Read · Brownian motion: from wandering to a measurable law · Results

Brownian motion: from wandering to a measurable law

Particles just large enough to see under a microscope must wander, Einstein argues, and how far they wander in a given time would let us count molecules. Read the argument, and open any step it leaves out.

The paper's results

Pages 550, 551, 553

The osmotic pressure of suspended particles

By the molecular-kinetic theory of heat, $n$ suspended particles held in a volume $V^*$ press on the wall that holds them with the osmotic pressure $p = \frac{RT}{V^*}\frac{n}{N}$ , as the same number of dissolved molecules would, provided the particles are far enough apart; classical thermodynamics expects no such force.

As printed

Nach dieser Theorie unterscheidet sich ein gelöstes Molekül von einem suspendierten Körper lediglich durch die Größe, und man sieht nicht ein, warum einer Anzahl suspendierter Körper nicht derselbe osmotische Druck entsprechen sollte, wie der nämlichen Anzahl gelöster Moleküle. Man wird anzunehmen haben, daß die suspendierten Körper infolge der Molekularbewegung der Flüssigkeit eine wenn auch sehr langsame ungeordnete Bewegung in der Flüssigkeit ausführen; werden sie durch die Wand verhindert, das Volumen $V^*$ zu verlassen, so werden sie auf die Wand Kräfte ausüben, ebenso wie gelöste Moleküle. Sind also $n$ suspendierte Körper im Volumen $V^*$ , also $n/V^* = \nu$ in der Volumeneinheit vorhanden, und sind benachbarte unter ihnen genügend weit voneinander entfernt, so wird ihnen ein osmotischer Druck $p$ entsprechen von der Größe: $p = \frac{R T}{V^*} \frac{n}{N} = \frac{R T}{N} \cdot \nu,$ Osmotic pressureGas constantAbsolute temperatureVolumeParticle countAvogadro's numberNumber densityExplain this equation wobei $N$ die Anzahl der in einem Gramm-Molekül enthaltenen wirklichen Moleküle bedeutet.
Durch diese Betrachtung ist gezeigt, daß die Existenz des osmotischen Druckes eine Konsequenz der molekularkinetischen Theorie der Wärme ist, und daß nach dieser Theorie gelöste Moleküle und suspendierte Körper von gleicher Anzahl sich in bezug auf osmotischen Druck bei großer Verdünnung vollkommen gleich verhalten.

Text on pages 550–551 of the German source, § 1, paragraph 3 · Text on page 553 of the German source, § 2, paragraph 8

A comparison, not a premise: Classical thermodynamics, as the paper reports it, expects no force on the wall, because the free energy seems not to depend on where the wall and the bodies are. The pressure is the molecular-kinetic theory's prediction, and the paper sets the two expectations side by side.

Decoder

$p$: The osmotic pressure on the wall that keeps the particles in $V^*$ ; modern texts write $\Pi$ .
$n$ , $V^*$: The number of suspended particles, and the part of the liquid's volume they are confined to.
$\nu$: The number of particles per unit volume, $n/V^*$ . In this paper $\nu$ is a number density, not a frequency; modern texts write $n$ .
$N$: The number of actual molecules in one gram-molecule, today Avogadro's constant $N_A$ .
$R/N$: The gas constant per molecule, today Boltzmann's constant $k_B$ . The paper writes the ratio and has no single letter for it.
"genügend weit voneinander entfernt": Far enough apart from one another: the law is for dilute suspensions, in which the particles do not act on one another.

Try it

BM-02: Switch between the molecular-kinetic mode and the classical expectation: what does each say the partition feels, and does the particles' size change it at the same number per volume?

BM-03: Count the positions 1000 particles can take in two volumes: does the free energy give the same pressure as the partition? Choose the preset “1000 particles (the osmotic-partition default)” there.

