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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.

Draft explanation, not yet reviewed

This is newly written explanation in modern notation, and its editorial review is pending. It is not the German source, an English translation, or a complete edition of the paper. The German source face holds a machine-drafted transcription with hand correction that no one has reviewed yet, and the facsimile face shows the pinned journal pages; an English translation is not ready. The headings name the part of the argument each passage discusses; they are not a list of the paper’s paragraphs.

New explanatory text authored with AI assistance. Mathematical and editorial review remains pending; none of these passages is presented as Einstein’s wording.

Einstein's letters change 1 of this paper's 18 formulas. 1 is already in his letters, and 16 stay in today's letters, with a note saying so.

First encounter · No algebra required

Do particles that wander in all directions ever get anywhere?

Imagine placing a microscopic particle in a drop of water and marking where it is after a few moments. Pushed at random by invisible water molecules, it is just as likely to move left as right. In the symmetric arithmetic example below, signed displacements cancel even though all four endpoints differ from the starting point.

Authored arithmetic examples: Displacements of −3, −1, +1, and +3 units (three steps left, one left, one right, three right). These are an authored educational case, not a measured dataset.

Step 1 · Explore the particle displacements

Drag a marker or select it and use ← / → arrow keys to shift its position.

1
2
3
4
Particle 1-3 units(three steps left)
Particle 2-1 units(one step left)
Particle 3+1 units(one step right)
Particle 4+3 units(three steps right)
Signed total0 unitsMean: 0.0 units
Mean distance, ignoring sign2.00 unitsProposal A (ignore sign)
Mean square5.00 squared unitsProposal B (square first)
Root mean square (RMS)2.236 units√(mean square)

Steps 2 and 3 · What a signed sum tells us

When we add the displacements algebraically, opposite directions cancel out:

(−3) + (−1) + (+1) + (+3) = 0 units

A signed total of zero tells us that the average endpoint has not shifted. It does not mean every particle remained at rest. In the authored example all four particles have nonzero displacements; after your edits the displayed totals describe your chosen endpoints.

Steps 4 and 5 · Two ways to keep track of distance

How do we keep track of how far particles wandered without opposite directions cancelling out? Both of these proposals are reasonable:

  • Proposal A (ignore the direction): Take the absolute value of each displacement. For our authored example (−3, −1, +1, +3), the absolute values are 3, 1, 1, 3 units, giving a mean absolute displacement of 2 units.
  • Proposal B (square each displacement): Multiplying any negative number by itself produces a positive number. The squared displacements are 9, 1, 1, 9 squared units, giving a mean square displacement of 5 squared units (and an RMS distance of √5 ≈ 2.236 units).

Steps 6 and 7 · What happens when every displacement doubles?

Suppose after a longer interval every particle has wandered twice as far (−6, −2, +2, +6 units):

Mean absolute displacement:(|−6| + |−2| + |+2| + |+6|) / 4 = 4 units (doubles from 2)
Mean square displacement:(36 + 4 + 4 + 36) / 4 = 20 squared units (quadruples from 5)

Notice that when displacement distances double, the mean square quadruples (2² = 4 times larger), and its square root (RMS = √20 ≈ 4.472 units) doubles exactly in proportion to distance.

Step 8 · Why the mean square has a simple additive rule

Both proposals measure spread. The mean square has a useful property when independent, zero-mean displacements are added. This is a pedagogical bridge, not the paper’s printed calculation.

First expand the square of a sum. This algebra holds without an independence assumption:

(Δx1 + Δx2)2 = Δx12 + 2 Δx1 Δx2 + Δx22

Now assume the displacements over the chosen time intervals are independent and each has zero mean. Independence makes the average product equal the product of the averages, so the cross term vanishes on averaging, not in every outcome. Equal finite step mean squares then add in proportion to the number of intervals. This coarse-grained assumption is not a claim about molecular motion at arbitrarily short times.

Absolute values have no such rule: the average absolute value of a sum is not, in general, the sum of the average absolute values. That is why the mean square is the measure that adds.

Step 9 · Mean absolute displacement is not a wrong answer

It measures spread too: the average distance from the start, ignoring direction. For the ideal Gaussian spread it also grows with the square root of time; along one coordinate it equals √(4Dt/π). The mean square is the one the argument uses because of what step 8 showed: for independent steps, mean squares add.

Why the square is useful: two steps, worked out

Take two successive independent steps, Δx1 and Δx2, each equally likely to be +1 or −1. The four possibilities are equally likely:

  • +1 then +1: total +2, squared 4
  • +1 then −1: total 0, squared 0
  • −1 then +1: total 0, squared 0
  • −1 then −1: total −2, squared 4

The average total is (2 + 0 + 0 − 2) / 4 = 0. The average square is (4 + 0 + 0 + 4) / 4 = 2, which is 1² + 1²: the squares of the two steps add. The cross terms, 2 Δx1 Δx2, are +2, −2, −2 and +2, and they average to zero.

Why do the cross terms vanish? Open the exact missing step

Step 10 · The bridge to the argument
New skill

keeping track of how far things went by squaring, so opposite directions stop cancelling.

Why useful in the paper

Section 5 says how far a particle typically wanders after a given time, and that statement is about the squared spread, not about a speed.

Continue with your choice of guidance
More guidance · Foundations

Review mean, variance, and root-mean-square displacement with worked algebraic examples in the Foundations library.

Open the lesson on mean, variance and RMS
Less guidance · Laboratory and paper

Test thousands of particles in the tracer-ensemble laboratory or jump straight to Einstein’s §5 displacement passage.

Introduction · A motion the theory requires

A qualification

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

What does Einstein say the molecular-kinetic theory requires of suspended bodies, how sure is he that this is the Brownian motion, and what would observing it decide?

The 2 printed paragraphs this passage explains
  • Introduction, paragraph 1: Einstein sets out to show that, on the molecular-kinetic theory of heat, bodies large enough to see in a microscope, suspended in a liquid, must move by amounts a microscope can detect. These motions may be the so-called Brownian molecular motion, but he finds the reports available to him too imprecise to judge.
  • Introduction, paragraph 2: If the motion and its laws can be observed, classical thermodynamics no longer holds exactly for spaces visible under a microscope, and the true size of atoms can be determined exactly; if the predicted motion is not found, that is a serious argument against the molecular-kinetic view of heat.

The paper opens with a claim about what a theory requires. The molecular-kinetic theory of heat takes the heat of a liquid to be the irregular motion of its molecules. From that theory, Einstein announces, it will be shown that bodies suspended in a liquid and large enough to see in a microscope must, because of the molecules' thermal motion, move by amounts large enough to be detected easily with a microscope.

