Re: [Phys-l] Gibbs paradox (redux)

[John Denker <jsd@av8n.com>]

On 06/26/2011 03:09 PM, Carl Mungan wrote:
> > > By dividing the glass of milk into 2 equal portions, I seem to
> > > have decreased the entropy of the system. This seems like a
> > > strange result: where is corresponding gain of entropy in the
> > > surroundings to ensure the 2nd law is not violated?
> >
> > Perhaps it would help to start with chocolate milk on one side and 
> > white milk laced with radioactive cesium on the other.  Remove the 
> > partition.  Observe the mixing.  Re-insert the partition.
> > Unmixing does not occur.
>
> I accept what you say for chocolate and white milk. Also for the
> cards.

:-)

> But I'm still stuck on what happens with just a glass of milk. My
> problem is that *macroscopically* I can't tell any difference between
> two half glasses of milk and one full glass of milk, in contrast to
> the case of chocolate and white where I can see the color gradations.
> Shouldn't there be some *macroscopically measurable* evidence of a
> change in entropy?

Again Carl gets the prize for asking the incisive question.

The short answer is no, you can't generally tell by macroscopic
measurements whether something irreversible is happening ...
but there is much more to the story than that.

I need to amend and extend the answer I gave to the question that 
was asked yesterday about the entropy of mixing.  Let's again use 
the deck-of-cards model system.  I'm pretty sure it is a faithful 
model of the milk.

a) Suppose I have three cards, namely the deuce, trey, and four of 
clubs.  I shuffle these three cards.  That gives 2.6 bits of entropy,
i.e. log2(3 factorial).

Let's call this the entropy of permutation of the hand.  For some
purposes we could consider this "the" entropy, but not for all purposes.
We must keep in mind that it is a conditional entropy, conditioned 
on knowing *which* three cards are in the hand.

b) If I shuffle a new deck and deal three cards at random, in some sense 
there is about 17 bits of entropy in that event, which breaks down as
14.4 bits associated with which cards got dealt, and the aforementioned
2.6 bits associated with the order in which they were dealt.

  Note the total -- i.e. 17 bits -- is explained as log2(52*51*50).

This tells us that the lion's share of the entropy is not the
entropy of permutation within the hand, but rather the entropy of
*which* cards were dealt into the hand.  I call this "the entropy 
of the deal".

  The entropy of the deal, by definition, does not include the
  entropy of permutation within the sample.

Now, for a sample of helium, the usual calculation for the entropy
rightly ignores the entropy of the deal because all helium atoms
are identical.  The entropy of the deal is zero.  We calculate
then entropy within the "hand" (within the sample) and rightly
call it quits at that point.

In contrast, for the milk, a proper accounting of the absolute entropy 
must account for the entropy of the deal.  If we don't do this, we are 
only calculating a conditional entropy, conditioned on knowing *which* 
globules got dealt into that sample of milk.

The formulas I gave yesterday did not include the entropy of the
deal, and therefore must be considered incomplete.

There are two scenarios:

a) We deal out two samples of milk.  In each sample, the lion's share
of the entropy is the entropy of the deal.  We don't peek at samples.
By that I mean we do not scrutinize or measure them in any way.
That is, we do not collect any information that would reduce the
entropy.

We pull out the partition.  The samples mix.  We put the partition
back in.  Unlike the scenario I considered yesterday, in this scenario 
the entropy does not increase.  Both samples were in a maximum entropy
to begin with, so the mixing could not possibly increase the entropy.
All we have done is re-deal the globules in a slightly complicated
and peculiar way.  Re-inserting the partition has no effect.

If we replace the samples of milk by hands of three cards, the total
entropy of the joint system would be 33.8 bits, i.e. log2(52*2*50*49*48*47).
This would be the entropy before and after mixing.  

b) Same as above, except that we peek at each of the samples before
mixing.  This reduces the entropy to zero temporarily.

Then we wait a little while, to allow permutation of the globules
within each sample.  The entropy rises until it equals the entropy
of permutation within the sample.  This is the entropy given by
my formulas yesterday.  For cards, this would be the aforementioned
2.6 bits per hand, for a system-wide total of 5.2 bits ... much less 
than the entropy in scenario (a).

If we now mix the two samples, the entropy goes up.  This is wildly
different from what happened in scenario (a).  This is the scenaroio
I considered yesterday.  For three-card hands, the entropy is now 
9.5 bits, i.e. log2(6 factorial).  It's not any larger than that, 
because we know *which* six cards are involved, because we peeked.  
Peeking zeroed out the entropy of the deal.

  FWIW I suspect that tagging one sample of milk with chocolate could 
  be considered a "weak peek" ... in the sense that it allows us to 
  look at the milk and get some partial information about it.

===================

Bottom line:  I reckon that the original question from yesterday
was to some degree underspecified.  As a result, my original 
answer was, alas, incomplete ... which I did not realize until 
the follow-up question called attention to it.

Depending on heretofore-unspecified details of how the milk
samples are handled before mixing, the entropy of mixing
could be zero or distinctly nonzero.  That's because the
entropy of deal the could be zero or could be completely 
dominant.