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So is charge fixed to a moving dielectric considered a current?
Yes.

Is the electric field of this moving charge (that is, current) in the direction of the current density vector?

No. In fact, Lorentz contraction causes the electric field from a moving charge to become weaker along the axis of motion and stronger in the transverse direction. The field becomes compressed along the axis of motion.

>>I'm familiar with the bunching of the electrostatic field lines on moving charge. I'm talking about the E aligned with the current density vector, not the coulombic E .

If you think carefully--very carefully--about this, it might seem that this gives rise to a situation that is not Lorentz invariant: as the charge passes by a "stationary" test charge, a stationary observer should expect the test charge to be subjected to a greater electostatic force than would be measured by an observer that is comoving with the moving charge. What's a test charge to do?

As it turns out, this difference in force is exactly compensated by the magnetic field that is observed by the stationary observer, created by the motion of the moving charge. This field is not observed by the comoving observer. They agree on the net motion of the test charge.

>>Correctly stated it is the changing magnetic field.

In other words, the magnetic field is the manifestation of effect of Lorentz contraction and time dilation upon an electric field. It's relativity you can play with at home.

Or is there no electric field of the charge in motion because it is mechanically forced to move?

The total electric field around the charge is invariant. Gauss's Law, once again.

>> The E field due to charge motion has zero divergence, just as the current density vector. This cancels out when gauss's law is applied. There's an "inny" and an "outy" on that vector over the closed surface.

There is no additional electric field that is created by the movement of the charge. Battery, black hole, or baseball bat, it doesn't matter what causes the charge to move. The divergence of the field equals the charge density, end of story.

>>The divergence definitely equals the charge density. The current density has zero divergence and the E field aligned with J also has zero divergence. If div J did not equal zero charge would be accumulating

I'm confused because there is no relative motion between the observer and apparatus.

Then that's the frame you use for that observer. The issue is that a moving observer will also have to observe a null result for that apparatus. It has to work in all possible reference frames. The different observers aren't entitled to disagree about the outcome of the experiment; either the interference fringes shift or they don't. They will, however, disagree about the size of the interferometer.

>>Agreed. But if I'm in the same frame then v = 0. You said the MM apparatus has measured the Lorentz contraction. I'm just trying to figure out how and contraction from what?

The E field due to charge motion has zero divergence, just as the current density vector.

There is no E field that is due to the motion of the charge. None whatsoever. I don't know what put that notion into your head.

You said the MM apparatus has measured the Lorentz contraction.

Yes. You measure the speed of light and time dilation in each frame with respect to the other. You predict from that a certain shifting of the interference fringes. You observe no such shift, so you can calculate how much the device must have shrunk to compensate.

I'm just trying to figure out how and contraction from what?

The contraction is due to the perspective from different inertial frames. A person looking at a yardstick from the side will see it as a long object, while a person looking at it from the end will see it as a short object. The Lorentz transformation is analogous to this effect of rotation, except that it describes a transformation between space and time, rather than the rotation of one space axis into another.