Re: Ohm's 'law' & minus sign

["John S. Denker" <jsd@MONMOUTH.COM>]

At 5:31 PM -0500 2/10/03, Bob LaMontagne wrote:

>> The form of Ohm's law relating current density and field does not
>> need the negative sign because the conventional current density
>> responds directly to the field.

Indeed!

Chuck Britton wrote:

> YES! This is EXACTLY what I'm trying to figure out.

What's left to figure out?
Bob's answer seems bullet-proof to me.
Clear and concise.  The microscopic law is
       E = rho j
and it would be perverse to rewrite it in a
way that would require a minus sign, and it
would be perverse to make the macroscopic law
nonparallel to the microscopic law.

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> This delta V is a SPACIAL delta rather than a temporal (final -
> initial) isn't it?

This is an electrostatic problem.
There are no temporal variables.

I suppose delta V could be considered a spatial
difference, but it is spatial in a quite
abstract sense.  In the approximation where
Kirchhoff's laws apply, geometry doesn't
matter; only topology matters.

Indeed in Kirchhoff's world, the "space" of
circuit schematics, there is the strongest
possible topology, the discrete topology;
any two nodes that aren't the same are
different, and that's all you can say about
them.

http://www.gap-system.org/~john/MT3822/Lectures/L11.html

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> ANOTHER use of delta is to mean an infinitesimal - or at least small
> - amount

That's a pun, nothing more.

The word "dove" can be a noun or a verb.  In
practice it should be obvious from context which
is intended.

-- In the expression "You give me an epsilon I'll give
you a delta" both epsilon and delta are variables,
usually infinitesimal variables.

-- In the expression delta V, delta is an operator.
It means a difference in V.  It does not mean delta
times V.  It does not imply or suggest that delta V
is small.