 8 strips of 0.5.

Each strip is as tall as the line at its own left edge, so each leaves a wedge uncovered above it. The 4 strips add to 0.75 and the 8 strips to 0.875, against the triangle's area of 1. Narrower strips leave smaller wedges, so the totals climb toward 1 and never past it. That limit is what the integral names.

What it does not show: these strips take the height at their left edge, which is one choice among several. Taking the right edge would overshoot 1 and climb down to it instead, and taking the middle of each strip would land much closer from the start. All three settle on the same 1, and that they agree is the fact the integral rests on. Nothing in either drawing demonstrates it.

Where this lesson stops

An integral is a total of height times width over a range, in the units of both. The widths and any endpoint term are part of the calculation, not decoration on the integral sign.

Explore the equations in this step

Read each operation, check its units and assumptions, or open the mathematical step behind it. These are modern teaching equations, not the equations as printed.

Explore the equation · Modern model notation

The classical allocation of energy to the spectrum

u_\nu = \frac{8\,\pi\,\nu^{2}}{c^{3}}\,k_B\,T

grows as squared, times and , over cubed.

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Each term and operation, in words

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The spectral energy density equals eight pi times the frequency squared over the speed of light cubed, times k B times the temperature.

Give every mode of the radiation the average energy that classical theory assigns, Boltzmann's constant times the temperature, and count the modes per unit volume per unit frequency: the density grows as the frequency squared without limit.

Model assumptions and every term’s meaning

Each radiation mode in equilibrium carries the classical average energy, Boltzmann's constant times the temperature.
These are modern teaching equations in SI notation; the paper writes L for c, and R/N for Boltzmann's constant.

Spectral energy density: The same quantity section 4 describes with Wien's law, here computed from classical theory. Read the prerequisite
Frequency: Squared: the number of modes per unit frequency grows with the frequency squared, so the density does too. Read the prerequisite
Speed of light: Cubed, from counting standing waves in a box; the paper writes it L. Read the prerequisite
Boltzmann's constant: With the temperature, the classical average energy per mode; R over N in the paper. Read the prerequisite
Temperature: The density is proportional to it at every frequency in this allocation. Read the prerequisite
No upper limit: Nothing in the formula stops the growth at high frequency, so the total energy over all frequencies diverges. That is the difficulty section 1 names, before any quantum is proposed. Read the prerequisite

Every term’s units were checked when this page was built. That checks the units, not the model. This equation is written for this edition in modern notation, not transcribed from the paper.

Explore the equation · Modern model notation

Add up the density below a cutoff

U\left(\nu_{c}\right) = \int_{0}^{\nu_{c}} u_\nu\,\mathrm{d}\nu

is the integral of over , from 0 to .

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Each term and operation, in words

No live values

This equation states a relation. No instrument on this page computes its terms.

Read the equation aloud in words

The energy density below cutoff frequency nu c is the integral of the spectral density from zero to nu c.

A density per unit frequency becomes an energy per unit volume only when it is added up over a range of frequencies. Here the range runs from zero to a cutoff.

Model assumptions and every term’s meaning

Each radiation mode in equilibrium carries the classical average energy, Boltzmann's constant times the temperature.
A cutoff frequency; the model counts only frequencies below it.
These are modern teaching equations in SI notation; the paper writes L for c, R/N for Boltzmann's constant, and β for B.

Add up over frequency: From zero to the cutoff: density times frequency width, summed. Read the prerequisite
What it asserts: The energy per volume is the area under the density curve up to the cutoff. Read the prerequisite
Cutoff frequency: The highest frequency the sum includes. Read the prerequisite
Spectral energy density: Radiation energy per cubic metre per unit of frequency. Read the prerequisite
Frequency: The frequency of the radiation. Read the prerequisite
Energy density below the cutoff: Energy per cubic metre in all frequencies up to the cutoff. Read the prerequisite

Every term’s units were checked when this page was built. That checks the units, not the model. This equation is written for this edition in modern notation, not transcribed from the paper.

