# Broken versus Non-Broken Time Reversal Symmetry: Irreversibility and Response
## 2 Standard Picture for Irreversible Relaxation to Equilibrium
For instance, taking \(\mathcal{A}=\{a,b\}\) and \(N=4\), the set of all possible phases contains 16 strings,  
\[\mathcal{M}=\left\{aaaa,aaab,aaba,abaa,baaa,aabb,bbaa,abba,bab,abab,baba,babb, bbbab,bbba,bbbb\right\}, \tag{1}\]  
and the possible macrostates are five:  
\[\
u(aaaa) =\left\{1,0\right\}, \tag{2}\] \[\
u(bbbb) =\left\{0,1\right\},\] (3) \[\
u(aaab) =\
u(aaba)=\
u(abaa)=\
u(baaa)=\left\{\frac{3}{4},\frac{1}{4} \right\},\] (4) \[\
u(bbba) =\
u(bbab)=\
u(babb)=\left\{\frac{1}{4},\frac{3}{4}\right\},\] (5) \[\
u(aabb) =\
u(bbaa)=\
u(abba)=\
u(baab)=\
u(abab)=\
u(abab)=\
u(baba)=\left\{ \frac{2}{4},\frac{2}{4}\right\}, \tag{6}\]two of which are given by a single arrangement, two are given by four distinct configurations, and the last one is achieved in six different ways. One can show that the number of different macrostates is not larger than \((N+1)^{L}\), cf. [7] Theorem 11.1.1, while the number of microstates equals \(L^{N}\). In particular, for sufficiently large \(N\), the number of strings corresponding to a given empirical distribution \(\
u\) is approximated by:  
\[W(\
u)\approx e^{NH(\
u)},\text{ with }H(\
u)=-\sum_{j=1}^{L}\
u_{j}\ln\
u_{j}. \tag{7}\]  
Note: like in the case of Equations (1), (3) and (5), this is what \"counting\" yields, i.e., it is the result of an enumeration excerise, with the aid of [PERSON]'s formula [8]. For instance, taking \(\mathcal{A}=\{a,b\}\) and strings of length \(N\), there are \(N!/K! (N-K)!\) strings with \(K\) symbols \(a\) and \(N-K\) symbols \(b\), while \(K=\
u_{a}N\) and \(N-K=\
u_{b}N\). Then, [PERSON]'s approximation yields:  
\[\ln\left[N!/K!(N-K)!\right]\approx N\left[\ln N-\
u_{a}\ln(\
u_{a}N)-\
u_{b} \ln(\
u_{b}N)\right]=-N\left[\
u_{a}\ln\
u_{a}+\
u_{b}\ln\
u_{b}\right],\]  
Large \(N\) implies that the macrostate \(\tilde{\
u}\) corresponding to the largest number of strings \(\widetilde{W}\) contains by far the largest fraction of the total set of microscopic phases of \(\mathcal{M}\), and dynamics \"freely jumping\" from string to string are _expected_ to spend most of their time in \(\tilde{\
u}\). Why should they remain localized in negligibly small fractions of the state space? This _vulgata_ relies on a kind of _counting_ argument: the microscopic phases corresponding to the different macrostates occupy hugely different fractions of the phase space. The differences are so vast that convergence to equilibrium appears unavoidable and requires no special dynamical properties or sophisticated probabilistic concepts. In this approach, counting states is thought to be more than enough to justify relaxation. To quantify these considerations, take all arrangements of \(10^{3}\times 10^{3}\) black and white pixels. \(\mathcal{M}\) contains \(2^{10^{6}}\) pictures, among which the remotely regular ones constitute a tiny fraction. The overwhelming majority are gray. If pictures are shown at a rate of 25 frames a second, and the movie goes through all of them before starting again, the period is \(T>10^{301020}\) years! As any picture appears just once in a time \(T\), the fraction of time in which one does not see noisy gray is ridiculously small. Many consider this such an exceedingly simple and compelling argument that they hardly understand why the debate is still open [3; 9], but, admittedly, it is far from simple to prove that specific dynamics are indeed like \"free jumping\". On the other hand, the dynamics of particles occur in continuous time in a continuous space, and not in discrete time and discrete space. Although not totally trivial, this issue is usually overlooked, and counting of states is readily replaced by the comparison of volumes in phase space. In other words, a macrostate, e.g., a given distribution of mass in space, corresponding to a larger portion of phase space than that of another mass distribution is considered realizable in \"more\" ways than the second. This approach can be criticized, but it is supported by some mathematical argument and, in the absence of better recipes, it is generally accepted.