OverUnity Research

Context:
Smudge's thread on the Marinov motor had been very instructive in leading us to questions about the vector potential, for example about a possible motor without an active magnetic field or about a possible induction by a spatial rather than temporal gradient of the vector potential.

The vector potential A is the equivalent for the magnetic field of what the (scalar) potential is for the electric field. The vector potential has its lines along the currents. If the conductor is linear, the lines will be oriented along a tube around the current. If it is a coil, they will be circles around it which, unlike the magnetic field confined to the interior of the coil, are established in all continuity from the interior to the exterior of a coil until infinity.

A temporal variation of this field is seen as an electric field E, which is another way of seeing the induction field created by a variation of magnetic field. E=-∂A/∂t.

The idea:
Starting from this, we can re-write E = - ∂A/∂x * ∂x/∂t = - v * ∂A/∂x. From this equation, we can derive that, with respect to an observer linked to the source of the vector potential, a charge moving at speed v along the x axis, will see an electric field proportional to the spatial gradient along x, of the vector potential. This is true whatever the space coordinate x, y or z. We have just assumed here that the gradient of A and the motion of the charge are only along x.

The question remains how one can verify this experimentally, namely to induce a current from a spatial gradient of the potential vector, which would be a new form of generator or transformer.
Smudge has proposed a method in his thread "Generator using a superconductor" (https://www.overunityresearch.com/index.php?topic=4382.0). Here is another one that I propose.

The potential vector is established along a current and depends on it, so we would need a circuit with a non constant current along the circuit. By "non-constant", we do not mean a current which varies in time, but a current which at a given moment, is not the same everywhere along the circuit, which in opposition to Kirchhoff's law seems difficult to obtain in the regime of quasi-stationary states.

Since A is proportional to I, the idea is that you just need a current gradient in a conductor. I think we can get this from a long capacitor. You can see the principle in view N#1 of the attached schematic. A long capacitor is formed by two rectangular plates fed by their opposite ends. The view is transverse to the plane of the capacitor.

Assume that current flows from the bottom plate to the top plate. No current can escape through the right end of the upper plate since it is open. And all current escapes through the left end since that is the only way out. And between the two? Well, the current will be evenly distributed from one plate to the other by the displacement current which is in fact the simple coulombic influence between plates. It should be noted that we are in a quasi-stationary regime. No propagation effect occurs along the plates. They remain equipotentials.

So along each plate the current will be maximum on the connection side, zero on the opposite side, and between the two of a value proportional to the distance from the open end. We will therefore have a current gradient along each plate, and therefore a gradient of the vector potential as well, which is what we are looking for.

However the sum of the currents along each plate will be constant all along. It will therefore be difficult to use the gradient of the potential vector of only one plate, because we will have the opposite influence of the other.
So we arrive at view N#2. The upper plate is now in the middle, sandwiched between two plates connected together by their opposite ends. The currents of the upper and lower plates oppose their effects, cancelling each other's A field, so we are left only with the middle plate having a current gradient from 0 at the right end, to its maximum value at the left, and so the same for the vector potential. This will be our inductor circuit.

Now that we have the device creating a vector potential with gradient, we have to set up the induced test circuit, view N#3. To do this we divide the thickness of the central plate in two, and we insert the folded green circuit. The charges must be at speed v to feel the spatial gradient of A of the brown plates and be accelerated by it, so we produce a DC current in the green test circuit.

Expected effect:
Here the spatial gradient of A is not time-constant, since it must be produced from an AC current for the capacitive effect to work, which complicates the experiment a bit.
But under these conditions, the DC current of the test circuit should be modulated and its AC component amplified, which is what the experiment is expected to show.

I have just posted the results of some calculations on my bench that show the induced effect is quite small.  That is for the A field from a pair of ring cores at saturation.  The effective surface current on the ring cores is huge compared to the practical currents we can provide in wires.  I can see what your experiment is meant to show but I suspect the actual induced voltage would be too small to readily detect.   Sorry!

Smudge

Quote from: Smudge on 2022.11.09, 17:08:14
I have just posted the results of some calculations on my bench that show the induced effect is quite small.  That is for the A field from a pair of ring cores at saturation.  The effective surface current on the ring cores is huge compared to the practical currents we can provide in wires.  I can see what your experiment is meant to show but I suspect the actual induced voltage would be too small to readily detect.   Sorry!

Smudge

E=- ∂A/∂t in classical induction is strictly equivalent to -v*∂A/∂x, so we can deduce that a time-varying current seen from the observer can also be seen as a spatial gradient by the charge, at a given instant.

If a charge in an induced circuit is influenced by E=- ∂A/∂t as seen from the observer at rest, then the moving charge following the influence of E at the current-related drift velocity v as seen from the observer, must see at each instant from its own frame of reference a spatial gradient of A, which must rotate to maintain itself.

The EMF being the same, I do not see why velocities corresponding to ordinary currents would not have significant effects in a spatial gradient of A created by another ordinary current.

It depends on what you mean by significant.  I haven't calculated the effect in your experiment so you may be right.  Let's leave it at that.

Smudge

Even if the induction was there. How would the induced current not itself also generate an A field which opposes the change of the source current? In other words, you would just have made a complicated transformer.

Quote from: broli on 2022.11.09, 20:51:35
Even if the induction was there. How would the induced current not itself also generate an A field which opposes the change of the source current? In other words, you would just have made a complicated transformer.

See the diagram in the attached file (already provided here: https://www.overunityresearch.com/index.php?topic=4382.msg102084#msg102084 )

It shows the orientation of the force, the speed of the charge and A.

The current of the test loop is of constant intensity along the conductor since v is constant, so the field A it produces is constant. It cannot therefore oppose the field A of the source which is not constant along the conductor, due to its gradient.

In the link above, I pointed out that the force is always opposite to the speed, so it is a braking force.

And Smudge replied with the braking uh breaking news:
"That braking force applies to a positive charge, in the case of a negative charge it becomes an accelerating force."

And to be honest, there's nothing to suggest that he's wrong. This would obviously be revolutionary, because while my test is only there to verify the basic principle of the effect of a gradient of A on a charge, the final idea is to use a gradient of A obtained from a permanent magnet.
In addition, the charge symmetry would be broken, which is also a novelty.
There are therefore many clues to continue.

I made the device and quickly did the experiment.
The attached file shows the copper strips making up the diagram attached in my reply #1, with their top and bottom faces before folding the strips in half, as well as the final device when tested (I know, it's a mess on the desk).
The insulation between the capacitor strips is made by a plastic film and the one between the inductor circuit and the test circuit by the paper strip provided with the copper strip coil.
The resistance in series with the DC supply to the circuit is 10 ohm. The large capacitor in the photo is used to decouple the DC supply.

