# Brownian motion: from wandering to a measurable law

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.

## 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 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.

[Open the foundation: Temperature and thermal energy](/foundations/temperature-thermal-energy/)

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.

[Open the foundation: Orders of magnitude](/foundations/orders-of-magnitude/)

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.

### Model limits

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.

## 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?

Take a liquid of total volume V. In part of it, a volume $V^*$, dissolve z gram-molecules of a non-electrolyte, a substance that does not split into ions, and separate $V^*$ 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 $V^*/z$ 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:

$$
pV^*=RTz
$$

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 $V^*$ 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.

[Open the foundation: Osmotic pressure and free energy](/foundations/free-energy-osmotic-pressure/)

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 $V^*$, they exert forces on it, just as dissolved molecules do.

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

$$
p=\frac{RT}{V^*}\frac{n}{N}=\frac{RT}{N}\,\nu
$$

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 $z = n/N$ 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.

[Open the foundation: Temperature and thermal energy](/foundations/temperature-thermal-energy/)

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.

### Model limits

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.

## 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?

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 $p_1, \ldots, p_l$ 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

$$
\frac{\partial p_\nu}{\partial t}=\varphi_\nu(p_1,\ldots,p_l)
$$

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

with the condition $\sum \frac{\partial \varphi_\nu}{\partial p_\nu} = 0$. Two of these letters are used elsewhere in the paper for other things: these $p_\nu$ are not the osmotic pressure p of §1, and these $\varphi_\nu$, 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 $p_\nu$. 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

$$
F=-\frac{R}{N}T\lg\int e^{-\frac{EN}{RT}}\,dp_1\ldots dp_l=-\frac{RT}{N}\lg B
$$

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.

[Open the foundation: Logarithms: turning products into sums](/foundations/logarithms/)

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 $V^*$ in which all the particles are held. (Particles, 'Teilchen', is his short word for dissolved molecules and suspended bodies alike.)

Put n particles in $V^*$, held there by a semipermeable wall, their total volume small compared with $V^*$. Where the wall stands limits the range of the integral B. Name the coordinates of the particles' centres of gravity $x_1, y_1, z_1$ through $x_n, y_n, z_n$, give each centre a tiny box inside $V^*$, and ask for the part of B that comes from states with every centre in its box. It has the form

$$
dB=dx_1\,dy_1\ldots dz_n\cdot J
$$

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 $V^*$, 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 $V^*$; its part of B is $dB'$ with a factor $J'$. Since the sizes are equal,

$$
\frac{dB}{dB'}=\frac{J}{J'}
$$

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

$$
\frac{dB}{B}=\frac{dB'}{B}
$$

d B over B equals d B prime over B.

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

[Open the foundation: Probability and independence](/foundations/probability-independence/)

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

$$
B=\int J\,dx_1\ldots dz_n=JV^{*n}
$$

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

$$
F=-\frac{RT}{N}\left\{\lg J+n\lg V^*\right\}
$$

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 $V^*$ grows:

$$
p=-\frac{\partial F}{\partial V^*}=\frac{RT}{V^*}\frac{n}{N}=\frac{RT}{N}\,\nu
$$

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.

[Open the foundation: Partial derivatives and held-fixed quantities](/foundations/partial-derivatives/)

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 $V^*$ is needed, and that follows from equal boxes being equally probable. The volume enters only through $V^{*n}$, and its logarithm, $n\lg V^*$, gives the pressure.

### Model limits

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.

## Zero average is not no movement

What can we measure when left and right cancel?

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.

$$
\langle x\rangle=0,\qquad\langle x^2\rangle>0
$$

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.

[Open the foundation: Mean, variance and RMS](/foundations/mean-variance-rms/)

### Model limits

The model expectation need not equal the mean of one small sample.

Net displacement is not total path length.

## Why the square grows with time

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

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.

$$
\left\langle\left(\sum_{i=1}^n\Delta_i\right)^2\right\rangle=\sum_{i=1}^n\langle\Delta_i^2\rangle=nl^2
$$

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

$$
t=n\tau,\qquad D=\frac{l^2}{2\tau},\qquad\langle x^2\rangle=2Dt
$$

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

[Open the foundation: Probability and independence](/foundations/probability-independence/)

### Model limits

Correlated steps, bias or an infinite second moment change the argument.

