Re: physical meaning of entropy

Juan R. González-Álvarez schrieb:
Arnold Neumaier wrote on Sat, 03 Jan 2009 02:19:09 +0000:

Juan R. Gonzalez-Alvarez schrieb:
Arnold Neumaier wrote on Wed, 29 Oct 2008 11:43:02 -0400:


I have downloaded a copy and will study it with care [#].

[#] In a faster scan I can see several interesting (e.g. about mixture
of water and oil) and serious remarks, but I can also see several
strong mistakes (you write about Gibbs potential and entropy is
plain wrong. Entropy *is* maximal in your cup of coffee example,
its convex character guaranties the evolution back to equilibrium
after applying any small perturbation over the system). It can be
proved that condition of minimum for G is a consequence *derived*
from the condition of maximum for S.
Please detail your argument to substantiate your claims. I don't believe
them.

Whether the entropy in a process is maximized or a free energy is
minimized depends on the boundary conditions. Entropy is maximal only
under conditions that guarantee that N,V,H are kept constant.

Using deDonder formulae

dS = d_iS + d_eS

The second law states that, with *independence* of boundaries, production
is non-negative

d_iS >= 0

The applicability of Gibbs function as potential is limited

G == U + pV - TS

dG = dU + pdV + Vdp - TdS - SdT

dG = dQ + Vdp - Td_iS - Td_eS - SdT

For a closed system, Td_eS = dQ

dG = Vdp - Td_iS - SdT

Using the second law

dG =< Vdp - SdT

if, and only if, p and T are constant then dG =< 0

G evolves to a minimum only when p and T are held constant for a closed
system.

The law (d_iS >= 0) is valid also when p and T are not constant or when
the system is open.

d_iS >= 0 only says that what is called called entropy production is
nonnegative. But this not imply that entropy is maximal. Indeed, the
local change of entropy is dS = d_iS + d_eS, so locally, entropy can
increase or decrease, and one cannot say anything in general about
the global entropy.

In particular, in conditions where P and T are constant, as in the cup
of coffee, it is known that the Gibbs free energy decreases, but no
assertion is possible about the behavior of the entropy.


But when stirring a cup of coffee, V and H are not controlled. Thus
there is no theory available to infer that entropy must increase.

In your book you state that "the cup of coffee ends up in a state of
minimal Gibbs energy and not in a state of maximal entropy!" This is plain
wrong.

First, the role of potential for G is derived from S as showed above.

Second, when the cup of coffee ends up in the final state there are no
flows d_eS = 0, thus

dS >= 0

dS is short for the time derivative of S. Thus, in the final state,
dS =0 and dG = 0, since there is no flow anymore. This just says
that equilibrium is reached. It says nothing about maximality or
minimality of S or G.

But since G decreased all the time, G ends up at a minimum
(according to the relaxed rigor of typical physicist arguments),
while since S has no determined direction, it can end up anywhere.


With the inequality valid only when there in internal inhomogeneities, for
instance fluctuations in local concentration or temperature.

Which can be excluded for the end state of the cup of coffee. Thus dS=0.


It is the non-negative character of dS that guarantizes that any
spontaneous fluctuation is compensated and the system returns to
equilibrium.

This is irrelevant to our discussion of the behavior of S and G.
Outside equilibrium, only d_iS >= 0, and one can deduce nothing about
S but an inequality for dG. At equilibrium, dS=0, and again nothing can
be concluded for S.


Moreover, your book considers less general thermodynamic systems
than the other references I already cited.
To substantiate your claim, please give an example of a thermodynamic
system not covered by Chapter 6 of my book.

What about matter in gravitational fields? Non-extensive systems?
Fluctuations? Dissipative structures? relativistic effects? What about the
Gibbs energy for a rotating system?...

Only nonextensive systems are not covered. But all the formulas you
quoted in your mail are not valid either in such systems.

All other systems you mentioned are covered by my concept of a
thermodynamic system. See Table 7.1 on p. 136.

The current description only refers to a fixed moment in time;
dynamical aspects are not covered but only hinted at on p. 63
and p.142. They will be treated in a subsequent volume that
I am currently preparing.


Arnold Neumaier

.

Relevant Pages

  • Re: physical meaning of entropy
    ... The applicability of Gibbs function as potential is limited ... But this not imply that entropy is maximal. ... In a single phase equilibrium state (as in the cup of coffee), ...
    (sci.physics.research)
  • Re: physical meaning of entropy
    ... perturbation theory) apply a so-called "Markovian approximation" to the ... are the kind of systems studied by Extended irreversible thermodynamics ... In my opinion, this is still local equilibrium, just with more fields. ... Entropy *is* maximal in your cup of coffee example, ...
    (sci.physics.research)
  • Re: From Maximal Entropy Random Walk to Quantum Thermodynamics
    ... I don't know what you refer as Jaynes maximum entropy theory - ... Then I answered how in stochastic processes is defined entropy ... limit we can only assume stationary probability density - equilibrium ...
    (sci.physics.research)
  • Re: Ludwig Boltzmann, entropy
    ... > The site you refer to above is the creation of Rob Swenson of ... What seems to have confused Swenson is the fact that ... Boltzmann's definition of the entropy of an equilibrium ...
    (talk.origins)