No effect seen on the scope when the current in the test circuit goes from 0 to 1A. Only the very weak AC signal, independent of the DC current, is seen, due to a probable residual capacitive coupling of the test circuit to the inductor.

I would also assume that if this induction effect depends on the speed of the electrons. You would rather use a material with high electron mobility property. I think this was mentioned before but Robert Murray does sell a conductive ink that contains graphene.

https://secure.workingink.co.uk/shop/working-ink-ink/working-ink-water-resistant-conductive-ink-emf-shielding-1l/

Not sure about the actual electron mobility properties but anything higher than coppers would give a significant boost no?

Also a while ago I came across these guys who sell carbon nanotubes in a yarn form:

https://dexmat.com/

Hi Broli,

Interesting your link to the nanotube wire, I didn't know it existed. I already experimented with carbon wire, very fragile and whose electron mobility is not better than in metals, but not with nanotube. So this would be a new field to test.

I agree with your point of view, and Smudge also expressed the same thing. The higher the mobility, the higher v is, and therefore the higher the electric field E will be seen from the mobile charge.

However E=-∂A/∂t or E=-v.∂A/∂x is the same thing said in two different ways depending on whether the observer is moving in the field or not.
This implies that a temporal variation of A is always accompanied by a non-uniformity in space. Indeed, if an electron at rest at position x sees a field E=-∂A/∂t at its position, at position x+dx not far from there, it would see a different field, the one that has varied during the time interval dt=c/dx where c is the propagation speed of the variation of the field in the medium and therefore almost the speed of light.

E=-∂A/∂t being the classical electric field induced by a magnetic field variation, which is obtained by a common variable current, then the same order of magnitude of current is also the cause of the non-uniformity of A in space. This implies that the drift velocity of electrons in an ordinary current should be more than sufficient to see the effect of spatial non-uniformity.

I therefore believe that the negative result has a more subtle cause. It seems that the A-field is a closed line that varies in time as a block, so that the intensity of A is always uniform along the loop, and that a temporal variation amounts to replacing a loop of a certain amplitude by a loop of a different amplitude, as for the magnetic field. In this case, and contrary to what I have just said, a charge following a conductor along this loop could never see a spatial non-uniformity. It remains to be seen precisely what happens when the conductor that determines the path of the charge does not follow this equipotential loop of A in such a way that we have spatial non-uniformity.
This implies not to limit oneself to x but to treat the problem with the 3 dimensions of space, which is quite annoying, especially since it is likely in this case that the spatial variations of A in one direction are exactly compensated by opposite variations :(...

Quote from: F6FLT on 2022.11.12, 08:50:03
.....................
This implies that a temporal variation of A is always accompanied by a non-uniformity in space.

Not true.  A spatially uniform A field can have a temporal variation.  I admit that to create that uniform field the source of the A field has to be at large distance (actually infinity for perfect uniformity) but science is full of situations where the distance is large enough that the non-uniformity can be ignored.

QuoteIndeed, if an electron at rest at position x sees a field E=-∂A/∂t at its position, at position x+dx not far from there, it would see a different field, the one that has varied during the time interval dt=c/dx where c is the propagation speed of the variation of the field in the medium and therefore almost the speed of light.

Again I disagree, you have got your dimensions mixed up, dt=c/dx is dimensionally wrong.   Perhaps you meant dt=dx/c.  If so I disagree with that as a general case.  That is only true if the x axis coincides with the direction of the distant source.  And in any case with dx small and c being large that apparent spatial variation is insignificant compared to the temporal one.

QuoteE=-∂A/∂t being the classical electric field induced by a magnetic field variation, which is obtained by a common variable current, then the same order of magnitude of current is also the cause of the non-uniformity of A in space. This implies that the drift velocity of electrons in an ordinary current should be more than sufficient to see the effect of spatial non-uniformity.

Perhaps you would reconsider that statement in light if my criticism above

QuoteI therefore believe that the negative result has a more subtle cause. It seems that the A-field is a closed line that varies in time as a block,

agreed for temporal variation

Quoteso that the intensity of A is always uniform along the loop,

not agreed for spatial variation

Quoteand that a temporal variation amounts to replacing a loop of a certain amplitude by a loop of a different amplitude, as for the magnetic field.

Not agreed.

Quote.......This implies not to limit oneself to x but to treat the problem with the 3 dimensions of space

My work looks at the 3D A field but then looks at loops lying in a plane within that 3D space.  I think that is OK.

Smudge

Quote from: Smudge on 2022.11.12, 10:58:08
A spatially uniform A field can have a temporal variation.

This is wrong in the sense that I used the word "uniformity". "Uniform" means "the same always or everywhere; not changing or varying".
By "non-uniformity in space" I only mean that the field has gradients in space, which is always the case in a sufficiently large 3D volume, whether the field is static or time-varying, but this is particularly obvious when it is time-varying.

Quote
Again I disagree, you have got your dimensions mixed up, dt=c/dx is dimensionally wrong.   Perhaps you meant dt=dx/c.

Of course, thanks for the correction.

Quote
  If so I disagree with that as a general case.  That is only true if the x axis coincides with the direction of the distant source.

That was my hypothesis that A is along x, that's what the device is designed to do, except in the small part of the hairpin bend

Quote
And in any case with dx small and c being large that apparent spatial variation is insignificant compared to the temporal one.
...

I think that, on the contrary, it is significant, as much as ∂A/∂t.  E=-∂A/∂t is seen by the observer at rest with respect to the circuit. But because of its motion, a charge moving in a circuit can no longer see the same A, A is not a Lorentz invariant. From its frame of reference it sees A' which depends on its velocity v, and it sees a spatial gradient which allows the EMF that drives the charge to be consistent with what the observer sees.
My idea is that the charge sees E = -∂A'/∂t - v.∂A'/∂x = -∂A/∂t. In a circuit classically induced by an AC current, v is not constant, and since A' = A(v) and there is acceleration, the calculation is no longer trivial. I am now trying to clarify this by calculation.

I checked the magnetic field along my test strip, and it remains the same, which means that the vector potential does too.
The expected spatial gradient of A is absent, so the null result is normal, and the idea of a FEM from such a gradient is not invalidated.

It remains to be seen why I could not obtain this gradient of A with the setup N°2 in the attached file of reply #1.

Quote from: F6FLT on 2022.11.13, 13:52:15
I checked the magnetic field along my test strip, and it remains the same, which means that the vector potential does too.
The expected spatial gradient of A is absent, so the null result is normal, and the idea of a FEM from such a gradient is not invalidated.

It remains to be seen why I could not obtain this gradient of A with the setup N°2 in the attached file of reply #1.

