# Brownian Motion

*Michael Fowler, U. Va.
*

See Applet here!

### Introduction: Jiggling Pollen Granules

In 1827 Robert Brown, a well-known botanist, was studying
sexual relations of plants, and in particular was interested in the particles
contained in grains of pollen. He began
with a plant (*Clarkia pulchella*) in
which he found the pollen grains were filled with oblong granules about 5
microns long. He noticed that these
granules were in constant motion, and satisfied himself that this motion was
not caused by currents in the fluid or evaporation. Smaller spherical grains, which at first he
took to be oblongs end-on, but later realized weren’t, had even more vigorous
motion. He thought at first that he was
looking at the plant equivalent of sperm$\u2014$they were
jiggling around because they were alive.
To check this, he did the same experiment with dead plants. There was just as much jiggling. Perhaps all organic matter, everything that
ever was alive, still contained some mysterious life force at this microscopic
level? Sure enough, he found the
movement in tiny fragments of fossilized wood!
But then he went on to find it in matter that never was alive$\u2014$tiny
particles of window glass, and even dust from a stone that had been part of the
Sphinx. The movement evidently had nothing to do with the substance ever being
alive or dead, much to Brown’s surprise.
So what was causing it? Perhaps
it *was* evaporation currents, or the
incident light energy, or just tiny unnoticed vibrations. But none of these explanations was very
satisfactory.

Half a century later, a new possible explanation emerged. The kinetic theory of heat developed by Maxwell, Boltzmann and others was gaining credence. If all the molecules in the fluid were indeed in vigorous motion, maybe these tiny granules were being moved around by this constant battering from all sides as the fluid molecules bounced off. But there was a problem with this explanation: didn’t it violate the second law of thermodynamics? It had been well established that energy always degrades, as friction slows movement kinetic energy goes to heat energy. This seemed to be the other way round$\u2014$the molecular battering was certainly disorganized heat energy, but when the granule moved it had evidently gained kinetic energy. Since many scientists regarded the second law as an absolute truth, they were very skeptical of this explanation.

In 1888, French experimentalist Léon Gouy investigated the movement in detail, finding it to be more lively in low viscosity liquids. He established that it was unaffected by intense illumination or by strong electromagnetic fields. Despite the second law, Guoy believed$\u2014$correctly$\u2014$the random motion was indeed generated by thermal molecular collisions.

It’s easy to see the Brownian movement, or Brownian motion (it’s called both) by looking through a microscope at tobacco smoke in air. There's a movie here.

### Einstein’s Theory: the Osmosis Analogy

In 1905, Einstein published a theoretical analysis of Brownian motion. He saw it as a crucial test of the kinetic theory, even of the atomic/molecular nature of matter. Previous discussions of the phenomenon had all been qualitative$\u2014$Einstein demonstrated that careful observation of Brownian motion could reveal the actual size of molecules, that is, find Avogadro’s number.

If the results of such experiments were consistent with other estimates of Avogadro’s number, based on unrelated phenomena such as gas viscosity measurements and van der Waal’s equation fit to isotherms of real gases, it would be a powerful argument for the kinetic theory. On the other hand, if there was real disagreement then the kinetic theory was in serious trouble.

Einstein’s approach was based on an analogy with osmosis:
recall that osmosis involves one substance (the solute) dissolved in another,
the solvent. The solute has larger
molecules. Now imagine a container
divided in two by a semipermeable membrane, meaning the solvent molecules can
pass through the tiny holes in it, but the solute molecules are too big to get
through. Suppose on one side of the
membrane there’s pure solvent, on the other side solvent plus solute (assumed
fairly dilute), and initially the pressure on the two sides of the membrane is
the same. The pressure on the side with
solute is generated by both solvent and solute molecules bouncing off the
membrane, so necessarily the rate at which *solvent*
molecules are hitting the membrane on this side is less than on the other
side. Some percentage of the solvent
molecules hitting the membrane pass through the little holes, so what will
happen is that more will pass through from the pure solvent side, and gradually
pressure will build up on the solvent + solute side until equilibrium is
reached, meaning equal numbers of solvent molecules going each way on average.

*Einstein’s insight was
that a liquid containing a large number of tiny identical particles, such as
those observed in Brownian motion, was really no different from a solvent
containing solute molecules*. True,
the Brownian particles were a lot bigger than molecules, but they were buzzing
around, and would therefore bounce off the walls of a container, generating
pressure. The formal analysis should be the same: the kinetic theory, with
equipartition of energy, predicted they would have kinetic energy 1.5*k _{B}T*. If the
concentration of particles varied spatially, they would flow to even it
out.

Here again he used an osmosis analogy: think of a cylindrical container, with a semipermeable membrane which is like a piston, free to move. The solute concentration is initially greater to the left of the piston.

From the previous discussion, it’s clear that the solvent will flow to the left, raising the pressure there, so the piston will move to the right. The solute molecules cannot cross the piston, so the piston will move until the solute concentrations on the two sides are equal.

