Brownian motion: from wandering to a measurable law: §5 · From displacement to molecular scale

Fractions and ratios

What does dividing one quantity by another tell you?

The fraction 6/2 is 3, because 2 fits into 6 three times; it also says that 6 is three times 2. Every fraction a/b works the same way. The bottom number can never be zero, since no count of zeros adds up to 6.

When the two quantities have different units, their ratio is a new quantity with a unit of its own: 12 micrometres travelled in 4 seconds is 3 micrometres per second. When they have the same unit, the units cancel and the ratio is a plain number: 2 particles out of 8 is 0.25 of the group.

Worked example: Fourteen metres in ten seconds

1. A steady walk covers 14 metres in 10 seconds.
2. Divide the distance by the time: 14 divided by 10 is 1.4, and the unit divides with it, so the rate is 1.4 metres per second.
3. After 60 seconds the same walk has covered 84 metres, and 84 divided by 60 is 1.4 again. The walk is steady, so the ratio does not depend on how long it is watched.
4. Metres and seconds are different units, so neither cancels and what comes out is a new quantity that keeps a unit of its own.
5. The drawing below is the other case: 2 particles out of 8 has the same unit above and below, they cancel, and 0.25 is left as a plain number.

Where this lesson stops

A ratio of two quantities in the same unit is a plain number. A ratio of different units, such as micrometres per second, is a new quantity and keeps its unit.

This lesson builds on

• Adding and averaging

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

A probability, a frequency and a single trial

What does a probability of 1/4 say about one try, and about a hundred?

Drop two tokens into a box, each landing in the left or the right half with equal chance. There are four outcomes: left and left, left and right, right and left, right and right. One of the four is both left, so the model probability of both left is 1/4.

Now drop them 100 times and count. Both might land left 23 times: a frequency of 23/100. Over more and more trials the frequency tends to settle near 1/4, but any particular count can differ from 25.

A single drop either lands both left or it does not. A probability of 1/4 is not a quarter of an outcome. It says how often, over many drops.

The light-quanta paper counts in exactly this way in §5. The chance that n independent points are all in a part v of a volume v₀ is (v/v₀)ⁿ, and the paper calls it a statistical probability: a frequency over many moments.

Worked example: Counting outcomes for two tokens and for four

1. Two tokens: 4 equally likely outcomes, 1 of them both left, so the probability is 1/4.
2. Four tokens: 2⁴ = 16 outcomes, 1 of them all left, so the probability is 1/16, which is (½)⁴.
3. A run of 100 drops of four tokens might show all left 5 times: a frequency of 5/100, near 1/16 (about 6 in 100) but not equal to it.
4. Any one drop shows all left or it does not. The 1/16 is about the long run.

Where this lesson stops

A probability counts favoured outcomes among equally likely ones; a frequency counts what happened. How a probability becomes an entropy is the lesson on entropy and the number of ways.

This lesson builds on

• Fractions and ratios

Powers of ten and physical units

How do you write a very small length without counting zeros?

The Brownian paper works its example, in §5, for particles 0.001 mm across. In metres that is 0.000001 m: easy to miscount by one zero, which is a factor of ten. Written as 10⁻⁶ m, the −6 says the decimal point has moved six places to the left, and that length has a name of its own, the micrometre (μm).

The unit goes through the arithmetic with the number. A micrometre is 10⁻⁶ m, so a square micrometre is 10⁻¹² m², not 10⁻⁶ m²: squaring a length squares its unit, power of ten included.

Worked example: Squaring six micrometres

1. 1 micrometre (1 μm) is 10⁻⁶ metres, or 0.000001 m.
2. The paper's displacement after one minute, about 6 μm, is 6 × 10⁻⁶ m.
3. Squaring it squares the number and the unit: (6 × 10⁻⁶ m)² = 36 × 10⁻¹² m² = 3.6 × 10⁻¹¹ m².

Where this lesson stops

A measurement is a number and a unit together. Squaring it squares both: (6 × 10⁻⁶ m)² is 3.6 × 10⁻¹¹ m².

This lesson builds on

• Fractions and ratios

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.

