Read · Brownian motion · Explanation preview

§3 · The diffusion of small suspended spheres

Detail

Letters

Show modern qualifications

Einstein's letters change 1 of this paper's 18 formulas. 17 stay in today's letters, with a note saying so.

JavaScript is off. The full explanation is shown. Each passage’s “Show every step here” disclosure works without scripts, and foundation links open complete reading pages. The embedded laboratory remains a static worked example.

A derivation, to an approximation

Why viscosity changes the spread

What fixes D for a small spherical tracer in a liquid?

The 9 printed paragraphs this passage explains

§3, paragraph 1: Suspended particles are spread irregularly through a liquid, and each is acted on by a force K that depends on position but not on time; for simplicity the force points along the X axis everywhere.
§3, paragraph 2: In equilibrium the number of particles per unit volume varies along x so that the free energy does not change under any small virtual displacement; working out the energy and entropy changes gives equation (1), in which the force K is held in balance by osmotic pressure. The passage that explains §3 states this balance directly; it does not follow Einstein's route through the variation of the free energy.
§3, paragraph 3: Einstein uses equation (1) to find the diffusion coefficient, reading the equilibrium as two opposed processes: the particles drifting under the force K, and diffusion driven by their irregular thermal motion.
§3, paragraph 4: For spheres of radius P in a liquid of viscosity k, the force K gives each particle the speed K divided by 6πkP, so that many particles times that speed cross unit area in unit time.
§3, footnote 1: The footnote refers to Kirchhoff's lectures on mechanics, lecture 26, section 4, for the drag on a sphere moving slowly through a liquid.
§3, paragraph 5: With D the diffusion coefficient, diffusion carries particles across unit area at a rate D times the fall of their number per unit volume along x; in dynamic equilibrium the drift and the diffusion cancel, which is equation (2).
§3, paragraph 6: Equations (1) and (2) together give D as RT/N divided by 6πkP: apart from universal constants and the absolute temperature, the diffusion coefficient depends only on the liquid's viscosity and the size of the particles.
§5, paragraph 1: Combining the diffusion coefficient found in §3 for small spheres of radius P with the displacement law of §4 and eliminating D gives λx in terms of the time, the temperature, the viscosity k and the radius P.
§5, paragraph 2: Taking N as 6 · 10²³ from the kinetic theory of gases, water at 17 °C as the liquid, and particles 0.001 mm across, λx for one second comes to 8 · 10⁻⁵ cm, or 0.8 micron, and about 6 microns in one minute.

The displacement law tells us what a given $D$ predicts. A separate model connects $D$ to a tracer’s physical surroundings. Let $b$ be mobility, so a small force $F$ produces mean drift $bF$ . Stokes drag for a sphere gives $b = 1/(6\pi\eta a)$ .

Open the foundation: Viscosity and Stokes drag

Let $c$ be number density. At isothermal balance the force density $cF$ balances the osmotic-pressure gradient. With ideal osmotic pressure $cRT/N$ , the drift flux $cbF$ becomes $b\,(RT/N)$ times the density gradient. Equating it with the opposite diffusive flux gives $D = bRT/N$ .

Open the foundation: Osmotic pressure and free energy

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.

Stokes–Einstein diffusivity is the gas constant times temperature, divided by six pi times viscosity times radius times Avogadro's number; equivalently, Boltzmann's constant times temperature, divided by six pi times viscosity times radius.

The equality kB = R/N relates the two forms. For a prediction using modern constants it is convenient. For an inference of N, using a value of kB derived from that same N would defeat the point.

Open the foundation: Temperature and thermal energy

Open the foundation: Counting what crosses a boundary

Show every step here: Why viscosity changes the spread

A tool for this step

Ratios and scaling

If one quantity doubles, what happens to the others?

Double the side of a square and its area grows four times. Double the side of a cube and its volume grows eight times. Length, area and volume scale as the first, second and third power of the side.

Physical laws scale in the same ways. Stokes's drag on a sphere is in proportion to its radius: double the radius, double the drag. So a particle's diffusion coefficient D is inversely proportional to its radius: double the radius, and D halves.

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

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

lambda x is the square root of 2 D t, and D is k B T over six pi eta a.

