22
L. Rondoni
and
D
(1)
x = v(t) D
(1)
v = −γv(t) − f
(x(t))
D
(2)
xx = 0 D
(2)
xv = 0 D
(2)
vx = 0 D
(2)
vv = σ
2
For Fokker-Planck
∂W
∂t
=
−
∂
∂x
D
(1)
x −
∂
∂v
D
(1)
v +
∂ 2
∂x 2 D
(2)
xx +
∂ 2
∂x∂v
D
(2)
xv +
∂ 2
∂v∂x
D
(2)
vx +
∂ 2
∂v 2 D
(2)
vv
W
= −
∂
∂x
vW −
∂
∂v
(−γv(t) − f
(x(t)))W +
∂ 2
∂v 2 σ
2 W
(1.66)
If σ
2
= γ
k B T
m
= γv
2
th and f
(x) = w
2
o x,
∂W
∂t
= −v
∂W
∂x
+ γ
W + v
∂W
∂v
+ w
2
o x
∂W
∂v
+ γv
2
th
∂
2 W
∂v 2
(1.67)
the solution for the stationary case is
W st (x, v) =
w o
2πv
2
th
e
−
1
2
v 2
v 2
th e
−
1
2
w 2
o x 2
v 2
th
=
mw o
2πk B T
e
−
mv 2 +mw 2
o x 2
2k B T
=
mw o
2πk B T
e
−
E
k B T
(1.68)
By integration, we obtain the stationary densities for the various observables:
W st (x) =
mw o
2πk B T
e
−
mw 2
o x 2
2k B T
+∞
−∞
e
−
mv 2
2k B T dv
(1.69)
Substituting
mv
2
k B T
=
u
2
s 2 ⇒ du = s
m
k B T
dv ⇒ dv =
k B T
ms 2 du
(1.70)
Therefore,
W st (x) =
mw o
2πk B T
e
−
mw 2
o x 2
2k B T
+∞
−∞
e
−
u 2
2s 2
k B T
ms 2 du =
mw o
2πk B T
e
−
mw 2
o x 2
2k B T
2πk B T
m
=
mw 2
o
2πk B T
e
−
mw 2
o x 2
2k B T
(1.71)
Similarly
W st (v) =
mw o
2πk B T
e
−
mv 2
2k B T
+∞
−∞
e
−
mw 2
o x 2
2k B T dx =
m
2πk B T
e
−
mv 2
2k B T
(1.72)