Bhattacharjee J.K., Bhattacharyya S. Non-Linear Dynamics Near and Far from Equilibrium
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8.3 The Correlation Function 243
We now expand C
ij k
as the sum of 8 terms and note that two of them vanish since
v
i
(x)v
j
(x)v
k
(x)cannot be expressed in terms of δ
ij
. Further, v
i
(x)v
j
(x)v
k
(x +
r) is the negative of v
i
(x +r)v
j
(x +r)v
k
(x) and consequently,
C
ij k
= 2[v
i
(x)v
j
(x)v
k
(x +r)+v
i
(x)v
k
(x)v
j
(x +r)
+v
k
(x)v
j
(x)v
i
(x +r)]
=−2(rC
+C)(δ
ij
n
k
+δ
ik
n
j
+δ
jk
n
i
) +6(rC
−C)n
i
n
j
n
k
(8.3.23)
where use has been made of Eqns.(8.3.17), (8.3.21) and (8.3.12) in arriving at the
last line. The different components of C
ij k
can be written down from Eq.(8.3.23)
as
C
rrr
=−12C
C
rtt
=−2(C +rC
)
C
trt
= C
ttt
=0
C
rtt
=
1
6
d
dr
(rC
rrr
) (8.3.24)
To proceed further, we need to use the Navier-Stokes equation. We start with the
two point function and take a time derivative to write
∂
t
v
i
(x)v
j
(x +r)=˙v
i
(x)v
j
(x +r)+v
i
(x)˙v
j
(x +r)
=−v
l
∂
l
v
i
(x)v
j
(x +r)−v
i
(x)v
l
(x +r)∂
l
v
j
(x +r)
+νv
j
(x +r)∇
2
v
i
(x)+νv
i
(x)∇
2
v
j
(x +r)
−v
j
(x +r)∂
i
P(x)−v
i
(x)∂
j
P(x +r) (8.3.25)
The term v
j
(x +r)∂
i
P(x) can be written as ∂
i
v
j
(x +r)P(x). The quantities
v
j
(x +r)P(x) are the components of a divergence free vector and since there
is no cetro-symmetric divergence free vector which is finite at the origin, we infer
v
j
(x +r)P(x)=0. Using translational invariance, whenever needed, we can
then write Eq.(8.3.25) as
∂
t
v
i
(x)v
j
(x +r)=−2∂
l
v
i
(x)v
l
(x)v
j
(x +r)
+2ν∇
2
v
i
(x)v
j
(x +r) (8.3.26)
We now use
v
i
(x)v
j
(x +r)=
1
3
v
2
δ
ij
−
1
2
C
ij
(8.3.27)
and
v
i
(x)v
l
(x)v
j
(x +r)=−
1
12
C
rrr
δ
il
n
j
+
1
12
r
2
C
rrr
+C
rrr
(δ
ij
n
l
+δljn
i
)
−
1
12
(rC
rrr
−C
rrr
)n
i
n
j
n
l
(8.3.28)
244 8 Turbulence
and work with i =j =l =r (i.e. the longitudinal component) to write
−
2
3
¯ −
1
2
C
rr
=
1
6r
4
∂
∂r
(r
4
C
rrr
) −
ν
r
4
∂
∂r
r
4
∂C
rr
∂r
(8.3.29)
where ¯ =∂
t
(
v
2
2
). For small values of r, We can safely drop C
rr
with respect to ¯
and we have the result,
C
rrr
=−
4
5
¯r +6ν
d
dr
C
rr
(8.3.30)
Being in the inertial range is equivalent to setting ν =0 and then
C
rrr
=−
4
5
¯r (8.3.31)
which is the result that would be anticipated on the basis of Kolmogorov’s picture
(Eq.(8.2.3)), with C
3
=−
4
5
. This is the only exact inertial range result in three
dimensional turbulence.
