Probability - University of Michigan



6.3 Diffusion and Brownian Motion.

Brownian motion is a probablistic model of certain types of motion. For example

1. Diffusion of particles in gases, e.g. diffusion of water molecules in air

2. Diffusion of particles in liquids, e.g. diffusion of sugar molecules in water

3. Heat conduction

4. Movement of stock prices

Brownian motion is a continuous version of a random walk which is a discrete model of certain types of motion which have a random aspect. So, we first take a look at random walks and then see how we get Brownian motion as the limit of random walks.

Random Walks. A random walk is a special type of Markov chain in which one gets from one state to another by adding a random amount. As an example we shall consider the motion of a water molecule in air. For simplicity we shall assume its motion is in one dimension which we take as the x axis.

Water molecules move fairly fast at room temperature. A typical speed is

v = 100 m/sec ( 0.06 miles/sec ( 200 miles/hr

It doesn't go very far until it hits a nitrogen or oxygen molecule in the air. A typical distance before it hits another molecule is

l = 5 ( 10-5 cm = mean free path = average distance between collisions

Let's suppose when it hits it rebounds in the opposite direction with the same speed v. Let

(x = = 2.5 ( 10-5 cm = ½ average distance between collisions

We assume that after the water molecule has gone a distance (x it has a 50% chance of striking another molecule in the air. If it does it rebounds and goes in the opposite direction with velocity v. If it doesn't, then the water molecule continues in the same direction with velocity v and goes another distance (x and then again has a 50% chance of a collision. So on the average it goes a distance 2((x) = l before it strikes another molecule. Let

(t = = = 2.5 ( 10-9 sec = time it takes to go a distance (x

We construct a Markov chain model of the water molcule's motion as follows. The states are the points on the x axis whose coordinates are multiple of (x, i.e the points j((x) where j is an integer, positive, negative or zero. We consider the position of the water molecule at times which are non-negative integer multiples of (t, i.e. n((t) where n is a positive integer of zero. Let

Un = displacement of water molecule during the time interval (n – 1)((t) < t < n((t)

We treat Un as a random variable and assume that there is a 50% chance the velocity during the time interval (n – 1)((t) < t < n((t) is either v or – v. So Pr(Un = (x) = ½ and Pr(Un = - (x) = ½. The Un all have the same distribution. We also assume that the Un are all independent of each other. Note that

E{Un} = mean of Un = 0

E{(Un)2} = (1/2)(- (x – 0)2 + (1/2)(- (x – 0)2 = ((x)2

(Un = standard deviation of Un = (x

We let

X(t) = the position of the water molecule at time t

Yn = X(n((t)) = the position of the water molecule at time n((t)

In particular,

X(t) = Yt/((t)

Y0 = X(0) = starting position of the water molecule at time t = 0

Y1 = X((t) = the position of the water molecule at time (t

Y2 = X(2((t)) = the position of the water molecule at time 2((t), etc

One has

Yn = Yn-1 + Un

i.e. the state at time n is obtained by adding Un to the state at time n – 1. It follows that

Y1 = Y0 + U1

Y2 = Y0 + U1 + U2

Yn = Y0 + U1 + U2 + ( + Un = Y0 + Vn

where

Vn = U1 + U2 + ( + Un = Yn - Y0 = X(n((t)) - X(0)

or

X(t) - X(0) = Vt/((t)

If n = t/((t) is large, then by the central limit theorem Vn is approximately normal with mean 0 and standard deviation

((Un) = ((x) = ((x) = (

where

( = =

For the water molecule this works out to be

( = = = 0.5 cm/(sec)1/2

So at time t the water molecule's displacement X(t) - X(0) is approximately normal with mean zero and standard deviation ( with ( = . So the density function of X(t) - X(0) is approximately

ft(x) = e-x2/(2(2t) = e-x2/(4Dt)

where

D = = = Diffusion coefficient

Note that

( =

For example, if t = 25 sec and ( = ½ cm/(sec)1/2 then f25(x) = e-x2/(2(2t) = e-x2/12.5.

( and D are macroscopic parameters that describe the motion of the water molecules or whatever particles one is modeling by diffusion. Most science and engineering books tend to use the diffusion coefficient D instead of (. However, sometimes ( is sometimes more convenient for hand computations using tables of the normal distribution. For the water molecule the diffusion coefficient is

D = = = 0.125 cm2/sec

Another way of saying that the displacement X(t) - X(0) at t is approximately normal with mean zero and standard deviation ( is to say that Z(t) = (X(t) – X(0)/(() is approximately a standard normal random variable.

Example 1. What is the approximate probability that a water molecule is a distance of at least 3 cm from its starting point after 25 sec. Assume ( = 0.5 cm/sec1/2.

We want Pr{ | X(25) – X(0) | > 3}. In this case Z = (X(25) – X(0)/((0.5)) = (X(25) - X(0)/2.5 is approximately a standard normal random variable. So

Pr{ | X(25) – X(0) | > 3} = Pr{ | | > 3/(2.5)} = Pr{ | Z | > 1.2} = 2 Pr{ Z > 1.2}

= 2(1 - Pr{ Z ( 1.2}) = 2(1 - ((1.2)) = 2(1 – 0.8849) = (2)(0.1151) = 0.2302

Here

((z) = Pr{Z ( z} =

is the cummulative distribution function of a standard normal random variable. It is available in mathematics software and tables are usually included in statistics books.

