40

5 Lecture 5: Simple kinetic theory of transport

phenomena

Summary∗ Transport phenomena are outlined. Fluxes are proportional to (−)gradients of the

density fields.∗ Transport coefficients can be estimated with the aid of elementary kinetic theory.∗ Maxwell used shear viscosity and the van der Waals equation of state to estimate

Avogadro’s constant and the molecular size for the first time.∗ If a mesoscopic object is placed in a gas, due to molecular bombardment, it jiggles

and wanders around = Brownian motion. Its displacement distance ∆ duringtime t is roughly ∆ ∝

√t.

∗ Langevin explained this behavior, starting from the Newton’s equation of motionwith a stochastic driving force.

∗ The trajectory of a Brownian particle after coarse-grained may be understood asa random walk, which is closely relate to the polymer conformation.

Key wordsTransport phenomena, flux, gradient, transport coefficient, shear viscosity, diffusion,Brownian motion, Langevin equation, random walk, random polymer

What you should be able to do∗ Understand how to handle the averages of vector components.∗ You should be able to follow Langevin’s logic and calculation.∗ Be able to demonstrate that a random walker after n unit steps is about

√n from

her starting point.

The derivation of the mean free path length

` =1√

2πnd2(5.1)

may not have been convincing. ` must be the mean traveling distance in t/averagenumber of collisions in t. The mean traveling distance in t may be represented by theroot-mean square velocity v times t. The collision occurs with the relative velocity

5. LECTURE 5: SIMPLE KINETIC THEORY OF TRANSPORT PHENOMENA41

w. Therefore, z = wπd2nt must be the total number of collisions (we follow the ideaof the sweep volume again). Hence,

` =vt

wπd2nt=

1√2πd2n

. (5.2)

This was the formula we guessed.

Maxwell used the idea of mean free path to compute the shear viscosity. Beforegoing to this calculation we discuss the general transport phenomena.

If a system is not far away from equilibrium and if there is a spatial nonuniformityin some physical quantity X, we can expect a flow of that physical quantity to reducethe nonuniformity. Thus, X must be transported from one point to another. This isgenerally called the transport phenomena.

grad X

X(r)

JX

Figure 5.1: Gentle nonuniformity causes transport phenomena. grad X points the direction ofincreasing X, so the flux driven by the gradient points in the −grad X direction.

Since we are interested in the spatially non-uniform systems, we describe the dis-tribution of X in the system as a field. That is, we define the density X(r) of thisquantity at around r and the distribution of X in the system is described by {X(r)}.The amount of flow of quantity X must be driven by the gradient of its density, soit is sensible to assume that the flux JX of X is proportional to grad X(r). Here,a flux is a vector pointing the direction of the flow, whose magnitude is the amountof the quantity going through the unit cross section per unit time (see Fig. 5.2). Wemay write J to be the product of the density of X and the speed of the flow.

J

A

X

Figure 5.2: The flux vector JX for the quantity X: its direction is the transport direction, and itsmagnitude is the flow rate: the quantity of X through the area A perpendicular to JX (convertedto the amount per unit area) per unit time.

JX = −L grad X(r), (5.3)

42

where L is a positive constant called the transport coefficient.If X is conserved, then

∂X(r)∂t

= −div JX(r), (5.4)

which reads a parabolic partial differential equation called diffusion equation:

∂X(r)∂t

= L∆X(r), (5.5)

where ∆ is the Laplacian.If you need a review of vector analysis, go to, e.g.,

http://www.yoono.org/ApplicableMath/ApplicableMath_files/AMI-2.pdf

Section 2.C.

Let X be a physical quantity transferred by molecules. Its density may be ex-pressed as

X(r) =

∑ri∈dτ(r) xi

dτ(r), (5.6)

where xi is the quantity carried by the ith molecule whose spatial location is ri.Here, the volume element dτ is very small from the macroscopic point of view, so itis actually huge from the molecular point of view. The law of large numbers tells usthat X(r) is not appreciably fluctuating, so we identify it with the density of X at(around) r.

l

v

r

−r

l

Figure 5.3: If a particle moves with velocity {v alon ght free path l, on the average the quantityaround r − l displaces that around r.

