OverUnity Research

https://phys.org/news/2025-03-generate-electricity-earth-rotational-energy.html

If we're at the center of the earth in a fixed (non-rotating) inertial frame of reference, we see any conductor on the earth's surface crossing the earth's magnetic field with a speed depending on its latitude. The conductor's charges are therefore subject to the Lorentz force, and if the conductor is properly oriented, a voltage will develop along its length.
However, this voltage causes a rearrangement of the charges and a new static configuration develops (extremely rapidly), cancelling out the electric field.
If we carry out the analysis from the conductor's frame of reference, i.e. rotating with the earth, it's identical: we don't see the Lorentz force since v=0 in our own frame of reference, but we do see an electric field E equivalent to that VxB seen from the center of the earth, with the same consequence of charge redistribution preventing us from circulating current.
After the well-known observation that no energy can be extracted from the Earth's magnetic field, the authors provide the key to getting around the problem, but I confess I'm having trouble following them.

"The extremely rapid and continuously ongoing field cancellation within any conductor appears to make it impossible to use the 𝐯×𝐁 force to generate electricity even for the m=0 components of Earth's field. However, Eq. (12) hides an implicit assumption that can be violated [2]. Since ∇x∇𝑉=0 always, Eq. (12) cannot be satisfied within a magnetically permeable object with a topology chosen to ensure ∇x(vxB)≠0 (13) within the object. The inequality of Eq. (13) holds, for example, in the interior region 𝑎<𝜌<𝑏 of a magnetically permeable conducting cylindrical shell of inner and outer radii 𝑎 and 𝑏 (with 𝜌 the radial coordinate in a cylindrical coordinate system centered in the shell), and long axis perpendicular to 𝐯 and 𝐁 (see Fig. 1)."

Why? If anyone understands, please enlighten us. For me, this is incompatible with relativity, where it can only be in the frame of reference where we see both the field B and the velocity V of the charges, that we can recover their energy.

In any case, with a 30 cm MnZn ferrite tube, they claim to produce a permanent voltage in the 15 µV range, eliminating possible temperature-related artifacts, and opening the door to possible scalability (otherwise, of course, the interest would remain purely theoretical).

Quote from: F6FLT on 2025.03.21, 17:31:44

https://www.youtube.com/watch?v=yZR95bXuMbQ

O0

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Quote from: F6FLT on 2025.03.21, 17:31:44
In any case, with a 30 cm MnZn ferrite tube, they claim to produce a permanent voltage in the 15 µV range, eliminating possible temperature-related artifacts, ...

I will believe it if that voltage varies sinusoidally as a function of the tube's angle.
It would make a novel compass.

 :)
If a copper wire generates a voltage on the earth, it cannot be measured or utilized. Because of the measurement or utilization, a closed loop will be formed, so that the total voltage = 0.
If the video says it correctly: a ferrite coil can slow down electrons.       https://www.youtube.com/watch?v=yZR95bXuMbQ
There is a voltage difference between the power generation coil that is decelerated and the electronic speed difference between the conventional measurement circuit without deceleration or the circuit using the circuit.

Quote from: panyuming on 2025.04.04, 13:19:26
If a copper wire generates a voltage on the earth, it cannot be measured or utilized. Because of the measurement or utilization, a closed loop will be formed, so that the total voltage = 0.

O0

Quote
If the video says it correctly: a ferrite coil can slow down electrons.       https://www.youtube.com/watch?v=yZR95bXuMbQ
There is a voltage difference between the power generation coil that is decelerated and the electronic speed difference between the conventional measurement circuit without deceleration or the circuit using the circuit.

What we do know is that a closed rigid circuit crossing a constant magnetic field produces no current. The reason is obvious, just consider that we have two half-circuits moving together and therefore cutting the same field, so their ends are at the same potential, consequently closing the circuit serves no purpose. This is true even if the observer on earth added an extra displacement to the circuit.

So what happens if we "slow down" the electrons somewhere with a ferrite? A ferrite contains charges and also moves with the conductor. The charges in the ferrite are subject to the Lorentz force. They are also redistributed. If it is an insulator, the redistribution consists of a displacement of the electron clouds in relation to the nuclei, under the effect of E=VxB, a displacement of charges which has an influence on the magnetic field and which, we can bet, cancelling out the delay we were hoping for.

This is best understood in terms of relativity. The magnetic field is a relativistic effect on the coulombic field. When all the charges are moving at the same time at the same speed, the overall effect on a rigid circuit can only be zero. It is only when one part of the circuit is mobile in relation to the other, by deformation or by sliding contacts as in the Faraday disc, that we can have an influence of electrons from one part on the other, creating the imbalance that allows a current to flow.
So in the end I think they're screwing up, and that a few microvolts could be created by various artefacts.

https://www.youtube.com/watch?v=jzHEmphBeFM

Quote from: panyuming on 2025.04.04, 21:15:55
https://www.youtube.com/watch?v=jzHEmphBeFM

According to this video, the voltage DOES vary as a function of the cylinder's orientation angle. I mentioned this  here (https://www.overunityresearch.com/index.php?topic=4764.msg115444#msg115444).

• Also, a full rod/cyliner does NOT work but a hollow one does.
• Furthermore, the choice of the material matters (the ones with lower magnetic Reynold's coefficients are better, e.g.: MnZn)
• A larger diameter hollow cylinder is supposed to generate a higher voltage.
• The generated voltage is supposed to be higher at lower geographical latitudes.
• Last but not least, thermoelectric and photoelectric effects have been ruled out.

P.S.
Electrons moving 30cm in a billionth of a second (1ns) is plausible but stretched, however moving this distance in a billionth of a billionth (sic!) of a second (1as) is FTL.