Your current gradient is probably minuscule. In an antenna you have the advantage of wavelengths to play with however at your low frequencies the capacitor is probably too short to exhibit any significant currents gradient and thus the current would appear to be uniform. So, you probably need to increase the frequencies and increase the length to match the wavelength.

I have offered a simple mechanical variation to confirm this idea: https://www.overunityresearch.com/index.php?topic=4391.msg102285#msg102285

Quote from: broli on 2022.11.16, 06:49:21
Your current gradient is probably minuscule. In an antenna you have the advantage of wavelengths to play with however at your low frequencies the capacitor is probably too short to exhibit any significant currents gradient and thus the current would appear to be uniform. So, you probably need to increase the frequencies and increase the length to match the wavelength.

I have offered a simple mechanical variation to confirm this idea: https://www.overunityresearch.com/index.php?topic=4391.msg102285#msg102285

The current gradient is not minuscule at all, it is maximum. If the current I in the capacitor is 1A, then the gradient is 1A along the length of the strip, since the current is zero on the open side of the center strip, and equal to I at the connected end.

The effect is therefore of the same order of magnitude as that obtained on an antenna where the gradient is obtained thanks to the signal propagation delay, but here we are in a quasi-stationary regime.

Again, if we consider that the spatial and temporal gradient are the same effect seen from an observer either at rest or in motion, then the same current that generates an EMF thanks to ∂A/∂t will also generate it by v.∂A/∂x where v is the drift velocity of the electrons generating ∂A/∂t. The effect must therefore be as significant as the classical induction by ∂A/∂t. The error is elsewhere.

Quote from: F6FLT on 2022.11.16, 08:09:49
The current gradient is not minuscule at all, it is maximum. If the current I in the capacitor is 1A, then the gradient is 1A along the length of the strip, since the current is zero on the open side of the center strip, and equal to I at the connected end.

The effect is therefore of the same order of magnitude as that obtained on an antenna where the gradient is obtained thanks to the signal propagation delay, but here we are in a quasi-stationary regime.

Again, if we consider that the spatial and temporal gradient are the same effect seen from an observer either at rest or in motion, then the same current that generates an EMF thanks to ∂A/∂t will also generate it by v.∂A/∂x where v is the drift velocity of the electrons generating ∂A/∂t. The effect must therefore be as significant as the classical induction by ∂A/∂t. The error is elsewhere.

This might be the case indeed. But then remains the obvious issue of having all plates so close to each other. The nonuniform A-fields will add up and all that remains is the "equivalent" uniform A-field of a straight wire. So any effect would remain very tiny. It's like having a loop sandwiched between two alternating current loops that are 180 degrees out of phase. The induction voltage will be non-existing.

To me the mechanical variant offers a very good and simple method to validate if this even has any merit. The problem can be solved theoretically quiet easily and can be compared to experimental data. I'm honestly a bit skpectical about this "convective" vector potential induction due to my own experience and experiments with it in the past but those were very crude and far from conclusive.

My english is going poor after covid so please bear with me .

Do you mean transformation of magnetic energy from the displacement of closed magnetic path into open ? Or rather this : https://www.youtube.com/watch?v=eH2TWPJiwEA

Quote from: broli on 2022.11.16, 10:44:11
This might be the case indeed. But then remains the obvious issue of having all plates so close to each other. The nonuniform A-fields will add up and all that remains is the "equivalent" uniform A-field of a straight wire. So any effect would remain very tiny.
It's like having a loop sandwiched between two alternating current loops that are 180 degrees out of phase. The induction voltage will be non-existing.

Look again at the diagram of reply #1, setup 2. The two external plates have opposite currents => A=0. There remains the middle plate with i=0 on the right and I max on the left => A with gradient.

And you can see in setup 3 that the test loop is not sandwiched between the powered plates, but inside the middle plate, which is at ground potential. There is no capacitive coupling allowing the AC signal to reach the test circuit. It is only sensitive to A.

Quote
To me the mechanical variant offers a very good and simple method to validate if this even has any merit. The problem can be solved theoretically quiet easily and can be compared to experimental data. I'm honestly a bit skpectical about this "convective" vector potential induction due to my own experience and experiments with it in the past but those were very crude and far from conclusive.

I'm skeptical too, but here I'm starting from conventional physics, and I would like to understand why I'm measuring a constant magnetic field along the strip when the theory says it shouldn't be.

I think I've found the cause of the negative result, and that's instructive.

The paradox is that, because the initial idea is good, the current remains constant along the central plate!
Indeed the expected current gradient of the central plate should cause an EMF along a parallel conductor. This is the original idea. But for the same reason it will cause the same effect on the electrons of the other two plates.
In other words, Lenz's law, which opposes an EMF generated by ∂A/∂t, applies in the same way against an EMF by v.∂A/∂x.

Seen from the point of view of magnetic fields rather than vector potential, we see on the attached file that I have redrawn, that we have two superimposed circuits, each constituting a loop with the current I/2 flowing in the opposite direction. But the surface of the circuit is not the same for both, because the displacement currents are spread in the opposite way along the plate for both circuits. So we have a mutual induction due to the surface delta that tends to equalise the currents everywhere in both circuits, causing us to lose the expected current gradient in the central conductor.

In my opinion this strongly confirms the equivalence of ∂A/∂t and v.∂A/∂x, but complicates the use of the second form. I will have to revise the setup.

We also discover something interesting. Spreading a displacement current along the plates of a capacitor can affect the surface of a magnetic circuit, with the known consequences on the flux induction through it. There may be something to experiment with here.

Quote from: F6FLT on 2022.11.17, 09:39:28
I think I've found the cause of the negative result, and that's instructive.

The paradox is that, because the initial idea is good, the current remains constant along the central plate!
Indeed the expected current gradient of the central plate should cause an EMF along a parallel conductor. This is the original idea. But for the same reason it will cause the same effect on the electrons of the other two plates.
In other words, Lenz's law, which opposes an EMF generated by ∂A/∂t, applies in the same way against an EMF by v.∂A/∂x.

Seen from the point of view of magnetic fields rather than vector potential, we see on the attached file that I have redrawn, that we have two superimposed circuits, each constituting a loop with the current I/2 flowing in the opposite direction. But the surface of the circuit is not the same for both, because the displacement currents are spread in the opposite way along the plate for both circuits. So we have a mutual induction due to the surface delta that tends to equalise the currents everywhere in both circuits, causing us to lose the expected current gradient in the central conductor.

In my opinion this strongly confirms the equivalence of ∂A/∂t and v.∂A/∂x, but complicates the use of the second form. I will have to revise the setup.

