Das Sh.P. Statistical Physics of Liquids at Freezing and Beyond
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A5.1 The microscopic-balance equations 261
where the prime in the sum implies that the α = β term is excluded. F
αβ
is the force on
the αth particle due to interaction with the βth particle. We can simplify the second term
with the formal manipulation
{δ(r − r
β
) − δ(r − r
α
)}F
i
αβ
=
&
1
0
ds
∂
∂s
δ(r − r
α
+ sr
αβ
)
'
F
i
αβ
=
⎧
⎨
⎩
1
0
ds
j
∂
∂sr
j
αβ
δ(r − r
α
+ sr
αβ
)
⎫
⎬
⎭
r
j
αβ
F
i
αβ
=
j
∇
j
&
1
0
ds δ(r − r
α
+ sr
αβ
)r
j
αβ
F
i
αβ
'
. (A5.1.4)
For an interaction potential that produces a central force field, i.e., F
αβ
is in the direction
of the vector r
αβ
, the second term on the RHS of (A5.1.3) reduces to the form
∇
j
&
1
0
ds δ(r − r
α
+ sr
αβ
)r
j
αβ
F
i
αβ
'
=−∇
j
αβ
ij
αβ
, (A5.1.5)
where we have defined
αβ
and
ij
αβ
as
αβ
=
1
0
ds δ(r − r
α
+ sr
αβ
), (A5.1.6)
ij
αβ
= (r
αβ
· F
αβ
)ˆr
i
αβ
ˆr
j
αβ
. (A5.1.7)
It is straightforward to show that
αβ
=
βα
. The quantity
ij
αβ
is also symmetric under
exchange of {α, β} as well as {i, j}. The balance eqn. (A5.1.3) for the momentum density
is obtained as
∂ ˆg
i
∂t
=−
j
∇
j
⎧
⎨
⎩
α
p
i
α
p
j
α
m
δ(r − r
α
) +
1
2
αβ
αβ
(ˆr
αβ
· F
αβ
)ˆr
i
αβ
ˆr
j
αβ
⎫
⎬
⎭
=−
j
∇
j
ˆσ
ij
, (A5.1.8)
where the stress-energy tensor ˆσ
ij
is
ˆσ
ij
(r, t) =
α
p
i
α
p
j
α
m
δ(r − r
α
) +
1
2
α,β
ij
αβ
αβ
. (A5.1.9)
262 Appendix to Chapter 5
The microscopic stress tensor ˆσ
ij
satisfies the symmetry ˆσ
ij
=ˆσ
ji
, signifying conservation
of angular momentum in the system. In a similar way we obtain the equation of motion for
the energy density e(x):
∂ ˆe
∂t
=
α
!
{∇
r
α
δ(r − r
α
)}· ˙r
α
e
α
+ δ(r −r
α
)˙e
α
"
. (A5.1.10)
To simplify the RHS of the above equation we first evaluate ˙e
α
as
˙e
α
=˙p
α
·
p
α
m
+
1
2
β
∂u(r
αβ
)
∂r
αβ
·
˙
r
αβ
=
β
F
αβ
·
p
α
m
−
1
2
F
αβ
·
˙
r
αβ
=
1
2
β
F
αβ
·
(
p
α
m
+
p
β
m
)
. (A5.1.11)
We have used the relations
˙
r
αβ
= (p
α
− p
β
)/m and the force on the particle at r
α
due to
the particle at r
β
is
F
αβ
=−∇
r
αβ
u(r
αβ
). (A5.1.12)
Using eqn. (A5.1.11) in eqn. (A5.1.10), we obtain
∂ ˆe
∂t
=
α
!
{∇
r
α
δ(r − r
α
)}· ˙r
α
e
α
+ δ(r − r
α
)˙e
α
"
=−∇
r
·
α
δ(r − r
α
)
p
α
m
e
α
+
1
2
α,β
(
p
α
m
+
p
β
m
)
· F
αβ
δ(r − r
α
)
=−∇
r
·
α
δ(r − r
α
)
p
α
m
e
α
+
1
4
α,β
(
p
α
m
+
p
β
m
)
·
!