If we are given that X(0) = ( then X(t) is normal with mean ( and standard deviation (, so its density is pt(x) = e-(x - ()2/(2(2t).

Brownian Motion. If we look at the motion of the water molecule in the limit that the mean free path l goes to zero and the velocity v goes to infinity in such a way that ( = stays constant the resulting motion is Brownian motion X(t). It has the following properties

1. X(s + t) – X(s) is normal with mean zero and standard deviation (t

2. If 0 < t0 < t1 < ( < tn then the displacements

X(t1) – X(t0)

X(t2) – X(t1)

…

X(tn) – X(tn-1)

are all independent

Above we focused on the motion of a single water molecule. Now suppose we have a bunch of water molecules. In particular, suppose we are given an initial density c0(x) of water molecules in the air. For example, we might have

(1) c0(x) =

We would like to find the density ct(x) of water molecules at a later time t. The initial density c0(x) acts like a probability distribution. If we let A = the p0(x) = c0(x)/A is the initial probablilty density function of X(0). Note that X(t) = X(0) + (X(t) – X(0)). X(0) and X(t) – X(0) are independent, so to get the probablity density function of X(t) we convolute the density function c0(x)/A of X(t) with the density function e-x2/(2(2t) of X(t) – X(0). Finally we multiply by A to get the density ct(x) of water molecules. Thus

ct(x) =

Using the formula (1) for c0(x) we have

ct(x) = (4 ( 10-3)

If we substitute z = (x - ()/(( ) the integral becomes

ct(x) = (4 ( 10-3) = (4 ( 10-3)

= (4 ( 10-3) [ ( - ( ]

Suppose t = 25. Then ( = (1/2)(5) = 2.5 cm so

ct(x) = (4 ( 10-3) [ ( - ( ]

A graph is at the right.

Problem 1. The diffusion coefficient of (ethyl) alcohol in air is D = 0.1 cm2/sec. Consider the diffusion of alcohol in one dimension.

a. Suppose an alcohol molecule starts at position x = 2 cm. What is the probability that its position 10 seconds later will be greater than or equal to x = 2.5 cm?

b. Suppose the density of alcohol to begin with is given by

c0(x) =

Find a formula for the density ct(x) of ammonia at time t.

Solution to a. We want Pr{ X(10) – X(0) > 0.5}. In this case  =  = = = . We know that X(10) – X(0) is normal with mean 0 and standard deviation ( So Z = (X(10) – X(0))/ is approximately a standard normal random variable. So

Pr{ X(10) – X(0) > 0.5} = Pr{ | > 0.5/} = Pr{Z > /4}

= 1 - Pr{Z ( /4}) = 1 - ((/4)) ( 1 – ((0.35)) ( 1 – 0.6368 = 0.3632

Brownian Motion with Drift. Let's look again at water molecules diffusing in air. So far we have been assuming that the air as a whole has no net motion. However, it may happen that the air as a whole is moving with a certain net velocity. Let

( = net velocity of the air

= drift rate

Then water molecules inherit this drift and the diffusion is superimposed on top of this drift. The result is that at time t the water molecule's displacement X(t) - X(0) is now approximately normal with mean (t and standard deviation ( . So the density function of X(t) - X(0) is approximately

ft(x) = e-(x - (t)2/(2(2t)

Example 2. Suppose a water molecule diffuses in air with ( = 0.5 cm/sec1/2. Suppose also the air has a drift of ( = 2 cm/sec. What is the probability that a water molecule will have traveled a distance of between 30 and 34 cm in the positive direction after 16 sec.

We want Pr{ 30 < X(16) – X(0) < 34}. In this case (t = (2 cm/sec)(16 sec) = 32 cm and (4 sec1/2) = 2 cm. We know that X(16) – X(0) is normal with mean (t and standard deviation (. So Z = (X(16) – X(0) – 32)/2 is approximately a standard normal random variable. So

Pr{ 30 < X(16) – X(0) < 34} = Pr{ - 2 < X(16) – X(0) – 32 < 2} = Pr{ -1 < < 1}

= Pr{- 1 < Z < 1} = 2 Pr{0 < Z < 1} = 2(Pr{Z < 1} – ½) = 2(((1) – ½)

= 2(0.8413 – 0.5) = (2)(0.3413) = 0.6826 ( 68%

Problem 2. Redo problem 1 if the air is drifting in the negative x direction at a rate of 0.05 cm/sec.

Appendix. Here are the values of some diffusion constants, D.

Diffusion in air:

Hydrogen, H2 1.3 cm2 / sec another source gave 0.74

Water vapor, H2O 0.22 cm2 / sec

Carbon monoxide, CO 0.21 cm2 / sec

Ammonia, NH3 0.2 cm2 / sec

Methane, CH4 0.2 cm2 / sec

Oxygen, O2 0.18 cm2 / sec

Nitrogen, N2 0.18 cm2 / sec

Hydrogen cyanide 0.17 cm2 / sec

Carbon dioxide, CO2 0.16 cm2 / sec

Mercury 0.12 cm2 / sec

Ethyl alcohol 0.1 cm2 / sec

Ethyl ether 0.079 cm2 / sec

Diffusion in water:

Alcohol 10 ( 10-6 cm2 / sec

Caffein 6 ( 10-6 cm2 / sec

Sugar 6 ( 10-6 cm2 / sec

Dyes 2 - 6 ( 10-6 cm2 / sec

Proteins 0.4 - 1 ( 10-6 cm2 / sec

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