We also discover something interesting. Spreading a displacement current along the plates of a capacitor can affect the surface of a magnetic circuit, with the known consequences on the flux induction through it. There may be something to experiment with here.

Gotta love a good synchronicity ;)
I came across this pic on my PC, saved from a forum years ago.  I can't recall exactly where and the original image seems to have disappeared.

It looks like it, but it's hard to know if it's related. We'll have to find out where it comes from

Quote from: F6FLT on 2022.11.17, 14:21:22
It looks like it, but it's hard to know if it's related. We'll have to find out where it comes from

Took way more work than I thought it would, but I did track it down :P
Turns out it came from this forum.  Was the profile pic of BEP (formerly WaveWatcher)

Quote from: Hakasays on 2022.11.17, 14:05:56
Gotta love a good synchronicity ;)
I came across this pic on my PC, saved from a forum years ago.  I can't recall exactly where and the original image seems to have disappeared.

;)

Mike

I come back with a new scheme that should solve the previous problem.
The principle is the same, it is still a question of generating a potential vector field with gradient from a current gradient.

But here the 2 plates of the capacitor follow the periphery of a disk or a cylinder. As the generator feeds the disc from the centre, only the outer plate will allow a linear current, which cannot induce an opposite current in the other plate as the current feeding it is radial.

(https://www.overunityresearch.com/index.php?action=dlattach;topic=4389.0;attach=46386;image)

I did the first tests. As the capacity is low, I must use high frequencies, so I made an HF type of setup otherwise the unwanted radiations disturb the measurements.

Everything is now coaxial 50 Ω, including the connection to the scope of the probe coil which allows to see the magnetic field around the circumference. It can be seen on the bottom left of the attached photo.

With the test-coil as shown in the photo, I get strange results. The magnetic field is visualised by holding the coil with a vertical axis and sliding it along the circular copper strip.
There is an inversion when the coil is between the two ends of the strip. There is a signal rise around the point of connection to ground, but the maximum is diametrically opposite the ends of the strip. This does not depend on the frequency. The level change between max and min is not linear.
If you turn the coil over, the level is higher. This is a sign that it is sensitive to the electric field in addition to the magnetic field, and that the two add or subtract depending on the position, hence the strange results.

So I screened the test coil, and you can't see any variation along the circular strip. The mystery remains.

I redid with a varriant today the experiment already indicated in reply #23:

(https://www.overunityresearch.com/index.php?action=dlattach;topic=4389.0;attach=46788;image)


An additional inductor is placed in series with the generator, in order to obtain a resonant circuit and therefore currents in phase with the voltage, and the ground connection is now at the center of the disk.
The inductor was placed directly at the output of the generator, therefore far from the device so as not to generate a disturbing magnetic field.

The frequency used were in the 300 KHz. The probe coil, on ferrite core, was shielded. All connections were coaxial (not shown on the diagram).
I checked that, as expected, the electric field at the surface of the plate 2 conductor is constant everywhere along the conductor, sign that it is an equipotential.

But once again, no variation of the magnetic field is detected around the conductor even though it is supposed to carry a current with a gradient (0 at the open end, and Imax at the energized end).
I don't understand this fact.

Quote from: F6FLT on 2023.01.04, 13:02:52
But once again, no variation of the magnetic field is detected around the conductor even though it is supposed to carry a current with a gradient (0 at the open end, and Imax at the energized end).
I don't understand this fact.

The gradient of any A field from a current does not create a magnetic field.  It is the curl function that does that and your experiment indicates that the A field everywhere does not have curl.  To better capture any effect of that A field gradient around your circle I would suggest having another circular conductor very close and parallel with yours, applying a current to it and looking for the voltage induction coming from E = -v* dA/dl where v is the electron drift velocity in this second wire. Note that this form of induction appears as a change of resistance (due to the presence of that A field) and it is a very small change in the resistance of that wire because the drift velocity is so low.  However it should be detectable and I would suggest having the current in the second wire at a slightly different frequency and look for the beat note between the two frequencies.  You will get some form of result but it may take some time to work out what it tells you.

Smudge

Quote from: Smudge on 2023.01.04, 15:08:20
The gradient of any A field from a current does not create a magnetic field.  It is the curl function that does that and your experiment indicates that the A field everywhere does not have curl.  To better capture any effect of that A field gradient around your circle I would suggest having another circular conductor very close and parallel with yours, applying a current to it and looking for the voltage induction coming from E = -v* dA/dl where v is the electron drift velocity in this second wire. Note that this form of induction appears as a change of resistance (due to the presence of that A field) and it is a very small change in the resistance of that wire because the drift velocity is so low.  However it should be detectable and I would suggest having the current in the second wire at a slightly different frequency and look for the beat note between the two frequencies.  You will get some form of result but it may take some time to work out what it tells you.

Smudge

I think you are talking about the wrong thing. The gradient of A does not create a magnetic field, I agree and I never claimed that.

The idea was explained in reply #1. The gradient of A is produced by a current gradient along the conductor. And a current creates a magnetic field. So the experiment done is to check that the magnetic field encircling the conductor is stronger where the current is supposed to be stronger.
As the current is assumed to be zero at the open end, and maximum at the connected end, the magnetic field it generates should also follow this increase (and A should be proportional to it so with spatial gradient).

I'm still a long way from trying to use the gradient of A. At the moment I'm only at the stage of producing it, and it doesn't work for some unknown reason.

@Smudge

I have found a possible explanation. The capacitively coupled plates 1 and 2 are also magnetically coupled. If, as expected, we have a current with a gradient along plate 2, we know that it is also variable in time and will therefore induce a current in plate 1 that will oppose the one coming from the generator. By opposing each other, the two currents along plates 1 and 2 will become uniform, destroying the gradient, while retaining their average value imposed by the looping of the circuit on the generator.

A possible workaround would be to segment plate 1, feeding it in star form with radial currents. Currently, plate 1 is supplied with radial currents, but as it is not segmented, a current can also flow along it.

Each segment of plate 1 being capacitively coupled to plate 2, allows current to be injected into it. But the current induced in plate 1 by the current along plate 2 will no longer be able to flow as plate 1 would be sliced.

@F6,
I sort of follow what you are saying here.  My only criticism is you talk of induced current where in fact it should be induced voltage (that of course "creates" the current that you mention as induced, but I think you mean creates the change of current there).  I was wrong in my previous post as I misread your post where you were looking for the expected spatial variations in the magnetic field, I read that as finding zero field.  Sorry about that.

Smudge

First step successful!
I segmented plate 1 (see photo, before putting plate 2 back in place around it). There are 8 segments. And now we can see the difference in the level of the magnetic field when we turn the probe-coil around.
It is therefore obvious that we have this current gradient along, and consequently, the potential vector A with a gradient too.

