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§2 · Osmotic pressure from the molecular-kinetic theory

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A derivation, to an approximation

How molecular theory gives the osmotic law without solving the motion

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

The 9 printed paragraphs this passage explains

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

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

Einstein describes the whole system, liquid, wall and particles, by state variables $p_1, \ldots, p_l$ that fix its momentary state completely, for example the coordinates and velocity components of all its atoms. How they change in time is given by equations of the form

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

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

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

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

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

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

and the integral is named B.

Open the foundation: Logarithms: turning products into sums

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

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

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

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

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

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

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

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

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

d B over B equals d B prime over B.

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

Open the foundation: Probability and independence

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

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

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

and so the free energy is

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

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

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

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

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

Open the foundation: Partial derivatives and held-fixed quantities

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

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

Letters, as the paper prints them. $p_1, \ldots, p_l$ are the state variables, l of them, which fix the state of the whole system: for example every atom's coordinates and velocity components. They are not the osmotic pressure p. $\varphi_\nu$ is the rate at which $p_\nu$ changes; it is not the φ of §4. ∂ marks a partial derivative, a rate of change with the other variables held fixed, and Σ a sum. T is the absolute temperature. Ē, printed with a bar, is the energy of the system; E is the energy as a function of the state variables. κ is a constant with 2κN = R, where R is the gas constant and N the number of real molecules in a gram-molecule, so 2κ is R/N. lg is the natural logarithm, today written ln. S is the entropy and F the free energy. B is an integral over all states. $V^*$ is the volume holding the particles, n their number, and $x_i, y_i, z_i$ the coordinates of the centre of the i-th particle.

The footnote first. The section takes Einstein's 1902 and 1903 papers as known, and says that neither they nor this section is needed to understand the paper's results. A reader may skip to §3 and lose none of them. What the section adds is an answer to a question: how can a theory of countless colliding molecules give van 't Hoff's simple law, for molecules and suspended bodies alike, without anyone solving their motions?
The system. Everything, liquid, wall and particles, is described by state variables $p_1, \ldots, p_l$ . How each one changes in time is given by an equation:

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

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

The condition $\sum \frac{\partial \varphi_\nu}{\partial p_\nu} = 0$ says that the motion neither crowds the possible states together nor spreads them apart; the equations of mechanics have this property when the forces do not depend on the velocities. Einstein's earlier theory needs it.
Entropy and free energy. The earlier theory gives the entropy S through the logarithm of an integral of $e^{-EN/RT}$ over every combination of the state variables that the conditions allow. States of higher energy E count for less, by that exponential factor. Writing 2κ as R/N, the free energy is:

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

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

The integral is named B. So F is minus RT/N times the logarithm of B: whatever makes B larger makes F smaller.
Hopeless in full, Einstein says: even with the molecular picture fixed in every detail, computing B would be too hard for an exact calculation of F.
Not needed in full. The pressure on the wall is the change of F when the wall moves, that is, when $V^*$ changes. So only how B depends on $V^*$ is needed.
The particles. n of them, dissolved molecules or suspended bodies, are held in $V^*$ by a semipermeable wall, and they take up little of $V^*$ . The wall's position sets the limits of the integral B.
Boxes. Give the centre of each particle a tiny box inside $V^*$ , of sides $dx_1, dy_1, dz_1$ for the first particle, and so on. The part of B from states with every centre in its box is the product of the box sizes times a factor J:

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

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

J collects everything else: the liquid's molecules, the velocities, the insides of the particles. It does not depend on the box sizes, nor on where the wall is.
Moving the boxes. Choose a second set of boxes, the same sizes, in other places inside $V^*$ . Their part of B is $dB'$ , with its own factor $J'$ . The box sizes are the same, so dividing one form by the other leaves:

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

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

What the parts mean. dB/B is the probability that, at a moment chosen at random, every centre is in its box. The footnoted 1903 paper supplies this reading.
The assumptions. The particles move independently of one another, to a sufficient approximation; the liquid is the same everywhere; no forces act on the particles. Then no place inside $V^*$ is preferred, and boxes of equal size are equally probable wherever they are:

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

d B over B equals d B prime over B.

