Chapter 8 Statistical Mechanics of Non-Equilibrium Systems
Ye Liang
August 27, 2018
8.1
The orientational correlation function ⟨u
z
(0)u
z
(t)⟩ indicates the rotational correlation of the molecules. In
a gas phase, the molecules are allowed to rotate freely. In the gure 8.5 in the text, the peak at ∼ 1 psec
indicates the molecule rotates back to its original direction, or has gone through a 2π rotation in ∼ 1 psec.
However, in a liquid phase the rotation of a molecule can be limited, for it may frequently collide with its
neighbors to keep its original direction.
8.2
In a perfect T = 0 crystal, the orientation of the molecules are frozen. Thus the orientational correlation
function should be
1
3
forever. However, if the temperature is above zero, molecules still have a chance to
rotate and ip with certain frequency. In such a situation, the orientational correlation function can slightly
decreases with time, and the speed of decreasing is related to the temperature. A sketch is shown in gure
1.
Figure 1: Orientational correlation function for a liquid.
8.3
∆
¯
A(t) =
¯
A(t) − ⟨A⟩ (1)
1
¯
A(t) =
dr
N
dp
N
A(t; r
N
, p
N
)F (r
N
, p
N
)
=
dr
N
dp
N
A(t; r
N
, p
N
)⟨A⟩
−1
f(r
N
, p
N
)A(r
N
, p
N
) (2)
where indeed A(0; r
N
, p
N
) = A(r
N
, p
N
).
⟨A⟩∆
¯
A(t) =⟨A⟩
dr
N
dp
N
A(t; r
N
, p
N
)⟨A⟩
−1
f(r
N
, p
N
)A(r
N
, p
N
) − ⟨A⟩
2
=
dr
N
dp
N
[A(t; r
N
, p
N
)A(r
N
, p
N
) − ⟨A⟩
2
]f(r
N
, p
N
)
=⟨A(t)A(0)⟩ − ⟨A⟩
2
=C(t). (3)
8.4
The rate equations are
dc
A
dt
= −k
BA
c
A
(t) + k
AB
c
B
(t)
dc
B
dt
= k
BA
c
A
(t) − k
AB
c
B
(t)
(4)
and the initial condition is c
A
(0) + c
B
(0) = c. Thus c
A
(t) + c
B
(t) = c. The rate equation reduces to
dc
A
dt
= −(k
BA
+ k
AB
)c
A
(t) + k
AB
c. (5)
The solution to this non-homogeneous linear dierential equation is
c
A
(t) =
k
AB
k
BA
+ k
AB
c +
c
A
(0) −
k
AB
k
BA
+ k
AB
c
e
−(k
BA
+k
AB
)t
. (6)
Because
⟨c
B
⟩
⟨c
A
⟩
=
k
BA
k
AB
(7)
⟨c
A
⟩ =
k
AB
k
BA
+ k
AB
c (8)
Therefore the solution (6) is
c
A
(t) = ⟨c
A
⟩ + [c
A
(0) − ⟨c
A
⟩]e
−(k
BA
+k
AB
)t
(9)
Therefore
∆c
A
(t) ≡ c
A
(t) − ⟨c
A
⟩ = [c
A
(0) − ⟨c
A
⟩]e
−(k
BA
+k
AB
)t
= ∆c
A
(0)e
−(k
BA
+k
AB
)t
. (10)
Let τ
−1
rxn
= k
BA
+ k
AB
, and the equation above becomes
∆c
A
(t) = ∆c
A
(0)e
−t/τ
rxn
. (11)
8.5
Since
H
A
(z) =
1, z < q
∗
,
0, z > q
∗
.
(12)
⟨H
A
⟩ =
∞
−∞
H
A
(z)p(z)dz =
q
∗
−∞
p(z)dz =
⟨c
A
⟩
⟨c
A
⟩ + ⟨c
B
⟩
= x
A
. (13)
Obviously H
2
A
(z) = H
A
(z). Thus ⟨H
2
A
⟩ = ⟨H
A
⟩ = x
A
.