Not only does it seem to work, but the negative result before segmentation is very encouraging. It shows that the current of plate 2 induced a current along plate 1, which opposed the gradient. This opposition can be seen as a consequence of Lenz's law, but also as an effect of the electric field along plate 1, created by the vector potential created by plate 2. Now that plate 1 can no longer neutralize the current of plate 2, we can start testing the effect of a gradient of the vector potential.

I may make a new larger setup, with a higher capacity and more segments, in order to use lower frequencies and also to improve the gradient by reducing the mutual influence of the different zones of plate 2. A return to a linear device is perhaps to be considered. I am at this stage of the reflections for the moment.

Second step not sucessfull
I added a conductive strip that goes back and forth around the one that creates the gradient of the vector potential.
I put a direct current through it, which also goes back and forth and should be modulated by the AC signal that generates the gradient of A, and differently depending on whether the alternation is positive or negative, this being due to the supposed acceleration or braking of electrons by the gradient of A.

Nothing of this kind is observed. Note that I made a differential connection of the scope in order to cancel the residual AC voltage that I measured until reaching the background noise, the measurement is therefore sensitive. But with a DC current up to 2.5 A, or without current, there is no difference.

Schematic diagram in the attached file :

The previous negative result prompted me to revise the theory. The reasoning up to now was done in the framework of Galilean kinematics, but electromagnetism is by nature relativistic, and only its framework allows a correct analysis.
I spent many painful hours working on it, even though the method was quite simple, but I still had to find it. I think I found the exact maths for a gradient of the vector potential through special relativity.

As a reminder:
We know that an electric field E is created by the time variation ∂A/∂t (which corresponds to the magnetic induction) and by a potential difference ∇φ.
The general case is E=-∂A/∂t-∇φ, where A is the vector potential and φ the scalar potential.

We assume a volume where φ=0 (no electric field deriving from a potential) and where A is oriented along the x axis, and has a spatial gradient ∇Ax. If a charge is moving at speed v along x, it should see a variation of A related to its position, hence an electric field by the virtue that E=-∂A/∂t=-∂A/∂x.∂x/∂t =-v.∂A/∂x.
This is the experiment that was proposed here and whose theory we want to verify.


In the framework of special relativity, A is a 4-vector to which the Lorentz transforms apply as a 4x4 matrix:

│φ'/c│     │ γ -γβ 0 0│ │φ/c│
│A'x │ =  │-γβ γ 0 0 │ │Ax │
│A'y │     │ 0  0  1  0│ │Ay │
│A'z │     │ 0  0  0  1│ │Az │

φ is the scalar potential, Ax, Ay, Az the values of the magnetic vector potential A on the 3 axes, β = v/c with v the velocity of the charge, and γ = 1/√(1-β²) is the Lorentz factor. A and φ are seen by the observer, A' and φ' by the charge.
In our case φ, Ay and Az are zero.


We simply apply the matrix calculation:

A'x = -γ.β.φ/c + γ.Ax = γ.Ax because φ=0
φ'/c = γ.φ/c - γ.β.Ax = - γ.β.Ax because φ=0

hence φ' = - γ.β.c.Ax = -γ.v.Ax
and ∇φ' = -γ.v.∇Ax

=> E' =-∂A'/∂t-∇φ' = -γ.∂Ax/∂t + γ.v.∇Ax = -γ.v.∂Ax/∂x + γ.v.∇Ax = 0

In other words: the electric field produced by the temporal variation that the charge sees due to the spatial gradient of the vector potential, is exactly compensated by the field from the gradient of a scalar potential that appears due to its movement!

The worst thing is that I had noticed this scalar potential 4 years ago and then completely forgotten about it :(. See https://www.overunityresearch.com/index.php?topic=2470.msg74654#msg74654 where I pointed out that
"We even see, which surprised me at first, that a scalar potential φ can appear in the charge referential when it moves in a place where there is only the vector potential. And vice versa".

The negative result of the experiment is therefore normal. It will be necessary to be more subtle to exploit the basic idea, for example by playing on the 3 space coordinates of A and perhaps even adding a scalar potential or making the speed of the charge variable...

@F6,

If the vector potential A is primary, then within a magnetic field, induced E from movement must come from how A changes as seen by the moving observer.  Then by your reasoning we should not obtain E = v X B, but we know that we do see that E so your reasoning seems flawed.  Perhaps the reason is for that flaw is the A field already comes from something that is moving relative to the observer, that velocity not being taken into account in the relativity argument.

Smudge   

@Smudge,

Very wise remark! I don't have the answer, we'll have to think about it seriously.

"The A field already comes from something that is moving relative to the observer", I agree. A comes from the coulombic field of the moving charges, those of the current which also creates B, and which is no longer isotropic because of the movement.
But this does not answer the problem because this same field A is continuous from the inside to the outside of a coil, so it should have the same effects inside and outside.

The calculation I made is when the charge moves along A and there is a gradient along A. Note that this does not change anything with a zero gradient, we will always have E=0.

Inside a coil, for my calculation to apply, the charge must rotate along A, i.e. in concentric circles inside the coil. A radial Lorentz force should therefore appear, and you are right, it should be obtained from the vector potential.

I also realise that this is the question already raised here: https://www.overunityresearch.com/index.php?topic=2470.msg74662#msg74662 where I proposed a device allowing to see a possible "Lorentz force" outside a magnetic field.

So I have to go back to the math :( to understand how to get the Lorentz force from A (https://en.wikipedia.org/wiki/Lorentz_force#Lorentz_force_in_terms_of_potentials) and why I found E=0 in the previous calculation.
So the original idea of E through the gradient of A is not necessarily dead yet! O0

@Smudge

I have some bad news. On Wikipedia, the starting formula for an A-dependent force is: F=q.[∇φ-∂A/∂t+Vx(∇xA)]

We recognise the one related to the potential difference ∇φ, the one related to the induction -∂A/∂t, and the Lorentz force Vx(∇xA) related to the speed.
We can thus see that the Lorentz force depends only on the rotational of A, which is B.

The relations between A and B being local, the question remains whether in our case the rotational of A is indeed zero. I got help from a physics forum.

Let's take the example of a long solenoid of radius R, where B is zero outside. The equipotentials of A are circles of radius r around the axis of the solenoid.
On a radial axis, the amplitude of A increases from 0 to R, then decreases in 1/r beyond R.
According to a contributor, a field that rotates tangentially along a circle and whose amplitude decreases in 1/r has a locally zero rotational. This is understandable, because A is a tangential vector to the equipotential circle, so the way the field evolves on the tangent matters and thus the gradient.