The spread of a diffusing particle grows as the square root of D times the time. So halving D does not halve the spread: it divides it by √2, about 1.41. The tempting first thought, that twice the radius means half as far, is out by that factor, and so is the same thought about twice the viscosity.

Scaling can also run away. Before 1905, giving every mode of radiation in a box its classical share of heat energy made the total grow as the cube of the highest frequency allowed: widen the range tenfold and the energy grows a thousandfold, without limit. That is the difficulty §1 of the light-quanta paper sets out.

Worked example: Double the radius: how far does the particle get?

A sphere of radius 0.5 μm in the paper's water at 17 °C spreads 0.7948 μm in 1 s (the paper's constants, set einstein-1905-brownian-printed).
Double the radius to 1 μm. $D = k_BT/(6\pi\eta a)$ halves.
λx = √(2Dt) is multiplied by √½ = 0.7071, so it becomes 0.7948 × 0.7071 = 0.5620 μm, not half of 0.7948.
Doubling the viscosity instead does exactly the same: D halves and λx is multiplied by 0.7071.

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: scaling by a factor

Choose a factor k and multiply a length by it. The readout shows what happens to an area, a volume and the spread of a diffusing particle, with the Brownian paper's 0.8 micrometres worked through.

Scale factor k

Type a factor and press Enter, or drag the slider from 0.5 to 5.

Length: × k = 2
Area: × k² = 4
Volume: × k³ = 8
Spread, k times as long: × √k = 1.414: 0.7948 μm in 1 s becomes 1.124 μm in 2 s
Spread, k times the radius: × 1/√k = 0.7071: 0.7948 μm becomes 0.5620 μm, and k times the viscosity does the same

What it shows, in words

Multiplying a length by k multiplies an area by k² and a volume by k³. A diffusing particle's spread grows as the square root of the time, so k times as long gives √k times the spread. Multiplying the radius or the viscosity by k divides the diffusion coefficient by k and the spread by √k. At k = 2 the spread changes by 1.414 or by 0.7071, not by 2 or by one half.

Where this lesson stops

This lesson stops at how a change scales. Why D depends on the radius and the viscosity in the first place is the lesson on viscosity and Stokes drag.

Name the quantities. $T$ is the absolute temperature, in kelvin. $\eta$ is the liquid’s dynamic viscosity, its resistance to flow, in pascal seconds. $a$ is the radius of the spherical tracer. $R$ is the molar gas constant, and $N$ is the number of molecules in a mole. Einstein prints the viscosity as $k$ and the radius as $P$ ; his $k$ is the viscosity, not Boltzmann’s constant.
$D$ is the diffusivity of the earlier passages: it says how fast the mean square grows, $\langle x^2\rangle = 2Dt$ . This passage asks what fixes $D$ . The route has two halves: how a tracer answers a steady push, and how the crowding of many tracers pushes back.
First half, the push. Suppose a small steady force $F$ acts on each tracer. In a viscous liquid the tracer soon moves at a steady mean drift speed in proportion to the force. Call the ratio of speed to force the mobility $b$ , so the drift speed is $bF$ .
Stokes’s law gives the drag on a sphere moving slowly at speed $v$ through a liquid as $6\pi\eta av$ . At steady drift the drag balances the push, $F = 6\pi\eta av$ , so $v = F/(6\pi\eta a)$ and the mobility is $b = 1/(6\pi\eta a)$ . Stokes’s law is brought in from the theory of fluids; the random-walk argument does not supply it.

A tool for this step

Viscosity and Stokes drag

How hard does a liquid hold back a small sphere drifting through it?

Stir honey and then water with the same spoon at the same speed. The honey pushes back far harder. Viscosity is the number that measures this. Water at room temperature has a viscosity of about 1 millipascal-second (0.001 Pa·s); honey's is thousands of times larger.

Now move a small sphere slowly through the liquid. The liquid pushes back with a force that grows in step with the speed: twice as fast, twice the drag. It also grows with the sphere's radius and with the viscosity. George Gabriel Stokes worked out the whole law in 1851:

F = 6\,\pi\,\eta\,a\,v

The drag F equals six pi times the viscosity eta, times the radius a, times the speed v.

F is the drag in newtons, η the viscosity, a the sphere's radius and v its speed. The factor 6π comes from the way the liquid flows around a sphere; another shape would have another factor.

