§4 · From random displacement to diffusion · Brownian motion: from wandering to a measurable law

§4 · From random displacement to diffusion

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.

Return to the measurement that still shows movement when left and right cancel. (included in this file)

Return to adding and averaging the four displacements. (included in this file)

Return to why leftward and rightward displacements can cancel. (included in this file)

Put every tracer’s starting point at zero and measure along one line, the $x$ -axis, with right as positive and left as negative. After the same observation time, each tracer’s displacement $x$ is a signed number: where it is now minus where it started.
The angle brackets stand for an average over the whole ensemble of tracers the model describes: $\langle x\rangle$ is the average of the signed displacement, and $\langle x^2\rangle$ is the average of its square, the mean square.
Take four displacements, in micrometres: −3, −1, +1, +3. As many went left as right, and by the same amounts, so they stand in for a centred distribution.
Their sum is (−3) + (−1) + (+1) + (+3) = 0: the two leftward numbers cancel the two rightward ones. Divide by four: the signed mean is zero.
The signed mean says where the centre of the group has moved, and here it has not moved at all. It says nothing about how far the members have spread out from that centre.
Square each displacement first. A negative number times itself is positive, so the squares are 9, 1, 1, 9 square micrometres, and now nothing can cancel.
The squares add to 20. Divide by four: the mean square is 5 square micrometres. It is positive although the signed mean was zero, which is the statement of the formula:

\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.

In the formula both averages belong to the model, over its whole ensemble. Four numbers drawn at random from such a model would give a signed mean near zero rather than exactly zero; these four were chosen to cancel exactly, so that the arithmetic stays easy to follow.
The mean square is measured in square micrometres, an area, not a length. To return to a length, take the square root: $\sqrt{5}$ is approximately 2.236, so the root-mean-square displacement, or RMS, is about 2.236 micrometres.
The mean distance, ignoring direction, is a third average: (3 + 1 + 1 + 3) / 4 = 2 micrometres, the mean absolute displacement. RMS and mean distance weigh the large displacements differently, so 2.236 and 2 answer different questions. They are not inconsistent answers.
Neither average is the length of the path a tracer wandered along. A tracer that moves 2 micrometres right and then 1 micrometre left ends 1 micrometre from its start, having travelled 3. The squares record where the tracers ended up, not how far they walked on the way.

Return to the distinction between signed mean, mean square and RMS. (included in this file)

Why the apparent speed depends on how you watch

Modern teaching equation, not the equation as printed.

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 spreading to a measurable distance

Modern teaching equation, not the equation as printed.

\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.

Assumptions

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

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.

Return to why the expected cross terms vanish. (included in this file)

Return to the independence and centring assumptions. (included in this file)

Name the pieces. A tracer takes $n$ steps, one in each interval of time $\tau$ . Its $i$ -th step is a signed displacement $\Delta_i$ , where $i$ counts the steps 1, 2, 3 and so on up to $n$ . Where it ends, measured from where it started, is the sum of its steps: $x = \Delta_1 + \Delta_2 + \ldots + \Delta_n$ .
The assumptions, stated once. Each step is centred: its average over many tracers is zero, $\langle\Delta_i\rangle = 0$ , so on average it goes neither left nor right. Every step has the same mean square, written $l^2$ , so $l$ is the root-mean-square length of one step. And the steps are independent: knowing one step tells you nothing about any other.
Start with two steps and call them $A$ and $B$ . The square of their sum is found by multiplying it out, $(A+B)(A+B)$ , which gives four products, two of which are the same:

(A+B)^2=A^2+2AB+B^2

The square of A plus B contains two squares and twice the cross product.

Average both sides over many tracers. The average of a sum is the sum of the averages, and a fixed number such as 2 can be taken outside an average, so $\langle(A+B)^2\rangle = \langle A^2\rangle + 2\langle AB\rangle + \langle B^2\rangle$ .
The middle term is where independence enters. For independent steps, the average of a product is the product of the averages: $\langle AB\rangle = \langle A\rangle\langle B\rangle$ . Both steps are centred, so both averages are zero, and $\langle AB\rangle = 0$ .
Check this with the simplest walk, where each step is +1 or −1 with equal chance. The four equally likely pairs (+1, +1), (+1, −1), (−1, +1) and (−1, −1) give the products +1, −1, −1, +1, whose average is zero. The squares $A^2$ and $B^2$ are always 1, so the mean square after two steps is 1 + 0 + 1 = 2.
Three steps work the same way. Multiplying out $(\Delta_1+\Delta_2+\Delta_3)^2$ gives three squares, $\Delta_1^2 + \Delta_2^2 + \Delta_3^2$ , and three cross terms, $2\Delta_1\Delta_2 + 2\Delta_1\Delta_3 + 2\Delta_2\Delta_3$ , one for each pair of different steps. Each cross term averages to zero, for the reason just given.
For $n$ steps there are $n$ squares and one cross term for every pair of different steps. Every cross term averages to zero. Each square averages to $l^2$ , so the $n$ squares together contribute $n$ times $l^2$ . In symbols, where $\sum_{i=1}^n$ means add up for $i$ from 1 to $n$ :

\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.

Now bring in time. Each step takes the interval $\tau$ , so after $n$ steps the elapsed time is $t = n\tau$ . Dividing both sides by $\tau$ , the number of steps is $n = t/\tau$ .
Replace $n$ in the mean square: $\langle x^2\rangle = nl^2 = (t/\tau)\,l^2$ . Regroup the same factors as $(l^2/\tau)\,t$ : a constant times the elapsed time.
Write the constant $l^2/\tau$ as 2 times $l^2/(2\tau)$ , and give $l^2/(2\tau)$ its own name, $D$ , the diffusivity of the walk. The factor 2 is a choice of convention: with it, this $D$ is the same coefficient that appears in the diffusion equation of the next passage. The three statements together:

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.

The mean square grows in proportion to time: four times the time gives four times the mean square. The RMS is its square root, and $\sqrt{4} = 2$ , so four times the time gives twice the RMS, not four times. In the ±1 walk, 4 steps give a mean square of 4 and an RMS of 2; 16 steps give a mean square of 16 and an RMS of 4.
Each assumption did work. If the steps were not centred, $\langle A\rangle\langle B\rangle$ would not vanish, and the cross terms would add a part growing like $n^2$ , the square of a steady drift. If the steps were not independent, $\langle AB\rangle$ need not factor at all. If the mean square of a step were infinite, there would be no $l^2$ to add.

Return to the step-count and elapsed-time comparison. (included in this file)

The mean square grows in proportion to time

Modern teaching equation, not the equation as printed.

\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 square of a sum

Modern teaching equation, not the equation as printed.

\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, not the equation as printed.

\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 diffusivity of a walk

Modern teaching equation, not the equation as printed.

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, not the equation as printed.

t = n\,\tau

t equals n tau.

n steps of duration tau take time t.

Assumptions

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

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.

