Bhattacharjee J.K., Bhattacharyya S. Non-Linear Dynamics Near and Far from Equilibrium
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3.2 Renormalization Group: General Framework 61
• i) elimination of the fast variables, φ(k) with
b
<k<
• ii) rescaling of k to k
=bk
• iii) rescaling of φ to φ
=φ/ζ
We now implement the steps for the Gaussian model. To do so it helps to write the
free energy in the Fourier space. This gives
F =
1
2
(m
2
+k
2
)φ(k)φ(−k)d
D
k (3.2.3)
and the partition function is
Z =
dφ(k)e
−
1
2
(m
2
+k
2
)φ(k)φ(−k)d
D
k
=
dφ(k)e
−
1
2
/b
k=0
(m
2
+k
2
)φ
<
(k)φ
<
(−k)d
D
k
×e
−
1
2
k=/b
(m
2
+k
2
)φ
>
(k)φ
>
(−k)d
D
k
=
dφ
<
(k) e
−
1
2
/b
k=0
(m
2
+k
2
)φ
<
(k)φ
<
(−k)d
D
k
×
dφ
>
(k) e
−
1
2
k=/b
(m
2
+k
2
)φ
>
(k)φ
>
(−k)d
D
k
= C
0
dφ
<
(k) e
−
1
2
/b
k=0
(m
2
+k
2
)φ
<
(k)φ
<
(−k)d
D
k
= C
0
dφ
<
(k
)e
−
1
2
k
=0
(m
2
+
k
2
b
2
)φ
<
(
k
b
)φ
<
(−
k
b
)
d
D
k
b
D
= C
0
dφ
<
(k
)e
−
1
2
k
=0
(m
2
+
k
2
b
2
)ζ
2
φ
<
(k
)φ
<
(−k
)
d
D
k
b
D
= C
0
dφ
<
(k
)e
−
1
2
k
=0
(m
2
+
k
2
ζ
2
b
2+D
)φ
<
(k
)φ
<
(−k
)d
D
k
(3.2.4)
with the definition
φ(k)=φ
<
(k) +φ
>
(k)
where
φ
<
(k) = φ(k) for 0 ≤k<
b
= 0 for
b
≤k<
and
φ
>
(k) = 0 for 0 ≤k<
b
= φ(k) for
b
≤k<.
62 3 The Renormalization Group
Now
m
2
=
ζ
2
m
2
b
D
(3.2.5)
and
ζ
2
=b
2+D
(3.2.6)
to keep the coefficient of k
2
φ(k)φ(−k) at 1/2. Thus the recursion relation for m
becomes
m
2
=b
2
m
2
(3.2.7)
Clearly, m
2
is relevant, since b
2
> 1 and for criticality m
2
=0 - this is consistent
with m
2
∼(T −T
c
). The rescaling of momentum scale by b, means that
ξ(m
) =ξ(m)/b (3.2.8)
or
ξ(m) =bξ(bm) =b
2
ξ(b
2
m) =..... =b
n
ξ(b
n
m) (3.2.9)
Choosing b
n
m =1
ξ(m) =
1
m
ξ(1)
= ξ
0
T −T
c
T
c
−1/2
(3.2.10)
giving the critical exponent ν =1/2.
Are terms like (∇
2
φ)
2
, [∇
2
(
∇φ)]
2
etc. in the free energy relevant? In momen-
tum space these terms contribute k
2n
φ(k)φ(−k) to the free energy density, with
n =2, 3.... For a term K
n
k
2n
φ(k)φ(−k) in Eq.(3.2.3) we get K
n
ζ
2
/b
2n+D
as the
coefficient of k
2n
φ
<
(k
)φ
<
(−k
) after the steps of the RGT and hence
K
n
=K
n
ζ
2
/b
2n+D
=K
n
b
2(1−n)
(3.2.11)
Clearly, for n ≥2, all couplings K
n
are irrelevant.
We now look at the couplings of the form u
n
φ
2n
d
D
x in F for n ≥2. In
momentum space this term in F has the form
u
n
φ(p
1
)φ(p
2
).......φ(−p
1
−p
2
....... −p
2n−1
)d
D
p
1
d
D
p
2
........d
D
p
2n−1
.