What if we were in the dissipation range? We anticipate that C
rrr
would vanish
faster than r for small r in the dissipation range and hence
C
rrr
=
1
15
¯
ν
r
2
(8.3.32)
for very small r. This is in accordance with Eq.(8.3.15) derived from a different
standpoint.
We end this section by noting the relation between the Fourier transform of
v
i
(
k)v
j
(−
k) and C
ij
(r)
C
ij
(r) = 2v
j
(x)
2
−2v
i
(x)v
j
(x +r)
= 2
d
D
k
(2π)
D
v
j
(
k)v
j
(−
k)−2
d
D
k
(2π)
D
e
i
k.x
v
j
(
k)v
j
(−
k)
= 2
d
D
k
(2π)
D
(1 −e
i
k.x
)v
j
(
k)v
j
(−
k) (8.3.33)
If C
ij
∝r
α
, then the dimensionally consistent v
j
(
k)v
j
(−
k) is proportional
to k
−α−D
.IfC
ij
(r) ∼r
2/3
as in the inertial range according to Kolmogorov,
v
j
(
k)v
j
(−
k)∼k
−11/3
. The resulting integral in Eq.(8.3.33) is convergent in both
the upper and lower ends of k. On the other hand if C
ij
(r) ∼r
2
as is true in the
dissipative range, then v
j
(
k)v
j
(−
k)would be proportional to k
−5
from a dimen-
sional argument. However, it is clear from Eq.(8.3.33) that v
j
(
k)v
j
(−
k)∼k
−5
does not give a convergent integral at the small k limit. This gives the indication
8.4 Randomly Stirred Model 245
that the momentum space correlation is scale dependent and we bring in a scale k
0
to write in the dissipation range
v
j
(
k)v
j
(−
k)=
C
0
k
n
e
−k/k
0
(8.3.34)
with n ≤5. For n =1, the Fourier transform gives
C
jj
(r) =
8π
(2π)
3
C
0
k
2
0
r
2
1 +k
2
0
r
2
c
0
π
2
k
2
0
r
2
(8.3.35)
when k
0
r 1. If we set k
0
equal to the dissipation scale s
−1
then
k
0
=
¯
1/4
ν
3/4
and comparing with Eq(8.3.32)
C
0
=
¯
15ν
π
2
ν
3
/2
¯
1/2
=
π
2
15
(ν ¯)
1/2
(8.3.36)
8.4 Randomly Stirred Model
It is customary to split the velocity field into a mean part and a fluctuating part as
v =u +
V (8.4.1)
where v=V and u is the fluctuation. Inserting this in the Navier-Stokes equation
(no external force) and taking an average we have
∂
V
∂t
+(
V.
∇)
V =−
∇P
0
+ν∇
2
V −(u.
∇)u (8.4.2)
where P
0
is the average part of the pressure term. Subtracting Eq(8.4.2) from the
full Navier-Stokes equation leads to
∂ u
∂t
+(u.
∇)u =−
∇P +ν∇
2
u −(
V.
∇)u
−(u.
∇)u +(u.
∇)u (8.4.3)
The mean flow as expressed in Eq.(8.4.2) is affected by the fluctuating flow through
the term (u.
∇)u. One of the standard techniques of obtaining the mean flow is
to make an Ansatz about (u.
∇)u, e.g Prandtl’s mixing length hypothesis, and
246 8 Turbulence
solving for
V from Eq.(8.4.2). It is then possible to make a consistency check from
Eq.(8.4.3). Our interest, here, however is to note that Eq.(8.4.3) can be written as
∂ u
∂t
+(u.
∇)u =−
∇P +ν∇
2
u +
f (8.4.4)
where
f is a force which expresses the interaction between the fluctuating and
the mean flow. There is a definite expression for this force as can be seen from
Eq.(8.4.3), however, one can take the view that since f is virtually impossible to
know from first principles we can replace it by a stochastic force (considering that
the velocity field u is random) with a prescribed correlation. We thus write the
correlator of f as
f
i
(x,t)f
j
(y,t)=C
(L)
ij
(x −y)δ(t −s) (8.4.5)
The delta function expresses the fact that the noise is temporally white. The spatial
function C
L
ij
(x −y) must be such that it must correctly portray the injection of
energy into the length scale L.