F
αβ
{δ(r − r
α
) − δ(r −r
β
)}
"
. (A5.1.13)
Now, using the same manipulations as in eqns. (A5.1.4)–(A5.1.6), the part
F
αβ
{δ(r −r
α
) −δ(r −r
β
)} in the second term on the RHS can be formally expressed as a
total derivative. This leads to the result
∂ ˆe
∂t
=−
i
∇
i
⎧
⎨
⎩
α
p
i
α
m
e
α
δ(r − r
α
) +
1
4
α,β
j
ij
αβ
p
j
α
m
+
p
j
β
m
αβ
⎫
⎬
⎭
.
(A5.1.14)
We therefore obtain the energy-balance equation as
∂ ˆe
∂t
=−∇·
ˆ
j
e
, (A5.1.15)
A5.1 The microscopic-balance equations 263
with the energy current
ˆ
j
e
being given by
ˆ
j
i
e
(r) =
α
e
α
p
i
α
m
δ(r − r
α
) +
1
4
α,β
j
ij
αβ
p
j
α
m
+
p
j
β
m
αβ
. (A5.1.16)
A5.1.1 The Euler equations
In order to determine the local equilibrium averages of the microscopic currents pre-
sented in eqn. (5.1.7), we consider the dynamics of the fluid in a frame moving with the
local velocity v(x, t) so that the fluid appears to be at rest in the vicinity of this point.
We consider the conserved densities {ˆρ
(r),
ˆ
g
(r), ˆe
(r)} and the corresponding currents
#
ˆ
g
(r), ˆσ
(r),
ˆ
j
e
(r)
$
in the locally moving frame in the fluid. The phase-space coordinates
of the particles in this frame are defined through the canonical transformation
p
α
= p
α
+ mv(r
α
) and r
α
= r
α
. (A5.1.17)
Now, using the above transformation rules in the definitions (5.1.1)–(5.1.3), we obtain the
relations between the conserved densities in the two frames:
ˆρ(r) =ˆρ
(r), (A5.1.18)
ˆ
g(r) =
ˆ
g
(r) +ˆρ
(r)v(r), (A5.1.19)
ˆe(r) =ˆe
(r) + v(r) ·
ˆ
g
(r) +
1
2
ˆρ
(r)v
2
(r). (A5.1.20)
Similarly, the transformation rules for the stress tensor ˆσ
ij
and the energy current
ˆ
j
e
are
obtained as
ˆσ
ij
(r) =ˆσ
ij
(r) +ˆg
j
(r)v
i
(r) +ˆg
j
(r)v
i
(r) +ˆρ
(r)v
i
(r)v
j
(r), (A5.1.21)
ˆ
j
e
(r) =
ˆ
j
e
(r) +
3
ˆe
(r) + v(r) ·
ˆ
g
(r) +
1
2
ˆρ
(r)v
2
(r)
4
v(r)
+
1
2
v
2
(r)
ˆ
g
(r) +ˆσ · v(r). (A5.1.22)
It is important to note that the transformation rule (A5.1.22) for the heat current
ˆ
j
e
holds
only for the local form of the heat current as described in eqn. (5.1.13), i.e., for short-
range potentials. Considering the transformation rules (A5.1.18)–(A5.1.20), the distribu-
tion function f
le
(t) in terms of the primed variables
N
≡
r
1
, p
1
,...,r
N
, p
N
is
obtained as
f
le
N
;t
= Q
−1
exp
−
dr β(r, t)
!
