The Statistical Nature of Diffusion
quick-and-dirty estimation of transport
coefficients
Cedric J. Gommes
February 20, 2014
Contents
1 Introduction
4
2 Diffusion is a random walk
5
3 Fick’s laws
8
4 Diffusion in some specific media
11
5 Diffusive transport of heat and momentum
19
6 Problems
6.1 Brownian motion . . . . . . . . . . . . . .
6.2 The size of a polymer in solution . . . . .
6.3 Dimensional analysis of gas-phase diffusion
6.4 Diffusion of O2 in air . . . . . . . . . . . .
6.5 Diffusion of O2 in water . . . . . . . . . .
6.6 So-called turbulent diffusion . . . . . . . .
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Figure 1: Albert Einstein (1879-1955) in 1904, about the time when he
was investigating Brownian motion. Einstein’s theoretical work on diffusion
notably enabled Jean Perrin to experimentally determine the value of Avogadro’s constant. This was tantamount to estimating the size of molecules,
at a time when the atomic nature of matter was still controversial.
Figure 2: Jean Perrin (1870-1942) is famous for his experimental investigations of Brownian motion. This enabled him to estimate kB via the
Stokes-Einstein relation, which eventually lead to an estimation of Avodagro’s constant. Incidentally, it is Perrin who suggested that the number of
molecules in a mole be named after Amedeo Avogadro.
2
Figure 3: Louis Bachelier (1870-1946) at age 15. Bachelier introduced the
idea of random walks to analyze stock market fluctuations in the context of
his PhD thesis ”Théorie de la Spéculation”. The approach of Bachelier is
now commonplace in financial mathematics.
Figure 4: Paul J. Flory (1910-1985) did not contribute to the theory of diffusion. He is a founding father of the physical chemistry of polymers (Nobel
prize for chemistry in 1974). The shape of a polymer in a solvent is very
similar to the path of a randomly diffusing molecule; the same mathematics
is useful for both domains.
3
”In this paper it will be shown that according to the molecular-kinetic theory of heat, bodies of microscopically-visible size suspended in a liquid will
perform movements of such magnitude that they can easily be observed in a
microscope, on account of the molecular motions of heat.”
(Albert Einstein, On the movement of small particles suspended in a stationary liquid demanded by the molecular-kinetic theory of heat, Annalen der
Physik, 1905)
1
Introduction
Quite often, people seem to have a wrong idea of the nature of diffusion.
In some way, they seem to understand that molecules diffuse down a concentration gradient in much the same way as they would run down a hill.
Perhaps, even, similar molecules would repel each other, which would lead
to the homogenisation of concentration in space. This would be a profound
misunderstanding of the nature of diffusion. Actually, the interaction of
similar molecules is quite rare in many instances where this concentration
homogenisation effect takes place1 .
In order to fully understand many properties of diffusive phenomena, and
in particular of diffusion-limited processes, it is necessary to have a deeper understanding of diffusion than this mere phenomenological description. What
diffusion really is, is the random unoriented motion of molecules under the effect of thermal agitation. The homogenisation of the concentrations is just a
statistical consequence of the random motion of a large number of molecules.
Let us make a back-of-the-envelope calculation to see how often a molecule
diffusing in air, say an odorant molecule, actually sees a similar molecule.
For this type of calculation it is always handy to have a few figures in mind.
Figures that are easily remembered about air in STP conditions is that the
mean free path of the molecules between two collisions is around λ ' 100
nm, and that their speed is about the same as the speed of sound, i.e., hui '
300 m/s. This means that a given air molecule collides with another one
every τ = λ/ hui ' 3 10−10 s. Now, olfaction is a very sensitive sense, and a
person can detect as little as 0.001 ppm of some odorant molecules. For such
a low concentration, an odorant molecules hits on another odorant molecule
1
Imagine gaseous molecules in a box: where do you expect to find the molecules if
they actually repelled each other? [Hint: think of an electromagnetic analogy with, say,
Coulomb interactions.]