⟨(δH
A
)
2
⟩ = ⟨H
2
A
⟩ − ⟨H
A
⟩
2
= x
A
− x
2
A
= x
A
(1 − x
A
) = x
A
x
B
. (14)
2
8.6
The notations are a little confusing in the text. By denition
n
A
(t) = H
A
[q(t)]. (15)
⟨n
A
⟩ = ⟨H
A
⟩, ⟨n
2
A
⟩ = ⟨H
A
⟩, ⟨(δn
A
)
2
⟩ = ⟨(δH
A
)
2
⟩. (16)
exp(−t/τ
rxn
) =⟨δn
A
(0)δn
A
(t)⟩/⟨(δn
A
)
2
⟩
=(x
A
x
B
)
−1
(⟨n
A
(0)n
A
(t)⟩ − ⟨n
A
⟩
2
)
=(x
A
x
B
)
−1
(⟨n
A
(0)n
A
(t)⟩ − x
2
A
). (17)
The time derivative gives
τ
−1
rxn
exp(−t/τ
rxn
) = −(x
A
x
B
)
−1
⟨n
A
(0) ˙n
A
(t)⟩ (18)
Because
⟨n
A
(0)n
A
(t)⟩ = ⟨n
A
(−t)n
A
(0)⟩ (19)
⟨n
A
(0) ˙n
A
(t)⟩ =
d
dt
⟨n
A
(0)n
A
(t)⟩ =
d
dt
⟨n
A
(−t)n
A
(0)⟩ = −⟨ ˙n
A
(−t)n
A
(0)⟩ = −⟨ ˙n
A
(0)n
A
(t)⟩. (20)
Thus
τ
−1
rxn
exp(−t/τ
rxn
) = (x
A
x
B
)
−1
⟨ ˙n
A
(0)n
A
(t)⟩ (21)
8.7
We need to understand the initial rate in a way of limit,
k
BA
(0) = lim
t→0
+
k
BA
(t) = lim
t→0
+
x
−1
A
⟨v(0)δ[q(0) − q
∗
]H
B
[q(t)]⟩. (22)
In the limit t → 0
+
, q(t) = q(0) + v(0)t. Notice that H
B
[q(t)] = H[q(t) − q
∗
], where H(x) is the Heaviside
function.
k
BA
(0) = lim
t→0
+
x
−1
A
⟨v(0)δ[q(0) − q
∗
]H[v(0)t + q(0) − q
∗
]⟩
= lim
t
→
0
+
x
−1
A
⟨v(0)δ[q(0) − q
∗
]H[v(0)t]⟩
=x
−1
A
⟨v(0)δ[q(0) − q
∗
]H[v(0)]⟩ (23)
Notice that the initial velocity and the initial coordinates are not correlated.
k
BA
(0) = x
−1
A
⟨v(0)H[v(0)]⟩⟨δ(q
0
− q
∗
)⟩. (24)
Since the distribution of v(0) is even, ⟨v(0)H[v(0)]⟩ =
1
2
⟨|v(0)|⟩ =
1
2
⟨|v|⟩. Thus
k
BA
(0) =
1
2x
A
⟨|v|⟩⟨δ(q − q
∗
)⟩. (25)
For the initial rate obtained from the transition state theory approximation, because
H
(TST)
B
[q(t)] = H[v(0)] (26)
k
(TST)
BA
= x
−1
A
⟨v(0)δ[q(0) − q
∗
]H[v(0)]⟩ (27)
which is identical to equation (23), the expression for k
BA
is exactly the same.
3
8.8
From the Exercise 8.7 we know k
(TST)
BA
is calculated assuming no trajectories recrossed the transition states
after a short time. However, in fact there might be trajectories that recross the transition state from B side
back to A side. Therefore in reality the reaction rate k
BA
should be lower than the approximation k
(TST)
BA
.
The shorter the time is, the closer these two are.
