A non-conservative E field is a vector function of position (and time, in the
general case). It is representable by continuous field lines, tangent to the
direction of the local E vector. If you choose a reference origin and consider
the line integral of this E field from this origin to some arbitrary position
(x,y,z), you do not produce a function of position. The value of your
integration to some point (x,y,z) will depend on the path taken. The line
integral is in no sense a function of position - any attempt to represent it as
such will be flawed and misleading. Be satisfied with the field line
representation of the E field - that IS a genuine (vector) function of position.
Bob Sciamanda (W3NLV)
Physics, Edinboro Univ of PA (em)
trebor@velocity.net http://www.velocity.net/~trebor
----- Original Message -----
From: "John S. Denker" <jsd@MONMOUTH.COM>
To: <PHYS-L@lists.nau.edu>
Sent: Saturday, April 26, 2003 8:16 PM
Subject: visualizing a non-potential
| Hi --
|
| I cobbled up an improved scheme for systematically
| portraying non-potentials, i.e. things that have a
| slope but no well-defined height.
What is the "thing" that has a "slope but no well-defined height"? a function,
or what?