 NASA has  explored generating electricity in space by "dangling" a wire through Earth's magnetic field. This concept is tied to experiments with electrodynamic tethers, notably the Tethered Satellite System (TSS) missions.
In one such experiment, during the TSS-1R mission in 1996 aboard the Space Shuttle Columbia, NASA deployed a satellite connected to a 19.7-kilometer (12.2-mile) conductive tether. The idea was to move the tether through Earth's magnetic field, inducing an electric current via electromagnetic induction—similar to how a generator works. The motion of a conductor through a magnetic field generates voltage, and in this case, the setup produced up to 3,500 volts and 0.5 amps of current. However, the experiment hit a snag when the tether unexpectedly snapped due to an electrical arc caused by the high current, ending the test prematurely.
The concept leverages the interaction between a long conductive wire and Earth's magnetic field to produce power, offering a potential method for spacecraft propulsion or electricity generation without fuel. While the TSS-1R mission didn't fully succeed, it demonstrated the principle and paved the way for further research into tether-based technologies. Earlier tests, like one in 1993 with a shorter 1,640-foot tether, also showed promise, generating 50-100 volts in a simpler setup. These efforts highlight NASA's innovative attempts to harness space's natural environment for power.

Having now educated myself for the term advection (which I had never used or known) compared to convection and diffusion (which I had known) I can now understand where this new principle comes from.  Treating magnetism moving within a permeable material acting like something (perhaps temperature or an impurity) advecting (moving in one direction) or diffusing within a fluid now makes sense.  Such magnetohydrodynamics requires a magnetic Reynold's number to act like the fluid Reynold's number in a real fluid.  So a lump of permeable material moving within a uniform magnetic field will have that field entering on one side and exiting on the other, hence moving within the material according to magnetohydrodynamic rules.  I can envisage the field within the material moving past a conductor embedded within the material at a rate that differs from the the field outside the material moving past the return path that contains the voltmeter, hence inducing a voltage reading.  What I cannot get to grips with is their vision of the field flowing radially through their rectangular closed circuit to reach the centre point of their cylinder.  I do not understand why they use the closed rectangular loop conductor within the ferrite for their math when the real closed loop that includes the voltmeter passes outside the ferrite.  In my view their internal path could be a single conductor, not two separate paths in parallel.

Smudge

Their statement that a voltage V induced into their rectagular closed loop will result in 0.5V being seen by the voltmeter is just plain wrong, and I am surprised that this has not been picked up.  It is zero.  So all their calculations are just rubbish.  However the fact that the external field moving relative to the ferrite is at a different velocity to the field moving within the ferrite yields an E=vXB voltage induction into a wire embedded along the tube that is different to the E=vXB cancelling induction in the external return wire with the voltmeter in series.  Hence that closed loop sees an induced voltage.

Using their Rm number for M60 ferrite I calculate 48 microvolts for a 20cm wire embedded in the ferrite at their earth location.  Not their measured 18 microvolts but its of the right order.  It should be a simple task to get a single wire running inside a ferrite rod and do the test.  It doesn't have to be a tube, all their math that led them to a tube is based on their false idea that you get half a closed loop induced voltage of you connect an external voltmeter across it, so they looked for induction into their internal closed loop.

Smudge 

Quote from: Smudge on 2025.04.06, 16:15:14
It should be a simple task to get a single wire running inside a ferrite rod and do the test. 

I should be simple if the ferrite rod does not need to be drilled axially, e.g. by making a hollow cylinder out of a bunch of ferrite rings stacked axially. 
Since spin-polarized electrons cannot jump these small air gaps between the rings, will these gaps inhibit the phenomenon ?

How does the length of the hollow ferrite rod affect the induced voltage, anyway ...in your opinion ?

Quote from: verpies on 2025.04.06, 17:23:27
I should be simple if the ferrite rod does not need to be drilled axially, e.g. by making a hollow cylinder out of a bunch of ferrite rings stacked axially. 
Since spin-polarized electrons cannot jump these small air gaps between the rings, will these gaps inhibit the phenomenon ?

I don't know how much the spin polarized electrons influence the "fluid" magnetic field velocity.  My guess is for the ferrite it has such a low conductivity that the velocity is more influenced by its dielectric property.  Doing the experiment with ferrite rings is so straightforward that it is worth doing anyway.  Incidentally Graham Gunderson built a low bandwith DC amplifier that gives a 'scope the ability to measure these microvolt DC levels.  He sent it to me, I passed it to Grum and he passed it to Itsu.  So Itsu has this facility.  For any measurement you have to null out DC offset using a control knob.  When you connect it to something using croc clips the thermocouple action between the croc clip and the item has a DC voltage just from the temperature difference at the instant of connection coming from your body temperature handling the croc clip, and that drifts as things stabilise.  For this experiment it needs a switch to make the connection to the circuit after the DC offset has been adjusted to zero.   

QuoteHow does the length of the hollow ferrite rod affect the induced voltage, anyway ...in your opinion ?

IMO the voltage is proportional to the length of the conductor within the rod.

Smudge

Quote from: Smudge on 2025.04.07, 07:37:30
Doing the experiment with ferrite rings is so straightforward that it is worth doing anyway. 

It would be better, maybe to use that small tube ? These were using often early.  :)

Just to clarify my position on this.  The first image below asks a question that is quite fundamental in regard to EM theory that maybe hasn't been asked before, and the second image is the result of the experiment that is the subject of this thread that seems to answer that question.

Smudge

A field is simply a scalar or vector (or tensor) associated with a position in space to model effects at that location.
A field increases or decreases at a given location, and/or changes direction if it is a vector. The projection of a spot of light moved along a wall is not associated with anything real moving along the wall, certainly not the light. It's the same with a magnetic field.
The notion of "field velocity" does not exist.  It's a convenient abuse of language, but it's not real at all; you mustn't confuse the map with the territory.

As the ferrite moves at constant speed in a constant field, there is no delay in the field inside, it is established and stable, and outside, the same.
A current generated by the Lorentz force can only be recovered in a circuit if there is a permanent imbalance in the mutual influence of all the charges in the circuit, resulting in a net force. By virtue of Stokes' theorem, the corollary of this is a variation in flux through the circuit, which we don't have here. So no current either. No current can be generated because, quite simply, there is no variation in flux through the circuit.