He does not claim to have explained a motion already seen. The motions to be treated, he writes, may be the same as the so-called Brownian molecular motion; but the information he could obtain about it was 'so ungenau, daß ich mir hierüber kein Urteil bilden konnte': so imprecise that he could form no judgment. The identification is left open. The paper predicts a motion and derives its laws; whether the reported motion is that one is a question for observation.

The second paragraph says what depends on the prediction, in both directions. If the motion, together with the laws the paper expects it to follow, can really be observed, then classical thermodynamics can no longer be regarded as exactly valid even for spaces that can be told apart in a microscope, and an exact determination of the true size of atoms becomes possible. If instead the prediction proves false, that would be a weighty argument against the molecular-kinetic view of heat.

Why would a visible motion limit thermodynamics? Classical thermodynamics describes a liquid in equilibrium at one temperature by a few quantities, such as its pressure and temperature, which then stay fixed. It has no place for a part of the liquid that keeps moving of its own accord, now one way and now another. On the molecular view such restless departures from the average are always present. For a body of ordinary size they are far too small to notice; the paper argues that for a grain about a thousandth of a millimetre across, the size §5 uses, they are large enough to see.

Why would it give the size of atoms? §§4 and 5 tie how far such a grain wanders in a given time to N, the number of real molecules in a gram-molecule. With N known, the mass of a single molecule is the mass of a gram-molecule divided by N, and its size can be estimated from the volume a gram-molecule fills. The introduction announces this; the later sections carry it out.

The road there: §1 argues that suspended bodies should exert osmotic pressure just as dissolved molecules do; §2 derives that from the molecular-kinetic theory; §3 turns it into a diffusion coefficient; §§4 and 5 give the spread of a grain over time and the displacement to look for.

Show every step here: What the paper sets out to show, and what would decide it

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Modern qualifications

Robert Brown described the irregular motion of small particles suspended in water in 1828, from observations made in 1827. Later students of the motion included Christian Wiener (1863) and Louis Georges Gouy, who argued in 1888 (Journal de Physique (2) 7, p. 561) that it comes from the thermal agitation of the liquid; Carl Nägeli had argued in 1879 that single molecular impacts are far too weak to move such particles visibly. The 1905 paper cites none of this literature, which fits the introduction's statement that the information available to Einstein was too imprecise for a judgment.

Einstein's next paper on the subject, 'Zur Theorie der Brownschen Bewegung' (Ann. d. Phys. 19, p. 371, 1906), opens by saying that after this paper appeared, Siedentopf told him that Gouy and other physicists had become convinced by direct observation that the Brownian motion comes from the irregular thermal motion of the liquid's molecules.

In 1905 the molecular-kinetic theory was in wide use, and it also had critics who treated atoms as a useful hypothesis rather than an established fact. The second paragraph speaks to that disagreement by naming an observation that could count against the theory as well as for it. Jean Perrin's measurements of 1908 and 1909 on suspended grains are later evidence, dated after this paper; they do not appear in it.

A modern lens: the motion the introduction predicts is what is now called a thermal fluctuation, and the spaces that can be told apart in a microscope are where such fluctuations become visible. The term is later usage; the paper speaks of classical thermodynamics ceasing to hold exactly.

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

Assumed here

  • The molecular-kinetic theory of heat: the heat of a liquid is the irregular motion of its molecules.
  • A body suspended in the liquid, large enough to see in a microscope, is struck by those molecules.

What this does not establish

  • The introduction states what the later sections derive; it derives nothing itself.
  • Einstein does not identify his motion with the Brownian motion; he says the reports available to him were too imprecise for a judgment.
  • Both consequences are conditional. The passage does not say the motion had been observed with its laws, nor that the existence of molecules was settled in 1905.

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

§1 · Osmotic pressure from suspended particles

A heuristic step, to an approximation

Why a suspended grain should press like a dissolved molecule

Should small bodies suspended in a liquid press on a wall that holds them back, as dissolved molecules do, and what would tell the two expectations apart?

The 3 printed paragraphs this passage explains
  • §1, paragraph 1: A dissolved substance held in part of a liquid by a wall that lets the solvent through presses on that wall with the osmotic pressure, which for a dilute solution obeys the gas law, pressure times volume equals R times T times the number of gram-molecules.
  • §1, paragraph 2: By classical thermodynamics, small suspended bodies held back by the same kind of wall should exert no force on it, because the free energy of the system seems not to depend on where the wall and the bodies are.
  • §1, paragraph 3: On the molecular-kinetic view a dissolved molecule differs from a suspended body only in size, so suspended bodies in slow irregular motion should press on the wall as dissolved molecules do, with osmotic pressure RT/N times their number per unit volume, N being the number of real molecules in a gram-molecule; §2 is to show that the theory leads there.

Take a liquid of total volume V. In part of it, a volume V∗V^*, dissolve z gram-molecules of a non-electrolyte, a substance that does not split into ions, and separate V∗V^* from the pure solvent by a wall that lets the solvent through but not the dissolved substance. The dissolved molecules push on that wall. The push on each unit of its area is the osmotic pressure p, and when V∗/zV^*/z is large enough, that is, when the solution is dilute, it obeys the law van 't Hoff found, which has the form of the gas law:

pV∗=RTzpV^*=RTz

p times V star equals R times T times z.

R is the gas constant and T the absolute temperature. Pressure is force per unit area: the force on the whole wall is p times the wall's area.

Now put small bodies suspended in the liquid into V∗V^* in place of the dissolved substance, bodies that also cannot pass through the wall. What does classical thermodynamics expect? Einstein states its answer together with its reason. At a fixed temperature the force on the wall follows from how the free energy of the system changes when the wall is moved. On the usual view the free energy depends on the total masses and kinds of the suspended substance, the liquid and the wall, and on pressure and temperature, but not on where the wall and the suspended bodies are. If moving the wall does not change the free energy, the wall feels no force. So, gravity aside, classical thermodynamics does not expect the suspended bodies to exert any force on the wall.

Einstein sets two effects aside so that the comparison is fair. Gravity, which would pull the bodies down, does not concern him here. The energy and entropy of the surfaces where the bodies meet the liquid (capillary forces) would also enter the free energy, but he assumes the moves considered do not change the size or nature of those surfaces, so they drop out.

The classical expectation is a coherent position, and for ordinary bodies it agrees with experience: a few pebbles held behind a sieve do not push on it measurably. The question is whether it still holds for bodies small enough to be jostled by the molecules of the liquid.

The molecular-kinetic theory of heat reaches a different view. On it, a dissolved molecule differs from a suspended body only in size (Einstein sets 'lediglich', only, in italics), and there is no reason why a number of suspended bodies should not give the same osmotic pressure as the same number of dissolved molecules. Jostled by the molecular motion of the liquid, the suspended bodies must perform an irregular motion in it, however slow; if the wall keeps them from leaving V∗V^*, they exert forces on it, just as dissolved molecules do.

