"On the other hand, I believe we can at least partially dispose of
"motional emf" by noting that we are trying to do the standard path
integral of E in a frame in which the path itself is moving. If we do
that integral in the proper frame we will find a background electric field
(v x B) which precisely accounts for the observed emf."
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I'm not sure what John is saying here. If, as if often the case,
different parts of the circuit have different velocities, then it seems to
me that no proper frame for the circuit exists. In J.D. Jackson's
"Classical Electrodynamics" he presents a line integral of E'ds, where E'
is the electric field at each circuit element in an inertial frame in
which that element, not the circuit, is momentarily at rest. It is clear
in his presentation that the line integral of E'ds and of E ds, where E is
the electric field in some particular reference frame, are not the same.
**********************
Nick Steph wrote:
"Here is where I get confused: Consider a conducting rod of length L,
moving perpendicularly through a UNIFORM magnetic field, B, at a constant
velocity, v. An observer in the lab will detect a potential difference
across the rod, PD = BLv. What about an observer on the rod? According
to special relativity, she must detect it also; but what does she consider
to be the cause?"
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If an electromagnetic field is observed from two distinct reference
frames, the E and B fields get mixed up in the transformation from one
frame to the other. For Nick's example, in the frame in which the rod is
moving the static B field transforms to a static B field and a static E
field in the frame moving with the rod. The observer on the rod considers
this E field as the cause of the charge separation in the rod. However,
she observes no potential difference across the rod because the rod is in
electrostatic equilibrium in her reference frame, so in this frame the
electric field E induces charges on the surface of the rod that result in
the net electric field inside the material of the conductor to be
everywhere zero. Thus the rod is an equipotential.