Return to the expansion at fixed position or fixed time. (included in this file)

Return to the two expansions and their remainders. (included in this file)

Name the pieces. $p(x,t)$ is the density of tracers at position $x$ and time $t$ : the chance of finding a tracer between $x$ and a slightly larger $x + dx$ is $p(x,t)\,dx$ . Einstein writes this density as $f(x, t)$ and counts particles rather than chances; the steps below are the same for both.
$\tau$ is a short interval of time. During it each tracer moves by a signed displacement $\Delta$ , and $\varphi(\Delta)$ is the probability density of that displacement: the chance of a displacement between $\Delta$ and $\Delta + d\Delta$ is $\varphi(\Delta)\,d\Delta$ .
Two facts about $\varphi$ are assumed. It is normalized: some displacement certainly happens, so all its chances add up to one, $\int_{-\infty}^{\infty}\varphi(\Delta)\,d\Delta = 1$ . And it is symmetric: a jump to the left is as likely as the same jump to the right, $\varphi(-\Delta) = \varphi(\Delta)$ .
To be at $x$ at time $t + \tau$ , a tracer must have been at $x - \Delta$ at time $t$ and then made the displacement $\Delta$ . The chance of that one route is the density at its start times the chance of the jump, $p(x - \Delta, t)\,\varphi(\Delta)\,d\Delta$ , because in this model the jump law is the same wherever the tracer starts. Adding over every possible $\Delta$ , from far left to far right, gives the transition relation:

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.

The integral sign $\int$ is that adding-up over every value of $\Delta$ . Both sides can now be simplified, using the assumption that $p$ changes little over one interval $\tau$ and over one jump $\Delta$ .
The left side, at fixed position. A short time $\tau$ later, the density is approximately its present value plus its rate of change in time, $\partial p/\partial t$ , multiplied by $\tau$ . The curly $\partial$ marks a partial derivative: a rate of change in one variable while the other is held fixed, here time with position held fixed.

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

Keep the density and its first time change on the left.

The right side, at fixed time. The density a distance $\Delta$ to the left of $x$ is approximately its value at $x$ , minus the slope $\partial p/\partial x$ times $\Delta$ , plus half the curvature $\partial^2p/\partial x^2$ times $\Delta^2$ . The slope term carries a minus sign because the step is to the left. The curvature term does not, because $(-\Delta)^2 = \Delta^2$ . Here $p$ and its derivatives are all taken at $x$ and $t$ .

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

Keep value, slope and curvature of the spatial density on the right.

Put the spatial expansion into the integral and split it into three integrals, one for each term. The density and its derivatives are taken at $x$ , so they do not change as $\Delta$ runs over its values, and each one comes outside its integral.
First term: $p$ times $\int\varphi\,d\Delta$ , which is $p$ times one, because $\varphi$ is normalized.
Second term: $-\partial p/\partial x$ times $\int\Delta\varphi\,d\Delta$ . For a symmetric $\varphi$ , each positive $\Delta$ is matched by the negative $-\Delta$ with the same weight, so this integral is zero. That is why there is no drift.
Third term: $\frac{1}{2}\,\partial^2p/\partial x^2$ times $\int\Delta^2\varphi\,d\Delta$ . This integral is the mean-square jump. It is positive, because $\Delta^2$ is never negative, so symmetry cannot remove it.
Set the two sides equal: $p + \tau\,\partial p/\partial t \approx p + \frac{1}{2}\,\frac{\partial^2p}{\partial x^2}\int\Delta^2\varphi\,d\Delta$ . Subtract $p$ from both sides; normalization is what makes the two $p$ terms cancel. Then divide both sides by $\tau$ : $\partial p/\partial t$ is approximately $\frac{1}{2\tau}\int\Delta^2\varphi\,d\Delta$ times $\partial^2p/\partial x^2$ .
The factor in front of the curvature depends only on the jump law and on $\tau$ , not on $x$ or $t$ . Give it a name, $D$ :

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.

It is the mean-square jump divided by $2\tau$ , the same form as the $l^2/(2\tau)$ of the previous passage. With that name, what remains is the diffusion equation:

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

The retained equation is the diffusion equation.

Read it term by term. Where the density has a peak, its curvature is negative, so $\partial p/\partial t$ is negative and the peak falls. Where it has a hollow, the curvature is positive and the hollow fills. $D$ sets how fast.
What was dropped. The next spatial term carries $\Delta^3$ and vanishes by symmetry, like the $\Delta$ term. The first nonzero term left out carries $\Delta^4$ , and on the left the term with $\tau^2$ was left out. Cutting the expansion there is an approximation. It becomes exact only in a limit where the jumps shrink with the time step so that $D$ stays finite, and the equation describes the density on scales larger than one jump, not individual collisions.

Return to where the jump chances add up to one and the lopsided terms cancel. (included in this file)

Return to the rates of change taken with position or time held fixed. (included in this file)

Return to what the resulting density equation predicts. (included in this file)

The diffusion equation

Modern teaching equation, not the equation as printed.

\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.

The diffusion coefficient from the jumps

Modern teaching equation, not the equation as printed.

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 density one interval later

Modern teaching equation, not the equation as printed.

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 density one jump away

Modern teaching equation, not the equation as printed.

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, not the equation as printed.

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.

Assumptions

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.

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.

Return to checking the bell curve: its total is one, and its spread is 2Dt. (included in this file)

Return to λx, the displacement after a time t. (included in this file)

Return to the bell curve for particles that all start at one point. (included in this file)

Return to the Gaussian's exponential factor. (included in this file)

Set the scene. Every tracer starts at $x = 0$ at time zero, on a line with no walls, with a constant, positive diffusivity $D$ and nothing pushing the tracers one way. Their density $p(x,t)$ then spreads by the diffusion equation of the previous passage. For every positive time $t$ , the solution is this bell-shaped curve, a Gaussian:

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.

Read its parts. $e$ is the base of natural exponentials, about 2.718, and $\pi$ is about 3.1416. The exponent $-x^2/(4Dt)$ is zero at $x = 0$ and more negative the farther $x$ is from the start, so the curve is highest at the start and falls away on both sides. The denominator $\sqrt{4\pi Dt}$ scales the whole curve so that its total is one, as the next steps check.
The curve depends on $x$ only through $x^2$ , so $p(-x,t) = p(x,t)$ : it is symmetric about the start.
Change variable to make the integrals standard ones. For positive $t$ , let $u = x/\sqrt{4Dt}$ . Then $x = \sqrt{4Dt}\,u$ , $x^2 = 4Dt\,u^2$ and $dx = \sqrt{4Dt}\,du$ , and the exponent becomes $-u^2$ .
Total probability. $\int p\,dx$ becomes $\frac{1}{\sqrt{4\pi Dt}}\int e^{-u^2}\sqrt{4Dt}\,du$ . The two square roots of $4Dt$ cancel, leaving $\frac{1}{\sqrt\pi}\int_{-\infty}^{\infty}e^{-u^2}\,du$ . The Gaussian integral $\int_{-\infty}^{\infty}e^{-u^2}\,du = \sqrt\pi$ is a known result, so the total is one.
Mean. The mean displacement is $\int x\,p\,dx$ . Because $p$ is symmetric, the product $x\,p$ is odd: its value at $-x$ is minus its value at $x$ . The negative side cancels the positive side, and the mean is zero.
Mean square. $\langle x^2\rangle = \int x^2p\,dx$ . Substitute $x^2 = 4Dt\,u^2$ and cancel the square roots as before; what remains is $4Dt$ times $\frac{1}{\sqrt\pi}\int u^2e^{-u^2}\,du$ .
That last integral is found by integration by parts. Write $u^2e^{-u^2}$ as $u$ times $ue^{-u^2}$ , and notice that $ue^{-u^2}$ is the derivative of $-\frac{1}{2}e^{-u^2}$ . Integration by parts moves the derivative across: $\int u^2e^{-u^2}\,du$ equals $-\frac{1}{2}ue^{-u^2}$ evaluated between far left and far right, plus $\frac{1}{2}\int e^{-u^2}\,du$ .
The first part is zero at both ends, because $e^{-u^2}$ falls to zero much faster than $u$ grows. So the integral is $\frac{1}{2}\sqrt\pi$ , and divided by $\sqrt\pi$ it gives one half:

\langle x^2\rangle=\frac{4Dt}{\sqrt\pi}\int_{-\infty}^{\infty}u^2e^{-u^2}\,du=2Dt

The second moment is four D t times the normalized Gaussian second-moment integral, giving two D t.