The rather surprising result is that if one assumes
equipartition of energy, *the pressure on
the piston from the solute on one side is the same as if those solute molecules
were freely moving in a vacuum*.

Their greatly reduced mean free path *doesn’t matter*: the pressure depends only on the concentration in
the immediate neighborhood of the piston, and the speed of the molecules. And, this is equally true if the solute
molecules are replaced by tiny but macroscopic spheres. At least, this is what Einstein asserted, and
he gave a formal proof based on an evaluation of the free energy, assuming a
dilute system (meaning interactions between the spherical granules could be
neglected).

So we can think of the little spheres as *moving freely through space* (!), and
although their paths will actually be very different, calculations of local
pressure based on this should be correct: the pressure on the walls from the
granules is therefore given by the ideal gas law, that is,

$$P={\scriptscriptstyle \frac{1}{3}}mn\overline{{v}^{2}}={\scriptscriptstyle \frac{2}{3}}n\overline{{\scriptscriptstyle \frac{1}{2}}m{v}^{2}}={\scriptscriptstyle \frac{2}{3}}n\text{\hspace{0.17em}}\overline{K.E.}={\scriptscriptstyle \frac{2}{3}}n\text{\hspace{0.17em}}W$$

where the average kinetic energy is written $W,$ and$\u2014$if the
kinetic theory is correct$\u2014$this
should equal $1.5{k}_{B}T.$ Of course, this is just the same as for a gas,
but the big difference is that these particles are *big enough to see*, so we can find the density $n$ just by counting! So if we can measure the pressure generated
by these particles, we can find $W$ and therefore Boltzmann's constant.

### An Atmosphere of Yellow Spheres

So how is this to be
done experimentally? As we shall see in
a moment, the first experiment used uniformly sized tiny *spheres* in place of granules. The first obvious thought is that if $\overline{{\scriptscriptstyle \frac{1}{2}}m{v}^{2}}$ is predicted to be $1.5{k}_{B}T,$ perhaps one can just measure the jiggling
velocity of the tiny sphere a few times and take an average. This, however,
misunderstands the nature of the motion: a molecule will bounce off the sphere
around 10^{20 }times per second, and although that only makes a tiny
difference to the sphere’s velocity, in one hundredth of a second the average
imbalance, $\sqrt{N},$ will be of order 10^{9}, enough to
make a change in a small sphere’s velocity. And all the successive changes are
completely random in direction, so it’s as hopeless as attempting to measure
the velocity of H_{2}S molecules in air by releasing a few and
measuring the time for the smell to reach the far end of a room.

A slightly less direct method of finding $\overline{{\scriptscriptstyle \frac{1}{2}}m{v}^{2}}$ is needed.
Now, it is well-known that in an isothermal atmosphere of an ideal gas
under gravity the density falls off exponentially with height, this is
established by balancing the gravitational force on a thin horizontal slice
against the pressure difference between top and bottom. (*Note*: if you don't recall this, see my
lecture here.) It occurred to the French experimentalist
Jean Perrin that this same argument should apply to a “gas” of tiny uniform
spheres in a fluid: their pressure is effectively generated by the Brownian
motion. In 1908, he chose gamboge, an
emulsion used for water color, which contains bright yellow spheres of various
sizes. By various ingenious tricks
described in his book he was able to separate out spheres all close to the same
size. He was able to measure the size,
he knew the density$\u2014$and that
of the “solvent”$\u2014$so he
could compute the gravitational pull. He
could also measure the decrease in density with height in isothermal
equilibrium.

The calculation is as follows: for a horizontal slice of unit area and thickness $dh,$ with $n$ spheres per unit volume, each of volume $\phi $ and density $\Delta ,$ in a liquid of density $\delta $ (I’m using Perrin’s notation here), the gravitational downward force on the slice is $ndh\phi \left(\Delta -\delta \right),$ this is balanced by the pressure difference:

$$-\Delta P=-{\scriptscriptstyle \frac{2}{3}}Wdn=ndh\phi \left(\Delta -\delta \right).$$

This is easily integrated to give the exponential vertical density profile.

$$\frac{dn}{dh}=-\frac{n\phi \left(\Delta -\delta \right)}{{\scriptscriptstyle \frac{2}{3}}W},\text{\hspace{1em}}n\left(h\right)=n\left(0\right)\mathrm{exp}\left(-\frac{\phi \left(\Delta -\delta \right)}{{\scriptscriptstyle \frac{2}{3}}W}h\right).$$

Perrin could establish by observation and measurement every term in this equation except $W,$ so this was a way of measuring $W,$ assuming of course the validity of the kinetic theory.

Now equating $W$ to $1.5{k}_{B}T$ gives a value for Boltzmann’s constant, and hence via the known gas constant $R={N}_{A}{k}_{B},$ a value for Avogadro’s number.

Perrin repeated the experiment with a wide variety of
different substances, the experiments were very challenging, his results for
Avogadro’s number were consistently between 5×10^{23} and 8×10^{23}. He remarked that for the largest granules
behaving like a perfect gas, *a gram
molecule would weigh 200,000 tons*!
The results were consistent with the other quite different ways of
finding Avogadro’s number, and these experiments convinced even the most
recalcitrant anti-atomic theory skeptics.
The kinetic theory was fully established.

