On the Motion of Particles Suspended in Liquids at Rest, as Required by the Molecular-Kinetic Theory of Heat

§ 4. On the irregular motion of particles suspended in a liquid and its relation to diffusion.Show the German source

We now turn to a closer investigation of the irregular motions which, produced by the molecular motion of heat, give rise to the diffusion investigated in the last section.Show the German source

It must evidently be assumed that each individual particle carries out a motion that is independent of the motion of all the other particles; the motions of one and the same particle in different time intervals are also to be regarded as mutually independent processes, as long as we think of these time intervals as not chosen too small.Show the German source

We introduce into the discussion a time interval $\tau$, which is to be very small compared with the observable time intervals, but still so large that the motions carried out by a particle in two successive time intervals $\tau$ are to be regarded as mutually independent events.Show the German source

Now let a total of $n$ suspended particles be present in a liquid.Show the German source In a time interval $\tau$ the $X$-coordinates of the individual particles will increase by $\Delta$, where $\Delta$ has a different (positive or negative) value for each particle.Show the German source A certain frequency law will hold for $\Delta$; the number $d n$ of particles that in the time interval $\tau$ undergo a displacement lying between $\Delta$ and $\Delta + d \Delta$ will be expressible by an equation of the formShow the German source $$d n = n \varphi(\Delta) d \Delta$$$n$Particle count$\phi$Jump density$\Delta$Displacement in one intervalExplain this equationShow the German source whereShow the German source $$\int_{-\infty}^{+\infty} \varphi(\Delta) d \Delta = 1$$$\phi$Jump density$\Delta$Displacement in one intervalExplain this equationShow the German source and $\varphi$ differs from zero only for very small values of $\Delta$ and satisfies the conditionShow the German source $$\varphi(\Delta) = \varphi(-\Delta)$$$\phi$Jump density$\Delta$Displacement in one intervalExplain this equationShow the German source

We now investigate how the diffusion coefficient depends on $\varphi$, again restricting ourselves to the case in which the number $\nu$ of particles per unit volume depends only on $x$ and $t$.Show the German source

Let $\nu = f(x, t)$ be the number of particles per unit volume; we calculate the distribution of the particles at the time $t + \tau$ from their distribution at the time $t$.Show the German source From the definition of the function $\varphi(\Delta)$ one easily obtains the number of particles that at the time $t + \tau$ are located between two planes perpendicular to the $X$-axis with the abscissas $x$ and $x + d x$.Show the German source One obtains:Show the German source $$f(x, t + \tau) d x = d x \cdot \int_{\Delta = -\infty}^{\Delta = +\infty} f(x + \Delta) \varphi(\Delta) d \Delta.$$$f$Number density$t$Time$\tau$Step interval$\Delta$Displacement in one interval$\phi$Jump densityExplain this equationShow the German source But since $\tau$ is very small, we can now set:Show the German source $$f(x, t + \tau) = f(x, t) + \tau \frac{\partial f}{\partial t}.$$$f$Number density$t$Time$\tau$Step intervalExplain this equationShow the German source Further, we expand $f(x + \Delta, t)$ in powers of $\Delta$:Show the German source $$f(x + \Delta, t) = f(x, t) + \Delta \frac{\partial f(x, t)}{\partial x} + \frac{\Delta^2}{2!} \frac{\partial^2 f(x, t)}{\partial x^2} \ldots \text{ in inf.}$$$f$Number density$\Delta$Displacement in one interval$t$TimeExplain this equationShow the German source We can carry out this expansion under the integral, since only very small values of $\Delta$ contribute anything to the latter.Show the German source We obtain:Show the German source $$\begin{aligned} f + \frac{\partial f}{\partial t} \cdot \tau = f \cdot \int_{-\infty}^{+\infty} \varphi(\Delta) d \Delta &+ \frac{\partial f}{\partial x} \int_{-\infty}^{+\infty} \Delta \varphi(\Delta) d \Delta \\ &+ \frac{\partial^2 f}{\partial x^2} \int_{-\infty}^{+\infty} \frac{\Delta^2}{2} \varphi(\Delta) d \Delta \ldots \end{aligned}$$$f$Number density$t$Time$\tau$Step interval$\phi$Jump density$\Delta$Displacement in one intervalExplain this equationShow the German source On the right-hand side, because $\varphi(x) = \varphi(-x)$, the second, fourth, etc. terms vanish, while of the first, third, fifth, etc. terms each successive one is very small compared with the one before.Show the German source From this equation we obtain, taking into account thatShow the German source $$\int_{-\infty}^{+\infty} \varphi(\Delta) d \Delta = 1,$$$\phi$Jump density$\Delta$Displacement in one intervalExplain this equationShow the German source and settingShow the German source $$\frac{1}{\tau} \int_{-\infty}^{+\infty} \frac{\Delta^2}{2} \varphi(\Delta) d \Delta = D$$$\tau$Step interval$\Delta$Displacement in one interval$\phi$Jump density$D$Diffusion coefficientExplain this equationShow the German source and taking into account only the first and third terms on the right-hand side:Show the German source $$\frac{\partial f}{\partial t} = D \frac{\partial^2 f}{\partial x^2}.$$$f$Number density$t$Time$D$Diffusion coefficientExplain this equationShow the German source

