For mn (which is generally true because = 0 and mems. ¹/2 while n >> 1), we obtain Pn (m) -exp(--m²/2n). (14) Taking x t

[answerhappygod]

For Mn Which Is Generally True Because 0 And Mems 2 While N 1 We Obtain Pn M Exp M 2n 14 Taking X T 1 (50.14 KiB) Viewed 41 times

For mn (which is generally true because = 0 and mems. ¹/2 while n >> 1), we obtain Pn (m) -exp(--m²/2n). (14) Taking x to be a continuous variable (and remembering that p. (m) = 0 either for even values of m or for odd values of m, so that in the distribution (14), Am = 2 and not 1), we may write this result in the Gaussian form: p(x) dx = •(-D.). where 2 √(2mn) 462 dx √(47Dt) exp D=1²/2t*. (15) (16) Fluctuations Later on we shall see that the quantity D introduced here is identical with the diffusion coefficient of the given system; eqn. (16) connects it with the (micro- scopic) quantities and r". To appreciate this connection, one has simply to note that the problem of Brownian motion can also be looked upon as a problem of "diffusion" of Brownian particles through the medium of the fluid; this point of view is also due to Einstein. However, before we embark upon these considera- tions, we would like to present here the results of an actual observation made on the Brownian motion of a spherical particle immersed in water; see Lee, Sears and Turcotte (1963). It was found that the 403 values of the net displacement Ar of the particle, observed after successive intervals of 2 seconds each, were distributed as follows: