2. Molecules in Motion

Kinetic Theory of Gases (microscopic viewpoint)

assumptions

(1) particles of mass m and diameter d ; ceaselessrandom motion

(2) dilute gas: d ¿ λ, λ = mean free path = averagedistance a particle travels between collisions

(3) no interactions between particles, except perfect-ly elastic collisions; particles are hard spheres (EK =E ′

K, ′= after collision)

(3) =⇒ EP ≡ 0 =⇒ E = EK1+EK2+·· ·EKN =∑Ni=1 EKi

pressure: particles bump into the walls of the contain-er =⇒ change of momentum =⇒ force

P = 1

3m

N

Vc2 (1)

where N is the number of particles and c is the root-

Adv PChem 2.1

mean-square velocity (r.m.s. speed)

velocity ~v = (vx, vy , vz)

v2 = v2x + v2

y + v2z

speed v = |~v | =√

v2x + v2

y + v2z

c = ⟨v2⟩1/2 =

(v2

1 + v22 +·· ·+ v2

N

N

)1/2

⟨EK

⟩= ⟨1

2mv2⟩= 1

2m

⟨v2⟩= 1

2mc2

(average kinetic energy of a particle)

c =(

2⟨

EK

⟩m

)1/2

Equation (1):

P = 1

3m

N

Vc2 = 1

3m

nNA

Vc2 = 1

3

nM

Vc2

Adv PChem 2.2

(M = molar mass of the gas) or

PV = 1

3nM c2 (2)

If the kinetic model correctly describes ideal gasesthen the ideal gas equation

PV = nRT

and (2) must be identical

PV = 1

3nM c2 != nRT

=⇒

c =√

3RT

M

must use SI units here!: m = mass of a particle in kg;M = molar mass of the gas in kg/mol; R = 8.314 J K−1

mol−1; c in m s−1

Adv PChem 2.3

example: r.m.s. speed of a nitrogen molecule, N2, atroom temperature, 25◦C, is 515 m s−1

c ∝p

T

the total energy E (internal energy U ) of an ideal gas:

U = E = N⟨

EK

⟩= 1

2N mc2 = 1

2nM c2 = 3

2nRT

⟨EK

⟩= 3

2kT equipartition theorem⟨

EK

⟩∝ T

distribution of molecular speeds: Maxwell distribu-tion f (v)

f (v)dv = N(v,v+dv)/N= fraction of molecules with aspeed in the range (v, v +dv) = probability of findinga molecule with a speed in the range (v, v +dv)

f (v)dv = 4π

(M

2πRT

)3/2

v2exp

(−M v2

2RT

)dv

Adv PChem 2.4

[Figure: Spherical shell in velocity space; Atkins 9th ed., Fig. 20.5]

qualitative behavior of the Maxwell distribution ofspeeds:

(1) Since xn exp(−x2) → 0 for x → ∞, we have thatf (v) → 0 for v →∞

(2) Since exp(−x2) → 1 for x → 0, we have that f (v) ∼v2 → 0 for v → 0

(3) The rate of the decrease of the exponential func-tion is governed by the factor M /(2RT ). Thereforelarge M or small T lead to rapid decrease, whilesmall M or large T lead to slow decrease.

Adv PChem 2.5

[Figure: Distribution of molecular speeds; Atkins 9th ed., Fig. 20.3]

The probability of finding a molecule with a speed be-tween 0 and ∞ is one:∫ ∞

0f (v)dv = 1

normalization

Adv PChem 2.6

Probability

discrete variable: u; takes values {u1,u2,u3, . . . }

P (ui ) = probability that u = ui

normalization:∑N

i=1 P (ui ) = 1 (N can be ∞)

mean value (average value, expectation value):

m = ⟨u⟩ =N∑

i=1

ui P (ui )

n-th moment:

⟨un⟩= N∑

i=1

uni P (ui )

variance (= spread of the probability):

σ2 = ⟨(∆u)2⟩

= ⟨(u −⟨u⟩)2⟩

= ⟨u2−2u⟨u⟩+⟨u⟩2⟩

Adv PChem 2.7

= ⟨u2⟩−2⟨u⟩⟨u⟩+⟨u⟩2

= ⟨u2⟩−⟨u⟩2 ≥ 0

expectation value of a function f (u):

⟨f (u)