Where this stops: Gravity is set aside, and so are the energy and entropy of the surfaces between bodies and liquid (capillary forces), on the assumption that the moves considered do not change those surfaces. §1 states the molecular-kinetic expectation; §2 derives it from the theory. Neither section says which view nature follows; that is for observation. The law is for great dilution. Crowded bodies, or bodies that act on one another, depart from it. Read the argument

Where this stops: The footnote to the heading says this section, and Einstein's earlier papers on the foundations of thermodynamics, are not needed to understand the paper's results. The passage explains the section for the reader who wants to know how the law is obtained. The argument finds only how the free energy depends on V*; it computes nothing else about the integral B. Independence, homogeneity and the absence of forces are assumptions. With particles that act on one another, or crowded ones, J would depend on their positions and the pressure would depart from the dilute law. Read the argument

Argument: §1 states a consequence of the molecular-kinetic theory, and §2 derives it from that theory
Model: Exact for independent particles at great dilution; an idealization beyond it
Evidence: A prediction, not a measurement
History: The law for dissolved substances was known; its extension to suspended bodies is introduced in this paper

Pages 555, 556

The diffusion coefficient of suspended spheres

Balancing the drift of spheres of radius $P$ , pulled by a force through a liquid of viscosity $k$ , against their diffusion gives $D = \frac{RT}{N}\frac{1}{6\pi kP}$ : apart from universal constants and the absolute temperature, the diffusion coefficient depends only on the liquid's viscosity and the particles' size, and the force does not appear in it.

As printed

Da dynamisches Gleichgewicht herrschen soll, so muß sein: $\frac{\nu K}{6 \pi k P} - D \frac{\partial \nu}{\partial x} = 0.$ (2)Number densityExternal force per particleDynamic viscosityParticle radiusDiffusion coefficientExplain this equation
Man erhält: $D = \frac{R T}{N} \frac{1}{6 \pi k P}.$ Diffusion coefficientGas constantAbsolute temperatureAvogadro's numberDynamic viscosityParticle radiusExplain this equation Der Diffusionskoeffizient der suspendierten Substanz hängt also außer von universellen Konstanten und der absoluten Temperatur nur vom Reibungskoeffizienten der Flüssigkeit und von der Größe der suspendierten Teilchen ab.

Text on page 555 of the German source, § 3, paragraph 7 · Text on pages 555–556 of the German source, § 3, paragraph 8

In modern notation

D = \frac{R\,T}{6\,\pi\,\eta\,a\,N_A}

D = \frac{k_B\,T}{6\,\pi\,\eta\,a}

Shown in modern letters; Einstein's are in §3 of the German source face.

Conditional: The formula is for spherical particles: the paper gives the drag for spheres of radius $P$ , and particles of another shape meet another drag.

Decoder

$D$: The diffusion coefficient: diffusion carries particles across unit area in unit time at a rate $D$ times the fall of their number per unit volume along $x$ .
$k$: The viscosity of the liquid, which the paper calls its coefficient of friction. It is not Boltzmann's constant; modern texts write $\eta$ .
$P$: The radius of a spherical particle; modern texts write $a$ .
$6\pi kP$: Stokes's drag on a slowly moving sphere: a force $K$ gives the sphere the speed $K/6\pi kP$ . The paper cites Kirchhoff's lectures on mechanics for it.
$K$: A force on each particle, introduced to set up the balance. It cancels, so the result holds without it.
$\frac{RT}{N}$: The gas constant per molecule times the temperature, today $k_B T$ .

Try it

BM-04: In the balance of drift against diffusion, change the applied force: does the diffusion coefficient change? Choose the preset “Einstein equilibrium balance (m = 1, F = 4.047 fN)” there.

Where this stops: Slip at the particle’s surface, its inertia, interactions between particles, unusual liquids, and gases are all left out. Stokes’s law and the osmotic-pressure law are brought in from outside; conservation alone does not give them. Read the argument

Common wrong turns

Doubling the viscosity halves the typical displacement.

Argument: A derivation from the osmotic law and Stokes's law
Model: Exact within the model of dilute, independent spheres obeying Stokes's law
Evidence: A derivation, not evidence
History: Introduced in this paper

Pages 557, 558

The diffusion equation, and its solution from a point

If each particle's jumps in successive short intervals $\tau$ are independent and symmetric, the number of particles per unit length $f(x,t)$ obeys the diffusion equation $\frac{\partial f}{\partial t} = D\frac{\partial^2 f}{\partial x^2}$ , with $D$ half the mean square jump per unit time; for particles that all start from one point its solution is the curve of random errors, widening with time.

