# Broken versus Non-Broken Time Reversal Symmetry: Irreversibility and Response
## 4 The Dissipation Function  
The above discussion can be framed in an abstract dynamical system setting. This section, apart from Section 4.1, summarizes with minor variations how this has been done in Refs. [23; 24; 25]. In Section 4.1, we introduce a novel approach to time dependent perturbations. Given a (sufficiently regular) manifold \(\mathcal{M}\), let \(S^{t}\) be the evolution operator which maps a point \(\Gamma\in\mathcal{M}\) into the point \(S^{t}\Gamma\) that is solution of  
\[\dot{\Gamma}=G(\Gamma)\,\qquad\text{with initial condition}\ \ \Gamma. \tag{14}\]  
Suppose that \(\mathcal{M}\) is endowed with a probability measure \(\mu_{0}\) that is absolutely continuous with respect to the Lebesgue measure, so that \(d\mu_{0}(\Gamma)=f_{0}(\Gamma)d\Gamma\), where \(f_{0}(\Gamma)\) is the probability density function. This density does not need to be related to the dynamics, but it may evolve with the dynamics, and turn into another density \(f_{t}\) after a time \(t\). The evolution rule for \(f_{t}\) will be given by the conservation of probability in phase space, i.e.,  
\[\mu_{t}(E)=\mu_{0}(S^{-t}E)\quad\text{or, equivalently, by}\quad\int_{E}f_{t}( \Gamma)\mathrm{d}\Gamma=\int_{S^{-t}E}f_{0}(\Gamma)\mathrm{d}\Gamma \tag{15}\]  
for any measurable set \(E\subset\mathcal{M}\). Given the phase space expansion rate \(\Lambda=\mathrm{div}G\), we define the _Dissipation Function_ as:  
\[\Omega^{f_{0}}(\Gamma)=-G(\Gamma)\cdot\left.\frac{d}{d\Gamma}\ln f_{0}\right| _{\Gamma}-\Lambda(\Gamma), \tag{16}\]which was first introduced in integral form in Ref. [26]. Note that the dissipation function is a phase function, and as such it does not depend explicitly on time. In its definition, the function \(f_{0}\) is used at all times, even though the probability density in phase space may change because of the dynamics. Solving the generalized Liouville Equation  
\[\frac{df}{dt}=-\Lambda f \tag{17}\]  
yields the evolution \(f_{t}\) of the initial distribution \(f_{0}\) which, using \(\Omega^{f_{0}}\), can be written as [24]:  
\[f_{t}(\Gamma)=\ f_{0}(\Gamma)\exp\left\{\Omega^{f_{0}}_{-t,0}(\Gamma)\right\}\, \tag{18}\]  
where for a phase function \(\mathcal{O}:\mathcal{M}\to R\), the subscripts denote integration along the trajectory:  
\[\mathcal{O}_{s,t}(\Gamma)=\int_{s}^{t}\mathcal{O}(S^{t}\Gamma)\mathrm{d}r, \tag{19}\]  
hence  
\[\Omega^{f_{0}}_{-t,0}(\Gamma)=\int_{-t}^{0}\Omega^{f_{0}}(S^{u}\Gamma)du. \tag{20}\]  
It follows that the dissipation function provides the necessary and sufficient condition for \(f_{0}\) to be invariant: \(\Omega^{f_{0}}(\Gamma)=0\) for all \(\Gamma\in\mathcal{M}\). If \(\Omega^{f_{0}}\) can be identified with the energy dissipation and it identically vanishes, then \(f_{0}\) and \(S^{t}\) can be considered an equilibrium distribution and dynamics, respectively. This happens indeed in known cases [23]. In order to understand the connection between \(\Omega^{f_{0}}\) and energy dissipation, consider the ensemble average of any observable \(\mathcal{O}\):  
\[\left\langle\mathcal{O}\right\rangle_{t}\equiv\int\mathcal{O}(\Gamma)f_{t}( \Gamma)d\Gamma. \tag{21}\]  
From the first-principles definition of the derivative and Equation (18), we obtain (see Appendix):  
\[\frac{d}{ds}\langle\mathcal{O}\rangle_{s}=\left\langle\mathcal{O}\cdot(\Omega ^{f_{r}}\circ S^{r-s})\right\rangle_{s} \tag{22}\]  
for any \(r\in[0,s]\).