So the mean square is $4Dt$ times one half, which is $2Dt$ : the same growth in proportion to time that the walk of independent steps gave. Its square root is a length, the RMS displacement along the axis, which Einstein writes $\lambda_x$ . Taking the nonnegative root:

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

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

These are statements about the ensemble. The curve does not say where one tracer will be; it gives the probability of an interval. The probability of a displacement between two positions $x_1$ and $x_2$ is the integral of $p$ from $x_1$ to $x_2$ , the area under the curve there.
For this curve, the interval from $-\lambda_x$ to $+\lambda_x$ holds about 68% of the probability (0.6827), and the interval from $-2\lambda_x$ to $+2\lambda_x$ about 95% (0.9545), at every positive time. Only the width changes as the curve spreads.
At $t = 0$ the formula has no meaning: $4Dt$ is zero, and the curve would be infinitely narrow and infinitely tall. The state itself is simple. Every tracer is at the start, so any interval containing the start has probability one and any interval that misses it has probability zero. No finite bell represents that, which is why the formula carries $(t>0)$ .
Last, check that the curve obeys the diffusion equation. Write $p$ as a factor $(4\pi Dt)^{-1/2}$ times $e^{-x^2/(4Dt)}$ , and take each rate of change with the other variable held fixed.
Time, with $x$ fixed. The factor contributes $-1/(2t)$ times $p$ , because the rate of change of $t^{-1/2}$ is $-\frac{1}{2}t^{-3/2}$ . The exponential contributes $x^2/(4Dt^2)$ times $p$ , because the rate of change of $-x^2/(4Dt)$ with $t$ is $+x^2/(4Dt^2)$ . By the product rule, $\partial p/\partial t = p\,(-1/(2t) + x^2/(4Dt^2))$ .
Position, with $t$ fixed. The first derivative is $\partial p/\partial x = p\,(-x/(2Dt))$ . Differentiate again with the product rule: $\partial^2p/\partial x^2 = p\,(x^2/(4D^2t^2)) + p\,(-1/(2Dt))$ .
Multiply that by $D$ : $p\,(x^2/(4Dt^2) - 1/(2t))$ , the same as the time derivative. So the curve satisfies $\partial p/\partial t = D\,\partial^2p/\partial x^2$ .

Return to the integration by parts that gives the one half. (included in this file)

Return to probabilities of intervals rather than heights. (included in this file)

The mean square from the Gaussian

Modern teaching equation, not the equation as printed.

\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 typical distance is the root of the mean square

Modern teaching equation, not the equation as printed.

\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.

Assumptions

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

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.

Foundations included in this file

Reading a graph

How does a curve show the relationship between two physical quantities?

Walk for a minute and note how far you have gone every ten seconds: 0 m at the start, 14 m after ten seconds, 28 m after twenty, and so on. The first drawing below puts time along the bottom and distance up the side, with a dot for each pair. That is a graph: each dot is one moment, read across for the time and up for the distance.

The dots of a steady walk fall on a straight line, and the steeper the line, the faster the walk. This one climbs 14 metres for every 10 seconds: 1.4 metres per second. A flat line would mean standing still, and a line that bends upward would mean speeding up.

The papers draw other pairs the same way. Along the bottom goes what you choose or wait for, such as a position or a time; up the side goes what you then find, such as how many particles sit there. A bell-shaped curve, high in the middle and falling away on both sides, says that most particles are near the centre and fewer are farther out.

Worked example: Four particles that average to zero

Four particles end at −3, −1, +1 and +3 micrometres from where they started.
The second drawing puts those positions along the bottom, with a bar one particle tall above each.
The bars stand two on each side of zero, at the same distances out. The picture is symmetric, so the average position is 0.
Yet no bar stands at zero: every particle moved. The graph shows at a glance what the average hides.

Where this lesson stops

Before reading a curve, read its two labels and their units, and ask whether its height is a count, a density or a running total.

This lesson builds on

A sign records direction

Return to the chapter

A letter stands for a quantity

What does a letter such as t mean in a formula?

Fill in the blanks: "After ___ seconds the particle had wandered about ___ micrometres." The Brownian paper fills them twice. After 1 second, about 0.8 micrometres; after 60 seconds, about 6.

Write t for the first blank and λx for the second, and the sentence becomes a relation between two quantities. The letter t keeps meaning the time elapsed, in seconds, while its value goes from 1 to 60. The letter λx keeps meaning how far the particle has wandered along one direction while its value grows from about 0.8 to about 6 micrometres.

A letter's meaning belongs to the place where it is used, not to the letter. In the light-quanta paper L is the speed of light in §§1 and 2, and in §9 it is the light energy absorbed. Each passage says what its letters mean, and so does every formula on this site, in its legend.

Worked example: The paper's two estimates, and where they come from

§5 of the Brownian paper prints λx = 8 × 10⁻⁵ cm = 0.8 micrometres for t = 1 second, and about 6 micrometres for one minute, for a particle 0.001 mm across in water at 17 °C.
Worked out again with the paper's own constants, R = 8.31 and N = 6 × 10²³ (the constant set einstein-1905-brownian-printed), the values are 0.7948 μm at t = 1 s and 6.1564 μm at t = 60 s.
With the modern Boltzmann constant in place of R/N (the exact value of the set modern-si-2019), and the same particle and water, they are 0.7935 μm and 6.1467 μm.
In every line t means the same thing, the time elapsed. Only its value changes, and λx changes with it.

Where this lesson stops

A letter names a quantity: its value can change while its meaning stays the same. The rule that turns t into λx is the subject of the Brownian paper's §4.

Return to the chapter

A sign records direction

How can movement add up to zero?

On the ruler below, the starting point is zero. A final mark three units to the right has displacement +3; one three units to the left has displacement −3. Adding the signed displacements gives zero. Adding the distances from the start gives six.

Displacement compares where something ends with where it started, whatever it did in between. A walker who goes three units right and comes back has displacement zero after walking six, which is the lower pair of arrows in the drawing.

Worked example: Two walkers, opposite directions, average zero

Two walkers both finish one unit from their start, one at −1 and the other at +1. The signed average is zero. The average distance is one.

Where this lesson stops

The sign names a direction relative to a chosen axis; it does not mean a negative distance.

This lesson builds on

Adding and averaging

Return to the chapter

Squares and square roots

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

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

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

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

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

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

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

Where this lesson stops

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

This lesson builds on

Adding and averaging

Return to the chapter

Adding and averaging

What does an average keep, and what does it lose?

Four jars hold 3, 1, 1 and 3 marbles. Pour them together and there are 8. Share the 8 equally between the four jars and each holds 2. Two is the average, although no jar held 2 to begin with.

The average keeps the total and the count and nothing else, so 3, 1, 1, 3 and 2, 2, 2, 2 have the same average. Listing every value twice doubles both the total and the count, and the average stays at 2.