ˆe
(r) − μ(r, t) ˆn
(r)
"
. (A5.1.23)
264 Appendix to Chapter 5
On averaging with respect to the distribution f
le
N
;t
in the local rest frame, it follows
directly from the p
α
→−p
α
antisymmetry (see the definitions (5.1.7) for the currents) that
1
g
2
le
= 0,
+
ˆ
j
e
,
le
= 0. (A5.1.24)
The average of the stress tensor σ
ij
, on the other hand, is nonzero in the local frame and is
linked to a time-dependent generalization of the corresponding thermodynamic pressure,
+
ˆσ
ij
,
le
= P(r, t)δ
ij
. (A5.1.25)
Similarly, the average of the energy density is linked to the internal energy density in the
fluid,
1
ˆe
2
le
= (r, t). (A5.1.26)
On substituting these relations into eqns. (A5.1.19), (A5.1.21), and (A5.1.22), respectively,
the averages of the microscopic currents are obtained as
1
ˆ
g
2
le
= ρv, (A5.1.27)
1
ˆσ
ij
2
le
= Pδ
ij
+ ρv
i
v
j
, (A5.1.28)
+
ˆ
j
e
,
le
=
+ρ
v
2
2
+ P
v, (A5.1.29)
for the pressure P(r, t ) and the internal energy density (r, t), respectively. On averaging
eqn. (A5.1.20) we obtain the relation
e(r, t) =
1
ˆe(r)
2
= (r, t) +
1
2
ρ(r, t)v
2
(r, t). (A5.1.30)
A5.1.2 The entropy-production rate
The entropy of the fluid in the local equilibrium state is defined as
S(t) =−
ln f
le
(
N
, t)
. (A5.1.31)
Using the expression (5.1.15) for f
le
, we obtain
ln f
le
(
N
, t) =−ln Q
l
−
dr β(r, t)
×
ˆe(r) − v(r, t) · ˆg(r) −
μ(r, t) −
1
2
mv
2
(r, t)
ˆn(r)
=−ln Q
l
−
dr
{a}
α
a
(r, t) ˆa(r), (A5.1.32)
A5.1 The microscopic-balance equations 265
where the set of local densities is given by ˆa ≡{ˆe,
ˆ
g, ˆn}. On substituting the expression
(5.1.15) on the RHS we obtain
S(t) = ln Q
l
(t) +
{a}
dr α
a
(r, t)a(r, t). (A5.1.33)
We define an entropy density s(r, t) whose spatial integral gives the total entropy S(t) of
the system as given by eqn. (A5.1.33):
s(r, t) = V
−1
ln Q
l
(t) +
{a}
α
a
(r, t)a(r, t). (A5.1.34)
The time rate of change of the entropy is obtained as
∂ S(t)
∂t
=
∂
∂t
ln Q
l
(t) +
{a}
dr
∂
∂t
[
α
a
(r, t)a(r, t)
]
. (A5.1.35)
Since the time dependence of Q
l
(t) comes through that of the α
l
after all the phase-space
variables have been integrated out, we write the first term on the RHS of eqn. (A5.1.35) in
terms of time derivatives of the α
a
. Hence we obtain
∂ S(t)
∂t
=
dr
{a}
δ ln Q
l
δα
a
∂α
a
(r, t)
∂t
+
∂
∂t
{
α
a
(r, t)a(r, t)
}
=
dr
{a}
−a(r, t)
∂α
a
∂t
+
∂
∂t
{
α
a
a(r, t)
}
=
dr
{a}
α
a
(r, t)
∂
∂t
a(r, t). (A5.1.36)
The LHS of the above equation is evaluated using the equation of motion for the slow vari-
ables. Let us first consider the Euler equations for reversible dynamics obtained from the
local equilibrium average of the microscopic balance equations as given in eqn. (5.1.21),
∂ S(t)
∂t
=−
dr
{a}
α
a
(r, t)∇ ·
+
ˆ
j
a
,
=
dr
{a}
!