8.9
Obviously,
dr
N
ρ(r, t) = N,
U[r
j
(t)]
dr
N
ρ(r, t) = 1 (28)
where U [r
j
(t)] indicates an innitesimal neighborhood of r
j
(t). These properties validate that ρ(r, t) has the
form
ρ(r, t) =
N
j=1
δ[r − r
j
(t)]. (29)
8.10
We can dene a quantity k
k(r, t) =
N
j=1
v
j
(t)δ[r − r
j
(t)] =
N
j=1
˙
r
j
(t)δ[r − r
j
(t)] (30)
Therefore
∂ρ(r, t)
∂t
=
N
j=1
∂
∂t
δ[r − r
j
(t)]
= −
N
j=1
∇δ[r − r
j
(t)] ·
d
dt
r
j
(t)
= −
N
j=1
˙
r
j
(t) · ∇δ[r − r
j
(t)]
= − ∇ ·
N
j=1
˙
r
j
δ[r − r
j
(t)]
= − ∇ · k(r, t). (31)
Comparing with the example of an activated process, H
A
[q] can be the density in the area q < q
∗
, and q is
a generalized coordinate. In one dimension,
˙
qδ
(
q
−
q
∗
)
is the divergence of the ux at the boundary
q
=
q
∗
.
Therefore the equation
˙
H
A
[q] = − ˙qδ(q − q
∗
) resembles the equation of continuity.
8.11
Suppose P (r
N
(t), r
N
(0)) is the joint congurational distribution at time t and time 0. By denition,
P
1
(r
1
(t), r
1
(0)) =
dr
2
(t) · · · dr
N
(t)
dr
2
(0) · · · dr
N
(0)P (r
N
(t), r
N
(0)) (32)
4
P
(t)
1
(r
1
(t)) =
dr
1
(0)P
1
(r
N
(t), r
N
(0)) (33)
P
(0)
1
(r
1
(0)) =
dr
1
(t)P
1
(r
N
(t), r
N
(0)) (34)
¯ρ(r(t), t, r(0), 0) = N P
1
(r
1
(t), r
1
(0)) (35)
¯ρ(r, t) = N P
(t)
1
(r) (36)
¯ρ(r, 0) = N P
(0)
1
(r) (37)
By denition,
P (r, t) = N
P
1
(r(t), r(0))
P
(0)
1
(0)
= N
¯ρ(r, t, 0, 0)
¯ρ(0, 0)
= N
⟨ρ(r, t)ρ(0, 0)⟩
¯ρ(0, 0)
. (38)
8.12
d
dt
∆R
2
(t) =
dr r
2
∂
∂t
P (r, t) =
dr r
2
D∇
2
P (r, t)
=r
2
D∇P (r, t)|
∞
−
drD∇r
2
∇P (r, t)
=r
2
D∇P (r, t)|
∞
− D∇r
2
P (r, t)|
∞
+
drD∇
2
r
2
P (r, t) (39)
Because the distribution should be bounded in nite space, itself as well as its gradient should vanish at
innity. Thus
d
dt
∆R
2
(t) =
drD∇
2
r
2
P (r, t) = 6
drDP (r, t). (40)
Since P (r, t) is normalized for any t, we have
d
dt
∆R
2
(t) = 6D. (41)
8.13
Since
d
dt
∆R
2
(t) = 2
t
0
dt⟨v(0) · v(t)⟩ (42)
and
⟨v(0) · v(t)⟩ ≈ ⟨v
2
⟩e
−t/τ
, (43)
d
dt
∆R
2
(t) ≈ 2
t
0
⟨v
2
⟩e
−t/τ
= −2⟨v
2
⟩τ
e
−t/τ
− 1
. (44)
∆R
2
(t) ≈ 2⟨v
2
⟩τ
2
(e
−t/τ
− 1) + 2⟨v
2
⟩τt. (45)
Given D ∼ 1 × 10
−5
cm
2
/s, ∆R
2
(t) = 6Dt,
3D = ⟨v
2
⟩τ. (46)
Because
⟨v
2
⟩ = 3k
B
T/m, (47)
τ =
mD
k
B
T
(48)
For a small molecule, m ∼ N
−1
A
× 1 kg. At room temperature T ∼ 300 K. Then τ ∼ 4 × 10
−13
s.
