X-Message-Number: 20857
Date: Mon, 13 Jan 2003 18:52:22 -0800 (PST)
From: Driven FromThePack <>
Subject: re: ettinger on heat death of the universe

From another forum, a poster replied to Mr Ettinger's
post on thermodynamics and the heat death of the
universe. Here are some excerpts from that post:


Quoting Mr Ettinger: "Actually, entropy is a concept
poorly understood by most people...." 

The reply:

....

This indicates a basic misunderstanding most people
have with entropy. This is because very often people
are not taught the connection between microscopic
states, and entropy associated with a macroscopic
state. You...are making a really fundamental mistake
based on the fact that you completely do not
understand what entropy really is. 

Now, Boltzmann tells us that when all configurations
are equally likely, S = k ln W where k is Boltzmann's
constant and W is the number of states accessible to
the system. 

[This] means that the more states that are accessible
to the system, the higher the entropy. Another way to
think of this is the broader the probability
distribution, the higher entropy is. 

.... a macroscopic example first. I have N books on my
shelf, which I can order in N! ways. Let's ask a
question. What is the entropy of the current
arrangement? Well, S = k ln W. What's W? Well, when I
say the current arrangement of my bookshelf, I am
applying a constraint. That is, I'm saying, my books
as currently ordered. In other words, there is 1
configuration possible. So the entropy is 0. In other
words, it doesn't matter how "disordered" my books
are; if I find the entropy of their current state,
it's 0. This is the same as the entropy of the books
if put in alphabetical order (assuming I have no
duplicates). On the other hand, what is the entropy of
all possible configurations? In this case, I can
arrange my bookshelf in N! possible ways, so the
entropy is k ln N! which is larger than 0 as long as N
> 1. 

Here's why this is important: talking about the
entropy of a specific configuration or state is
[wrong]. It tells you nothing. One state is not
inherently more ordered or less ordered than another. 

....

so how do we evaluate entropy? Well, we can do that
because we never actually know the EXACT state of a
system. If we know that we have 2 liters of helium gas
at 298 K, we can call that our "state" of the system.
But factually, we don't know the exact microscopic
state of the system. We don't know where each gas
molecule is. And frankly, we don't care. Because
talking about the entropy of a single configuration
doesn't make any sense; it's zero. When we talk about
the entropy of our system, we are talking about all
possible configurations which are accessible to the
system. Now what does the second law say? It says that
over time, you expect your system to maximize its
probability distribution. That is, over time, you
would expect your system to try and explore as much of
configuration space as it can. You would *not* expect,
for instance, that if you had a 2 L tank that your gas
would stay on one side of the tank for an arbitrarily
long time, leaving one liter of vacuum on the other
side. 

Now, imagine that our gas molecules in the two liter
tank had M possible configurations, with an entropy of
S = k ln M. Now, how could we make the entropy
decrease? There's only one way--we have to decrease
the number of possible configurations it can visit.
And in order to do that, we need to impose some
restrictions on our system that would prevent it from
entering some of those configurations. Once you impose
those restrictions, though, you are no longer in an
equilibrium state. So the state of highest entropy is
always the equilibrium state. 

Now what you claim above is that you start in one
configuration, which you say is "highest entropy".
That you eventually must move from this state,
implying entropy decreases. This is completely wrong,
because you're mixing up microscopic configurations
with macroscopic descriptions of entropy. 

....


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