We want to explore in this model the consequences of stationarity, i.e. the fact
that the correlation function u
i
(x,t)u
j
(y,t) is independent of time. To do this
we need to set the derivative of u
i
(x,t)u
j
(y,t) equal to zero. We note that
u
i
(x,t +δt) = u
i
(x,t)+
t+δt
t
(−u
l
∂
l
u
i
(x,s) −∂
i
P +ν∇
2
u
i
)ds
+
t+δt
t
f
i
(x,s)ds
= u
i
(t) +[−u
l
∂
l
u
i
(x,t)−∂
i
P +ν∇
2
u
i
]δt
+
t+δt
t
f
i
(x,s)ds +O((δt)
2
) (8.4.6)
We note that f
i
is O((δt)
−1/2
) because of Eq.(8.4.5). The stationarity condition
then takes the form (upto O(δt))
0 =−u
l
(x,t)∂
l
u
i
(x,t)u
i
(y,t)−∂
i
P(x, t)u
i
(x,t)
+νu
i
(y,t)∇
2
u
i
(x,t)+(x ↔y)
+2u
i
(t)
t+δt
t
f
i
(s)ds+
1
δt
t+δt
t
f
i
(x,s)f
i
(y,s
)dsds
(8.4.7)
8.4 Randomly Stirred Model 247
Now
1
δt
t+δt
t
f
i
(x,s)f
i
(y,s
)dsds
=
1
δt
t+δt
t
C
(L)
ii
(x −y)δ(s −s
)
1
δt
t+δt
t
TrC
(L)
(x −y)ds
= TrC
(L)
(x −y) (8.4.8)
t+δt
t
u
i
(x,t)f
i
(y,s)ds =0 (8.4.9)
because s is always greater than t and hence by causality the relation vanishes.
Because of translation invariance and divergence free condition
∂
i
P(x, t)u
i
(x,t)=0 (8.4.10)
u
i
(x,t)∇
2
u
i
(y,t)=−∂
j
u
i
(x,t)∂
j
u
i
(y,t) (8.4.11)
u
j
(x,t)∂
j
u
i
(x,t)u
i
(y,t)+x ↔y =−
1
4
∇
x
(u
i
(x,t)−u
i
(y,t))
×(u(x,t)−u(y, t))
2
(8.4.12)
With the above identities, Eq.(8.4.7) becomes
1
2
TrC
(L)
(x −y) =−
1
4
∇
x
(u
i
(x,t)−u
i
(y,t))[u(x, t) −u(y, t)]
2
+ν∂
i
u
i
(x,t)∂
j
u
j
(x,t) (8.4.13)
We now can take the limits ν →0 and x →y in twodifferentorders. If ν is held fixed
and the limit x →y taken, then the first term on the left hand side of Eq.(8.4.13)
vanishes and we have,
1
2
TrC
(L)
(0) = lim
x→y
ν→0
ν(∂
i
u
j
(x)
2
=¯ (8.4.14)
If we take the limit of ν →0 first and then consider x →y
−
1
4
∇
x
(u
i
(x,t)−u
i
(y,t))[u(x, t) −u(y, t)]
2
=¯ (8.4.15)
We now invoke isotropy to write,
−
4
D +2
¯[δ
ij
x
k
+δ
jk
x
i
+δ
ki
x
j
]=[u
i
(x,t)−u
i
(
0,t)][u
j
(x,t)−u
j
(
0,t)]
×[u
k
(x,t)−u
k
(
0,t)] (8.4.16)
248 8 Turbulence
from which it follows that
[{u(x, t)−u(
0,t)}.
ˆ
x]
3
=−
12
D(D +2)
¯r (8.4.17)
The above result is valid for D ≥3 and in D =3, gives the usual result
S
3
=−
4
5
¯r.