With n suspended bodies in V∗V^*, so that there are ν=n/V∗\nu = n/V^* of them in each unit of volume, and with neighbouring bodies far enough apart, the osmotic pressure should be

p=RTV∗nN=RTN νp=\frac{RT}{V^*}\frac{n}{N}=\frac{RT}{N}\,\nu

p equals R T over V star, times n over N, which equals R T over N, times nu.

where N is the number of real molecules in a gram-molecule. The first form is van 't Hoff's law with z=n/Nz = n/N gram-molecules; the second says that the pressure depends on the number of bodies per unit volume and on the temperature, and not on their size or mass.

The two views disagree about something that can be looked for. If suspended bodies exert osmotic pressure, then wherever their number per unit volume varies, so does the pressure, and it pushes them toward thinner regions; §3 shows that this appears as diffusion, and §§4 and 5 predict how far a grain wanders in a given time. On the classical expectation there is no osmotic pressure to drive such a spread. Observing the predicted wandering, with the predicted size, would count for the molecular-kinetic view, and its absence against it. First, §2 shows that the molecular-kinetic theory really leads to the extended law.

Show every step here: Why a suspended grain should press like a dissolved molecule

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Modern qualifications

Van 't Hoff stated the law for dilute solutions in 1887 (Zeitschrift für physikalische Chemie 1, p. 481), drawing on Wilhelm Pfeffer's measurements of 1877 with membranes that pass water but not sugar (Osmotische Untersuchungen). The paper takes the law as known and cites neither.

The expectation Einstein attributes to classical thermodynamics is his own characterization of 'the usual view' (der üblichen Auffassung); the paper names no one who held it. The paper does not call that view mistaken. It says the molecular-kinetic theory leads to a different one, and §2 shows how.

A modern lens: R/N is what is now written kBk_B, Boltzmann's constant, so the extended law reads p=νkBTp = \nu k_B T, the ideal-gas law for particles of any size. The paper's N is what is now called Avogadro's number; since 2019 it has an exact defined value, and in 1905 it was known only roughly. §5 of this paper proposes a way to measure it.

Later evidence: Jean Perrin's measurements of 1908 and 1909 found that the grains of a suspension at rest thin out with height as a gas of very heavy molecules would, which is what the extended law predicts once gravity is put back. That is later evidence, dated after this paper, and not a premise of it.

Assumptions and limits: Why a suspended grain should press like a dissolved molecule

Assumed here

  • Van 't Hoff's law for a dilute solution of a non-electrolyte held behind a wall that passes the solvent only: pV* = RTz, for large enough V*/z.
  • On the molecular-kinetic theory of heat, a dissolved molecule differs from a suspended body only in size.
  • The suspended bodies are few enough that neighbouring bodies are far apart.

What this does not establish

  • 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.

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

§2 · Osmotic pressure from the molecular-kinetic theory

A derivation, to an approximation

How molecular theory gives the osmotic law without solving the motion

How can a theory of countless moving molecules give the osmotic pressure of dissolved molecules and suspended bodies without anyone solving their motion?

The 9 printed paragraphs this passage explains
  • §2, footnote 1: The footnote says that this section assumes Einstein's earlier papers on the foundations of thermodynamics, of 1902 and 1903, and that neither they nor the section itself is needed to understand the paper's results.
  • §2, paragraph 1: Using his earlier statistical theory of heat, Einstein writes the entropy of a system whose state is fixed by its state variables as an integral over all their allowed values, and from it the free energy, minus RT/N times the logarithm of that integral, B.
  • §2, paragraph 2: A liquid of volume V holds n dissolved molecules or suspended bodies kept by a semipermeable wall within a part V* of it; where the wall stands changes the limits of the integral B, and the particles together take up little of V*.
  • §2, paragraph 3: Computing B exactly would be hopeless even with the whole molecular picture fixed, but all that is needed is how the free energy depends on the volume V* in which the particles are held.
  • §2, paragraph 4: Split the integral according to where each particle's centre lies: for small boxes around the centres, all inside V*, the contribution is the product of the box sizes times a factor J that does not depend on the wall; boxes of the same sizes placed elsewhere give a factor J′.
  • §2, paragraph 5: The share of B from a set of boxes is the probability of finding the particles' centres in them at a moment chosen at random; if the particles move independently, the liquid is uniform and no forces act, equal boxes are equally probable, so J equals J′.
  • §2, footnote 2: The footnote cites Einstein's 1903 paper in Annalen der Physik, volume 11, page 170.
  • §2, paragraph 6: So J depends neither on V* nor on where the particles are; integrating gives B as J times V* to the power n, the free energy follows, and its change with V* is the osmotic pressure, RT/N times the number of particles per unit volume.
  • §2, paragraph 7: This shows that osmotic pressure follows from the molecular-kinetic theory of heat, and that on this theory equal numbers of dissolved molecules and suspended bodies behave exactly alike as regards osmotic pressure at great dilution.

A footnote to the heading says what this section assumes and what it is for. It takes as known Einstein's papers on the foundations of thermodynamics (Ann. d. Phys. 9, p. 417, 1902; 11, p. 170, 1903), and says that neither those papers nor this section is needed to understand the results of the present paper. The section is still worth reading, because it answers the question §1 leaves open: how can a theory of countless colliding molecules give a law as simple as van 't Hoff's, for dissolved molecules and suspended bodies alike, without anyone solving their motions?

Einstein describes the whole system, liquid, wall and particles, by state variables p1,…,plp_1, \ldots, p_l that fix its momentary state completely, for example the coordinates and velocity components of all its atoms. How they change in time is given by equations of the form

∂pν∂t=φν(p1,…,pl)\frac{\partial p_\nu}{\partial t}=\varphi_\nu(p_1,\ldots,p_l)

The rate of change of p nu with time equals phi nu, a function of p 1 through p l.

with the condition ∑∂φν∂pν=0\sum \frac{\partial \varphi_\nu}{\partial p_\nu} = 0. Two of these letters are used elsewhere in the paper for other things: these pνp_\nu are not the osmotic pressure p of §1, and these φν\varphi_\nu, the rates of change of the state variables, are not the φ of §4.

For such a system his earlier theory gives the entropy S as an expression containing the logarithm, printed lg and meaning the natural logarithm, of an integral taken over every combination of the state variables that the conditions of the problem allow. In it T is the absolute temperature, Ē (printed with a bar) the energy of the system, and E the energy as a function of the pνp_\nu. The constant is printed as 2κ, and Einstein ties κ to N by 2κN = R, so 2κ is R/N. For the free energy F he obtains

F=−RNTlg⁡∫e−ENRT dp1…dpl=−RTNlg⁡BF=-\frac{R}{N}T\lg\int e^{-\frac{EN}{RT}}\,dp_1\ldots dp_l=-\frac{RT}{N}\lg B

F equals minus R over N times T times the logarithm of the integral of e to the minus E N over R T, over d p 1 through d p l, which equals minus R T over N times the logarithm of B.

and the integral is named B.