Suppose a particle moves along a free path l with a velocity v to r. A flux isthe density of the quantity carried by the flow × the mean velocity. The moleculecarry on the average X(r − l) to r and displace X(r). Therefore, X(r − l)−X(r)is carried by the velocity v. Thus, we may write

JX(r) = 〈[X(r − `)−X(r)]v〉 (5.7)

where the average is over all the cases of l and v (but they are parallel). Since |l| istiny, we may expand X(r − `) as

X(r − l) = X(r)− l · grad X + · · · . (5.8)

5. LECTURE 5: SIMPLE KINETIC THEORY OF TRANSPORT PHENOMENA43

or in terms of components

X(r − l) = X(r)−∑

i

li∂X

∂ri

+ · · · . (5.9)

Thus, (5.7) can be rewritten as

JX(r) = −〈v[l · grad X(r)]〉. (5.10)

To compute this average, consider for a vector A

〈v[l ·A]〉 = 〈v∑

i

liAi〉 (5.11)

Note that v and l are parallel, and that the components of v (therefor those of l aswell) are statistically independent, so we can approximately estimate

〈vi`j〉 '1

3v`δij, (5.12)

where v is the average speed of the particle, and ` is the mean free path. Thus,generally, we obtain

JX(r) = −1

3v` gradX(r). (5.13)

Suppose we have a shear flow with the velocity V in the x-direction and thevelocity gradient in the z-direction as shown in Fig. 5.4. To understand the decay ofthis velocity gradient we study the transport of the x-component of the momentum.Due to exchange of particles between positions with different z-coordinates, largerVx (or larger momentum density) and smaller Vx layers mix and the gradient in thez direction diminishes. This is the effect of shear viscosity.

V r( )l

z = 0

z

Figure 5.4: Shear flow

44

To apply the general formula (5.13) to shear flow, we must identify what X is. It isalready clear that the x-component of the momentum density

X =∑dτ

mvx/dτ, (5.14)

soX(r) = nmV (r) (5.15)

is the right density to study. Therefore, (5.13) reads

J = −1

3vlnm

∂Vx

∂z, (5.16)

where J is the z-component of the ‘x-component momentum flux’.43 Shear viscosityη is defined by

J = −η∂Vx/∂z, (5.17)

Comparing this with (5.16), we get the shear viscosity η:

η =1

3mnvl. (5.18)

With the already obtained estimate of ` (4.31) and v =√

8kBT/πm, we ob-tain44

η =2

3d2

√mkBT

π3. (5.19)

This is independent of the density n as noted by Maxwell. We generally expect thatthe viscosity increases with density, but in gases, higher densities imply shorter freepaths or a shorter mixing distance (actually the mean free path length is ∝ 1/n)and the expected density effect is cancelled. Also notice that the viscosity increaseswith temperature. Although this is contrary to the liquid behavior, it is easy tounderstand because higher temperatures imply better mixing.

To obtain the heat conductivity λ, X must be the energy contained in the volumeelement: X =

∑dτ mv2/2/dτ , so X(r) = 3nkBT (r)/2. Since J = −λ∇T defines

the heat conductivity λ, we obtain

λ =1

2nkB`v. (5.20)

43In a more advanced course, we use a tensor.44If we assume that the particle mass, the cross section (d2) and the particle thermal velocity

are only relevant quantities, dimensional analysis gives essentially this result. Even if we try to takethe density of the gas into account, it automatically drops out of the formula. This independencewas a bit of surprise. It is a good occasion to learn rudiments of dimensional analysis.

5. LECTURE 5: SIMPLE KINETIC THEORY OF TRANSPORT PHENOMENA45

For self-diffusion (i.e., isotope diffusion) X =∑

dτ 1/dτ , so X(r) = n(r). J =−D∇n implies

D =1

3v` =

`2

3τ, (5.21)

where τ is the mean free time τ = `/v.Notice that η = nmD, η/λ = 2m/3kB and λ/D = 3nkB/2. The last relation

tells us λ = cV D, where cV is the specific heat per molecule of gas under constantvolume. Again, we should note that these relations do not tell us anything about themicroscopic properties of the gas particles.

The following YouTube video about diffusion contains some interesting episodes:http://www.youtube.com/watch?v=H7QsDs8ZRMI

Most of the demos in this video is under strong influence of gravity, so you must becritical.

http://lessons.harveyproject.org/development/general/diffusion/diffnomemb/diffnomemb.html

The following simulation gives a nice bridge between diffusion and Brownian motions:http://www.chm.davidson.edu/vce/kineticmoleculartheory/diffusion.html

To establish the reality of atoms, we wish to determine the number of particlesN and their sizes d. Even if you could determine the mean-free path length we candetermine only the combination Nd2.

In 187345 van der Waals (1837-1923) proposed his equation of state of imperfectgases:

P (V − V0) = NkBT − α

V(1− V0/V ). (5.22)

V V0

free v

olum

e

Figure 5.5: The idea of van der Waals.

His basic idea is as follows (see Fig. 5.5): Since molecules are not points but havevolumes, they cannot run everywhere they wish (at least they must avoid each other).However, if we collect all the volumes of the molecules at a corner of the container,then, the centers of mass of the molecules can freely move around in the ‘free volume.’Therefore, if we ignore the attractive interactions, the ‘hard-core’ gas must look like

45Maxwell’s A Treatise on Electricity and Magnetism was published this year.