Figsure 2 shows the mean square displacement versus time with reduced units.
5
Figure 2: Diusion curve of particles in Exercise 8.13.
8.14
The perturbation is
∆
H
=
−
i
f
i
A
i
(0)
.
(49)
¯
A
j
(
t
) =
Tr
e
−β(H +∆H )
A
j
(t)
Tre
−β(H +∆H )
=
1
Tre
−βH
Tr
e
−βH
A
j
(t) − (β∆H )A
j
(t) + A
j
(t)
Tr[e
−βH
(β∆H )]
Tre
−βH
+ O[(β∆H )
2
]
=⟨A
j
⟩ − β[⟨∆H A
j
(t)⟩ − ⟨A
j
⟩⟨∆H ⟩] + O[(β∆H )
2
] (50)
∆
¯
A
j
(t) =
¯
A
j
(t) − ⟨A
j
⟩ = β
i
f
i
⟨A
i
(0)A
j
(t)⟩ −
i
f
i
⟨A
j
⟩⟨A
i
⟩
+ O(f
2
)
=β
i
f
i
⟨δA
i
(0)δA
j
(t)⟩ + O(f
2
). (51)
8.15
Since it is proved that
∆n(r, t) = β
dr
′
Φ(r
′
)⟨δρ(r
′
, 0)δρ(r, t)⟩ (52)
According to Fick’s law
∂
∂t
n(r, t) = D∇
2
n(r, t), (53)
6
∂
∂t
n(r, t) =
∂
∂t
∆n(r, t) = β
dr
′
Φ(r
′
)
∂
∂t
⟨δρ(r
′
, 0)δρ(r, t)⟩ (54)
D∇
2
n(r, t) = Dβ
dr
′
Φ(r
′
)∇
2
⟨δρ(r
′
, 0)δρ(r, t)⟩ (55)
Since Φ(r) is arbitraty external eld, for any r
′
∂
∂t
⟨δρ(r
′
, 0)δρ(r, t)⟩ = D∇
2
⟨δρ(r
′
, 0)δρ(r, t)⟩ (56)
Thus
∂C(r, t)
∂t
= D∇
2
C(r, t). (57)
8.16
First we calculate the response function at t > 0,
χ(t) = −β
d
dt
⟨δA(0)δA(t)⟩ = βτ
−1
⟨(δA)
2
⟩e
−t/τ
. (58)
Then we start with
∆
¯
A(t) = f
t
2
t
1
dt
′
χ(t − t
′
). (59)
For the case t < t
1
, because χ(t) = 0 when t < 0, ∆
¯
A(t) = 0.
For the case t
1
< t < t
2
,
∆
¯
A(t) = f
t
t
1
dt
′
χ(t − t
′
) = f βτ
−1
⟨(δA)
2
⟩
t
t
1
dt
′
e
−(t−t
′
)/τ
= fβ⟨(δA)
2
⟩
1 − e
−(t−t
1
)/τ
(60)
For the case t > t
2
,
∆
¯
A(t) = f
t
2
t
1
dt
′
χ(t − t
′
) = f β⟨(δA)
2
⟩
e
−(t−t
2
)/τ
− e
−(t−t
1
)/τ
(61)
gradually fades to zero.
A demonstration of the deviation of A under dierent τ is shown in gure 3. From the gure it can be seen
that if τ ≪ t
2
− t
1
, the system will be driven immediately following the square perturbation; if τ ≫ t
2
− t
1
,
the system will be perturbed and will restore slowly. τ = t
2
− t
1
should resemble a critical damping.