A PhD-level contributor used the software "Maxima (https://maxima.sourceforge.io)" to confirm this. ∇xA is indeed zero. I don't master "Maxima" to check it too, but it seems that no Lorentz-like force is to be expected outside B.
The software Maxima (https://maxima.sourceforge.io), which manipulates symbolic expressions, could be very useful to us, it will always be something positive to take away from this story.

Quote from: F6FLT on 2023.01.30, 11:29:52
@Smudge

I have some bad news. On Wikipedia, the starting formula for an A-dependent force is: F=q.[∇φ-∂A/∂t+Vx(∇xA)]

We recognise the one related to the potential difference ∇φ, the one related to the induction -∂A/∂t, and the Lorentz force Vx(∇xA) related to the speed.
We can thus see that the Lorentz force depends only on the rotational of A, which is B.

OK that is the Lorentz force but we are looking for another force that can not be called the Lorentz force.

QuoteThe relations between A and B being local, the question remains whether in our case the rotational of A is indeed zero. I got help from a physics forum.

Let's take the example of a long solenoid of radius R, where B is zero outside. The equipotentials of A are circles of radius r around the axis of the solenoid.
On a radial axis, the amplitude of A increases from 0 to R, then decreases in 1/r beyond R.
According to a contributor, a field that rotates tangentially along a circle and whose amplitude decreases in 1/r has a locally zero rotational. This is understandable, because A is a tangential vector to the equipotential circle, so the way the field evolves on the tangent matters and thus the gradient.

A PhD-level contributor used the software "Maxima (https://maxima.sourceforge.io)" to confirm this.  ∇xA is indeed zero. I don't master "Maxima" to check it too, but it seems that no Lorentz-like force is to be expected outside B.

That is all confirming that outside B ∇xA is zero.  That is all text book stuff.  That does not tell us that there can't be another force non-Lorentz term there that does not involve ∇xA.

Smudge

Quote from: Smudge on 2023.01.30, 16:47:20
OK that is the Lorentz force but we are looking for another force that can not be called the Lorentz force.
That is all confirming that outside B ∇xA is zero.  That is all text book stuff. 

We saw that around a solenoid ∇xA=0 because of a particular decay of the field (in 1/R). But the " book stuff" do not say whether A would not have a nonzero rotational in some more complex topologies. Certainly this would generate a magnetic field, but not necessarily around the current that generates A, may be somewhere in space. The book stuff are of no help on this point.

Quote
That does not tell us that there can't be another force non-Lorentz term there that does not involve ∇xA.

The other forces are those related to ∇φ and ∂A/∂t, and we have seen that in a gradient of A, ∇φ cancels the ∂A/∂t that arises from ∂A/∂x.
Again one can imagine more complex topologies so that this is not the case.

@F6,

You have missed the point I tried to make in reply #33 repeated here

QuoteIf the vector potential A is primary, then within a magnetic field, induced E from movement must come from how A changes as seen by the moving observer.  Then by your reasoning we should not obtain E = v X B, but we know that we do see that E so your reasoning seems flawed.

The A field must be non-uniform (in a special way we call curl and you call rotational) for us to recognize a B field.  You have considered a particular non-uniformity where the longitudinal E field given by E_x=-v_x*dA_x/dx is countered by a spatial non-uniform scalar electric potential φ derived from relativity theory.  If that is true then the transverse E field given by E_y=-v_x*dA_y/dx should also be countered by a φ derived from relativity theory.  That transverse E field is the one given by E=vXB that we know to be there, so clearly in that case the relativity argument no longer applies.  That is my dilemma, do we pick and chose which EM laws where we must or must not apply relativity?  I also have difficulty in understanding that Lorentz transverse force, how does an electron moving along the x direction know that the A_y vector it is seeing is changing amplitude along the y direction as it does not get to the y direction?  What are the A carriers that the electron collides with or interacts with?

Perhaps that dilemma is even more evident when we consider Ampere's law applied to two parallel (infinitely) long lines of current.  We find that the force on one line can be derived from Lorentz E=vXB where B is the magnetic field from current flowing in the other line, and that agrees with the force derived by Ampere.  So now we have two known force laws where we cant apply your relativity argument.  But in this case there is a relativity argument that applies to the Coulomb force law between the conduction electrons and ions in both wires.  I have long argued elsewhere that when we have two sources of electric charge, one positive and the other negative, the carriers of their respective E fields can not annihilate each other, if they did we would not have the laws of vector addition that apply in space.  What does annihilate is their effect on a test charge in space if two carriers arrive there at the same time.  Thus neutral material like our two copper wires does not radiate electric fields, but the carriers from the electrons and ions do radiate so the space around the wires has such carriers, call them sub-photons, virtual particles whatever, as part of the huge number density of particles that make up our active aether.  It is then possible to see how the Coulomb carriers from a moving line of electrons as seen by electrons moving in the other line inherit a Coulomb force that is different from that which would have applied if the electrons were stationary, and would have been cancelled by the Coulomb forces from the ions if the electrons were stationary.  That non-cancellation yields the B field we recognize as magnetic, and has a value v/c² times the (otherwise) cancelled Coulomb E field.

With those perceptions in mind I am wary of accepting your suggestion that there is no longitudinal force avaiable to us from movement through a non-curl A field.

Smudge

Quote from: Smudge on 2023.01.31, 11:39:35
...The A field must be non-uniform (in a special way we call curl and you call rotational) for us to recognize a B field.
...

I will respond later on the other points and only make a quick response here.

Sorry for the misuse of the term "rotational" instead of "curl". In French, the "curl" is called "rotationnel", and the automatic translators translate "rotationnel" into "rotational" instead of "curl", including Google (https://fr-m-wikipedia-org.translate.goog/wiki/Rotationnel?_x_tr_sl=auto&_x_tr_tl=en) :), so it didn't shock me.
I see that you understood that we were talking about the same thing, I'll be more careful next time.

Quote from: Smudge on 2023.01.31, 11:39:35
@F6,

You have missed the point I tried to make in reply #33 repeated hereThe A field must be non-uniform (in a special way we call curl and you call rotational) for us to recognize a B field.  You have considered a particular non-uniformity where the longitudinal E field given by E_x=-v_x*dA_x/dx is countered by a spatial non-uniform scalar electric potential φ derived from relativity theory.  If that is true then the transverse E field given by E_y=-v_x*dA_y/dx should also be countered by a φ derived from relativity theory.  That transverse E field is the one given by E=vXB that we know to be there, so clearly in that case the relativity argument no longer applies.  That is my dilemma, do we pick and chose which EM laws where we must or must not apply relativity?"

