deletions | additions       diff --git a/Gibbs.tex b/Gibbs.tex  index a1b16ed..40353d8 100644  --- a/Gibbs.tex  +++ b/Gibbs.tex 
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\subsubsection{Resolution according to Jaynes}  In \citet{1992mebm.book.....S}, REF?,  Jaynes says that the entropy increase has to be treated more ``subjectively''. Entropy production is not absolute: if {\it we} cannot distinguish the properties of two mixing gases, then there is no entropy increase and no work required to un-mix them. If we can distinguish the gases, then this is no longer true. To repeat, if the particles are experimentally indistinguishable for whatever reason, Gibbs' paradox is resolved.\footnote{In the quantum realm, this indistinguishability may be true as a matter of principle, rather than being due to an insufficiently refined experimental capability.} What this suggests to me is absurd: that a colour-blind person will calculate zero entropy generation when a box of green balls mix with a box of red balls. But a normally-sighted person calculates some entropy increase. Is Jaynes saying that they can both be right? I don't get it -- we can use the mixing to do work (see section~\ref{sec:workout}), and surely no-one can argue about that?  In \citet{1992mebm.book.....S}, Jaynes tells a slightly different story (originally due to Gibbs). When the gases are identical, our definitions of ``reversible'' and ``original state'' change: we have double standards. In the different-gases case, we want to separate all molecules originally in $V_1$ and put them back into $V_1$, and the same for $V_2$; but in the same-gases case we are happy to just reinsert the diaphragm. ``Reversible'' applies to thermodynamic variables we can measure, not to microscopic states. The work in statistical mechanics is to recover the same (family of) microstates, i.e. the original separation. What about gases which are identical except in magnetic moment, which does not come into play? $S=\ln W$ needs us to define an ``equivalence class'' $C$, the members of which belong to the same macrostate. But choice of $C$ depends on what we're measuring about the macrostate. Section 5, mixing isotopes: Ar2 soluble in Whifnium. No $\Delta S$ before, so there should be no $\Delta S$ unless there are observable consequences, e.g. work extractable. Can do this with Whifnium: work extractable depends on human information. Identical (even microscopically) physical processes can be assigned different entropy depending on knowledge -- knowledge lets us extract more work. Entropy not a physical property of microstate (as energy is), it is an anthropomorphic quantity.  %%--------------------------------------------------------------------------------------  \subsection{Goal} 
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\begin{enumerate}  \item For {\it distinguishable} particles, mixing two volume-$\frac{V}{2}$ boxes of $\frac{N}{2}$ ideal-gas particles each incurs an entropy increase of $\Delta S = N\ln2$.  \item The maximum work extractable from {\it this} process is therefore $W_{\rm max} = NT\ln2$. (This is also the minimum work required to recover the original state.)  \item Here's an idea. idea {\it *Update*: wrong: need to carry out mixing process isothermally and reversibly}.  The total number of particles $N$ is fixed, but not necessarily the final volume. If we set $v^{\rm final} < V^{\rm init}_1 + V^{\rm init}_2$, then the system will end in a pressurised state; this pressure can do work. \begin{itemize}  \item Does it cost work to do this?  \item Ignoring this, I calculated (I think) the entropy change for mixing to a valume $\eta V < V$. I found $\Delta S \approx N\ln2\eta$, which means the theoretical work is $W\approx NT\ln2\eta$.