46

an ideal gas with a reduced volume:

P (V − V0) = NkBT. (5.23)

The remaining part of the van der Waals equation is to take care of the attractiveintermolecular forces. Thus, from V0 ' bNπd3/6, where b is a geometrical constantof order unity, we can estimate the size of the molecules. Now, we know Nd2 andNd3, so we can estimate N and d. The method gives an estimate of Avogadro’sconstant NA ' (4 ∼ 6)× 1023.46

Since molecules are incessantly moving around vigorously, small objects shouldnot be able to sit still in fluids even in equilibrium. Indeed, we know Brownianmotions. Let us watch some examples:

Nanoparticles in water:http://www.youtube.com/watch?v=cDcprgWiQEY&feature=topics

Simulationshttp://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=24.msg158#msg158

http://www.youtube.com/watch?v=PtYP8uoN0lk&feature=topics

The Brownian motion was discovered in the summer of 1827.47,48 We are usuallygiven an impression that he simply observed the “Brownian motion.” However, hedid a very careful and thorough research to establish the universal nature of themotion.

Since the particles came from living cells, initially he thought that it was a vi-tal phenomenon. Removing the effect of advection, evaporation, etc., carefully, hetested many flowers. Then, he tested old pollens in the British Museum (he wasthe (founding) director of the Botanical Division), and still found active particles.He conjectured that this was an organic effect, testing even coal with no exceptionfound. This suggested him that not only vital but organic nature of the specimens

46〈〈Definition of Avogadro’s constant〉〉 This is defined as the number of atoms in a 0.012kg of 12C. The latest Avogadro constant value is due to B. Andreas et al., “Determination of theAvogadro constant by counting the atoms in a 28Si crystal,” Phys. Rev. Lett., 106, 030801 (2011).Cf. P. Beker, “History and progress in the accurate determination of the Avogadro constant,” RepProg Phys 64 1945 (2001).

47Beethoven died in March; Democratic party was founded.48The work was published the next year, but was communicated to Stokes very soon in August,

1827. See P Pearle, B Collett, K Bart, D Bilderback, D Newman, and S Samuels, “What Brownsaw and you can too,” Am. J. Phys. 78, 1278 (2010).

5. LECTURE 5: SIMPLE KINETIC THEORY OF TRANSPORT PHENOMENA47

were irrelevant. He then tested numerous inorganic specimens (including a piece ofSphinx; he also roasted his specimens).

Thus, he really deserves the name of the motion. As to Mr. Brown himself, see1.4.4.

Curiously enough, there was no work published on the Brownian motion between1831 and 1857, but the phenomenon was well known. From 1850s new experimentalstudies began by Gouy (1854-1926) and others. The established facts included (youwould find them very easy to understand in terms of molecular bombardment onmesoscopic particles):(1) Its trajectory is quite erratic without any tangent lines anywhere.(2) Two Brownian particles are statistically independent even when they come withintheir diameters.(3) Smaller particles move more vigorously.(4) The higher the temperature, the more vigorous the Brownian motion.(5) The smaller the viscosity of the fluid medium, the more vigorous the motion.(6) The motion never dies out.etc.

In the 1860s there were experimentalists who clearly recognized that the motionwas due to the impact of water molecules. Even Poincare (1854-1912) mentioned thismotion in 1900, but somehow no founding fathers of kinetic theory and statisticalmechanics paid any attention to Brownian motion.

Due to the bombardment of water molecules, the Brownian particle executes azigzag motion, and on the average it is displaced as seen in Fig. 5.6 (ref 1 =https://commons.wikimedia.org/wiki/File:500_frames_of_brownian_motion_integrated.ogv?uselang=de).

This site exhibits how the (Gaussian) distribution is accumulated.

As you see the salient feature of the Brownian displacement ∆x is

〈∆x2〉 ∝ t, (5.24)

where 〈 〉 is the ensemble average (you repeat the experiment again and again ordo many experiments simultaneously, and average the results) t is time, and theproportionality constant is related to (proportional to) the diffusion constant.

48

Figure 5.6: Displacement of Brownian particles [From ref1]

Figure 5.7: The left are four sample paths and their average is on the right. [Courtesy of Prof.Nishizaka of Gakushuin Univ.]