§3 of the Brownian paper runs the law backwards. A steady force K on a sphere of radius P, in a liquid whose viscosity the paper writes k, makes the sphere drift at the speed K/(6πkP). There k is the viscosity, not Boltzmann's constant, and P is the radius, not a pressure. For the law itself the paper refers to Kirchhoff's lectures on mechanics.

The law holds only in a regime. The flow must be slow and smooth, so that the liquid's own inertia plays no part; that is true for a micrometre sphere moving a few micrometres a second. The sphere must also be much larger than the liquid's molecules, which are about 0.3 nanometres across, so that the liquid acts as a smooth continuum. In a gas, where a molecule travels about 70 nanometres between collisions, a small particle partly slips through, and the drag is smaller than Stokes's law says.

Worked example: The drag on a half-micrometre sphere in water

Take a sphere of radius a = 0.5 μm = 5 × 10⁻⁷ m, moving at v = 1 μm/s = 10⁻⁶ m/s through water, with η = 10⁻³ Pa·s.
Multiply: 6π × 10⁻³ × 5 × 10⁻⁷ × 10⁻⁶ = 6π × 5 × 10⁻¹⁶, about 9.4 × 10⁻¹⁵ newtons.
Twice the speed gives twice the drag, 1.9 × 10⁻¹⁴ N. A sphere of twice the radius at the same speed feels twice the drag too.
Read backwards, as §3 does: a steady force of 9.4 × 10⁻¹⁵ N keeps this sphere drifting at 1 μm/s. The speed per unit force, 1/(6πηa), is what the paper needs next.

The flow here is very slow in the sense that matters: the liquid's inertia is less than a millionth of its viscous resistance (the ratio is 1000 kg/m³ × 10⁻⁶ m/s × 5 × 10⁻⁷ m ÷ 10⁻³ Pa·s = 5 × 10⁻⁷), so Stokes's law applies with room to spare.

Sinking spheres: drag against weight

A grain a little denser than water sinks at the speed where the drag holding it back equals its weight less the water's buoyancy. The excess weight grows with the cube of the radius and the drag at a given speed only with the radius, so the sinking speed grows with the square of the radius, v = 2Δρga²/(9η): ten times the radius, a hundred times as fast. Here the grains are 1200 kg/m³, 1.2 times as dense as water, in water at 1 mPa·s.

Grains of 1200 kg/m³ sinking through water. The 0.5 μm row is the lesson's sphere; times are rounded.
Radius, μm	Drag at 1 μm/s, 10⁻¹⁵ N	Sinks, μm/s	Time to sink 1 mm
0.25	4.71	0.0273	10 h
0.5	9.42	0.109	2.5 h
1	18.8	0.436	38 min
2.5	47.1	2.72	6.1 min
5	94.2	10.9	1.5 min

Every row is far inside Stokes's regime: the water's inertia is between 7 × 10⁻⁹ and 5 × 10⁻⁵ of its viscous resistance. The half-micrometre sphere takes hours to sink a millimetre. In one second, by the estimate in §5 of the Brownian paper, it wanders about 0.8 μm, several times as far as it sinks. That is why such spheres stay suspended long enough to watch.

What it shows, in words

The drag on a sphere at a given speed doubles when the radius doubles, but the weight that makes it sink grows eightfold, so larger grains sink much faster. A grain of 0.25 μm radius takes about 10 hours to sink a millimetre through water; one of 5 μm takes about a minute and a half. The half-micrometre sphere of the lesson feels a drag of 9.42 × 10⁻¹⁵ newtons at 1 μm/s and sinks at about 0.1 μm each second.

Where this lesson stops

This lesson stops at the drag on one slow sphere. Why the same 6πηa also sets how fast a crowd of such spheres spreads by diffusion is the subject of §§3 and 5 of the Brownian paper.

Second half, the crowding. Let $c$ be the number of tracers per unit volume, which Einstein writes $\nu$ . Few, small tracers in a liquid exert an osmotic pressure, written here Π, of the same form as an ideal gas’s pressure: Π = cRT/N. Dividing $R$ by $N$ turns the gas constant per mole into a constant per molecule.
Let the density vary along $x$ . A gradient is the rate at which a quantity changes with position. With $T$ held fixed, only $c$ varies in cRT/N, so the constant factor comes outside: the gradient of Π is $RT/N$ times the gradient of $c$ .
In a steady balance, the force on the tracers in a thin slab is held by the difference in osmotic pressure across it. Per unit volume that reads: $cF$ equals the gradient of Π, which is $RT/N$ times the gradient of $c$ . This osmotic law, like Stokes’s, is brought in from outside; conservation of tracers alone does not give it.

