§1 · Osmotic pressure from suspended particles · Brownian motion: from wandering to a measurable law

§1 · Osmotic pressure from suspended particles

Why a suspended grain should press like a dissolved molecule

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

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

pV^*=RTz

p times V star equals R times T times z.

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

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

Return to the classical expectation for suspended bodies. (included in this file)

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

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

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

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

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

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

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

Return to where the pressure depends on temperature and number, not size. (included in this file)

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

Letters, as the paper prints them. V is the total volume of the liquid and $V^*$ the part of it behind the wall. z is the number of gram-molecules dissolved in $V^*$ ; a gram-molecule is what is now called a mole. p is the osmotic pressure, R the gas constant, T the absolute temperature. n is the number of suspended bodies, a count. ν is their number per unit volume, $\nu = n/V^*$ , and not a frequency. N is the number of real molecules in a gram-molecule, now called Avogadro's number. Three letters, three different numbers: n counts the bodies, ν counts them per unit volume, N counts the molecules in a gram-molecule.

The wall. It is semipermeable: the solvent, say water, passes through it; the dissolved substance does not. The solution is on one side, pure solvent on the other.
Osmotic pressure. The dissolved molecules push on the wall. The push on each unit of area is the osmotic pressure p. Pressure is force divided by area, so a wall of area A feels the force $pA$ .
Van 't Hoff's law. For a dilute solution, one in which $V^*/z$ is large, the pressure obeys the same law as an ideal gas of as many molecules in the same volume:

pV^*=RTz

p times V star equals R times T times z.

Reading it. Doubling the amount dissolved, z, doubles p. Doubling the volume $V^*$ at fixed z halves p. Raising the absolute temperature T raises p in proportion.
A worked case, with today's value of the gas constant, $R \approx 8.314$ joule per kelvin per gram-molecule: one gram-molecule in 22.4 litres at 273 K gives $p = RTz/V^* \approx 8.314 \times 273 / 0.0224 \approx 101\,300$ pascal, about one atmosphere, the pressure one gram-molecule of gas exerts in that volume at 0 °C.
Classical thermodynamics, for suspended bodies. Replace the dissolved substance by small bodies that float in the liquid and also cannot pass the wall. At a fixed temperature the force on the wall is found from the free energy F of the whole system: if moving the wall a little changes F, there is a force; if not, there is none.
On the usual view, F depends on the total masses and kinds of the suspended substance, the liquid and the wall, and on pressure and temperature. It does not depend on where the wall is, or where the bodies are. So moving the wall leaves F unchanged, and classical thermodynamics expects no force on the wall from the suspended bodies.
What is set aside. Gravity, which would make the bodies sink. And the energy and entropy of the surfaces between bodies and liquid, the capillary forces, removed by assuming that the moves considered do not change those surfaces. Without these set-asides the classical expectation would carry extra forces that have nothing to do with the question.

Return to the classical expectation for suspended bodies. (included in this file)

Why the classical expectation is reasonable. For ordinary bodies it matches experience: a few pebbles held behind a sieve do not press on it measurably. It is a coherent position, and §1 treats it as one: the question it asks is whether the expectation still holds for bodies small enough to be jostled by molecules.
The molecular-kinetic view. A dissolved molecule and a suspended body differ only in size. Both are struck by the liquid's molecules and move irregularly, the body far more slowly. A body kept in by the wall strikes it now and then, and so pushes on it, as a dissolved molecule does.
So equally many bodies and molecules should give the same pressure. Take n bodies in $V^*$ . They amount to $z = n/N$ gram-molecules' worth of particles, because a gram-molecule holds N particles. Put that z into van 't Hoff's law and divide by $V^*$ ; since $\nu = n/V^*$ , the last step replaces $n/V^*$ by ν:

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

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

Reading the result. The pressure is RT/N times ν, the number of bodies per unit volume. Their size and mass do not appear. A body a thousand times bigger than a molecule presses no harder, as long as there are equally many per unit volume.
The condition. Neighbouring bodies must be far enough apart, the same condition as a dilute solution. Crowded bodies, or bodies that act on one another, depart from the law.
For scale, with today's value of R/N, $1.38\times10^{-23}$ joule per kelvin, at T = 290 K, the 17 °C of the paper's §5 example, the factor RT/N is about $4.0\times10^{-21}$ joule. A suspension with $10^{16}$ grains per cubic metre, ten million per cubic millimetre, would press with about $4\times10^{-5}$ pascal: far too little to measure directly. The paper's test lies in the motion, in §§4 and 5.