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

1. The displacements −3, −1, +1, +3 have squares 9, 1, 1, 9. Their average is 5, so the RMS is √5, about 2.236.
2. Double each displacement: −6, −2, +2, +6. The squares are 36, 4, 4, 36, and their average is 20.
3. 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

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

1. One morning the four jars hold 3, 1, 1 and 3 marbles. Add them: 8. Divide by the four jars: an average of 2.
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.
3. A third morning every jar holds 2, which is the average itself. Add them: 8. Divide by four: 2.
4. 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.
5. 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.

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:

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:

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.

This lesson builds on

• Counting what crosses a boundary
• Squares and square roots

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.

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

1. Spread the probability evenly over 4 μm. The density is 1 divided by 4 μm, or 0.25 per micrometre.
2. The area over the whole 4 μm is 0.25 per μm × 4 μm = 1.
3. A 1 μm piece has probability 0.25 × 1 = 0.25, and a 2 μm piece has 0.5.
4. 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

Uncertainty, evidence and inference

What can a set of measurements tell you, and what can it not?

Some questions the data cannot answer, however carefully you measure. If all you know of a rectangle is that its area is 12 square centimetres, its sides could be 3 and 4, 2 and 6, or 1 and 12. The area fixes their product, not each side; to get one side you must measure the other.

The Brownian paper meets the same limit. In its §3 the diffusion coefficient is D = RT/(N · 6πkP), with k the liquid's viscosity and P the particle's radius. So watching particles spread gives D, and with the temperature and viscosity known, D fixes the product of N and P, not each of them: a smaller particle and a larger N give the same spreading. To get N, the radius has to be measured some other way.

An estimate comes from a finite sample, and a different sample gives a different number. Two questions follow: whether the method lands on the right value on average, which is its bias, and how widely its answers scatter from sample to sample. Fitting a drift from the same displacements uses up part of the information and leaves less to estimate the scatter.

Uncertainty comes in four kinds, and each needs its own statement. Sampling uncertainty is the scatter from one sample to the next. Measurement uncertainty is the error in each reading, such as where the microscope places a particle. Model uncertainty is the chance that the formula itself does not fit these particles. Numerical uncertainty is the error of the computation. A confidence interval is not an interval-arithmetic enclosure, which is guaranteed to contain the exact result of a calculation, and a numerical error bar, which bounds the arithmetic, is not a measurement uncertainty.

A 95 per cent confidence interval comes from a method that, used on many samples, catches the true value in 95 of every 100 of them, provided its assumptions hold. It is not a 95 per cent chance that this one interval contains the value. A plot of many repeated intervals, with the misses left in, shows what the promise means.

For a diffusion coefficient the promise has an exact form. With M independent displacements in d coordinates there are q = dM squared displacements, and their sum divided by 2Dt has a known spread from sample to sample: the chi-square distribution with q degrees of freedom. Its 2.5 and 97.5 per cent points give the 95 per cent interval. In a synthetic example with 50 displacements in two coordinates, so q = 100, and an estimate of 0.43 μm²/s, those points are 74.2219 and 129.5612:

The interval runs from 43 divided by 129.5612 to 43 divided by 74.2219, that is from 0.3319 to 0.5793 square micrometres per second.

Inverting an estimate adds a bias of its own. The molecular number is found by dividing by the estimated D, and although that estimate is right on average, its reciprocal is not: with q = 100 degrees of freedom the average of the inverted estimate is too large by the factor q/(q − 2) = 1.020408, about two per cent. The interval itself survives inversion, with its ends swapped, as the worked example shows.

Reading positions adds noise of its own. If each position is read with an error of spread σ in each coordinate, each measured step gains 2σ² in its mean square per coordinate, and two neighbouring steps share one reading, so their errors have a covariance of −σ². Neighbouring steps are then no longer independent, and cutting one path into overlapping windows does not make independent samples.

An interval that treats the radius, viscosity and temperature as exact holds only on that condition. Allowing for their uncertainty needs to know how each was measured and how often its own interval misses. Without that, a plus-or-minus has no stated coverage and is not a confidence statement.

