Re: physical meaning of entropy

Juan R. González-Álvarez schrieb:
Arnold Neumaier wrote on Thu, 29 Jan 2009 00:00:50 +0100:

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.

(snip)

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.

(snip)

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.

I will remark that

dG = Vdp - Td_iS - SdT (for closed system)

dS = d_iS + Td_eS (for any system)

In an equilibrium state dp = dT = d_eS = 0, thus

dG = - Td_iS =< 0

dS = d_iS >= 0

The step from these two inequalities to you conclusion:

G is minimum and S is maximum at equilibrium.

is logically unfounded.

In a single phase equilibrium state (as in the cup of coffee),
dG=0 and dS=0, since equilibrium is stable and changes nothing.
This is stronger than your inequalities, but only asserts that
G and S do not change (of course, else equilibrium were not
yet reached). The equations therefore contain no other information.

Thus from consideration of equilibrium alone, no assertion about
being minimal or maximal (which is a statement about comparing the
state to nonequilibrium states with the same boundary condition)
can be obtained.


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.

It is at the root of your confusion. In your reply you *deleted* a
fundamental part of my message that was:

"Gibbs theory of stability is based in the virtual variations of second
order: (delta^2)S < 0."

Unfortunately, this is not universally valid close to equilibrium,
but holds only in a closed and isolated system, which the cup of
coffee is not.

For the cup of coffee, we have instead (delta^2)G > 0, with exactly
the same derivation as for (delta^2)S < 0 in the closed and isolated
case, and with the same implications in case of a system at constant
T and P.


I will illustrate application of Gibbs theory to your "cup of coffee".

The system is in an equilibrium state with T_eq, U_eq, P_eq, V_eq,
mu_i,eq, and N_i,eq for each chemical specie i. Consider a thermal virtual
process to a nonequilibrium state (S_eq --> S). A Taylor expansion of S
around equilbrium gives

S = S_eq + deltaS + 1/2 delta^2S + �·�·�·

where

deltaS = (part S / part U)_eq delta U

delta^2S = (part^2 S / part U^2)_eq (delta U)^2

Here you use the equation of state in the form of S=S(U,V,N).
You can easily see that your formula only holds at constant V and N.
But nothing keeps V constant in the cup of coffee example.
Instead, P is kept constant.


At equilibrium (part S / part U)_eq = 0 and

S = S_eq + 1/2 delta^2S + �·�·�·

Since equilibrium entropy is maximum, it follows that (delta^2S < 0) so

delta^2S = - (C_V (delta T)^2 / T^2) < 0

This is the well-known Gibbs condition for thermal stability.

Gibbs condition for *entropy* explains why your "cup of coffee" remains at
constant T at equilibrium. Any thermal fluctuation (S_eq --> S) driving
the system away from equilibrium will decrease entropy so that spontaneous
entropy increasing processes (S --> S_eq) will drive the system back to
the state of equilibrium.

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.

Table in Chapter 7 contains an interesting collection of variables. It
covers gravitational fields and rotating systems but fluctuations and
dissipative structures are not covered in any part.

Fluctuations are covered in Theorem 6.3.3 and its subsequent discussion.

Dissipative structures are intrinsically time-dependent, hence cannot
be treated in a static thermodynamic framework as that considered in
my book. They will be covered in a future volume II.


Also I cannot find any section discussing relativistic effects in Chapter
6. What do you prefer, scalar coldness \beta and Lorentz covariance or
four-coldness \beta^\mu and invariance?

The current description only refers to a fixed moment in time;

Where does your book deal with stationary states where S(t) /= S_eq?

At a fixed moment, they are still characterized by the theory given in
Chapter 6, using the correspondence of variables given in Chapter 7.
But discussing stationaity requires a dynamical formulation, which will
be treated via master equations ans stochastic differential equations,
the subject of a future Volume II.

One cannot do everything in one book. But the formulation given is given
with foresight to what is needed for Volume II, and hence is completely
general.

Thus you'd be right in claiming that my book is incomplete, but not in
claiming that it is not general enough.


Arnold Neumaier

.

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