4
100
15
100
0
5
0
100
0
t=30
%
x(t)
t=10
%
10
−5
100
0
%
−10
−15
−20
t=2
%
20
0
10
20
30 40 50 60 70
time (nb of jumps)
80
90 100
t=90
0
−20 −15 −10
−5
0
x
5
10
15
20
Figure 5: Left: 50 realizations of discrete random walks, whereby each
walker jumps left x → x − 1 or right x → x + 1 with equal probability. The
time is the number of jumps. Right: distributions of the distance traveled
after various times, also shown as dashed lines on the left.
only once in a billion collisions.
Therefore, the odorant molecule is constantly colliding with air molecules
(every nanosecond) but a collision with another odorant molecule occurs
only every second or so. In other words : in many situations, diffusing
molecules barely see each other! However, this does not prevent the
concentration of the diffusing molecule to equalize, eventually.
2
Diffusion is a random walk
The simplest diffusion model is the one-dimensional random walk. Consider
a walker, initially at position x(0) = 0. Each time the clock ticks a second,
you toss a coin and have the walker move to the right or to the left with
equal probability p = 1/2. Examples of random walks are shown in Figure
5.
The situation of Figure 5 can be considered as a crude mathematical
model of what happens to a coumarin molecule diffusing in air. As we have
shown in the Introduction, all the collisions have to be taken into account.
This means that most jumps are due to a collision of the coumarin with an
5
air molecule. The clock is set to tick every τ ' 10−10 seconds, and the length
of the jump is set to be λ ' 100 nm.
The concentration equalization property of diffusion is clear in the distribution of distances shown on the right of the figure. Initially, all molecules are
at x = 0, and the statistical distribution of the distances spreads with time.
To make the spreading analysis quantitative one has to calculate the standard
deviation of the distributions, which is the square root of the variance. This
is a measure of the breadth of any statistical distribution. Mathematically,
the variance σ 2 of a random variable X is defined as
σ 2 = (X − hXi)2
(1)
where the brackets h.i stand for the average value. If you look carefully at
Eq. 1, you will realize that the variance is simply the average value of the
squared deviation from the mean.
In order to calculate the spreading with time of the distributions, we have
first to calculate the average value hXi (t) as a function of time. To formalize
the random walk, we may write it as
X(t + ∆t) = X(t) + (t)
where is a random number
+∆x with probability p = 1/2
(t) =
−∆x with probability p = 1/2
(2)
(3)
and X(0) = 0. The latter equation simply means that all the random walks
start at the origin of the x’s.
With that formalism, we find immediately that hX(t + 1)i = hX(t)i
because h(t)i = 0. This means that the average distance traveled by a
random walker is always zero. This has a simple explanation: whatever move
the walker can make in the positive x direction, it is exactly counterbalanced
by an equally probable move in the negative x direction. With that result in
mind, the variance of the distribution becomes simply σ 2 (t) = hX 2 (t)i. To
calculate it, consider that
X 2 (t + ∆t) = X 2 (t) + 2 (t) + 2X(t)(t)
(4)
which naturally implies that
hX 2 (t + ∆t)i = hX 2 (t)i + (∆x)2
6
(5)
because h2 (t)i = (∆x)2 and hX(t)(t)i = 0. A beautiful consequence of Eq.
5 is the following
(∆x)2
σ 2 (t) =
t
(6)
∆t
This is a key result. The variance of the distance traveled by a diffusing
molecule is proportional to the time. This means that the linear size of the
region of space explored by a diffusing molecule increase like the square root
of time. This is often considered as being the signature of a diffusive process.
For further purposes, it is useful at this stage to derive the exact analytical
expression of the probability Pt (x) of finding a diffusing particle at position
x at time t, if it was at x = 0 at time t = 0. This is tantamount to deriving
the analytical expression of the histograms on the right side of Figure 5. The
position x reached by the molecules after n = t/(∆t) jumps can be written
as follows
x
= +1 + 1 − 1 − 1{z. . . + 1 − 1 − 1}
∆x |
(7)
n=t/(∆t) terms
where each one of the +1 or −1 term corresponds to a jump of the molecule
to the right of to the left. Let n+ be the number of jumps to the right and
n− be the number of jumps to the left. Obviously, these numbers satisfy
n = n+ + n− . Now, every term in Eq. 7 has the probability 1/2 of being +1
and probability 1/2 of being −1, so the probability of any given sequence of
+1 and −1 of total length n is exactly 1/2n . However, it has to be noticed
that the order in which the +1 and −1 appear in the sequence is irrelevant
to the final position of the molecule. You can can jump now to the right or
later, it does not matter as long as the total number of jumps n+ and n− are
unchanged. The total number of ways of making a total of n+ jumps to the
right (and n− = n − n+ jumps to the left) out of a series of n jumps is
n!
n
(8)
=
n+
n+ !(n − n+ )!