Return to where the pressure depends on temperature and number, not size. (included in this file)

What tells the views apart. On the molecular-kinetic view, wherever the number per unit volume varies, the pressure varies, and it pushes the bodies toward thinner regions: §3 shows that this is diffusion, and §§4 and 5 predict how far a grain wanders in a given time. On the classical expectation there is no osmotic pressure to drive such a spread.
So observing the predicted wandering, with its predicted size, would count for the molecular-kinetic view, and its absence against it. §1 only states the expectation; §2 shows that the molecular-kinetic theory really leads to it.

Assumptions

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

Limits

Gravity is set aside, and so are the energy and entropy of the surfaces between bodies and liquid (capillary forces), on the assumption that the moves considered do not change those surfaces.
§1 states the molecular-kinetic expectation; §2 derives it from the theory. Neither section says which view nature follows; that is for observation.
The law is for great dilution. Crowded bodies, or bodies that act on one another, depart from it.

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. Draw a line along the bottom of a page for time and one up the side for distance, and put 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.
Mark those positions along the bottom and draw 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

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: One in four, two in eight

In a group of 4 particles, 1 moves to the right: the fraction is 1/4 = 0.25.
In a group of 8, 2 move to the right: the fraction is 2/8 = 0.25.
Doubling both the number that moved and the size of the group leaves the fraction at 0.25.

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

Return to the chapter

A sign records direction

How can movement add up to zero?

Put the starting point at zero on a ruler. 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.

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

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 micrometre (1 μm) is 10⁻⁶ metres, or 0.000001 m.
The paper's displacement after one minute, about 6 μm, is 6 × 10⁻⁶ m.
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

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: Adding four numbers and dividing by four

Add 3 + 1 + 1 + 3 to get 8.
Count the values: there are four.
Divide 8 by 4 to get 2.
List the values twice: 16 divided by 8 is still 2.

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

Entropy and temperature

What does temperature have to do with entropy?

Add a small amount of heat to a body, slowly and reversibly, and keep its volume fixed so that it does no work. Its energy rises by that amount, and its entropy rises by the energy divided by its absolute temperature. So one joule raises the entropy of a body at 100 kelvin by 0.01 joule per kelvin, and of a body at 1000 kelvin by only 0.001: the same energy counts for more in a cold body. In symbols, with dE a small gain of energy and T the temperature:

dS=\frac{dE}{T},\qquad\left(\frac{\partial S}{\partial E}\right)_V=\frac{1}{T}

At fixed volume and the stated constraints, entropy change is energy change divided by absolute temperature.

Read the other way round, 1/T is how steeply a body's entropy climbs as its energy grows.

The subscript V matters. If the volume can change, the body can do work while it takes in energy, and the relation gains another term. The derivative describes one kind of change, at fixed volume, not every process.

The relation also says why bodies in contact end at one temperature. Share a fixed total of energy between two bodies, and the total entropy is greatest when moving a little energy from one to the other no longer changes it: when their entropy slopes, and so their temperatures, are equal. §3 of the light-quanta paper uses the same relation for radiation, setting its entropy slope equal to 1/T.

Return to the constraints on the entropy derivative. (included in this file)

Worked example: Three joules into a reservoir held at 300 kelvin

A reservoir so large that its temperature stays at 300 kelvin receives 3 joules of heat, slowly.
Its entropy rises by 3 J ÷ 300 K = 0.01 joule per kelvin.
The same 3 joules given to a reservoir at 150 kelvin would raise its entropy by 0.02 joule per kelvin, twice as much.
A small body warms as it takes in heat, so its temperature changes on the way. Its entropy change is found by adding dE/T over the warming, not by dividing by one temperature.

The relation fixes how entropy changes, not its absolute value: two entropy formulas that differ by a constant have the same slope everywhere, and a separate condition is needed to fix the constant. For an entropy per unit volume, that constant multiplied by two different volumes can change the difference between two states.

Where this lesson stops

Before using 1/T as an entropy slope, say what is held fixed. Before comparing two entropies, say whether they are totals or per unit volume, and what fixes the constant.

This lesson builds on

Partial derivatives and held-fixed quantities
integration

Return to the chapter

Osmotic pressure and free energy

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

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

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

p = \frac{RT}{N}\,\nu

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

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

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

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

Worked example: A sugar solution and a suspension of grains

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

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

Where this lesson stops

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

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

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

Quantities and units

Can a number of metres be compared with a number of seconds?

"The particle moved 6" says nothing until the unit is added: 6 micrometres, or 6 seconds? A physical quantity is a number times a unit, and the number alone changes when the unit does. Six micrometres is 0.006 millimetres: the same length, written with another unit.

Quantities of different kinds cannot be added or compared: 3 metres plus 2 seconds has no meaning. So in a correct equation every term, on both sides, carries the same units. Checking that is the quickest test a formula can pass or fail.

Take the Brownian diffusion coefficient, $D = k_BT/(6\pi\eta a)$ . Boltzmann's constant is in joules per kelvin, T in kelvins, the viscosity η in pascal-seconds and the radius a in metres. The units work out to square metres per second, which is what a diffusion coefficient must be. The worked example below goes through it.

Units also tell apart quantities that share a letter. The light paper's $\rho_\nu$ is an energy per unit volume per unit of frequency, joules per cubic metre per hertz, not per unit of wavelength: two different quantities with different units. And its L is the speed of light in §§1 and 2 but an absorbed energy in §9. A unit check starts from what the quantity is, not from its letter.

Worked example: Checking that D comes out in square metres per second

$k_BT$ : joules per kelvin times kelvins is joules, which are newton-metres.
$6\pi\eta a$ : 6π has no units, and pascal-seconds times metres is (newtons per square metre) × seconds × metres, which is newton-seconds per metre.
Divide: newton-metres divided by newton-seconds per metre is square metres per second.
With numbers: 1.380649 × 10⁻²³ × 290.15 ÷ (6π × 1.35 × 10⁻³ × 5 × 10⁻⁷) = 3.15 × 10⁻¹³ m²/s. That is the modern Boltzmann constant (set modern-si-2019) with the paper's water at 17 °C and its particle.

Where this lesson stops

This lesson stops at checking units. Converting the papers' own units into SI units is a separate step, with its own conventions.

This lesson builds on

Powers of ten and physical units
Fractions and ratios

Return to the chapter

Ratios and scaling

If one quantity doubles, what happens to the others?

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

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

\lambda_x=\sqrt{2Dt},\qquad D=\frac{k_BT}{6\pi\eta a}

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

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

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

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

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

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

Return to the chapter

Temperature and thermal energy

What does a temperature say about the jostling of molecules and grains?

The air in a room at 20 °C is a crowd of molecules moving in every direction and colliding billions of times a second. Warm the room and, on average, they move faster. The kinetic theory of heat reads temperature this way: through the average energy of motion of the particles, taken over very many of them.

For any particle in equilibrium with its surroundings, the average energy of motion along each direction is the same:

\left\langle \frac{1}{2} m v_x^2 \right\rangle = \frac{1}{2} k_B T

The average of one half m v x squared equals one half k B T.

Here m is the particle's mass, $v_x$ its speed along x, T the absolute temperature, and $k_B$ = 1.38 × 10⁻²³ joules per kelvin is Boltzmann's constant. It equals R/N, the gas constant shared out per molecule. The Brownian paper writes RT/N where a modern text writes $k_BT$ ; the paper's own letter k means the viscosity, not this constant.

