Re: Entropy: quenched disorder is entropy

[David Bowman <David_Bowman@GEORGETOWNCOLLEGE.EDU>]

Regarding John Denker's comments:

> Today in several messages, David Bowman and Leigh Palmer have been
> advocating the notion that quenched disorder should not be counted as
> entropy;  for instance:
> > > The ordered
> > > pack has exactly the same entropy as the disordered pack,
>
> Alas, that isn't right.

Oh?

> In fact, it is a standard result from the thermodynamics of computation
> that quenched disorder is disorder, and it counts as entropy.

Clearly, disorder is (by tautology) disorder whether it is quenched or
not.  Depending on just what is meant quenched disorder can have *an*
entropy (in some info-theoretic sense) associated with it.  But this
entropy is not thermodynamic entropy.

> In particular, we can draw a very strict analogy:
>         sorted card deck <==> blank tape for Turing machine
>         shuffled card deck <==> "dirty" tape containing random 1s and 0s

I fail to see the analogy, and how the theory of the thermodynamic cost
of computation has anything to do with whether or not an entropy measure
associated with an amount of quenched disorder contributes to the actual
physical thermodynamic entropy for the system of interest.

> and beautiful classic results show that the thermodynamic cost of erasing
> the dirty tape is on the order of kT per bit, whereas blank tape doesn't
> need to be erased, and any tape with a known (or easily computable) pattern
> can be erased at arbitrarily small cost (much less than kT per bit).

So what?  (The result may be both beautiful and classic but it appears
irrelevant to me.)

>  ....   On the contrary,
> detailed calculations show that it doesn't matter that the tape is not part
> of an ergodic system.

I believe it definitely does matter whether the underlying dynamics is
ergodic.  I don't see how the calculations you mention can make it not
matter.

>  Similarly, it doesn't matter that you cannot afford
> to exhibit the entire ensemble of tapes.  The dirty tape is entropic -- get
> used to it.

So you say.

> The correct physical argument goes something like this:  Imagine that bits
> are represented by the position of a particle in a double-well
> potential.  The left well represents zero, and the right well represents
> one.  Imagine now a machine to erase a large number of such bits.  If it
> knows the bit pattern, it knows where to find each particle, so it can grab
> each particle that needs to be moved and isentropically move it.  OTOH if
> it doesn't know which ones need to be moved, it has to guess, and each time
> it guesses wrong the particle will dissipate something like kT of
> energy.  You can visualize trying to scoop up the particle with a scoop big
> enough to encompass the zero-well and the one-well;  if the scoop is that
> big the particle will go "clunk" when it gets scooped.  For details, please
> refer to the scintillatingly clear and scholarly papers by Charlie Bennett
> and Rolf Landauer.

With all due respect to the interesting work of Landauer, et al.
regarding the thermodynamic cost of computation, what does this have to
do with whether or not the entropy difference between the total
microscopic entropy consistent with a macrostate description and the
entropy of just the dynamically accessible microstates consistent with
that macrostate description actually counts as thermodynamic entropy?

> ---
> A reversible computer has the interesting property that the thermodynamic
> cost of doing the computation is only the cost of writing down the answer
> -- and writing on blank tape is costless.  Writing on dirty tape is
> costly.  Blank tape is more valuable than dirty tape.

That's nice.  How is it relevant?

> These ideas are central to the work of Seth Lloyd (which as you may recall
> was the original subject of this thread).

It's true that those ideas are very important for Lloyd's ongoing work
(everything from his introduction of his thermodynamic depth-type measure
of complexity to his recent work mentioned in the news article), but
these ideas (thermodynamic cost of computation), and Lloyd's work do not
appear to me to address the current relevant issue.

Please explain more fully so I can understand why you think the entropy
associated with quenched disorder (i.e. essentially the logarithm of the
number of separate dynamical domains in phase space that are effectively
dynamically disconnected from each other over a time scale relevant to
experimental equilibration time scales, such that all the microstates in
such domains are consistent with the system's macrostate description)
actually does count as contributing to the thermodynamic entropy of a
system.  For some reason I can't figure out why arguments about which
steps in a computation do and do not have a thermodynamic cost have
anything to do with whether or not the entropy of a quenched system
really is or is not the same as that of a well-equilibrated/annealed one.

David Bowman
David_Bowman@georgetowncollege.edu