When you consider that Physical Review Research is a peer-reviewed journal, it's frightening to see the level of many physicists today. I'm afraid the days of the Amperes and Faradays are over.
Fortunately, thanks to AI, we can remain hopeful. I had a very interesting discussion on the anomalous Hall effect, associated with the skin effect, with Deepseek. Deepseek is much more pragmatic than ChatGPT, with lots of practical information and order-of-magnitude calculations. I'd recommend it. I look forward to the time when AI will be able to make discoveries.

That said, experiments takes precedence over theory. So if the current they're measuring isn't just an artefact, even if their theory is bogus, that doesn't mean their setup isn't worth exploring.

Quote from: F6FLT on 2025.04.08, 19:27:52
A field is simply a scalar or vector (or tensor) associated with a position in space to model effects at that location.

I would refine this statement like this: A field is a spatial distribution of scalars or vectors (or tensors) associated with various locations in space that model effects at these locations.
These effects usually result in forces, which makes most fields "the field of forces" ...or "force fields", if you like.
Normally, forces accelerate things (cause motion) unless prevented by other forces.

Such definitions of fields usually do not concern themselves with the mechanistic or relativistic causes of these effects (the effects are assumed to just happen in space with a semblance of an explanation given about their cause [mostly heuristic], but their directions and magnitudes are meticulously defined).  This leads to mathematical (quantitative and qualitative) correctness and conceptual ignorance.  A common occurrence nowadays.

It is important to note that observer's motions can create the "illusion" of such fields of forces.  For example if the observer starts spinning, he can rightly claim that a field of forces suddenly appeared outside of him and that field has accelerated everything around him to spin/move with a velocity that is perpendicular to the radius of his rotation and proportional in magnitude to the radius of his observation.  He could make a similar claim if he started to move linearly albeit without the dependence of motions' magnitudes on the distance of his observation.  The neglect of observer's motion always causes the attribution of this motion to other objects. That's classical relativity that assumes 1D time and 3D space.

Increasing the dimensionality of time to 3 allows for much more complex relativity ...in which the forces can vary in space in more complex ways then the two manners which I just described above ...but most people's brains balk at such possibility.

Quote from: F6FLT on 2025.04.08, 19:27:52
A field is simply a scalar or vector (or tensor) associated with a position in space to model effects at that location.
A field increases or decreases at a given location, and/or changes direction if it is a vector. The projection of a spot of light moved along a wall is not associated with anything real moving along the wall, certainly not the light. It's the same with a magnetic field.
The notion of "field velocity" does not exist.  It's a convenient abuse of language, but it's not real at all; you mustn't confuse the map with the territory.

I fully understand that, but I make no excuse for using E = v X B for a conductor moving through a B field for an observer who is stationary, and the same formula for an observer moving with the conductor where the E field now appears to come from a B field that is moving relative to the observer.   

QuoteAs the ferrite moves at constant speed in a constant field, there is no delay in the field inside, it is established and stable, and outside, the same.
A current generated by the Lorentz force can only be recovered in a circuit if there is a permanent imbalance in the mutual influence of all the charges in the circuit, resulting in a net force. By virtue of Stokes' theorem, the corollary of this is a variation in flux through the circuit, which we don't have here. So no current either. No current can be generated because, quite simply, there is no variation in flux through the circuit.

Yes, that is all textbook stuff.  But since Stokes' is just vector math, and our concept of vector fields is a simplification of the real situation of trillions of atomic orbiting electrons in one region of space influencing trillions of conduction electrons in another region of space, is it not possible that Stokes' would not apply in certain cases?  The results of the experiment we are looking at here suggests this.  Are you certain that a conductor lying along the axis of a cylinder of plasma obeying magnetohydrodynamic rules moving through a uniform magnetic field will endure the same induced E field as one lying outside the cylinder?  Has anyone ever done the experiment?

Smudge

Quote from: Smudge on 2025.04.09, 07:33:49
... the real situation of trillions of atomic ...

More like sextillions...

Quote from: Smudge on 2025.04.09, 07:33:49
...orbiting electrons in one region of space influencing trillions of conduction electrons in another region of space,

Sounds like a lot of motion that we ordinarily neglect.
...do you remember what I wrote about neglected motions ?

Here is a FEMM run of a Ferrite tube in a uniform magnetic field.  Shown there are two conductors crossing that field, one inside the tube and one outside the tube.  I show the B field magnitudes given by FEMM for the two positions.  The field inside the tube is virtually zero.  Now if the tube and the conductors move together in the x direction will there be a difference in the v X B electric field seen by the conductors?  If the zero B field at the inner conductor results in zero v X B electric field on that inner conductor then the answer is a resounding yes.  So we can construct a rectangular closed loop that extracts energy from that movement.  Admittedly FEMM is only a 2D program so it simulates something that is infinitely long in the z dimension.  But we know intuitively that a long tube (length greater than diameter) will exhibit its magnetic screening properties around the centre point so we can still construct a rectangular loop that doesn't extend near the ends.  Worth trying the experiment.

Smudge 

Anyone up for trying this experiment?  The voltmeter has to register 10's of microvolts DC so maybe not just a simple meter.  A more complex experiment uses multiple turns in the loop (say 100 then you are in the millivolts range).  I am surprised this has not been done before, but then maybe it has and I am ignorant of it.

Smudge

Edit.  For the Princeton data B=45uT and v=354m/S, if the zero B field inside the tube genuinely results in zero vXB induction there then for a 10cm length of ferrite tube the voltmeter would read 1.6mV DC from induction in the outside wire!!!  If this were real surely this type of induction would have been noted before.  Methinks the E=vXB that appears in every textbook on EM theory may need some qualifications.       

What would be if we take a  magnet pipe,instead ferrite one  ?   :-\

Quote from: chief kolbacict on 2025.04.10, 11:24:20
What would be if we take a  magnet pipe,instead ferrite one  ?   :-\

Since the relative permeability of a magnet is close to 1 it will not divert the earth's field around the pipe so there will be zero voltage created in the loop.

