1 Introduction to Nonequilibrium Statistical Physics and Its Foundations
15
Consider now the squared distance travelled by a particle in a time t:
(x(t) − x 0 )
2
=
t
0
v(t 1 )dt 1
t
0
v(t 2 )dt 2 =
t
0
t
0
v(t 1 )v(t 2 )dt 1 dt 2
(1.32)
Again, averaging over all realizations, one obtains
E[(x(t) − x 0 )
2
] =
t
0
t
0
E[v(t 1 )v(t 2 )]dt 1 dt 2
=
t
0
dt 1
t
0
dt 2
q
2γ
e
−γ|t 1 −t 2 |
− e
−γ(t 1 +t 2 )
+ v
2
0 e
−γ(t 1 +t 2 )
=
t
0
dt 1
t
0
dt 2
q
2γ
e
−γ|t 1 −t 2 |
+
v
2
0 −
q
2γ
e
−γ(t 1 +t 2 )
(1.33)
where
t
0
t
0
e
−γ(t 1 +t 2 ) dt 1 dt 2 =
t
0
e
−γt 1 dt 1
t
0
e
−γt 2 dt 2 =
1
γ 2 (e
−γt
− 1)
2
(1.34)
and
t
0
t
0
e
−γ|t 1 −t 2 | dt 1 dt 2 =
t
0
⎡
⎣
t 1
0
e
−γ(t 1 −t 2 ) dt 2 +
t
t 1
e
γ(t 1 −t 2 ) dt 2
⎤
⎦ dt 1
=
t
0
e
−γt 1
e
γt 1 − 1
γ
+ e
γt 1
e
−γt 1 − e
−γt
γ
dt 1
=
t
0
2
γ
−
1
γ
e
−γt 1 −
1
γ
e
γt 1 e
−γt
dt 1
=
2
γ
t +
1
γ 2 (e
−γt
− 1) −
1
γ 2 e
−γt
(e
γt
− 1)
=
2
γ
t +
2
γ 2 (e
−γt
− 1)
(1.35)
Eventually, one can write: