1 Introduction to Nonequilibrium Statistical Physics and Its Foundations
37
I (ω) =
k B T
mπ
Re
˜
γ(ω)
, Re
˜
γ(ω)
≥ 0
(1.145)
which mimics Eq. (1.100). Here, we require Re
˜
γ(ω)
≥ 0, because the power spectrum cannot be negative. To obtain this result, one may extend the definition of γ to
negative times, setting γ(−t) = γ(t) = γ(t), for t ≥ 0, and then try the following
assumtpion:
E [(t 1 ))(t 2 )] =
k B T
m
γ(t 1 − t 2 )
(1.146)
Considering that ˜
γ(ω) =
∞
0 dtγ(t) exp(−iωt)/2π, the result is:
I (ω) =
1
2π
∞
−∞
E [(0))(t)] e
−iωt dt =
k B T
2πm
0
−∞
γ(t)e
−iωt dt +
∞
0
γ(t)e
−iωt dt
=
k B T
2πm
˜
γ(−ω) + ˜
γ(ω)
=
k B T
mπ
Re
˜
γ(ω)
(1.147)
where the last equality holds if ˜
γ(−ω) = ˜
γ(ω)
∗ . Also, taking γ(t) = γδ(t), with
γ ∈ R, one falls back in the original situation with δ-correlated noise and obtains:
I =
γk B T
πm
(1.148)
To generalize the mobility, observe first that it can be cast in the following form:
μ
=
D
k B T = lim t→∞
E
(x(t)−x(0)) 2
2tk B T
= lim t→∞
1
2tk B T
t
0
dt 1
t
0
dt 2 E [v(t 1 )v(t 2 )]
(1.149)
= lim t→∞
1
2tk B T
t
0
dt 1
t
0
dt 2 E [v(0)v(t 2 − t 1 )] = lim t→∞
1
k B T t
t
0
(t − s)E [v(0)v(s)] ds
(1.150)
=
1
k B T lim t→∞
t
0
E [v(0)v(s)] ds
(1.151)
where the first term of line (1.150) holds because we consider a steady state, while
the second term of line (1.150) and line (1.151) can be derived with a little bit of
algebra, provided the following is satisfied:
∞
0
φ v (s)ds
< ∞, and
t
0
sφ v (s)ds = o(t) for t → ∞
(1.152)
To illustrate what this means, consider a positive φ v , although in general it fluctuates
between positive and negative values. In that case, the second condition in Eq. (1.152)