Explore the equation · Modern model notation

The total grows as the cutoff cubed

U\left(\nu_{c}\right) = \frac{8\,\pi\,k_B\,T}{3\,c^{3}}\,\nu_{c}^{3}

is eight pi times times over three cubed, times cubed.

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Explain this equation

Each term and operation, in words

No live values

This equation states a relation. No instrument on this page computes its terms.

Read the equation aloud in words

The energy density below the cutoff frequency is eight pi times Boltzmann's constant times the temperature, divided by three times the speed of light cubed, times the cutoff frequency cubed.

Integrating the classical density, which grows as the frequency squared, gives a total that grows as the cutoff cubed. Doubling the cutoff multiplies it by eight, and with no cutoff it has no limit.

Model assumptions and every term’s meaning

Each radiation mode in equilibrium carries the classical average energy, Boltzmann's constant times the temperature.
A cutoff frequency; the model counts only frequencies below it.
These are modern teaching equations in SI notation; the paper writes L for c, R/N for Boltzmann's constant, and β for B.

The coefficient: Fixed by the temperature alone. Read the prerequisite
Cutoff cubed: Double the cutoff and this is eight times larger. Read the prerequisite
Three c cubed: The three comes from integrating frequency squared. Read the prerequisite
c cubed: From counting standing waves in a box. Read the prerequisite
8π times Boltzmann's constant times T: From the classical density's coefficient. Read the prerequisite
Coefficient times cutoff cubed: Finite for any cutoff, unbounded as the cutoff grows. Read the prerequisite
What it asserts: A finite window gives a finite total, but no window is singled out, and the full total is infinite. Read the prerequisite
Boltzmann's constant: With the temperature, the classical average energy per mode; R over N in the paper. Read the prerequisite
Cutoff frequency: The highest frequency the sum includes. Read the prerequisite
Speed of light: The paper writes it L. Read the prerequisite
Temperature: The temperature of the radiation in equilibrium. Read the prerequisite
Energy density below the cutoff: Energy per cubic metre in all frequencies up to the cutoff. Read the prerequisite

Every term’s units were checked when this page was built. That checks the units, not the model. This equation is written for this edition in modern notation, not transcribed from the paper.

Modern qualifications

$k_B$ and $c$ are modern explanatory notation. The paper writes $R/N$ for the thermal constant and $L$ for the speed of light. Historical names and later terminology are not additional evidence for the model.

Assumptions and limits: A finite window cannot cure an infinite total

Assumed here

Classical equilibrium assigns mean energy k_B T to each radiation mode.
The number of modes per unit volume and per unit frequency grows as the square of the frequency.

What this does not establish

The allocation is a classical model, not the spectrum measured across all frequencies.
A finite cutoff is where the calculation stops, not a property of light.

Source context: German source · English · Interlinear gloss · Facsimile

An assumption

Two assumptions under the equilibrium of section 1

What does the equilibrium of §1 assume about the electrons and about the radiation?

The 2 printed paragraphs this passage explains

§1, footnote 1: The footnote says this assumption amounts to supposing that gas molecules and electrons have equal mean kinetic energies at the same temperature, which Drude assumed when he derived the ratio of heat conduction to electrical conduction in metals.
§1, footnote 3: The footnote defines radiation as disordered as possible: when the electric field at a point is expanded in a long Fourier series, the probabilities of the amplitudes and phases are independent of one another. The radiation here comes closer to that the more each pair of them depends on emission and absorption by particular groups of resonators.

§1 puts gas molecules, free electrons and bound electrons in a box with mirror walls, and lets the free electrons collide like the molecules of a gas. A footnote says what that amounts to: at equilibrium, gas molecules and electrons have equal mean kinetic energies. Drude, Einstein notes, used the same assumption to derive the ratio of heat conduction to electrical conduction in metals.