Even if the molecular picture were fixed in every detail, Einstein says, computing B would be so hard that an exact calculation of F is hardly conceivable. But the pressure needs only how F depends on the volume V∗V^* in which all the particles are held. (Particles, 'Teilchen', is his short word for dissolved molecules and suspended bodies alike.)

Put n particles in V∗V^*, held there by a semipermeable wall, their total volume small compared with V∗V^*. Where the wall stands limits the range of the integral B. Name the coordinates of the particles' centres of gravity x1,y1,z1x_1, y_1, z_1 through xn,yn,znx_n, y_n, z_n, give each centre a tiny box inside V∗V^*, and ask for the part of B that comes from states with every centre in its box. It has the form

dB=dx1 dy1…dzn⋅JdB=dx_1\,dy_1\ldots dz_n\cdot J

d B equals d x 1, d y 1, and so on up to d z n, times J.

where the factor J does not depend on the box sizes, nor on V∗V^*, that is, on where the wall is. J does not depend on where the boxes are either. Take a second set of boxes, of the same sizes, in other places inside V∗V^*; its part of B is dB′dB' with a factor J′J'. Since the sizes are equal,

dBdB′=JJ′\frac{dB}{dB'}=\frac{J}{J'}

d B over d B prime equals J over J prime.

Einstein's earlier theory gives these parts a meaning: dB/B is the probability that, at a moment chosen at random, the centres are in the given boxes. If the particles move independently of one another, to a sufficient approximation, the liquid is homogeneous and no forces act on the particles, then equal boxes are equally probable wherever they are, so

dBB=dB′B\frac{dB}{B}=\frac{dB'}{B}

d B over B equals d B prime over B.

and with the previous equation, J=J′J = J'.

So J depends neither on V∗V^* nor on where the particles are. Integrating over all positions of the n centres, each ranging over the volume V∗V^*, gives

B=∫J dx1…dzn=JV∗nB=\int J\,dx_1\ldots dz_n=JV^{*n}

B equals the integral of J over d x 1 through d z n, which equals J times V star to the power n.

and so the free energy is

F=−RTN{lg⁡J+nlg⁡V∗}F=-\frac{RT}{N}\left\{\lg J+n\lg V^*\right\}

F equals minus R T over N, times the logarithm of J plus n times the logarithm of V star.

The pressure on the wall is minus the rate at which F changes as V∗V^* grows:

p=−∂F∂V∗=RTV∗nN=RTN νp=-\frac{\partial F}{\partial V^*}=\frac{RT}{V^*}\frac{n}{N}=\frac{RT}{N}\,\nu

p equals minus the partial derivative of F with respect to V star, which equals R T over V star times n over N, which equals R T over N times nu.

This shows, Einstein concludes, that osmotic pressure is a consequence of the molecular-kinetic theory of heat, and that on this theory equal numbers of dissolved molecules and suspended bodies behave exactly alike as regards osmotic pressure at great dilution. The question of how the theory avoids solving every molecular motion has a plain answer: the hard part of B, the factor J, is never computed. Only its independence of V∗V^* is needed, and that follows from equal boxes being equally probable. The volume enters only through V∗nV^{*n}, and its logarithm, nlg⁡V∗n\lg V^*, gives the pressure.

Show every step here: How molecular theory gives the osmotic law without solving the motion

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Modern qualifications

The footnote cites Einstein's 'Kinetische Theorie des Wärmegleichgewichtes und des zweiten Hauptsatzes der Thermodynamik' (Ann. d. Phys. 9, p. 417, 1902) and 'Eine Theorie der Grundlagen der Thermodynamik' (11, p. 170, 1903); the second footnote cites the 1903 paper for the probability reading of dB/B. A third paper of the series, 'Zur allgemeinen molekularen Theorie der Wärme' (14, p. 354, 1904), is not cited here. J. Willard Gibbs set out a closely related formulation in Elementary Principles in Statistical Mechanics (1902).

A modern lens: the condition ∑∂φν∂pν=0\sum \frac{\partial \varphi_\nu}{\partial p_\nu} = 0 is the property later texts discuss under the name of Liouville's theorem; Einstein does not use the name. In modern notation 2κ = R/N is kBk_B, B is, up to a constant factor, the classical partition function, and F=−RTNlg⁡BF = -\frac{RT}{N}\lg B is F=−kBTln⁡ZF = -k_BT\ln Z.

A modern lens on the limits: for particles that act on one another, J depends on their positions, and the osmotic pressure picks up corrections in powers of the number density, the virial expansion of later statistical mechanics. §2 excludes them by its stated assumptions, and §1's condition that neighbours be far apart says the same.

Assumptions and limits: How molecular theory gives the osmotic law without solving the motion

Assumed here

  • Einstein's statistical theory of heat of 1902 and 1903: for a system whose state variables change by equations whose rates satisfy the sum condition of §2, the entropy and the free energy are given by the logarithm of an integral over all its states.
  • The particles move independently of one another to a sufficient approximation, the liquid is homogeneous, and no forces act on the particles.
  • The total volume of the particles is small compared with the volume V* that holds them.

What this does not establish

  • 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.

Earlier step: Why a suspended grain should press like a dissolved molecule

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

§4 · From random displacement to diffusion

A definition

Zero average is not no movement

What can we measure when left and right cancel?

The 2 printed paragraphs this passage explains
  • §4, paragraph 9: The displacements in any time t are distributed like random errors; what matters is how the constant in the exponent depends on D, and it gives the root of the mean square displacement along X, λx, as the square root of 2Dt.
  • §4, paragraph 10: The mean displacement therefore grows as the square root of the time, and the root mean square of the total displacement in space is λx times the square root of 3.

A signed mean answers where the ensemble’s centre has moved. It does not answer how far its members have wandered. For a centred distribution, rightward and leftward contributions balance even while the distribution broadens.

⟨x⟩=0,⟨x2⟩>0\langle x\rangle=0,\qquad\langle x^2\rangle>0

The model’s mean displacement can be zero while its mean-square displacement is positive.

To retain the movement, square each displacement before averaging. Taking the square root of that mean square returns a length: the root-mean-square displacement, or RMS. A mean absolute displacement is a different observable, and neither is the length of a wandering trajectory.