Relativity must be applied all the time, everywhere. We will see that it can also be applied in the case of the transverse Lorentz force.
The relativistic transforms achieve only one thing: the taking into account of the contraction of lengths and the dilation of time when we change reference frame, direct consequence of the fact that we are in a 4D space and the invariance of the space-time interval, making the space-time coordinates dependent on relative motions, see here (https://www.overunityresearch.com/index.php?topic=4370.msg103700#msg103700).

I didn't miss anything. In the case I have dealt with, the velocity vector of the charge and A are collinear.
In the case of the Lorentz force, the velocity vector of the charge and A are perpendicular.
In the first case we have length contraction, not in the second case. This is what you missed. Moving towards Alpha Centauri at almost speed c will not make its diameter smaller.
The two cases are totally different, there is no reason for a potential to be seen by a charge when it moves perpendicular to A.

Let's check this with the maths. Let's take the matrix from my answer #32 and apply it to the case of a charge moving inside a solenoid on a radial axis x, through a constant magnetic flux directed along y. Then A constitutes circles in the x-z plane, concentric around y. At any point on x, A is therefore directed along z. Ax=Ay=0. φ=0 because we have no scalar potential in the reference frame at rest, so all that remains from the matrix calculation is A'z = Az. Same field A because no length contraction. And no scalar potential appears, contrary to the other case.

Since the force on a charge is given by F=q.[∇φ-∂A/∂t+Vx(∇xA)], we are left here with F=q.Vx(∇xA) where A reduces to Az, the classical Coulomb force, everything is consistent.

QuoteI also have difficulty in understanding that Lorentz transverse force, how does an electron moving along the x direction know that the A_y vector it is seeing is changing amplitude along the y direction as it does not get to the y direction?  What are the A carriers that the electron collides with or interacts with?

I think this misunderstanding is a consequence of the oversimplification of the charge when we think of it as a point. A charge is not point-like. This approximation is certainly very valid in most cases where one seeks to know its influence at a distance thanks to the fields, or conversely the effects of the fields on it, but it is irrelevant when it comes to knowing a transverse influence at the exact location of the charge. A charge is always extended, the electron has a non-zero classical radius. It is therefore not surprising that a field gradient is felt by it whatever its direction.

Quote
Perhaps that dilemma is even more evident when we consider Ampere's law applied to two parallel (infinitely) long lines of current.  We find that the force on one line can be derived from Lorentz E=vXB where B is the magnetic field from current flowing in the other line, and that agrees with the force derived by Ampere.  So now we have two known force laws where we cant apply your relativity argument.
...

As in the other case, of course we can and must apply relativity. The case of Ampere wires treated by relativity is perfectly known and consistent with classical electromagnetism. Paul Bickerstaff's course deals with it clearly in chapter 14.8.3, extract here (http://exvacuo.free.fr/div/Sciences/Cours/R.Paul%20Bickerstaff%20-%20Claustrophobic%20physics%20ch.%2014.8.1.pdf). Full course here (http://exvacuo.free.fr/div/Sciences/Cours/R.Paul%20Bickerstaff%20-%20Claustrophobic%20physics.pdf).

If one can imagine a physical sub-universe of "sub-photons, virtual particles whatever", this is by no means necessary for the calculation of the effects, the application of relativity to the coulomb field of charges or their related potentials is sufficient.

Quote from: F6FLT on 2023.02.03, 13:07:05
Relativity must be applied all the time, everywhere. We will see that it can also be applied in the case of the transverse Lorentz force.
The relativistic transforms achieve only one thing: the taking into account of the contraction of lengths and the dilation of time when we change reference frame, direct consequence of the fact that we are in a 4D space and the invariance of the space-time interval, making the space-time coordinates dependent on relative motions, see here (https://www.overunityresearch.com/index.php?topic=4370.msg103700#msg103700).

I didn't miss anything. In the case I have dealt with, the velocity vector of the charge and A are collinear.
In the case of the Lorentz force, the velocity vector of the charge and A are perpendicular.

Only for some trajectories (the one you chose below).

QuoteIn the first case we have length contraction, not in the second case. This is what you missed. Moving towards Alpha Centauri at almost speed c will not make its diameter smaller.
The two cases are totally different, there is no reason for a potential to be seen by a charge when it moves perpendicular to A.

Let's check this with the maths. Let's take the matrix from my answer #32 and apply it to the case of a charge moving inside a solenoid on a radial axis x, through a constant magnetic flux directed along y. Then A constitutes circles in the x-z plane, concentric around y. At any point on x, A is therefore directed along z. Ax=Ay=0. φ=0 because we have no scalar potential in the reference frame at rest, so all that remains from the matrix calculation is A'z = Az. Same field A because no length contraction. And no scalar potential appears, contrary to the other case.

Since the force on a charge is given by F=q.[∇φ-∂A/∂t+Vx(∇xA)], we are left here with F=q.Vx(∇xA) where A reduces to Az, the classical Coulomb force, everything is consistent.

Now take a trajectory that is parallel to your radial x axis.  Ax and Ay are not zero and dAx/dx is not zero.  You will claim that there will be a non-zero scalar potential negating any Ex field from Ex=-V*dAx/dx.  Will that also apply to Ey=-V*dAy/dx? By determining the force at increments along a trajectory directly from the A field using a finite element program I can reproduce that F=q.Vx(∇xA) Lorentz force, if I introduce a "correction" to take account of this relativity effect it does not reproduce correctly.  I note that the 4 vector that you used is the case for momentum and energy, and certainly qA is considered a momentum and qφ an energy, so I understand why you went down this route.  When the 4 vector is applied to mass momentum which is related to V and energy is related to V² then length contraction and time dilation would be expected to have an effect.  But here the momentum is not related to velocity, so is it correct to apply that relativity correction?  Also the A field comes from many distant charges that have velocity, and it can be argued that the A field already has a V/c relativity connection to those velocities in that the A field from a current element is V/c times the E field from the moving charges making up that current.  Is it right to apply a relativity argument twice?     

QuoteI think this misunderstanding is a consequence of the oversimplification of the charge when we think of it as a point. A charge is not point-like. This approximation is certainly very valid in most cases where one seeks to know its influence at a distance thanks to the fields, or conversely the effects of the fields on it, but it is irrelevant when it comes to knowing a transverse influence at the exact location of the charge. A charge is always extended, the electron has a non-zero classical radius. It is therefore not surprising that a field gradient is felt by it whatever its direction.

But by saying that you have imbued the electron with some form of intelligence in that by sensing the A field over an area it can deduce whether or not the variations offer a field with or without curl.  I can't see that.