That equality is what fixes the mean energy of each straight-line vibration of a resonator. A resonator whose average energy were higher or lower would, through the collisions, keep giving energy to the gas or taking it, so equilibrium needs:

E=\frac{R}{N}T

The mean energy E of one resonator vibration equals R over N times the temperature T.

The second assumption belongs to Planck's relation between a resonator and the radiation around it, which §1 uses. Planck derived it for radiation that is 'as disordered as possible', and a footnote defines the phrase. At one point of the space, expand one component Z of the electric field over a long time T in a Fourier series of sine waves with amplitudes A and phases α. Repeat the expansion from starting moments chosen at random: the amplitudes and phases come out differently each time, with a statistical probability for each combination.

Z=\sum_{\nu=1}^{\infty}A_\nu\sin\left(2\pi\nu\frac{t}{T}+\alpha_\nu\right)

Z equals the sum over nu from one to infinity of A nu times the sine of two pi nu t over T plus alpha nu.

The radiation is as disordered as possible when that probability splits into separate factors, one for each amplitude and one for each phase:

f(A_1,A_2,\ldots,\alpha_1,\alpha_2,\ldots)=F_1(A_1)F_2(A_2)\ldots f_1(\alpha_1)f_2(\alpha_2)\ldots

The joint probability density of all the amplitudes and phases equals a product of separate factors, one for each amplitude and one for each phase.

Then the value of any one amplitude or phase tells nothing about the others. Einstein adds that the radiation in his box comes closer to this the more each pair of an amplitude and its phase depends on emission and absorption by its own particular group of resonators. Neither assumption is derived in §1. The footnotes state them, and the equilibrium, with the unbounded total energy it leads to at the end of §1, rests on both.

Open the foundation: Probability and independence

Show every step here: Two assumptions under the equilibrium of section 1

Letters: E is the mean energy of one straight-line vibration of a resonator (the paper writes it with a bar), R the gas constant, N the number of real molecules in a gram-equivalent and T the absolute temperature. In the footnote, Z is one component of the electric field at a point, T there is a long stretch of time, A the amplitudes and α the phases of the Fourier series.

The first assumption: in the box, free electrons collide with gas molecules the way gas molecules collide with each other.
Its content, from the footnote: at a common temperature the mean kinetic energy of an electron equals the mean kinetic energy of a gas molecule. Drude had made the same assumption to derive the ratio of the thermal to the electrical conductivity of metals.
A resonator's electron shares in that equality. Kinetic theory gives a gas molecule the mean kinetic energy of its forward motion as (3/2)(R/N)T, so the resonator electron has that mean kinetic energy too.
Split the electron's motion into three vibrations along lines at right angles. Each gets a third of the kinetic energy: (1/3) × (3/2)(R/N)T = (1/2)(R/N)T.
In a vibration the time averages of kinetic and potential energy are equal, so each vibration's mean energy is twice its kinetic part: 2 × (1/2)(R/N)T = (R/N)T. That is (R/N)T ÷ (3/2)(R/N)T = 2/3 of the kinetic energy of a free monatomic gas molecule, as the paper says.
If a resonator had more than that on average, collisions would drain the excess into the gas; if less, collisions would feed it. Equilibrium therefore needs exactly:

E=\frac{R}{N}T

The mean energy E of one resonator vibration equals R over N times the temperature T.

Worked value, with today's value of R/N, $1.38\times10^{-23}$ joule per kelvin, which the paper does not use: at 300 kelvin, $\overline{E}=1.38\times10^{-23}\times300\approx4.14\times10^{-21}$ joule. The frequency does not enter, so a resonator of any frequency has the same mean energy.
The second assumption comes with Planck's relation, which §1 borrows to connect a resonator to the radiation. Planck derived it for radiation that is as disordered as possible, and the footnote says what that means.
Pick a point in the box and one component Z of the electric field there. Over a time T much longer than every period of vibration that matters, write Z as a sum of sine waves. (Here T is that stretch of time, not the temperature.)

