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Re: TdS is not dQ or d(anything)



Before I get creamed, I repeat with typos all (hopefully) corrected:

----- Original Message -----
From: "John S. Denker" <jsd@MONMOUTH.COM>
To: <PHYS-L@lists.nau.edu>
Sent: Tuesday, May 13, 2003 5:47 PM
Subject: 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 qualification "essentially" is troubling - what does it mean?

|
| >>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.

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.

Bob Sciamanda (W3NLV)
Physics, Edinboro Univ of PA (em)
trebor@velocity.net
http://www.velocity.net/~trebor