Re: TdS is not dQ or d(anything)

["John S. Denker" <jsd@MONMOUTH.COM>]

On 05/13/2003 08:36 PM, Bob Sciamanda wrote:
| >>The exterior derivative of the 0-form f is f_{x}dx+f_{y}dy+f_{z}dz
| >>(f_{x} denotes the partial of f with respect to x.
> grad f would be the vector f_{x} [i] + f_{y} [j] +  f_z [k]
> and dr is the vector dx[i] = dy[j] + dz[k]
>
> so that the above 0-form "f_{x}dx+f_{y}dy+f_{z}dz" is clearly grad f (dot) dr
>
> or are you now defining dx, dy and dz to be the unit vectors [i] [j] and [k] ???
>
> This would indeed be a cruel usurpation and corruption  of well established
> notation.

Surprise!  dx, dy, and dz are precisely the
basis one-forms in (x, y, z) space.

This is not only well established notation, it
is the only possible logical notation.  It is
an interesting exercise to choose something else
such as FOO_x, FOO_y, and FOO_z as the basis
one-forms, and then prove that dx = FOO_x, so
dx is equal to a basis one-form whether you
wanted it or not.  Hint:  start from
   df = f_{x}FOO_x + f_{y}FOO_y + f_{z}FOO_z
and take f(x) = x.

This and other properties of one-forms are
discussed at
  http://www.monmouth.com/~jsd/physics/thermo-forms.htm#sec-def-forms

Differential forms are not something that
just crawled out from under a rock.

====================

I wrote:
>>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.

On 05/13/2003 04:17 PM, Bob LaMontagne wrote:
>
> I'm still missing how dP is a state function.
> If I give your machine two
> different sets of input vectors, I get two different
> scalar dp values. How do
> you reconcile this lack of uniqueness with a state function?

The point is that the MACHINE is a well-defined
state function.

Let's forget about thermo for a moment, and
let's forget about one-forms.  Let's talk
about plain old vector fields.  In particular,
imagine pressure as a function of position
in (x,y,z) space.  The pressure gradient is
a vector field.  I hope you agree that there
is no lack of uniqueness in this vector field.
There is a perfectly real vector at each
(x,y,z) point.

If somebody wanted to make trouble, they could
say that the vector is merely a list of three
numbers, and the numerical values depend on
the choice of basis, so the vector is not
really unique.  But that's not how we think
of the physics.  We think of the vector as
being more real than its components.  The
vector is a machine which, given a basis, will
tell you its projections on the basis vectors,
i.e. its components.  The components are non-
unique, because they depend on the basis, but
we attach physical reality to the vector
nevertheless.

The pressure gradient is a vector field.  Now
it turns out that there are two different kinds
of vectors, leading to two perfectly good ways
of representing the pressure gradient.
  1) It can be represented by _pointy vectors_,
little critters with a tip and a tail, one at
each point in the field.  This should be
familiar to all.
  2) It can also be represented by contours,
i.e. shells of constant pressure.  This is
the one-form representation.  Closely-spaced
shells represent a large-magnitude one-form.
One-forms are vectors, they're just less
familiar than ye olde pointy vectors.

If you believe that the pointy vector field
representing the pressure gradient is unique
and well-defined, you ought to believe that
the one-form field representing the same
pressure gradient is equally unique and well-
defined.

Given a nice Cartesian metric, in any basis
the three numbers representing the pointy
vector are numerically equal to the three
numbers representing the one-form.

Returning to thermo:  Let's not leave behind
all our physical and geometrical intuition
when we start doing thermo.  Thermo is weird,
but it's not so weird that we have to forget
everything we know about vectors.

The point I've been trying to make for a couple
of days now is that one-forms are vectors.
They are as real as the more-familiar pointy
vectors.  Row vectors versus column vectors.

Sure, the one-form gives different projections
(more properly called contractions) if you
look at it from different angles, but so what?
Pointy vectors do the same, and that doesn't
make them any less real or any less unique or
any less well defined.

Row vectors are just as real as column vectors.
Bras are just as real as kets.