### Langevin’s Theory

In 1908, Langevin gave a more direct treatment of Brownian
motion. He focused on following one
particle as it jiggled around. We’ll
follow him in *restricting the motion to
one dimension*$\u2014$assuming
the molecular collisions driving the motion are completely random, the motions
in the three directions are uncorrelated, so can be treated separately and
added. Finally, we’ll neglect gravity
and any other external force fields.

Let’s assume, then, that we’re tracking a small spherical object, of mass $m$ and radius $a.$ It will experience a viscous drag force $-6\pi a\eta v$ (Stokes’ formula). We’ll denote the random thermal molecular collision force by $X,$ which clearly averages to zero.

Assuming the equipartition of energy applies also to the kinetic energy of our sphere (and remember we're working in one dimension only),

$$\overline{m{\left(\frac{dx}{dt}\right)}^{2}}={k}_{B}T$$

where the average is over a long time.

The equation of motion $ma=F$ is:

$$m\frac{{d}^{2}x}{d{t}^{2}}=-6\pi a\eta \frac{dx}{dt}+X.$$

To integrate this equation, we begin by multiplying throughout by $x:$

$$mx\frac{{d}^{2}x}{d{t}^{2}}=-6\pi a\eta x\frac{dx}{dt}+Xx$$

which can be written

$$m\frac{d}{dt}\left(x\frac{dx}{dt}\right)-m{\left(\frac{dx}{dt}\right)}^{2}=-3\pi a\eta \frac{d}{dt}{x}^{2}+Xx.$$

To eliminate the unknown random force, we average over a long time:

$$m\overline{\frac{d}{dt}\left(x\frac{dx}{dt}\right)}-m\overline{{\left(\frac{dx}{dt}\right)}^{2}}=-3\pi a\eta \overline{\frac{d}{dt}{x}^{2}}+\overline{Xx}.$$

Since $X$ is random, $\overline{Xx}=0.$ Also, $m\overline{{\left(dx/dt\right)}^{2}}={k}_{B}T,$ and $x\left(dx/dt\right)={\scriptscriptstyle \frac{1}{2}}d\left({x}^{2}\right)/dt.$

The operations of averaging and taking the time derivative commute, so we can write the equation:

$$\frac{m}{2}\frac{{d}^{2}}{d{t}^{2}}\overline{{x}^{2}}+3\pi a\eta \frac{d}{dt}\overline{{x}^{2}}={k}_{B}T.$$

The trick to solving this equation is to write $\frac{d}{dt}\overline{{x}^{2}}=y,$

So the equation becomes:

$$\frac{dy}{dt}+\frac{6\pi a\eta}{m}y=\frac{2{k}_{B}T}{m},$$

with solution:

$$y\left(t\right)=\frac{{k}_{B}T}{3\pi a\eta}+C\mathrm{exp}\left(-\frac{6\pi a\eta}{m}t\right).$$

For the actual systems examined experimentally, the exponential term dies off in far less than a microsecond, so for a particle beginning at the origin:

$$\frac{d}{dt}\overline{{x}^{2}}\left(t\right)=y\left(t\right)=\frac{{k}_{B}T}{3\pi a\eta},$$

therefore

$$\overline{{x}^{2}}\left(t\right)=\frac{{k}_{B}T}{3\pi a\eta}t.$$ $$

Hence by doing multiple experiments and averaging, Boltzmann’s constant ${k}_{B}$ can be found, and from that Avogadro’s number, as before.

Note that the constancy of $\left(d/dt\right)\overline{{x}^{2}}\left(t\right)$ also appears in discussions of random molecular motion and the random path$\u2014$this is all the same thing.

** Exercises**:

Estimate the decay time of the exponential term in the integrated expression for $y\left(t\right)$ above. You’ll need to find the viscosity of water, and estimate the sphere size as a few microns.

Estimate how rapidly density of yellow spheres drops with height in Perrin’s “atmosphere”.

Notice the average distance traveled in the last equation above depends on the kinetic energy, the size, and the viscosity. This means a tiny lead sphere would diffuse the same distance, on average, as a tiny sphere of oil of the same size. But isn’t the lead moving a lot more slowly, since it has the same average kinetic energy? Explain.

## Reference Books

For Brown’s work, see *A
Source Book in Physics*, W. F. Magie, Harvard, 1963, page 251, where several
pages from the original pamphlet are reproduced.

Albert Einstein: *Investigations
on the Theory of the Brownian Movement*,
Dover, New York, 1956.

Jean Perrin: *Brownian
Motion and Molecular Reality*, Dover, New York, 2005.

Langevin’s paper in translation: *Am*. *J*. *Phys*. **65** (11), November 1997, 1079.

Wolfgang Pauli: *Pauli
lectures on Physics Volume 4, Statistical Mechanics*. Dover, New York, 2000,
page 64.