As printed

Wir erhalten aus dieser Gleichung, indem wir berücksichtigen, daß $\int_{-\infty}^{+\infty} \varphi(\Delta) d \Delta = 1,$ Jump densityDisplacement in one intervalExplain this equation und indem wir $\frac{1}{\tau} \int_{-\infty}^{+\infty} \frac{\Delta^2}{2} \varphi(\Delta) d \Delta = D$ Step intervalDisplacement in one intervalJump densityDiffusion coefficientExplain this equation setzen und nur das erste und dritte Glied der rechten Seite berücksichtigen: $\frac{\partial f}{\partial t} = D \frac{\partial^2 f}{\partial x^2}.$ (1)Number densityTimeDiffusion coefficientExplain this equation
Dies ist die bekannte Differentialgleichung der Diffusion, und man erkennt, daß $D$ der Diffusionskoeffizient ist.
Das Problem, welches mit dem Problem der Diffusion von einem Punkte aus (unter Vernachlässigung der Wechselwirkung der diffundierenden Teilchen) übereinstimmt, ist nun mathematisch vollkommen bestimmt; seine Lösung ist: $f(x, t) = \frac{n}{\sqrt{4 \pi D}} \frac{e^{-\frac{x^2}{4 D t}}}{\sqrt{t}}.$ Number densityTimeParticle countDiffusion coefficientExplain this equation

Text on pages 557–558 of the German source, § 4, paragraph 6 · Text on page 558 of the German source, § 4, paragraph 9 · Text on page 558 of the German source, § 4, paragraph 10

In modern notation

D = \frac{1}{2\,\tau}\,\int_{-\infty}^{\infty} \Delta^{2}\,\varphi\left(\Delta\right)\,\mathrm{d}\Delta

\frac{\partial p}{\partial t} = D\,\frac{\partial^{2} p}{\partial x^{2}}

Shown in modern letters; Einstein's are in §4 of the German source face.

Premise: The paper takes $\varphi$ to be the same for $\Delta$ and $-\Delta$ , and different from zero only for very small $\Delta$ . Jumps biased in one direction would add a drift that the equation does not contain.

Decoder

$\tau$: A time interval very short beside the intervals one can observe, yet long enough that a particle's motions in two successive intervals are independent.
$\Delta$ , $\varphi(\Delta)$: A particle's jump along $x$ in the time $\tau$ , and the probability density of that jump, which is the same for $\Delta$ and $-\Delta$ .
$f(x,t)$: The number of particles per unit length at $x$ and time $t$ . For the solution each particle is measured from its own starting point.
$\frac{1}{\tau}\int\frac{\Delta^2}{2}\varphi(\Delta)\,d\Delta = D$: The paper's definition of $D$ from the jumps. Of the whole distribution of jumps, only this moment survives in the equation.
$\frac{n}{\sqrt{4\pi D}}\frac{e^{-x^2/4Dt}}{\sqrt{t}}$: The bell curve of random errors. Its width grows as $\sqrt{t}$ , and the area under it stays $n$ , the number of particles.

Try it

BM-05: Add up many coin-flip steps: which feature of the step sets how fast the bell curve widens? Choose the preset “Coin to bell (the central limit theorem)” there.

BM-06: Start every tracer at one point and let time run: how does the width of the distribution grow? Choose the preset “The starting point” there.

Where this stops: Cutting the expansion after the second term is an approximation; it becomes exact only in a limit where the jumps shrink with the time step. Over very short times a particle’s motion is not independent from one moment to the next, so the diffusion law is not a picture of individual collisions. Read the argument

Where this stops: A finite closed box has a different long-time distribution. At t = 0 the distribution is a point mass, not an ordinary density. Coordinate RMS is not the three-dimensional RMS distance. Read the argument

Argument: A derivation from stated assumptions about the jumps
Model: An approximation, exact in the limit of small jumps and short intervals
Evidence: A derivation, not evidence
History: The diffusion equation was known, as the paper says; deriving it from the particles' irregular jumps is introduced in this paper

Pages 558, 559

The displacement grows as the square root of the time

The displacements in a time $t$ are distributed like random errors, and the root of the mean square displacement along one axis is $\lambda_x = \sqrt{2Dt}$ , so it grows as the square root of the time, not in proportion to it; for the total displacement in space the same root is $\lambda_x\sqrt{3}$ .