⟩= N∑i=1

f (ui )P (ui )

example: u with u ∈ {0,1,2,3, . . . }

P (u = i ) = mi

i !exp(−m) Poisson distribution

⟨u⟩ = m⟨(∆u)2⟩=σ2 = m⟨un⟩= n∑

k=0

S(n,k)mk

S(n,k) Stirling numbers of the second kind

S(n,k) = 1

k !

k∑j=0

(−1)k− j

(k

j

)j n

Adv PChem 2.8

continuous variable x; takes values in [a,b], where acan be −∞ and b can be +∞P (x) probability density; P (x)dx = probability that xtakes a value in the interval (x, x +dx)

normalization:

∫ b

aP (x)dx = 1

mean value (average value, expectation value):

⟨x⟩ =∫ b

axP (x)dx

n-th moment:

⟨xn⟩= ∫ b

axnP (x)dx

variance:

σ2 = ⟨(∆x)2⟩= ⟨

x2⟩−⟨x⟩2

Adv PChem 2.9

⟨f (x)

⟩= ∫ b

af (x)P (x)dx

example x ∈ (−∞,+∞)

P (x) = 1√2πσ2

exp

(−(x −m)2

2σ2

)Gaussian distribution

⟨x⟩ = m⟨(∆x)2⟩=σ2

⟨(∆x)n⟩={

0 n ≥ 1, odd

(n −1)!!σn n ≥ 2, even

(n −1)!! = 1 ·3 ·5 · · · · · (n −1)

If one knows the probability or probability density,then all moments of the random variable can be de-termined (some may be infinite)

the converse is not true: the probability or probabilitydensity is in general not uniquely determined by itsmoments

Adv PChem 2.10

the r.m.s speed represents a typical speed of the mol-ecules in the gas

a second way to define a typical speed is the averagespeed:

c =⟨v⟩= ∫ ∞

0v f (v)dv

=(

8RT

πM

)1/2

=√

8

3πc ≈ 0.921c

R

M= NAk

NAm= k

m

c =(

8kT

πm

)1/2

a third way is the most probable speed, location ofthe peak of the Maxwell distribution:

f ′(c∗) = 0

c∗ =(

2RT

M

)1/2

c∗ =√

2

3c ≈ 0.816c

Adv PChem 2.11

c∗ < c < c

[Figure: Characteristic speeds; Atkins 9th ed., Fig. 20.6]

collision frequency

determine how often a specific particle collides withother particles in the gas

since this is a representative particle, we can assumethat it moves with the average speed c

we can replace the moving collision partners by sta-tionary particles, if we replace the average speed c

Adv PChem 2.12

by the relative average speed crel:

c =(

8kT

πm

)1/2

, crel =(

8kT

πµ

)1/2

1

µ≡ 1

mA+ 1

mB, µ = reduced mass

for identical particles

1

µ= 2

m=⇒ µ= m

2

crel =p

2c

[Figure: Collosion tube; Atkins 9th ed., Fig. 20.8]

number of stationary particles inside the collisiontube: N σ0 · crel∆t , N = N /V number density, σ0 =

Adv PChem 2.13

πd 2 = elastic collision cross-section, for collisions be-tween identical particles, d = 2r

collision frequency z = average number of collisionsof a particle per unit time

z = N σ0 · crel∆t

∆t=N σ0crel

z =p2N σ0c

z =p2

N

Vσ0c

z =p2

P

kTσ0c; (PV = nRT = N

NART = N kT )

mean free path λ = the average distance a moleculetravels between two successive collisions

λ= c · 1

z= cp

2N σ0c= 1p

2(N /V )σ0

= kTp2Pσ0

collisions with walls and surfaces

ZW = Pp2πmkT

Adv PChem 2.14

Transport

transfer of matter, energy, charge, momentum, etc.,from one place to another

flux ~J = (Jx, Jy , Jz)

Ji =amount of property in direction i

area ⊥ to direction i · time interval

flux of matter

x

zy

transport: driven by spatial gradients: experiments=⇒

Jx =−const · d property

dx

i.e., flux ∝ gradient, if gradient is not too large: linearnonequilibrium thermodynamics

Adv PChem 2.15

diffusion

matter: transport process = diffusion, gradient = con-centration

Jx(x) =−DdN (x)

dxFick’s first law

D diffusion coefficient, [D] = m2

s

for liquids, customary units: [D] = cm2

s

for small molecules in water at 25◦C: D ∼ 10−5 cm2 s−1

[Figure: Matter flux; Atkins 9th ed., Fig. 20.9]Adv PChem 2.16

energy transport: rate of thermal conduction

Jx(x) =−κdT (x)

dxFourier’s law

κ coefficient of thermal conductivity, [κ] = J

K m s

momentum transport: viscosity

Newton: “Viscosity is a lack of slipperiness betweenadjacent layers of fluid.”