Therefore, the probability of having n+ jumps to the right in a random walk
of length n is
1
n
(9)
n+ 2n
7
which is the well-known binomial (Bernoulli) distribution B(n, p = 1/2).
In order to put this result in a slightly more usable form, we shall use
the central limit theorem, which states that a binomial distribution B(n, p)
converges for large values of n to a normal (Gaussian) distribution with mean
np and variance np(1 − p). This leads to
1 (n+ − n/2)2
(10)
P ' exp −
2
n/4
where the ' sign means that a normalizing constant has been ignored.
Furthermore, we shall now express n+ in terms of x. For that purpose,
we notice that the position of the molecule can be written as
x
= n+ − n−
∆x
(11)
which can be combined with n+ + n− = n to yield
n+ =
1
x n+
2
∆x
Together with n = t/∆t, the final expression becomes
x2
P ' exp −
2[(∆x)2 /∆t]t
(12)
(13)
which is valid for large values of t.
3
Fick’s laws
How does that random walk analysis compare with the classical approach
based on Fick’s first law? They are of course equivalent, as we now show.
Fick’s first law is the usual way in which diffusion is introduced: the
flux JD of molecules per unit area and unit time across a given surface is
negatively proportional to the gradient of concentration across that surface.
Mathematically, this reads
JD = −D∇c
(14)
where bold-faced characters are vectorial quantities2 .
2
We use the notations
8
Let us consider again our random walk analysis of diffusion. Imagine a
one-dimensional world that is discretized into boxes of size ∆x. Every time
∆t, each molecule in all the boxes can jump either to the box on their right
or to the box on their left. Consider two adjacent boxes: the one on the left
has nlef t molecules in it and the one on the right has nright molecules. The
net flux of molecules across the surface between them is therefore
n
nright lef t
−
/∆t
(15)
J=
2
2
where the factors 1/2 account for the fact that each molecule can jump in
both directions, so that only half the molecules in the two boxes cross that
surface. We can put that equation in a more familiar form as
(∆x)2 cright − clef t
J =−
(16)
2∆t
∆x
where we have introduced the concentrations c = n/∆x, and the term between square brackets is the gradient in the limit where ∆x → 0 . Comparing
with Eq. 14 immediately suggests
D=
(∆x)2
2∆t
(17)
To check is everything is really consistent, we now use Fick’s laws to
determine the probability of finding a molecule at a given position, if it was
initially at another position (equivalent of Eq. 13). For that purpose, we need
a second equation in addition to Eq. 14, which translates mass conservation
in mathematical terms. This reads
∂t c + ∇ · J = 0
(18)
When combining the two equations, and assuming that the diffusion coefficient is space-independent leads to
∂t c = D∆c
• ∇ for the gradient;
• ∇· for the divergence;
• ∆x = ∇ · (∇x) for the Laplacian of a scalar quantity;
9
(19)
which is sometimes referred to as Fick’s second law.
Let us consider again a one-dimensional domain. In this case Fick’s second law writes
2
c
(20)
∂t c = D∂xx
In the situation we want to analyze, there are initially a large number, say
N molecules at x = 0 and we are interested in determining how the concentration evolves according to Eq. 20.
This is a nice mathematical problem that can be considerably simplified
using dimensional analysis. The solution of Eq. 20 that we are looking for is
of the type
c = f (x, t, D, N )
(21)
where f is an unknown function. Let us apply Buckingham’s Π theorem:
there is a total of 5 variables and 3 dimensions (number of molecules, distance
and time). There are therefore only two dimensionless numbers for this
problem. A possibility is 3
√
x
c Dt
(22)
π1 = √
π0 =
N
Dt
The unknown solution to our problem cannot but take the form
N
x
c(x, t) = √ f √
Dt
Dt
(23)
Introducing this particular function in Eq. 20, one finds that the function
f (ξ) has to satisfy the following equation
1
ξ df
d2 f
f+
+ 2 =0
2
2 dξ
dξ
(24)
Note that the task of solving this equation is considerably simpler than the
one of solving Eq. 20. The dimensional analysis has enabled us to transform
a partial differential equation (PDE) into an ordinary differential equation
(ODE). The solution is
2
−ξ
f = A exp
(25)
4
3
Remember that this is a one-dimensional problem so a concentration has dimensions
mole/m.
