8.8 Langevin dynamics simulation 249
subject to the following conditions.
- As the average effect of the collisions is already absorbed in the friction, the
expectation value of the random force should vanish:
〈R(t)〉=0. (8.145)
- The values ofRare taken to be uncorrelated:
〈R(t)R(t+τ)〉= 0 forτ>0. (8.146)
- The values ofRare distributed according to a Gaussian:
P[R(t)]=( 2 π〈R^2 〉)−^1 /^2 exp(−R^2 / 2 〈R^2 〉). (8.147)
All these assumptions can be summarised in the following prescription for the
probability for a set of random forces to occur betweent 0 andt 1 :
P[Ri(t)]t 0 <t<t 1 ∝exp
(
−
1
2 q
∫t 1
t 0
dtR^2 i(t)
)
(8.148)
withqa constant to be determined.
Below we consider the numerical integration of the equations of motion for the
heavy particles, and in that case it is convenient to assume that the random force is
constant over each time step: at stepn, the value of the random force isRn. For this
case, the correlation function for theRnreads
〈RnRm〉=
∫
dRndRn+ 1 ···dRmexp(− 1 / 2 q
∑m
l=nR
2
∫ lt)RnRm
dRndRn+ 1 ···dRmexp(− 1 / 2 q
∑m
l=nR
2
lt)
(8.149)
which yields the value 0 forn=m, in accordance with the previous assumptions.
Forn=mwe find the valueq/t, so we arrive at
〈RnRm〉=
q
t
δnm. (8.150)
For the continuum caset→ 0 (8.150) converges to theδ-distribution function
〈R(t)R(t+τ)〉=qδ(τ). (8.151)
We now return to the continuum form of the Langevin equation(8.144). This
can be solved analytically and the result is
v(t)=v( 0 )exp(−γt/m)+
1
m
∫t
0
exp[−(t−τ)γ/m]R(τ )dτ. (8.152)
Because the expectation value ofRvanishes we obtain
〈v(t)〉=v( 0 )exp(−γt/m) (8.153)
which is to be expected for a particle subject to a friction force proportional and
opposite to the velocity.