velocity gradient =⇒ transfer of momentum~p = (px, py , pz) = m~v = (mvx,mvy ,mvz)

Adv PChem 2.17

[Figure: Transfer of momentum in laminar flow; Atkins 9th ed., Fig. 20.10]

Jz,px(z) =−ηdvx(z)

dz

η viscosity (coefficient), [η] = kg

ms

1 Poise = 1 P = 1 g cm−1 s−1

if η is independent of v , then the fluid is a Newtonianfluid

kinetic theory: transport parameters

D = 1

3λc

κ= 1

3λcC V [A]

η= 1

3λcM [A]

λ↘ as P ↗ =⇒ D ↘ as P ↗c ↗ as T ↗ =⇒ D ↗ as T ↗

Adv PChem 2.18

λ↗ as σ0 ↘ =⇒ D ↗ as σ0 ↘

λ∝ P−1 and P ∝ [A] =⇒ κ is independent of P

λ∝ P−1 and P ∝ [A] =⇒ η is independent of P

c ∝pT =⇒ η∝p

T ; viscosity of a gas increases withtemperature

as T ↗, molecules travel faster =⇒ greater flux ofmomentum

liquids: η↘ as T ↗ (typically)

liquids have attractive intermolecular forces, and η

decreases with increasing T

T / ◦C η/ P

water 0 0.0178920 0.0100540 0.00653

100 0.00282blood 18 0.0475glycerin 20 14.99air 0 0.000171

Adv PChem 2.19

How does the concentration change due to the diffu-sive flux?

non-stationary system: non-equilibrium state

∂c(x, t )

∂t=−∂Jx(x, t )

∂xcontinuity equation

The continuity equation holds for any type of flux.

In three dimensions:

∂c(x, y, z, t )

∂t=−∂Jx(x, y, z, t )

∂x−∂Jy(x, y, z, t )

∂y−∂Jz(x, y, z, t )

∂z

Consider only one-dimensional systems in this class.

∂c(x, t )

∂t=−∂Jx(x, t )

∂x

Need to close the equation; need a constitutiveequation:

Fick’s first law: Jx =−Dx∂c

∂x

Adv PChem 2.20

Drop subscript x, since we consider only one-dimensional systems:

∂c(x, t )

∂t=−∂J (x, t )

∂x, J =−D

∂c

∂x

Combine the two equations:

∂c(x, t )

∂t= ∂

∂xD∂c(x, t )

∂x

If D is independent of c and x, then

∂c(x, t )

∂t= D

∂2c(x, t )

∂x2 Fick’s second law

diffusion equation

point source: at t = 0 all the solute particles are lo-cated in a small region of width ∆x around x = 0: con-centration c0

c(x, t ) = c0∆xp4πDt

exp

(− x2

4Dt

), t > 0

Adv PChem 2.21

solution of the diffusion equation with D = 10−5 cm2/sand c0∆x = 1 mol

-0.02 -0.01 0.01 0.02

20

40

60

80

Figure 1: t = 1 s, x in cm

-0.2 -0.1 0.1 0.2

2

4

6

8

Figure 2: t = 100 s, x in cm

Adv PChem 2.22

-1 -0.5 0.5 1

0.2

0.4

0.6

0.8

1

1.2

1.4

Figure 3: t = 3600 s, x in cm

N (x, t ) = number of particles in (x, x +dx) = c(x, t )dx(one-dimensional system)

total number of particles N = c0∆x

P (x, t )dx = N (x, t )

N= c(x, t )dx

c0∆x

= probability of finding a particle in (x, x +dx)

P (x, t ) = 1p4πDt

exp

(− x2

4Dt

)

Gaussian distribution

Adv PChem 2.23

P (x, t ) obeys of course the diffusion equation

∂P (x, t )

∂t= D

∂2P (x, t )

∂x2

integral table:

∫ ∞

0exp(−q2x2)d x =

pπ

2q, q > 0

∫ ∞

−∞P (x, t )d x = 1p

4πDt

∫ ∞

−∞exp

(− x2

4Dt

)d x = 1

⟨x(t )⟩ =∫ ∞

−∞xP (x, t )d x = 0

integral table:

∫ ∞

0x2n exp(−px2)d x = (2n −1)!!