10
where A is an undetermined constant. Its value can be obtained by expressing
that
Z +∞
c(x, t)dx = N
(26)
−∞
The solution takes finally the form
N
c(x, t) = √
exp
4πDt
−x2
4Dt
(27)
which is mathematically identical to Eq. 13 provided D takes the value given
in Eq. 17.
4
Diffusion in some specific media
Diffusion in gases
There exist very accurate theories for the diffusion coefficients in gases. The
most famous is probably the Chapman-Enskog theory, which leads to the
following expression of the binary diffusion coefficient of species A in species
B
1/2
1
1
1
3/2
+
(28)
DAB = 0.0018583 T
2
MA MB
P σAB ΩAB
where the diffusion coefficient is expressed in cm2 /s, the temperature T is
expressed in Kelvin, the molar masses MA/B are in g/mole, the collision
diameter σAB is in Å, and the collision integral ΩAB is a dimensionless factor
close to one.
Our purpose here is not to reproduce the calculations that lead to Eq.
28, but rather to explain its very particular dependence on the temperature,
pressure and molar masses, based on Eq. 17.
Remember that ∆x is the length of each random jump in our simple
model of diffusion, and that ∆t is the time between jumps. In the context
of gases ∆x has to be identified with the mean free path λ, corresponding
to the distance a molecule travels before it hits on another one. A simple
estimation of λ is based on the sketch in Figure 13, in which all molecules are
assumed to be spherical with diameter d. A collision happens whenever the
distance between two molecules becomes smaller that d. Consider that only
one molecule is moving; the others are randomly distributed in space with
11
d
v
2d
lambda
Figure 6: Mean free path estimation in gases.
mean free path λ (nm)
700
600
Helium
Air
Cyclohexane
500
400
300
200
100
0
200
400
600
800
1000
Temperature (K)
Figure 7: Values of the mean free path of three molecules, estimated from
Eq. 29 at P = 1 bar, with σ = 2.57 Å(Helium), σ = 3.62 Å(Air), and
σ = 6.14 Å(Cyclohexane).
12
concentration c = P/kB T (ideal gas law) but they do not move. With this
assumption, the distance λ that the moving molecule has to travel before it
hits another one is such that there should be on average one molecule in the
cylinder of diameter 2d and height λ. This leads to the values
kB T
(29)
P πd2
This average value does naturally not change if all the molecules are moving.
The other quantity left to be determined is the time between two collisions, which can be calculated as ∆t = λ/v if the velocity v is known. Of
course the velocity is a fluctuating quantity so only an average value can be
estimated. An easy way to do that is to remember that the average kinetic
energy of a molecule satisfies the equipartition theorem, i.e.
λ=
3
1
M v 2 ' RG T
(30)
2
2
where M is the molar mass. When writing it in this way, it is the average
kinetic energy of a mole that is calculated so that RG has to be used instead
of Boltzmann’s constant kB .
Putting all this together, one finds
r
√
3
1 1
1 (∆x)2
3/2
=
(kB T )
(31)
D=
2 ∆t
2π
m P d2
which is our back-of-the-envelope estimate of the diffusion coefficient. Of
course, this expression is inaccurate in several respects. Notably, it assumes
that all the molecules are identical, so that it would strictly apply to selfdiffusion. However, the temperature, pressure, and mass dependence of D
in Eq. 31 is consistent with the one of Chapman-Enskog (Eq. 28). In the
case of a mixture of gases, the mass m that appears in the equation is an
”average” mass. It so happens that the right ”average” mass to use is the
reduced mass
1
1
1
=
+
(32)
m
mA mB
which is quite common when analyzing two-body interactions.