A tool for this step

Osmotic pressure and free energy

Why should grains floating in water push on a wall the way dissolved molecules do?

Put sugar water on one side of a membrane that lets water through but not sugar, and pure water on the other. Water flows toward the sugar until the sugar side stands higher: the sugar pushes outward on the membrane. That push on each square metre is the osmotic pressure.

For a dilute solution van 't Hoff found in 1887 that the pressure is the one an ideal gas of the same particles would exert in the same volume. With ν particles in each cubic metre at absolute temperature T, R the gas constant and N the number of molecules in a gram-molecule:

p = \frac{R\,T}{N}\,\nu

The osmotic pressure p equals R T over N, times the number of particles per unit volume, nu.

The paper's §1 asks why a grain floating in water, large enough to see under a microscope, should be any different. Classical thermodynamics expected no force at all from suspended grains: the free energy of the system did not seem to depend on where the wall or the grains were. On the kinetic theory of heat, the paper answers, a dissolved molecule and a suspended grain differ only in size. The grains are jostled into a slow, irregular motion, and a wall that keeps them in must be pushed as a wall that keeps molecules in is pushed. So n grains in a volume V*, ν = n/V* in each unit of volume, press on it with p = (RT/N)ν.

§2 checks this inside the kinetic theory with free energy, F = E − TS: the energy minus the temperature times the entropy. At a fixed temperature a system settles where F can no longer decrease. Moving the wall that confines the grains by a small step changes F, and requiring that change to vanish gives the same pressure, (RT/N)ν.

The law holds for dilute suspensions and solutions, where the particles are far apart and do not act on one another. The paper states it for a large volume per gram-molecule; crowded particles depart from it.

Worked example: A sugar solution and a suspension of grains

Dissolve 0.1 gram-molecule of sugar in each litre: 100 in each cubic metre, so ν = 100 × 6.02 × 10²³ = 6.02 × 10²⁵ molecules per cubic metre.
At T = 293 K, RT/N = 8.314 × 293 ÷ 6.02 × 10²³ = 4.05 × 10⁻²¹ joules. Multiply by ν: p ≈ 2.4 × 10⁵ Pa, about 2.4 atmospheres, enough to hold up a column of water nearly 25 metres tall.
Now a suspension of grains, a million in each cubic millimetre: ν = 10¹⁵ per cubic metre. The same law gives p = 4.05 × 10⁻²¹ × 10¹⁵ ≈ 4 × 10⁻⁶ Pa.
Each grain counts exactly as much as one sugar molecule. The pressure is tiny only because there are fewer grains, by a factor of about 6 × 10¹⁰.

While the particles are few and far apart, the pressure depends on their number and the temperature, not on their size or mass. That is what lets §3 treat a grain like a molecule and weigh its jostling against a force.

Behind the partition: same count, same push

A wall that lets the liquid through but not the particles feels a pressure from them, p = νk_BT, while they are dilute. Only the number per cubic metre and the temperature enter, so a grain pushes exactly as much as a sugar molecule. The table is at 293 K.

What is behind the wall and how many per m³, the share of the volume they fill, and the pressure
Behind the wall	Share filled	Pressure, Pa
Sugar, 0.1 gram-molecule per litre: 6.02 × 10²⁵ per m³	dilute	2.4 × 10⁵
Grains of 0.5 μm radius, a million per mm³: 10¹⁵ per m³	5 × 10⁻⁴	4.0 × 10⁻⁶
Grains of 0.05 μm radius, a million per mm³: 10¹⁵ per m³	5 × 10⁻⁷	4.0 × 10⁻⁶
Grains of 0.5 μm radius, a billion per mm³: 10¹⁸ per m³	0.5	not given: too crowded for the dilute law

The two grain sizes at a million per cubic millimetre push equally, although one is a thousand times the volume of the other. The last row is refused rather than computed: grains filling half the volume press on one another, and p = νk_BT holds only when they do not. The osmotic partition laboratory lets you change the count, the volume and the radius and see which of them the pressure depends on.