A path simulated from a chosen N is useful for checking that a method recovers the N it was given. It is not evidence about N in nature. Since 2019 the SI fixes Avogadro's number by definition, so recovering it from modern data checks consistency. Measuring N, as Perrin did from 1908, needs inputs measured independently of it.

Worked example: Inverting an interval, and watching its endpoints swap

1. Take a made-up relation, N = 12/D, with D measured as 4. The estimate is N = 12/4 = 3.
2. Suppose the method gives D between 3 and 6. A larger D means a smaller N, so N runs from 12/6 = 2 to 12/3 = 4: the ends swap over.
3. Now suppose N = 12/(aD), with a an unknown scale. The same D fits many pairs of a and N, and no interval for N follows until a is measured separately.
4. Declaring a = 2 halves both the estimate and its interval without changing a single measurement. It changes an assumption, not the data.
5. If three inputs each come from a method that misses at most once in 100 uses, and D's method at most twice, the combination misses at most 5 times in 100, even if the misses are related. Without the stated miss rates, this arithmetic guarantees nothing.

Where this lesson stops

Say what the data can tell apart, which method and model you used, what you held exact, and whether the result recovers a value you put in, estimates it independently, or checks consistency.

This lesson builds on

• Density is not probability
• Mean, variance and RMS
• The Gaussian and its width
• Probability and independence
• Squares and square roots
• A probability, a frequency and a single trial

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.

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

1. In one second, 7 particles cross into the stretch and 5 cross out. The count inside rises by 2.
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

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:

1. Within one width of the start, between −1 µm and +1 µm, end about 68 per cent of the particles.
2. Within two widths, ±2 µm, end about 95 per cent; within three, about 99.7 per cent.
3. 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 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.

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.

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.

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:

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

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:

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:

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

1. 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.
2. Four strips 1 µm wide, each taking the height at its left edge: 0 + 0.125 + 0.25 + 0.375 = 0.75.
3. Eight strips 0.5 µm wide: the heights add to 1.75, and times 0.5 that is 0.875.
4. Sixteen strips 0.25 µm wide: 0.9375. The shortfall halves each time the strips do: 0.25, then 0.125, then 0.0625.
5. 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

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.

Add the M signed displacements and divide by M.

Square each displacement before adding and dividing by M.

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

1. For −3, −1, +1, +3 the signed sum is 0, so the mean is 0.
2. The distances, ignoring sign, are 3, 1, 1, 3; their mean is 2.
3. The squares are 9, 1, 1, 9; their mean is 5.
4. 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.
5. 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

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?

1. The four outcomes, right-right, right-left, left-right and left-left, are equally likely: one chance in four each.
2. They leave you 2 m to the right, back at the start, back at the start, or 2 m to the left.
3. 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 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:

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

1. Now let the second flip always copy the first. Only right-right and left-left can happen, one chance in two each.
2. The product of the two steps is +1 every time, so the mixed term no longer averages zero.
3. 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

Ratios and scaling

If one quantity doubles, what happens to the others?

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

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

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

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

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

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

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

Where this lesson stops

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

This lesson builds on

• Fractions and ratios
• Squares and square roots

Two measurements, two unknowns

Can a measured spreading rate tell you how big the particle is?

In §5 of the Brownian paper the diffusion coefficient of a suspended sphere is $D = \frac{RT}{N}\cdot\frac{1}{6\pi\eta a}$. Einstein writes k for the viscosity and P for the radius; this lesson uses η and a. At a fixed temperature and viscosity, a measured D fixes one combination of the two unknowns, the product of the molecular number N and the radius a.

N times a equals R T divided by six pi eta D.

Every pair with that product fits the measurement equally well. Drawn in a plane with the radius along one axis and the molecular number along the other, the pairs form a curve. Double the radius and halve N, and the product, and with it D, is exactly what it was. Displacements alone cannot say which point on the curve is the true one, and measuring D more carefully does not change that.

Einstein's dissertation measured something else: how much a dissolved substance thickens water. The volume taken up by the dissolved particles, as a fraction of the whole, is $\varphi = \frac{4}{3}\pi a^3 N n$, where n is the number of moles dissolved per unit volume. The viscosity relation therefore fixes a different combination, $a^3 N$. Its curve in the same plane has another shape, and the two curves cross.

a cubed times N equals three times the quantity eta star over eta minus one, divided by four pi c n.