Diffusion in porous solids
Diffusion in porous solids is an important matter for many technologies and
industrials processes as well as natural phenomena. Examples can be found
13
theta
c1
c2
L
Figure 8: Overly simplified slab of porous material.
in heterogeneous catalysis, separation processes, electrochemistry, the weathering of rocks, etc.
When diffusion is taking place in the pores of a solid, several cases have
to be considered. The simplest case is when the pore size is much larger than
the mean free path of the molecules. In this case, diffusion takes place in the
pores just in the same way as in a free gas. Only, the overall space where
the molecules diffuse is limited by impermeable walls. In many cases, the
diffusion coefficient is modeled as follows
D=
Dm
τ2
(33)
where is the porosity of the material and τ is a tortuosity factor. The
tortuosity is the ratio of the actual distance between two points in the pore
space of the material and the straight-line distance between them.
The rationale behind Eq. 33 is the following. Imagine a porous material
made up of a large number of tortuous pores, and a slab of thickness L is
mentally cut out of it such as in Fig. 8. The cross section of the pores is
Ω and they all have an angle θ with the direction perpendicular to the slab.
The length of each pore is therefore L0 = L/cos(θ). Let there be n such pores
per unit surface of the slab. The porosity is easily found to be
= nΩ
14
L0
L
(34)
If a concentration difference ∆c in any molecule is set across the slab, each
pore will individually contribute to transport a molar flow of
∆c
j = Dm Ω 0
(35)
L
The total macroscopic molar flow is obtained by summing over all pores,
which introduces simply a factor n in Eq. 35. Combining Eqs. 35 and 34
leads to
Dm ∆c
(36)
J= 0
(L /L)2 L
where L0 /L is nothing but the tortuosity τ .4
In the case where the pore size dpore is smaller than the mean free path of
the molecules, the molecular diffusion coefficient is irrelevant. Indeed, Dm is
calculated from the collision of the diffusing molecules with other molecules,
but these collisions are less frequent than the collision with the pore wall in
the limit d λ. This is regime is called Knudsen diffusion. In this case, the
random walk is obtained by the molecule hitting the wall and bouncing back
randomly. The corresponding diffusion coefficient is obtained in the same
way as for the molecular diffusion, but setting the length of the jump to the
value dpore . This leads to the
r
kB T
DK ∼ dpore
(37)
m
Diffusion in liquids
Diffusion in liquids is considerably more difficult to analyze that diffusion
in gases, mostly because the successive collisions that a diffusing molecule
undergoes are not independent from one another. A breakthrough in that
domain was obtained by Einstein, in the context of his work on Brownian
motion. That breakthrough is the celebrated Einstein’s relation that relates
the diffusion coefficient of a small macroscopic particles to their mobility.
The latter quantity is defined as follows. When you apply a given force F to
a small particle in a fluid at rest, its velocity will progressively increase and
reach a stable value V . The mobility is defined as
V
(38)
µ=
F
4
Can you imagine an argument, based on the diffusion time of a molecule, to justify
why the tortuosity enters Eq 33 with an exponent 2?
15
1
300 K
6000 K
107 K
Probability
0.8
0.6
0.4
0.2
0
−3
10
0
10
Energy (eV)
3
10
Figure 9: Boltzmann distribution for three different temperatures. 300 K is
the typical temperature on the surface of Earth: all states with energy around
1 meV are excited thermally; 6000 K is the temperature at the surface of the
sun: atoms and molecules are ionized because the thermal energy reaches 1
eV; 107 K is the estimated temperature at the center of the sun: the thermal
energy reaches the keV, which can trigger thermonuclear reactions.
This concept is more general than a particle in a fluid. It applies also, e.g.,
to charge carriers in semi-conductor crystals.
To relate the concept of mobility to the diffusion coefficients, imagine a
suspension of small solid particles in a fluid, which are subject to Brownian
motion, i.e. to diffusion. If you exert a force on the particles, the concentration profile will be modified because the particles will accumulate in the
regions where the force field pushes them. This is a familiar observation with
suspensions that settle under the effect of gravity. Einstein’s insight was that
in the equilibrium state of the suspension, the distribution of the particles
should obey Boltzmann’s distribution
E
finding a particle in
(39)
Probability
' exp
a state with energy E
kB T
In the present context, the energy is the potential energy corresponding to
the force field imposed on the particles.