Why don't we put down permanent magnet near our device? It will  give field much more stronger than earth's magnetic field. And it will be  moving with our device together as well.  :)

Quote from: chief kolbacict on 2025.04.11, 08:46:01
Why don't we put down permanent magnet near our device? It will  give field much more stronger than earth's magnetic field. And it will be  moving with our device together as well.  :)

Since the conductor will not be moving with respect to the magnet it will not get induction from the magnet's field.  But you could create the situation where the PM nulls the earth's field on one half of the closed loop and not on the other half.  Then ask whether that null takes away the movement induction on that half?  That is similar to the ferrite tube case.  All our teachings and our gut feelings are that we will not get free energy in this way.

I am reminded of being called into a laboratory about 40 years ago where a military thermal imager was being developed.  This used a miniature Stirling engine to cool the IR sensor and the device suffered from noise from the sensor associated with the Stirling pump frequency, and no amount of fiddling could get rid of it.  I took a small fridge magnet, held it near the sensor to null out the earth's field to demontrate that the stainless steel spring in the Stirling engine was the culprit.  This was supposed to be non-magnetic but it clearly wasn't.  So nulling away the earth's field does work for some situations.

Smudge     

Quote from: Smudge on 2025.04.09, 19:16:50
Here is a FEMM run of a Ferrite tube in a uniform magnetic field.  Shown there are two conductors crossing that field, one inside the tube and one outside the tube.  I show the B field magnitudes given by FEMM for the two positions.  The field inside the tube is virtually zero.  Now if the tube and the conductors move together in the x direction will there be a difference in the v X B electric field seen by the conductors?  If the zero B field at the inner conductor results in zero v X B electric field on that inner conductor then the answer is a resounding yes. 
<Snip>
Smudge

Hi Smudge,
Interesting. I appreciate the FEMM. And agree for a static condition. However when there is motion, left to right, on the diagram, field lines move from one side of the tube to the other. Since, as we are taught, these lines are continuous and remain unbroken, they must cross the conductor which is inside the tube. This infers that both conductors will have equal lines "cut" for a distance traveled in a uniform field, would it not?
bi

Quote from: bistander on 2025.04.11, 21:16:58
Hi Smudge,
Interesting. I appreciate the FEMM. And agree for a static condition. However when there is motion, left to right, on the diagram, field lines move from one side of the tube to the other. Since, as we are taught, these lines are continuous and remain unbroken, they must cross the conductor which is inside the tube. This infers that both conductors will have equal lines "cut" for a distance traveled in a uniform field, would it not?
bi

Yes you can consider lines entering the moving tube on one side and leaving on the other side, but no lines appear inside the tube where the field remains at zero.  FEMM has the capability of giving you the vector potential A field that lies along the z dimension parallel to the conductors.  When you examine that and use E=-dA/dt instead of flux cutting E=vXB you find that both conductors endure the same E field so the closed loop induction is zero.  My take is that the flux cutting rule cannot be used under all situations but E=-vXB is more fundametal and can be used.

The first image below shows a red line where FEMM has given a chart of the A field along the line.  The second image shows two charts superimposed for the initial case and for a 10mm movement of the tube and conductors.  It is seen that both conductors obtain the same change of A field.

Smudge

If our wire put inside a tube from superconductor instead ferrite? There will be not magnetic field there  at all.   ;) in deed.

Quote from: Smudge on 2025.04.12, 09:24:52
Yes you can consider lines entering the moving tube on one side and leaving on the other side, but no lines appear inside the tube where the field remains at zero.  FEMM has the capability of giving you the vector potential A field that lies along the z dimension parallel to the conductors.  When you examine that and use E=-dA/dt instead of flux cutting E=vXB you find that both conductors endure the same E field so the closed loop induction is zero.  My take is that the flux cutting rule cannot be used under all situations but E=-vXB is more fundametal and can be used.

The first image below shows a red line where FEMM has given a chart of the A field along the line.  The second image shows two charts superimposed for the initial case and for a 10mm movement of the tube and conductors.  It is seen that both conductors obtain the same change of A field.

Smudge

Thanks Smudge,
My take is that the moving tube cuts lines, so that which is inside the tube cuts lines. It seems to work in this case even though FEMM never shows flux inside the tube. I do appreciate your explanation.

Flux cutting is E=-vXB. E=-dA/dt is the more fundamental, right?
bi

Quote from: Smudge on 2025.04.09, 07:33:49
But since Stokes' is just vector math, and our concept of vector fields is a simplification of the real situation of trillions of atomic orbiting electrons in one region of space influencing trillions of conduction electrons in another region of space, is it not possible that Stokes' would not apply in certain cases?

A vector field is first and foremost a model. In the case of the magnetic field, the B field allows us to characterise space locally rather than using the billions of billions of equations describing each interaction at a distance with a delay, of each charge with each other.

If anyone has a better model, don't hesitate to provide it.

You can always imagine that since it's a model, and since it would be a simplification, phenomena could deviate from it. Even if we don't assert that something is true because it hasn't been shown to be false, we assume it, which is simply the argument from ignorance, and therefore an irrelevant argument.

Even if the experiment we're talking about produces a current, we don't even need to assume that the field model or Stockes' theorem wouldn't apply, as long as we don't have experimental confirmation that not only is there a real current produced but that no field model would explain it, for example by adding the field of a source we hadn't thought of.
On the other hand, since no experiment has yet shown a deviation from field theory, we can say that it is highly improbable that this would be the case here, and that it is up to those who are convinced of the contrary to demonstrate it.

Quote from: Smudge on 2025.04.12, 09:24:52
...When you examine that and use E=-dA/dt instead of flux cutting E=vXB you find that both conductors endure the same E field so the closed loop induction is zero.  My take is that the flux cutting rule cannot be used under all situations but E=-vXB is more fundametal and can be used.
...

The flux cutting rule is generally false, except when applied in the near-field approximation, i.e. when field propagation times can be neglected. Here, it is correct.  Sometimes it's also difficult to identify, as in the case of the Faraday disk.