Worked example: Three mornings with the same average

One morning the four jars hold 3, 1, 1 and 3 marbles. Add them: 8. Divide by the four jars: an average of 2.
Another morning they hold 8, 0, 0 and 0. Add them: 8 again. Divide by four: 2 again, and now three jars are empty.
A third morning every jar holds 2, which is the average itself. Add them: 8. Divide by four: 2.
Write any of those three lists out twice over. The total doubles to 16 and the count doubles to 8 jars, and 16 divided by 8 is still 2.
Four descriptions and one average. It keeps the total and the count, and the differences between the mornings are what it leaves behind.

Where this lesson stops

An average is the total shared equally among the observations, and two very different lists can have the same one.

Return to the chapter

Rates of change and derivatives

How can a body have a speed at a single instant?

A speed needs two readings of position and the time between them: divide the distance covered by the time taken, and you have the average speed over that interval. An instant has no duration, so that recipe cannot be used at an instant as it stands.

So shorten the interval and watch what happens. A ball rolls so that after t seconds it has gone 3t² metres: 3 m after one second, 12 m after two. Start at the one-second mark and average over shorter and shorter stretches:

Over the next second it goes from 3 m to 12 m: 9 m in 1 s, an average of 9 m/s.
Over the next tenth of a second it reaches 3.63 m: 0.63 m in 0.1 s, an average of 6.3 m/s.
Over the next hundredth it reaches 3.0603 m: 0.0603 m in 0.01 s, an average of 6.03 m/s.
Over the next thousandth it reaches 3.006003 m: 0.006003 m in 0.001 s, an average of 6.003 m/s.

The averages close in on 6 m/s. That number is the derivative: the ball's speed at the instant t = 1 s. The interval is never set to zero, which would mean dividing zero distance by zero time. It only shrinks, and the averages settle.

In symbols, with Δt for the short interval, the same recipe reads:

\frac{dx}{dt} = \lim_{\Delta t \to 0} \frac{x(t+\Delta t) - x(t)}{\Delta t}

d x by d t is the value that the change in position divided by the change in time settles on as delta t shrinks towards zero.

A derivative carries units: those of the quantity that changes, divided by those of the quantity you vary. Position in metres, varied over time in seconds, gives metres per second. On a graph of position against time, the derivative at a point is the steepness of the curve there.

Einstein's Brownian paper needs rates like this for a concentration of particles, which changes along a tube and in time at once. With two things changing, a rate has to say which one it follows, so the paper writes a curly ∂ and holds the other fixed:

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

The partial derivative of f with respect to t equals D times the second partial derivative of f with respect to x.

The lesson on partial derivatives, listed at the end of this page, reads that equation one piece at a time.

Worked example: Why the averages settle on 6 m/s

The list of averages can be explained, not only watched. Start at any time t and let the interval be Δt. In that interval the ball moves 3(t + Δt)² − 3t² metres. Since (t + Δt)² is t² + 2tΔt + (Δt)², that distance is:

3(t+\Delta t)^2 - 3t^2 = 6t\,\Delta t + 3(\Delta t)^2

Three times t plus delta t, squared, minus three t squared, equals six t delta t plus three delta t squared.

Divide that distance by the time Δt to get the average speed over the interval:

\frac{6t\,\Delta t + 3(\Delta t)^2}{\Delta t} = 6t + 3\,\Delta t

Six t delta t plus three delta t squared, divided by delta t, equals six t plus three delta t.

The 3Δt term shrinks with the interval and leaves 6t metres per second. At t = 1 s that is 6 m/s. The extra 3, 0.3, 0.03 and 0.003 in the list above are that 3Δt term, for Δt of 1, 0.1, 0.01 and 0.001 seconds.

Where this lesson stops

A derivative is the value an average rate settles on as the interval shrinks, in units of one quantity per unit of the other. The interval shrinks but is never set to zero, so nothing is ever divided by zero.

This lesson builds on

Functions and graphs

Return to the chapter

What the diffusion coefficient means

What changes when D changes?

Put a drop of dye into a long tube of still water. At first the dye sits in one spot; an hour later it has spread along the tube, thinner and wider. The diffusion coefficient D says how fast that spreading goes.

The rule it follows is local. Look at one spot. If the dye is thicker on both sides than at the spot itself, a dip, dye drifts in from both sides and the spot fills. If the spot is a peak, thicker than its neighbours, dye drifts out and it thins. The more sharply the dip or peak is curved, the faster it changes, and D sets the pace. With p for how much dye sits per metre of tube at place x and time t:

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

The time rate of change of density is D times its spatial curvature.

The right-hand side is D times the curvature, which is positive at a dip and negative at a peak. For one particle instead of a drop of dye, p is the probability per metre of finding it at x. Where the tube ends matters too: a closed tube keeps all its dye, and after a long time spreads it evenly, not in the ever-widening bell of an endless tube.

The same rule can be read as a flow. The amount crossing a point each second runs from thick to thin, faster where the density falls more steeply:

J=-D\frac{\partial p}{\partial x}

Fick’s constitutive law sends flux down the density gradient.

Combining the flux law with conservation gives the diffusion equation for constant D. The flux law itself is a model premise in this route, not something conservation alone proves.

Worked example: Doubling the diffusion coefficient, then quadrupling the time

Suppose the dye's typical distance from where it started is 1 mm after one minute. Double D, keeping everything else the same: the mean square doubles, so the typical distance becomes about 1.4 mm, the square root of 2 times as far, not twice. Keep the original D and wait four minutes instead of one: the typical distance doubles, to 2 mm. Neither comparison says where any single particle goes.

Where this lesson stops

D is measured in square metres per second. Before using it, say where the tube ends and what the model assumes, since both change the answer.

Return to the chapter

Density is not probability

What does the height of a probability curve mean?

Mark where 100 particles end up on a ruler divided into 2 μm bins. If 30 land in one bin, that bin holds a probability of about 0.3, and the density there is 0.3 divided by 2 μm, or 0.15 per micrometre. Measured in metres the same density is 150,000 per metre, while the bin still holds 0.3: the height depends on the unit of length, and the probability does not.

P(a\le X\le b)=\int_a^b p(x)\,dx

The probability of the interval from a to b is the integral of the density over that interval.

Under a smooth curve, any single exact position has probability zero, since an interval of zero width has no area. At the starting instant of the Brownian problem every particle sits at one point. No curve can draw that: the whole probability, 1, sits on the single point, and a curve appears only once the particles have begun to spread.

Worked example: A flat density across four micrometres

Spread the probability evenly over 4 μm. The density is 1 divided by 4 μm, or 0.25 per micrometre.
The area over the whole 4 μm is 0.25 per μm × 4 μm = 1.
A 1 μm piece has probability 0.25 × 1 = 0.25, and a 2 μm piece has 0.5.
Squeeze the same probability into 2 μm and the height doubles to 0.5 per micrometre, while the total stays 1. The height was never a probability.

Where this lesson stops

The height of the curve is a density, a probability per unit length. Only the area over an interval is a probability.

This lesson builds on

Adding and averaging

Return to the chapter

Exponential change

What kind of change does an exponential describe?

Suppose a quantity changes at a rate proportional to its own size. A radioactive sample loses the same fraction of its atoms every hour; a beam of light loses the same fraction of its intensity in every millimetre of absorbing glass. Take 1,000 atoms that lose half their number each hour:

After one hour 500 remain, after two 250, after three 125.
The losses shrink as the sample does: 500 in the first hour, 250 in the second, 125 in the third.
Each hour multiplies the count by the same factor, one half. Equal steps, equal factors: that is exponential change.