16
Figure 10: Ludwig Boltzmann (1844-1906) is one of the founding fathers of
statistical mechanics. His celebrated equation S = kB ln(Ω), which gives a
statistical meaning to entropy, is written on his grave in Vienna.
From the point of view of diffusion, the concentration profile that is eventually reached results from a balance between the force exerted on the particles, which would tend to concentrate them all at the bottom of the solution,
and diffusion which would spread them uniformly over the entire solution.
Comparing this analysis with Boltzmann’s distribution should enable us to
relate the diffusion coefficient to the mobility.
The analysis proceeds as follows. We shall first express that the total flux
of particles Jp is the sum of a diffusive contribution, and of a drift contribution
resulting from the applied force. Next, we shall express that the total flux is
zero at equilibrium, which will provide us with an equation that characterizes
the equilibrium distribution. Take the The diffusive contribution reads
JDif f usion = −D
dc
dz
(40)
The drift contribution is
JDrif t = cµF
(41)
because the force F is positive upwards. At equilibrium, the sum of two
contributions is zeros, which leads to a first order differential equation that
is solved into
Z z
c(z)
µ
= exp
F dz
(42)
c(0)
D 0
17
Figure 11: George Gabriel Stokes (1819-1903) was an Irishman who gave
his name to the fundamental equation of fluid mechanics: the Navier-Stokes
equation. He held the Lucasian chair of mathematics at Cambridge, a position held long before him by none other than Isaac Newton, and long after
him by Paul Dirac and Stephen Hawking.
Realizing that the integral of F dz is the negative of the potential energy, Eq.
42 is seen to have exactly the same form as Boltzmann’s distribution (Eq.
39). This leads to the identification
D = µkB T
(43)
which is known as Einstein’s relation. If you go through the derivation again,
you will realize that no assumption whatever has been made about the nature
of the diffusing species. Equation 43 is quite general, and it applies also
beyond the particular case of diffusion in liquids, which we shall consider
now.
To particularize Einstein’s relation to the diffusion in liquids, one needs
only to find an expression of the mobility µ that incorporates the properties of
the liquid and of the diffusing molecule. The determination of the mobility of
a spherical particle suspended in a viscous fluid is a classical mathematical
problem solved by Stokes. What limits the velocity of the particle is the
viscous forces exerted by the fluid. Using dimensional analysis, based on the
dimensional variables µ, particle radius R, fluid viscosity η, one can easily
find that the mobility should scale as µ ∼ 1/(Rη). A more precise analysis,
based on Stokes’ solution leads to
1
µ=
(44)
6πηR
18
Putting this result together with Einstein’s relation, one finds
D=
kB T
6πηR
(45)
for a particle diffusing in a viscous fluid.
Equation 45 is strictly valid only for macroscopic objects for which Stokes
relation applies and many authors have tries to generalize the equation to
smaller solutes. The most common equation is the Wilke-Chang correlation,
which is often used for quick estimations. It reads
7.4 10−8 (φM2 )1/2 T
D=
ηV10.6
(46)
where D is the diffusion coefficient of the solute 1 in solvent 2, in cm2 /s ; M2
is the molar weight of the solvent in Daltons; T is the temperature; η is the
viscosity of the solvent in centiPoise; V1 is the molar volume of the solute
in cm3 /mol. The empirical constant φ is 1 for most organic solvents, 1.5 for
alcohols, and 2.6 for water.
5
Diffusive transport of heat and momentum
Diffusive phenomena are not limited to the transport of matter. Imagine the
case of colliding molecules in a gas, and consider what happens to a molecule
that is much faster than all the others. When that molecules collides with a
slower molecule, the former will necessarily transfer some of its energy to a
degree of freedom of the latter. Say, it starts spinning rapidly and the other
slows down. Because the subsequent random trajectories of these molecules
are uncorrelated, this is a process by which energy is transported. It should
be no surprise to you that this process is described by the equations that are
mathematically identical to Fick’s equations. The equation that governs the
transport of heat bears the name of Joseph Fourier; it is written
JH = −k∇T
(47)
where JH is the heat flux and k is the heat conductivity. In a resting medium
with a constant k, the conversation of energy takes the following form
cV ∂t T = k∆T
19
(48)
Figure 12: Joseph Fourier (1768-1830) is mostly famous for the Fourier
series, by which any function is written as an infinite sum of periodic components. It is not so well-known that Fourier developed this mathematical
technique to solve the heat equation that also bears his name. In addition
to his achievements as a scientist, Fourier had also an illustrious career as a
civil servant. He was notably governor of Lower Egypt during Bonaparte’s
1798 military expedition.