The average is the point. One molecule's energy of motion changes at every collision, from almost nothing to several times the average. The temperature fixes how energy is shared out over many particles. It is not the energy of any one of them.

Nothing in the rule mentions size. A grain ten billion times heavier than a molecule has the same average energy of motion, so it moves more slowly, by the square root of the mass ratio. That is why a suspended grain belongs to the same thermal story as a dissolved molecule, as §1 of the paper insists.

Worked example: A nitrogen molecule and a half-micrometre grain at 20 °C

At T = 293 K, $k_BT$ = 1.381 × 10⁻²³ × 293 = 4.05 × 10⁻²¹ J, so the average energy of motion along one direction is half of that, 2.02 × 10⁻²¹ J.
A nitrogen molecule has a mass of 4.65 × 10⁻²⁶ kg. Setting $\frac{1}{2} m v_x^2$ equal to $\frac{1}{2} k_BT$ gives a typical speed along one direction of √(4.05 × 10⁻²¹ ÷ 4.65 × 10⁻²⁶), about 295 metres a second.
A grain of radius 0.5 μm and density 1200 kg/m³ has a mass of 6.3 × 10⁻¹⁶ kg, about 1.4 × 10¹⁰ times the molecule's. The same average energy gives it about 2.5 millimetres a second.
In water that motion is turned about within some 70 nanoseconds: the grain's mass divided by its drag coefficient 6πηa. That is far too brief to follow, so what a microscope sees is the net result of the zigzag.

This is why the Brownian paper asks for a displacement over a time rather than a speed: the thermal speed is real, but it turns about far faster than any observer can watch.

Where this lesson stops

This lesson stops at the average energy of motion. How the jostling of many molecules makes a grain wander, and how far, is §4 of the Brownian paper.

This lesson builds on

Energy of motion and inertia
Powers of ten and physical units
Squares and square roots

Return to the chapter

Energy of motion and inertia

If a body moves at the same speed but carries less energy of motion, what has changed?

A force that pushes a body through a distance does work on it, and the work becomes the body's energy of motion, its kinetic energy. That is separate from energy stored inside the body, as heat or in its chemistry.

For one body at everyday speeds, doubling the speed multiplies the energy of motion by four. At a fixed speed, doubling the mass doubles it. The mass-energy paper's title asks about a body's Trägheit, its inertia: its resistance to a change in its motion, which is measured by pushing it, not by weighing it.

K=\frac{1}{2}mv^2

In Newtonian mechanics, kinetic energy is one half times inertial mass times speed squared.

This formula holds at speeds small compared with the speed of light, and it is not exact close to that speed. So the mass-energy argument works with the exact expression and reads the mass from its low-speed limit, where it takes this form.

Worked example: Two bodies at the same speed, different masses

Two bodies move at 2 metres per second.
The 3 kg body has ½ × 3 × 2² = 6 joules of energy of motion; the 2 kg body has ½ × 2 × 2² = 4 joules.
The speed is the same, and the energies differ by 2 joules because the masses differ by 1 kg.

K_0-K_1=\frac{1}{2}(m_0-m_1)v^2

At the same low speed, the kinetic-energy drop equals one half times the mass decrease times the squared speed.

Here the two masses are given, so the example proves nothing about mass and energy. The 1905 argument runs the other way: it computes the energy difference from the light the body gives off, finds that it has this form, and reads the mass difference from the coefficient of ½v².

Where this lesson stops

At the same low speed, less energy of motion means less mass. The comparison gives only a difference in mass, never the total energy stored inside the body.

This lesson builds on

Squares and square roots
Fractions and ratios

Return to the chapter

Static laboratory examples

bm-02

This laboratory is not included in this file. The chapter's authored foundation examples remain available above. Open bm-02 online

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.
A. Einstein, Does the inertia of a body depend upon its energy content? Annalen der Physik (4), 18, 639–641 (1905). External 1923 Perrett–Jeffery translation, electronically transcribed by John Walker; its notation was modernized. A reference for this explanatory preview, not this edition’s reviewed translation or pinned facsimile.