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Author Topic: New generator from a spatial gradient of the vector potential and current  (Read 6980 times)
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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". 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.



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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.



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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!

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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.


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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.

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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.
   
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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.

« Last Edit: 2022-11-11, 10:20:27 by F6FLT »


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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.


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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/
   
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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 :(...




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.....................
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.
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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.
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.

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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.

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

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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,
agreed for temporal variation
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so that the intensity of A is always uniform along the loop,
not agreed for spatial variation
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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.
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.

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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.



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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.



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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
   
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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.


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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.
   
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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
   
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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.


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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.



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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.



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


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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)


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Buy me a beer
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


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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.





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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.



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