The ring model by Mark Kac was formulated precisely to understand the influ- ence of a Stoßzahlansatz on a reversible dynamics. For an excellent summary of the properties of the Kac model, we refer to appendix A in the paper [13]. The Kac ring model consists of a ring on whichN equidistant sites are placed. At each site, one has two possible states: the site can be filled by a white ball or by a black ball. The segments of the ring between each site play an important role in the dynamics: to mimic a collision, a segment can either be colour-changing or colour-preserving, i.e., it can scatter a ball from one colour to another. The total dynamics is obtained by performing a clockwise rotation of the ring, changing the colour of a ball if it passes a colour-changing scatterer. Write k < N as the number of colour-changing scatterers. Write the number of white and black balls at timetasNw(t) andNb(t) respectively,
and writenw(t) andnb(t) as the number of white and black balls in front of a
colour-changing scatterer. Then, we have the obvious equations of motion (
Nw(t+ 1) =Nw(t)−nw(t) +nb(t),
white ball colour-changing segment direction of rotation
t = t
1 direction of rotationt = t
2Figure 5.1: An example of a realization of a (part of the) Kac ring at two consecutive timest1 < t2. The upper black ball is in front of a colour-changing segment/scatterer, so
after the next time step, it will become white and, as all other balls, shift downwards one position due to the clockwise rotation. We see how the lower black ball on the left shifts out of the picture, and how a white ball enters again at the top.
and the conservation laws (
Nw(t) +Nb(t) =N,
nw(t) +nb(t) =k. ∀
t
This model has the typical properties of a Hamiltonian dynamics:
• Isolated and deterministic: the ring and the scatterers are isolated and the original dynamics is perfectly deterministic.
• Reversibility: if we reverse the direction of rotation at timet, then the system neatly travels back to its initial state (cfr. the reversibility of the Hamiltonian dynamics2, see section 1.2.2).
• Periodicity: the dynamics is strictly periodic with period at most 2N (cfr. the Poincar´e recurrences), while the phase space contains 2N con-
figurations.
Let us now solve the equations of motion. We assume that we start from an initial state with all white balls, Nw(0) =N, Nb(0) = 0.
Solution using the Stoßzahlansatz
The equivalent of Boltzmann’s Stoßzahlansatz for this model would be that the colour of a ball is uncorrelated with the property of having a colour-changing
2The direction of motion, let’s write it p ∈ {+1,−1} with +1 for clockwise and −1
for counter-clockwise, should be included as a dynamical variable to be able to define a kinematic time-reversal operationπin the sense of equation (1.3).
scatterer in front of it.3 In other words, n
w(t) should be proportional to
Nw(t) and nb(t) to Nb(t). Combined with the conservation equations, that
Stoßzahlansatz gives ( nw(t) =µNw(t), nb(t) =µNb(t). µ= k N, Inserting the Stoßzahlansatz in the equations of motion yields:
(
Nw(t+ 1) =Nw(t) +µ[Nb(t)−Nw(t)],
Nb(t+ 1) =Nb(t) +µ[Nw(t)−Nb(t)].
The average colour of the system
m(t) =Nw(t)−Nb(t) N
satisfies the following equation of motion, obtained by subtracting the equa- tions for Nw(t) andNb(t):
m(t+ 1) = (1−2µ)m(t). By recursion, we get:
m(t) = (1−2µ)tm(0).
For µ <1/2, we have a monotonic approach to the stationary value m= 0, forµ >1/2, there is an oscillatory approach to this value.
The variablesNw(t) andNb(t) play the role of macroscopic variables. We can
associate a Boltzmann entropy
SB(Nw, Nb) = ln
N Nw
to them since the binomial coefficient gives the number of microscopic config- urations with Nw white balls (andNb =N−Nw black balls). This entropy
is maximal when Nw = Nb =N/2, which corresponds to an average colour
m = 0. The Stoßzahlansatz predicts an exponential relaxation towards that equilibrium state, precisely in the line of thoughts of Boltzmann, see section 2.4.3.
We now wish to clarify how the Stoßzahlansatz reproduces the microscopic dy- namicstypically, and what is meant by that statement. It is clear that also here the Stoßzahlansatz solution cannot be exact: the periodicity and reversibility of the original dynamics are lost. On the level of macroscopic variables, only irreversibility and exponential relaxation to equilibrium is found.
3The role of the velocity is played by the colour, the role of the collisions by the passage
over a scatterer. The colour after passing a scatterer (velocity after collision) should be uncorrelated to the colour before (the velocity before the collision). Thus, it does not matter if a colour-changing scatterer is in front of a ball in the Stoßzahl approximation.
Ensemble approach to the Kac ring
Let us look at the model again from a formal perspective. We write ηi ∈ {+1,−1}, the colour of the ball at sitei∈ {1, . . . , N}, and interpret
(
ηi= +1, when the ball at siteiis white,
ηi=−1, when the ball at siteiis black.
The properties of the ring are encoded in the scattering variablesǫi∈ {+1,−1}
with (
ǫi= +1, when the segment between sitesiandi+ 1 is colour-preserving,
ǫi=−1, when the segment between sitesiandi+ 1 is colour-changing.
The equation of motion then becomes
ηi(t) =ǫi−1ηi−1(t−1),
which can be recursed to
ηi(t) =ǫi−1ǫi−2. . . ǫi−tηi−t(0), (5.1)
where all the subtractions in the indices are understood to be modulo N. This is the explicit solution of the microscopic equations of motion. Our macroscopic observable of interest, the average colour, can be expressed in terms of the variablesηi as follows:
m(t) = Nw(t)−Nb(t) N = 1 N N X i=1 ηi(t) = 1 N N X i=1 ǫi−1ǫi−2. . . ǫi−tηi−t(0).
The distribution of the scatterers determines the dynamical properties of the system, so we cannot continue the calculation until we know more about that distribution. The question that arises is if we are interested in every possi- ble single realization of the distribution of ǫi’s, and the answer is clearly: no.
The natural approach at this point would be to follow the idea’s of Boltz- mann and Gibbs and to introduce an ensemble of Kac ring’s, all consisting of N sites, all with their own distribution of k colour-changing scatterers, and to ask what one would typically expect if one ring is chosen at ran- dom from this ensemble. In other words, if we fix k and N, with N ≫ t, will we typically see that m(t) ∼ (1−2µ)tm(0)? The answer is yes: we
consider ǫ = (ǫ1, . . . , ǫN) ∈ {+1,−1}N with the additional constraint that
P
iǫi=N−2k, i.e., that there arekcolour-changing scatterers.
After some work, Kac proves that up to corrections of order O(N−1/2), see
[67, p.100] and [127, p.401], that for almost all ǫ
withµ=k/N. For large systems and appropriate timescales, Kac suggests to think ofN∼1023 andt∼106 s, we get back the Stoßzahlansatz solution
m(t) = (1−2µ)tm(0), t≪N. (5.2) Kac goes one step further4and also shows that it is not necessary to constrain the total fraction of scatterers to µ. If at each segment, we independently put a colour-changing segment with probabilityµand a colour-preserving seg- ment with probability 1−µ, then Kac proves that again the Stoßzahlansatz solution is typically obtained. In that sense, the Stoßzahlansatz consists of replacing the “real” dynamics by an effective dynamics in which a particle is scattered independently at each segment with probabilityµ and preserved with probability 1−µ.