Under the transformations described above this term becomes
u
n
ζ
2n
b
(2n−1)D
φ
<
(p
1
)φ
<
(p
2
).......φ
<
(−p
1
−p
2
....... −p
2n−1
)d
D
p
1
d
D
p
2
........d
D
p
2n−1
.
3.2 Renormalization Group: General Framework 63
Thus the coupling constant u
n
goes to u
n
, where
u
n
=
ζ
2n
b
(2n−1)D
u
n
=
b
2n+nD
b
(
2n −1)D
u
n
= b
2n+(1−n)D
u
n
(3.2.12)
The exponent is positive if (this makes u
n
relevant)
D<
2n
n −1
(3.2.13)
For n =2, u
2
is relevant for D<4, while for n =3, u
3
is relevant for D<3 and
so on. Consequently, if we are interested in critical phenomena in D =3, the only
relevant perturbation to the Gaussian model of Eq.(3.2.1) is a term
φ
4
d
D
x. (it
is easy to check that terms of the form
(
∇φ
2
)
2
d
D
x etc. where derivatives are
included will be irrelevant). This leads to the standard model for treating critical
phenomena
F =
d
D
x
1
2
m
2
φ
2
+
1
2
(
∇φ)
2
+
λ
4
φ
4
(3.2.14)
The performance of the RGT on the Hamiltonian F , leads to renormalized m
and λ. For dimensions D>4, λ is irrelevant and the Gaussian model holds. For
dimensions D<4, the φ
4
term is relevant and a new fixed point is reached, where
a first order in λ calculation yields
ν =
1
2
+
12
(3.2.15)
We will not provide a full derivation of Eq.(3.2.15) which can be found in a large
number of texts, butwill content ourselves with outlining how the mode elimination
is done. This is done perturbatively by writing
φ =φ
<
+φ
>
(3.2.16)
Where φ
<
contains wavenumbers from 0 to /b and φ
>
contains wavenumbers
from /b to
Z =
D[φ
<
]D[φ
>
]e
−
d
D
x[
m
2
2
(φ
<
+φ
>
)
2
+
1
2
(
∇φ
<
+
∇φ
>
)
2
+
λ
4
(φ
<
+φ
>
)
4
]
=
dφ
<
(k)dφ
>
(k) exp[−
1
2
d
D
k(k
2
+m
2
)φ
<
(k)φ
<
(−k)]
× exp[−
λ
4
d
D
k
1
d
D
k
2
d
D
k
3
φ
<
(k
1
)φ
<
(k
2
)φ
<
(k
3
)φ
<
(−k
1
−k
2
−k
3
)]
× exp[−
1
2
d
D
k(k
2
+m
2
)φ
>
(k)φ
>
(−k)]×exp
−
λ
4
d
D
k
1
d
D
k
2
d
D
k
3
64 3 The Renormalization Group
×{4φ
<
(k
1
)φ
<
(k
2
)φ
<
(k
3
)φ
>
(−k
1
−k
2
−k
3
) +4φ
<
(k
1
)φ
>
(k
2
)
× φ
>
(k
3
)φ
>
(−k
1
−k
2
−k
3
) +6φ
<
(k
1
)φ
<
(k
2
)φ
>
(k
3
)
× φ
>
(−k
1
−k
2
−k
3
) +φ
>
(k
1
)φ
>
(k
2
)φ
>
(k
3
)φ
<
(−k
1
−k
2
−k
3
)}
=
dφ
<
(k)e
−F
<
dφ
>
(k)e
−F
0
>
e
−λF
I
(φ
<
,φ
>
)
(3.2.17)
The strategy is to separate the Hamiltonian F
<
which contains only the fields
φ
<
from the rest where φ
>
appears. In the part which contains φ
>
, we split the
Hamiltonian into the Gaussian part F
0
>
and a part λF
I
(φ
<
,φ
>
), which involves
both φ
<
and φ
>
or φ
>
alone in a non-quadratic form. The perturbative calculation
proceeds by expanding
e
−λF
I
=1 −λF
I
+
λ
2
2
F
2
I
+..... (3.2.18)
One now performs the integration over the fields φ
>
, and exponentiating the result
ends up with a contribution of the form exp(−
˜
F
<
(φ
<
)). This casts Eq.(3.2.18) in
the form
Z =
D[φ
<
]exp
−(F
<
+
˜
F
<
)
Rescaling of k and φ paves the way to the recursion relations.
