Foundation lesson

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:

Within one width of the start, between −1 µm and +1 µm, end about 68 per cent of the particles.
Within two widths, ±2 µm, end about 95 per cent; within three, about 99.7 per cent.
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\left(x,\,t\right) = \frac{1}{\sqrt{4\,\pi\,D\,t}}\,\exp\left(-\frac{x^{2}}{4\,D\,t}\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^{2}\,\exp\left(-u^{2}\right)\,\mathrm{d}u = \frac{1}{2}

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.

The bell, and what each width is worth

One curve, with a rule drawn at one, two and three widths either side of the start. The bands are nested: the second contains the first.

One width is the root mean square displacement, √(2Dt). It is a label too long to sit inside a drawing, where text does not reflow on a narrow screen.

Within one width either way, between −1 and +1: about 68 per cent of the chance.
Within two widths either way, between −2 and +2: about 95 per cent of the chance.
Within three widths either way, between −3 and +3: about 99.73 per cent of the chance.

So about 32 per cent, roughly one particle in three, ends more than one width from where it started. A width is a typical distance, never a limit.

What it shows, in words

A bell curve centred on the start, with a rule drawn at one, two and three widths either side. The band within one width either way holds about 68 per cent of the chance; within two widths about 95 per cent; within three about 99.73. The bands are nested, so the 95 already includes the 68, and each further width adds less than the one before it.

The curve is the same shape to the left of the start as to the right, because the displacement enters the formula squared. About 32 per cent of the chance lies outside one width altogether, which is the part a reader most often forgets: the width says where particles typically end, and roughly one in three ends farther out than that.

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.

Going further: How far a sample's width can stray

The bell curve says where one particle is likely to end. It also says how far a whole sample's width can stray from the model's. Divide each measured displacement by √(2Dt) and it becomes a draw from the standard bell curve, whose mean square is 1. The sum of q such squares follows a curve of its own, the chi-square distribution with q degrees of freedom, whose mean is q.

With q = 100 squared displacements, 95 samples in 100 have a mean square between 0.742 and 1.296 times 2Dt. Their widths, the square roots, lie between 0.86 and 1.14 times the true width.
With q = 4, the same 95 samples in 100 run from 0.121 to 2.79 times 2Dt: a mean square about an eighth of the true one, or nearly three times it.
The spread narrows as q grows, but it is never zero for a finite sample.

Turned around, the same two points of the chi-square curve make an interval for D from a single sample. With q = 100, and $D_{\mathrm{est}}$ for the estimate:

\left[\frac{100\,D_{\mathrm{est}}}{129.56},\ \frac{100\,D_{\mathrm{est}}}{74.22}\right]

From one hundred D est over one hundred twenty-nine point five six, to one hundred D est over seventy-four point two two.

Used on sample after sample, this recipe catches the true D in 95 of every 100 of them. The promise is about the recipe, repeated. It says nothing more about any one interval once it has been drawn.

All of this assumes the displacements follow the bell curve and are independent of one another. Reading errors and overlapping windows break the second assumption, and then q is not the number of independent squares, so the interval's promise no longer holds as stated.

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