Clas Blomberg - Physics of life-Elsevier Science (2007)

Chapter 21. Brownian motion: first description 219
where v 0 is an initial velocity. A particle that at the onset had a certain velocity v 0 , continues in
the same direction, with a declining velocity in a timescale given by the damping time 1/g.
One gets a solution of (21.4) which expresses the velocity as an integral over the random
force component:
t
∫
vt () v t
( t )
0e g
1
e g t )
Firr( t)
dt
0
(21.7)
This can be used for further development, and, in particular, to calculate correlation functions.
These are given by averages of products of two integrals containing random force
components. This leads to an integral over a correlation function of the irregular force, thus
the d-function. That integral falls out simply, and one gets the desired result:
kT
( vt () vt ()) 2 B
m
(21.8)
This is important and shows the consistency in the explicit choice of the proportionality
factor for the irregular force given by the delta-function. It means that the average value of
the kinetic energy of the Brownian particle in equilibrium is the same as the average energy of
all other particles, which is what the classical statistical thermodynamics requires.
One can continue with expressions for the position x(t) as function of time. The velocity
is of course the time derivative of the position, and the previous equation becomes:
d2x
dx
F t
dt
g 2
dt
irr ()
(21.9)
The average of the position can be calculated easily:
⎛ v ⎞
xt () x 0
( t
0 1
g
e )
⎝⎜
g ⎠⎟
(21.10)
where x 0 is the starting position and v 0 the initial velocity. The second term signifies the
average distance (v 0 /g) travelled by the particle along the direction of the velocity.
Again, we are interested in the correlation function, the average of the square of the displacement
from the initial position at a time t. We can use the above expression for velocity,
which again leads to products of integrals with the irregular force component and d-functions,
which according to their definition leads to simple integrals. The final result is:
⎛ 1 ⎞ kT
[ xt ( ) x( 0)]
2
⎛ ⎞
B
t
⎝⎜
g ⎠⎟
⎝⎜
m ⎠⎟
(21.11)