Smudge

Quote from: F6FLT on 2023.01.04, 13:02:52
I redid with a varriant today the experiment already indicated in reply #23:

(https://www.overunityresearch.com/index.php?action=dlattach;topic=4389.0;attach=46788;image)


An additional inductor is placed in series with the generator, in order to obtain a resonant circuit and therefore currents in phase with the voltage, and the ground connection is now at the center of the disk.
The inductor was placed directly at the output of the generator, therefore far from the device so as not to generate a disturbing magnetic field.

The frequency used were in the 300 KHz. The probe coil, on ferrite core, was shielded. All connections were coaxial (not shown on the diagram).
I checked that, as expected, the electric field at the surface of the plate 2 conductor is constant everywhere along the conductor, sign that it is an equipotential.

But once again, no variation of the magnetic field is detected around the conductor even though it is supposed to carry a current with a gradient (0 at the open end, and Imax at the energized end).
I don't understand this fact.

As an inverted version of the previous test, what do you think would be the result on the periphery if you were to have an RF microwave source passing through the center?
Would it be DC mixed with AC?  (DC component due to the waves traveling uni-directionally, AC coming from the microwave and 60hz AC inter-oscillation)

Quote from: Hakasays on 2023.02.05, 01:04:25
As an inverted version of the previous test, what do you think would be the result on the periphery if you were to have an RF microwave source passing through the center?
Would it be DC mixed with AC?  (DC component due to the waves traveling uni-directionally, AC coming from the microwave and 60hz AC inter-oscillation)

This was conceivable, but we saw by using the relativistic reference frame change, that the charge will not see the expected electric field, so the reverse operation will not work either.

However, devices with a non-constant current along a conductor (while remaining in quasi-stationary regimes, otherwise in RF it is trivial), could bring new effects, I have not seen any experimentation in this area.

Quote from: Smudge on 2023.02.04, 15:15:19
Only for some trajectories (the one you chose below).
Now take a trajectory that is parallel to your radial x axis.  Ax and Ay are not zero and dAx/dx is not zero.  You will claim that there will be a non-zero scalar potential negating any Ex field from Ex=-V*dAx/dx.  Will that also apply to Ey=-V*dAy/dx? By determining the force at increments along a trajectory directly from the A field using a finite element program I can reproduce that F=q.Vx(∇xA) Lorentz force, if I introduce a "correction" to take account of this relativity effect it does not reproduce correctly.  I note that the 4 vector that you used is the case for momentum and energy, and certainly qA is considered a momentum and qφ an energy, so I understand why you went down this route.  When the 4 vector is applied to mass momentum which is related to V and energy is related to V² then length contraction and time dilation would be expected to have an effect.  But here the momentum is not related to velocity, so is it correct to apply that relativity correction?

The matrix in reply #32 shows us what a charge moving along an x-axis at speed v sees of the 4-vector potential A:

│φ'/c│ │ γ -γβ 0 0│ │φ/c│
│A'x │ = │-γβ γ 0│ │Ax │
│A'y │ │ 0 0 1 0│ │Ay │
│A'z │ │ 0 0 0 1│ │Az │

So :
φ'/c = γ.φ/c - γβ.Ax
A'x = -γβ.φ/c + γ.Ax
A'y = Ay
A'z = Az

In the absence of a scalar potential difference, we are left with :
φ'/c = -γβ.Ax
A'x = γ.Ax
A'y = Ay
A'z = Az

This is remarkably simple and true all the time, no matter what the choice of x-axis is, as long as the charge moves well along x.
If the x-axis is offset from the solenoid axis, these 4 equations still apply.

In my previous example, the axis of the solenoid was y, so Ay=0 since A is in circles in the x-z plane that cuts the solenoid. I guess it is Az that in my context you see non-zero, and that is correct. Az is no longer zero. We are left with:
φ'/c = -γβ.Ax
A'x = γ.Ax
A'y = Ay = 0
A'z = Az

But we see that Az not zero changes nothing, no scalar potential arises because of Az or its gradient. The potential appears only when the velocity of the charge is along x and it depends only on the component of A on x and the velocity on x.

The x-axis is chosen for simplicity, that of the velocity v supposed to be linear. This choice is legitimate.
If we do not do it, it is also correct, but we would have to decompose the vector v on the 3 axes, use its projection on each axis and then take it into account in γ and β in 3 matrix calculations to be done for the 3 components, finally we would have to add respectively the 4 values found.

Quote
Also the A field comes from many distant charges that have velocity, and it can be argued that the A field already has a V/c relativity connection to those velocities in that the A field from a current element is V/c times the E field from the moving charges making up that current.  Is it right to apply a relativity argument twice?     
But by saying that you have imbued the electron with some form of intelligence in that by sensing the A field over an area it can deduce whether or not the variations offer a field with or without curl.  I can't see that.

Smudge

The field A is a local property in space, seen from an observer. It is created by the distant charges which are at the source of the field. It is also a relativistic effect of their motion related to their speed v' with respect to the observer. The relativistic effect is not taken twice, it is taken only once, but once for the velocity v' of the source charges with respect to the observer, which gives us the field A, and another time for the test charges influenced by A and moving at velocity v with respect to the observer. A is only an intermediate.
In the examples we have taken, we assume that A is known. It could be calculated in the same way from the field A0 of each charge in its own reference frame (which reduces to the scalar potential), transforming it by our matrix into the field A seen by the observer who sees these charges moving at speed v'.

One could also treat the case using the velocity v" of the test charges with respect to the source charges, without using the intermediate A-field, but this would certainly be more complex and tedious, the use of the field being precisely made to avoid having to deal with the direct interactions between charges. This use of fields is not specific to relativity, it is the same thing in classical electromagnetism.

Its a spinning sphere, metal (inductive conductor)
Or dielectric (ionic charge carrier)

Smaller spheres with greater charge density produce larger spatial fields

Spheres inside of spheres can warp space time with relatively low energy input
(before you ask i'll give you the simplified version, briefly)
theres a type of radio-sensing warp drive that can basically take itself to the source of a far away signal
by folding spacetime. Consists of a radio isotope and a 3-way transceiver. which consists of 3 smaller spheres in the center. Each has a coil in it, Wired in a closed series loop.
On the center coil is place a small radioisotope oscillator.
The other two act as stimulated receivers, which can sense the distance to the radio source, as a geometrical property related to the distance between the small spheres. Curiously, when subjected to multitudes of frequencies at close range, the spheres vibrate in a way that changes the center of gravity and it rolls like orbi

Anyways, you want a single sphere and you can rotate on 1, 2 , or 3 gimbals
through an electric field.
or in the case of a conductive sphere, through a magnetic field
Then measure the field at distance x, where the vector of x is any direction away from the sphere

@Sm0ky2

What exactly are you referring to?
Do you have a link, diagrams?