The function e to the power x, with e about 2.718, is the one whose rate of change equals its own value everywhere, which is why it describes this kind of change. In symbols, with N₀ atoms at the start and a constant k that sets how fast they decay (for halving every hour, k is about 0.693 per hour):

\begin{gathered} \frac{dN}{dt} = -kN, \\ \text{so}\quad N(t) = N_0\,e^{-kt} \end{gathered}

If the rate of change of N is minus k times N, then N at time t is N nought times e to the minus k t.

Whatever sits in the exponent must be a pure number with no units. The reason is the series that defines e to the power x:

e^{x} = 1 + x + \frac{x^{2}}{2} + \frac{x^{3}}{6} + \ldots

e to the x equals one plus x plus x squared over two plus x cubed over six, and so on.

If x were a length, the series would add a length to an area to a volume, which has no meaning. In the decay law, k is a rate per hour and t is in hours, so kt is a pure number.

Two exponentials from the 1905 papers pass that check. Wien's radiation law, which the light-quanta paper uses from §4 on, contains βν/T: β times a frequency is a temperature, so dividing by the temperature T leaves a pure number. The formula below writes e to the power −βν/T as exp(−βν/T). exp(u) is another way of writing e to the power u, used when the power is long.

\rho = \alpha\,\nu^{3} e^{-\beta \nu / T}

Rho equals alpha nu cubed times e to the minus beta nu over T.

The Brownian paper's §4 solution contains x²/4Dt: x² is an area, and a diffusion coefficient D in square metres per second times a time t in seconds is also an area, so their ratio is a pure number.

f(x,t) = \frac{n}{\sqrt{4\pi D}}\,\frac{e^{-x^{2}/4Dt}}{\sqrt{t}}

f of x and t equals n over the square root of four pi D, times e to the minus x squared over four D t, divided by the square root of t.

Worked example: The bell curve at its centre and one width out

The Brownian curve's shape comes from its exponential factor, e raised to −x²/4Dt. Follow that factor outward from the centre:

At the centre, x = 0, the exponent is 0 and e⁰ = 1: the curve is at its peak.
At x = √(2Dt), one root-mean-square width out, the exponent is −2Dt/4Dt = −1/2, and e raised to −1/2 is about 0.607 of the peak.
At x = √(4Dt), the exponent is −1, and e raised to −1 is about 0.368 of the peak.
Doubling the time doubles 4Dt, so each of these heights is reached √2 times farther from the centre.

None of these steps depends on the units chosen for x, t and D, as long as they agree with one another. That is what a pure-number exponent guarantees.

Where this lesson stops

Whatever sits in an exponent must be a pure number. Check it by cancelling the units, as with βν/T and x²/4Dt.

This lesson builds on

Rates of change and derivatives
logarithms

Return to the chapter

Counting what crosses a boundary

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

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

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

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

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

Worked example: Seven particles in, five out

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

Where this lesson stops

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

This lesson builds on

Adding and averaging

Return to the chapter

Functions and graphs

How does a graph record the way one quantity depends on another?

A bath fills from a tap. After one minute it holds 10 litres, after two 20, after three 30. For each time there is exactly one amount of water, and that pairing, one output for each input, is a function. Plot the pairs with time along the bottom and litres up the side and they fall on a straight line that climbs 10 litres for every minute. The slope, in litres per minute, is how fast the bath fills.

In the papers the pairs are physical: the mean square displacement of a Brownian particle for each elapsed time, or the concentration for each position along a tube. A function says how one thing responds when another changes.

A graph puts the input along the horizontal axis and the output up the vertical one, so each point is one pair. Both axes carry units, and so does everything read off the graph: the slope of mean square displacement against time is in square micrometres per second, and the area under a probability density is a plain probability.

A formula such as ⟨x²⟩ = 2Dt describes an average over many particles. One particle's squared displacement scatters widely around the line; the average over many lies close to it. The Brownian paper predicts the average, and ends by hoping that a researcher will soon decide the question by observation.

Worked example: Mean square displacement plotted against time

\langle x^2\rangle = 2Dt

The mean square displacement equals two times the diffusion coefficient times the time.

Take D = 0.5 square micrometres per second, so 2D = 1 square micrometre per second.
At t = 1 s the mean square displacement is 1 µm²; at t = 4 s it is 4 µm²; at t = 9 s, 9 µm².
Plotted against time, these points lie on a straight line through the origin whose slope is 2D, 1 µm² per second.
Their square roots, the root-mean-square displacements, are 1, 2 and 3 µm: four times the time gives twice the distance.

Where this lesson stops

A graph is a record of paired quantities, each with its units. A slope read off it has units too, and a slope stated without them explains nothing.

This lesson builds on

Reading a graph

Return to the chapter

The Gaussian and its width

After a time t, where is a spreading particle likely to be?

Release many particles at one point in still water and look again after a time t. They have spread out: most are still near the start, fewer are farther away, and as many went left as right. Count how many ended up at each distance and the counts make a bell-shaped curve, the Gaussian. One number sets its width, √(2Dt), the root-mean-square distance. For D = 0.5 µm² per second and t = 1 s the width is 1 µm, and:

Within one width of the start, between −1 µm and +1 µm, end about 68 per cent of the particles.
Within two widths, ±2 µm, end about 95 per cent; within three, about 99.7 per cent.
So about one particle in three ends more than one width from where it started, and about one in twenty more than two widths away.

Wait four times as long and the width doubles, because it grows as the square root of the time. The curve itself, with p the probability per micrometre of finding a particle at displacement x after time t, is:

p(x,t)=\frac{1}{\sqrt{4\pi Dt}}\exp\left(-\frac{x^2}{4Dt}\right)

p of x and t equals one over the square root of four pi D t, times e to the minus x squared over four D t.

The factor in front keeps the total probability at one as the curve spreads: a wider bell has to be a lower one. The exponential makes the curve fall away on both sides, the same way to the left as to the right, because x appears squared.

Worked example: Checking the curve: total probability one, and mean square 2Dt

In the formula above, x is the displacement, t the time since the start and D the diffusion coefficient. Two things need checking: that the total probability under the curve is 1, and that the mean square displacement is 2Dt. One change of variable turns both into standard integrals.

u=\frac{x}{\sqrt{4Dt}},\qquad dx=\sqrt{4Dt}\,du

Let u equal x over the square root of four D t. Then d x equals the square root of four D t, times d u.

\begin{aligned}\int_{-\infty}^{\infty}p(x,t)\,dx&=\frac{1}{\sqrt{\pi}}\int_{-\infty}^{\infty}e^{-u^2}\,du\\&=1\end{aligned}

The total probability becomes one over the square root of pi, times the integral of e to the minus u squared, and that equals one.

The last step needs the integral of e to the minus u squared to be √π. One way to see it is to square the integral, which makes it an integral over a whole plane, and then add the plane up in rings around the origin: a ring of radius r and width dr has area 2πr dr. This check is a modern one; the paper does not show it.

\begin{aligned}I^2&=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(u^2+v^2)}\,du\,dv\\&=\int_{0}^{\infty}e^{-r^2}\,2\pi r\,dr=\pi\end{aligned}

I squared is the integral of e to the minus u squared plus v squared over the whole plane, which equals the integral from zero to infinity of e to the minus r squared times two pi r d r, which is pi.

