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Re: TdS is not dQ or d(anything)



Bob LaMontagne wrote:

> I'm confused. Unless there is some redefinition of dV going on,
>
> dv = (dv/dT) dT + (dv/dP) dP,
>
> where the first term in parenthesis is evaluated at constant P and the
> second at constant T. At a particular point, dT and dP are arbitrary,
> which means dv at a particular point is not unique - so how can it be a
> state function?

I agree that dV is no less (and no more!) arbitrary
than dT and dP. I view all three as functions of state.

So the question is, what is the basis for asserting
that dT and/or dP is arbitrary.

As mentioned in a previous note, I think any
"redefinition" going on is very, very slight.
Some people seem to think that dP is the change
in P due to a small step in some unspecified
direction. I prefer to think of dP as a machine
which will tell you the exact change in P if you
give it an exact description of the step. The
output of the machine is arbitrary (unless/until
the input is specified) but the machine itself is
not arbitrary at all.

I don't feel that I'm redefining anything, just
shifting attention from the output of the machine
to the machine itself.

I find the resulting decrease in arbitrariness to
be tremendously powerful. For starters, it means
we can draw pictures of dE and dT and dP.
http://www.monmouth.com/~jsd/physics/thermo-forms.htm

By way of analogy: You could consider any matrix to
be a machine that says: if you give me an input
vector, I'll give you an output vector. The output
of this machine is arbitrary unless/until you specify
the input vector. That that doesn't mean the matrix
itself is arbitrary!

This analogy is very, very tight. If state space
is two-dimensional (e.g. spanned by T and V) then
dP at each point in that space can be represented
by non-square matrix (one row and two columns) that
says: give me an input vector (specifying the
direction and magnitude of a step in state space)
and I'll give you a scalar which is the change in
P that results from that step.