2(2p)n

√π

p, p > 0

(2n −1)!! = 1 ·3 ·5 · · ·2n −1

∆x = x −⟨x⟩⟨x(t )2⟩= ⟨

(∆x(t ))2⟩= ∫ ∞

−∞x2P (x, t )d x

= 2∫ ∞

0x2 1p

4πDtexp

(− x2

4Dt

)d x

Adv PChem 2.24

= 2Dt

diffusion:⟨

x(t )2⟩= 2D t in one dimension⟨x(t )2⟩= 4Dt in two dimensions⟨x(t )2⟩= 6Dt in three dimensions

if Dx = D y = Dz = D

one dimension: root mean square distance

d =√⟨

x(t )2⟩=p2Dt

convective flow: streaming fluid, carries particleswith it, fluid velocity ~v(x, y, z, t )

convective flux: Jx = cvx

Consider a one-dimensional system and vx(x, t ) = v ,i.e., constant velocity.

J = cv

∂c(x, t )

∂t=−∂J (x, t )

∂x

Adv PChem 2.25

∂c(x, t )

∂t=−v

∂c(x, t )

∂x

convection equation

Particles in a fluid flow are generally also subject todiffusive motion:

∂c(x, t )

∂t=−v

∂c(x, t )

∂x+D

∂2c(x, t )

∂x2

diffusion-convection equation

convection: particle velocity v =⇒ x(t ) = v · t , withx(0) = 0

diffusion:⟨

x(t )2⟩= 2Dt

convection: x(t ) ∝ t

diffusion: xrms(t ) = ⟨x(t )2⟩1/2 ∝p

t

“Mesoscopic” picture of diffusion

consider a one-dimensional lattice, lattice spacing ∆x

Adv PChem 2.26

solid; in fluids, ∆x ≈λ mean free path

Simple random walk: hopping transport, activatedtransport

particle hops a distance ∆x during the time ∆t (= τ

average collision time in fluids)

ii−1 i+1

q p

∆x

at each time step the particle jumps to the right withprobability p and to the left with probability q = 1−p;particle does not stay in place; the particle has nomemory, the jumps are statistically independent

consider unbiased random walk

p = q = 1

2

no systematic force acts on the particle

time t = n∆t , n = 0,1,2, . . . ; position x = i∆x, i =. . . ,−2,−1,0,+1,+2, . . . (i ∈Z)

Adv PChem 2.27

statistical formulation: large collection of indepen-dent particles

P (x, t ) = P (i ,n)

evolution equation

P (i ,n +1) = qP (i +1,n)+pP (i −1,n)

= 1

2P (i +1,n)+ 1

2P (i −1,n)

Solution: Assume the particle starts at x = 0, i.e., i = 0

n = nR +nL

nR = number of jumps to the right

nL = number of jumps to the left

x = i∆x = (nR −nL

)∆x

n = nR +nL, i = nR −nL

add the two equation

n + i = 2nR

Adv PChem 2.28

nR = 1

2(n + i )

nL = n −nR = 1

2(n − i )

Since the jumps are independent, the probability toobserve a walk with nR steps to the right and nL stepsto the left is

p ·p · · · · ·p︸ ︷︷ ︸nR

·q ·q · · · · ·q︸ ︷︷ ︸nL

= pnR qnL

1

2· 1

2· · · · · 1

2︸ ︷︷ ︸nR

· 1

2· 1

2· · · · · 1

2︸ ︷︷ ︸nL

=(

1

2

)nR(

1

2

)nL

=(

1

2

)n

number of different ways of taking n steps, such thatnR are to the right and nL are to the left:

Wn(nR) = n!

nR !nL!

Consequently

P (i ,n) = n!

nR !nL!

(1

2

)n

Adv PChem 2.29

= n![12(n + i )

]![

12(n − i )

]!

(1

2

)n

binomial distribution

(p +q)n =∞∑

k=0

n!

k !(n −k)!pk qn−k

⟨i ⟩n = mn =∞∑

i=−∞i P (i ,n) = 0 (by symmetry)⟨

(∆i )2⟩n =σ2

n = n

consider the continuum limit of the simple randomwalk:

t = n∆t , x = i∆x

∆t → 0, ∆x → 0

P (i ,n +1) = 1

2P (i +1,n)+ 1

2P (i −1,n)

P (i∆x, (n +1)∆t ) = 1

2P ((i +1)∆x,n∆t )+ 1

2P ((i −1)∆x,n∆t )