The energy absorbed is
Abs = −
∞
−∞
dt
˙
f(t)
¯
A(t) =
∞
−∞
dt
˙
¯
A(t)f(t) = f
t
2
t
1
dt
˙
¯
A(t)
=f[
¯
A(t
2
) −
¯
A(t
1
)] = f [∆
¯
A(t
2
) − ∆
¯
A(t
1
)]
=f
2
β⟨(δA)
2
⟩
1 − e
−(t
2
−t
1
)/τ
. (62)
8.17
If n = 0,
1
T
T
0
dt e
inωt
=
1
T
T
0
dt =
T
T
= 1. (63)
If n ̸= 0,
1
T
T
0
dt e
inωt
=
1
inωT
e
inωT
− 1
=
2e
inωT /2
nωT
sin(nωT/2). (64)
7
Figure 3: Response to a square pulse in Exercise 8.16.
lim
ωT →∞
1
T
T
0
dt e
−nωt
= lim
ωT →∞
2e
inωT /2
nωT
sin(nωT/2)
⩽ lim
ωT →∞
2
nωT
= 0. (65)
Thus when n ̸= 0, ωT → ∞,
1
T
T
0
dt e
inωt
→ 0. (66)
8.18
First expand f(t) =
1
2
f
ω
e
−iωt
+ f
∗
ω
e
iωt
,
abs(ω) =
1
T
T
0
dt
iω
2
f
ω
e
−iωt
− f
∗
ω
e
iωt
⟨A⟩ +
∞
−∞
dt
′
χ(t
′
)f(t − t
′
) + O(f
2
)
=
1
T
T
0
dt
iω
2
f
ω
e
−iωt
− f
∗
ω
e
iωt
⟨A⟩ +
∞
−∞
dt
′
χ(t
′
)
1
2
f
ω
e
−iω( t−t
′
)
+ f
∗
ω
e
iω (t−t
′
)
+ O(f
2
)
(67)
Notice Chandler’s book has a typo in abs equation: a redundant minus sign. Use the result in Exercise 8.17,
and suppose ωT → ∞.
abs(ω) =
1
T
T
0
dt
iω
2
f
ω
e
−iωt
− f
∗
ω
e
iωt
∞
−∞
dt
′
χ(t
′
)
1
2
f
ω
e
−iω (t−t
′
)
+ f
∗
ω
e
iω (t−t
′
)
+ O(f
2
)
=
iω
4
∞
−∞
dt
′
χ(t
′
)
f
ω
f
∗
ω
e
−iωt
′
− f
∗
ω
f
ω
e
iωt
′
+ O(f
3
)
=
iω
4
|f
ω
|
2
∞
−∞
dt
′
χ(t
′
)
e
−iωt
′
− e
iωt
′
+ O(f
3
)
=
ω
2
|f
ω
|
2
∞
−∞
dtχ(t) sin(ωt) + O(f
3
). (68)
8
8.19
Suppose A(t) obeyed simple harmonic oscillator dynamics,
d
2
A(t)
dt
2
= −ω
2
0
A(t) (69)
The real solution to this dierential equation is
A(t) = C sin(ω
0
t) + D cos(ω
0
t) (70)
˙
A(t) = Cω
0
cos(ω
0
t) − Dω
0
sin(ω
0
t) (71)
A(0) = D,
˙
A(0) = Cω
0
(72)
Once the initial condition C and D given, the evolution of the system is determined. However, required by
statistical mechanics,
⟨A
˙
A⟩ = ⟨DCω
0
⟩ = 0 (73)
Thus
⟨CD⟩ = 0. (74)
The distribution of C and D should be even for equilibrium, thus ⟨C⟩ = ⟨D⟩ = 0. Thus ⟨δCδD⟩ = 0.
