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Re: [Phys-l] thermo differential and extensive/intensive variables

On 06/15/2007 09:22 AM, Stefan Jeglinski wrote:

Here's my attempt to formulate thermodynamics without the usual

I was reading this, but missed a link to your partial derivatives page (maybe it's there) you also cited. So obviously, my example is doubly wrong:

E(S,T) -> dE = (dE/dS)(const T) dS + (dE/dT)(const S) dT [1]
= (!T)dS + (!Cv)dT [2]

(! means "not")


Both [1] and [2] are correct as written (including the "!").

So it might seem that the most general way of approaching thermo problems is to assume that all variables are always included, for example E = E(S,V,N,P,T), and an equation of state (eg PV=NkT). I haven't looked at this yet, but I don't recall thermo texts with this approach. At first glance it would appear that this approach would not lead to elegant relationships, but this may be a premature conclusion.

As a point of /advanced/ mathematics, there is nothing "wrong" with
writing E = E(S,V,N,P,T) plus some collection of constraints,
although you will typically need more than just the equation of
state; a typical problem will have boundary conditions, too.

OTOH I don't recommend it. Instead I recommend using a complete
but not overcomplete set of variables. That is, as a matter of
convenience and elegance, you should find a set of variables
that is
-- sufficient to span the space (i.e. complete), yet
-- linearly independent (i.e. not overcomplete).

In particular, equation [1] above will not be correct if the
variables T and S are linearly dependent.

The reason I was reviewing this was to reconsider the simple heating/cooling of water (for example) without resorting to the dE = d'Q + d'W formulation, as per previous discussions here. To use J Denker's terminology, uncramped vs cramped (if I followed it correctly). But I'm finding the going tough. The standard treatment is simply "you heat the water, d'Q = CvdT, and if you want the entropy change, solve TdS = CvdT."

That paragraph is easily debugged.

1) Interesting part: The words talk about energy, which is correct ...
so make the symbols agree with the words. Just get rid of every mention
of "Q" and "d'", and then the rest is correct. Specifically:
"you heat the water, d'Q = CvdT"
becomes "you heat the water, dE = Cv dT".

It's just that simple. Thermodynamics is not nearly so complicated and
ugly as the thermo books make it seem.

2) Nitpick: It's obvious from context, but we ought to make it explicit
that we are working under conditions of constant volume.

In general there is another term in the expansion of dE, but it vanishes
under conditions of constant V:


Take-home lessons:
*) In uncramped thermodynamics, there is no such thing as Q. If you
are tempted to talk about Q, find something else (perhaps E or S) to
talk about instead.
*) There is absolutely positively no such thing as d'.
-- There is a d, which can be rigorously formalized as the exterior
derivative operator. AFAICT, anything worth differentiating can be
differentiated using d.
-- In contrast, when the books write d' it is like saying
"we know this is wrong but we're going to do it anyway".

See also next note.