So $J = J'$ : J does not depend on where the boxes are, and, from before, not on $V^*$ either.
Adding up the boxes. To get all of B, let each centre range over the whole of $V^*$ . Each centre contributes a factor $V^*$ , and there are n centres:

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

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

A worked case. Two particles in a box: doubling the box doubles the room for each centre, so the number of ways to place both grows by 2 × 2 = 4. With 10 particles, doubling the volume multiplies B by $2^{10} = 1024$ . In general B grows as $V^*$ to the power n.
Taking the logarithm. The logarithm turns a product into a sum and a power into a multiple: $\lg(JV^{*n}) = \lg J + n\lg V^*$ . So:

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

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

A tool for this step

Logarithms: turning products into sums

Why does entropy need a logarithm?

Adding is easier than multiplying, and a logarithm turns one into the other. The natural logarithm of 2 is about 0.693, and of 4 about 1.386. Add them and you get 2.079, which is the natural logarithm of 8, the product of 2 and 4.

The natural logarithm ln x answers one question: to what power must e, about 2.718, be raised to give x? Raised to 0.693, e gives 2, and raised to 1.386 it gives 4. Multiplying two powers of e adds the powers, so raised to 2.079 it gives 8. Written for any numbers, the logarithm of a product is the sum of the logarithms, and a power becomes a multiple:

\ln\left(A\,B\right) = \ln\left(A\right) + \ln\left(B\right)

\ln\left(f^{n}\right) = n\,\ln\left(f\right)

The log of A times B is log A plus log B, and the log of f to the power n is n times log f.

Entropy needs exactly this. In §5 of the light-quanta paper, Einstein takes two independent systems. Their entropies add, S = S₁ + S₂, while the probabilities of their states multiply, W = W₁W₂. If entropy is some function φ of probability, then φ(W₁W₂) = φ(W₁) + φ(W₂), and he concludes that φ is a logarithm:

S - S_{0} = \frac{R}{N}\,\ln\left(W\right)

S minus S nought equals R over N times the natural log of W.

R is the gas constant and N the number of molecules in a mole, so R/N is what is now called Boltzmann's constant. The step from φ(W₁W₂) = φ(W₁) + φ(W₂) to the logarithm needs φ to be a smooth function, which Einstein takes for granted.

Einstein prints lg for the natural logarithm, as German physics did in 1905. Today lg usually means the base-10 logarithm and ln the natural one. So a printed lg 2 means ln 2, about 0.693, not log₁₀ 2, about 0.301.

Worked example: Ten molecules found in half the volume

In §5 Einstein asks how probable it is that n molecules, each moving independently through a volume v₀, are all found at one moment in a part v of it. Each molecule is in that part with probability v/v₀, and independent probabilities multiply:

W = \left(\frac{v}{v_{0}}\right)^{n}

W equals v over v nought, to the power n.

Take n = 10 molecules and v = v₀/2. Then W = (1/2)¹⁰ = 1/1024, about 0.000977.
The logarithm turns the tenth power into ten times: ln W = 10 × ln(1/2) = 10 × (−0.693) = −6.93.
So the entropy of the state with every molecule in one half is lower by 6.93 × R/N: 6.93 times Boltzmann's constant.

For a mole of gas, n = N, and the same formula gives S − S₀ = R ln(v/v₀). That is the entropy change thermodynamics gives for an ideal gas compressed from v₀ to v at constant temperature.

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Try it: turning multiplication into addition

In §5 of the light-quanta paper, Einstein takes two independent systems. The probabilities of their states multiply, W = W₁ · W₂, while their entropies add, S = S₁ + S₂. A logarithm turns the one into the other, which is why he concludes that entropy is a constant times ln W, plus a constant. The table checks the rule on three pairs of numbers.

Notation display toggle:

Switches the table between the symbol printed in 1905, lg, and the modern ln.

1905 historical notation vs Modern ISO standard

In 1905 German scientific printing (including Annalen der Physik), the symbol lg denoted the natural logarithm (base e).

In modern ISO 80000-2 notation, ln denotes the natural logarithm, while lg is reserved for the common base-10 logarithm (log₁₀).

1905 printed “lg 2”: 0.693147, the natural logarithm ln 2.

Modern ISO “lg 2” (log₁₀ 2): 0.301030, the common base-10 logarithm.

ln(W₁ · W₂) = ln(W₁) + ln(W₂), checked on three pairs
W₁ and W₂	ln(W₁) + ln(W₂)	ln(W₁ · W₂)
2 and 4 product 8	0.693147 + 1.386294 = 2.079442	ln 8 = 2.079442
3 and 5 product 15	1.098612 + 1.609438 = 2.708050	ln 15 = 2.708050
10 and 10 product 100	2.302585 + 2.302585 = 4.605170	ln 100 = 4.605170

What it shows, in words

For each of three pairs of numbers, the logarithm of the product equals the sum of the two logarithms. That is the property entropy needs: the probabilities of independent systems multiply and their entropies add. In the 1905 papers the logarithm is printed lg and means the natural logarithm, ln.