Here $\eta^*$ is the viscosity of the solution, η that of the water, and c the coefficient of φ in the viscosity law: 1 as the dissertation printed it in 1906, and 5/2 after Einstein's correction of 1911. This lesson takes the relation as given: deriving it, and the history of its correction, belong to the dissertation. Dividing the second combination by the first leaves $a^2$ alone, so the crossing gives the radius, and then either relation gives N.

a squared equals three times the quantity eta star over eta minus one, over four pi c n, times six pi eta D over R T.

A crossing is not yet an answer. Two curves that cross at a shallow angle fix their crossing poorly, so a small error in either measurement moves it a long way. Each measurement has an uncertainty, which widens its curve into a band, and the honest result is the region where the two bands overlap, not a point. Two measurements that rest on the same calibration can cross and still add nothing new. And each relation holds only within its own domain: dilute suspensions, spheres much larger than the molecules of the liquid.

Worked example: Why solving for the radius gives a curve

1. The question: can a measured spreading rate tell you how big the particle is?
2. What is given: the measured D, the temperature T, the viscosity η, and the relation $D = RT/(6\pi\eta a N)$.
3. The first thought, and a reasonable one: solve the relation for a. The algebra looks as if it allows that.
4. The decisive step: solving gives $a = RT/(6\pi\eta N D)$, the radius in terms of N. Every choice of N gives its own a. Double a and halve N, and $6\pi\eta a N$, and with it D, is exactly what it was. The answer is a curve, not a number.
5. The limitation: a second relation that depends on a and N in another way is needed. How well it fixes the pair depends on the angle at which the two curves cross and on the two measurements being independent, and what comes out is a region, not a point.

Where this lesson stops

One measurement of one combination of two quantities leaves a curve, not a point.

This lesson builds on

• Ratios and scaling
• Uncertainty, evidence and inference
• functions-graphs

The units of 1905

How do Einstein's numbers, printed in centimetres, grams, ergs and the old electrical units, become SI units?

Einstein's papers measure length in centimetres, mass in grams and time in seconds, the CGS system. Energy is in ergs, and one erg is 10⁻⁷ joules. Viscosity is in grams per centimetre per second, the unit now called the poise, and one poise is 0.1 pascal-seconds. These mechanical factors are exact, because they follow from 1 cm = 0.01 m and 1 g = 0.001 kg.

To convert, multiply by a ratio equal to one. Since 1 poise is 0.1 pascal-seconds, the ratio (0.1 Pa s)/(1 P) equals one, and multiplying by it changes the unit without changing the viscosity. The Brownian paper gives water's viscosity as 1,35 · 10⁻², with no unit name; it is in poise, so the viscosity is 1.35 × 10⁻³ Pa s. The paper calls this viscosity k, which is not Boltzmann's constant.

1.35 times ten to the minus two poise, times 0.1 pascal-seconds per poise, equals 1.35 times ten to the minus three pascal-seconds.

The same ratio works for compound units. The gas constant R = 8.31 × 10⁷ erg per mole per kelvin is 8.31 joules per mole per kelvin. Neither the light paper nor the Brownian paper prints this number; the edition supplies the standard value of 1905 as an editorial input and labels it that way.

Electricity had two CGS systems. The electrostatic system builds its charge unit, the statcoulomb, from the force between charges. The electromagnetic system builds its current unit from the force between currents; its units are the abcoulomb and the abvolt. One abcoulomb is 10 coulombs, one abvolt is 10⁻⁸ volts, one statvolt is 299.792458 volts, and one gauss is 10⁻⁴ tesla. Planck's 1901 elementary charge, 4,69 · 10⁻¹⁰ statcoulombs, is 1.56 × 10⁻¹⁹ coulombs.

These electrical correspondences are conventional, not exact. They assume the magnetic constant $\mu_0$ is exactly 4π × 10⁻⁷ henry per metre. Since the 2019 revision of the SI it is a measured quantity, equal to that value within about one part in a billion. The site marks every electrical conversion conventional.