As for using A instead of B, we've already done quite a bit of work on this, remember, and it turns out that the electromagnetic results agree, since what links them is not a physical theory but a purely logical mathematical relationship.
The only problem we have is when we limit ourselves to 3D Euclidean space when we live in a 4D universe.
"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".
https://www.overunityresearch.com/index.php?topic=4389.msg103488#msg103488

When we use the 4-vector, many surprises disappear!

https://www.youtube.com/watch?v=Qod24255i-0

Quote from: F6FLT on 2025.04.08, 19:41:26
That said, experiments takes precedence over theory. So if the current they're measuring isn't just an artefact, even if their theory is bogus, that doesn't mean their setup isn't worth exploring.

I agree as it could show something that has been missed in our theories.  An experiment conducted at the equator would involve using a vertical assembly, but why not do the equivalent at a much higher rotational speed using a laboratory magnet?  The composite image below shows first a variation of the Faraday homopolar generator where the disc is moved from the pole to the equator.  The magnet plus disc (annulus) is rotated.  The magnet is a conducting one and we would expect to get a measurable voltage across the brushes. In the second image the meter rotates with the magnet and we would expect a zero voltage.  The third image includes the ferrite tube, and if this shows a voltage then we need to find out why.

Smudge

@Smudge

What I can say is that in the middle image, we don't have zero. There is a current if one of the contact points is towards the middle of the cylinder (from which no field line emerges), and the other towards one end, i.e. there is a field gradient between the two.
This is what I had demonstrated experimentally, and which we had already discussed here: https://www.overunityresearch.com/index.php?topic=3738
(https://www.overunityresearch.com/index.php?action=dlattach;topic=3738.0;attach=31330;image) (https://www.overunityresearch.com/index.php?topic=3738)

The only difference was that the rotating shaft was not a magnet, but ferromagnetic and magnetized at a distance by a magnet.

Your case 3 with ferrite should therefore also show an EMF.

Quote from: F6FLT on 2025.04.18, 10:54:50
@Smudge

What I can say is that in the middle image, we don't have zero. There is a current if one of the contact points is towards the middle of the cylinder (from which no field line emerges), and the other towards one end, i.e. there is a field gradient between the two.
This is what I had demonstrated experimentally, and which we had already discussed here: https://www.overunityresearch.com/index.php?topic=3738
(https://www.overunityresearch.com/index.php?action=dlattach;topic=3738.0;attach=31330;image) (https://www.overunityresearch.com/index.php?topic=3738)

The only difference was that the rotating shaft was not a magnet, but ferromagnetic and magnetized at a distance by a magnet.

Your case 3 with ferrite should therefore also show an EMF.

Sorry I did not make it clear that in the bottom two images the meter is connected to the magnet and revolves with it.  I have edited the images to make them more like the experiment under discussion.

Smudge

@Smudge

Sorry, I misinterpreted your diagrams. But I don't see what the ferrite is for. An electric field from a dB/dt cannot be masked. On the other hand, as the whole device is rigid and rotates in a constant field, it is invariant to rotation, so no force is to be expected on the electrons.

A few days ago I came across a revolutionary theory of electromagnetism, treating charges in the same way as masses through relativity, based on Weyl geometry : https://iopscience.iop.org/article/10.1088/1742-6596/2987/1/012001 .
Electrons thus follow geodesics according to the principle of least action. In this theory, the Lorentz force becomes a fictitious force, an idea you had imagined in the Coler/Unruh thread. This theory may therefore be of interest to you, especially as it also reactivates the idea of the ether, which the quantum vacuum actually becomes and whose charge is a compression. It explains the Aharonov-Bohm effect.

I haven't yet had time to look at it all in detail. Unfortunately, this theory is difficult and requires a level of mathematics that I don't have. But it seems simple and elegant on the principles, electromagnetism becomes a geometrical theory like relativity. It would be interesting to try to use it qualitatively on our problems (quantitatively seems out of reach to me).

A few extracts:  "The Lorentz force law is derived from the same metrics as a geodesic equation. The charge density is shown to obey a covariant wave equation, which indicates that charge density is a field, which propagates at the speed of light. This viewpoint promotes the wave-picture of the electron."
"As the electrodynamic force, i.e. the Lorentz force can be related directly to the metrical structure of spacetime, it directly leads to the explanation of the Zitterbewegung phenomenon and quantum mechanical waves as well."
"Charge is therefore to be understood as a local compression of the metric in the spacetime, which relates to longitudinal waves as described in [12]."

Quote from: F6FLT on 2025.04.18, 19:52:58
@Smudge

Sorry, I misinterpreted your diagrams. But I don't see what the ferrite is for. An electric field from a dB/dt cannot be masked. On the other hand, as the whole device is rigid and rotates in a constant field, it is invariant to rotation, so no force is to be expected on the electrons.

This is simply the earth's field experiment discussed here that gave 18uV reading, replicated using a magnet in place of the earth.  We get greater field and greater rotation speed, but it's the same experiment.  Big disadvantage is the meter revolving with the magnet is somewhat difficult to read.  A variation where the meter is stationary is shown in the images below.  This uses plastic annuli attached to the magnet carrying a radial conductor.  We know we will get a voltage from image 1 and zero from image 2, but what will we get from image 3?  You will say zero but the earth's field experiment did not yield zero.  It strikes me that this experiment where we can vary the rotation speed is more likely to find answers than another earth's field one.

QuoteA few days ago I came across a revolutionary theory of electromagnetism, treating charges in the same way as masses through relativity, based on Weyl geometry : https://iopscience.iop.org/article/10.1088/1742-6596/2987/1/012001 .
Electrons thus follow geodesics according to the principle of least action. In this theory, the Lorentz force becomes a fictitious force, an idea you had imagined in the Coler/Unruh thread. This theory may therefore be of interest to you, especially as it also reactivates the idea of the ether, which the quantum vacuum actually becomes and whose charge is a compression. It explains the Aharonov-Bohm effect.