3.3 Dynamics of Model A
When dealing with dynamics there is the additional variable which is time and when
differentiating between fast and slow components of the Fourier field φ(x,t),we
need to think about the double Fourier transform
φ(x,t) =
1
(2π)
D/2
1
√
2π
φ(
k, ω)e
i(
k.x−ωt)
d
D
kdω (3.3.1)
The fast variables are now the high momentum, high frequency components and
the slow variables are the low frequency, low momentum components. For the
momentum variable, as before, we introduce the quantity b(> 1) in terms of which,
we define the fast range as
b
<k< and the slow range is 0 <k<
b
. For the
frequency variable, we need to define a similar variable ω
b
. If the full frequency
range is 0 to , then the slow variable range is 0 <ω<
ω
b
and the fast variable
range is
ω
b
<ω<. The parameter ω
b
is the additional one for the dynamics and
its dependence on b is specified as
ω
b
∝b
z
(3.3.2)
3.3 Dynamics of Model A 65
where z is called the dynamic scaling exponent which was defined from a slightly
different standpoint in Eq.(2.4.27). A considerable part of the effort in dynamic
critical phenomena revolves around the calculation of the exponent z. To explain
how the RG works in this context, we consider the trivial problem of the linear
part of model A (Eq.(2.4.19)) where by directly integrating the linear equation, we
already saw in Chapter 2, that z =2. The equation of motion for φ in the k, ω
space (Fourier transform of Eq.(2.4.19)) is
−
iω
+k
2
+m
2
φ(
k, ω)=
N(
k, ω)
(3.3.3)
The first step is equivalent to the first step in the statics (previous section)- we need
to find the effective equation of motion for the slow variable φ
<
(
k, ω). For the
linear equation of motion in Eq.(3.3.3), this is trivially done - the equation for φ
<
is
−
iω
/χ
+1
φ
<
(
k, ω)=
N
<
/χ
(3.3.4)
where χ =(k
2
+m
2
)
−1
is the static susceptibility. The next few steps are,
• i) rescale momentum k →k
=bk, so that the range of k
is back to 0 →λ
• ii) rescale frequency ω →ω
=b
z
ω, so that the range of ω
is back to 0 →
• iii) rescale field φ
<
(k, ω) to φ
=φ/ζ with ζ such that it is consistent with the
statics.
This gives the transformation law for /χ. The above rescalings lead to
−
iω
b
z
/χ
+1
φ
<
=
N
<
/ζ
/χ
(3.3.5)
To keep the scale factor of φ consistent with statics,
ζ =b
1+
D
2
+z
(3.3.6)
χ
=χb
−2
(3.3.7)
The above equation of motion becomes
−
iω
/χ
+1
φ
<
=
N
<
/χ
(3.3.8)
where
=
b
z
b
2
=b
z−2
(3.3.9)
N
=Nb
z
/ζ =Nb
z−2
(3.3.10)
66 3 The Renormalization Group
The relaxation rate renormalizes as shown in Eq.(3.3.9) and the fixed point corre-
sponds to
z =2. (3.3.11)
as expected.