This pedagogical walk does not describe fixed physical jumps in a liquid.

## From a step law to a density law

How can random individual steps produce a deterministic equation?

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

$$
p(x,t+\tau)=\int_{-\infty}^{\infty}p(x-\Delta,t)\varphi(\Delta)\,d\Delta
$$

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.

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

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

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

The retained equation is the diffusion equation.

[Open the foundation: Approximating a curve near a point](/foundations/taylor-expansion/)

### Model limits

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.

## What the spreading curve predicts

How does the density law become a measurable displacement?

$$
p(x,t)=\frac{e^{-x^2/(4Dt)}}{\sqrt{4\pi Dt}}\quad(t>0)
$$

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.

$$
\lambda_x=\sqrt{\langle x^2\rangle}=\sqrt{2Dt}
$$

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.

[Open the foundation: The Gaussian and its width](/foundations/gaussian-distributions/)

### Model limits

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.

## Why viscosity changes the spread

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

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](/foundations/viscosity-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](/foundations/free-energy-osmotic-pressure/)

$$
D=\frac{RT}{6\pi\eta aN}=\frac{k_BT}{6\pi\eta a}
$$

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](/foundations/temperature-thermal-energy/)

[Open the foundation: Counting what crosses a boundary](/foundations/flux-continuity/)

### Model limits

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.

## What would let us count molecules?

Which additional measurements turn displacement into an estimate of N?

$$
D=\frac{\langle x^2\rangle}{2t},\qquad N=\frac{RTt}{3\pi\eta a\langle x^2\rangle}
$$

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.

[Open the foundation: Two measurements, two unknowns](/foundations/two-measurements-two-unknowns/)

[Open the foundation: Mean, variance and RMS](/foundations/mean-variance-rms/)

[Open the foundation: Uncertainty, evidence and inference](/foundations/error-and-inference/)

[Open the foundation: The units of 1905](/foundations/unit-system-1905/)

### Model limits

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.

## Why the apparent speed depends on how you watch

Modern teaching equation; review pending.

$$
v_{\mathrm{app}} := \frac{\lambda_x}{t}
$$

Apparent coordinate speed equals the model coordinate root mean square displacement divided by the observation interval.

This quotient depends on the observation interval. It is not instantaneous physical velocity. At zero interval the quotient is undefined, even though the displacement is zero.


## From a measured spread to the number of molecules

Modern teaching equation; review pending.

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

Avogadro's number N equals the gas constant times the temperature times the observation time, divided by the product of three pi, the viscosity, the particle radius and the mean square displacement.

With temperature, viscosity and radius measured independently, the observed mean square gives Avogadro's number. The radius has to come from somewhere else: the spread alone cannot separate it from N.


## The diffusion equation

Modern teaching equation; review pending.

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

The rate of change of the probability density with time equals D times its second derivative with respect to position.

Where the density curves upward it grows, and where it curves downward it falls, at a rate set by D. A peak therefore flattens and spreads.


## Read D off the observations

Modern teaching equation; review pending.

$$
D = \frac{\left\langle x^{2}\right\rangle}{2\,t}
$$

D equals the mean square displacement over two t.

Turn the relation around: a measured mean square over a known interval gives D.


## The same diffusivity, written with the gas constant

Modern teaching equation; review pending.

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

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.


## Resistance to motion controls spreading

Modern teaching equation; review pending.

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

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.


## The mean square from the Gaussian

Modern teaching equation; review pending.

$$
\left\langle x^{2}\right\rangle = \frac{4\,D\,t}{\sqrt{\pi}}\,\int_{-\infty}^{\infty} u^{2}\,\exp\left(-u^{2}\right)\,\mathrm{d}u = 2\,D\,t
$$

The mean of x squared equals four D t over root pi, times the integral from minus infinity to infinity of u squared e to the minus u squared, d u, which equals two D t.