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

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Modern qualifications

The four numbers are this page’s own example. Einstein’s observable is the mean square of the displacement along one axis, λx2\lambda_x^2; its square root, λx\lambda_x, is the RMS used here.

Assumptions and limits: Zero average is not no movement

Assumed here

  • All displacements use the same origin, axis, units and observation interval.

What this does not establish

  • The model expectation need not equal the mean of one small sample.
  • Net displacement is not total path length.
What is getting in the way?
  • An unfamiliar word or symbol

    Two averages carry this passage. ⟨x⟩ is the average of the displacements themselves, each with its sign: plus for a tracer that ended to the right of where it started, minus for one that ended to the left. ⟨x²⟩ is the average of their squares, and a square is never negative.

    Open the explanation that addresses this

  • An algebraic move

    Squaring before averaging is the move that keeps movement when signs cancel. The signed mean of −3, −1, +1, +3 is zero; the mean of those squares is not.

    Open the explanation that addresses this

  • The physical reason for a step

    Nothing in the liquid prefers left to right. Molecules strike each tracer from every side, so in the model a step to the right is exactly as likely as the same step to the left, and over the whole ensemble the signed steps balance. The strikes never stop, so every tracer keeps moving; only the average of the signed displacements stays at the start.

    Open the explanation that addresses this

  • The connection to the picture

    Watch the cloud of tracers rather than any one of them. Its centre stays on the starting line: that centre is ⟨x⟩, and it stays at zero. Its edges move outward: how wide the cloud is, measured as √⟨x²⟩, is the quantity that grows.

    Open the explanation that addresses this

  • The purpose of the calculation

    The calculation answers how far members of the ensemble have wandered, not where their centre has moved. That is why a zero signed mean is not a claim that nothing moved.

    Open the explanation that addresses this

  • Too much at once

    One example, four numbers. Four tracers end at −3, −1, +1 and +3. Their signed average is 0. Square them to get 9, 1, 1 and 9; the average of those is 5, and its square root, about 2.24, is how far a typical tracer has gone. That is the whole passage.

    Open the explanation that addresses this

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

A derivation

Why the square grows with time

What permits us to add the contributions of many random steps?

The 3 printed paragraphs this passage explains
  • §4, paragraph 2: Each particle is assumed to move independently of all the others, and the motions of one particle in different time intervals count as independent too, as long as the intervals are not taken too small.
  • §4, paragraph 3: He introduces a time interval τ, very small compared with the intervals one can observe, but long enough that a particle's motions in two successive intervals are independent events.
  • §4, paragraph 10: The mean displacement therefore grows as the square root of the time, and the root mean square of the total displacement in space is λx times the square root of 3.

Write the displacement after n steps as the sum of their increments. Expanding its square exposes both the squared increments and their cross terms. Independence factors each expected cross term into the product of two means; centring makes that product zero.

⟨(∑i=1nΔi)2⟩=∑i=1n⟨Δi2⟩=n l2\left\langle \left(\sum_{i=1}^{n} \Delta_{i}\right)^{2}\right\rangle = \sum_{i=1}^{n} \left\langle \Delta_{i}^{2}\right\rangle = n\,l^{2}

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

The mean square of the sum of independent centred increments is the sum of their mean squares.

t=n τt = n\,\tau
D=l22 τD = \frac{l^{2}}{2\,\tau}
⟨x2⟩=2 D t\left\langle x^{2}\right\rangle = 2\,D\,t

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

Elapsed time is n tau; defining D as ell squared over two tau gives mean-square displacement two D t.

Show every step here: Why the square grows with time

Read every step on this section’s own page, where they are part of the page.

Why the mean square is tractable

Choose a transition to open its explanation. Without JavaScript, open the worked bridge below; it contains the same four explanations.

  1. Expand the square
  2. Average each contribution
  3. Why do the cross terms vanish?
  4. Add equal mean squares
Read all four transitions here (also works without JavaScript)
Expand the square

Pedagogical bridge, not the paper’s printed calculation

Expand the square

This is newly authored explanatory prose, not reviewed German text or a translation. The two-step calculation below illustrates the pairwise argument for any finite number of steps.

Changed subexpression: the total mean-square expression. The outline marks it on both sides.

Before

⟨(A+B)2⟩\htmlData{expression-id=totalMeanSquare}{\boxed{\left\langle \left(A + B\right)^{2}\right\rangle}}

After

⟨A2+2 A B+B2⟩\htmlData{expression-id=totalMeanSquare}{\boxed{\left\langle A^{2} + 2\,A\,B + B^{2}\right\rangle}}

Rule: Expand algebraic terms: use the stated equality without changing its assumptions.

Open the mathematical tool behind this step →

Expand (A + B)² as A² + 2AB + B² before averaging. The 2AB term records how the two steps combine.

In fewer words

Two added steps produce three kinds of term, not just two squares.

Show every step in this transition

This algebra holds for every pair of step values, even correlated or biased ones. No probability assumption has been used yet.

What this step assumes

No new independence or centring premise is used at this step. The averages are assumed to exist.

Treat increments over the chosen intervals as independent, centred, and having the same finite second moment. In a Brownian model this is a coarse-grained assumption, not a claim about arbitrarily short times.

Two signed steps: see every possible outcome

Each step is +1 or −1 in arbitrary step units. Rows have equal probability within each selected model. These are exact finite teaching distributions, not observations or simulated Brownian paths.

Independent steps

All four pairs are equally likely. The cross products cancel only in the average, not in each outcome.

4 equally likely outcomes. Squared columns use squared step units.
ABA + B(A + B)²A² + B²2AB
-1-1-2422
1-1002-2
-11002-2
112422

Mean square: 2. Mean cross contribution: 0. Signed mean: 0. Mean absolute displacement: 1.

Always the same direction

Both individual steps still have zero mean and mean square one. Removing independence makes the cross contribution positive.

2 equally likely outcomes. Squared columns use squared step units.
ABA + B(A + B)²A² + B²2AB
-1-1-2422
112422

Mean square: 4. Mean cross contribution: 2. Signed mean: 0. Mean absolute displacement: 2.

Always opposite directions

The same individual step averages now produce complete cancellation of the total displacement. Zero means alone are not enough.

2 equally likely outcomes. Squared columns use squared step units.
ABA + B(A + B)²A² + B²2AB
-11002-2
1-1002-2

Mean square: 0. Mean cross contribution: -2. Signed mean: 0. Mean absolute displacement: 0.

From two steps to many

For n steps, expand the square of their sum. There are n individual squared terms and cross terms for distinct pairs. Every cross average vanishes under independence and zero mean; equal step mean squares then give n times the one-step mean square. With t = nτ and D defined as the one-step mean square divided by 2τ, the result is 2Dt.