P (i∆x, (n +1)∆t )−P (i∆x,n∆t ) =1

2P ((i +1)∆x,n∆t )+ 1

2P ((i −1)∆x,n∆t )−P (i∆x,n∆t )

Adv PChem 2.30

P (i∆x, (n +1)∆t )−P (i∆x,n∆t ) =1

2[P ((i +1)∆x,n∆t )+P ((i −1)∆x,n∆t )−2P (i∆x,n∆t )]

P (i∆x, (n +1)∆t )−P (i∆x,n∆t )

∆t=

1

2∆t[P ((i +1)∆x,n∆t )+P ((i −1)∆x,n∆t )−2P (i∆x,n∆t )]

P (x, t +∆t )−P (x, t )

∆t=

1

2∆t[P (x +∆x, t )+P (x −∆x, t )−2P (x, t )]

∂P (x, t )

∂t= 1

2∆t

[P (x, t )+ ∂P

∂x(x, t )∆x + 1

2

∂2P

∂x2 (x, t )(∆x)2+·· ·

+P (x, t )+ ∂P

∂x(x, t )(−∆x)+ 1

2

∂2P

∂x2 (x, t )(−∆x)2+·· ·−2P (x, t )

]∂P (x, t )

∂t= (∆x)2

2∆t

∂2P

∂x2 (x, t )

(∆x)2

2∆t→ 02

0??

obtain a nontrivial result if ∆x → 0 and ∆t → 0 such

Adv PChem 2.31

that

(∆x)2

2∆t= const= D

∂P (x, t )

∂t= D

∂2P (x, t )

∂x2 diffusion equation

one can show that the binomial distribution approach-es the Gaussian distribution in the continuum limit

Note: the diffusion equation can be obtained eitherfrom thermodynamics

∂c(x, t )

∂t=−∂J (x, t )

∂xcontinuity equation

J (x, t ) =−D∂c(x, t )

∂xconstitutive equation; Fick’s 1st law

or from a mesoscopic description, simple randomwalk

transport equations need to have a solid foundation,either macroscopic (thermodyanics) or mesoscopic(microscopic)

Adv PChem 2.32

diffusion equation: unrealistic feature of infinitely fastpropagation

c(x, t ) = c0∆xp4πDt

exp

(− x2

4Dt

)

no matter how small t and how large x, the concen-tration c will be nonzero, though exponentially small

reason: lack of inertia of Brownian particles; their di-rection of motion in successive time intervals is un-correlated

consequences: (i) particles move with infinite veloc-ity; there is some probability, though exponential-ly small, that a particle will travel an infinite dis-tance from its current position in a small but nonzeroamount of time. (ii) motion of the particles is unpre-dictable even on the smallest time scales.

in most applications, infinitely fast propagation is nota concern, since the density is exponentially small,i.e., zero for all practical purposes

diffusion equation is adequate for aqueous solutions

Adv PChem 2.33

if infinitely fast propagation is a concern, there areseveral remedies:

(i) thermodynamics: replace Fick’s first law by theCattaneo equation

∂c(x, t )

∂t=−∂J (x, t )

∂xcontinuity equation

τ∂J (x, t )

∂t+ J (x, t ) =−D

∂c(x, t )

∂xconstitutive eq.= Cattaneo eq.

eliminating J we obtain the telegraph equation in-stead of the diffusion equation

τ∂2c(x, t )

∂t 2 + ∂c(x, t )

∂t= D

∂2c(x, t )

∂x2 telegraph equation

The solution of the telegraph equation has the prop-erty that

c(x, t ) = 0 for |x| >√

D

τt

and it converges to the solution of the diffusion equa-tion for τ→ 0

Adv PChem 2.34

(ii) mescoscopic description: replace the simple ran-dom walk by a persistent random walk, where a par-ticle takes steps of length ∆x and duration ∆t and theparticle continues in its previous direction with prob-ability α= 1−µ∆t and reverses direction with proba-bility β = 1−α = µ∆t ; continuum limit −→ telegraphequation

different mescoscopic description

consider a particle of mass m, to which a force ~F isapplied, moving in a fluid (examples: sedimentation,F = gravity; centrifuge, F = centrifugal force)

m~a = ~F + ~Ft

Newton’s second law; ~Ft force from huge number ofcollisions with fluid particles

one dimension:

mdv

dt= F +Ft

Adv PChem 2.35

decompose the random force Ft into its systematicpart, the friction force, and its fluctuating part

Ft =⟨Ft

⟩+ξt⟨Ft

⟩=− f v⟨ξt

⟩= 0

fluctuating force ξt varies very rapidly compared to v ,it is statistically independent of v , and has very shortmemory,

