My question stems from a confusion of the definition of entropy for the Fluctuation Theorem (see the comments too).

Normally, entropy is as an average over different ensembles. In the Canonical Ensemble,

\begin{align} e^{-{\frac {F}{kT}}} &= \int \ldots \int {\frac {1}{h^{n}C}}e^{\frac {-E}{kT}}\,dp_{1}\ldots dq_{n} \\ S &= - \frac{\partial F}{\partial T} \end{align}

This is for an n-particle system, so clearly, we are integrating over all possible system states.

Now assuming identical particles, if I prepare a single system in a micro-state $ \Gamma_0 = (\vec{q}, \vec{p})_{t=0} $ that need not be at equilibrium, then I could possibly compute a different kind of entropy over the particles in this system.

For example, imagine I chose a microstate that had a phase space density $ \rho(q, p) $. Meaning each particle is chosen from $ \rho $.

And use this (or approximations of this) to compute an entropy with the Boltzmann/Shannon formula $ S = - \int dq~dp~\rho \log \rho $. This seems like it could inherently be different from the standard ensemble entropy which computes entropy over the p.d.f. $ \rho(\Gamma) $.

Is there such a notion for micro-state entropy? In what cases is it equivalent to macro-state entropy? How should it properly be computed?

  • $\begingroup$ In your example, you have full knowledge over the state of the system and thus, the Shannon (information) entropy is zero. Or is there some other type of uncertainty in your question (for example at a time $t \neq 0$)? $\endgroup$ – Julian Helfferich Jul 31 '17 at 10:02
  • $\begingroup$ @JulianHelfferich Good point. That was an accident. I just picked a p.d.f. without thinking too much about it in particular. I was thinking of how I usually compute entropy when simulating. I usually bin the particles on a discrete grid and then compute. I edited it $\endgroup$ – aidan.plenert.macdonald Jul 31 '17 at 17:35

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