"so that the above O-form . . ."
should read:
"so that the above exterior derivative of the 0-form f . . ."
Bob Sciamanda (W3NLV)
Physics, Edinboro Univ of PA (em)
trebor@velocity.net http://www.velocity.net/~trebor
----- Original Message -----
From: "Bob Sciamanda" <trebor@VELOCITY.NET>
To: <PHYS-L@lists.nau.edu>
Sent: Tuesday, May 13, 2003 8:44 PM
Subject: 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