−∇ · {α
a
(r, t)j
a
}+j
a
· ∇α
a
(r, t)
"
, (A5.1.37)
where
ˆ
j
a
le
≡ j
a
are the local currents obtained in eqns. (5.1.22)–(5.1.24). If we define
an entropy density s(r, t) whose spatial integral gives the total entropy S(t) of the system,
then the above equation transforms into the form
dr
⎡
⎣
∂s
∂t
+∇·
⎧
⎨
⎩
{a}
α
a
j
a
⎫
⎬
⎭
⎤
⎦
=
dr
{a}
{j
a
· ∇α
a
(r, t)}. (A5.1.38)
266 Appendix to Chapter 5
For the set of conserved densities ˆa ≡{ˆe,
ˆ
g, ˆn}, the corresponding α
a
are given by
#
β, −βv, −β
(
μ −
1
2
mv
2
)$
and the currents j
a
are listed in eqns. (5.1.22)–(5.1.24).Using
these results, we obtain
{a}
α
a
j
a
= β(e + P)v
i
− βv
j
(Pδ
ij
+ ρv
i
v
j
) − β
μ −
1
2
mv
2
nv
i
=
3
βe −βv
j
g
j
− β
μ −
1
2
mv
2
n
4
v
i
=
⎧
⎨
⎩
{a}
α
a
(r, t)a(r, t)
⎫
⎬
⎭
v
i
(r, t)
={s(r, t) − V
−1
ln Q
l
(t)}v
i
(r, t), (A5.1.39)
wherewehaveusedeqn. (A5.1.34) in reaching the last equality. On substituting the above
result into eqn. (A5.1.38), we obtain
dr
∂s
∂t
+∇· (sv)
=
dr
{a}
{j
a
·∇α
a
(r, t)}+
dr ∇ · {V
−1
ln Q
l
v}.
(A5.1.40)
Finally, we evaluate the first term on the RHS of eqn. (A5.1.40) explicitly:
{a}
{j
a
· ∇α
a
(r, t)}=
i
(e + P)v
i
∇
i
β
−
i, j
(Pδ
ij
+ ρv
i
v
j
)∇
i
(−βv
j
) − nv
i
∇
i
3
β
μ −
1
2
mv
2
4
= v · {( + P − nμ)∇β + (∇P − n ∇μ)β}−∇·(β Pv)
= v · {Ts∇β + (∇P − n ∇μ)β}−∇· (β Pv)
= βv · {∇P − s ∇T −n ∇μ}−∇· (β Pv). (A5.1.41)
In the hydrodynamic regime all the local functionals T (r, t), μ(r, t), P(r, t), and v(r, t)
are the same functionals of the conserved densities of mass, momentum, and energy as in
the respective cases in equilibrium. Therefore, using the thermodynamic identity, i.e., the
Gibbs–Duhem relation
VdP − sdT − μdN =
0 (A5.1.42)
in this case, the quantity within the curly brackets on the RHS of eqn. (A5.1.41) is equal to
zero. We define the entropy-production rate in the fluid as R
E
,
R
E
= T
d
dt
s(r, t)dr, (A5.1.43)
A5.1 The microscopic-balance equations 267
where the integration extends over the volume of the system. Using the results from
(A5.1.41), we obtain
R
E
= T
dr
∂s
∂t
+∇· (sv)
= T
dr ∇ ·
v
EQ
V
−1
(A5.1.44)
where
EQ
= ln Q
l
−β PV. The RHS, which is expressed as a volume integral of the diver-
gence, can be reduced to a surface integral, which is assumed to vanish when the system is
not externally perturbed. Note that, for the equilibrium limit, f
l
(
N
) becomes the grand-
canonical distribution and the corresponding partition function satisfies β PV = ln Q
l
(see
eqn. (1.2.40)), making
EQ
= 0 identically zero. Therefore, we have for reversible dynam-
ics that the entropy density satisfies the continuity equation
∂s
∂t
+∇· (sv) = 0 (A5.1.45)
with current sv. The net production of entropy of the isolated system with reversible
dynamics is equal to zero. In the present situation we have completely reversible dynamics
within the local-equilibrium approximation starting from the microscopic-level description
in terms of Newton’s equations.
