I don't know. Perhaps you are right. My understanding of the theory is limited, but please read the introduction here:
http://www.springerlink.com/content/q6634pp556m08500/fulltext.pdf Even Feynman and Maxwell say there is no e-field. Harry ----- Original Message ----- From: "Stephen A. Lawrence" <[email protected]> Date: Tuesday, September 15, 2009 11:46 am Subject: Re: [Vo]:The Electric Field Outside a Stationary Resistive Wire Carrying a Constant Current > > > Harry Veeder wrote: > > Maxwell's theory needs the field concept. The theory says and > > electric force can not be present without an electric field. > > > > If we follow Maxwell's theory to the letter, it says there will > be no > > electric field outside a current carrying wire. > > I don't know what you're talking about here. If by "Maxwell's theory" > you mean Maxwell's four equations, as they are normally written, > and as > they are embedded in the model of special relativity (which is how > thisis normally applied), then what you said is simply false. > > If there is a current flowing through a resistive wire, then there > is an > E field within the wire directed parallel to the wire. That E > field is > what drives the current, and its value is proportional to \rho*I where > \rho is the resistivity per unit length of the wire. The curl of > the E > field within the wire and near the surface of the wire is zero, since > > Del x E = -dB/dt > > in rationalized CGS units. > > Since the curl is zero, if the field points along the wire just > withinthe wire, it must also point along the wire just outside the > wire. > Otherwise you'd get a nonzero integral of the E vector around a small > loop which is partly inside the wire and partly outside the wire, > whichwould imply the curl was nonzero. > > For points near the wire, that field runs parallel to the wire. This > field is independent of the presence or absence of a charge outside > thewire. > > Arguments straight out of Purcell, based directly on Maxwell's > equationsand the Lorentz force law, lead to the conclusion that a > point charge > located close to the wire will also induce a local charge on the wire, > which will result in a local field which is perpendicular to the > surfaceof the wire. This field vanishes if we remove the external > charge. > But did you perhaps mean something else by "Maxwell's theory? > > (Incidentally I said "As embedded in the model of SR" because without > that extra bit of icing you have no way of transforming the equations > from one frame of reference to another, and no answer to the > question of > what happens when moving a uniform velocity.) > > > > Consequently, the > > theory leads one to expect an electric force is absent as well. > > > > Weber's theory is not built on the field concept, so this curious > > expectation does not arise. > > > > My analysis is based on reading of this preface to the book > suggested> by Taylor J. Smith. > > http://www.ifi.unicamp.br/~assis/Preface-Webers-Electrodynamics.pdf > > > > > > Harry > > > > ----- Original Message ----- From: "Stephen A. Lawrence" > > <[email protected]> Date: Monday, September 14, 2009 6:18 pm Subject: > > Re: [Vo]:The Electric Field Outside a Stationary Resistive Wire > > Carrying a Constant Current > > > >> > >> Harry Veeder wrote: > >>> fyi Harry > >>> > >>> Foundations of Physics © Plenum Publishing Corporation 1999 > >>> 10.1023/A:1018874523513 > >>> > >>> The Electric Field Outside a Stationary Resistive Wire Carrying a > >>> Constant Current > >>> > >>> A. K. T. Assis, W. A. Rodrigues Jr. and A. J. Mania > >>> > >>> Abstract We present the opinion of some authors who believe > >> there is > >>> no force between a stationary charge and a stationary resistive > >>> wire carrying a constant current. > >> That's stated a lot, but it's just sloppiness. Anyone who knows > >> electronics realizes it's not really true. > >> > >> A good conductor carrying small current has *nearly* zero voltage > >> drop along any small length, and calling the drop "zero" is usually > >> "good enough". But really the voltage drop along any segment is > >> equal to I*Rwhere R is the resistance of that segment. > >> > >> When the voltage drop along a (resistive) wire is nonzero, then > >> *any* path which leads from a higher voltage point on the wire > to a > >> lower voltage point on the same wire must traverse the same exact > >> potential change, which means that there must be an electric field > >> *outside* the wire, running parallel to the wire. > >> > >> This is well known but, as I said, usually neglected, because it's > >> usually too small to matter in real-world problems. > >> > >> The fact that there's an "image charge" induced in the wire as > >> well, which consequently must be having its effect on the charge > >> sitting outside the wire, is certainly the case and could even be > >> called "obvious", but it's not something I ever thought of until I > >> saw it mentioned in the abstract. :-) > >> > >> > >>> We show that this force is different from zero and present its > >>> main components: the force due to the charges induced in the wire > >>> by the test charge and a force proportional to the current in the > >>> resistive wire. We also discuss briefly a component of the force > >>> proportional to the square of the current which should exist > >>> according to some models and another component due to the > >>> acceleration of the conduction electrons in a curved wire > >>> carrying a dc current (centripetal acceleration). Finally, we > >>> analyze experiments showing the existence of the > >> electric> field proportional to the current in resistive wires. > >>> complete paper available here: > >>> http://www.springerlink.com/content/q6634pp556m08500/fulltext.html > >>> > >>> > >>> > >> > > > > > >