δA(t) = δ C sin(ω
0
t) + δD cos(ω
0
t), δA(0) = δ D (75)
⟨δA(0)δA(t)⟩ =⟨δD[δC sin(ω
0
t) + δD cos(ω
0
t)]⟩
=⟨(δD)
2
cos(ω
0
t) + δCδD sin(ω
0
t)⟩
=⟨(δD)
2
⟩ cos(ω
0
t) + ⟨δCδD⟩ sin(ω
0
t)
=⟨(δD)
2
⟩ cos(ω
0
t) = ⟨(δA(0))
2
⟩ cos(ω
0
t). (76)
8.20
The model described in section 8.8 in the text is an oscillator coupled to a random harmonic bath. The
target is to describe the dynamics of the oscillator. The Hamitonian of the system is
H = H
0
(x) − xf + H
b
(y
1
, · · · , y
N
) (77)
where
H
0
=
1
2
m ˙x
2
+ V (x), f =
i
c
i
y
i
(78)
Here y
i
are the normal modes of the harmonic bath. Because the bath is purely harmonic, the evolution and
response to evolving x is exactly linear. Thus the evolution of f can be written as
f(t) = f
b
(t) +
∞
−∞
dt
′
χ
b
(t − t
′
)x(t
′
) (79)
where
χ
b
(t − t
′
) =
−β
dC
b
(t − t
′
)
d(t − t
′
)
, t > t
′
0, t < t
′
.
(80)
From the Hamiltonian the equation of state is
m¨x(t) = f
0
[x(t)] + f
b
(t) +
∞
−∞
dt
′
χ
b
(t − t
′
)x(t
′
) (81)
9
where f
0
comes from H
0
f
0
[x] = −
dV
dx
. (82)
Plug in the equation (80), and notice that the time origin is 0,
m¨x(t) = f
0
[x(t)] + f
b
(t) − β
t
0
dt
′
C
′
b
(t − t
′
)x(t
′
) (83)
Integrated by part,
t
−∞
dt
′
C
′
b
(t − t
′
)x(t
′
) = − C
b
(t − t
′
)x(t
′
)
t
0
+
t
0
C
b
(t − t
′
) ˙x(t
′
)dt
′
= − C
b
(0)x(t) + C
b
(t)x(0) +
t
0
C
b
(t − t
′
) ˙x(t
′
)dt
′
(84)
Thus
m¨x(t) =f
0
[x(t)] + f
b
(t) + βC
b
(0)x(t) − βC
b
(t)x(0) − β
t
0
C
b
(t − t
′
) ˙x(t
′
)dt
′
={f
0
[x(t)] + βC
b
(0)x(t)} + [f
b
(t) − βC
b
(t)x(0)] − β
t
0
C
b
(t − t
′
) ˙x(t
′
)dt
′
(85)
Dene
¯
V (x) = V (x) −
1
2
βC
b
(0)x
2
(t) (86)
δf(t) = f
b
(t) − βC
b
(t)x(0) (87)
and
¯
f[x(t)] = −
d
¯
V
dx
= −
dV
dx
+ βC
b
(0)x(t). (88)
Thus
m¨x(t) =
¯
f[x(t)] + δf(t) − β
t
0
dt
′
C
b
(t − t
′
) ˙x(t
′
). (89)
Notice that the distribution of f
b
(t) is Gaussian with mean value βC
b
(t)x(0) and variance C
b
(t − t
′
).