Closely following Paul Langevin’s argument49, let us demonstrate indeed 〈∆x2〉 ∝t. Let us try to describe the motion of a Brownian particle classical mechanically.Let x be its position vector, m its mass. Newton’s equation of motion requires theforces acting on the particle. Since the particle is being hit ‘randomly,’ we expect arandom force X (whose direction and magnitude change incessantly and erratically)acting upon the particle. If the Brownian particle moves at a constant velocity v,then it would be hit by more particles of the medium on its front than on its back(imagine running in the rain). Therefore, it is natural to expect a force opposingthe motion whose magnitude is proportional to the speed. Therefore, the equation

49Paul Langevin, “Sur la theorie du mouvement brownien,” C. R. Acad. Sci. Paris 146, 530-533(1908). A translation may be found in D. S. Lemons and A. Gythiel, “Paul Langevin’s 1908 paper“On the Theory of Brownian Motion” [“Sur la theorie du mouvement brownien,” C. R. Acad. Sci.(Paris) 146, 530-533 (1908)],” Am. J. Phys., 65, 1079 (1997).

5. LECTURE 5: SIMPLE KINETIC THEORY OF TRANSPORT PHENOMENA49

of motion reads

md2x

dt2= −ζ

dx

dt+ X. (5.25)

Let us try to make an equation for x2 by scalar-multiplying x to this equation.Since

x · d2x

dt2=

d

dt

(x

dx

dt

)−

(dx

dt

)2

=d

dt

(1

2

dx2

dt

)−

(dx

dt

)2

(5.26)

Therefore, we have

m

2

d2x2

dt2−m

(dx

dt

)2

= −ζ

2

dx2

dt+ X · x. (5.27)

Let us ‘ensemble-average’ this equation; that is, we prepare many such Brownianparticles and average the equations for them. Let us denote this averaging procedureby 〈 〉. Since averaging procedure is linear and time-independent, we can exchangethe order of differentiation and averaging, we have

m

2

d2〈x2〉dt2

−m

⟨(dx

dt

)2⟩

= −ζ

2

d〈x2〉dt

+ 〈X · x〉. (5.28)

Langevin says, “The average value of the term X ·x is evidently null by reason of theirregularity of the complementary forces X.” Thanks to the equipartition of kineticenergy in equilibrium

1

2m

⟨(dx

dt

)2⟩

=3

2kBT, (5.29)

where kB is the Boltzmann constant and T is the absolute temperature.If we introduce

z =d〈x2〉

dt, (5.30)

(5.28) readsm

2

dz

dt+

ζ

2z = 3kBT. (5.31)

This implies after a sufficiently long time50, z = 6kBT/ζ, or

〈x2〉 =6kBT

ζt. (5.32)

50which is actually a very short relaxation time: τ ' m/ζ.

50

That is, the absolute value of the displacement during time t is proportional to√

t.See Fig. 5.7.

As we have seen, due to random bombardment by fluid particles a Brownianparticle executes an erratic motion. Let ∆xi be the i-th total displacement betweentime (i− 1)τ and iτ .

Figure 5.8: Actual observation results of a latex particle trajectory for 3.3 sec. Left: every1/8000 sec; Right: every 1/30 sec. [Courtesy of Prof Nishizaka of Gakushuin U]

Then, we may model the movement of the particle by a random walk. After n steps(after nτ):

x(t) = ∆x1 + ∆x2 + · · ·+ ∆xn, (5.33)

where t = nτ .Let us compute the mean square displacement:

〈x2〉 =∑

i

〈∆x2i 〉+ 2

∑i<j

〈∆xi ·∆xj〉 (5.34)

Since the movement of the Brownian motion is uniform (e.g., the initial and thefinal stages of the motions are the same on the average), so we may expect 〈∆x2

1〉 =〈∆x2

2〉 = · · ·. Since there is no systematic direction to move into, 〈∆xi〉 = 0. Since∆xi are totally random (statistically independent), we expect that the average 〈∆xi ·∆xj〉 = 0 for i 6= j. Therefore, (5.34) implies

〈x2〉 = n〈∆x2i 〉 ∝ t (5.35)

5. LECTURE 5: SIMPLE KINETIC THEORY OF TRANSPORT PHENOMENA51

This is consistent with (5.32).

We can consider a random walk on a lattice. Let `i be the ith step of the walk.Starting from the origin, a random walk of n steps on a lattice may be definedas

R(n) = `1 + `2 + · · ·+ `n, (5.36)

where `i are chosen from the bond vectors, and R(n) is the position of the walk aftern steps. If the lattice spacing is unity, then the consideration in ?? tells us

〈R(n)2〉 = n. (5.37)

We may interpret the trajectory of a random walk as a conformation of a polymerconsisting of n monomers (without any steric interactions among monomer units ex-cept perhaps for bond angle constraints). Then, R(n) is the end-to-end vector of thepolymer chain, and the root-mean square end-to-end distance satisfies (5.37).