The important thing to note in this derivation is the anomaly that the operator
lim
x→y,ν→0
∇
x
.[u(x) −u(y))[u(x)−u(y)]
2
]
is non-zero. Had u(x) been a differentiable field then this term would have been
zero.
This kind of anomaly occurs for arbitrary correlation functions
n
F
n
(u(x
n
))
where F
n
(u(x
n
)) are arbitrary functions of u(x
n
) without any derivatives. We work
out the case of n =2 in detail. The stationarity condition for a two point function
is
∂
t
F
1
(u(x
1
))F
2
(u(x
2
))=0 (8.4.18)
Noting that ∂
t
F (u(x)) =(∂
t
u).δF (u) by the chain rule we have
∂
t
F
1
(u(x
1
))F
2
(u(x
2
))+x
1
↔x
2
=0
or
0 =∂
t
u
j
(x
1
)∂
j
F
1
(u(x
1
))F
2
(u(x
2
))+x
1
↔x
2
= [−∂
l
(u
l
(x
1
)u
j
(x
1
) −∂
j
P(x
1
) +ν∇
2
u
j
(x
1
)]∂
j
F
1
(u(x
1
))F
2
(u(x
2
))
+f
j
∂
j
F
1
(u(x
1
))F
2
(u(x
2
))+(x
1
↔x
2
) (8.4.19)
The first two terms represent correlation functions which are well defined in the
inviscid limit. The viscosity involving term which would naively vanish in the
ν →0 limit is actually nonzero due to the dissipative anomaly. Using translational
invariance, we can write
ν∇
2
∂
j
F
1
[u(x
1
)]F
2
[u(x
2
)] = ν∂
kx
1
[∂
j
F
1
[u(x
1
)∂
k
u
j
(x
1
)]F
2
(u(x
2
))
−ν∂
k
u
j
(x
1
)∂
kx
1
∂
j
F
1
[u(x
1
)]F
2
[u(x
2
)]
= ν∂
kx
1
∂
k
F
1
[u(x
1
)F
2
[u(x
2
)]
−ν∂
k
u
j
(x
1
)∂
k
u
l
(x
1
)∂
j
∂
l
F
1
[u(x
1
)]F
2
[u(x
2
)]
=−ν∂
k
F
1
[u(x
1
)]∂
k
F
2
[u(x
2
)]
−
lj
(x
1
)∂
l
∂
j
F
1
[u(x
1
)]F
2
[u(x
2
)]
=−ν∇
k
x
1
∇
k
x
2
F
1
[u(x
1
)]F
2
[u(x
2
)]
−
lj
(x
1
)∂
l
∂
j
F
1
[u(x
1
)]F
2
[u(x
2
)] (8.4.20)
8.4 Randomly Stirred Model 249
where
lj
=ν∂
k
u
j
(x)∂
k
u
l
(x) (8.4.21)
In the inviscid limit, the correlation F
1
[u(x
1
)]F
2
[u(x
2
)] is finite and hence
the derivatives are finite at ν =0. Consequently, the first term on the R.H.S of
Eq.(8.4.20) vanishes in the inviscid limit and we have
lim
ν→0
ν∇
2
u
j
(x
1
)∂
j
F
1
[u(x
1
)]F
2
[u(x
2
)] = −
lj
(x
1
)∂
l
∂
j
F
1
[u(x
1
)]F
2
[u(x
2
)]
(8.4.22)
Finally we need to evaluate the correlation function involving the stochastic force.
