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Re: TdS is not dQ or d(anything)



On 05/13/2003 04:44 PM, Bob Sciamanda wrote:

A 0-form on 3-space (x,y,z) is a smooth map f(x,y,z) of
3-space onto the real numbers.

How is this different from a scalar field - a scalar function of space?

It's not. Scalars and 0-forms are essentially synonymous.

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.

How is this different from GRAD f (dot ) dr ?

The exterior derivative of a scalar field (f) can
be, and generally should be, defined to be the same
thing as (grad f). There is no "dot dr" involved.

The object so defined is a one-form, which you can
think of as a row-vector, in this case a field of
row-vectors. If you wish to interpret your (grad f)
as a column-vector, and if you have a metric handy,
you can convert one to the other.

What is this new language adding to good old fashioned vector analysis?

1) Old-fashioned vectors (which I like to call pointy
vectors) are necessary for representing displacements
but not sufficient for representing things like
gradients WHEN YOU DON'T HAVE A METRIC HANDY. Thermo
is a good example of not having a metric. It's hard
to make sense of the length of dV or the dot product
dV dot dS. Any such things would be unlikely to
survive a coordinate transformation.

If the gradient is formulated as a one-form, you
can do everything you need to do without a metric.
Everything is automatically invariant under coordinate
transformations.

2) The new language allows us to replace a vague
thing dU (the differential of U) with a non-vague
thing dU (the exterior derivative of U, written
with a boldface d when possible). The latter is
a function of state. We can draw pictures of it.

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

By the way, I want to thank folks (especially Bob
and Bob) for asking exceptionally clear questions.
It's been great fun to figure this stuff out and
(attempt to) explain it.