What it shows, in words

At 293 kelvin, sugar dissolved at 0.1 gram-molecule per litre pushes on a partition with about 2.4 × 10⁵ pascals. A million grains per cubic millimetre push with about 4.0 × 10⁻⁶ pascals, whether their radius is 0.5 or 0.05 micrometres: the pressure counts particles and does not weigh or measure them. At a billion grains of 0.5 micrometres per cubic millimetre they would fill half the volume, and the dilute law no longer applies.

Where this lesson stops

This lesson stops at the pressure of a dilute suspension. How that pressure, balanced against a force on each grain, fixes how fast the grains spread is §3 of the Brownian paper.

Now count flows. A flux is the number of tracers crossing unit area in unit time. Tracers drifting at speed $bF$ with density $c$ carry the drift flux $c\,bF$ . Multiply the balance of the last step by $b$ : the drift flux is $b\,(RT/N)$ times the gradient of $c$ .
Fick’s law gives the diffusive flux: tracers spread from crowded to sparse regions, and the flux is $-D$ times the gradient of $c$ , with the minus sign because the flow runs down the gradient.
In the balance nothing flows overall, so the two fluxes add to zero: $b\,(RT/N)$ times the gradient, minus $D$ times the gradient, is zero. Divide by the gradient, which is not zero where the density varies: $D = bRT/N$ .
Substitute the Stokes mobility $b = 1/(6\pi\eta a)$ : $D = (RT/N) \times 1/(6\pi\eta a)$ , which is $RT$ divided by $6\pi\eta aN$ . Einstein found this in §3 and quotes it at the start of §5 as $D = \frac{RT}{N}\frac{1}{6\pi kP}$ , with his $k$ for $\eta$ and his $P$ for $a$ .
Boltzmann’s constant $k_B$ equals $R/N$ , the gas constant per molecule. Replacing $R/N$ by $k_B$ gives the second form. Both are one relation:

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.

Which form to use depends on the question. To predict a spread from modern constants, the $k_B$ form is convenient. To infer $N$ from a spread, it defeats the point: a value of $k_B$ derived from a known $N$ would put the answer into the data. The $R$ form keeps $N$ separate, because $R$ can be measured on gases without counting molecules.

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, under half a width either way: 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: 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.

Einstein’s own numbers, from §5: water at 17 °C, so $T$ = 290.15 K; the viscosity printed as $k$ = $1{,}35\cdot10^{-2}$ , a bare number that in the centimetre-gram-second units of the time means poise, and is $1.35\times10^{-3}$ Pa·s; a particle diameter printed as 0,001 mm, so $a$ = 0.0005 mm = $5\times10^{-7}$ m; and $N$ = $6\times10^{23}$ from the kinetic theory of gases. The paper does not print $R$ ; this edition supplies $R$ = 8.31 J/(mol·K).
The denominator: 6π × ( $1.35\times10^{-3}$ Pa·s) × ( $5\times10^{-7}$ m) = $1.272\times10^{-8}$ kg/s. The numerator: $RT/N$ = 8.31 × 290.15 / ( $6\times10^{23}$ ) = $4.019\times10^{-21}$ J. Their quotient is $D$ = $3.16\times10^{-13}$ $\mathrm{m}^2/\mathrm{s}$ , since a joule divided by a kilogram per second is a square metre per second.
With $\lambda_x = \sqrt{2Dt}$ from §4: at $t$ = 1 s the mean square is $2Dt$ = $6.32\times10^{-13}$ $\mathrm{m}^2$ , whose square root is $7.95\times10^{-7}$ m, about 0.8 μm. Einstein prints “ $8\cdot10^{-5}$ cm = 0,8 Mikron”. At $t$ = 60 s the mean square is 60 times larger, so $\lambda_x$ is $\sqrt{60}$ , about 7.75, times larger: 6.16 μm, which Einstein gives as about 6 micrometres in one minute.
Now double the viscosity, holding $T$ , $a$ and the constants fixed. $\eta$ sits in the denominator, so $D$ halves, to $1.58\times10^{-13}$ $\mathrm{m}^2/\mathrm{s}$ , and the mean square at a fixed time halves with it. The RMS is a square root, so it changes by a factor $1/\sqrt{2}$ , about 0.707: at 1 s it falls from 0.795 μm to 0.562 μm. Doubling the viscosity divides the RMS displacement by $\sqrt{2}$ , not by two.