I haven't yet had time to look at it all in detail. Unfortunately, this theory is difficult and requires a level of mathematics that I don't have. But it seems simple and elegant on the principles, electromagnetism becomes a geometrical theory like relativity. It would be interesting to try to use it qualitatively on our problems (quantitatively seems out of reach to me).

A few extracts:  "The Lorentz force law is derived from the same metrics as a geodesic equation. The charge density is shown to obey a covariant wave equation, which indicates that charge density is a field, which propagates at the speed of light. This viewpoint promotes the wave-picture of the electron."
"As the electrodynamic force, i.e. the Lorentz force can be related directly to the metrical structure of spacetime, it directly leads to the explanation of the Zitterbewegung phenomenon and quantum mechanical waves as well."
"Charge is therefore to be understood as a local compression of the metric in the spacetime, which relates to longitudinal waves as described in [12]."

Thank you for that, I will study it with interest.

Smudge

Quote from: Smudge on 2025.04.19, 10:45:15
...
Big disadvantage is the meter revolving with the magnet is somewhat difficult to read.  A variation where the meter is stationary is shown in the images below.  This uses plastic annuli attached to the magnet carrying a radial conductor.  We know we will get a voltage from image 1 and zero from image 2, but what will we get from image 3?
...

There's a simple way to carry out the experiment: replace the voltmeter by a capacitor, rigidly connected on one side, and with a light contact on the other, with a sort of ratchet. The speed of rotation is slowly increased, and once a certain threshold is reached, the centrifugal force on the capacitor causes it to disconnect. We can then stop the rotation and quietly measure the voltage across the capacitor.
I did this experiment in the context of the Faraday disk, with a capacitor placed radially and rotating with the disk, in order to verify the absence of voltage in the rotating frame of reference, which the experiment did indeed show.

In the present case, I'm not inclined to do so. For one thing, the experiment is very different from that of the terrestrial field, because the B field cannot be uniform throughout space. On the other hand, if the theory were true, i.e. that speed would have an influence, we'd have measurements that would also depend on the speed of the earth around the sun, or the speed of rotation of our galaxy, which are much greater than the linear speed at the earth's surface due to its angular velocity.

In F=q.VxB, V is the velocity of the charge relative to the observer who sees B, not relative to the source of the field. The observer at the earth's surface will see F=0 because V=0, ferrite or not.

If the effect is real, it can't be a question of the conductor's speed, but perhaps of acceleration. We have two of these: gravity acceleration and centrifugal acceleration. It's possible that the positioning of the device in relation to the accelerations and the B field, and its nature thanks to the ferrite, could produce an imbalance somewhere on a closed circuit, resulting in a non-zero net force on the electrons, knowing that the energy could only be taken from centrifugal acceleration, slowing down the earth. I think that's where we should be looking.

Quote from: F6FLT on 2025.04.20, 10:50:44
....the experiment is very different from that of the terrestrial field, because the B field cannot be uniform throughout space.

I disagree.  The earth's field experiment is very much like the one I describe, you have the obsever rotating with the magnet.  The non-uniformity of the B field would also apply to the Faraday disc experiment, the field is not uniform across the disc, but that does induce according to F=q.VxB even when the disc moves with the magnet.

QuoteOn the other hand, if the theory were true, i.e. that speed would have an influence, we'd have measurements that would also depend on the speed of the earth around the sun, or the speed of rotation of our galaxy, which are much greater than the linear speed at the earth's surface due to its angular velocity.

Again I disagree, we don't consider those velocities in the Faraday disc experiments.

QuoteIn F=q.VxB, V is the velocity of the charge relative to the observer who sees B, not relative to the source of the field. The observer at the earth's surface will see F=0 because V=0, ferrite or not.

If you apply that argument to the Faraday experiment you are completely wrong.  The earth's field experiment is just something larger than the Faraday one.  By your reasoning, if we had a giant disc at the pole and an observer who was not rotating with the earth (but was translating with the earth around the sun etc.) had a voltmeter connected to brushes on that disc would not measure anything.  I think he would as it is just a giant Faraday disc experiment.

I stand by my perception that a radial conductor at the equator is just a larger version of the radial conductor in my experiment and it will inherit an induced E field.  The reason the closed circuit sees zero voltage is the two radial conductor's voltages cancel out.  The non-zero result of the actual earth's field experiment suggests that having ferrite around one of the conductors negates this cancellation.  Surely this is something worth investigation to see if it is true.

QuoteIf the effect is real, it can't be a question of the conductor's speed, but perhaps of acceleration. We have two of these: gravity acceleration and centrifugal acceleration. It's possible that the positioning of the device in relation to the accelerations and the B field, and its nature thanks to the ferrite, could produce an imbalance somewhere on a closed circuit, resulting in a non-zero net force on the electrons, knowing that the energy could only be taken from centrifugal acceleration, slowing down the earth. I think that's where we should be looking.

OK but I think that is far more speculative than my approach.  In my simple mind if a conductor is within a magnetic field that is a constant value and somehow Nature has has arranged that its movement through that field can induce a voltage, then perhaps Nature also has a way of doing something to some magnetic material that is moving, something we don't know yet.

Smudge

Maybe we should change the planet so that there is enough energy for everyone?

Quote from: Smudge on 2025.04.20, 16:04:13
I disagree.  The earth's field experiment is very much like the one I describe, you have the obsever rotating with the magnet.  The non-uniformity of the B field would also apply to the Faraday disc experiment, the field is not uniform across the disc, but that does induce according to F=q.VxB even when the disc moves with the magnet.
Again I disagree, we don't consider those velocities in the Faraday disc experiments.

If you apply that argument to the Faraday experiment you are completely wrong.  The earth's field experiment is just something larger than the Faraday one....

You disagree because you haven't interpreted my comments correctly.
The question of whether the magnet is rotating or not is irrelevant as long as the magnetic field remains the same, which is the case when a cylindrical magnet rotates around its magnetic axis.
The only question is: who is the observer who sees the charges at speed V?
Obviously it's the voltmeter connected to the sliding contacts that are at rest. It sees the electrons spinning and therefore sees the Lorentz force acting on them. If the voltmeter turns with them, it won't measure anything, which is what my experiment with the radial capacitor confirmed.