The mean absolute displacement is not a wrong answer. It also measures spreading, but absolute value does not distribute over addition, so it lacks this simple additive calculation. Neither absolute displacement nor RMS displacement is total path length.

Modern lens: where the independence model stops

Modern lens: inertia introduces a short-time regime in which successive motions need not be independent. Momentum relaxation time is a later dynamical interpretation, not a premise used in this teaching derivation.

Compare the paper’s step-distribution route (explanation preview) →

Average each contribution

Pedagogical bridge, not the paper’s printed calculation

Average each contribution

This is newly authored explanatory prose, not reviewed German text or a translation. The two-step calculation below illustrates the pairwise argument for any finite number of steps.

Changed subexpression: the total mean-square expression. The outline marks it on both sides.

Before

⟨A2+2 A B+B2⟩\htmlData{expression-id=totalMeanSquare}{\boxed{\left\langle A^{2} + 2\,A\,B + B^{2}\right\rangle}}

After

⟨A2⟩+(⟨2 A B⟩)+⟨B2⟩\htmlData{expression-id=totalMeanSquare}{\boxed{\begin{aligned}& \left\langle A^{2}\right\rangle \\ &{}+ \left(\left\langle 2\,A\,B\right\rangle\right) \\ &{}+ \left\langle B^{2}\right\rangle\end{aligned}}}

Rule: Linearity of averaging: average the sum by adding the averages. Independence is not required for this move.

Open the mathematical tool behind this step →

Apply linearity of expectation to the two squared terms and the cross term. Linearity does not require independence.

In fewer words

Averaging a sum is the same as adding its averages.

Show every step in this transition

Finite second moments ensure these averages exist. Expectation distributes over addition and a fixed multiplier: ⟨A² + 2AB + B²⟩ = ⟨A²⟩ + 2⟨AB⟩ + ⟨B²⟩.

What this step assumes

No new independence or centring premise is used at this step. The averages are assumed to exist.

Treat increments over the chosen intervals as independent, centred, and having the same finite second moment. In a Brownian model this is a coarse-grained assumption, not a claim about arbitrarily short times.

Two signed steps: see every possible outcome

Each step is +1 or −1 in arbitrary step units. Rows have equal probability within each selected model. These are exact finite teaching distributions, not observations or simulated Brownian paths.

Independent steps

All four pairs are equally likely. The cross products cancel only in the average, not in each outcome.

4 equally likely outcomes. Squared columns use squared step units.
ABA + B(A + B)²A² + B²2AB
-1-1-2422
1-1002-2
-11002-2
112422

Mean square: 2. Mean cross contribution: 0. Signed mean: 0. Mean absolute displacement: 1.

Always the same direction

Both individual steps still have zero mean and mean square one. Removing independence makes the cross contribution positive.

2 equally likely outcomes. Squared columns use squared step units.
ABA + B(A + B)²A² + B²2AB
-1-1-2422
112422

Mean square: 4. Mean cross contribution: 2. Signed mean: 0. Mean absolute displacement: 2.

Always opposite directions

The same individual step averages now produce complete cancellation of the total displacement. Zero means alone are not enough.

2 equally likely outcomes. Squared columns use squared step units.
ABA + B(A + B)²A² + B²2AB
-11002-2
1-1002-2

Mean square: 0. Mean cross contribution: -2. Signed mean: 0. Mean absolute displacement: 0.

From two steps to many

For n steps, expand the square of their sum. There are n individual squared terms and cross terms for distinct pairs. Every cross average vanishes under independence and zero mean; equal step mean squares then give n times the one-step mean square. With t = nτ and D defined as the one-step mean square divided by 2τ, the result is 2Dt.

The mean absolute displacement is not a wrong answer. It also measures spreading, but absolute value does not distribute over addition, so it lacks this simple additive calculation. Neither absolute displacement nor RMS displacement is total path length.

Modern lens: where the independence model stops

Modern lens: inertia introduces a short-time regime in which successive motions need not be independent. Momentum relaxation time is a later dynamical interpretation, not a premise used in this teaching derivation.

Compare the paper’s step-distribution route (explanation preview) →

Why do the cross terms vanish?

Pedagogical bridge, not the paper’s printed calculation

Why do the cross terms vanish?

This is newly authored explanatory prose, not reviewed German text or a translation. The two-step calculation below illustrates the pairwise argument for any finite number of steps.

The move: cross terms average away, not individual displacements.

Changed subexpression: the cross-term average. The outline marks it on both sides.

Before

⟨A2⟩+(⟨2 A B⟩)+⟨B2⟩\begin{aligned}& \left\langle A^{2}\right\rangle \\ &{}+ \htmlData{expression-id=crossTerm}{\boxed{\left(\left\langle 2\,A\,B\right\rangle\right)}} \\ &{}+ \left\langle B^{2}\right\rangle\end{aligned}

After

⟨A2⟩+(0)+⟨B2⟩\left\langle A^{2}\right\rangle + \htmlData{expression-id=crossTerm}{\boxed{\left(0\right)}} + \left\langle B^{2}\right\rangle

Rule: For independent, zero-mean quantities, the mean of their product is the product of their means, hence zero.

Open the mathematical tool behind this step →

Independence gives ⟨AB⟩ = ⟨A⟩⟨B⟩. The two means are zero, so the average cross term is zero. It is not zero in each individual outcome.

In fewer words

Independent, unbiased steps have no average cross contribution.

Show every step in this transition

These are two separate premises. Zero marginal means alone do not force ⟨AB⟩ to vanish: when B always equals A, both means are zero but AB is always positive. Pairwise uncorrelated zero-mean steps would also suffice; independence is the stronger premise used here.

What this step assumes

  • The increments A and B are independent.
  • Each increment has mean zero.

Treat increments over the chosen intervals as independent, centred, and having the same finite second moment. In a Brownian model this is a coarse-grained assumption, not a claim about arbitrarily short times.

Two signed steps: see every possible outcome

Each step is +1 or −1 in arbitrary step units. Rows have equal probability within each selected model. These are exact finite teaching distributions, not observations or simulated Brownian paths.

Independent steps

All four pairs are equally likely. The cross products cancel only in the average, not in each outcome.

4 equally likely outcomes. Squared columns use squared step units.
ABA + B(A + B)²A² + B²2AB
-1-1-2422
1-1002-2
-11002-2
112422

Mean square: 2. Mean cross contribution: 0. Signed mean: 0. Mean absolute displacement: 1.

Always the same direction

Both individual steps still have zero mean and mean square one. Removing independence makes the cross contribution positive.

2 equally likely outcomes. Squared columns use squared step units.
ABA + B(A + B)²A² + B²2AB
-1-1-2422
112422

Mean square: 4. Mean cross contribution: 2. Signed mean: 0. Mean absolute displacement: 2.