For the mean square, the same substitution leaves the integral of u² times e to the minus u squared. Integrating by parts once, the end terms vanish at both infinities, and half the integral above remains:

\frac{1}{\sqrt\pi}\int_{-\infty}^{\infty}u^2e^{-u^2}\,du=\frac12

The normalized integral of u squared times e to the minus u squared is one half.

Substituting back, the mean square is 4Dt times one half, which is 2Dt: the value the Brownian paper's §4 gives for the mean of the squared displacement, λx² = 2Dt.

Where this lesson stops

The bell curve says where many particles end up. Any one particle can end anywhere, and about one in three ends more than one width from its start.

This lesson builds on

Adding continuously
Mean, variance and RMS

Return to the chapter

Adding continuously

How do you add up something that varies continuously?

Chance spread along a line is measured per unit length: a density. Where the density is the same all along a stretch, the chance of the stretch is the density times its length. A particle equally likely to be anywhere in a 4 µm stretch has density 0.25 per micrometre, and the chance of finding it in the first 1.5 µm is 0.25 × 1.5 = 0.375.

Where the density changes from place to place, cut the line into narrow strips. Each strip contributes its height times its width, as if the density were flat across it. Add the strips, then cut them narrower and add again. For a smooth curve the totals settle on one value, and that value is the integral:

\int_a^b p(x)\,dx = \lim_{\Delta x \to 0} \sum_i p(x_i)\,\Delta x

The integral of p of x from a to b is the value that the sum of p of x i times delta x settles on as delta x shrinks.

The units multiply. A probability density is a probability per metre, so a strip of it, times a width in metres, is a plain probability. For a density the whole area must be 1, because the particle has to be somewhere. The Brownian paper's §4 writes exactly that for the chances of a jump Δ, ∫φ(Δ) dΔ = 1, and the light-quanta paper's §3 adds entropy over every frequency, S = v∫φ dν.

One more tool recurs in the papers: integration by parts. It is the product rule for derivatives added up over an interval, and it moves a derivative from one factor to the other at the price of an endpoint term:

\int_a^b u\frac{dv}{dx}\,dx = [uv]_a^b - \int_a^b v\frac{du}{dx}\,dx

The integral of u times the derivative of v equals u v at b minus u v at a, minus the integral of v times the derivative of u.

The endpoint term, uv at b minus uv at a, is part of the answer. It disappears only when uv is zero at both ends, as it is for a density that falls to zero far out on both sides.

Worked example: A rising density, added up in narrower strips

A particle's density rises steadily across a 4 µm stretch, from 0 at the left end to 0.5 per micrometre at the right. The whole area is a triangle, ½ × 4 × 0.5 = 1.
Four strips 1 µm wide, each taking the height at its left edge: 0 + 0.125 + 0.25 + 0.375 = 0.75.
Eight strips 0.5 µm wide: the heights add to 1.75, and times 0.5 that is 0.875.
Sixteen strips 0.25 µm wide: 0.9375. The shortfall halves each time the strips do: 0.25, then 0.125, then 0.0625.
The totals settle on 1, the area of the triangle. That settled value is the integral.

A flat density gives the same total for every cut, which is why the uniform stretch above needed no strips. A curved one, such as the Brownian bell curve, does, and the integral is the value the totals settle on.

Where this lesson stops

An integral is a total of height times width over a range, in the units of both. The widths and any endpoint term are part of the calculation, not decoration on the integral sign.

This lesson builds on

Density is not probability

Return to the chapter

Mean, variance and RMS

If particles wander both ways, which average shows how far they got?

Four particles move −3, −1, +1 and +3 micrometres. Their average position has not changed, yet they have spread out. Three averages answer three different questions.

\langle x\rangle=\frac{1}{M}\sum_{i=1}^{M}x_i

Add the M signed displacements and divide by M.

\langle x^2\rangle=\frac{1}{M}\sum_{i=1}^{M}x_i^2

Square each displacement before adding and dividing by M.

\operatorname{Var}(x)=\langle x^2\rangle-\langle x\rangle^2

Variance is the mean square minus the square of the mean.

RMS means root mean square: average the squares, then take the square root. The standard deviation is the square root of the variance, so it measures spread around the mean, where the RMS measures distance from zero. The two are equal only when the mean is zero. The average of the distances, ignoring sign, is a third statistic and gives a different number.

Worked example: Four displacements, four different averages

For −3, −1, +1, +3 the signed sum is 0, so the mean is 0.
The distances, ignoring sign, are 3, 1, 1, 3; their mean is 2.
The squares are 9, 1, 1, 9; their mean is 5.
The RMS is the square root of 5, about 2.236. With a mean of 0, the variance is also 5 and the standard deviation is also about 2.236.
A small random sample need not have a signed mean of exactly zero, even when the model's mean is zero.

Where this lesson stops

Say which average you mean. Squaring then averaging, averaging then squaring, and averaging the distances give three different numbers.

This lesson builds on

A sign records direction
Squares and square roots

Return to the chapter

Partial derivatives and held-fixed quantities

When a quantity depends on two things, what does its rate of change mean?

The concentration of particles in a tube depends on where you look, x, and when you look, t. Ask how fast the concentration changes and there are two different answers: how it changes as you move along the tube at one instant, and how it changes at one place as time passes.

A partial derivative picks one of them. ∂f/∂x is the rate of change along the tube with the time t held fixed, like comparing points in one photograph. ∂f/∂t is the rate of change in time at a fixed position x, like watching one spot under the microscope. The curly ∂ says that something else is being held fixed, and a subscript names it:

\left(\frac{\partial f}{\partial x}\right)_t \qquad \left(\frac{\partial f}{\partial t}\right)_x

The partial derivative of f with respect to x at fixed t, and the partial derivative of f with respect to t at fixed x.

Thermodynamics depends on the same care. How much a gas's volume changes as the pressure changes depends on whether its temperature T is held fixed, with heat flowing in and out, or its entropy S is held fixed, with the gas insulated. The two derivatives have different values, and ∂V/∂p written alone does not say which is meant:

\left(\frac{\partial V}{\partial p}\right)_T \quad \text{versus} \quad \left(\frac{\partial V}{\partial p}\right)_S

The partial derivative of V with respect to p at fixed temperature T, versus the same derivative at fixed entropy S.

In §3 of the light-quanta paper, radiation fills a fixed volume and φ is its entropy per unit volume and per unit interval of frequency, a function of the energy density ρ and of ν. Varying ρ with the frequency ν held fixed, Einstein finds that the derivative is 1/T:

\left(\frac{\partial \varphi}{\partial \rho}\right)_{\nu} = \frac{1}{T}

The partial derivative of phi with respect to rho, at fixed frequency nu, equals one over T.

In §6 of the relativity paper, Maxwell's equations are rewritten in the coordinates ξ, η, ζ, τ of a moving system. A derivative with respect to x taken with y, z and t held fixed then becomes a combination of a derivative with respect to ξ and one with respect to τ, because holding the resting time t fixed does not hold the moving time τ fixed.

Worked example: The same diffusion profile, differentiated two ways

Take the spreading profile of the Brownian paper's §4, written for a single particle:

f(x,t) = \frac{1}{\sqrt{4\pi D t}}\,e^{-x^2/4Dt}

f of x and t equals one over the square root of four pi D t, times e to the minus x squared over four D t.