To do this we note that
f
i
(x)K(u(y))=
d
3
zC
ij
(x −z)
δK(u(y))
δf
j
(z)
(8.4.23)
for any functional of G, when we integrate by parts and use the fact that the force
is Gaussian. To evaluate
δK(u(y))
δf
j
(z)
, we really need to evaluate
δu
k
(y,t )
δf
j
(x,s)
in the equal
time limit t =s. This derivative is zero if t<s. Starting from the Navier-Stokes
equation and considering a differential δu
k
caused by a differential δf
j
,wehave
∂
∂t
+L(u(y))
δ(u
k
(y, t))
δf
j
(x, s)
=δ
jk
δ(y −x)δ(t −s) (8.4.24)
where L is some differential operator depending on u. From Eq.(8.4.24), it is clear
that
δu
k
(y, t)
δf
j
(x, s)
=θ(t −s)G
jk
(y, x|t,s) (8.4.25)
where G
jk
(y, x|t,s) is the solution of
∂
∂t
+L(u(y))
G
jk
(y, x|t,s)=0
with the initial condition that at t =s,
G
jk
(y, x|t,s)=δ
jk
δ(y −x).
250 8 Turbulence
Taking the equal time limit Eq.(8.4.25) and setting θ(t →s)=
1
2
,wehave
lim
t→s
δu
k
(y, t)
δf
j
(x, s)
=
1
2
δ
jk
δ(x −y) (8.4.26)
From Eq.(8.4.23), we then have
f
i
(x)K(u(y))=
d
3
zC
ij
(x −z)
∂K(y)
∂u
k
∂u
K
(y)
∂f
j
(z)
=
d
3
z
∂K(y)
∂u
k
C
ij
(x −z)
1
2
δ
jk
δ(y −z)
=
1
2
C
ij
(x −y)
∂K(y)
∂u
k
(8.4.27)
Putting the information of the Eqns.(8.4.27) and (8.4.22) into the stationarity
condition of Eq.(8.4.19), we have after a little rearrangement
{(u.
∇)u
j
+
∇p}(x)
∂
∂u
j
+
ij
∂
2
∂u
i
∂u
j
F
1
[u(x
1
)]F
2
[u(x
2
)]
=
1
2
C
ij
( x
1
−x
2
)
∂
2
∂u
i
∂u
j
F
1
[u(x
1
)]F
2
[u(x
2
)]
(8.4.28)
For the n-point correlation function
n
F
n
(u(x
n
)), we need to replace F
1
[u(x
1
)]
F
2
[u(x
2
)] by
n
F
n
(u(x
n
)).
We end this section by exhibiting the long distance and short distance correla-
tions for a driven diffusion equation, i.e. a diffusing scalar field φ(x,t) driven by
a stochastic noise f, such that
∂φ
∂t
=D
1
∇
2
φ +f (8.4.29)
In momentum space, we have
∂
∂t
φ(k,t)+D
1
k
2
φ(k,t)=f(k) (8.4.30)
with the solution
φ(k,t)=
dt
e
−D
1
k
2
(t−t
)
f(k,t
) (8.4.31)
8.4 Randomly Stirred Model 251
The two point correlation function is given by
φ(k,t
1
)φ(−k, t
2
)=
t
1
−∞
dt
t
2
−∞
dt
e
−D
1
k
2
(t
1
−t
)
×e
−D
1
k
2
(t
2
−t
)
f(k,t
)f (−k, t
)
=
dt
dt
C(k)δ(t
−t
)e
−D
1
k
2
(t
1
+t
2
)
e
D
1
k
2
(t
+t
)
=
t
2
−∞
C(k)e
−D
1
k
2
(t
1
+t
2
)
e
2D
1
k
2
t
=
C(k)
2D
1
k
2
e
−D
1
k
2
|t
1
−t
2
|
(8.4.32)
If we are to form the spatial correlation
φ(x
1
,t
1
)φ(x
2
,t
2
)=
d
D
k
(2π)
D
C(k)
2D
1
k
2
e
−D
1
k
2
|t
1
−t
2
|
e
i
k.(x
1
−x
2
)
(8.4.33)
Equal time correlation function is
φ(x
1
, t)φ(x
2
,t)=
d
D
k
(2π)
D
C(k)
2D
1
k
2
e
i
k.(x
1
−x
2
)
If large |x
1
−x
2
)| behaviour is required, then we look for the k-part alone and
φ(x
1
, t)φ(x
2
,t)∼
C(0)
2D
1
d
D
k
(2π)
D
e
i
k.(x
1
−x
2
)
k
2
|x
1
−x
2
|
2−D
(8.4.34)
Now suppose we want to extract the small distance behaviour from Eq.(8.4.33).