8.21
(i) Since
m
d
2
dt
2
x(t) =
¯
f[x(t)] + δf(t) − β
t
0
dt
′
C
b
(t − t
′
)
d
dt
′
x(t
′
)
= − m¯ω
2
x(t) + δf(t) − β
t
0
dt
′
C
b
(t − t
′
)
d
dt
′
x(t
′
) (90)
multiply by x(0) and take the average on both side
d
2
dt
2
⟨x(0)x(t)⟩ = −¯ω
2
⟨x(0)x(t)⟩ −
β
m
t
0
dt
′
C
b
(t − t
′
)
d
dt
′
⟨x(0)x(t)⟩. (91)
(ii) The Laplace transform gives
s
2
˜
C(s) − s⟨x
2
(0)⟩ − ⟨x(0) ˙x(0)⟩ = −¯ω
2
˜
C(s) −
β
m
˜
C
b
(s)
s
˜
C(s) − ⟨x
2
(0)⟩
(92)
10
Required by statistical mechanics, ⟨x(0) ˙x(0)⟩ = 0. Then
˜
C(s) =
s⟨x
2
⟩ +
β
m
˜
C
b
(s)⟨x
2
⟩
s
2
+ ¯ω
2
+ s
β
m
˜
C
b
(s)
=
s +
β
m
˜
C
b
(s)
s
2
+ ¯ω
2
+ s
β
m
˜
C
b
(s)
⟨x
2
⟩. (93)
(iii) Assume
C
b
(t) = C
b
(0)e
−t/τ
. (94)
Consequently,
˜
C
b
(s) =
C
b
(0)
s + τ
−1
. (95)
The cosine Fourier transform of ⟨x(0)x(t)⟩ is
ˆ
C(ω) =
∞
0
cos(ωt)⟨x(0)x(t)⟩ =
1
2
[
˜
C(iω) +
˜
C(−iω)]
=
1
2
iω +
β
m
˜
C
b
(iω)
−ω
2
+ ¯ω
2
+ iω
β
m
˜
C
b
(iω)
+
−iω +
β
m
˜
C
b
(−iω)
−ω
2
+ ¯ω
2
− iω
β
m
˜
C
b
(−iω)
⟨x
2
⟩
=
1
2
iω +
βC
b
(0)
m(iω + τ
−1
)
−ω
2
+ ¯ω
2
+ iω
βC
b
(0)
m
(
iω
+
τ
−1
)
+
−iω +
βC
b
(0)
m(−iω + τ
−1
)
−ω
2
+ ¯ω
2
− iω
βC
b
(0)
m
(
−
iω
+
τ
−1
)
⟨x
2
⟩. (96)
The function at ω = ¯ω should be a peak
ˆ
C(¯ω) =
1
2
i¯ω +
βC
b
(0)
m(i¯ω + τ
−1
)
i¯ω
βC
b
(0)
m(i¯ω + τ
−1
)
+
−i¯ω +
βC
b
(0)
m(−i¯ω + τ
−1
)
−i¯ω
βC
b
(0)
m(−i¯ω + τ
−1
)
⟨x
2
⟩
=
1
2
i¯ωm(i¯ω + τ
−1
) + βC
b
(0)
i¯ωβC
b
(0)
+
−i¯ωm(−i¯ω + τ
−1
) + βC
b
(0)
−i¯ωβC
b
(0)
⟨x
2
⟩
=
1
2
2i¯ωmτ
−1
i¯ωβC
b
(0)
⟨x
2
⟩
=
m⟨x
2
⟩
βτC
b
(0)
, (97)
which indicates a strong absorption.
8.22
The corresponding sketches are shown in gure 4. Plot (a) shows a smooth decay, while plot (b) shows a
pattern of periodicity. This is because in the solid state, the particles may oscillate with a certain period
T . Plot (c) shows the decay in ⟨v
(
0)v
2
(t)⟩. Because ⟨v
2
(t)⟩ ̸= 0, when T goes to innity, ⟨v
2
(0)v
2
(t)⟩ →
⟨v
2
(0)⟩⟨v
2
(t)⟩ = ⟨v
2
⟩
2
nonzero. Plot (d) shows the direction of velocity coupling, exactly the three times as
that is shown in gure 1.
11
Figure 4: The sketches in Exercise 8.22.
12
8.23
Since
D =
1
3
∞
0
dt⟨v(0) · v(t)⟩ ≈
1
3
∞
0
dt⟨v
2
⟩e
−t/τ
=
1
3
⟨v
2
⟩τ. (98)
τ ≈
3D
⟨v
2
⟩
(99)
At the time right after turning down the external electric eld, the ensemble average of v
2
can be describes
by the temperature
⟨v
2
⟩ =
3k
B
T
m
. (100)
Thus
τ ≈
mD
k
B
T
. (101)
8.24
At room temperature T ≈ 300 K, β ≈ 2.4 × 10
20
J
−1
,
D ≈
1
2πβση
∼ 1 × 10
−10
m
2
/s. (102)
8.25
Because
∆R
2
(t) = 6Dt (103)
Consider the particles doing a Brownian motion. For t = 5 psec, ∆R
2
= 3 Å. Since r
i
(t) =
|r
i
(t) − r
i
(0)|
2
,
r
i
(t)/Å satises χ
2
(3) distribution. F
χ
2
(3)
(5) = 0.828. Thus 17.2% particles have moved more than 5 Å.
