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Brownian motion · The statistical mechanics derivation

Why counting positions
gives the pressure law.

How can a vast microscopic problem, with every solvent molecule and every suspended particle in motion, yield a law as simple as Π = n k_B T without solving any molecular equations of motion?

Read the statistical argument and open its derivation steps →

BM-03 · Statistical mechanics derivation

The configuration integral laboratory

Configuration integral & free energy, host calculation

Predict before deriving

Doubling the volume available to two independent particles multiplies the number of position arrangements by 2, 4, or 8?

Presets and parameters
Particle placement model
Notation

Whole integer Np ≥ 1, positive volume ratio V/V₀, reference volume V₀ and temperature T.

Derivation step:

Step 1: One Particle in Accessible Volume V*Position options are proportional to the accessible room.V* (2 × V₀)particle 1 (x₁, y₁, z₁)Integral over (x₁, y₁, z₁)B₁ = ∫ dx₁dy₁dz₁ = V*Factor ratio: 2
One accepted calculation, in explicit units
Particle count2
Volume expansion ratio V/V₀2
Position arrangement factor4
Free-energy difference ΔF-5.61085e-21 J
Ideal pressure p4.0473725e-9 Pa
Locked-cluster pressure2.0236863e-9 Pa
Volume-independent factor Jsymbolic (cancels in derivative)
Momentum integralssymbolic (cancels in derivative)
Free-energy offset F₀symbolic (cancels in derivative)

Editorial commentary on Einstein §2

R0 (Overview): Counting where independent particles can be, not how they move, gives the pressure. Doubling the room doubles each particle's options.

R1 (Physical reasoning): Under independence, N_p particles explore volume V with a state-variable integral proportional to V^{N_p}. The free energy contains -k_B T ln(V^{N_p}) = -N_p k_B T ln V. Differentiating with respect to volume yields the pressure p = -dF/dV = N_p k_B T / V.

R2 (Mathematical structure): One particle has position options proportional to V. Two independent particles have options proportional to V * V = V^2. For N_p independent particles the factor is V^{N_p}. Taking the logarithm turns this product into the sum N_p ln V. The volume derivative removes the volume-independent factor J and the constant offset F_0, leaving p = N_p k_B T / V.

R3 (Historical notation & counterexample): Einstein's printed §2 notation uses B for the configuration integral, J for the volume-independent factor, lg for the natural logarithm, n for the particle count, V* for volume, and 2 kappa N = R for the gas constant relation. If the particles are locked into a single rigid cluster, the spatial arrangements grow only as V/V_0 rather than (V/V_0)^{N_p}, yielding pressure p = k_B T / V. Pressure counts independently placed units, not constituents.

What this model assumes

• Independence of particle positions (the §2 premise).

• Dilution (no excluded volume interactions).

• Uniform potential inside the accessible volume.

• Thermal equilibrium.

• The volume-independent factor J does not depend on volume V.

Not modeled: interactions between particles; excluded volume effects; external potential fields; non-ideal solutions; quantum statistics; momentum integrals beyond their cancellation in the derivative; molecular dynamics or collision trajectories; cluster formation or breakup kinetics.

Show the calculation owner and source identity

Configuration volume term, factor ratio and locked cluster pressure are computed by src/physics/reference/diffusion/routeA.ts.

Live terms bind particleCount, volume, temperature, and freeEnergy.

Compare two setups side by side in your reading. Each laboratory has its own settings, stepwise state and accepted results.

Open the derivation

From available volume to free energy

Einstein's §2 writes the free energy using the logarithm of an integral over all state variables (the configuration integral B). Under the premises of independent particle positions and dilution, the spatial positions of N_p particles contribute a clean volume factor V^{N_p}.

B=∫⋯∫dx1…dzn=V∗nJ,F=−2κTlg⁡B=−2κTnlg⁡V∗−2κTlg⁡J+constB=\int\cdots\int dx_1\dots dz_n=V^{*n}J,\qquad F=-2\kappa T\lg B=-2\kappa Tn\lg V^*-2\kappa T\lg J+\text{const}

In modern notation, 2κN = R gives 2κ = k_B, the particle count n is written N_p, and the volume V* is V:

F=−NpkBTln⁡V−kBTln⁡J+F0F=-N_p k_B T\ln V - k_B T\ln J + F_0

Why the complicated molecular factor J drops out

The factor J represents the integral over all solvent positions and interactions for fixed tracer coordinates. Under Einstein's three premises (homogeneous fluid, dilute independent particles, no external forces), shifting the particle positions does not change the probability of finding solvent molecules in any region. Thus J is independent of V, and differentiating with respect to volume leaves zero:

p=−∂F∂V=NpkBTV=nkBTp=-\frac{\partial F}{\partial V}=\frac{N_p k_BT}{V}=n k_BT

The locked-cluster counterexample: what is actually counted?

If the particles are rigidly locked together into a single cluster, the spatial arrangements grow only like V (or V/V_0) rather than (V/V_0)^{N_p}. The resulting osmotic pressure is that of one independent unit:

plocked=kBTVp_{\text{locked}}=\frac{k_B T}{V}

This demonstrates that osmotic pressure counts independently placed units, not constituents.

Explore the tracer ensemble →From random steps to diffusion