Z=\sum_{\nu=1}^{\infty}A_\nu\sin\left(2\pi\nu\frac{t}{T}+\alpha_\nu\right)

Z equals the sum over nu from one to infinity of A nu times the sine of two pi nu t over T plus alpha nu.

Each term has an amplitude $A_\nu$ , zero or more, and a phase $\alpha_\nu$ between 0 and 2π. Start the stretch of time at a different moment, chosen at random, and the amplitudes and phases come out different.
Doing that many times gives a statistical probability for each combination of amplitudes and phases, written as a density f of all of them together.
The disorder condition: f is a product of separate factors, one for each amplitude and one for each phase.

f(A_1,A_2,\ldots,\alpha_1,\alpha_2,\ldots)=F_1(A_1)F_2(A_2)\ldots f_1(\alpha_1)f_2(\alpha_2)\ldots

The joint probability density of all the amplitudes and phases equals a product of separate factors, one for each amplitude and one for each phase.

What the product means: learning one amplitude or phase changes nothing about the chances of any other. Worked case: if the first amplitude lies above some value in half of the expansions, and the first phase lies in the first half of its circle in half of them, then disorder gives both at once in 1/2 × 1/2 = 1/4 of the expansions.
When the condition holds only roughly: Einstein says it holds the better, the more each pair of an amplitude and its phase comes from the emission and absorption of its own particular group of resonators.
Neither assumption is proved in §1. They are the ground under the equilibrium from which §1 draws its conclusion: that on this picture the radiation energy grows without limit as higher frequencies are allowed.

A tool for this step

Temperature and thermal energy

What does a temperature say about the jostling of molecules and grains?

The air in a room at 20 °C is a crowd of molecules moving in every direction and colliding billions of times a second. Warm the room and, on average, they move faster. The kinetic theory of heat reads temperature this way: through the average energy of motion of the particles, taken over very many of them.

For any particle in equilibrium with its surroundings, the average energy of motion along each direction is the same:

\left\langle \frac{1}{2}\,m\,v_x^{2}\right\rangle = \frac{1}{2}\,k_B\,T

The average of one half m v x squared equals one half k B T.

Here m is the particle's mass, $v_x$ its speed along x, T the absolute temperature, and $k_B$ = 1.38 × 10⁻²³ joules per kelvin is Boltzmann's constant. It equals R/N, the gas constant shared out per molecule. The Brownian paper writes RT/N where a modern text writes $k_BT$ ; the paper's own letter k means the viscosity, not this constant.

The average is the point. One molecule's energy of motion changes at every collision, from almost nothing to several times the average. The temperature fixes how energy is shared out over many particles. It is not the energy of any one of them.

Nothing in the rule mentions size. A grain ten billion times heavier than a molecule has the same average energy of motion, so it moves more slowly, by the square root of the mass ratio. That is why a suspended grain belongs to the same thermal story as a dissolved molecule, as §1 of the paper insists.

Worked example: A nitrogen molecule and a half-micrometre grain at 20 °C

At T = 293 K, $k_BT$ = 1.381 × 10⁻²³ × 293 = 4.05 × 10⁻²¹ J, so the average energy of motion along one direction is half of that, 2.02 × 10⁻²¹ J.
A nitrogen molecule has a mass of 4.65 × 10⁻²⁶ kg. Setting $\frac{1}{2} m v_x^2$ equal to $\frac{1}{2} k_BT$ gives a typical speed along one direction of √(4.05 × 10⁻²¹ ÷ 4.65 × 10⁻²⁶), about 295 metres a second.
A grain of radius 0.5 μm and density 1200 kg/m³ has a mass of 6.3 × 10⁻¹⁶ kg, about 1.4 × 10¹⁰ times the molecule's. The same average energy gives it about 2.5 millimetres a second.
In water that motion is turned about within some 70 nanoseconds: the grain's mass divided by its drag coefficient 6πηa. That is far too brief to follow, so what a microscope sees is the net result of the zigzag.