We then expand the exponential and have
φ(x
1
, t)φ(x
2
,t)=
d
D
k
(2π)
D
C(k)
2D
1
k
2
1 −D
1
k
2
|t
1
−t
2
|−
k
2
2D
(x
1
−x
2
)
2
=
d
D
k
(2π)
D
C(k)
2D
1
k
2
−
d
D
k
(2π)
D
C(k)
2D
1
×
D
1
|t
1
−t
2
|+
(x
1
−x
2
)
2
2D
=
d
D
k
(2π)
D
C(k)
2D
1
k
2
−
C(r =0)
2D
1
D
1
|t
1
−t
2
|+
(x
1
−x
2
)
2
2D
(8.4.35)
The first term is a diverging constant and hence the scaling shown in the second
term is exhibited only by the difference variable φ(x
1
,t)−φ(x
2
,t). This is the
typical turbulence related problem - the presence of a blowing up constant mode
which may be eliminated only by considering field differences.
252 8 Turbulence
8.5 Advection of a Passive Scalar
We have seen that the central problem in the theory of homogeneous isotropic
turbulence is the determination of the scaling exponents ζ
n
associated with the
structure functions as shown in Eq.(8.2.7). These exponents differ from the Kol-
mogorov assertion which was based on purely dynamical grounds. To obtain these
non-trivial exponents, we wrote down the most general relation between various
correlation functions in Eq.(8.4.28). However, apart from n =3, where it is possi-
ble to obtain an exact answer as shown in Eq.(8.4.17), it is still not practical to use
these equations. In this section we will treat a much simpler problem and explicitly
show how the anomalous scaling arises. This is the problem of advection of a scalar
field φ by a turbulent velocity field v. The equation of motion is,
∂φ
∂t
+(v.
∇)φ =κ∇
2
φ +f (8.5.1)
where κ is the diffusion coefficient and random force f has the correlation
f(x,t)f(y,s)=C(
x −y
L
)δ(t −s) (8.5.2)
The random velocity field has the correlation
v
α
(x,t)v
β
(y,s)=D
αβ
(x −y)δ(t −s) (8.5.3)
with ∂
α
D
αβ
=0. An explicit form for D
αβ
that we choose is
D
αβ
=D
0
δ
αβ
−d
αβ
(8.5.4)
and
d
αβ
=D
0
(D +ζ −1)δ
αβ
−ζ
r
α
r
β
r
2
r
ζ
(8.5.5)
where ζ is a parameter which is less than 2. A more rigorous form for D
αβ
(r) is
D
αβ
(r) =d
0
d
D
k
(2π)
D
e
i
k.r
(k
2
+m
2
)
(D+ξ)/2
δ
αβ
−
k
α
k
β
k
2
(8.5.6)
We will work with the form in Eq.(8.5.5) which is more convenient. The para-
meter ζ fixes the naive dimension under the rescaling x →µx and t →µ
z
t. From
Eq.(8.5.5), d
αβ
(µr) =µ
ζ
d
αβ
(r) and hence from Eq.(8.5.3), the scaling of the
velocity field by the factor µ
ζ
2
−
z
2
. From Eq.(8.5.2), we note that f rescales as
1/
√
T . In, Eq.(8.5.1), comparing the terms on the L.H.S, we see that T rescales as
µ
2−ζ
and comparing
∂φ
∂t
with f, we have φ rescaling as µ
1−ζ/2
. We note that the
Kolmogorov velocity field satisfies (δv
r
)
2
∼r
2/3
∼r
4/3
/τ
r
. This is to be com-
pared with (δv
r
)
2
∼r
ζ
δ(t) in the above model. Hence in the above ζ =4 /3is