Prigogine's theorem of minimum entropy production (Pentcho Valev)

RUMYM at BGEARN.ACAD.BG RUMYM at BGEARN.ACAD.BG
Tue Nov 4 13:57:29 EST 1997


In reply to Bryant Fujimoto

Maybe it is not very suitable to present so special problems on this
forum. On the other hand, nonequilibrium thermodynamics is regarded as
very promising for the future theoretical biology. I addressed this
message to you, in the btk group, a few months ago. You were tired and did
not answer, but Herbert Sauro replied and confirmed my conclusion. As you
can see, the problem is still existing, and while nobody should care about
my personal problems, everybody should care about scientific ones.

That was my message:
Bryant, I do not find your answers satisfactory. Of course, I do not blame
you - nonequilibrium thermo is so complicated and abstruse that maybe one
can only use some final results - verifying all mathematical technicalities
is too difficult. Still I think that any theory - even the most respectable
one - should be verified in detail from time to time. That is what I am
going to do - I hope nobody would find this obtruding - maybe I will even
receive some help.
   Let me describe an interesting paradox I have recently found in textbooks.
Maybe I am wrong, maybe the textbooks are wrong, but still the problem should
somehow be settled.
   Consider the following process (the notation is somewhat strange, due to
e-mail restrictions):

   S1  -v1->  S2  -v2->  S3                             /1/

S1 is converted into S2 (at a rate v1), and S2 is converted into S3 (at a rate
v2).
   In a steady state,

   v1 = v2 = v                                          /2/

The affinity is

   A = mu(S1) - mu(S3)                                  /3/

Also, it is said in textbooks that, in a steady state,

  J = LA = dC(S2)/dt = 0                                /4/

where C(S2) is the concentration of S2.
   The condition /4/ is essential for the derivation of Prigogine's
theorem of minimum entropy production:

   B = JA = LA^2                                        /5/

where B = T(dSi/dt) is the entropy production. Therefore,

   dB/dA = 2LA = 2J = 0                                 /6/

i.e. the conditions of steady state and minimum entropy production are
equivalent.
   In my opinion, /4/ is wrong. The flow, J, corresponding to A, is not
dC(S2)/dt, but

   J = LA = v                                           /7/

Therefore,

   dB/dA = 2LA = 2J = 2v                                /8/

This means that, in a steady state, the entropy production is not minimal.
Rather, it is minimal at equilibrium (v = 0), what is, of course, a
triviality.
   I am waiting for someone to refute or confirm my analysis.

Pentcho



HERBERT SAURO' s reply was:
>   Consider the following process (the notation is somewhat strange, due to
>e-mail restrictions):
>
>   S1  -v1->  S2  -v2->  S3                             /1/
>
>S1 is converted into S2 (at a rate v1), and S2 is converted into S3 (at a rate
>v2).
>   In a steady state,
>
>   v1 = v2 = v                                          /2/
>
>The affinity is
>
>   A = mu(S1) - mu(S3)                                  /3/
>
>Also, it is said in textbooks that, in a steady state,
>
>  J = LA = dC(S2)/dt = 0                                /4/
>
>where C(S2) is the concentration of S2.
>   The condition /4/ is essential for the derivation of Prigogine's
>theorem of minimum entropy production:
>
>   B = JA = LA^2                                        /5/
>
>where B = T(dSi/dt) is the entropy production. Therefore,
>
>   dB/dA = 2LA = 2J = 0                                 /6/
>
>i.e. the conditions of steady state and minimum entropy production are
>equivalent.
>   In my opinion, /4/ is wrong. The flow, J, corresponding to A, is not
>dC(S2)/dt, but
>
>   J = LA = v                                           /7/
>
I think the problem lies in the way J is interpreted and I think that
depends on how the problem is cast. One of the problems I've had with
non-equil thermo is that in it's attempt to be so general the physical
interpretation of the quantities is missed. I've seen in books and
papers different meanings given to J, somtimes dC/dt other times v, I
can only assume that the differences are either errors on the part of
the authors or it's a problem dependent matter. Most of the references
state that Js are rates of reaction so /4/ is wrong in that sense but
this only appears to make the problem worse. If the flows are indeed the
rates then one ends up with the problem you cite, ie the rates do not
vanish at steady state, therefore the steady state is not the state of
minimum entropy production. I think it is more likely that we are doing
something very wrong here, but having never understood non-equil thermo
fully anyway, I don't know what.

Here's a sentence that's confused me:

'If X1, the input or driving force, is held constant, but no restriction
is placed on X2, the flow J1 will continue until X2 achieves a value
adequate to bring it to a halt' - fair enough, I get that, even if it an
odd way to describe the approach to steady state, but the next bit:

'Thereafter J1 will remain zero' - what, zero? - 'and thus X2 will
remain constant so long as X1 and L coeffieients retain their original
values.'

If the Js are indeed rates of reaction then how can J1 remain zero -
they seem to be talking about equilibrium not steady state.

I would love to see a concrete problem such as the one you give above
cast in non-equil thermo so I can see what symbols mean what. Anybody
like to offer?

>Therefore,
>
>   dB/dA = 2LA = 2J = 2v                                /8/
>
>This means that, in a steady state, the entropy production is not minimal.
>Rather, it is minimal at equilibrium (v = 0), what is, of course, a
>triviality.
>
>Pentcho

Herbert
--
Herbert M Sauro
email:     HSauro at fssc.demon.co.uk
Telephone: 01974 282428

Again, I am sorry for presenting so many technicalities. Nobody of course
should feel forced to go through this - maybe only you Bryant, when you are
not so tired, will understand how fatal Prigogine's mistake is and what
sad conclusions should be drawn.

Pentcho



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