brownian motion

The equation explained

N=tλx2⋅RT3πkP.N = \frac{t}{\lambda_x^2} \cdot \frac{R T}{3 \pi k P}.
In wordsThe number of molecules in a gram-molecule equals the elapsed time divided by the square of the observed displacement, times the gas constant times the absolute temperature, divided by three pi times the coefficient of friction of the liquid times the radius of a particle.OverviewRead the relation backwards and it becomes a method: watch the particles, measure how far they wander in a known time, and the number of molecules in a gram-molecule follows from quantities a laboratory can measure.Full explanationThe paper turns the relation round. In the preceding form the number of molecules was an input taken from the kinetic theory of gases and the displacement was predicted; here the displacement is what is measured and the number is what comes out. Every other quantity on the right is available to an experimenter: the time of the observation, the temperature, the viscosity of the liquid, and the radius of the particles. The quantity this yields is an estimate obtained from a measurement, not the same thing as the number assumed a paragraph earlier, and the edition binds it as such. The paper ends on the strength of it, with the wish that a researcher may soon succeed in deciding the question raised here, which it calls important for the theory of heat, and the date line Bern, May 1905.Every stepλx=t⋅RTN13πkP\lambda_x = \sqrt{t} \cdot \sqrt{\frac{R T}{N} \frac{1}{3 \pi k P}}The relation of section 5, with the displacement on the left.λx2=t⋅RTN13πkP\lambda_x^2 = t \cdot \frac{R T}{N} \frac{1}{3 \pi k P}Square both sides. Both are positive, so nothing is lost.N=tλx2⋅RT3πkP.N = \frac{t}{\lambda_x^2} \cdot \frac{R T}{3 \pi k P} .Solve for the number: the printed line. Reading it with the figures of the preceding paragraph returns 6 times ten to the twenty-third, as it must, since those figures were obtained from the same relation.Also

Read it in the paper