Always opposite directions

The same individual step averages now produce complete cancellation of the total displacement. Zero means alone are not enough.

2 equally likely outcomes. Squared columns use squared step units.
ABA + B(A + B)²A² + B²2AB
-11002-2
1-1002-2

Mean square: 0. Mean cross contribution: -2. Signed mean: 0. Mean absolute displacement: 0.

From two steps to many

For n steps, expand the square of their sum. There are n individual squared terms and cross terms for distinct pairs. Every cross average vanishes under independence and zero mean; equal step mean squares then give n times the one-step mean square. With t = nτ and D defined as the one-step mean square divided by 2τ, the result is 2Dt.

The mean absolute displacement is not a wrong answer. It also measures spreading, but absolute value does not distribute over addition, so it lacks this simple additive calculation. Neither absolute displacement nor RMS displacement is total path length.

Modern lens: where the independence model stops

Modern lens: inertia introduces a short-time regime in which successive motions need not be independent. Momentum relaxation time is a later dynamical interpretation, not a premise used in this teaching derivation.

Compare the paper’s step-distribution route (explanation preview) →

Add equal mean squares

Pedagogical bridge, not the paper’s printed calculation

Add equal mean squares

This is newly authored explanatory prose, not reviewed German text or a translation. The two-step calculation below illustrates the pairwise argument for any finite number of steps.

Changed subexpression: the total mean-square expression. The outline marks it on both sides.

Before

⟨A2⟩+(0)+⟨B2⟩\htmlData{expression-id=totalMeanSquare}{\boxed{\left\langle A^{2}\right\rangle + \left(0\right) + \left\langle B^{2}\right\rangle}}

After

2 l2\htmlData{expression-id=totalMeanSquare}{\boxed{2\,l^{2}}}

Rule: Substitution: use the stated equality without changing its assumptions.

Open the mathematical tool behind this step →

Here l² denotes the mean square of one step. The two remaining contributions are l² + l² = 2l².

In fewer words

Two equal contributions give twice one contribution.

Show every step in this transition

For n increments the expansion contains n square terms and one ordered cross term for every i ≠ j. The same independence and centring argument eliminates each cross average. If every step has mean square l², the result is nl².

What this step assumes

  • Each increment has the same finite mean square, called l² in this worked bridge.

Treat increments over the chosen intervals as independent, centred, and having the same finite second moment. In a Brownian model this is a coarse-grained assumption, not a claim about arbitrarily short times.

Two signed steps: see every possible outcome

Each step is +1 or −1 in arbitrary step units. Rows have equal probability within each selected model. These are exact finite teaching distributions, not observations or simulated Brownian paths.

Independent steps

All four pairs are equally likely. The cross products cancel only in the average, not in each outcome.

4 equally likely outcomes. Squared columns use squared step units.
ABA + B(A + B)²A² + B²2AB
-1-1-2422
1-1002-2
-11002-2
112422

Mean square: 2. Mean cross contribution: 0. Signed mean: 0. Mean absolute displacement: 1.

Always the same direction

Both individual steps still have zero mean and mean square one. Removing independence makes the cross contribution positive.

2 equally likely outcomes. Squared columns use squared step units.
ABA + B(A + B)²A² + B²2AB
-1-1-2422
112422

Mean square: 4. Mean cross contribution: 2. Signed mean: 0. Mean absolute displacement: 2.

Always opposite directions

The same individual step averages now produce complete cancellation of the total displacement. Zero means alone are not enough.

2 equally likely outcomes. Squared columns use squared step units.
ABA + B(A + B)²A² + B²2AB
-11002-2
1-1002-2

Mean square: 0. Mean cross contribution: -2. Signed mean: 0. Mean absolute displacement: 0.

From two steps to many

For n steps, expand the square of their sum. There are n individual squared terms and cross terms for distinct pairs. Every cross average vanishes under independence and zero mean; equal step mean squares then give n times the one-step mean square. With t = nτ and D defined as the one-step mean square divided by 2τ, the result is 2Dt.

The mean absolute displacement is not a wrong answer. It also measures spreading, but absolute value does not distribute over addition, so it lacks this simple additive calculation. Neither absolute displacement nor RMS displacement is total path length.

Modern lens: where the independence model stops

Modern lens: inertia introduces a short-time regime in which successive motions need not be independent. Momentum relaxation time is a later dynamical interpretation, not a premise used in this teaching derivation.

Compare the paper’s step-distribution route (explanation preview) →

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 a reviewed transcription.

Open them on this section’s own page, where they are part of the page.

Modern qualifications

Einstein’s independence holds over his interval τ\tau, not at every instant: over very short times a particle’s velocity carries over from one moment to the next, which is why τ\tau must not be too small. Paul Langevin’s equation of 1908, which keeps the particle’s inertia, shows where that crossover lies. A simulation that agrees with 2Dt2Dt checks the arithmetic of the model, not the existence of molecules.

Assumptions and limits: Why the square grows with time

Assumed here

  • Increments are independent, have zero mean, and share a finite mean square ℓ².
  • One step corresponds to a declared interval τ.

What this does not establish

  • Correlated steps, bias or an infinite second moment change the argument.
  • This pedagogical walk does not describe fixed physical jumps in a liquid.

Earlier step: Zero average is not no movement

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

A derivation, to an approximation

From a step law to a density law

How can random individual steps produce a deterministic equation?

The 6 printed paragraphs this passage explains
  • §4, paragraph 1: Einstein now examines more closely the irregular motions, brought about by the thermal motion of the molecules, that give rise to the diffusion of §3.
  • §4, paragraph 3: He introduces a time interval τ, very small compared with the intervals one can observe, but long enough that a particle's motions in two successive intervals are independent events.
  • §4, paragraph 4: In each interval τ every particle's x coordinate changes by some amount Δ, positive or negative; the number of particles with a change between Δ and Δ + dΔ is n φ(Δ) dΔ, where φ adds up to one over all Δ, is non-zero only for very small Δ, and is the same for Δ and −Δ.
  • §4, paragraph 5: He asks how the diffusion coefficient depends on φ, again restricting himself to a number of particles per unit volume that depends only on x and t.
  • §4, paragraph 6: Calling that number f(x, t), he expresses the particles present at time t + τ through those present at time t, expands for small τ and small Δ, drops the terms that vanish by symmetry or are very small, and, calling the mean square jump divided by twice τ the quantity D, obtains equation (1): the rate of change of f in time equals D times its second derivative in x.
  • §4, paragraph 7: This is the known differential equation of diffusion, and D is recognised as the diffusion coefficient.

