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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