Annus MirabilisEinstein in 1905
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brownian motion

The equation explained

D=RTN16πkP.D = \frac{R T}{N} \frac{1}{6 \pi k P}.
In wordsThe diffusion coefficient is the gas constant times the absolute temperature over the number of molecules in a gram-molecule, times one over six pi times the friction coefficient of the liquid times the radius of a particle.OverviewThe result of the section. How fast a suspended substance spreads is settled by the temperature, the friction of the liquid, the size of its particles, and the number of molecules in a gram-molecule. The force that was used to find it has gone: the answer holds whether or not any force is acting.Full explanationThe two conditions found for dynamic equilibrium, (1) and (2), are enough to compute the diffusion coefficient. Equation (1) says Kν=RTN∂ν∂xK \nu = \frac{R T}{N} \frac{\partial \nu}{\partial x}, so wherever νK\nu K appears in equation (2) it may be replaced. Every term then carries the factor ∂ν∂x\frac{\partial \nu}{\partial x}, which cancels: the answer does not depend on how the particles happen to be distributed, nor on the force, both of which were scaffolding. What is left is the printed value. The sentence that closes the section on the next page states what has been shown: the diffusion coefficient of the suspended substance depends, apart from universal constants and the absolute temperature, only on the friction coefficient of the liquid and on the size of the suspended particles. Two of those quantities can be measured with a microscope and a thermometer, and one of them, NN, was not known with precision, which is why the paper can later turn the relation round.Every step−Kν+RTN∂ν∂x=0- K \nu + \frac{R T}{N} \frac{\partial \nu}{\partial x} = 0Equation (1), the equilibrium of the force against the crowding.Kν=RTN∂ν∂xK \nu = \frac{R T}{N} \frac{\partial \nu}{\partial x}The same, with the force term on one side.νK6πkP−D∂ν∂x=0\frac{\nu K}{6 \pi k P} - D \frac{\partial \nu}{\partial x} = 0Equation (2), the equilibrium of the two flows.16πkPRTN∂ν∂x−D∂ν∂x=0\frac{1}{6 \pi k P} \frac{R T}{N} \frac{\partial \nu}{\partial x} - D \frac{\partial \nu}{\partial x} = 0Equation (1) put into equation (2) in place of νK\nu K.(RTN16πkP−D)∂ν∂x=0\left( \frac{R T}{N} \frac{1}{6 \pi k P} - D \right) \frac{\partial \nu}{\partial x} = 0The common factor taken out. Wherever the number density varies at all, the bracket must vanish.D=RTN16πkPD = \frac{R T}{N} \frac{1}{6 \pi k P}The printed result. The force and the distribution have both dropped out, and what remains can be measured.Historian's marginThe section reaches the diffusion coefficient by holding a thermodynamic equilibrium beside a mechanical one, and the force KK appears in both and survives in neither. The resistance law it uses is the one for a slowly moving sphere, which the footnote at the foot of this same page, p. 555, refers to Kirchhoff's Vorlesungen über Mechanik, 26. Vorlesung § 4, hanging it on the velocity K/(6 π k P) earlier on the page; its kk is the friction coefficient of the liquid, not Boltzmann's constant.Also

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