182 6 Systems Far from Equilibrium
Writing Eq.(6.7.6) as
1
k
2
+
1
λ
∂
φ
(k
)
∂t
=
δF
δ
φ
(−k
)
+
ξ
(
k
/b, b
z
t
) (6.7.9)
we have
=b
y+z−2−2ζ
At the fixed point
z =2 +2ζ −y (6.7.10)
This evolution is absolutely general. If we work at the ordinary critical point, then T
does not change under the scale tranformations, i.e. y =0. This makes z =2 +2ζ .
In this case, we have the structure factor given by
S(k,L)=k
−2+η
˜g(kL) =L
2−η
g(kL).
This implies
S(k,t) =t
2−η
z
g(kt
1/z
)
and with this choice Eq.(6.7.4) leads to ζ =1 −
η
2
. The dynamic exponent becomes
z =2 +2ζ =4 −η which is the model B answer. Since does not renormalize at
any order and that is the basis for our derivation, z =4 −η is an exact answer
independent of the order of perturbation theory.
If we work at the zero temperature fixed point which is the relevant one for the
problem of quench, then ζ =D/2 as seen from Eq.(6.7.4). This yields
z =2 +D −y (6.7.11)
The exponent y which describes the scaling of the Hamiltonian (hence the energy)
is determined by the excitations in the system. For the scalar order parameter (i.e.
n =1) the excitations are the domain walls of dimension D −1, while for the vector
order parameter (i.e. n ≥2), the excitations are spin waves with dimension D −2.
Hence y =D −1 for n =1 and y =D −2 for n ≥2. Consequently,
z = 3 for n =1
z = 4 for n ≥2
(6.7.12)
The derivation requires no assumptions about the details of the domain growth
kinetics and the central result (Eq.(6.7.11)) is of general validity. It is not restricted
to the O(n) model, but is valid for any symmetry group. One needs only to insert
the appropriate value of η.