The light paper's photoelectric check in §8 uses the electromagnetic system. It takes E = 9,6 · 10³ for the charge of one gram-equivalent of a monovalent ion, so a potential Π comes out in abvolts, and the paper multiplies by 10⁻⁸ to get volts. It prints the result as about 4,3 volts. On the result line the factor is set as 10⁷; the edition records this as a misprint, because the sentence four lines earlier gives 10⁻⁸. The check can also be run per electron, with Planck's charge in statcoulombs and the result in statvolts, and it gives 4.31 volts; the edition keeps that as a documented alternative and follows the page.

The paper's R, E and J are amounts per gram-equivalent, a mole of monovalent ions. Dividing by N gives the amount per molecule, R/N, or per electron, ε. With the paper's E and the N it prints in §2, 6,17 · 10²³, the charge on one ion is 9.6 × 10³ ÷ 6.17 × 10²³ = 1.56 × 10⁻²⁰ abcoulombs, which is 1.56 × 10⁻¹⁹ coulombs. The paper does not print this quotient.

The relativity paper writes its fields in Gaussian units, where electric and magnetic fields share one unit. That is why its transformation reads β(Y − (v/V)N), with the dimensionless ratio v/V; here N is a magnetic field component, not Avogadro's number. SI measures the electric field in volts per metre and the magnetic field in tesla, so the same law reads $\gamma(E_y - vB_z)$, with the speed itself multiplying the magnetic field. One statvolt per centimetre is 2.998 × 10⁴ volts per metre.

One abvolt corresponds to ten to the minus eight volts, one abcoulomb to ten coulombs, one statvolt to 299.792458 volts, and one gauss to ten to the minus four tesla, all by convention.

Worked example: The photoelectric check, converted to volts

1. The inputs: β = 4,866 · 10⁻¹¹ and ν = 1,03 · 10¹⁵ per second are printed in §8, and so is E = 9,6 · 10³. R = 8.31 × 10⁷ erg per mole per kelvin is the edition's editorial input, because the paper does not print it.
2. With P′ = 0, Π = Rβν/E = 8.31 × 10⁷ × 4.866 × 10⁻¹¹ × 1.03 × 10¹⁵ ÷ 9.6 × 10³ = 4.34 × 10⁸ abvolts.
3. Multiply by the ratio 10⁻⁸ volts per abvolt: 4.34 volts. The paper prints about 4,3 volts.
4. In §9 the conversion runs the other way. An energy per gram-equivalent divided by E is a potential: the printed Rβν = 6,4 · 10¹² erg gives 6.4 × 10¹² ÷ 9.6 × 10³ = 6.67 × 10⁸ abvolts, or 6.67 volts.
5. Stark's ionization voltage of 10 volts is 10⁹ abvolts, and 10⁹ × 9.6 × 10³ = 9.6 × 10¹² erg per gram-equivalent, the upper bound for J that §9 prints.
6. E itself, 9.6 × 10³ abcoulombs, is 9.6 × 10⁴ coulombs per gram-equivalent. Beside it, the modern Faraday constant is 96 485 coulombs per mole (set modern-si-2019); Einstein's value is 0.50 percent below it.

Where this lesson stops

A conversion factor is a ratio equal to one. This lesson converts units only: it never turns a historical constant's value into a modern one, and it treats every electrical correspondence as a convention.

This lesson builds on

• Powers of ten and physical units
• Fractions and ratios
• quantities-units

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

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-07: 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-07 online

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

Settings (canonical units)

M

50

T

293.15

a

5e-7

calibrationScale

1

constantSetId

scenario-bm07-hidden-number

coverage

0.95

coverageTrials

0

d

2

dt

1

estimator

independent-increment-known-zero-drift

eta

0.001

generatorEta

0.001

generatorRadius

5e-7

generatorT

293.15

inputCoverage

0.99

intervalKind

conditional

observationSet

synthetic

radiusError

0.1

radiusKnown

false

seed

1905

temperatureError

0.005

viscosityError

0.02

Evaluator: source:sha256:d7c143066043c23f1cba39a975b3928834a1275abe090374b8853e99570ecfc5