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Re: visualizing a non-potential



On 04/27/2003 05:38 PM, Bob Sciamanda wrote:
| a) Given a potential, equipotential lines can always be
| constructed. They represent the exterior derivative
| of the potential.

Does the exterior derivative of the potential generate the vector E field
or the equipotentials?

It depends.... Here are some true facts that
may or may not answer the question:

1) The gradient of the potential is a vector.

2) The exterior derivative of the potential is a
one-form.

3) If you have a metric, the distinction between
the two is minor. One is the dual of the other.

If you have a nice Cartesian metric, the distinction
between them is practically imperceptible. The
components of one are numerically equal to the
components of the other.

4) As to assertion that the exterior derivative
generates the equipotential lines, I'm not sure
"generates" is exactly the right verb, but I
agree with the basic sentiment. The contour
lines are the natural representation of the
exterior derivative.

A slightly technical point: You know the
distinction between a number and a numeral.
-- I consider the one-form to be highly
abstract, like a number.
-- I consider the contour lines to be a
concrete representation thereof, like a
numeral.

5) In contrast, the natural representation for
the gradient (a vector field) is a bunch of
arrows. They are oriented perpendicular to
the contour lines.

6) If cases (e.g. thermodynamics) where you don't
have a metric, I don't know how to define a
gradient, and I wouldn't know what to do with
a gradient if you gave me one. The exterior
derivative is the only game in town AFAIK.

You can see how this works by looking at the
first-order term in the multi-dimensional
Taylor series:

delta Phi = (d Phi | delta x)

where

Phi = some potential
d Phi = the exterior derivative of Phi
(A | B) = the contraction of a one-form A
with a vector B
i.e. the number of times the arrow representing B
pierces the contours representing A.

Contraction of a one-form with a vector is purely
geometrical and can be performed just fine without
a metric.

In contrast, the dot product of one vector with
another cannot be performed without a metric.
This includes dotting delta x (a vector) with the
gradient (a vector, if it exists at all).

In thermodynamics, there are various potentials
including E, S, P, V, et cetera. In general
AFAIK there is no such thing as the "gradient" of
the pressure. But the exterior derivative dP is
a well-defined (exact) one-form. And VdP is a
well-defined (non-exact) one-form. As good
geometric objects should, they exist quite
independent of what variables, if any, you have
chosen to consider the "independent" variables.

If this doesn't answer the question, please ask
again.