So in the case of the Faraday disc, obviously the speed of the earth plays no role, because the observer is on earth. But in the context of this thread, where the "observer", the voltmeter, moves with the circuit, but is claimed to be measuring a current, then the speed of the whole through the galaxy should also have an effect. Obviously this is all wrong.
The voltmeter has speed zero in relation to the charges in the circuit, the Lorentz force is zero on all the charges it sees, so it cannot measure a current. The presence of the ferrite doesn't change anything because the Lorentz force is zero. Your remark 'the conductors negates this cancellation' is irrelevant, since there is no force on the electrons, so there is nothing to cancel, everything is already zero.

The Lorentz force on electrons in a circuit involving current can only be invoked when there is relative velocity between charges! In the end, these authors' paper makes no sense in terms of theory.

Secondly, when I spoke of a 'uniform field', I was simply referring to the experimental context of the experiment you proposed: when you rotate a magnet, it's almost impossible to avoid small variations in B, which could cause artefacts of a few µV in the measurements, whereas the Earth's magnetic field is considerably more uniform and stable. I wasn't referring to the cut-off flux.

If we talk about this, then it is clear that there is a variation in flux in the case of the Faraday disc. The flux through the disc is constant, but not through the circuit, because the rotation of the disc tends to change the circuit, and therefore the flux, by moving the charges participating in the current. In other words, in the case of the Faraday disc, Stokes' theorem is also valid. All this is known and calculated for a long time.
In the case of a rigid circuit moving with the earth, there is no variation in flux through the circuit. By Stokes' theorem, there is no current in the circuit either, which is perfectly in line with the absence of Lorentz force.
If there is a current, its cause really has nothing to do with the Lorentz force.

Quote from: F6FLT on 2025.04.21, 15:18:45
You disagree because you haven't interpreted my comments correctly.

I think the disgreement is because we see things differently.

QuoteThe question of whether the magnet is rotating or not is irrelevant as long as the magnetic field remains the same, which is the case when a cylindrical magnet rotates around its magnetic axis.
The only question is: who is the observer who sees the charges at speed V?

And there is where we see things differenty with regard to the reference frame of the observer and what that means.

QuoteObviously it's the voltmeter connected to the sliding contacts that are at rest.

No disageement there.

QuoteIt sees the electrons spinning and therefore sees the Lorentz force acting on them.

When you say the electrons spinning I think you mean it sees the electrons in the moving disc and the spin is the disc movement.  I don't see it quite like that.  The voltmeter sees a voltage, and that voltage is because the electrons in the disc are enduring the Lorentz force due to their movement through the magnet's field.  That force is real, it doesn't disappear when the voltmeter is removed and it doesn't disappear if the voltmeter moves with the disc.

QuoteIf the voltmeter turns with them, it won't measure anything.......The voltmeter has speed zero in relation to the charges in the circuit, the Lorentz force is zero on all the charges it sees, so it cannot measure a current.

The Lorentz force on all the electrons in the disc, and now on all the electrons in the voltmeter circuit, can't just disappear.  They are still there.  The reason the voltmeter measures zero is because the Lorentz force on the voltmeter and its connecting wires induces voltage there in exact opposition to that from the disc.  Moving the voltmeter onto the disc has simply added more electrons there.  All those electrons are moving through the magnetic field and they all endure the force.

QuoteThe presence of the ferrite doesn't change anything because the Lorentz force is zero. Your remark 'the conductors negates this cancellation' is irrelevant, since there is no force on the electrons, so there is nothing to cancel, everything is already zero.

And there is our disagreement, in your mind moving the voltmeter onto the disc has made the Lorentz force disappear within the voltmeter's world so the voltmeter thinks there is zero force everywhere.  But we know different, we know that there is that movement which is really a relative movement between electrons in the disc (which now includes the voltmeter circuit electrons) and some of the electrons in the magnet, the forces are still there.

QuoteThe Lorentz force on electrons in a circuit involving current can only be invoked when there is relative velocity between charges!

Yes and the relative velocities inducing current are between electrons in the disc (and anything we put on the disc) and some electrons in the magnet.  I stand by my position that electrons in a closed circuit on the disc will endure the Lorentz force that will normally only drive electrons to one side of the circuit, will not drive electrons around the circuit.  If part of that closed circuit is surrounded by high mu material, that part can be shielded from the B field, hence the Lorentz force expressed by F=e.vxB no longer applies, but F=-e.dA/dt does still apply.  Are we absolutely sure that adding ferrite to the circuit will not allow the closed circuit current to appear?  Are there conditions where the F=-e.dA/dt does not account for the missing Lorentz F=e.vxB.  Has anybody ever done the experiment?     

Smudge

Further to my previous message (post#40) I will change my position slightly.  If we have a closed circuit on the Faraday disc like the rectangular one I showed previously (or on the annulus at the equatorial mid position of the magnet) there will be no current around that circuit because the forces on each conduction electron is zero.  The force is zero not because the Lorentz force is zero, but because each electron also endures a Coulomb force that negates the Lorentz one.  The Lorentz force from the "flux cutting" movement drives electrons within the closed circuit such that one side has a surfeit of electrons and the other side has a deficit.  That redistribution of charge creates the Coulomb E field that everywhere in the conductor negates the Lorentz E field.  I was wrong to claim that the reason for zero current is the two long arms of the rectangular circuit each having an induced identical voltage induction from the Lorentz force that cancel out (although in fact they do) and IMO F6 was wrong to claim that the Lorentz force disappears.  The question remains, can putting one of the radial conductor arms inside a ferrite tube then allow a current to appear?  The earth's field experiment suggests it can.