As printed

Die Häufigkeitsverteilung der in einer beliebigen Zeit $t$ erfolgten Lagenänderungen ist also dieselbe wie die der zufälligen Fehler, was zu vermuten war. Von Bedeutung aber ist, wie die Konstante im Exponenten mit dem Diffusionskoeffizienten zusammenhängt. Wir berechnen nun mit Hilfe dieser Gleichung die Verrückung $\lambda_x$ in Richtung der $X$ -Achse, welche ein Teilchen im Mittel erfährt, oder — genauer ausgedrückt — die Wurzel aus dem arithmetischen Mittel der Quadrate der Verrückungen in Richtung der $X$ -Achse; es ist: $\lambda_x = \sqrt{\overline{x^2}} = \sqrt{2 D t}.$ Coordinate RMS displacementMean-square displacement (one dimension)Diffusion coefficientObservation intervalExplain this equation
Die mittlere Verschiebung ist also proportional der Quadratwurzel aus der Zeit. Man kann leicht zeigen, daß die Wurzel aus dem Mittelwert der Quadrate der Gesamtverschiebungen der Teilchen den Wert $\lambda_x \sqrt{3}$ besitzt.

Text on pages 558–559 of the German source, § 4, paragraph 11 · Text on page 559 of the German source, § 4, paragraph 12

In modern notation

\lambda_x = \sqrt{\left\langle x^{2}\right\rangle} = \sqrt{2\,D\,t}

Shown in modern letters; Einstein's are in §4 of the German source face.

Approximation: The law is for times long beside the short interval $\tau$ of §4. Over shorter times the motion is not a series of independent jumps, and $\sqrt{2Dt}$ does not describe it.

Decoder

$\lambda_x$: The root of the mean of the squares of the displacements along $x$ . The paper first calls it the displacement a particle undergoes on average, then says more precisely what it is.
$\overline{x^2}$: The mean of the squared displacements along $x$ . Squaring counts moves in both directions without letting them cancel.
$\sqrt{t}$: Four times the time gives twice the displacement. A speed read off as displacement over time therefore depends on the interval chosen.
$\lambda_x\sqrt{3}$: The same root for the total displacement in space: the three axes each contribute a mean square of $2Dt$ .

Try it

BM-01: Shorten the observation interval: what happens to the apparent speed, and which statistic stays fixed? Choose the preset “The velocity trap (Exner's mistake)” there.

BM-06: After one second, take the interval from minus to plus $\lambda_x$ : what share of the tracers lies inside it? Choose the preset “One second (modern golden)” there.

Where this stops: The model expectation need not equal the mean of one small sample. Net displacement is not total path length. Read the argument

Common wrong turns

Argument: A consequence of the solution from a point
Model: Exact within the diffusion model
Evidence: A prediction that observation can test
History: Introduced in this paper

Page 559

How far a particle moves in a second

Putting §3's diffusion coefficient into §4's law gives $\lambda_x = \sqrt{t}\sqrt{\frac{RT}{N}\frac{1}{3\pi kP}}$ , and with $N = 6\cdot10^{23}$ , water at 17 °C and particles 0.001 mm across, the paper finds 0.8 micron in one second and about 6 microns in one minute.