8.26
Adopted from the ocial solution manual. (a) Solving the dierential equations we get
∆c
1
(t) = A
1
e
−λ
1
t
+ B
1
e
−λ
2
t
(104)
∆c
2
(t) = A
2
e
−λ
1
t
+ B
2
e
−λ
2
t
(105)
where
λ
1,2
=
1
2
[(k
31
+ k
13
) + (k
32
+ k
23
)] ±
1
2
[(k
31
+ k
13
) − (k
32
+ k
23
)]
2
+ k
13
k
23
. (106)
(b)
λ
1
≈ k
13
+ k
23
= τ
−1
transient
(107)
λ
2
≈
k
31
k
23
+ k
13
k
32
k
13
+ k
23
= τ
−1
rxn
. (108)
τ
−1
rxn
≈ k
31
k
32
k
13
+ k
23
+ k
32
k
13
k
13
+ k
23
≈ k
31
+ k
32
≪ k
13
+ k
23
≈ τ
−1
transient
. (109)
Therefore the relaxation is dominated by τ
rxn
.
(c) As shown above, the faster transient decay occurs on a time scale of (k
13
+ k
23
)
−1
≈ k
−
1
13
or k
−
1
23
.
13
(d) The two decay rates are analogous to the two rates in the reactive ux description:
τ
mol
≈ τ
transient
≪ τ
rxn
. (110)
The connection can be made by imagining preparing the system at the transition state, i.e., in state 3. Then
the decay into states 1 and 2, c
1
(t) and c
2
(t), follow the two decay rates, one much faster than the other.
But c
1
(t) ∝ ⟨δn
A
(0)δn
A
(t)⟩ by the regression hypothesis, and the time derivative of ⟩δn
A
(0)δn
A
(t)⟩ is just
the ux in the reactive ux picture. In particular, τ
−1
rxn
is on the order of k
31
or k
32
, and is the plateau value
for the reaction rate.
(e) This is similar to the transition state theory idea where e
−βQ
is the probability o getting to the transition
states 3, and D ∝ 1/η is the rate to cross the barrier 3 once there. As we showed previously,
k
(TST)
∝ ⟨|v|⟩⟨δ(q − q
∗
)⟩. (111)
8.27
From transition state theory, we still have
k
12
=
1
2x
2
⟨|v|⟩⟨δ[q(0) − q
∗
]⟩ (112)
because when t → 0, the trajectories starting from region 3 do not have a chance to cross the barrier at q
∗
.
8.28
(a) The RMS velocity of an argon atom in the vapor is
v
g
=
3k
B
T
m
= 4.3 × 10
2
m/s. (113)
(b) The RMS velocity of an argon atom in the solution is the same as that in (a)
v
l
= v
g
=
3k
B
T
m
= 4.3 × 10
2
m/s. (114)
(c) Because in the gas phase, the spaces of the particles are larger than l = 10 Å, the motion is in the inertial
regime
t =
l
v
g
= 2.3 × 10
−12
s = 2.3 psec. (115)
(d) Because in the solution, the particles are eected by the solvent molecules, the motion is in the diusion
regime
t =
l
2
6
D
= 1.67 × 10
−9
s = 167 psec. (116)
(e) In the solution,
τ
relax
=
mD
k
B
T
= 1.6 × 10
−14
s = 1.6 × 10
−2
psec. (117)
(f) When η
2
= 2η
1
, according to Stokes’ law D ∝ η
−1
,
t
2
= 2t
1
= 333 psec. (118)
The velocity is not inuenced since the temperature does not change.
14