A tool for this step

Counting what crosses a boundary

If particles keep crossing the edges of a region, what decides whether it fills up or empties?

Take a short stretch of a tube. If more particles cross into it than cross out, the number inside goes up. The flux J counts the net number crossing a surface per unit area and per second, with rightward crossings positive; the concentration c counts particles per unit volume. They measure different things, and the equation below connects them.

\frac{\partial c}{\partial t} = -\frac{\partial J}{\partial x}

The rate at which the concentration changes equals minus the rate at which the flux changes along x.

The minus sign is the bookkeeping. If the rightward flux is larger at the right end of the stretch than at the left, more particles leave on the right than arrive on the left, and the concentration falls.

Worked example: Seven particles in, five out

In one second, 7 particles cross into the stretch and 5 cross out. The count inside rises by 2.
Next second, 5 cross in and 5 cross out. The count is unchanged, although 10 particles crossed.

Where this lesson stops

The contents change by the net crossing. A steady count does not mean the particles have stopped, only that as many leave as arrive.

A tool for this step

Squares and square roots

If the mean square grows four times, why does the typical distance only double?

Both +3 and −3 have square 9, so squaring removes the sign, and it weights a large distance more than a small one. A square root returns to the original unit: the square root of a square micrometre is a micrometre.

The root mean square, or RMS, uses both steps: square each displacement, average the squares, then take the square root. The result is a typical distance from the start, in the original unit.

\sqrt{4q}=2\sqrt q\quad(q\ge 0)

The square root of four q is twice the square root of q, for nonnegative q.

Worked example: Doubling every displacement, and what the RMS does

The displacements −3, −1, +1, +3 have squares 9, 1, 1, 9. Their average is 5, so the RMS is √5, about 2.236.
Double each displacement: −6, −2, +2, +6. The squares are 36, 4, 4, 36, and their average is 20.
The mean square went from 5 to 20, four times as large. The RMS went from about 2.236 to √20, about 4.472: twice as large.

Where this lesson stops

Four times the mean square is twice the RMS. That is why, in the Brownian paper, waiting four times as long doubles the typical distance.

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 notationEinstein's letters

The same diffusivity, written with the gas constant

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

D = \frac{R\,T}{6\,\pi\,k\,P\,N}

is times over six pi times , and .

Select a term or operation. In the formula, Down enters an operation, Up returns to its parent, and Left/Right move between siblings. Escape clears selection. Tab leaves the formula.

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

D equals R T over six pi eta a N sub A.

Einstein's form: the gas constant per molecule, R over N, stands where the modern form writes Boltzmann's constant. A larger N would make each molecule's kick smaller and the spreading slower.

Model assumptions and every term’s meaning

A dilute suspension of spheres, each moving slowly through a Newtonian liquid (Stokes drag).
Avogadro's number is the quantity the paper sets out to measure.
These are modern teaching equations in SI notation, not a transcription of the printed paper. The paper writes k for the viscosity.

Six pi eta a N sub A: Stokes's drag coefficient for one sphere, times Avogadro's number. Read the prerequisite
R times T: Thermal energy per mole. Read the prerequisite
Thermal energy over drag: Kicks drive spreading; drag resists it. Read the prerequisite
What it asserts: The spreading rate is fixed by temperature, viscosity, size and the number of molecules in a mole. Read the prerequisite
Avogadro's number: Molecules per mole: what the measurement is meant to find. Read the prerequisite
Diffusion coefficient: How fast the mean square displacement grows: half its rate of growth. Not a speed. Read the prerequisite
Gas constant: Energy per mole per kelvin. Read the prerequisite
Particle radius: The radius of the suspended sphere, not its diameter. Read the prerequisite
Absolute temperature: Temperature in kelvin. Read the prerequisite
Viscosity: The liquid's resistance to shear. The paper writes it k, which is not Boltzmann's constant. 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

Resistance to motion controls spreading

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

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

is thermal energy divided by the viscous drag coefficient.

Select a term or operation. In the formula, Down enters an operation, Up returns to its parent, and Left/Right move between siblings. Escape clears selection. Tab leaves the formula.

Each term and operation, in words

No live values

Symbolic here. Open the linked laboratory for a worked example and live values.

