New generator from a spatial gradient of the vector potential and current

[F6FLT]

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




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.

[Smudge]

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

[F6FLT]

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.

[F6FLT]

@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.

[Smudge]

@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

[F6FLT]

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.

[F6FLT]

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 :

[F6FLT]

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...

[Smudge]

@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   

[F6FLT]

@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 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!

[F6FLT]

@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" 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, which manipulates symbolic expressions, could be very useful to us, it will always be something positive to take away from this story.

[Smudge]

@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.

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" 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

[F6FLT]

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.

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.

[Smudge]

@F6,

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

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.

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

[F6FLT]

...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 , 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.

[F6FLT]

@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.

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.

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?

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.

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. Full course here.

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.

[Smudge]

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.

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).

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.

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?     

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.

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

[Hakasays]

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




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)

[F6FLT]

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.

[F6FLT]

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.

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.

[Sm0ky2]

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

[F6FLT]

@Sm0ky2

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