Irreversible growth of average energy

How does the energy spreading cause growth of the average energy? It is a result of both the diffusion term $D_{{\mbox{\tiny E}}}$ being energy-dependent, and of the extra drift (bias) due to transformation from $\Omega$ to $E$ (see Fig. 2.2b). In the billiard context, the former effect is easy to visualize: we can imagine that particles which happen to have a slightly increased kinetic energy will hit the walls more often, thus diffusing faster in energy. We will see in Chapter 4 that the dependence is $D_{{\mbox{\tiny E}}} \propto E^{3/2}$, independent of $d$. The latter effect is less intuitive. Considering that $g(E) \propto E^{d/2 - 1}$ for billiards it is not hard to show that the ratio of the latter to the former effect is $(d-2)/3$, independent of energy. Therefore which effect dominates depends on $d$ the system dimensionality.

We compute the rate of change of average energy, substituting (2.21) and integrating by parts,

$$\langle \dot{E}(t) \rangle \; \equiv \; \int dE \, E \, \dot{\rho}(E,t)$$

$$= \dot{x} \int dE \left. \frac{\partial E}{\partial x}\right\vert _... ...ox{\tiny E}}} \, g \frac{\partial }{\partial E} \left( \frac{\rho}{g} \right) .$$ (2.22)

The first term is the reversible heating (easily verified using the microcanonical case $\rho(E) = \delta(E-E_0)$). The second is irreversible since it relies on diffusion; it can be written (by substituting for $D_{{\mbox{\tiny E}}}$) as

$$\dot{Q}_{{\mbox{\tiny irrev}}}(t) \; = \; \mu V^2 .$$ (2.23)

$\mu(\omega)$ is a friction coefficient, named so because he agency moving $x$ has to do work against a nonzero average force. We have ohmic dissipation, corresponding to an average force $\langle{\mathcal{F}}\rangle = - \mu \dot{x}$ which is proportional to velocity. The value of $\mu(\omega)$ depends on the energy distribution $\rho$ at time $t$, and has the general expression

$$\mu(\omega) \; = \; -\frac{1}{2} \int_0^\infty dE \,\tilde... ...ox{\tiny E}}}(\omega)) \hspace{0.4in} \mbox{general $\rho$} .$$ (2.24)

It is interesting that Koonin and Randrup derived an equivalent expression (Eq. (2.23) of [120]), without explicitly considering energy spreading. If we start in a particular choice of ensemble (initial $\rho_0$), and diffusion has not yet caused $\rho$ to differ much from $\rho_0$, then $\mu(\omega)$ therefore takes special forms:

$$\mu_{{\mbox{\tiny E}}}(\omega) = \frac{1}{2} \frac{1}{g(E)} \frac{\partial }{\partial E} (g(E) \tilde{C}_{{\mbox{\tiny E}}}(\omega)) \hspace{1.5in} \mbox{microcanonical},$$ (2.25)

$$\mu(\omega) = \frac{g(E_{{\mbox{\tiny F}}})}{2\Omega_{{\mbox{\tiny F}}}} \tilde{C}_{{\mbox{\tiny F}}}(\omega) \hspace{2.0in} \mbox{Fermi distribution},$$ (2.26)

where the subscript ${\mbox{\tiny F}}$ means evaluation at the Fermi energy $E_{{\mbox{\tiny F}}}$. The Fermi distribution has a constant phase-space density of $\Omega_{{\mbox{\tiny F}}}^{-1}$, normalised to represent a single particle. For an application to many non-interacting fermions, see Section 2.2.4. The case of the canonical distribution, relevant to a thermal gas of non-interacting classical particles, is covered by Cohen (Section 4 of [46]) and I do not discuss it here. After long times, if the system remains isolated, it can be that $\rho$ is very different from $\rho_0$, and the special forms (2.25) or (2.26) will have to be replaced by (2.24).

What I have presented is only the dissipation appearing to lowest order in $V$ (Jarzynski [107] discusses this by way of an expansion in powers of a `slowness parameter' $\propto V$). There will be higher-order terms in the friction which become more relevant as $V$ approaches for example the limit (2.12). There are other ways that the classical spreading picture can break down. For instance for $t$ less than the correlation time $\tau_{{\mbox{\tiny cl}}}$ the energy spreading should be much less than $E$ itself. Also for this same $t$ the average energy increase should remain small compared to the spreading. These set additional upper limits on $V$ [46].