E-Field vs B-Field in SR

Trifis

Macroscopically the wire seems to possess a pure B-Field. However I really doubt it that it is possible to cancel out all the leftover E-Fields of the current (electrons) with positive ions.

I will try to examine the problem again adding some more electrons to my initial example, while avoiding any complicated interactions, but I fear that it could hardly yield any satisfying results...

DaleSpam

I suspect that you probably have it backwards. In all likelyhood if there is a microscopic region where the invariant is positive it would be near the nucleus where the electrons couldn't cancel out the field of the protons. Remember, the invariant is the same in all reference frames, but it generally varies from place to place and from time to time.

It won't work unless you include some positive charge distribution also.

Also, don't forget that there is no such thing as a classical point charge, classical EM is runs into some irritating problems if you use point charges rather than charge distributions. Otherwise you have to do a full QED analysis, in which case you don't have a classical point charge either.

pervect

A superconducting ring with no net charge but carrying a current (or the non-superconducting equivalent with normal wire and a battery) is a good simple example. In the lab frame B is nonzero, and E is zero.

Trifis

Yes that is well understood. What I meant to say is that the ideal wire of magnetostatics is not possible even with a theoretical assumptions when analyzed in microscopic level.

Yes I would have added some positive charges or distributions if you like, but the whole attempt would be pointless since the flow of electrons must be at least constant in order to give the observer a chance to detect a pure B-Field.

Am I right to think that something like that could not be analyzed within the scope of SR, since the reference frame of the charges would have to accelerate (circular orbit) ?

DaleSpam

I think you are thinking of a EM field as something monolithic that covers all space as a single object. A field has a different value at each point. The fact that there may be some points where the invariant is negative does not change the fact that there are definitely points where the invariant is positive. At those points there is no reference frame where you will ever see an E field only.

SR can handle acceleration just fine:
http://math.ucr.edu/home/baez/physic...eleration.html

Trifis

In order to detect a pure magnetic field at a specific point the invariant has to be 2B² not just positive right? How could that be possible?


"In SR it is still possible to use co-ordinate systems corresponding to accelerating or rotating frames of reference just as it is possible to solve ordinary mechanics problems in curvilinear co-ordinate systems. This is done by introducing a metric tensor. The formalism is very similar to that of many general relativity problems but it is still special relativity so long as the space-time is constrained to be flat and Minkowskian."

That is extremely interesting! Have you any books or papers, which elaborate that, to suggest?

DaleSpam

Where are you getting the factor of 2 from? As long as it is positive in any frame there exists some frame where E=0.

I like the first chapter here:
http://arxiv.org/abs/gr-qc/9712019/

Trifis

I think you are referring to the EM Lorentz scalar F_μνF^μν=2(Β²-[itex]\frac{Ε^2}{c^2}[/itex])

DaleSpam

I was referring to half that quantity, but both quantities are invariant. The point remains that if it is positive in any frame then there exists some frame where E=0. Do you understand that now?

Trifis

Hmmm so theoretically we can assume that there are somewhere some observers that happen to detect no E-Field. I was just wondering if we could examine analytically such a bizarre configuration.

DaleSpam

I don't know what would make you use the word "bizarre", but yes, you can do it analytically for simple current distributions (straight wires and circular loops). For more complicated geometries you need to do it numerically.

PAllen

An obvious point: there are pure, macroscopic B fields, but they are dipole plus higher moments. There is no B analog of a coulomb field without monopoles.

Muphrid

This is...I recognize it's the common view, but let me provide an alternative perspective on this.

We already know that the fundamental object representing the EM field is the Faraday tensor, [itex]F_{\mu \nu}[/itex]. This object can be called a bivector field--a field of oriented planes. The six components represent the six planes in a 3+1D spacetime: tx, ty, tz, yz, zx, xy. The first three, of course, correspond to the electric field; the other three correspond to the magnetic field. The full EM tensor is just a linear combination of these basis planes.

This should also be suggestive: maybe we're wrong to consider the magnetic field a vector field. After all, the yz, zx, and xy planes exist in ordinary 3D space already.

With that in mind, consider the magnetic field outside a straight current carrying wire and, in your mind, imagine instead the planes perpendicular to those field lines. What do you get? You get planes extending radially outward from this wire in a manner similar to a Coulomb field. The difference is that there is a translational symmetry along the direction of the wire.

But, remember the electric field can be interpreted as planes also, just with one of the directions in the plane being the t direction. In this way, a stationary charge looks exactly like the current-carrying wire. It just goes in the t direction instead of a spatial direction. And the electric field (as planes) looks exactly like the magnetic field (as planes).

This is why the straight current-carrying wire is as fundamental to magnetic fields as the stationary point charge is to electric fields. And while in most circles it's probably good enough to say there are no magnetic monopoles, I think this connection between the magnetic field from a wire and the electric field from a point charge is too significant to ignore.