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Re: Question 07/02 CURRENT IN A WIRE



I think everybody's got it pretty much figured
out now. Maybe a picture will help nail the
lid on it, and perhaps help when explaining it to
students.

Fact #1: A beam of plain old electrons will repel
another beam of plain old electrons. This is
obvious in the frame comoving with the electrons,
and what's true in one frame must be true in all.

Fact #2: An electrically netural current-carrying
wire will attract a similar parallel wire.

Analysis: The electrons in one wire are strongly
repelled by the electrons in the other wire, just
as in case (1). But to first order this effect is
cancelled by the attraction to the positive metal
ions. What's left is a "small difference between
large numbers" problem that must be analyzed with
a bit of care.

Please refer to
http://www.monmouth.com/~jsd/physics/gif48/wire-bivector.gif

We analyze one of the wires. The other wire will
be just the same. The wire runs in the z direction,
right-to-left in the picture. The lab frame's time
axis (t) runs vertically in the picture. The radial
direction r is perpendicular to t and z in real life,
and appears to run across the figure at an angle
because of perspective.

The electric field bivector of the positive metal
ions is shown in green. It is in the r /\ t
direction, and in fact it is proportional to
(1/r) /\ t.

The "primed" reference frame is comoving with the
electrons. The electric field bivector of the
electrons goes like (1/r) /\ t', shown in red.
And it has a minus sign, because the electrons
are negatively charged.

Observe that the t' axis is tilted relative to the
t axis by an angle proportional to the electrons'
speed (relative to the ions).

The total electromagnetic field is the superposition
of these two contributions. Add the bivectors
edge-to-edge just as you add vectors tip-to-tail,
with due regard to sign. The resultant is shown in
gray. It is in the r /\ z direction, and in fact
it goes like (1/r) /\ z, with additional factors
proportional to the density and speed of the carriers.
The exact answer is (2/r) /\ I, where I is the
current, as discussed in
http://www.monmouth.com/~jsd/physics/straight-wire.htm

Note that all three bivectors in the figure are
frame-independent physical objects. Everybody in
every frame agrees that they are what they are.

The way you get into trouble is if people in different
frames start aguing about frame-dependent notions
such as whether such-and-such bivector is purely
electrical or purely magnetic. The "electric" and
"magnetic" pieces are merely _components_ of the
real, physical electromagnetic bivector, and as
always, the decomposition into components is
frame-dependent (and irrelevant to the physics).

==============

Pedagogical hint: It is annoyingly difficult
to draw the figure on a chalkboard with sufficient
detail and clarity that students can follow it.
Even the simplest "3D visual aids" give better
results: grab a couple of pieces of card-stock
and hold them edge to edge at an acute angle as
shown in the figure. One card represents the
electromagnetic field bivector of the metal ions,
the other represents the EM field of the electrons,
and you can wave your hand over the "mouth"
they form and identify it as the resultant.

=========================

Trebor directed us to the Peters AJP paper:

"In what frame is a current-carrying conductor neutral?", AJP 53 (12), pg
1165, Dec 1985.

I haven't read the paper, but I suspect I can guess
the answer:

Glance at the figure.
-- To zeroth order, the two E-field bivectors cancel.
-- To first order in v/c (the velocity of the carriers)
the resultant is in the purely magnetic direction in the
lab frame.
-- To second order, we see that the resultant is
slightly tilted in the lab frame.

I don't know much, but I know a symmetry argument when I
see one. It's fairly obvious that there is a frame where
the resultant is 100% spacelike (i.e. purely magnetic) and
it's not the lab frame.

=========

Homework puzzle: Can you fly past the wire at some special
speed such that the field thereof becomes purely ELECTRICAL?

Huge hint: Convince yourself that X /\ (X~) is a
relativistic invariant, for any physical object X.
In the case where X=p, the momentum 4-vector, the
invariant is just the rest-mass squared. What about
X=F, the electromagnetic field bivector?

This posting is the position of the writer, not that of Thomas Bowdler,
Anthony Comstock, or William Hays.

This posting is the position of the writer, not that of SUNY-BSC, NAU or the AAPT.