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Re: [Phys-L] work versus mechanical transfer of energy



On 01/06/2016 05:02 PM, Diego Saravia wrote:

You have U(S,V) and its partial derivatives T and -P

You assume that you have in all points in that path U S V T and P defined,
they are continuos and derivable.

Do you have an example of a irreversible process in that case?

___________ ___________
| X X |
| E1 X X E2 |
| X ~~~> X |
| S1 X X S2 |
| X <~~~ X |
| T1 X X T2 |
|___________X X___________|


Two huge chunks of metal. E1 is known as a function of S1,
differentiable and well behaved in all ways. Ditto for E2.
The chunks are black on the side marked X, but shiny and
thermally insulated elsewhere. Black-body radiation is
emitted by block 1 and absorbed by block 2, and vice versa.
The geometry is such that negligible radiation escapes,
so the two-block system is closed.

The ambient pressure is zero /and/ the blocks have constant
volume, so the P dV terms are negligible squared. Assume
that the energy and entropy stored in the radiation
field at any given moment are negligible.

Assume WLoG that initially T2 > T1. Then over time block
1 will gain energy from block 2. This continues until the
two blocks asymptotically approach a common temperature.
The whole process is grossly irreversible. Because P and
V are negligible, the path through thermodynamic state
space is obvious and ultra-simple. The analysis is as
simple as could be.

Here's another version, where the radiation has been replaced
by a filament with some small but nonzero thermal conductivity.
Assume the energy and entropy stored in the filament at any
given moment are negligible.
___________ ___________
| | | |
| E1 | | E2 |
| | | |
| S1 | | S2 |
| |~~~~~~~~~~| |
| T1 | | T2 |
|___________| |___________|


I do not understand the taylor expansion thing.

I sympathize. The whole "Taylor" thing strikes me as both
unnecessarily complicated and unnecessarily unsophisticated.

I assume that equation is a
differential equation along a path.

Actually, if you do it right, the differential equation is
valid at each point in thermodynamic state space. You can
imagine it as a point along the path, but the differential
equation doesn't actually care what the overall path is;
the derivatives are defined in a local neighborhood.

It takes a fair bit of sophistication to fully formalize
things in this local, path-agnostic form ... but the
resulting equations are the same as what you would get
via any other approach, e.g. things like
dE = T dS - P dV [1a]

One crucial step is realizing that dE is a /vector/, in
particular the gradient vector in some abstract high-
dimensional space. You absolutely must not think of it
as some "infinitesimal". Mathematicians have tried for
350 years to make infinitesimals work, without success.
However, circa 1900 Cartan figured out how to make
exterior derivatives work. This allows us to write
equations such as [1a] and actually make sense of them.

When I say this approach reproduces all the equations of
classical thermodynamics, I should say that it reproduces
only the /correct/ ones. There are a bunch of equations
you find in thermo books that never made any sense. It
is however easy to live without them and/or replace them
with things that do make sense.