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Microcanonical averages and trajectory averages

In the preceding sections $F(x)$ and $\tilde{C}_{{\mbox{\tiny E}}}(\omega)$ were defined in terms of the microcanonical average over initial conditions,

$\displaystyle \langle \cdots \rangle_{{\mbox{\tiny E}}} \; \equiv \;
\int d{\mathbf q} d{\mathbf p} \; f({\mathbf q},{\mathbf p},0)
(\cdots) ,$     (2.27)
$\displaystyle f({\mathbf q},{\mathbf p},0) \; = \;
\frac{1}{g(E)} \delta({\mathcal{H}}({\mathbf q},{\mathbf p},x) - E)$     (2.28)

The assumption was made that the change in $x$ over the correlation time $\tau_{{\mbox{\tiny cl}}}$ is insignificant. This means that the `frozen' Hamiltonian (at fixed $x$) can be used, so the distribution is unchanging in time $f({\mathbf q},{\mathbf p},t) = f({\mathbf q},{\mathbf p},0)$. All resulting averages are constant in time (or, if they involve multiple times, they are functions of time differences only), and the choice of initial time $t=0$ is arbitrary.

However, these averages need not be taken using an average over phase space. By ergodicity [79], they are equal to time averages over a single trajectory. Namely, the conservative force (2.3) can be written

F(x) \; = \;
- \lim_{T\rightarrow\infty}
\frac{1}{T} \; \... \mathcal{H}}{\partial x}({\mathbf q}(t),{\mathbf p}(t),x) ,
\end{displaymath} (2.29)

seen to be the mean force due to motion of a single particle in the system. Similarly, the auto-correlation is written
C_{{\mbox{\tiny E}}}(\tau) \; = \; \lim_{T\rightarrow\infty...
... \int_{0}^{T} dt \; {\mathcal{F}}(t) {\mathcal{F}}(t + \tau) .
\end{displaymath} (2.30)

There is now an issue of convergence: the number of independent samples of any quantity along a trajectory is $N \approx T/\tau_{{\mbox{\tiny cl}}}$, and the fractional error of the estimate of the above quantities converges slowly as $\sim N^{-1/2}$. Therefore very long trajectories are required to get good estimates. However, this is often easier than performing the $2d{-}1$ dimensional integral over the energy shell which would be required for the explicit evaluation of the phase-space average, especially since the integrand in (2.30) already involves propagation forward in time. This technique of evaluation of a multi-dimensional integral using a random sample of points taken from the distribution function is called Monte Carlo integration [161]. However, in practice, rather than compute $C_{{\mbox{\tiny E}}}(\tau)$ and take the Fourier transform, $\tilde{C}_{{\mbox{\tiny E}}}(\omega)$ is most efficiently estimated directly from the Fourier transform of ${\mathcal{F}}(t)$. This approach is discussed in detail in Appendix B.

In an identical fashion to that shown in Fig. 2.2a, Eq.(2.30) describes $C_{{\mbox{\tiny E}}}(\tau)$ as the projection of the function ${\mathcal{F}}(t_1){\mathcal{F}}(t_2)$ onto the $\tau = t_2-t_1$ axis. However only a single trajectory is involved, so ${\mathcal{F}}(t)$ is noisy, and an average over the time axis $t' = \mbox{\small$\frac{1}{2}$}(t_2 + t_1)$ is required. In this figure, one can imagine the `box' of allowed $t_1$,$t_2$ values now bounded by $[0,T]$. The average converges in the limit $T \rightarrow \infty$.

An instructive convergence issue arises if we extend this single-trajectory estimate to the noise intensity $\nu_{{\mbox{\tiny E}}}$. Naive use of (2.14) and (2.30) would give

\nu_{{\mbox{\tiny E}}}\; = \; \lim_{T\rightarrow\infty}
... {\mathcal{F}}(t + \tau)
\hspace{.5in} \mbox{non-convergent}.
\end{displaymath} (2.31)

This does not converge because the $\tau$ integral involves an infinite number of similarly-sized contributions (the integrand is similar in size for all $t$, $\tau$). Averaging over any finite $T$ cannot remove this divergence. Another option is to look back to Eq.(2.8) which is responsible for the appearance of $\nu_{{\mbox{\tiny E}}}$ in the spreading rate. So I could define
\nu_{{\mbox{\tiny E}}}
= \lim_{T\rightarrow\infty}
...1) {\mathcal{F}}(t_2)
\hspace{.2in} \mbox{nearly convergent},
\end{displaymath} (2.32)

which is the same integral (2.31) with different limits. This is simply proportional to the energy variance $\delta E^2$ after time $T$ for the single trajectory involved, divided by $T$. The term inside the square brackets is simply a random walk (on timescales $\gg \tau_{{\mbox{\tiny cl}}}$), giving a Gaussian distribution whose variance grows linearly. Therefore this estimate of $\nu_{{\mbox{\tiny E}}}$ will not converge, rather it will wander for all $T \rightarrow \infty$, taking values with a $\chi^2$ distribution whose mean is the correct $\nu_{{\mbox{\tiny E}}}$. In effect this reproduces exactly the stochastic energy spreading whose variance is desired! However, it is `more convergent' than (2.31). A convergent estimate can only be created by limiting the $\tau$ integration further, to give
\nu_{{\mbox{\tiny E}}}\; = \; \lim_{t_0\rightarrow\infty}
...}(t) {\mathcal{F}}(t + \tau)
\hspace{.5in} \mbox{convergent},
\end{displaymath} (2.33)

which will converge once $T \gg t_0 \gg \tau_{{\mbox{\tiny cl}}}$. In essence the convergence arises because the limits do not allow the number of independent $\tau_{{\mbox{\tiny cl}}}$-squared-sized `patches' of the integrand to grow as fast as $N^2$, where $N$ is defined as above. The nearly-convergent case (2.32) corresponds to exactly $N^2$ growth.

The above considerations will not relate directly to the numerical method of finding $\nu_{{\mbox{\tiny E}}}$ (which will be via $\tilde{C}_{{\mbox{\tiny E}}}(\omega\rightarrow 0)$). However they serve to warn and provide intuition about convergence when a single trajectory is used for estimation.

next up previous
Next: Review of the linear Up: Review of classical dissipation Previous: Irreversible growth of average
Alex Barnett 2001-10-03