# The Gaussian and its width

The chance of finding a diffusing particle at a displacement x follows a bell curve whose mean square is 2Dt. About 68 per cent of the chance lies within one root-mean-square width, √(2Dt), of the start.

Written for this edition, not translated from Einstein.

## After a time t, where is a spreading particle likely to be?

Release many particles at one point in still water and look again after a time t. They have spread out: most are still near the start, fewer are farther away, and as many went left as right. Count how many ended up at each distance and the counts make a bell-shaped curve, the Gaussian. One number sets its width, √(2Dt), the root-mean-square distance. For D = 0.5 µm² per second and t = 1 s the width is 1 µm, and:

1. Within one width of the start, between −1 µm and +1 µm, end about 68 per cent of the particles.
2. Within two widths, ±2 µm, end about 95 per cent; within three, about 99.7 per cent.
3. So about one particle in three ends more than one width from where it started, and about one in twenty more than two widths away.

Wait four times as long and the width doubles, because it grows as the square root of the time. The curve itself, with p the probability per micrometre of finding a particle at displacement x after time t, is:

$$
p(x,t)=\frac{1}{\sqrt{4\pi Dt}}\exp\left(-\frac{x^2}{4Dt}\right)
$$

p of x and t equals one over the square root of four pi D t, times e to the minus x squared over four D t.

The factor in front keeps the total probability at one as the curve spreads: a wider bell has to be a lower one. The exponential makes the curve fall away on both sides, the same way to the left as to the right, because x appears squared.

## Worked example: Checking the curve: total probability one, and mean square 2Dt

In the formula above, x is the displacement, t the time since the start and D the diffusion coefficient. Two things need checking: that the total probability under the curve is 1, and that the mean square displacement is 2Dt. One change of variable turns both into standard integrals.

$$
u=\frac{x}{\sqrt{4Dt}},\qquad dx=\sqrt{4Dt}\,du
$$

Let u equal x over the square root of four D t. Then d x equals the square root of four D t, times d u.

$$
\begin{aligned}\int_{-\infty}^{\infty}p(x,t)\,dx&=\frac{1}{\sqrt{\pi}}\int_{-\infty}^{\infty}e^{-u^2}\,du\\&=1\end{aligned}
$$

The total probability becomes one over the square root of pi, times the integral of e to the minus u squared, and that equals one.

The last step needs the integral of e to the minus u squared to be √π. One way to see it is to square the integral, which makes it an integral over a whole plane, and then add the plane up in rings around the origin: a ring of radius r and width dr has area 2πr dr. This check is a modern one; the paper does not show it.

$$
\begin{aligned}I^2&=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(u^2+v^2)}\,du\,dv\\&=\int_{0}^{\infty}e^{-r^2}\,2\pi r\,dr=\pi\end{aligned}
$$

I squared is the integral of e to the minus u squared plus v squared over the whole plane, which equals the integral from zero to infinity of e to the minus r squared times two pi r d r, which is pi.

For the mean square, the same substitution leaves the integral of u² times e to the minus u squared. Integrating by parts once, the end terms vanish at both infinities, and half the integral above remains:

$$
\frac{1}{\sqrt\pi}\int_{-\infty}^{\infty}u^2e^{-u^2}\,du=\frac12
$$

The normalized integral of u squared times e to the minus u squared is one half.

Substituting back, the mean square is 4Dt times one half, which is 2Dt: the value the Brownian paper's §4 gives for the mean of the squared displacement, λx² = 2Dt.

## Where this lesson stops

The bell curve says where many particles end up. Any one particle can end anywhere, and about one in three ends more than one width from its start.

## This lesson builds on

- [Adding continuously](/foundations/integration/)
- [Mean, variance and RMS](/foundations/mean-variance-rms/)