where cV is the heat capacity. The analogy with Fick’s second law is complete
by writing this as
∂t T = α∆T
(49)
where α = k/cV is the heat diffusivity. 5
Diffusive transport also applies to the transport of momentum, and this
is the nature of viscosity. In the case of a pure sheared flow, the viscosity is
defined by Newton’s law
σxy = µ∂y vx
(50)
which relates the mechanical shear stress σxy to the sheer strain ∂y vx by a
proportionality relation. The proportionality constant is the dynamic viscosity µ. The fluids that obey Eq. 50 are referred to as being Newtonian. The
conservation of momentum in Newtonian fluids with constant viscosity leads
to the celebrated Navier-Stokes equations
1
∂t v + (v · ∇) v = − ∇P + ν∆v
ρ
5
What is the dimension of the heat diffusivity α ?
20
(51)
where ρ is the fluid density and ν = µ/ρ is the kinematic viscosity.
Although the Navier-Stokes equation may seem more complicated that
Fick’s second law, it has to be stressed that their physical basis is the same.
Only, in the case of momentum transport it would make no sense to assume
that the medium is at rest. Therefore a convective transport term (v · ∇) v
had to be included in the equation.
The similarity between the Navier-Stokes, Fick’s and Fourier’s equation
is made clearer by considering the more general case of mass and heat conservation in a flowing medium. In this case, Fick’s second law becomes
∂t c + (v · ∇) c = D∆c
(52)
∂t T + (v · ∇) T = α∆c
(53)
and Fourier’s equation is
We leave it to a next chapter to study the role of the convective term.
21
6
6.1
Problems
Brownian motion
The equation
R2 (t) = 6Dt
(54)
was derived by Einstein in his famous 1905 paper on Brownian motion. In
this equation, R2 (t) is average value of the squared distance of a diffusing
molecule from its starting point. This equation holds for diffusion in a threedimensional space. Use the same methods as in Sec. 2 to derive this equation.
Hint: Decompose the 3D random walk into three independent one-dimensional
random walks.
6.2
The size of a polymer in solution
In a neutral solvent (a so-called θ-solvent) the conformation of a linear polymer made up of N monomers can be modeled as a random walk of length
N . Consider the following model
x(n + 1) = x(n) + lx (n)
y(n + 1) = y(n) + ly (n)
z(n + 1) = z(n) + lz (n)
(55)
where x(n), y(n) and z(n) are the Cartesian coordinates of subunit n of the
polymer, l is the length of the bonds, and x,y,z are independent random
numbers that take value +1 or −1 with equal probability. Calculate the
average squared distance between the two ends of the polymer chain as a
function of N .
6.3
Dimensional analysis of gas-phase diffusion
Use dimensional analysis to find the general expression for the diffusion coefficient in a gas as a function of temperature T , molecular mass M , pressure
P , and a molecular size d. Verify that Eq. 31 has the right functional form.
Hint: remember that the only physical meaning of the temperature T is
via the thermal energy kB T .
22
8
7
6
5
4
3
2
1
0
15
5
10
5
0
0
−5
−5
Figure 13: The conformation of a linear polymer in a θ-solvent can be
modeled as a random walk. The two ends of the chain are shown with
circles.
6.4
Diffusion of O2 in air
Use the Chapman-Enskog relation to estimate the diffusion coefficient of O2
in air at 25 ◦ C and 1 atm.
6.5
Diffusion of O2 in water
Estimate the diffusion coefficient of O2 is water at 25 ◦ C using Stokes-Einstein
relation and the Wilke-Chang correlation. Compare the results with the
experimental value D = 1.8 10−5 cm2 /s.
6.6
So-called turbulent diffusion
23