As printed

In § 3 haben wir für den Diffusionskoeffizienten $D$ eines in einer Flüssigkeit in Form von kleinen Kugeln vom Radius $P$ suspendierten Stoffes den Wert gefunden: $D = \frac{R T}{N} \frac{1}{6 \pi k P}.$ Diffusion coefficientGas constantAbsolute temperatureAvogadro's numberDynamic viscosityParticle radiusExplain this equation Ferner fanden wir in § 4 für den Mittelwert der Verschiebungen der Teilchen in Richtung der $X$ -Achse in der Zeit $t$ : $\lambda_x = \sqrt{2 D t}.$ Coordinate RMS displacementDiffusion coefficientObservation intervalExplain this equation Durch Eliminieren von $D$ erhalten wir: $\lambda_x = \sqrt{t} \cdot \sqrt{\frac{R T}{N} \frac{1}{3 \pi k P}}.$ Coordinate RMS displacementObservation intervalGas constantAbsolute temperatureAvogadro's numberDynamic viscosityParticle radiusExplain this equation Diese Gleichung läßt erkennen, wie $\lambda_x$ von $T$ , $k$ und $P$ abhängen muß.
Wir wollen berechnen, wie groß $\lambda_x$ für eine Sekunde ist, wenn $N$ gemäß den Resultaten der kinetischen Gastheorie $6 \cdot 10^{23}$ gesetzt wird; es sei als Flüssigkeit Wasser von 17° C. gewählt ( $k = 1{,}35 \cdot 10^{-2}$ ) und der Teilchendurchmesser sei 0,001 mm. Man erhält: $\lambda_x = 8 \cdot 10^{-5} \, \text{cm} = 0{,}8 \, \text{Mikron}.$ Coordinate RMS displacementExplain this equation Die mittlere Verschiebung in 1 Min. wäre also ca. 6 Mikron.

Text on page 559 of the German source, § 5, paragraph 1 · Text on page 559 of the German source, § 5, paragraph 2

In modern notation

\lambda_x = \sqrt{2\,D\,t}

D = \frac{R\,T}{6\,\pi\,\eta\,a\,N_A}

Shown in modern letters; Einstein's are in §4 of the German source face.

Premise: The numbers rest on a value of $N$ taken from the kinetic theory of gases. The calculation shows what size of motion to look for; it measures nothing.
Approximation: The printed values are rounded, and the second is marked "ca.", about.
Conditional: The paper writes "wäre", would be: it states what the theory predicts and reports no observation.

Decoder

$3\pi kP$: Half of Stokes's $6\pi kP$ : the 2 in $\sqrt{2Dt}$ cancels a factor 2 of the drag.
$k = 1{,}35\cdot10^{-2}$: The viscosity of water at 17 °C, printed as a bare number in centimetre-gram-second units (poise). In SI units it is $1.35\times10^{-3}$ Pa·s.
0,001 mm: The particles' diameter. The radius $P$ in the formula is half of it.
$N = 6\cdot10^{23}$: The number of molecules in a gram-molecule as the kinetic theory of gases gave it, taken here as an input.
Mikron: The micron, a thousandth of a millimeter; today the micrometer.

Try it

BM-01: The laboratory opens on the paper's case, water at 17 °C and a particle of radius 0.5 micron: read $\lambda_x$ after one second and after sixty, and compare them with the printed 0.8 and 6 microns.

BM-01: Take the printed 0.001 mm as the radius instead of the diameter: how far does $\lambda_x$ move from 0.8 micron? Choose the preset “The radius probe” there.

BM-01: Double the viscosity: does $\lambda_x$ halve? Choose the preset “The viscosity comparison (common random numbers)” there.

Where this stops: A synthetic run generated from an assumed N is not independent evidence for N. Without the radius, the diffusivity fixes only the product aN. Measurement noise, drift, exposure and finite sampling require separate treatment. Read the argument

Common wrong turns

Doubling the viscosity halves the typical displacement.

Argument: A worked example of the combined law
Model: Exact within the model; the inputs are the paper's round values
Evidence: A prediction of the size of the motion, not a measurement
History: Introduced in this paper

Page 560

Determining N from the displacement

Turned around, the same relation gives $N = \frac{t}{\lambda_x^2}\cdot\frac{RT}{3\pi kP}$ , so a measured displacement, with the temperature, the viscosity and the particles' radius, would determine the number of molecules in a gram-molecule; the paper ends by hoping that a researcher will soon decide the question it has raised.

As printed

Umgekehrt läßt sich die gefundene Beziehung zur Bestimmung von $N$ benutzen. Man erhält: $N = \frac{t}{\lambda_x^2} \cdot \frac{R T}{3 \pi k P}.$ Estimated Avogadro numberObservation intervalCoordinate RMS displacementGas constantAbsolute temperatureDynamic viscosityParticle radiusExplain this equation
Möge es bald einem Forscher gelingen, die hier aufgeworfene, für die Theorie der Wärme wichtige Frage zu entscheiden!