This is the well-known differential equation of diffusion, and one sees that $D$ is the diffusion coefficient.Show the German source

Another important consideration can be attached to this development.Show the German source We have assumed that the individual particles were all referred to the same coordinate system.Show the German source This is not necessary, however, since the motions of the individual particles are independent of one another.Show the German source We shall now refer the motion of each particle to a coordinate system whose origin coincides with the position of the center of gravity of the particle in question at the time $t = 0$, with the difference that $f(x, t) d x$ now denotes the number of particles whose $X$-coordinate has increased from the time $t = 0$ to the time $t = t$ by an amount lying between $x$ and $x + d x$.Show the German source In this case too, then, the function $f$ changes according to equation (1).Show the German source Further, for $x \gtrless 0$ and $t = 0$ we must evidently haveShow the German source $$f(x, t) = 0 \quad \text{und} \quad \int_{-\infty}^{+\infty} f(x, t) d x = n$$$f$Number density$t$Time$n$Particle countExplain this equationShow the German source The problem, which coincides with the problem of diffusion from a single point (neglecting the interaction of the diffusing particles), is now mathematically completely determined; its solution is:Show the German source $$f(x, t) = \frac{n}{\sqrt{4 \pi D}} \frac{e^{-\frac{x^2}{4 D t}}}{\sqrt{t}}.$$$f$Number density$t$Time$n$Particle count$D$Diffusion coefficientExplain this equationShow the German source

The frequency distribution of the changes of position that have taken place in an arbitrary time $t$ is therefore the same as that of random errors, as was to be expected.Show the German source Of importance, however, is how the constant in the exponent is connected with the diffusion coefficient.Show the German source With the help of this equation we now calculate the displacement $\lambda_x$ in the direction of the $X$-axis that a particle undergoes on average, or — more precisely expressed — the square root of the arithmetic mean of the squares of the displacements in the direction of the $X$-axis; it is:Show the German source $$\lambda_x = \sqrt{\overline{x^2}} = \sqrt{2 D t}.$$${\lambda }_x$Coordinate RMS displacement$\overline{x^2}$Mean-square displacement (one dimension)$D$Diffusion coefficient$t$Observation intervalExplain this equationShow the German source

The mean displacement is therefore proportional to the square root of the time.Show the German source It can easily be shown that the square root of the mean of the squares of the total displacements of the particles has the value $\lambda_x \sqrt{3}$.Show the German source

Bern, May 1905.Show the German source

(Received 11 May 1905.)Show the German source