Dissipative effects
The dissipative effects due to the irreversible transport in the fluid are included in the
dynamical equations in a phenomenological manner. To infer about the form of the dissi-
pative terms, we extend the currents in the equations for the dynamics. Let us first define
the total current as a sum of reversible and irreversible parts, denoted by superscripts R and
D, respectively:
σ
ij
= Pδ
ij
+ ρv
i
v
j
+ σ
D
ij
≡ σ
R
ij
+ σ
D
ij
, (A5.1.46)
j
e
=
+ρ
v
2
2
+ P
v + j
D
e
≡ j
R
e
+ j
D
e
. (A5.1.47)
There is no dissipative part in the mass current g, which is itself a conserved quantity.
The continuity equation remains unaltered at the coarse-grained level. The corresponding
equations of motion for the coarse-grained or averaged densities ˆa(r, t) are given by
∂ρ
∂t
+∇· g = 0, (A5.1.48)
∂g
i
∂t
+
j
∇
j
σ
ij
= 0, (A5.1.49)
∂e
∂t
+∇· j
e
= 0. (A5.1.50)
268 Appendix to Chapter 5
Now, using these equations of motion in eqn. (A5.1.36), we obtain
∂ S(t)
∂t
=
dr
{a}
α
a
(r, t)
∂
∂t
a(r, t)
=−
dr
⎡
⎣
−β
μ −
1
2
mv
2
∇ · g −
i, j
(βv
i
)∇
j
σ
R
ij
+ β ∇ · j
R
e
⎤
⎦
−
dr
⎡
⎣
−
i, j
(βv
i
)∇
j
σ
D
ij
+ β ∇ · j
D
e
⎤
⎦
. (A5.1.51)
Now the first integral on the RHS has already been computed in the steps between eqns.
(A5.1.36) and (A5.1.45). Using this, we obtain
d
dt
s(r, t)dr ≡
dr
∂s
∂t
+∇· (sv)
=
dr
⎡
⎣
i, j
(βv
i
)∇
j
σ
D
ij
− β ∇ · j
D
e
⎤
⎦
. (A5.1.52)
The entropy-production rate R
E
defined in eqn. (A5.1.43) in the presence of the dissipative
effects is given by
R
E
=−
dr
⎡
⎣
i, j
(∇
j
v
i
)σ
D
ij
+
∇T
T
· j
D
e
⎤
⎦
, (A5.1.53)
ignoring terms O(v
3
) and terms of higher order in gradients of T or v. Now, in order that
R
E
≥ 0 in the irreversible process, the dissipative currents must have the following forms
to ensure positive definiteness:
σ
D
ij
=−η
ijkl
∇
k
v
l
, (A5.1.54)
!
j
D
e
"
i
=−λ
ij
∇
j
T, (A5.1.55)
where the positive constants η
ijkl
and λ
ij
represent the viscosity and the thermal-
conductivity tensors, respectively. Note that no cross term is present in the relation above
linking the dissipative part of the stress tensor to ∇T or the heat current to the ∇
i
v
j
.This
is linked to the fundamental property of the equations of motion, since the dissipative part
breaks the time-reversal symmetry of the equations of motion.
A5.2 The second fluctuation–dissipation relation 269
A5.2 The second fluctuation–dissipation relation
Let us consider the linear Langevin equation for the variable
ˆ
φ,
∂
∂t
ˆ
φ
i
(t) + i K
(d)
ij
ˆ
φ
j
+
t
0
d
¯
t K
(d)
ij
ˆ
φ
j
(
¯
t) = f
i
(t), (A5.2.1)
where, as discussed in Section 5.3, the full memory function K
ij
has been divided into a
static and a dynamic part, denoted with superscripts (s) and (d), respectively,
K
ij
(z) = K
(s)
ij
+ K
(d)
ij
(z). (A5.2.2)
Here we obtain a relation between the dynamic part of the memory kernel K
(d)
ij
and the
correlation of the noise f
i
(Mazenko, 2006). To obtain this so-called second fluctuation–
dissipation relation we make use of the following identity for the solution
ˆ
φ
i
of the Langevin
equation:
ˆ
φ
i
(t) = C
ik
(t)S
−1
kl
ˆ
φ
l
+
t
0
ds C
ik
(t − s)S
−1
kl
f
l
(s). (A5.2.3)
We establish the result (A5.2.3) by starting from the Laplace transform of the Langevin
equation (A5.2.1) in the form (5.3.16). On the other hand, from eqn. (5.3.23) it follows
directly that
zδ
lj
− K
lj
= S
lm
C
−1
mj
(z). (A5.2.4)
Using the above relation to eliminate the memory function K
ij
from (5.3.16), we obtain
S
lm
C
−1
mj
(z)
ˆ
φ
j
(z) =[zδ
lj
− K
lj
]
ˆ
φ
j
(z) =
ˆ
φ
l
+if
l
(z). (A5.2.5)
On multiplying the above equation by C
ik
S
−1
kl
from the left and summing over the repeated
indices l and k, we obtain the result
ˆ
φ
i
(z) = C
ik
(z)S
−1
kl
{
ˆ
φ
l
+if
l
(s)}. (A5.2.6)
On taking the inverse Laplace transform of (A5.2.6) the result (A5.2.3) follows.Now,in
order to establish the second FDT (5.3.48), we rewrite the Langevin equation (A5.2.1) in
the form
H
i
(t) +
t
0
d
¯
t K
(d)
ij
(t −
¯
t)
ˆ
φ
j
(
¯
t) = f
i
(t), (A5.2.7)
where we define H
i
(t) in terms of the static part K
(s)
ij
of the memory function,
H
i
(t) =
∂
ˆ
φ
i
∂t
+i K
(s)
ij
ˆ
φ
j
. (A5.2.8)
We multiply this equation by
ˆ
φ
j
and take the equilibrium average to obtain the relation
H
i
(t)
ˆ
φ
j
+
t
0
ds K
(d)
ik
(t − s)C
kj
(s) = 0, (A5.2.9)
270 Appendix to Chapter 5
since
ˆ
φ
j
is orthogonal to the noise f
i
(t). Next, using time translational invariance of
equilibrium averages, we obtain that
H
i
(t)
ˆ
φ
j
≡H
i
ˆ
φ
j
(−t), (A5.2.10)
where H
i
denotes the value of H
i
(t) at t = 0. From the Langevin equation (A5.2.7) it
follows that H
i
= f
i
. Now we compute the RHS of (A5.2.10) by using (A5.2.3) for
ˆ
φ
j
(−t)
and obtain
H
i
(t)
ˆ
φ
j
=
-
f
i
3
C
jk
(−t)S
−1
kl
ˆ
φ
l
+
−t
0
ds C
jk
(−t − s)S
−1
kl
f
l
(s)
4.
. (A5.2.11)
The contribution from the first term in the curly brackets on the RHS gives zero since
f
i
ˆ
φ
l
=0. With a change of variables s →−s in the second term on the RHS, the above
relation reduces to
H
i
(t)
ˆ
φ
j
=−
t
0
dsf
i
f
l
(−s)C
jk
(s −t )S
−1
kl
=−
t
0
dy f
i
(t − y) f
l
C
jk
(−y)S
−1
kl
. (A5.2.12)
In obtaining the last equality we have made a change of variable s → t − y and used the
time translational invariance of the equilibrium correlation functions
f
i
f
l
(−s)=f
i
(s) f
l
. (A5.2.13)
Similarly, on applying the same time translational invariance for the equilibrium correlation
of the slow modes we have
C
jk
(−y) = C
kj
(y). (A5.2.14)
The equal-time correlation matrix S satisfies
S
−1
kl
= S
−1
lk
. (A5.2.15)
Using these two results, (A5.2.12) reduces to the form
H
i
(t)
ˆ
φ
j
=−
t
0
dy f
i
(t − y) f
l
S
−1
lk
C
kj
(y). (A5.2.16)
On comparing the relations (A5.2.9) and (A5.2.16) we obtain the so-called second
fluctuation–dissipation relation between the noise correlation and the dynamic part of the
memory function,
f
i
(t) f
j
(t
)=K
(d)
ik
(t − t
)S
kj
. (A5.2.17)