Let φ(Δ)\varphi(\Delta) be the probability density for a displacement Δ\Delta during τ\tau. To end at xx, a tracer must start at x−Δx - \Delta and then make that displacement. Adding over all possible increments gives the transition relation, written here in modern notation.

p(x, t+τ)=∫−∞∞p(x−Δ, t) φ(Δ) dΔ\begin{aligned}& p\left(x,\,t + \tau\right) \\ &\qquad \mathrel{=} \int_{-\infty}^{\infty} p\left(x - \Delta,\,t\right)\,\varphi\left(\Delta\right)\,\mathrm{d}\Delta\end{aligned}

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

The density one interval later is the integral, over every jump, of the density one jump away now, times how likely that jump is.

Expand to first order in time and second order in displacement. Normalization cancels the zeroth-order term. Symmetry removes the first moment. The second moment remains.

D=12 τ ∫−∞∞Δ2 φ(Δ) dΔD = \frac{1}{2\,\tau}\,\int_{-\infty}^{\infty} \Delta^{2}\,\varphi\left(\Delta\right)\,\mathrm{d}\Delta

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

D is half the mean-square step divided by the step interval.

∂p∂t=D ∂2p∂x2\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.

The retained equation is the diffusion equation.

Show every step here: From a step law to a density law

Read every step on this section’s own page, where they are part of the page.

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 a reviewed transcription.

Open them on this section’s own page, where they are part of the page.

Modern qualifications

Einstein takes an interval τ\tau very short compared with the times we observe, yet long enough that a particle’s motions in two successive intervals are independent. That double condition, not a formal limit τ→0\tau \to 0, is what the expansion needs: the jumps must shrink with τ\tau so that DD stays finite. Einstein writes the density as f(x,t)f(x, t), the number of particles per unit volume, and the jump law as φ(Δ)\varphi(\Delta), as here.

Assumptions and limits: From a step law to a density law

Assumed here

  • The transition density is normalized, symmetric and has finite second moment.
  • The density changes little over one jump, so a short expansion describes it on scales larger than a jump.

What this does not establish

  • 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.

Earlier step: Why the square grows with time

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

A derivation

What the spreading curve predicts

How does the density law become a measurable displacement?

The 2 printed paragraphs this passage explains
  • §4, paragraph 8: Because the particles move independently, each can be measured from its own starting point at time zero; f then counts particles by how far they have moved, it still obeys equation (1), all start at one point, and the solution is the spreading curve of diffusion from a point.
  • §4, paragraph 9: The displacements in any time t are distributed like random errors; what matters is how the constant in the exponent depends on D, and it gives the root of the mean square displacement along X, λx, as the square root of 2Dt.
p(x,t)=e−x2/(4Dt)4πDt(t>0)p(x,t)=\frac{e^{-x^2/(4Dt)}}{\sqrt{4\pi Dt}}\quad(t>0)

The point-source solution is the normalized Gaussian density for positive time.

Its symmetry gives zero mean. Its second moment is 2Dt. These are ensemble statements: the curve assigns probabilities to intervals, not destinations to individual particles.

λx=⟨x2⟩=2 D t\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.

Coordinate RMS displacement is the square root of two D t.

The probability of finding a displacement between a and b is the integral of this density over that interval. At time zero, an interval containing the starting point has probability one; no finite bell represents that state.

Show every step here: What the spreading curve predicts

Read every step on this section’s own page, where they are part of the page.

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 a reviewed transcription.

Open them on this section’s own page, where they are part of the page.

Modern qualifications

Einstein writes the solution for nn particles starting together as f(x,t)=n4πDe−x2/4Dttf(x, t) = \frac{n}{\sqrt{4\pi D}}\frac{e^{-x^2/4Dt}}{\sqrt{t}}, then gives the displacement λx=2Dt\lambda_x = \sqrt{2Dt}. Three independent coordinates would give a total mean square of 6Dt6Dt; the paper asks about one coordinate, hence 2Dt2Dt.

Assumptions and limits: What the spreading curve predicts

Assumed here

  • D is constant and positive, the line is unbounded, nothing pushes the particles one way, and they all start at one point.

What this does not establish

  • 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.

Earlier step: From a step law to a density law

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

§5 · From displacement to molecular scale

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 DD predicts. A separate model connects DD to a tracer’s physical surroundings. Let bb be mobility, so a small force FF produces mean drift bFbF. Stokes drag for a sphere gives b=1/(6πηa)b = 1/(6\pi\eta a).

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

D=R T6 π η a NAD = \frac{R\,T}{6\,\pi\,\eta\,a\,N_A}
D=kB T6 π η aD = \frac{k_B\,T}{6\,\pi\,\eta\,a}

Shown in modern letters; Einstein's are in §5 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.

Show every step here: Why viscosity changes the spread

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Modern qualifications

Einstein reaches this relation in §§2–3 through osmotic pressure and Stokes’s law, writing kk for the viscosity and PP for the radius: D=RTN16πkPD = \frac{RT}{N}\frac{1}{6\pi kP}. His kk 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.

A derivation, to an approximation

What would let us count molecules?

Which additional measurements turn displacement into an estimate of N?

The 3 printed paragraphs this passage explains
  • §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.
  • §5, paragraph 3: Turned around, the same relation can be used to determine N from observed displacements.
  • §5, paragraph 4: Einstein closes by hoping that a researcher will soon succeed in deciding the question raised here, which he calls important for the theory of heat.
D=⟨x2⟩2 tD = \frac{\left\langle x^{2}\right\rangle}{2\,t}
N=R T t3 π η a ⟨x2⟩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.

Diffusivity is the mean-square displacement divided by twice the time; the molecular number is the gas constant times temperature times time, divided by three pi times viscosity times radius times the mean-square displacement.

The first expression refers to the model mean square, or to an estimate obtained from an appropriate sample. The second is an inversion under the stated physical assumptions. An estimate needs uncertainty and checks of those assumptions; rearranging symbols does not remove experimental error.

Without an independent radius, the same D can result from many pairs of a and N. The data then select a compatible family rather than a unique molecular number. The existing synthetic laboratories explore this relationship but do not supply a historical measurement.

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Modern qualifications

Einstein did not estimate NN. He took N=6×1023N = 6 \times 10^{23} from the kinetic theory of gases and predicted a displacement of about 0.8 μm in one second for particles 0.001 mm across in water at 17 °C, closing with the hope that someone would soon test it. Jean Perrin’s measurements from 1908 onward did, and gave values of NN near 7×10237 \times 10^{23}. This page supplies no dataset of its own.

Assumptions and limits: What would let us count molecules?

Assumed here

  • A measured mean-square displacement along one axis, or how fast it grows with time.
  • Separately measured R, T, η and particle radius a, for few small spheres in a liquid.

What this does not establish

  • 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.

Earlier step: Why viscosity changes the spread

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.

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