We now turn to the complete model A, when the equation of motion is defined
by Eq.(2.4.2) with F given by Eq.(2.4.1). We work with n =1 i.e. a scalar order
parameter. In Fourier space, the equation of motion reads
−
iω
+k
2
+m
2
φ(k,ω) =−λ
φ( p
1
,ω
1
)φ( p
2
,ω
2
)
φ(
k −p
1
−p
2
,ω−ω
1
−ω
2
) +
N(
k, ω)
(3.3.12)
The slow variable k<
b
, ω<
b
z
satisfies
−
iω
+k
2
+m
2
φ
<
(k, ω) =−λ
φ
<
( p
1
,ω
1
)φ
<
( p
2
,ω
2
)
× φ
<
(
k −p
1
−p
2
,ω−ω
1
−ω
2
)
− 3λ
φ
<
( p
1
,ω
1
)φ
>
( p
2
,ω
2
)
× φ
>
(
k −p
1
−p
2
,ω−ω
1
−ω
2
)
− 3λ
φ
>
( p
1
,ω
1
)φ
<
( p
2
,ω
2
)
× φ
<
(
k −p
1
−p
2
,ω−ω
1
−ω
2
)
− λ
φ
>
( p
1
,ω
1
)φ
>
( p
2
,ω
2
)
× φ
>
(
k −p
1
−p
2
,ω−ω
1
−ω
2
) +
N
<
(3.3.13)
The fast variable φ
>
satisfies a similar equation with the understanding that
b
<
k<and
b
z
<ω<.Wehave
−
iω
+k
2
+m
2
φ
>
(k, ω) =−λ
φ
>
( p
1
,ω
1
)φ
>
( p
2
,ω
2
)
× φ
>
(
k −p
1
−p
2
,ω−ω
1
−ω
2
)
− 3λ
φ
<
( p
1
,ω
1
)φ
>
( p
2
,ω
2
)
× φ
>
(
k −p
1
−p
2
,ω−ω
1
−ω
2
)
− 3λ
φ
<
( p
1
,ω
1
)φ
<
( p
2
,ω
2
)
× φ
>
(
k −p
1
−p
2
,ω−ω
1
−ω
2
)
3.3 Dynamics of Model A 67
− λ
φ
<
( p
1
,ω
1
)φ
<
( p
2
,ω
2
)
× φ
<
(
k −p
1
−p
2
,ω−ω
1
−ω
2
) +
N
>
(3.3.14)
The strategy is to solve for φ
>
from Eq.(3.3.14) in terms of φ
<
and substitute into
Eq.(3.3.13) to get the equation of motion for φ
<
. The way to solve for φ
>
from
Eq.(3.3.14) is by perturbation theory. So one expands,
φ
>
=φ
(0)
>
+λφ
(1)
>
+λ
2
φ
(2)
>
+...... (3.3.15)
and at the different orders
−
iω
+k
2
+m
2
φ
(0)
>
(k, ω) =
N
>
(k, ω)
(3.3.16)
−
iω
+k
2
+m
2
φ
(1)
>
(k, ω) =−3
φ
<
( p
1
,ω
1
)φ
(0)
>
( p
2
,ω
2
)
× φ
(0)
>
(
k −p
1
−p
2
,ω−ω
1
−ω
2
)
− 3
φ
(0)
>
( p
1
,ω
1
)φ
<
( p
2
,ω
2
)
× φ
<
(
k −p
1
−p
2
,ω−ω
1
−ω
2
)
−
φ
(0)
>
( p
1
,ω
1
)φ
(0)
>
( p
2
,ω
2
)
× φ
(0)
>
(
k −p
1
−p
2
,ω−ω
1
−ω
2
)
−
φ
<
( p
1
,ω
1
)φ
<
( p
2
,ω
2
)
× φ
<
(
k −p
1
−p
2
,ω−ω
1
−ω
2
)
(3.3.17)
Inserting the expansion of Eq.(3.3.15) into Eq.(3.3.13), we have
−
iω
+k
2
+m
2
φ
<
(
k, ω) =−λ
φ
<
( p
1
,ω
1
)φ
<
( p
2
,ω
2
)
× φ
<
(
k −p
1
−p
2
,ω−ω
1
−ω
2
)
− 3λ
φ
<
( p
1
,ω
1
)φ
(0)
>
( p
2
,ω
2
)
× φ
(0)
>
(
k −p
1
−p
2
,ω−ω
1
−ω
2
)
− 3λ
φ
<
( p
1
,ω
1
)φ
<
( p
2
,ω
2
)
× φ
(0)
>
(
k −p
1
−p
2
,ω−ω
1
−ω
2
)
68 3 The Renormalization Group
