76
L. Rondoni
while they are not known for the field particle. In this model, the field particle does
not depend on the test particle, and its equations of motion can be readily solved:
x 1 (t) = x 10 cos
√
kt +
v 10
√
k
sin
√
kt
(1.269)
v 1 (t) = −
√
kx 10 sin
√
kt + v 10 cos
√
kt
(1.270)
Substituting x 1 (t) in the equations of motion of the test particle, and solving, one
obtains:
v(t) =
x 10
k + γ 2
γ cos
√
kt +
√
k sin
√
kt − γe
−γt
=
x 10
k + γ 2 f (t)
(1.271)
where the last equality defines the function f .
In the case in which the initial condition x 10 is not known, the dynamics of the test
particle remains also unknown. One may however know the probability distribution of
its values. For instance, suppose x 10 takes values in {−1, 0, 1}, with equal probability,
1/3. Then we may compute the average of v, obtaining:
E[v(t)] =
1
3
(−1 + 0 + 1)
1
k + γ 2
γ cos
√
kt +
√
k sin
√
kt − γe
−γt
= 0
(1.272)
For the velocity autocorrelation function, note that:
v(t)v(t
) =
x 10
k + γ 2
2
f (t) f (t
)
(1.273)
which leads to
E[v(t)v(t
)] =
1
3
(1 + 0 + 1)
1
k + γ 2
2
f (t) f (t
) =
2
3
1
k + γ 2
2
f (t) f (t
)
(1.274)
In summary, we have averaged the solution of the equations of motion of the test
particle over the initial conditions weighted with their probability distribution. This
can be interpreted as an average over an ensemble of identical and independent
test particles, under the assumption that the dynamics of both test particles and
field particles is deterministic and given by the corresponding equations of motion.
This is the picture of Eq. (1.14), with N = 1. Clearly, there is a difference between
trajectories that start from different initial conditions, and also a difference between
the single trajectories and their averages. The same happens in the Brownian motion,
of which this exercise is but an exaggeration. Indeed, one may legitimately maintain
that for small particles in a liquid at room temperature, pressure etc, the classical
mechanics description is accurate, the only difficulty being related to knowledge of
the interaction potentials, and the huge number of degrees of freedom to consider. But
this is a difficulty only in case quantitative explicit calculations are to be performed.
L. Rondoni
while they are not known for the field particle. In this model, the field particle does
not depend on the test particle, and its equations of motion can be readily solved:
x 1 (t) = x 10 cos
√
kt +
v 10
√
k
sin
√
kt
(1.269)
v 1 (t) = −
√
kx 10 sin
√
kt + v 10 cos
√
kt
(1.270)
Substituting x 1 (t) in the equations of motion of the test particle, and solving, one
obtains:
v(t) =
x 10
k + γ 2
γ cos
√
kt +
√
k sin
√
kt − γe
−γt
=
x 10
k + γ 2 f (t)
(1.271)
where the last equality defines the function f .
In the case in which the initial condition x 10 is not known, the dynamics of the test
particle remains also unknown. One may however know the probability distribution of
its values. For instance, suppose x 10 takes values in {−1, 0, 1}, with equal probability,
1/3. Then we may compute the average of v, obtaining:
E[v(t)] =
1
3
(−1 + 0 + 1)
1
k + γ 2
γ cos
√
kt +
√
k sin
√
kt − γe
−γt
= 0
(1.272)
For the velocity autocorrelation function, note that:
v(t)v(t
) =
x 10
k + γ 2
2
f (t) f (t
)
(1.273)
which leads to
E[v(t)v(t
)] =
1
3
(1 + 0 + 1)
1
k + γ 2
2
f (t) f (t
) =
2
3
1
k + γ 2
2
f (t) f (t
)
(1.274)
In summary, we have averaged the solution of the equations of motion of the test
particle over the initial conditions weighted with their probability distribution. This
can be interpreted as an average over an ensemble of identical and independent
test particles, under the assumption that the dynamics of both test particles and
field particles is deterministic and given by the corresponding equations of motion.
This is the picture of Eq. (1.14), with N = 1. Clearly, there is a difference between
trajectories that start from different initial conditions, and also a difference between
the single trajectories and their averages. The same happens in the Brownian motion,
of which this exercise is but an exaggeration. Indeed, one may legitimately maintain
that for small particles in a liquid at room temperature, pressure etc, the classical
mechanics description is accurate, the only difficulty being related to knowledge of
the interaction potentials, and the huge number of degrees of freedom to consider. But
this is a difficulty only in case quantitative explicit calculations are to be performed.