Changing to the unitless u moves every unit into the factor in front, and leaves an integral worth the square root of pi over two. The mean square is 2Dt.


## The diffusion coefficient from the jumps

Modern teaching equation; review pending.

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

D equals one over two tau times the integral, from minus infinity to infinity, of Delta squared times phi of Delta, d Delta.

D is the mean squared jump in one short interval, divided by twice that interval. Nothing about the fluid appears here yet; that comes with Stokes' law.


## The mean square grows in proportion to time

Modern teaching equation; review pending.

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

The mean square displacement equals two D t.

The squares of independent steps add, so the mean square grows in proportion to time, not to its square.


## The density one interval later

Modern teaching equation; review pending.

$$
p\left(x,\,t + \tau\right) = \int_{-\infty}^{\infty} p\left(x - \Delta,\,t\right)\,\varphi\left(\Delta\right)\,\mathrm{d}\Delta
$$

p of x and t plus tau equals the integral, over every jump Delta from minus infinity to infinity, of p of x minus Delta and t, times phi of Delta.

A particle found at x after one more interval tau was somewhere else a jump earlier. Add up every jump Delta that could have brought it: the density at x minus Delta now, times how likely a jump of that size is.


## The typical distance is the root of the mean square

Modern teaching equation; review pending.

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

Lambda x is the square root of the mean square displacement, which is the square root of two D t.

The root of the mean square is a distance in metres. It grows as the square root of time: four times as long, twice as far.


## From spreading to a measurable distance

Modern teaching equation; review pending.

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

The model coordinate root mean square displacement equals the square root of two times the diffusion coefficient times the observation interval.

Squaring measures spread without cancellation between directions. Taking the positive square root turns squared distance back into a distance. Four times the observation interval gives twice the model RMS, not four times.


## The square of a sum

Modern teaching equation; review pending.

$$
\left(A + B\right)^{2} = A^{2} + 2\,A\,B + B^{2}
$$

A plus B, squared, equals A squared plus two A B plus B squared.

The cross term 2AB is what averages to zero when the next step is independent of the ones before it, which is why the mean squares simply add.


## The mean square of n independent steps

Modern teaching equation; review pending.

$$
\left\langle \left(\sum_{i=1}^{n} \Delta_{i}\right)^{2}\right\rangle = \sum_{i=1}^{n} \left\langle \Delta_{i}^{2}\right\rangle = n\,l^{2}
$$

The mean of the square of the sum, for i from 1 to n, of Delta sub i, equals the sum, for i from 1 to n, of the mean of Delta sub i squared, which equals n l squared.

Square the sum of n steps and average. Every cross term between two different steps averages to zero, because the steps are independent and centred. Only the n squared steps remain, each contributing l squared.


## The density one jump away

Modern teaching equation; review pending.

$$
p\left(x - \Delta,\,t\right) \approx p - \Delta\,\frac{\partial p}{\partial x} + \frac{\Delta^{2}}{2}\,\frac{\partial^{2} p}{\partial x^{2}}
$$

p at x minus Delta and t is approximately p minus Delta times the partial derivative of p with respect to x, plus Delta squared over two times the second partial derivative of p with respect to x.

Expanding the density about x in the jump Delta keeps its value, its slope and its curvature. Averaged over the jump density, the slope term cancels and the curvature term leaves the diffusion coefficient.


## The density a short interval later

Modern teaching equation; review pending.

$$
p\left(x,\,t + \tau\right) \approx p\left(x,\,t\right) + \tau\,\frac{\partial p}{\partial t}
$$

p of x and t plus tau is approximately p of x and t, plus tau times the partial derivative of p with respect to t.

Over a short interval tau, the density at a fixed place changes by about tau times its rate of change there. Terms in tau squared are dropped.


## The diffusivity of a walk

Modern teaching equation; review pending.

$$
D = \frac{l^{2}}{2\,\tau}
$$

D equals l squared over two tau.

A walk of steps of length l every tau spreads like diffusion with this D.


## Time counts steps

Modern teaching equation; review pending.

$$
t = n\,\tau
$$

t equals n tau.

n steps of duration tau take time t.