Read the equation aloud in words

The diffusion coefficient equals the Boltzmann constant times the absolute temperature, divided by six times pi times the dynamic viscosity times the particle radius.

Thermal energy competes with viscous drag. Doubling viscosity halves the diffusion coefficient while reducing RMS displacement only by the square root of two. Radius means radius, not diameter.

Model assumptions and every term’s meaning

Dilute, approximately spherical tracers with no-slip Stokes drag in a homogeneous Newtonian liquid.
Wall corrections, interactions, inertia, and observation noise are not included.
The modern SI 2019 constant set is used; this is not a historical inversion exercise.

An ideal model: This equation assumes the dilute spherical tracer and Stokes-drag model; dimensional consistency alone does not establish those assumptions. Read the prerequisite
Diffusion coefficient: A squared-distance-per-time coefficient. This value comes from the accepted model calculation. Read the prerequisite
Why divide by drag?: Increasing the denominator reduces diffusivity at fixed thermal energy. This is a model dependence, not a rule that every larger quantity must reduce motion. Read the prerequisite
Thermal energy scale: Boltzmann constant times absolute temperature supplies energy per particle. Read the prerequisite
A known modern constant: The SI 2019 constant set supplies this value. It must not be treated as independent historical evidence when trying to infer molecular number. Read the prerequisite
Absolute temperature: Use kelvin. The laboratory holds viscosity as an independently supplied parameter. Read the prerequisite
The drag coefficient: Six pi times viscosity times radius is Stokes drag per speed for an isolated sphere in the assumed low-Reynolds regime. Read the prerequisite
Dynamic viscosity: Viscosity is in the denominator. Doubling it at fixed temperature and radius halves D, not the RMS displacement. Read the prerequisite
Particle radius: The input is a radius, not a diameter. Confusing them changes the predicted coefficient by a factor of two. 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

Einstein reaches this relation in §§2–3 through osmotic pressure and Stokes’s law, writing $k$ for the viscosity and $P$ for the radius: $D = \frac{RT}{N}\frac{1}{6\pi kP}$ . His $k$ is not Boltzmann’s constant. William Sutherland published the same relation independently in 1905, and Marian Smoluchowski reached the displacement law by another route in 1906. The laboratories use modern SI constants and say so.

Assumptions and limits: Why viscosity changes the spread

Assumed here

Few, small, spherical particles in a uniform liquid, with Stokes’s drag and ideal osmotic pressure.
R, the gas constant, is known separately; N is the number of molecules in a mole.

What this does not establish

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.

Earlier step: What the spreading curve predicts

Source context: Annalen der Physik (4), 17, 549–560 (1905), §§1–5. Bibliographic pointer; this preview is not a source transcription or translation.

Explanation

Einstein's displacement argument: the typical distance grows with the square root of time, not with time itself. Every explanation stays on the page when reading-only is on.

Static worked case

For radius 0.5 μm, viscosity 1.35×10⁻³ Pa·s, and T = 290.15 K, the RMS displacement is about 0.8 μm in one second. This static worked case stays in the markup; loading the live ensemble does not replace it.

Try the displacement argument yourself

Once open, the tracer ensemble stays open while you change the detail or read a lesson. It runs a new trial only when you apply its settings; opening an explanation never starts one.

Open the tracer ensemble in this reading

Open the tracer ensemble on its own page, where its worked example is part of the page.

Notes and laboratory

Page 554 of Annalen der Physik, volume 17, where §3 begins. — §3 begins on page 554 of Annalen der Physik, volume 17. Read it in Einstein’s German

Symbols in §3

Diffusion coefficient
Gas constant
Absolute temperature
Dynamic viscosity
Particle radius
Avogadro's number
Boltzmann constant

Keep the tracer ensemble in view

Read in another form

Download the full explanation as Markdown · The same explanation as data (JSON) · Browse the foundation library

Read this offline

Each chapter is one HTML file that opens in any browser without a connection. It holds the explanation at every level of detail, the lessons it links to, its source references and the numbers worked out when the site was built. It has no notes of yours and no running simulations. It is an explanation, not an edition of the paper.

Download §3 · The diffusion of small suspended spheres HTML, 637 KB

Links to online sources still need a connection, and the interactive plots are not included. Every chapter you can keep

Original journal scan