With regard to Lorentz E=vxB against E=-dA/dt, I reported doing some FEMM simulations where the latter approach showed identical values along both radial conductors.  That used the 2D program where my uniform field came from two magnetic slabs that in the 2D program go to +- infinity in the z direction.  The A field there is nothing like the 3D A field that forms circles around the typical rod magnet used in the Faraday experiment.  The circles are also concentric to the rotation axis.  Each part of the radial conductor is therefore moving along a constant A field so it would seem E=-dA/dt can't apply.  But the screening effect of the ferrite tube will apply, the inner conductor there is in virtually zero B field which suggests Lorentz can't apply there.   I think an experiment using magnet and rotating part like the Faraday one is worth trying.  Incidentally the 3D axisymmetric version of FEMM can give useful data for this problem on the assumption that the magnet axis and the rotation axis are the same (which isn't quite true in the earth's field experiment).

Smudge

Quote from: Smudge on 2025.04.22, 16:22:35
...
I stand by my position that electrons in a closed circuit on the disc will endure the Lorentz force that will normally only drive electrons to one side of the circuit
...

The Lorentz force doesn't exist in the electron's frame of reference, since its velocity is zero relative to itself. So if you don't specify the frame of reference in which you're doing the analysis, your reasoning is irrelevant.

F = q.VxB is only in the frame of reference of the observer who sees charges at speeds V in a field B. In its own frame of reference, the electron sees only an electric field E and the force exerted on it is F=q.E. Since B is constant and uniform in the case of the earth's rotation, so is the E field seen from the electron (since E and B are linked by E=VxB).

E is conservative so the integral of E around a closed loop is zero. Same null result whatever the choice of observer, whatever the case, whether we analyze with B or E. End of story.

Ferrite is useless because an electric field from VxB is not shieldable. The reason is obvious: no configuration of static charges can mask the field of moving charges, those at the origin of the magnetic field, due to the non-spherical symmetry of the coulombic field of moving charges seen by charges at rest. The +/- charge fields can no longer be balanced. This is exactly the same problem as when B varies with time.

Quote from: F6FLT on 2025.04.25, 11:29:15
Ferrite is useless because an electric field from VxB is not shieldable. The reason is obvious: no configuration of static charges can mask the field of moving charges, those at the origin of the magnetic field

OK we have moving charges creating a magnetic field.  Let's assume that movement is the magnetically aligned orbital motion of many electrons around nuclei in a PM.  Now we have ferrite moving through the B field from the PM at some velocity V.  That is not a configuration of static charges.  All the charges in the ferrite have that velocity V.  Also there are many electrons having orbital velocity added to V, in particular the ones that give the ferrite its magnetic characteristic.  We know that those orbits can get aligned within the ferrite due to the B field within which the ferrite is moving.  And with a certain ferrite geometry (e.g. tube) the B field is masked to be virtually zero in the centre.  If B=0 then VxB must also be zero.  I do not see how this fits your obvious reason.

I can accept that using VxB for an electric field may not apply to all situations and there is a more fundamental approach that is likely to involve movement through the A field, not the B field.  That I am trying to look at in regard to the earth's field experiment.  There we have the A field from the earth that runs east-west, and that has a high value (100 weber/m at the equator if my memory is correct).   The closed circuit of the experiment is being transported at earth's surface velocity along the A field, a giant circle A field.  The electrons in the Faraday disc are transported along a circular A field and they obtain a radial force, and the same applies to the earth's field.  If the A field is the primary one to be used, we discover that electrons in the rotating Faraday disc move along a constant value circular A field.  An observer on the disc would see the A field as a rotating vector.  We have discussed this before and for the E field from VxB to match that from the A field we must use |E|=2ωA where ω is the angular rotation rate.  This also applies to the earth's field where the earth is rotating at ω=7.272E-5.  If we have a conductor within a ferrite tube, will the A field there have a different value from that of a conductor without the tube?  Playing around with FEMM, both as a planar problem and an axisymmetric one it appears that the ferrite tube could affect the A field.  In which case we could have a means to examine the earth's field experiment in more detail using 3D finite element programs.

Smudge

Quote from: Smudge on 2025.04.27, 09:12:46
OK we have moving charges creating a magnetic field.  Let's assume that movement is the magnetically aligned orbital motion of many electrons around nuclei in a PM.  Now we have ferrite moving through the B field from the PM at some velocity V.  That is not a configuration of static charges.  All the charges in the ferrite have that velocity V.
...

Once again, you forgot to mention the reference frame. You talk about speed as if it were absolute!
If I move with the ferrite, the Lorentz force is zero because V is zero. Would the force on the electrons depend on who's looking at them?
Of course there's a force on the electrons, since a force doesn't depend on the reference frame and we have an E=VxB field seen from the charges, where V is their velocity relative to the observer who sees the B field. If there's a force in one frame of reference, there's a force in all.
Now, if we, the observer, want to use this force, therein lies the problem, since energy depends on the frame of reference. By moving at the same speed as the charges, we are immersed in the same E field as the charges, so there can be no more field gradients or potential differences that would enable us to obtain a current. We are in the same referential.

The magnetic field is simply a question of speed relativity. If I'm moving relative to a fixed charge, I see a current, and therefore a magnetic field. If it's the charge that's moving in relation to me, same thing. There's nothing to distinguish a charge at rest from a charge at speed v. It's just a question of point of view, and it's only the relative speed that can create an exploitable effect.

It's the same problem in mechanics. If you see a car passing by you at speed V, you think you can recuperate its kinetic energy, for example by capturing it with a spring that will stretch until the car stops, at which point the spring is blocked. The energy has passed into the spring. But if you go at the same speed as the car, you can't recover any of its energy. With electrons, it's the same.
We have to stop drowning ourselves in experimental details and special cases, when the great principles of nature are perfectly clear on the situation.

Researchers using NASA's Hubble Space Telescope have discovered the magnetar called SGR 0501+4516 is traversing our galaxy from an unknown place of origin.
https://science.nasa.gov/missions/hubble/nasas-hubble-tracks-a-roaming-magnetar-of-unknown-origin/ (https://science.nasa.gov/missions/hubble/nasas-hubble-tracks-a-roaming-magnetar-of-unknown-origin/)
It will soon. The dream of lovers of the OU will come true. In every piece of metal on the earth, a huge EMF will be induced.  :)