Balancing directional drag against random spreading.

Push every suspended particle the same way with a small steady force and they drift and pile up, while their random jostling spreads them back out. When the two balance, the strength of the jostling is fixed by the temperature and the drag of the liquid alone, whatever the force was.

Section 3 imagines a force K acting along x on every suspended particle. In equilibrium the force is balanced by osmotic pressure: Kν = ∂p/∂x, with p = (RT/N)ν, so the density falls off exponentially over a length (RT/N)/K. Einstein then reads the same equilibrium as two opposite processes. The force drives each sphere at K/(6πkP), carrying νK/(6πkP) particles through unit area each second, and diffusion carries D ∂ν/∂x back; for the two to cancel everywhere, D = (RT/N)/(6πkP). The force does not appear in the result. The instrument takes a sphere of radius 0.5 μm in water at 20 °C, with viscosity 1.002 mPa·s, and a force of 4.05 fN, chosen so that the decay length k_BT/F is 1 μm. The drag gives a drift of 0.429 μm/s and D = 0.429 μm²/s, and the density in a 10 μm box relaxes towards that 1 μm exponential. With the jostling switched off, Nägeli's objection, only the drift is left and the particles pile against the wall; with it doubled, the profile stretches to 2 μm, twice what osmotic balance allows.

Take unit cross-section and a force K on each particle along x. The suspended particles exert an osmotic pressure p = (RT/N)ν, like a gas of ν particles per unit volume. For a thin slab to stay at rest, the force on its particles, Kν per unit volume, must be matched by the change of pressure across it: Kν = dp/dx = (RT/N) dν/dx. So ν changes by the same fraction, K/(RT/N), over every unit of length, which makes the profile an exponential with decay length (RT/N)/K = k_BT/K. With k_BT = 1.381 × 10⁻²³ × 293.15 = 4.047 × 10⁻²¹ J and K = 4.047 × 10⁻¹⁵ N, that length is 10⁻⁶ m, 1 μm. Now the moving picture. Stokes's law gives a sphere of radius P, in a liquid of viscosity k, the speed K/(6πkP) under the force K. Here 6πkP = 6 × 3.1416 × 0.001002 × 0.5 × 10⁻⁶ = 9.44 × 10⁻⁹ kg/s, so the drift speed is 4.047 × 10⁻¹⁵/(9.44 × 10⁻⁹) = 4.29 × 10⁻⁷ m/s, or 0.429 μm/s, and ν such spheres per unit volume carry νK/(6πkP) across unit area each second. Diffusion carries D dν/dx the other way. In equilibrium the two cancel: νK/(6πkP) = D dν/dx. Put in dν/dx = νK/(k_BT) from the pressure balance: νK/(6πkP) = DνK/(k_BT). Both ν and K cancel, leaving D = k_BT/(6πkP) = 4.047 × 10⁻²¹/(9.44 × 10⁻⁹) = 4.29 × 10⁻¹³ m²/s. The same D follows from any force at all, which is why Einstein can use an imagined one. The flow around the sphere is extremely slow, with a Reynolds number of 2 × 10⁻⁷, so Stokes's law applies. In the box the density starts uniform and, after 1 s, the drift still outweighs the returning diffusion, so it is still settling towards the exponential.

Einstein wrote K for the force, ν for the number of particles per unit volume, k for the viscosity and P for the radius. The force is a device of the argument, and it drops out of the result. The same relation, D = (RT/N)/(6πkP), was found independently by Sutherland in 1905 and is often called the Stokes–Einstein relation. Nägeli had argued in 1879 that molecular impacts are far too small to move a visible particle; the branch with the jostling switched off shows what the balance would look like without them, drift alone with nothing to hold the particles off the wall. Stokes's law for the drag on a sphere dates from 1851.