Hold t fixed and differentiate with respect to x. Only the exponent depends on x, and the derivative of −x²/4Dt with respect to x is −x/2Dt:

\left(\frac{\partial f}{\partial x}\right)_t = -\frac{x}{2Dt}\,f(x,t)

The partial derivative of f with respect to x at fixed t equals minus x over two D t, times f.

That is the slope of the profile in one snapshot: downhill to the right of the centre, uphill to the left. Now hold x fixed and differentiate with respect to t. Both the factor in front and the exponent depend on t:

\left(\frac{\partial f}{\partial t}\right)_x = \left(-\frac{1}{2t} + \frac{x^2}{4Dt^2}\right) f(x,t)

The partial derivative of f with respect to t at fixed x equals minus one over two t, plus x squared over four D t squared, all times f.

That is what a probe fixed at one place records. Near the centre, where x² is less than 2Dt, the bracket is negative: the concentration falls as the cloud spreads away. Farther out it is positive: the concentration rises as the cloud arrives. Differentiate the first result once more with respect to x and multiply by D, and you get the second. That is the §4 equation ∂f/∂t = D ∂²f/∂x², a rate at one place equal to a curvature at one time.

Where this lesson stops

A partial derivative is not defined until you have said which quantities are held fixed.

This lesson builds on

Rates of change and derivatives

Return to the chapter

Probability and independence

When you square a sum of random steps, why do the mixed terms average to zero?

Flip a coin twice and take one step for each flip: a metre to the right for heads, a metre to the left for tails. How far from the start do you end up, on average?

The four outcomes, right-right, right-left, left-right and left-left, are equally likely: one chance in four each.
They leave you 2 m to the right, back at the start, back at the start, or 2 m to the left.
Square those distances so that left and right count alike: 4, 0, 0 and 4. Their average is 2, one square metre for each step.

Combining the two steps added nothing beyond one square metre each. In symbols, with Δ₁ and Δ₂ for the two steps, squaring the sum gives each step squared and a mixed term, twice their product:

(\Delta_1 + \Delta_2)^2 = \Delta_1^2 + 2\Delta_1\Delta_2 + \Delta_2^2

Delta one plus delta two, squared, equals delta one squared plus two delta one delta two plus delta two squared.

In the coin walk the product of the two steps is +1 for right-right and left-left and −1 for the other two, so on average it is zero and the mixed term adds nothing. Two conditions make that happen. Independence: learning the first step does not change the chances for the second, so the chance of a pair is the product of the two chances. Centring: each step averages zero, left as likely as right. Then the average of the product is the product of the averages, zero times zero:

\langle \Delta_1 \Delta_2 \rangle = \langle \Delta_1 \rangle \langle \Delta_2 \rangle = 0

The average of delta one times delta two equals the average of delta one times the average of delta two, which is zero.

Both conditions matter. If each step drifts, averaging +1, the product of the averages is 1, not 0. If the second step tends to copy the first, the product is positive more often than negative. The Brownian paper's §4 assumes that a particle's motions in successive intervals are independent, as long as the intervals are not chosen too small.

Worked example: What changes when the second step copies the first

Now let the second flip always copy the first. Only right-right and left-left can happen, one chance in two each.
The product of the two steps is +1 every time, so the mixed term no longer averages zero.
You always end 2 m from the start, so the squared distance is 4 every time: the average is 4, not 2. Independence removed that extra 2.

Where this lesson stops

Before dropping a mixed term, ask two things: does learning one step change the chances for the next, and does each step average zero?

This lesson builds on

Adding and averaging
A sign records direction

Return to the chapter

From steps to spread

Why does typical displacement grow as the square root of the step count?

A walker flips a coin before every step: heads, one metre to the right; tails, one metre to the left. After four steps, how far from the start is the walker likely to be? Not four metres: the steps partly cancel. Count the walks, and square each final distance so that left and right count alike:

After one step the walker is 1 m away, every time. The squared distance is 1.
After two steps there are four equally likely walks, ending 2 m right, at the start, at the start, or 2 m left. The squared distances 4, 0, 0 and 4 average 2.
After four steps there are sixteen equally likely walks. Two end 4 m out, eight end 2 m out and six end at the start. The squared distances average (2 × 16 + 8 × 4) ÷ 16 = 4.

The average squared distance grows by one square metre per step: after n steps it is n. The typical distance, its square root, grows only as the square root of n: 1 m after one step, about 1.4 m after two, 2 m after four, 10 m after a hundred. To get twice as far takes four times as many steps.

In symbols, call the steps Δ₁, Δ₂ and so on, and the position after n steps X, their sum:

X_n=\Delta_1+\ldots+\Delta_n

Position after n steps is the sum of the n signed increments.

Suppose each step averages zero and has the same mean square l², where l is the typical size of one step (1 m in the coin walk). Squaring the sum gives the n squared steps and many products of two different steps. When the steps are independent and each averages zero, every such product averages zero, as the lesson on independence shows for two coin steps. Only the squares are left to add up:

\langle X_n^2\rangle=nl^2,\qquad \sqrt{\langle X_n^2\rangle}=l\sqrt n

The mean square is n times ell squared, and RMS displacement is ell times the square root of n.

Worked example: A hundred steps, then four hundred

A hundred steps of typical size 1 m leave a typical distance of 10 m from the start, because the square root of 100 is 10. Four hundred steps leave 20 m: four times the steps, twice the distance. No single walker has to end exactly 10 or 20 m away; these are typical sizes over many walkers. Nor is the distance from the start the length of the path: to end 10 m from home, the walker has walked 100 m.

Where this lesson stops

The coin walk is a picture, not a claim that molecules push a particle in fixed jumps. The Brownian paper needs only that the particle's moves in successive intervals are independent of one another.

This lesson builds on

Probability and independence
Mean, variance and RMS

Return to the chapter

Approximating a curve near a point

Why can a curve be replaced, close to one point, by a few simple terms?

What is 2.1 squared, without a calculator? Start from what you know: 2 squared is 4. Picture a square 2 wide growing to 2.1 wide. It gains two strips, each 2 long and 0.1 thick, which add 0.4, and a small corner, 0.1 by 0.1, which adds 0.01. So 2.1 squared is 4.41. The first correction, 0.4, did almost all the work; the second, 0.01, is forty times smaller.

The same recipe works for any smooth curve, close to a point where you know it. Start with its value there. Add its slope times the step Δ: the slope, or first derivative, is the change in output per unit of input. Then add half its curvature times Δ²: the curvature, or second derivative, says how fast the slope itself changes. Further terms carry higher powers of Δ, so they are small only when Δ is:

\begin{aligned}f(x+\Delta)&=f(x)+\Delta f'(x)\\&\quad+\frac{\Delta^2}{2}f''(x)+\ldots\end{aligned}

The value a step delta away equals the value, plus delta times the slope, plus half of delta squared times the curvature, plus further terms.

The Brownian paper does exactly this in §4. It expands the number of particles at x + Δ in powers of the jump Δ and keeps the terms up to Δ², because only very small jumps contribute and each further surviving term is much smaller than the one before. The Δ term averages away, since jumps to the left and to the right are equally likely, and the Δ² term becomes the diffusion coefficient. A real particle does not make independent jumps over arbitrarily short times, so the expansion holds for times long enough for successive jumps to be independent and short enough for each jump to be small.

Worked example: What the dropped term costs

(x+\Delta)^2=x^2+2x\Delta+\Delta^2

For the square function, the local expansion is exact through the quadratic term.