Where this lesson stops

A logarithm turns products into sums, which is why it connects probabilities that multiply to entropies that add. In the 1905 papers, lg means the natural logarithm.

The pressure. Pressure is minus the rate at which F changes with the volume. lg J does not change with $V^*$ . The rate of change of $n\lg V^*$ with $V^*$ is n divided by $V^*$ . The two minus signs cancel, and $n/V^*$ is ν, the number per unit volume:

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

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

A tool for this step

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

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}{2\,D\,t}\,f\left(x,\,t\right)

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}{2\,t} + \frac{x^{2}}{4\,D\,t^{2}}\right)\,f\left(x,\,t\right)

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.

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Try it: hold time fixed, or hold position fixed

In §4 of the Brownian paper, f(x, t) is the number of particles per unit volume at a place x along a tube, at a time t. “How fast does the concentration change?” then has two answers, and which one you mean depends on what you keep still: the moment, or the place.

Keep one of them still:

One moment: t fixed, ∂f/∂x

Freeze the moment and compare neighbouring places, as in a single photograph of the tube. ∂f/∂x says how steeply the concentration changes along it: particles per µm³ for each µm, which is particles per µm⁴.

Particles drift down that slope. The flux, the number crossing each square micrometre in a second, is J = −D ∂f/∂x: Fick’s first law.

The same question in thermodynamics

Squeeze a gas and its pressure rises, but by how much depends on what else you keep fixed:

Three derivatives of pressure, and what each holds fixed
Held fixed	Derivative	What happens
Temperature T (isothermal)	(∂p / ∂V)_T	Heat flows in or out to keep the temperature constant.
Entropy S: no heat exchanged (adiabatic)	(∂p / ∂V)_S	The gas warms as it is squeezed, so its pressure climbs faster than at fixed temperature.
Volume V (isochoric)	(∂p / ∂T)_V	In a rigid vessel the pressure rises as the gas is heated.

What it shows, in words

∂f/∂x compares places at one moment; ∂f/∂t follows one place through time. They answer different questions and carry different units. In thermodynamics a subscript names what is held fixed, so (∂p/∂V)_T and (∂p/∂V)_S are different numbers for the same gas.

Where this lesson stops

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

The conclusion, as Einstein states it: osmotic pressure is a consequence of the molecular-kinetic theory of heat, and at great dilution equal numbers of dissolved molecules and suspended bodies behave exactly alike as regards osmotic pressure.
How no molecular motion had to be solved: the factor J, which holds all the hard physics, was never computed. Its independence of $V^*$ was enough, and that came from equal boxes being equally probable. Nothing about a particle's size or mass entered; only its number, the temperature and the volume did.
Where it would fail: if the particles act on one another, or crowd each other, some placements are more probable than others, J depends on the positions, and the pressure departs from the dilute law. Those are exactly the assumptions Einstein states.

A tool for this step

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:

\left(\Delta_{1} + \Delta_{2}\right)^{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?

Modern qualifications

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

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

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

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

Assumed here

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

What this does not establish

The footnote to the heading says this section, and Einstein's earlier papers on the foundations of thermodynamics, are not needed to understand the paper's results. The passage explains the section for the reader who wants to know how the law is obtained.
The argument finds only how the free energy depends on V*; it computes nothing else about the integral B.
Independence, homogeneity and the absence of forces are assumptions. With particles that act on one another, or crowded ones, J would depend on their positions and the pressure would depart from the dilute law.

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

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

Explanation

Einstein's displacement argument: the typical distance grows with the square root of time, not with time itself. Every explanation stays on the page when reading-only is on.

Static worked case

For radius 0.5 μm, viscosity 1.35×10⁻³ Pa·s, and T = 290.15 K, the RMS displacement is about 0.8 μm in one second. This static worked case stays in the markup; loading the live ensemble does not replace it.

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Notes and laboratory

Page 551 of Annalen der Physik, volume 17, where §2 begins. — §2 begins on page 551 of Annalen der Physik, volume 17. Read it in Einstein’s German

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