⟨ξtξt ′

⟩= 0 for t 6= t ′

mdv

dt= F − f v +ξt Langevin equation

md⟨v⟩

dt= F − f ⟨v⟩+⟨

ξt

⟩m

d⟨v⟩dt

= F − f ⟨v⟩+0

d⟨v⟩dt

=− f

m⟨v⟩+ F

m

Adv PChem 2.36

The solution of the general first order linear ODE

dy

dt=−h(t )y(t )+ g (t ), y(0) = y0

is given by

y(t ) = exp[−H(t )]

{∫ t

0g (s)exp[H(s)]ds + y0

}H(t ) =

∫ t

0h(s)ds

So with v(0) = 0

⟨v(t )⟩ = F

f

[1−exp

(− f t

m

)]t →∞ : ⟨v(t )⟩→ v∞ = F

fterminal velocity or drift velocity

relaxation time: τ= m

f: ⟨v(τ)⟩ = F

f

[1−e−1]

spherical protein with molar mass M = 6×105g/mol in

Adv PChem 2.37

water: τ≈ 10−11s; cell with radius r = 10−3cm: τ≈ 20µs

v∞ = 1

fF =µF, µ= mobility

for an ion in an electric field, F = zeE

v∞ =µzeE = uE

u = ionic mobility

Stokes law: frictional coefficient f for a spherical par-ticle of radius a in a fluid of viscosity η:

f = 6πηa

µ= 1

6πηa

u = ez

6πηa

“diffusive” motion: F = 0

mdv

dt=− f v +ξt

Adv PChem 2.38

x(t ) = x(0)+∫ t

0v(s)ds, assume x(0) = 0

⟨x(t )⟩ =∫ t

0⟨v(s)⟩ds

⟨v(s)⟩ = 0 for F = 0

⟨x(t )⟩ = 0⟨(∆x(t ))2⟩= ⟨

x(t )2⟩=?

md2x

dt 2 =− fdx

dt+ξt

mxx =− f xx +xξt

m

[d(xx)

dt− x2

]=− f xx +xξt

md(xx)

dt= mx2− f xx +xξt

m

⟨d(xx)

dt

⟩= ⟨

mx2⟩− f ⟨xx⟩+⟨xξt

⟩EK =

1

2mv2 = 1

2mx2

m

⟨d(xx)

dt

⟩= ⟨

2EK(t )⟩− f ⟨xx⟩+⟨

xξt

⟩Adv PChem 2.39

⟨xξt

⟩= ⟨x⟩⟨ξt

⟩, x(t ) and ξt independent

m

⟨d(xx)

dt

⟩= ⟨

2EK(t )⟩− f ⟨xx⟩+⟨x⟩⟨ξt

⟩⟨

EK(t )⟩= 1

2kT equipartition theorem

m

⟨d(xx)

dt

⟩= kT − f ⟨xx⟩+0

d⟨xx⟩dt

=− f

m⟨xx⟩+ kT

m

with ⟨xx⟩(0) = 0 since x(0) = 0

⟨xx⟩ = kT

f

[1−exp

(− f

mt

)]

xx = 1

2

dx2

dt

1

2

d⟨

x2⟩dt

= kT

f

[1−exp

(− f

mt

)]⟨

x2(t )⟩= ∫ t

0

2kT

f

[1−exp

(− f

mt ′

)]dt ′

⟨x2(t )

⟩= 2kT

f

{t − m

f

[1−exp

(− f

mt

)]}

Adv PChem 2.40

two cases:Case 1: t ¿ τ= m/ f

exp

(− f

mt

)= 1− f

mt + 1

2

(f

m

)2

t 2+·· ·

⟨x2(t )

⟩= 2kT

f

{t − m

f

[1−1+ f

mt − 1

2

(f

m

)2

t 2+·· ·]}

⟨x2(t )

⟩= 2kT

f

{1

2

f

mt 2+·· ·

}⟨

x2(t )⟩= kT

mt 2

x ∝ t (ballistic motion for short times)

v =√

kT

m, thermal velocity

Case 2: t À τ= m/ f

exp

(− f

mt

)u 0

⟨x2(t )

⟩= 2kT

ft

!= 2Dt (diffusive motion for long times)

Adv PChem 2.41

D = kT

fEinstein relation

example of a fluctuation-dissipation theorem

combine with Stokes law

D = kT

6πηa

Adv PChem 2.42