This is why the Brownian paper asks for a displacement over a time rather than a speed: the thermal speed is real, but it turns about far faster than any observer can watch.

JavaScript is off, so the controls below cannot change anything. The construction shows its first setting, and its closing paragraph, “What it shows, in words”, describes the rest.

Try it: one spread of speeds, two scales

At any moment, the speeds of particles along one axis are spread in a bell curve. The temperature sets its width, and the width depends on the particle's mass. Label the same curve for a molecule or for a grain.

Label the axis for

At 293 K, a nitrogen molecule moves along one axis with a spread of about 295 m/s. The curve does not change; only the numbers on its axis do, by a factor of about 120 000.

Shaded in the middle, within half a width either way (−148 to 148 m/s): the particles that have, at that moment, less than a quarter of the average energy of motion along the axis, about 38 per cent of them.
Outlined at the two ends, beyond two widths (past −590 or 590 m/s): those with more than four times the average, about 5 per cent.

Collisions move each particle about the whole curve.

What it shows, in words

At 293 kelvin a nitrogen molecule's speed along one axis is spread with a width of about 295 metres a second, and a half-micrometre grain's with a width of about 2.5 millimetres a second. Measured in its own width the spread is the same bell curve for both, with the same average energy of motion along the axis, 2.02 × 10⁻²¹ joules. At any moment about 38 per cent of the particles have less than a quarter of that average and about 5 per cent more than four times it. The temperature fixes the spread, not the energy of any one particle.

Where this lesson stops

This lesson stops at the average energy of motion. How the jostling of many molecules makes a grain wander, and how far, is §4 of the Brownian paper.

Modern qualifications

Drude's electron theory of metals (Annalen der Physik, 1900) treated the conduction electrons as a gas sharing the thermal energy of the metal, which is the use the Drude footnote points to. For Planck's relation §1 cites Ann. d. Phys. 1, p. 99, 1900; that paper derives the relation under a statistical condition on the radiation that Planck called natural radiation, and the footnote examined here is Einstein's own statement of the condition in terms of Fourier amplitudes and phases. On the plate (p. 135) the second α in the sentence after the product is set as an italic x, a misprint the sense of the sentence corrects. Whether cavity radiation meets the condition is a question §1 does not take up.

Assumptions and limits: Two assumptions under the equilibrium of section 1

Assumed here

Free electrons collide with the gas molecules as gas molecules collide with each other.
At equilibrium, gas molecules and electrons have equal mean kinetic energies (the footnote to the setup of §1).
Planck's relation between a resonator and the radiation around it holds for radiation that is as disordered as possible (the footnote that defines it).

What this does not establish

The passage explains what the footnotes state; it does not test whether real radiation in a cavity meets the disorder condition.
Einstein cites Drude's theory of metals in support; the passage does not assess that theory.

Earlier step: A finite window cannot cure an infinite total

Source context: German source · English · Interlinear gloss · Facsimile

References

A. Einstein, Über einen die Erzeugung und Verwandlung des Lichtes betreffenden heuristischen Gesichtspunkt. Annalen der Physik (4), 17, 132–148 (1905).

A. Einstein, On the motion of particles suspended in liquids at rest required by the molecular-kinetic theory of heat. Annalen der Physik (4), 17, 549–560 (1905), §§1–5. Bibliographic pointer; this preview is not a source transcription or translation.

A. Einstein, Zur Elektrodynamik bewegter Körper. Annalen der Physik (4), 17, 891–921 (1905).

A. Einstein, Does the inertia of a body depend upon its energy content? Annalen der Physik (4), 18, 639–641 (1905). External 1923 Perrett–Jeffery translation, electronically transcribed by John Walker; its notation was modernized. A reference for this explanatory preview, not this edition’s translation or its pinned facsimile.

R. A. Millikan, A Direct Photoelectric Determination of Planck's "h". Physical Review (2), 7 (3), 355–388 (1916).

26 foundation readings sit behind this argument.

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