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

§4 · From random displacement to diffusion

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 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

§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