Keep only the value and the slope term, 4 + 2 × 2 × Δ, and see what dropping Δ² costs as the step grows:

Δ = 0.1: 4.4 instead of 4.41. The error, 0.01, is about 0.2 per cent.
Δ = 0.5: 6 instead of 6.25. The error, 0.25, is 4 per cent.
Δ = 2: 12 instead of 16. The error, 4, is a quarter of the answer.
Halving the step quarters the error, because the dropped term is Δ².

Where this lesson stops

An expansion is a claim about size. Say which terms you keep, and why the ones you drop are smaller: here, because they carry higher powers of a small step.

This lesson builds on

Squares and square roots

Return to the chapter

Static laboratory examples

bm-01: a built-in worked result

Static host-calculated example, not a running experiment or an observation. Canonical units and owner-supplied statuses are retained. Open bm-01 online

Build-time scalar readouts
Quantity	Value	Unit	Status
temperature	293.15	K	value
viscosity	0.001	Pa s	value
particleRadius	5e-7	m	value
observationInterval	1	s	value
boltzmannConstant	1.380649e-23	J/K	value
diffusionCoefficient	4.294395645549615e-13	m2/s	value
rmsDisplacement1d	9.267573194261392e-7	m	value
modelSecondMoment	8.588791291099231e-13	m2	value
modelMeanNorm	7.394453567811659e-7	m	value
modelRmsNorm	9.267573194261392e-7	m	value
sampleMean	2.9236813725820092e-9	m	value
sampleMeanAbsolute	7.780756267413605e-7	m	value
sampleMeanSquare	9.722317299024293e-13	m2	value
sampleRms	9.8601811844531e-7	m	value
sampleMeanNorm	7.780756267413605e-7	m	value
sampleMeanSquareNorm	9.722317299024293e-13	m2	value
sampleRmsNorm	9.8601811844531e-7	m	value
modelApparentSpeed	9.267573194261392e-7	m/s	value
sampledApparentSpeed	9.8601811844531e-7	m/s	value
underflow	0	1	value
overflow	0	1	value
recordingDraws	1200000	1	value
reusedRecording	0	1	value
ensembleSize	400	1	value
signedMean	2.9236813725820092e-9	m	value
meanSquare	9.722317299024293e-13	m2	value
lambdaX1s	9.267573194261392e-7	m	value
lambdaX60s	0.00000717863132822653	m	value
signedMeanLowerBand	-1.524759866588756e-7	m	value
signedMeanUpperBand	1.524759866588756e-7	m	value
meanSquareLowerBand	6.729893316990064e-13	m2	value
meanSquareUpperBand	1.0728827253311408e-12	m2	value
kolmogorovDistance	0.03627333580623167	1	value

18 array outputs are not reproduced. This table does not claim to reproduce the interactive plot.

Settings (canonical units)

H: 10
M: 400
T: 293.15
a: 5e-7
axis: 0
d: 1
eta: 0.001
h: 0.02
interval: 1
seed: 1905
statistic: rms

Evaluator: source:sha256:6e9e4cb91ff4a2493d2b641284d3697eec06b64a62f056de23ddc501ecb7aa67

bm-05: a built-in worked result

Static host-calculated example, not a running experiment or an observation. Canonical units and owner-supplied statuses are retained. Open bm-05 online

Build-time scalar readouts
Quantity	Value	Unit	Status
stepMean	0	m	value
stepSecondMoment	2.5e-13	m2	value
stepFourthMoment	6.25e-26	m4	value
diffusionCoefficient	1.25e-12	m2/s	value
stepKurtosis	-2	1	value
sampleMean	4.999999999999988e-10	m	value
sampleMeanSquare	9.865e-13	m2	value
sampleRms	9.932270636667127e-7	m	value
modelMeanSquare	1e-12	m2	value
modelRms	0.000001	m	value
elapsedTime	0.4	s	value
samplingTerm	0.04359157733881077	1	value
walkerCount	2000	1	value
sumKurtosis	-0.5	1	value
kolmogorovDistance	0.192	1	value
shapeTerm	0.1875	1	value
agreementBound	0.23109157733881078	1	value
withinBound	1	1	value
underflow	0	1	value
overflow	0	1	value
coinDenominator	16	1	value
biasedDiffusion	No numeric value	m2/s	outside-domain
biasedMean	9.999999999999997e-8	m	value
biasedDrift	9.999999999999995e-7	m/s	value
biasedCenteredDiffusion	1.2e-12	m2/s	value
cauchyDiffusion	No numeric value	m2/s	outside-domain
continuumLimit	No numeric value	m2/s	outside-domain
recordingDraws	800000	1	value
requestDraws	800000	1	value
replayedDraws	0	1	value
reusedRecording	0	1	value
retainedBytes	144160	1	value

20 array outputs are not reproduced. This table does not claim to reproduce the interactive plot.

Settings (canonical units)

bias: 0.6
kernel: coin
n: 4
runSteps: 400
seed: 1905
stepRms: 5e-7
tau: 0.1
walkers: 2000

Evaluator: source:sha256:3887735794dc6c231ef169876a981f8900999052e1f3a0ecc112841dfaf42329

bm-06: a built-in worked result

Static host-calculated example, not a running experiment or an observation. Canonical units and owner-supplied statuses are retained. Open bm-06 online

Build-time scalar readouts
Quantity	Value	Unit	Status
diffusionCoefficient	4.294395645549615e-13	m2/s	value
activeDiffusionCoefficient	4.294395645549615e-13	m2/s	value
rmsDisplacement1d	9.267573194261392e-7	m	value
meanSquareDisplacement1d	8.588791291099231e-13	m2	value
intervalProbability	0.7194261506398167	1	value
meanRadius2d	0.0000011615180502973971	m	value
rmsRadius2d	0.0000013106327701609808	m	value
mostLikelyRadius2d	9.267573194261391e-7	m	value
meanRadius3d	0.0000014788907135623319	m	value
rmsRadius3d	0.0000016051907635324123	m	value
gridDensity	No numeric value	1/m	not-applicable
cellMasses	No numeric value	1	not-applicable
cellProbabilities	No numeric value	1	not-applicable
maxCellMassDifference	No numeric value	1	not-applicable
wallContact	No numeric value	1	not-applicable
stabilityRatio	No numeric value	1	not-applicable
gridTimeStep	No numeric value	s	not-applicable

4 array outputs are not reproduced. This table does not claim to reproduce the interactive plot.

Settings (canonical units)

T: 293.15
a: 5e-7
copiedDiffusivityInstanceId
copiedDiffusivityRunId
copiedDiffusivitySnapshotVersion: 0
copiedDiffusivityValue: 0
dx: 1e-7
eta: 0.001
gridEnabled: false
lower: -0.000001
n: 101
steps: 250
t: 1
upper: 0.000001

Evaluator: source:sha256:c649ff3318872a0745312cb8c36ff143e63024ccba89f85fd8eb4b24ec660b3c

Source references

A. Einstein, Über einen die Erzeugung und Verwandlung des Lichtes betreffenden heuristischen Gesichtspunkt. Annalen der Physik (4), 17, 132–148 (1905).
A. Einstein, On the motion of particles suspended in liquids at rest required by the molecular-kinetic theory of heat. Annalen der Physik (4), 17, 549–560 (1905), §§1–5. Bibliographic pointer; this preview is not a source transcription or translation.