Text on page 560 of the German source, § 5, paragraph 3 · Text on page 560 of the German source, § 5, paragraph 4

In modern notation

N = \frac{R\,T\,t}{3\,\pi\,\eta\,a\,\left\langle x^{2}\right\rangle}

Shown in modern letters; Einstein's are in §5 of the German source face.

Conditional: The paper proposes this determination and reports no measurement. It closes by hoping that a researcher will soon decide the question it has raised.

Decoder

$N$: Here the number to be found, an estimate of Avogadro's constant $N_A$ from observed displacements, and no longer an input as in the previous example.
$\frac{t}{\lambda_x^2}$: What the observations supply: the time divided by the mean square displacement along one axis, which is $\frac{1}{2D}$ .
$\frac{RT}{3\pi kP}$: What the inference needs besides the displacement: the temperature, the viscosity and the radius, each known by other means.
"Möge es bald einem Forscher gelingen": May a researcher soon succeed: the paper leaves the test to observation and reports none.

Try it

BM-07: Leave the radius undeclared: which values of $N$ fit the same displacements? Choose the preset “Identifiability family, radius undeclared” there.

BM-07: Load Perrin's 1909 observations, later evidence the paper could not use: what value of $N$ do they give, and how sure is it? Choose the preset “Perrin 1909 historical set” there.

Common wrong turns

Argument: The previous result turned around, proposed as a measurement
Model: Exact within the model, given independent values of the temperature, the viscosity and the radius
Evidence: A proposed test, not evidence
History: Proposed in this paper; carried out later, among others by Perrin in 1909

Introduction · A motion the theory requires

What the paper sets out to show, and what would decide it

On the molecular-kinetic theory of heat, bodies just large enough to see in a microscope must move by amounts a microscope can detect. Einstein thinks this may be the Brownian motion, but says the reports he could obtain were too imprecise for him to judge. Observing the motion and its laws would mean classical thermodynamics is not exactly valid for spaces a microscope can resolve and would allow the size of atoms to be determined; a failed prediction would weigh heavily against the molecular view.

Read the argument

§1 · Osmotic pressure from suspended particles

Why a suspended grain should press like a dissolved molecule

Classical thermodynamics, as Einstein characterizes it, expects no force on the wall from suspended bodies, because the free energy seems not to depend on where they are. On the molecular-kinetic view a dissolved molecule differs from a suspended body only in size, so equally many of either, far enough apart, should exert the same osmotic pressure, RT/N times their number per unit volume.

Read the argument

§2 · Osmotic pressure from the molecular-kinetic theory

How molecular theory gives the osmotic law without solving the motion

Einstein writes the free energy as F = -(RT/N) lg B, where B is an integral over every state of the system. He cannot compute B, but he needs only how it depends on the volume V* the particles are held in. If they move independently to a sufficient approximation, in a homogeneous liquid, with no forces on them, B is a factor J that does not depend on V* times V* to the power n, and the change of F with V* gives p = (RT/N)ν for dissolved molecules and suspended bodies alike.

Read the argument

§3 · The diffusion of small suspended spheres

Why viscosity changes the spread

Combining ideal osmotic pressure with mobility and diffusive balance relates diffusivity to temperature, viscosity and radius.

Read the argument

§4 · From random displacement to diffusion

Zero average is not no movement

Signed displacements can cancel. Their squares still record movement.

Read the argument

Why the square grows with time

Independent centred steps add mean squares; the root therefore grows as the square root of time.

Read the argument

From a step law to a density law

Normalization preserves probability; symmetry removes drift; the second moment supplies diffusion.

Read the argument

What the spreading curve predicts

For an unbounded point source, the Gaussian has mean square 2Dt and coordinate RMS √(2Dt).

Read the argument

§5 · From displacement to molecular scale

What would let us count molecules?

Displacement constrains diffusivity; independent radius, viscosity, temperature and gas-constant information are needed to infer molecular number.

Read the argument