Re: function of state vs exact differentials

From: Alfred Einstead (whopkins_at_csd.uwm.edu)
Date: 06/30/04 

Date: 30 Jun 2004 11:24:17 -0700

"Anja" <anja@no.spam.com> wrote:
> if X is a function of state then is dX an exact differential?
> Is it actually a biimplication?

That's not the correct question. An exact differential form W
is always of the form dX for some function, by definition of
"exact".

The question you're trying to ask is whether a *closed*
differential form W is always exact, W = dX. W is closed
if dW = 0. For 1-forms, this boils down to the question
of whether and when (for instance) W = A dx + B dy with
dA/dy - dB/dx = 0 implies that A = dX/dx and B = dX/dy.
For 2-forms, for instance, with W = A dy^dz + B dz^dx + C dx^dy
the question is whether and when dA/dx + dB/dy + dC/dz = 0
implies that A = dX/dx, B = dX/dy and C = dX/dz.

If the variables (x,y) range over some domain S in which the
following is true:
           every path P from any points (x0,y0) to (x1,y1)
           in S can be deformed continuously to any other
           path P' from (x0,y0) to (x1,y1)
then it is true that dA/dy - dB/dx = 0 implies A = dX/dx,B=dX/dy.
A similar condition holds for the second example cited for
the 2-form W = A dy^dz + B dz^dx + C dx^dy.

In the case you're interested in, if the system's state is
given by the coordinates (q1,...,qn,p1,...,pn) for some n > 0,
then a 1-form W is given by
          W = Q1 d(q1) + ... + Qn d(qn) + P1 d(p1) + ... + Pn d(pn).
The domain S over which (q1,...,qn,p1,...,pn) should have the
above-mentioned path property. Then it will be true that a
closed differential form W
                dW = 0, that is:
                d(Qi)/d(qj) = d(Qj)/d(qi); i,j = 1,...,n
                d(Qi)/d(pj) = d(Pj)/d(qi); i,j = 1,...,n
                d(Pi)/d(pj) = d(Pj)/d(pi); i,j = 1,...,n
implies W is exact, W = dX, or
                Qi = dX/d(qi), Pi = dX/d(pi); i = 1,...,n.