− λ
φ
(0)
>
( p
1
,ω
1
)φ
(0)
>
( p
2
,ω
2
)
× φ
(0)
>
(
k −p
1
−p
2
,ω−ω
1
−ω
2
)
− 3λ
2
φ
<
( p
1
,ω
1
)φ
(0)
>
( p
2
,ω
2
)
× φ
(1)
>
(
k −p
1
−p
2
,ω−ω
1
−ω
2
)
− 3λ
2
φ
<
( p
1
,ω
1
)φ
(1)
>
( p
2
,ω
2
)
× φ
(0)
>
(
k −p
1
−p
2
,ω−ω
1
−ω
2
)
− 3λ
2
φ
<
( p
1
,ω
1
)φ
<
( p
2
,ω
2
)
× φ
(1)
>
(
k −p
1
−p
2
,ω−ω
1
−ω
2
)
− λ
2
φ
(1)
>
( p
1
,ω
1
)φ
(0)
>
( p
2
,ω
2
)
× φ
(0)
>
(
k −p
1
−p
2
,ω−ω
1
−ω
2
)
− λ
2
φ
(0)
>
( p
1
,ω
1
)φ
(1)
>
( p
2
,ω
2
)
× φ
(0)
>
(
k −p
1
−p
2
,ω−ω
1
−ω
2
)
− λ
2
φ
(0)
>
( p
1
,ω
1
)φ
(0)
>
( p
2
,ω
2
)
× φ
(1)
>
(
k −p
1
−p
2
,ω−ω
1
−ω
2
)
+ O(λ
3
) +
N
<
(3.3.18)
The solution to Eq.(3.3.16)is
φ
(0)
>
=GN
>
/ (3.3.19)
where
G
−1
=−
iω
+k
2
+m
2
(3.3.20)
The solution to Eq.(3.3.17) is
φ
(1)
>
(
k, ω) =−G(
k, ω)
3
φ
(0)
>
( p
1
,ω
1
)φ
(0)
>
( p
2
,ω
2
)
× φ
<
(
k −p
1
−p
2
,ω−ω
1
−ω
2
)
+ 3
φ
<
( p
1
,ω
1
)φ
<
( p
2
,ω
2
)
× φ
(0)
>
(
k −p
1
−p
2
,ω−ω
1
−ω
2
)
+
φ
(0)
>
( p
1
,ω
1
)φ
(0)
>
( p
2
,ω
2
)
× φ
(0)
>
(
k −p
1
−p
2
,ω−ω
1
−ω
2
)
+
φ
<
( p
1
,ω
1
)φ
<
( p
2
,ω
2
)
× φ
<
(
k −p
1
−p
2
,ω−ω
1
−ω
2
)
(3.3.21)
3.3 Dynamics of Model A 69
The thing to note is that φ
(0)
>
and φ
(1)
>
are determined by the noise (see Eq.(3.3.19))
and hence to determine the equation of motion satisfied by φ
<
, we will have to
average over the noise which can be done by using Eq.(2.4.2) in Fourier space
N
>
(
k
1
,ω
1
)N
>
(
k
2
,ω
2
)=2δ(
k
1
+
k
2
)δ(ω
1
+ω
2
)
We will be looking at the O(λ) term in Eq.(3.3.18) in some detail to establish the
procedure,
=
3λ
φ
<
( p
1
,ω
1
)φ
(0)
>
( p
2
,ω
2
)φ
(0)
>
(
k −p
1
−p
2
,ω−ω
1
−ω
2
)
=
3λ
φ
<
( p
1
,ω
1
)G( p
2
,ω
2
)N( p
2
,ω
2
)G(
k −p
1
−p
2
,ω−ω
1
−ω
2
)
× N(
k −p
1
−p
2
,ω−ω
1
−ω
2
)
−2
= 3λ
φ
<
( p
1
,ω
1
)G( p
2
,ω
2
)G(
k −p
1
−p
2
,ω−ω
1
−ω
2
)
×N(p
2
,ω
2
)N(
k −p
1
−p
2
,ω−ω
1
−ω
2
)
−2
= 3λ
φ
<
( p
1
,ω
1
)G( p
2
,ω
2
)G(
k −p
1
−p
2
,ω−ω
1
−ω
2
)
× 2δ(
k −p
1
)δ(ω −ω
1
)/
2
= 6
λ
φ
<
(
k, ω)
p,ω
1
1
[−
iω
1
+p
2
+m
2
]
1
[
iω
1
+p
2
+m
2
]
= 6
λ
φ
<
(
k, ω)
d
D
p
(2π)
D
∞
−∞
dω
1
2π
1
(p
2
+m
2
)
2
+
ω
2
1
2
= 3λφ
<
(
k, ω)
d
D
p
(2π)
D
1
p
2
+m
2
(3.3.22)
Similarly, the third term on the R.H.S of Eq.(3.3.18) is
= 3λ
φ
<
( p
1
,ω
1
)φ
<
( p
2
,ω
2
