I fail to see how this affects the argument. We are dealing with systems where the distance between the source and the charge is so small that the propagation time is negligible.
This argument was just related to your comment: "But our A fields come from spinning or orbiting electrons". A potential as well as a field is a tool to avoid to discuss the question of the source and of the propagation of an effect from a source.
If we deal with the potential vector, the question of the source is off topic.
Explain why it will no longer see the same A. You quote the Lorentz transform which invokes beta and gamma but since we are dealing with electron velocities that are small compared to c, beta tends to zero and gamma tends to unity. So IMO we can forget the Lorentz transform, it has negligible effect and to all intents and purposes the moving electron sees the unmodified A.
Electromagnetism effects are relativistic effects. It is a common mistake to think that the low electron velocity could be negligible while it is the only cause of the magnetic field. I found the best explanation is the book "
Claustrophobic physics" from Paul Bickerstaff.
As the question comes up regularly, I have attached an extract from his chapter on currents, with the question of the force between 2 parallel wires carrying a current.
I quote him:
"
So the magnetic field experienced by the test charge is correctly understood as arising from an electric force in another frame due to a line of net charge density γ.β².λ0 [...]. The relative strength of the magnetic and electric force is thus of the order of β² where β is determined by the drift velocity of the electrons. As remarked above, this is of the order of just a few mm/s and so
β ≈ 10-11
for typical current bearing wires. The magnetic force they exert is therefore just 10-22 times that of an electric force due to just the positive (or negative) charges. That the magnetic force is observed at all (and is appreciable) is due to the enormous amount of charge carried by a number of electrons of the order of the Avogadro's number".
In other words, the electrical force is really a huge force and the large number of charges makes visible the smallest relativistic effect. Without the natural balance of positive and negative charges in nature, this force would prevail everywhere.
The electrical effects come from the separation of charges.
Magnetic effects come from relative speeds between charges, the relativistic effects cancelling the apparent balance of the charges, in fact the equilibrium of their electric field.
Inside the solenoid the equipotentials of A are also concentric circles but they strengthen proportionally with distance. This also applies to the A field in the Faraday disc. As you well know the force on any radially moving electron there is not radial, it is always transverse to the movement.For E=-dA/dt the E is along A. If you examine the components of (v.del)A the E is along A.
My interrogation was general. I wrote dA/dx by simplification but a spatial gradient is dA/dl where A and l are vectors. And we know there is no field gradient along A, so the question of the gradient direction arises, and this question doesn't depend on the speed of the charge that moves in.
A and dA/dt are independent of the load, they are evaluated by the observer.
So the spatial gradient of A should be as well. Let's not forget that the electric field is a property related to the place. If we assume that the spatial gradient gives rise to an electric field, this field cannot depend on the test charge that is placed in it. If charges at different speeds see different fields, this is understandable, it is related to relativity, but in any case the observer should only see a single electric field!
The inconsistency is due to the fact that I have already mentionned, that of calculating the spatial gradient in the referential of the charge instead of that of the observer, and mixing it with dA/dt or VxB which are relative to the observer's referential.
Can you give the spatial gradient of A, viewed in the referential frame of the observer ?
IMO the Marinov generator is just that.I look forward to seeing it.
Smudge
The Marinov generator is just that but perhaps there is simpler. In the Marinov generator there is a spatial A gradient, not a constant A. In the Faraday disk, A is constant, spatially and temporally. Just keep A and remove B in the Faraday disk, and you have a possible homopolar generator working on A, and not on a gradient of A, which is different from Marinov. I don't know yet the result. See my next post.
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Topic: Marinov Generator (Read 54660 times)
