integrability in thermodynamics

[rge11x]

Hello

Hannay ("Carnot and the fields formulation of elementary
thermodynamics," Am. J. Phys, v74, Feb 2006 pp134-140) writes the 1st
and 2nd laws as d^w'+d^q'= 0 (1st) and d^(q'/T) = 0 (2nd). Here the
prime ' means a co-vector and ^ is the wedge product.

In both cases it is true locally in a simple domain (one that can be
deformed to a single point) of the thermodynamic variables that w'+q' =
dE and q'/T = dS but why is this true "globally"? In other words, what
assures that the domain of thermodynamic variables is always simple; is
this a separate assumption, the "(-1)th law"? What happens when it is
not true?

Thanks
Robert

[Norbert Dragon]

* rge11x wrote:

> In both cases it is true locally in a simple domain (one that can be

^^^^^^
simply connected

> deformed to a single point) of the thermodynamic variables that w'+q' =
> dE and q'/T = dS but why is this true "globally"? In other words, what
> assures that the domain of thermodynamic variables is always simple; is
> this a separate assumption, the "(-1)th law"? What happens when it is
> not true?

This cannot happen. Each domain D which is not simply connected defines
a simply connected, universal covering space C. If a thermodynamic
potential turned out to be a multivalued function of D then this
would prove C to be the real domain of the thermodynamics variables.

--
Superstition brings Bad Luck

www.itp.uni-hannover.de/~dragon

[rge11x]

If I understand you correctly you seem to imply that since the
thermodynamic potentials must obviously be single-valued, therefore the
domain of the variables must be singly connected.

Here is what bothering me: Consider a thermodynamic system of three
variables that has two phases. Is it possible that the domain of one of
the phases is a doughnut and the other phase is the complement of the
doughnut? I have never heard of such case but if that were possible then
the integrability would also require either the proof or the assumption
that the integral of any closed loop of w'+q' and that of q'/T be equal
to zero for loops confined within either of the phases. Or can we just
say that the variable domain is the whole space and we are integrating
across the phase boundary but then don't we need some argument to
support that this can be done since the functional form of the dependent
variables changes with the phase?

thanks

[Norbert Dragon]

* rge11x wrote:

> If I understand you correctly you seem to imply that since the
> thermodynamic potentials must obviously be single-valued, therefore the
> domain of the variables must be singly connected.

Exactly. If the thermodynamic potential f is multivalued, i.e. if
at x it has values f_1(x), f_2(x), ..., depending on which path
Gamma_1, Gamma_2, ... from 0 to x you chose to integrate changes of f,
then you must not identify the endpoints of the different paths but
distinguish x_1, which you reached via Gamma_1, from x_2, which you
reached via Gamma_2. Then the potential is a function f with
f_1 = f(x_1) and f_2 = f(x_2).

As simple, though unrealistic example: How did Magellan know that he
returned to Lisbon and not to a different Lisbon'? On the sphere, he
could have contracted his closed path to smaller and smaller paths
to the point Lisbon. But not on a torus. There he would have to
investigate all properties of Lisbon'. If one of them depended on the
winding number, this would prove Lisbon' to be different from Lisbon.