)φ
(0)
>
(
k −p
1
−p
2
,ω−ω
1
−ω
2
)
= 3λ
φ
<
( p
1
,ω
1
)φ
<
( p
2
,ω
2
)GN
>
−1
= 0 (3.3.23)
The fourth term on the R.H.S of Eq.(3.3.18) does not contribute either. Correct to
O(λ) in Eq.(3.3.18), we have, on using the results of Eqns.(3.3.22) and (3.3.23)
−
iω
+k
2
+m
2
φ
<
=−λφ
<
φ
<
φ
<
where
m
2
=m
2
+3λφ
<
(
k, ω)
d
D
p
(2π)
D
1
p
2
+m
2
(3.3.24)
70 3 The Renormalization Group
The dynamics is not changed. It is the statics that gets modified and Eq.(3.3.24)
is exactly what a one-loop statics calculation would give
if carried out in detail in
§(3.2). Similarly, the fourth, fifth and sixth terms on the R.H.S of Eq.(3.3.18)
contribute to the statics. These terms renormalize the coupling constant λ. (The
sixth is not important as it gives a reducible contribution).
To find out the effect on the dynamics, we need to find a contribution to the
equation of motion that can renormalize the coefficient
−1
of the −iω term on
the L.H.S of Eq.(3.3.18). To do so, we need to consider the last three terms on the
R.H.S and pick out the terms that are linear in φ
<
. Further, we are interested in
terms which have the structure (k,ω)φ
<
(k, ω), since the coefficient (k,ω) can
then modify the relaxation rate . Let us now look at the eighth term on the R.H.S
of Eq.(3.3.18) and substitute for φ
(1)
>
by the first term on the R.H.S of Eq.(3.3.21).
The term becomes (after the averaging)
3λ
2
G( p
1
,ω
1
)φ
<
( q
1
,ω
1
)φ
(0)
>
( q
2
,ω
2
)
× φ
(0)
>
( p
1
−q
1
−q
2
,ω
1
−ω
1
−ω
2
)φ
(0)
>
( p
2
,ω
2
)
× φ
(0)
>
(
k −p
1
−p
2
,ω−ω
1
−ω
2
)
In factoring the above average to get the contribution that has the above structure
(k, ω)φ
<
(k, ω) requires the φ
(0)
>
coming from Eq.(3.3.21), to be correlated with
φ
(0)
>
coming from Eq.(3.3.18). This takes the above term to
6λ
2
G( p
1
,ω
1
)φ
<
( q
1
,ω
1
)φ
(0)
>
( q
2
,ω
2
)φ
(0)
>
( p
2
,ω
2
)
×φ
(0)
>
( p
1
−q
1
−q
2
,ω−ω
1
−ω
2
)φ
(0)
>
(
k −p
1
−p
2
,ω−ω
1
−ω
2
)
=6λ
2
G( p
1
,ω
1
)φ
<
( q
1
,ω
1
)
2
1
ω
2
2
2
+(p
2
2
+m
2
)
2
×
2
1
(ω−ω
1
−ω
2
)
2
2
+[(
k −p
1
−p
2
)
2
+m
2
]
2
δ(q
1
−
k)δ(ω
1
−ω)
Noting that
2
dω
2π
1
ω
2
2
+β
2
=
1
β
and that
2
dω
2π
1
[−
iω
+β
1
]
1
[−
i(ω−ω
)
+β
2
]
=
1
[−
iω
+β
1
+β
2
]